Part I: Getting started
August 1, 2026 · View on GitHub
kimi-k3-in-c
A 2.78-trillion-parameter model. One CPU. 8 GB of RAM.
Kimi K3 inference in portable C99.
No BLAS. No framework. No GPU.
| 2.78T parameters |
1.56 TB checkpoint on disk |
8.24 GB peak RSS, measured |
176 KB the whole engine |
0 GPUs |
$ ./bin/k3 ~/k3model --trunk ~/k3trunk --preset laptop \
--tok ~/k3model --prompt "The capital of France is" --gen 8 --incremental
--- generated text ---
Paris.",
+ "The Eiffel
----------------------
8 tokens in 261.5 s, 32.69 s/token average
PEAK RSS for the whole run: 8.24 GB
Slow, and answering correctly, in 8.24 GB, from a checkpoint of 1.56 TB. This is a base model, so what follows " Paris." is a continuation rather than a reply; there is no chat template. Give it more memory and the answer does not change, only the clock:
$ ./bin/k3 ~/k3model --trunk ~/k3trunk --preset server \
--tok ~/k3model --prompt "def fibonacci(n):" --gen 28 --incremental
--- generated text ---
if n <= 1:
return n
else:
return fibonacci(n-1) + fibonacci
----------------------
28 tokens in 299.3 s, 10.69 s/token average
PEAK RSS for the whole run: 127.92 GB
Every figure in this document comes from the measurement output in
docs/data/.
The dense trunk stays in memory to whatever depth you choose and streams the rest; the 1.45 TB of routed experts are never resident, and are multiplied straight out of their packed 4-bit form. The consequence is that the same model runs in 8 GB and in 224 GB and produces byte-identical output at every budget between.
Four decisions about where bytes live take it from a cluster to a laptop, and the answer at the bottom is the same as the answer at the top:
Part II builds every box in both diagrams from scratch, one component at a time.
Contents
- Requirements
- Quick start: clone, build and verify in about a minute, with no model
- Full setup: the whole path to generated text
- Usage
- Choosing a preset
- Reading the run report
- Common questions
- The problem: a model that does not fit
- The four reductions
- The machine, and what it assumes
- The codebase
- Five invariants
- 1. Reading a 1.56 TB checkpoint from its headers
- 2. The config reader that refuses to guess
- 3. The tokenizer, byte for byte
- 4. Reduction one: the experts already ship at half a byte
- 5. Kernels with a floating point contract
- 6. Reduction two: KDA, attention with a memory that never grows
- 7. Reduction three: MLA, one latent instead of ninety-six heads
- 8. Attention residuals: layers that look back
- 9. Picking 16 experts of 896
- 10. Packing the trunk: 93 layers, one read each
- 11. Reduction four: streaming the trunk turns a floor into a dial
- 12. An LRU cache for the experts
- 13. How big should that cache be? Ask the trace
- The gate ladder
- A tiny oracle first
- Proving it on the full checkpoint
- The first tokens
- Sustained generation: text in, text out
- The memory ladder: 8 GB to 224 GB
- The cache that was not participating
- Allocation beats capacity
- Measuring the measurement
- Storage is the whole game
- Why the trunk is not quantised
Part I: Getting started
Requirements
The gate is storage: the checkpoint is 1.56 TB. Everything else is ordinary.
| OS | Linux, x86-64 | uses O_DIRECT, posix_memalign, getrusage |
| CPU | AVX2 + FMA | AVX-512 unnecessary. make portable targets generic AVX2 |
| RAM | 8 GB and up | every preset works; more memory is faster, never different |
| Storage | ~1.7 TB free | 1.56 TB checkpoint + 109 GB packed trunk, ideally on fast local disk |
| Toolchain | GCC ≥ 9 or Clang ≥ 10 | GNU make, or CMake |
| Python | 3.9+ | for the download, pack and analysis tools; not for make test |
The tokenizer and config reader are portable C99 and build anywhere. Without a checkpoint you can still do everything in Quick start.
Quick start
Clone, build and run the entire test suite. No checkpoint, no network, no Python. The whole thing takes about a minute.
git clone https://github.com/FareedKhan-dev/kimi-k3-in-c.git
cd kimi-k3-in-c
make -j # seconds. Seven C files, a compiler and OpenMP
make test # under a minute
It ends like this, or it failed:
GATE 1 teacher forcing : 32/32 positions match tf_pred
generated span : 20/20 <- must be exact
GATE 2 greedy decode : 20/20 generated tokens match full_ids
GATE 3 incremental : 20/20 generated tokens match full_ids <- KV cache + carried KDA state
VERDICT: ENGINE MATCHES THE REFERENCE EXACTLY
ALL WEIGHTLESS TESTS PASSED
That is the whole engine: every kernel, the streaming cache, the safetensors reader, the config reader, the tokenizer, and an end-to-end oracle over a 13-layer model built with the same tensor graph as the released one, checked against a PyTorch reference from fixtures committed to the repository.
One published measurement also replays on the spot, from a trace recorded during a full 93-layer run (this one needs Python 3.9+ and numpy):
python3 tools/sim_cache.py tests/fixtures/expert_trace.bin
100,096 expert requests, reprinting the capacity table in
expert-cache-capacity.txt.
Full setup
Six steps from an empty directory to generated text. Only step 4 is slow.
./scripts/k3-doctor.sh can be run at any point. It checks the toolchain, sizes your RAM
to a preset, measures your storage, and prints the exact command to run next.
Step 0. clone
git clone https://github.com/FareedKhan-dev/kimi-k3-in-c.git
cd kimi-k3-in-c
About 45 MB, most of it the diagrams and the test fixtures.
Step 1. check the machine
./scripts/k3-doctor.sh
Takes about a minute, because it measures your disk the way the engine reads it. It exits non-zero if the machine cannot run the model at all.
Step 2. build
make -j
Seconds. The only dependencies are a C99 compiler, libm and OpenMP. CMake works too:
cmake -B build && cmake --build build -j && ctest --test-dir build
Step 3. verify before downloading anything
make test
This is worth doing before committing to a 1.56 TB download: it proves the engine matches its reference on a model with the same tensor graph, and it needs nothing but the repository.
Step 4. fetch the checkpoint
1.56 TB, so hours rather than minutes. Get a token from huggingface.co/settings/tokens:
export HF_TOKEN=hf_your_token_here # read from the environment, never echoed
./scripts/download-model.sh ~/k3model # resumable, re-run to continue
The script finishes by verifying the shard count, the exact byte total, and then every individual per-shard size against the published figures:
verifying…
shards : 96 (expect 96)
bytes : 1560936091448 (expect 1560936091448)
shards : all 96 match their published sizes individually
RESULT : byte-exact match
A partial download does not fail loudly; it produces wrong tokens. Treat a FAIL here as
a stop. Checking per shard also turns "re-download 1.56 terabytes" into "re-download this
one 17 gigabyte file", and it catches the one case a total cannot: two shards wrong in
opposite directions by the same amount.
Step 5. pack the trunk
./scripts/pack-trunk.sh ~/k3model ~/k3trunk
About four minutes, once. It rewrites the 93 dense layers into one 109 GB file where layer L lives at a known offset and can be read in a single call. This is what turns the memory requirement into a dial. Put the output on the fastest disk you have.
Step 6. run
./bin/k3 ~/k3model --trunk ~/k3trunk --preset workstation \
--tok ~/k3model --prompt "The capital of France is" --gen 8 --incremental
The tokenizer ships with the checkpoint, which is why --tok points at the model
directory.
Where everything ends up
kimi-k3-in-c/ ~45 MB source, docs, images, and bin/k3
~/k3model/ 1.56 TB 96 shards · config.json · tiktoken.model · tokenizer_config.json
~/k3trunk/ 109 GB trunk.bin · trunk.json, on the fastest disk you have
The first token of any run loads every pinned layer from disk, about 108 GB at the
server preset, so it takes far longer than the steady rate. That cost is paid once per
run, not once per token.
Usage
Synopsis
k3 <model_dir> [prompt] [memory] [generation] [diagnostics]
<model_dir> is the directory holding the .safetensors shards. It is required for any
run, but --help, --version and --list-presets work without it:
./bin/k3 --help
./bin/k3 --version
./bin/k3 --list-presets
Prompt options
Exactly one of these is required. Passing none, or more than one, is a usage error (exit 2).
| flag | argument | |
|---|---|---|
--prompt | TEXT | tokenize TEXT and run it. Requires --tok. |
--prompt-file | PATH | tokenize the file's bytes. Requires --tok. Preferred for anything non-ASCII: the shell re-encodes argv, whereas a file is read verbatim |
--ids | 1,2,3 | token ids directly. No tokenizer is loaded at all, so this works on a machine with no tokenizer files. The reproducible channel the tests use |
# text in
./bin/k3 ~/k3model --tok ~/k3model --prompt "The capital of France is" ...
# text in, from a file. Use this for CJK, emoji, accents
printf 'La capitale de la France est' > /tmp/p.txt
./bin/k3 ~/k3model --tok ~/k3model --prompt-file /tmp/p.txt ...
# ids in, ids out, no tokenizer needed
./bin/k3 ~/k3model --ids 1008,10484,318,15383,387 ...
Memory options
| flag | argument | default | |
|---|---|---|---|
--preset | NAME | none | laptop · desktop · workstation · server · max. Sets both budgets below |
--trunk | DIR | off | the packed trunk directory from step 5. This is what enables streaming. Without it the trunk loads fully resident, around 113.5 GB |
--trunk-gb | X | 16 | budget for pinned layers plus the streaming ring |
--cache-gb | X | 64 | budget for the routed-expert LRU cache |
--preset and the two -gb flags set the same two numbers, so a preset is just a
shorthand. Order matters if you mix them: a later flag wins, so
--preset server --cache-gb 40 gives you the server trunk budget with a 40 GB cache.
--presetwithout--trunkdoes nothing useful. Every preset assumes the trunk is streamed. Omit--trunkand the engine loads all 113.5 GB resident regardless of the budget you asked for.
Generation options
| flag | argument | default | |
|---|---|---|---|
--gen | N | 8 | tokens to generate. Ceiling 4096; prompts may be up to 32768 tokens |
--incremental | none | off | carry the KV cache and the recurrent state between tokens instead of re-running the whole prefix |
--tok | DIR | none | directory holding tiktoken.model and tokenizer_config.json |
Pass --incremental for any generation of length. Without it every step re-runs the
entire prefix, which is O(T²); with it, step 0 pays for the prompt and every later step
costs the same fixed amount. Both paths are gated on producing identical tokens, so this
is a pure speed choice.
Diagnostic options
| flag | argument | |
|---|---|---|
--config | PATH | model config; defaults to <model_dir>/config.json |
--layers | N | bind only the first N layers, for partial shard sets |
--out | FILE | JSON results (default k3_run.json) |
--dump-logits | PATH | float32 logits for the first step, for elementwise comparison |
--dump-cache-trace | DIR | writes expert_hist.json and expert_trace.bin, which tools/sim_cache.py replays |
Exit codes
Scripts can rely on these.
0 | success |
1 | a tensor failed to bind, or a forward pass failed |
2 | usage error, or a config that could not be read with confidence; the engine declines to guess |
4 | the run finished, but at least one routed expert failed to load, so the emitted ids are unsound. Distinct from 1 because the process otherwise succeeded, and it is the code that catches silent numerical corruption |
Environment variables
| variable | used by | |
|---|---|---|
HF_TOKEN | download-model.sh | HuggingFace token, read from the environment and never echoed |
OMP_NUM_THREADS | the engine | thread count, defaulting to all cores |
K3_TOK_FILES | tokenizer tools and CI | directory holding tiktoken.model, when it is not in a default location |
K3_MODEL_DIR | tools/budget.py | checkpoint directory, when not given as an argument |
Worked examples
# Smallest possible run, the 8 GB floor.
./bin/k3 ~/k3model --trunk ~/k3trunk --preset laptop \
--tok ~/k3model --prompt "Hello! My name is" --gen 16 --incremental
# Fastest per gigabyte. Pins 90 of 93 trunk layers.
./bin/k3 ~/k3model --trunk ~/k3trunk --preset server \
--tok ~/k3model --prompt "def fibonacci(n):" --gen 28 --incremental
# Hand-tuned split instead of a preset: everything to the trunk.
./bin/k3 ~/k3model --trunk ~/k3trunk --trunk-gb 110 --cache-gb 13 \
--tok ~/k3model --prompt-file prompt.txt --gen 32 --incremental
# Reproducible: ids in, ids out, no tokenizer, JSON results.
./bin/k3 ~/k3model --trunk ~/k3trunk --preset desktop \
--ids 1008,10484,318,15383,387 --gen 8 --incremental --out run.json
# Capture a cache trace, then replay it offline at any capacity.
./bin/k3 ~/k3model --trunk ~/k3trunk --preset workstation \
--ids 1008,10484,318,15383,387 --gen 8 --incremental \
--dump-cache-trace /tmp/trace
python3 tools/sim_cache.py /tmp/trace/expert_trace.bin
# Elementwise logit comparison against the PyTorch reference.
./bin/k3 ~/k3model --trunk ~/k3trunk --preset server \
--ids 3,4,5,6,7 --gen 1 --dump-logits /tmp/c_logits.bin
python3 tools/cmp_logits.py /tmp/c_logits.bin ref_logits.json
# Partial shard set: bind only the first 8 layers.
./bin/k3 ~/k3model --trunk ~/k3trunk --layers 8 \
--ids 1,2,3 --gen 1
# Under a hard memory ceiling, which is how the ladder was measured.
systemd-run --scope --user -q -p MemoryMax=8G -p MemorySwapMax=0 \
./bin/k3 ~/k3model --trunk ~/k3trunk --trunk-gb 2.5 --cache-gb 0.5 \
--ids 1008,10484,318,15383,387 --gen 8 --incremental
Choosing a preset
$ ./bin/k3 --list-presets
presets (trunk / expert-cache, in GB):
laptop 3.0 / 1.0 8.2 GB peak RSS. The floor. Runs, slowly.
desktop 16.0 / 10.0 31.9 GB peak RSS.
workstation 60.0 / 30.0 95.5 GB peak RSS; the expert cache starts to matter here.
server 110.0 / 13.0 ~128 GB peak RSS; 90 of 93 trunk layers pinned. Fastest.
max 110.0 / 109.0 ~224 GB peak RSS; trunk pinned and a large expert cache.
All presets stream the trunk, so they need --trunk <packed_dir>.
Run scripts/k3-doctor.sh to see which one this machine fits.
The boundaries come from the measured ladder, and the doctor keys on MemAvailable rather
than MemTotal:
if [ "$AVAIL_GB" -ge 192 ]; then PRESET=server; EXPECT="~19-21 s/token"
elif [ "$AVAIL_GB" -ge 96 ]; then PRESET=workstation; EXPECT="~24 s/token"
elif [ "$AVAIL_GB" -ge 32 ]; then PRESET=desktop; EXPECT="~28-31 s/token"
elif [ "$AVAIL_GB" -ge 10 ]; then PRESET=laptop; EXPECT="~32 s/token"
else PRESET=""; fi
Two things worth knowing before you pick:
maxis not faster thanserverin these measurements. The extra 96 GB buys nothing outside the noise floor.- Give the trunk memory before the expert cache. At a fixed 128 GB budget that was worth 1.69×. Allocation beats capacity has the data.
Reading the run report
The engine prints a memory plan, then a line per generated token, then a summary. Abridged
from a workstation run:
cache [final step]
requests : 1472 hits 1472 (100.00%) misses 0 evictions 729
TRUE resident hit rate 50.48%
I/O share of wall clock: 71.1% (trunk 62.4 s + experts 34.2 s of 135.8 s)
trunk [final]
pinned 48/93 layers, ring 1 slots
read 368.65 GB in 62.40 s (5908 MB/s)
PEAK RSS for the whole run: 94.74 GB <- quote this, not the plan
Three numbers carry the meaning:
TRUE resident hit rate: experts served from RAM. The rawhitscounter also counts experts the prefetcher pulled off disk moments earlier, so it reads 100% at every cache size; the resident figure is printed beneath it.I/O share of wall clock: whole-run disk time against total, measured between 41% and 61% across the ladder.PEAK RSS: fromgetrusage, after the run. This is the memory figure; the up-front plan runs slightly above it.
Common questions
Memory sits near 113 GB even at a small preset. --trunk was omitted. Without a packed
trunk directory the whole trunk is loaded resident; every preset assumes streaming.
A non-ASCII prompt tokenizes oddly. The shell re-encodes argv, so the engine receives
different bytes than you typed. Put the prompt in a file and use --prompt-file, which is
read verbatim.
--prompt/--prompt-file need --tok DIR. The tokenizer ships with the checkpoint, so
add --tok ~/k3model. The engine exits rather than guessing where the vocabulary lives. To
skip the tokenizer entirely, pass token ids with --ids.
Throughput is well below the table. Almost always storage. python3 tools/devbw.py <file-on-that-disk> measures the disk the way the engine reads it, using large random
O_DIRECT reads at queue depth 1 and 16, which dd does not. Network volumes run several
times slower than local NVMe; keep ~/k3trunk local.
The run refused to start over the KV cache. Context costs about 2.37 MB per position
regardless of budget, and the engine computes that up front rather than discovering it an
hour in. Shorten the request, or drop --incremental, which carries no KV cache at all.
Is the whole 1.56 TB needed? For generation, yes. For development, no: make test
needs nothing at all, and --layers N runs against partial shard sets.
macOS, Windows, WSL? The engine targets Linux. The tokenizer and config reader are portable C99 and are built portably in CI.
Part II: How it works
The problem: a model that does not fit
Kimi K3 has 2.78 trillion parameters and is 1.56 terabytes as shipped. No consumer machine can hold it, and waiting for better hardware does not help, because the wall is not speed, it is capacity.
But it is a mixture of experts, so only 16 of its 896 experts per layer fire for any given token and the rest sit asleep on disk. Keep the always-on part in memory, stream the sleeping experts, and it fits in 8.24 gigabytes on one CPU with no GPU.
The naive requirement is the one every parameter count implies.
So 5.56 terabytes is the number to beat.
Kimi K3 has 93 layers. Layer 0 is a plain dense feed-forward layer, so the other 92 layers route, and each of those picks the top 16 experts out of 896.
About 104 billion parameters are active for any given token, out of 2.78 trillion, which is 3.7 percent. The other 96.3 percent still has to exist somewhere reachable, but it does not have to be in RAM.
Counting the actual bytes on disk rather than guessing:
=== shard census: what the 1.56 TB actually is ===
shards : 96
total bytes : 1560936091448 (1.56 TB)
--- routed experts (the part that is streamed, never resident) ---
experts total : 82,432 (896 routed x 92 MoE layers)
bytes per expert : 17,547,264 exactly
= 33,030,144 params x 0.53125 bytes
= 0.5 bytes/nibble + 1/32 byte for the shared E8M0 scale
routed expert set : 82,432 x 17,547,264 = 1.447 TB
There are 82,432 routed experts, each occupying exactly 17,547,264 bytes. Together they are 1.447 terabytes, which is 93 percent of the entire checkpoint. Everything else (attention projections, routers, norms, embeddings) is the remaining 7 percent.
That census is the whole strategy in one picture. If those 1.447 terabytes can be reachable but never resident, the memory problem shrinks by more than an order of magnitude before a single kernel is written.
What is left is 56,743,648,000 parameters, or 113.49 gigabytes at bfloat16. Of that, 108.81 GB is the per-layer dense trunk and 4.70 GB is the embedding table plus the output head.
The four reductions
- 5,560 GB: every parameter at bfloat16, where we start.
- 1,560 GB: the checkpoint as shipped, because the experts already arrive at half a byte per weight.
- 113.49 GB: what has to be resident once routing means the experts never load.
- 8.24 GB: what is measured, once the trunk is streamed instead of held.
End to end that is a 675× reduction from the bfloat16 model and 189× from the shipped checkpoint. Nothing is approximated and no weight is dropped: the output at the bottom of that ladder is byte for byte the output at the top. The chart at the top of this document is that ledger drawn to scale.
The machine, and what it assumes
Every measurement here comes from one workstation: a two-socket AMD EPYC 7763 with 124 cores and no SMT, 228 GB of RAM, and 3.2 TB of NVMe. It also has four NVIDIA L40 GPUs, which sat completely idle for the entire campaign, because this engine has no GPU path.
--- ISA (note: AVX2 present, AVX-512 ABSENT) ---
avx avx2 fma sse4_2
--- memory ---
Mem: 228Gi 5.1Gi 207Gi 3.1Mi 18Gi 223Gi
MemTotal: 239308464 kB
MemAvailable: 233961008 kB
Hugepagesize: 2048 kB
There is no AVX-512. The engine needs AVX2 and FMA and nothing more, the instruction set on any desktop CPU from the last decade.
The storage numbers matter more than the CPU numbers, and one runs against expectation.
--- storage bandwidth, measured ---
O_DIRECT cold : 3.2 GB/s (dd bs=4M iflag=direct after drop_caches)
buffered warm : 2.3 GB/s
engine, trunk : 5373-6064 MB/s sustained during runs
NOTE O_DIRECT is FASTER than buffered here. That is the opposite of the usual
expectation, and it is why the engine opens the trunk O_DIRECT.
Reading with O_DIRECT, bypassing the page cache entirely, is faster here than
reading through it. That single measurement decided the whole I/O design.
One piece of hygiene, because a loaded machine is easy to measure badly:
--- measurement hygiene ---
unattended-upgrades: STOPPED and DISABLED before measurement (was using ~63% of a
core during the smoke run).
apt-daily.timer and apt-daily-upgrade.timer: DISABLED
A background package updater eating most of a core moves a timing by more than most optimisations do, so it goes off before anything is measured.
How much memory does the engine actually need? Multiplying config values by hand gives the
wrong answer in an instructive way, which is why tools/budget.py exists:
# Streamable only if ROUTED. The 2 SHARED experts sit in the same namespace and
# are NOT streamable, which is where hand arithmetic goes wrong.
def classify(name: str) -> str:
if ".block_sparse_moe.experts." in name:
return "routed_expert" # streamable: only 16 of 896 per token
if ".block_sparse_moe.shared_expert" in name:
return "shared_expert" # RESIDENT: runs on every token
if ".self_attn." in name:
return "attention" # resident
if "embed_tokens" in name or "lm_head" in name:
return "embedding" # resident
return "other" # norms, router gates, biases: resident
The two shared experts run on every token, so they belong in the resident set even though their tensor names sit next to the routed experts. Getting that wrong makes the floor look smaller than it is, which is the worst direction to be wrong in.
The codebase
Six C files compiled into one binary. No BLAS, no PyTorch, no ONNX runtime, no GPU library. The only dependencies are libm and OpenMP.
include/k3/
k3.h # the public header: config, weights, every kernel prototype
k3_cfg.h # config reader, header-only, refuses to substitute defaults
src/
core/k3_ops.c # every numeric kernel: RMSNorm, KDA, MLA, MoE, MXFP4 matmul
io/k3_st.c # safetensors reader, hand-written JSON scan, O_DIRECT reads
io/k3_load.c # locating one expert's bytes inside a shard
io/k3_trunk.c # streaming the dense trunk, pinned prefix plus a ring slot
cache/k3_cache.c # the routed-expert LRU cache and its batch prefetch
model/k3_bind.c # binding checkpoint tensor names to kernel arguments
tokenizer/k3_tok.h# byte-level BPE loaded from tiktoken.model
cli/k3_run.c # the k3 binary: memory plan, decode loop, reporting
tools/ # python: pack the trunk, replay the cache, verify against torch
benchmarks/ # the cgroup memory ladder and the split sweep
tests/ # fixtures, the tiny oracle, the 93-layer conformance run
CFLAGS = -O3 -std=gnu99 -Wall -Wextra -Wpointer-arith -Wshadow -Wvla \
-march=native -fopenmp -ffp-contract=off
LDFLAGS = -lm -fopenmp
The flag that looks unusual is -ffp-contract=off. By default a compiler may fuse a
multiply and an add into a single FMA, which changes the rounding. That is normally good.
Here it is a problem, because the scalar path, the OpenMP path and the AVX2 path must
produce bit-identical results, so that a performance change can never quietly become
an accuracy change.
1. build, warnings are failures
-> clean build, no diagnostics
test_ops 97784 bytes
k3_model 89392 bytes
k3_run 179736 bytes
The whole inference engine is 179,736 bytes, a 176 kilobyte binary whose job is to run a 1.56 terabyte model.
One check crosses machines. The tokenizer was run on the same input file under Windows and under Linux:
=== cross-platform tokenizer determinism ===
Linux gcc 13.3.0 x86_64
Windows gcc 16.1.0 x86_64
input src/k3.h (24,499 bytes) -> 6,862 ids
result IDENTICAL id streams
(a naive md5 of stdout DIFFERS by one byte: Windows text-mode stdout writes the
trailing newline as CRLF. That is the pipe, not the tokenizer.)
Two compilers on two operating systems produce the same 6,862 token ids from the same 24,499 bytes. The md5 sums differ by exactly one byte, and the reason is the line ending the shell added, not anything the tokenizer did.
Five invariants
The public header opens with five invariants that must hold. Each one is a place where a plausible-looking implementation produces a model that runs, emits fluent text, and is wrong, with no crash and no NaN to warn you.
A_logis indexed per head, not per channel. The checkpoint shipshead_dimfloats but only the firstnum_headsare meaningful; the rest are padding.- The UT-transform inverse is
(I + Akk)^-1. The sign is not a convention. Aqkretains its diagonal andAkkdoes not.- MLA uses NoPE, yet the 64 rope dimensions still exist and are still cached. Only the rotation is absent; dropping the slots changes the head width.
- The MoE routing bias steers selection only. The combining weights come from the unbiased sigmoid scores.
Each is ticked off below as its component arrives.
1. Reading a 1.56 TB checkpoint from its headers
The checkpoint is 96 safetensors files. The format is deliberately simple, which is what makes it possible to treat 1.56 terabytes as an index rather than as data.
Every file starts with an 8-byte little-endian length, then that many bytes of JSON describing every tensor, then the raw tensor bytes back to back. Nothing is compressed and nothing is interleaved.
No JSON library is used. The header can be tens of megabytes and only four fields per tensor are wanted, so the reader scans it directly.
/* Walk the header once, copy nothing we do not need. `p` sits just past the
* opening quote of the tensor name. */
static const char *st_scan_entry(const char *p, const char *end, K3Tensor *t)
{
const char *q = memchr(p, '"', (size_t)(end - p));
if (!q || (size_t)(q - p) >= sizeof t->name) return NULL;
memcpy(t->name, p, (size_t)(q - p));
t->name[q - p] = '\0';
const char *d = st_find_key(q, end, "dtype");
if (!d) return NULL;
t->dtype = st_dtype_code(d);
const char *s = st_find_key(q, end, "shape");
if (!s) return NULL;
t->rank = 0;
t->nelem = 1;
for (const char *c = s; c < end && *c != ']'; c++) {
if (*c >= '0' && *c <= '9') {
long v = strtol(c, (char **)&c, 10);
if (t->rank >= K3_ST_MAXRANK) return NULL;
t->shape[t->rank++] = v;
t->nelem *= (size_t)v;
}
}
/* offsets are RELATIVE to the start of the data section */
const char *o = st_find_key(q, end, "data_offsets");
if (!o) return NULL;
t->off = (size_t)strtoull(o, (char **)&o, 10);
while (o < end && (*o < '0' || *o > '9')) o++;
t->nbytes = (size_t)strtoull(o, (char **)&o, 10) - t->off;
return o;
}
Every tensor goes into a hash table keyed by a hash of its name. The choice of hash is not arbitrary.
/* Names are long and share deep prefixes
* ("language_model.model.layers.N.block_sparse_moe.experts.M...."), so the hash must
* mix every byte; a prefix-only or length-only hash would pile every expert of a
* layer into one bucket. */
static uint64_t fnv1a(const char *s)
{
uint64_t h = 1469598103934665603ull;
while (*s) { h ^= (unsigned char)*s++; h *= 1099511628211ull; }
return h;
}
Half a million tensor names that all begin with the same forty characters is a genuinely hostile input for a hash function. FNV-1a mixes on every byte, so the expert index at the end of the name still moves the result.
Reading a tensor afterwards has one wrinkle: O_DIRECT requires the offset and the length
to be multiples of the block size, and a tensor starts wherever the previous one ended.
int64_t k3_st_read_aligned(const K3St *s, int shard, int64_t off, int64_t nbytes,
void *buf, int64_t bufcap, int64_t *payload_off)
{
/* widen outward to the enclosing aligned window */
const int64_t lo = off & ~(int64_t)(K3_ST_ALIGN - 1);
const int64_t hi = (off + nbytes + K3_ST_ALIGN - 1) & ~(int64_t)(K3_ST_ALIGN - 1);
const int64_t len = hi - lo;
const int64_t pad = off - lo;
if (len > bufcap) return 0;
if (payload_off) *payload_off = pad;
int64_t got = 0;
while (got < len) {
ssize_t r = pread(dfd, (char *)buf + got, (size_t)(len - got), (off_t)(lo + got));
if (r <= 0) break; /* the last window may run past EOF */
got += r;
}
return got >= pad + nbytes ? nbytes : (got > pad ? got - pad : 0);
}
Note the break rather than a failure on a short read: the final aligned window of a shard
extends past the end of the file, which is expected, so the return value checks that the
payload was covered rather than that the whole window was.
indexed 497220 tensors from 96 shards in 0.27 s
Half a million tensors indexed in about a quarter of a second. This is what makes everything afterwards possible: the engine never reads a shard it does not need, so the 1.56 terabytes on disk is a catalogue, not a working set.
A parser that agrees with itself proves nothing, so the index is dumped and re-parsed independently in Python, comparing dtype, shape, offsets, and the widened float bit patterns.
# Bit patterns, not tolerances: widening bf16 to f32 is lossless.
c_bits = np.asarray(c_values[name], dtype=np.float32).view(np.uint32)
p_bits = ref.astype(np.float32).view(np.uint32)
if not np.array_equal(c_bits, p_bits):
bad = int(np.count_nonzero(c_bits != p_bits))
fail(f"{name}: {bad} of {c_bits.size} float32 bit patterns differ")
=== shard verification ===
shards: 96
bytes: 1560936091448
expected: 1560936091448
RESULT: EXACT MATCH
2. The config reader that refuses to guess
The model's dimensions come from the checkpoint's own config.json, and this is the first
place invariant four can silently bite.
Kimi K3 alternates two attention mechanisms. Most layers use one, every fourth uses the other, and the last two are both the second kind so the final layer always does global attention. The config lists those layers explicitly, and the list is one-based.
--- every value below is READ from the checkpoint, not assumed ---
config: config.json (nested shape) | hidden=7168 layers=93 vocab=163840
| 24 MLA + 69 KDA | experts 896 top16 shared2 | latent=3584
--- KDA/MLA layer map (ONE-based, from full_attn_layers) ---
full_attn_layers (24, all MLA): 4,8,12,16,20,24,28,32,36,40,44,48,52,56,60,64,68,72,76,80,84,88,92,93
note 92 AND 93 are both MLA - the report (2.1) places an extra Gated MLA layer
at the end of the backbone so the final layer always does global attention.
kda_layers (69): every other layer.
Every one of those numbers is read from the file; none is compiled in. They land in one struct, which is the entire model on one screen:
typedef struct {
int hidden; /* 7168 */
int n_layers; /* 93 */
int vocab; /* 163840 */
float rms_eps; /* 1e-5 */
/* Kimi Delta Attention. 69 of the 93 layers. */
int kda_heads; /* 96 */
int kda_head_dim; /* 128, and d_k == d_v */
int conv_k; /* 4, depthwise, causal, SiLU fused */
float gate_lb; /* -5.0, the decay lower bound */
/* Gated MLA. 24 of the 93 layers. */
int n_heads; /* 96 */
int q_lora; /* 1536 */
int kv_lora; /* 512 */
int qk_nope; /* 128 */
int qk_rope; /* 64, PRESENT BUT NEVER ROTATED */
int v_head; /* 128 */
int mla_out_gate; /* 1 */
/* Stable LatentMoE. 92 of the 93 layers. */
int n_experts; /* 896 */
int topk; /* 16 */
int n_shared; /* 2, full width, added UNWEIGHTED */
int latent; /* 3584, the routed-expert width */
int moe_inter; /* 3072 */
float routed_scale; /* 1.0 */
int moe_renorm; /* 1 */
int latent_norm; /* 1, RMSNorm on the AGGREGATE, not per expert */
/* the single dense layer, layer 0 */
int first_dense; /* 1 */
int dense_inter; /* 33792 */
int attn_res_block; /* 12. Boundaries fire when layer_idx % this == 0. */
float situ_b1; /* 4.0 */
float situ_b2; /* 25.0 */
int n_full_attn; /* 24 */
int *full_attn; /* ONE-BASED layer indices */
} K3Cfg;
That struct is the contract between the checkpoint and every kernel. If it is right, the model is Kimi K3. If any field is wrong, the model is something else that still speaks English.
Consider what a permissive reader would do. The released config nests its fields one level
deeper than a fixture does, so a reader that only knows the flat shape finds nothing it
recognises. If it then fills in defaults, two things happen: the SiTU betas get 4.0 and
25.0, which are the correct values, so nothing looks wrong, and full_attn_layers
comes back empty, so all 93 layers run as KDA and the 24 global-attention layers vanish.
The model loads, streams, decodes, and produces grammatical English from an architecture
that is not Kimi K3.
/* An absent field is an ERROR, never a default. Missing names are accumulated so
* the message lists all of them at once. */
static int cfg_req_int(jval root, const char *key, int *out,
const char **missing, int *nmissing)
{
jval v = json_get(root, key);
if (v.type != JSON_NUM) { /* absent OR the wrong type */
if (*nmissing < K3_CFG_MAXMISS) missing[(*nmissing)++] = key;
return 0;
}
*out = (int)v.num;
return 1;
}
[no_layermap]
k3_cfg: no_layermap.json is missing 1 required field(s):
full_attn_layers
refusing to substitute defaults: a config this reader cannot
fully understand would silently produce a DIFFERENT model.
ok correctly rejected no_layermap.json
[bad_layer_index]
k3_cfg: bad_layer_index.json full_attn_layers[2] = 999 is outside 1..93
(the list is ONE-based)
ok correctly rejected bad_layer_index.json
A config reader is about a hundred and fifty lines of the most boring code in the project, and it is one of exactly two places that can hand you a different model without telling you.
3. The tokenizer, byte for byte
The other one is the tokenizer. Kimi K3 uses a byte-level BPE with 163,584 ranks plus 256
special tokens, shipped as a tiktoken.model file.
The loader reads that file straight into the vendored BPE structures. It rests on three assumptions, each of which produces a tokenizer working perfectly on ASCII and diverging on everything else:
- The merge keys are bytes, not code points. A key that happens to decode as valid UTF-8 must still be treated as its raw bytes.
- Ranks come from the file. They are not derived from frequency at load time.
- The added-token block is appended after the ranks, so an added token's id is 163,584 plus its index, not its position in a merged table.
The test compares the C tokenizer against the Python tiktoken library case by case,
through files rather than command-line arguments:
oracle : tiktoken 0.13.0
method : token-for-token comparison; every case passed through a FILE, never argv
(argv is re-encoded to the active code page on Windows and would compare
different bytes on every non-ASCII case)
PASS han only 2 ids
PASS japanese 6 ids
PASS korean 5 ids
PASS cyrillic 4 ids
PASS arabic 7 ids
PASS emoji zwj 5 ids
PASS code python 11 ids
PASS json 19 ids
tokenizer parity: 45/45 cases match
Then whole files are pushed through and decoded back:
roundtrip: 48353 bytes -> 14797 ids -> 48353 bytes : PASS <- k3_ops.c
roundtrip: 24499 bytes -> 6862 ids -> 24499 bytes : PASS <- k3.h
roundtrip: 201775 bytes -> 52671 ids -> 201775 bytes : PASS <- REPORT.md
roundtrip: 53444 bytes -> 12145 ids -> 53444 bytes : PASS <- modeling_kimi_k3.py
Two hundred kilobytes of markdown becomes 52,671 token ids and comes back as exactly the same two hundred kilobytes. Every later claim about identical output rests on the tokenizer being deterministic.
/* Greedily merge the lowest-rank adjacent pair. Everything here is BYTES. */
static int tok_encode_piece(const Tok *t, const unsigned char *p, int n, int *out)
{
int parts[K3_TOK_MAXPIECE + 1], np = n + 1;
for (int i = 0; i <= n; i++) parts[i] = i; /* byte boundaries */
for (;;) {
int best = -1, bestrank = INT_MAX;
for (int i = 0; i + 2 < np; i++) {
const int r = tok_rank(t, p + parts[i], parts[i + 2] - parts[i]);
if (r >= 0 && r < bestrank) { bestrank = r; best = i; }
}
if (best < 0) break; /* no mergeable pair left */
memmove(&parts[best + 1], &parts[best + 2],
(size_t)(np - best - 2) * sizeof(int));
np--;
}
for (int i = 0; i + 1 < np; i++)
out[i] = tok_rank(t, p + parts[i], parts[i + 1] - parts[i]);
return np - 1;
}
The loop keeps a list of slice boundaries and repeatedly joins whichever adjacent pair has the lowest rank, which is what makes the result independent of any tie-breaking order.
4. Reduction one: the experts already ship at half a byte
The first of the four reductions, and the largest single one.
The routed experts do not ship at bfloat16. They ship in MXFP4, a microscaling 4-bit float format. Each weight is a 4-bit nibble indexing a 16-entry table, and every group of 32 consecutive weights shares one 8-bit exponent.
Half a byte plus one thirty-second of a byte is 0.53125 bytes per weight, and one expert has 33,030,144 parameters, so one expert is exactly 17,547,264 bytes. That matches the census to the byte.
The format is checked against the released checkpoint rather than against documentation:
{
"note": "Kimi K3 MXFP4 bytes from the released checkpoint. w = E2M1[nibble] * 2^(scale - 127), one scale per 32 elements.",
"source": "language_model.model.layers.1.block_sparse_moe.experts.0.w1",
"rows": 64, "packed_cols": 1792, "scale_cols": 112,
"logical_width": 3584, "group_size": 32,
"e2m1_lut": [0.0, 0.5, 1.0, 1.5, 2.0, 3.0, 4.0, 6.0,
-0.0, -0.5, -1.0, -1.5, -2.0, -3.0, -4.0, -6.0]
}
Both halves of the decode are lookup tables, and building them is the only setup the format needs.
/* E2M1: sign, two exponent bits, one mantissa bit. Sixteen values in total. */
static const float K3_E2M1[16] = {
0.0f, 0.5f, 1.0f, 1.5f, 2.0f, 3.0f, 4.0f, 6.0f,
-0.0f, -0.5f, -1.0f, -1.5f, -2.0f, -3.0f, -4.0f, -6.0f
};
/* Byte -> its two weights, so the inner loop does one lookup, not two shifts. */
static void k3_pair_init(void)
{
for (int b = 0; b < 256; b++) {
K3_E2M1_PAIR[b][0] = K3_E2M1[b & 0x0F]; /* low nibble = EVEN element */
K3_E2M1_PAIR[b][1] = K3_E2M1[b >> 4]; /* high nibble = ODD element */
}
}
/* Scale byte -> power of two. 255 is NaN by spec, mapped to 0 to contain damage. */
static void k3_e8m0_init(void)
{
for (int b = 0; b < 256; b++)
K3_E8M0[b] = (b == 255) ? 0.0f : ldexpf(1.0f, b - 127);
}
Now the nibble order, which is a convention no statistic can check.
"expected_swapped_nibbles": {
"note": "what you get if the low nibble is treated as the ODD element.
Statistics are identical; positions are wrong."
}
Every mean, every standard deviation, every histogram of the swapped version is identical to the correct one, because it is the same multiset of numbers. Only the positions differ. A verification that checks distributions would pass a matrix with every adjacent pair of weights transposed.
The obvious way to use these weights is to decode them into floats and then do a normal matrix multiply. Pricing that:
One expert at 17.55 MB becomes 132 MB once expanded to float32. Each token touches 16 experts across 92 layers (1,472 experts), so decoding them all would mean writing out 194 gigabytes per token of pure format conversion, before a single multiply-accumulate.
So nothing is ever dequantised. The matrix multiply reads packed nibbles directly.
/* y[rows] = x[in] . W[rows][in], W stored as MXFP4. Nothing is dequantized. */
void k3_matmul_mxfp4(float *y, const float *x, const unsigned char *packed,
const unsigned char *scales, int in, int rows, int group)
{
const int ngroup = (in + group - 1) / group;
const int rowbytes = (in + 1) / 2; /* two nibbles per byte */
#pragma omp parallel for schedule(static)
for (int o = 0; o < rows; o++) {
const unsigned char *pb = packed + (size_t)o * rowbytes;
const unsigned char *sb = scales + (size_t)o * ngroup;
double acc = 0.0; /* double, always */
for (int g = 0; g < ngroup; g++) {
const float s = K3_E8M0[sb[g]];
if (s == 0.0f) continue; /* a NaN scale zeroes the group */
const int i0 = g * group;
const int n = (i0 + group <= in) ? group : (in - i0);
float wf[64];
/* low nibble is the EVEN weight, high nibble the ODD one */
const int half = n / 2;
for (int k = 0; k < half; k++) {
const unsigned char b = pb[(i0 / 2) + k];
wf[2 * k] = K3_E2M1_PAIR[b][0]; /* low nibble -> even index */
wf[2 * k + 1] = K3_E2M1_PAIR[b][1]; /* high nibble -> odd index */
}
if (n & 1) wf[n - 1] = K3_E2M1_PAIR[pb[(i0 / 2) + half]][0];
double part = 0.0;
for (int k = 0; k < n; k++) part += (double)x[i0 + k] * (double)wf[k];
acc += part * (double)s;
}
y[o] = (float)acc;
}
}
Two details are worth pausing on. if (s == 0.0f) continue handles a scale byte of 255,
which the MXFP4 specification defines as NaN; mapping it to zero means one corrupt byte
kills one group of 32 weights instead of turning an entire row into NaN and poisoning
every downstream layer. And if (n & 1) handles a group with an odd number of elements.
with a group size of 32 that can never happen on this checkpoint, and it is written
anyway. That is the difference between code that works and code that is correct, and it
costs one line.
PASS mxfp4 64 rows x 3584 elems, EXACT on released checkpoint bytes
Not "within tolerance", but exact. Both sides read identical bytes off identical weights, so there is nothing that could legitimately differ, and the test is written with zero tolerance to say so.
That is reduction one. The experts arrive at 0.53125 bytes per weight instead of 2, which takes 5.45 TB of expert weights down to 1.447 TB, and never expanding them saves another 194 GB of memory traffic per token.
5. Kernels with a floating point contract
Every code path must produce the same bits. RMSNorm is the most used kernel in the model.
void k3_rmsnorm(float *out, const float *x, const float *w, int n, float eps)
{
double ss = 0.0; /* double, not float */
for (int i = 0; i < n; i++) ss += (double)x[i] * (double)x[i];
const float inv = (float)(1.0 / sqrt(ss / (double)n + (double)eps));
for (int i = 0; i < n; i++) out[i] = x[i] * inv * (w ? w[i] : 1.0f);
}
Two details are load-bearing: the accumulator is a double even though every input and output is a float, and epsilon goes inside the square root rather than outside.
/* Four accumulators partitioned by i % 4. This pins the summation order, so a
* 4-wide vector loop adds the same numbers in the same sequence. */
static float k3_dot4(const float *x, const float *w, int n)
{
double a0 = 0.0, a1 = 0.0, a2 = 0.0, a3 = 0.0;
int i = 0;
for (; i + 3 < n; i += 4) {
a0 += (double)x[i] * (double)w[i];
a1 += (double)x[i + 1] * (double)w[i + 1];
a2 += (double)x[i + 2] * (double)w[i + 2];
a3 += (double)x[i + 3] * (double)w[i + 3];
}
for (; i < n; i++) a0 += (double)x[i] * (double)w[i];
return (float)((a0 + a1) + (a2 + a3)); /* the parentheses are the contract */
}
A bfloat16 value is the top 16 bits of a float32 with the bottom 16 dropped, so widening is a shift left by 16 and it is exact. That is why the trunk can be streamed in its shipped precision with no accuracy question at all, a point that becomes important much later.
void k3_matmul_bf16(float *y, const float *x, const uint16_t *W, int in, int out)
{
#pragma omp parallel for schedule(static) if (out > 64)
for (int o = 0; o < out; o++) {
const uint16_t *row = W + (size_t)o * in;
int i = 0;
double acc;
#if defined(__AVX2__)
{
__m256d v = _mm256_setzero_pd();
for (; i + 3 < in; i += 4) {
/* bf16 -> f32 is a 16-bit shift: no table, no rounding */
const __m128i h = _mm_loadl_epi64((const __m128i *)(row + i));
const __m128i b32 = _mm_slli_epi32(_mm_cvtepu16_epi32(h), 16);
const __m256d wd = _mm256_cvtps_pd(_mm_castsi128_ps(b32));
const __m256d xd = _mm256_cvtps_pd(_mm_loadu_ps(x + i));
v = _mm256_add_pd(v, _mm256_mul_pd(wd, xd)); /* NOT fmadd */
}
double a[4];
_mm256_storeu_pd(a, v);
acc = (a[0] + a[1]) + (a[2] + a[3]);
}
#else
{
double a0 = 0.0, a1 = 0.0, a2 = 0.0, a3 = 0.0;
for (; i + 3 < in; i += 4) {
a0 += (double)k3_bf16f(row[i ]) * (double)x[i ];
a1 += (double)k3_bf16f(row[i + 1]) * (double)x[i + 1];
a2 += (double)k3_bf16f(row[i + 2]) * (double)x[i + 2];
a3 += (double)k3_bf16f(row[i + 3]) * (double)x[i + 3];
}
acc = (a0 + a1) + (a2 + a3);
}
#endif
for (; i < in; i++) acc += (double)k3_bf16f(row[i]) * (double)x[i];
y[o] = (float)acc;
}
}
Both branches accumulate into four doubles, both partition by i % 4, and both reduce as
(a0 + a1) + (a2 + a3). The vector path is the scalar path with the same additions
performed in the same order, four at a time.
Note the multiply-add. _mm256_add_pd of an _mm256_mul_pd is deliberately not
_mm256_fmadd_pd. A fused multiply-add rounds once instead of twice and is therefore
more accurate, which is exactly the problem: it would give a different answer from
the scalar loop, and a hardware capability must never change the output.
Proving the AVX2 and scalar paths agree cannot be done with a tolerance check, because a tolerance check happily passes a kernel that quietly reassociated its sum. So the benchmark hashes the output instead: FNV-1a over the exact bit pattern of every float in the result, built twice, once with AVX2 and once without, then diff the digests.
tolerance: atol=1.0e-05 rtol=1.0e-04 (from MANIFEST.json)
PASS rmsnorm n=384 worst=0.01x tol
PASS situ_glu n=48 worst=0.00x tol
bound check |out|=100.000 must be <= b1*b2=100.0 : ok
PASS kda_decay H=4 D=16 tok=4 max|dg|=4.768e-07 max|dalpha|=1.788e-07
PASS mla n=768 worst=0.08x tol
H=4 qh=32 (nope 24 + rope 8) v=16 kv_lora=32 scale=0.176777
PASS mxfp4 64 rows x 3584 elems, EXACT on released checkpoint bytes
PASS matmul_bf16 n=129 bit-identical to k3_matmul
22 passed, 0 failed, 0 skipped
The worst case across all 22 kernels is 8 percent of the allowed tolerance, and two are exact rather than merely close.
Benchmarked at the model's own dimensions on the smallest configuration, the split is about 36 seconds of trunk reading, 11 seconds of expert reading and 10 seconds of arithmetic. Eighty percent of a token is waiting for a disk, which is why the second half of this document is about I/O and not about kernels.
6. Reduction two: KDA, attention with a memory that never grows
Sixty-nine of the 93 layers use Kimi Delta Attention, and its property that matters here is easy to state: its memory does not grow with context length.
A standard attention layer stores a key and a value for every token it has seen, so its cache grows linearly forever. KDA instead keeps one fixed-size matrix per head and updates it in place as tokens arrive.
Ninety-six heads times a 128×128 matrix per head is the entire memory of a KDA layer, at any sequence length. Across all 93 layers that is 626.25 megabytes, whether you feed it ten tokens or a million.
That plot is the argument for the whole design. The flat line is KDA. The rising line is what the other 24 layers cost, and it crosses this machine's memory somewhere around 100,000 tokens. If all 93 layers behaved like it, the model would not fit at any context length worth having.
First, the projections go through a short depthwise causal convolution of width four with SiLU fused in.
/* Causal depthwise conv with SiLU fused. State is carried across calls. */
void k3_shortconv(float *y, const float *x, const float *w, float *state,
int channels, int k, int T)
{
const int hist = k - 1; /* guard on hist, not on buf */
float *buf = hist ? (float *)malloc((size_t)hist * sizeof(float)) : NULL;
if (hist && !buf) k3_fatal_oom("ShortConv history", (size_t)hist * sizeof(float));
for (int c = 0; c < channels; c++) {
if (hist) {
if (state) memcpy(buf, state + (size_t)c * hist, (size_t)hist * sizeof(float));
else memset(buf, 0, (size_t)hist * sizeof(float));
}
for (int t = 0; t < T; t++) {
const float cur = x[(size_t)t * channels + c];
float acc = w[(size_t)c * k + hist] * cur; /* taps run oldest to newest */
for (int j = 0; j < hist; j++)
acc += w[(size_t)c * k + j] * buf[j];
for (int j = 0; j + 1 < hist; j++) buf[j] = buf[j + 1];
if (hist > 0) buf[hist - 1] = cur;
y[(size_t)t * channels + c] = acc * sigmoidf_(acc); /* SiLU, fused */
}
if (state && hist) memcpy(state + (size_t)c * hist, buf, (size_t)hist * sizeof(float));
}
free(buf);
}
Note the guard on hist rather than on buf: with a kernel width of one there is no
history at all, malloc(0) is allowed to return NULL, and a check on the pointer would
silently skip the entire convolution and leave the output untouched.
Then the queries and keys are L2-normalised, a sum of squares, not a mean of squares, which is a different function that looks nearly identical in code.
Then the decay gate, which is invariant one.
void k3_kda_decay(float *g, float *alpha, const float *z, const float *A_log,
const float *dt_bias, int H, int D, float lb)
{
for (int h = 0; h < H; h++) {
const float a = expf(A_log[h]); /* PER HEAD, not per channel */
for (int d = 0; d < D; d++) {
const int i = h * D + d;
const float u = a * (z[i] + dt_bias[i]);
const float gi = lb * sigmoidf_(u); /* in (lb, 0] */
g[i] = gi;
alpha[i] = expf(gi); /* in (e^lb, 1] */
}
}
}
The gate lower bound is −5, so alpha lands between e^-5 and 1. Near one means this key
channel keeps almost all of its history; near e^-5 means it forgets almost everything,
per channel and per token.
void k3_kda_step(float *S, float *o, const float *q, const float *k,
const float *v, const float *alpha, float beta, int dk, int dv)
{
/* 1. decay: scale ROW i of S by alpha[i], per key channel */
for (int i = 0; i < dk; i++) {
float *row = S + (size_t)i * dv;
const float a = alpha[i];
for (int j = 0; j < dv; j++) row[j] *= a;
}
/* 2. read the state along k: u = S^T k */
float *u = (float *)calloc((size_t)dv, sizeof(float));
if (!u) k3_fatal_oom("KDA recurrence temporary", (size_t)dv * sizeof(float));
for (int i = 0; i < dk; i++) {
const float ki = k[i];
if (ki == 0.0f) continue;
const float *row = S + (size_t)i * dv;
for (int j = 0; j < dv; j++) u[j] += ki * row[j];
}
/* 3. rank-one delta write: (v - u) is the prediction error */
for (int i = 0; i < dk; i++) {
const float ki = k[i];
if (ki == 0.0f) continue;
float *row = S + (size_t)i * dv;
for (int j = 0; j < dv; j++) row[j] += ki * beta * (v[j] - u[j]);
}
/* 4. output from the ALREADY UPDATED state: o = S^T q */
for (int j = 0; j < dv; j++) o[j] = 0.0f;
for (int i = 0; i < dk; i++) {
const float qi = q[i];
if (qi == 0.0f) continue;
const float *row = S + (size_t)i * dv;
for (int j = 0; j < dv; j++) o[j] += qi * row[j];
}
free(u);
}
That calloc happens after step one has already scaled the state, so an early return
on allocation failure would leave the recurrent matrix permanently decayed but never
updated. Every subsequent token would then be computed from a state that is quietly wrong,
with nothing to indicate it. That is why the failure path aborts instead of returning.
void k3_kda_layer(float *out, const float *x, const K3KdaW *w, const K3Cfg *c,
int T, float *state, float *scratch)
{
const int E = c->hidden, H = c->kda_heads, D = c->kda_head_dim;
const int P = H * D, K = c->conv_k, hist = K - 1;
float *q = scratch; float *k = q + (size_t)T * P;
float *v = k + (size_t)T * P; float *z = v + (size_t)T * P;
float *al = z + (size_t)T * P; float *bt = al + (size_t)T * P;
float *o = bt + (size_t)T * H; float *gb = o + (size_t)T * P;
float *wr = gb + P; float *fa = wr + P;
/* 1. projections */
for (int t = 0; t < T; t++) {
const float *xt = x + (size_t)t * E;
k3_mmw(q + (size_t)t * P, xt, w->q, w->wdt, E, P);
k3_mmw(k + (size_t)t * P, xt, w->k, w->wdt, E, P);
k3_mmw(v + (size_t)t * P, xt, w->v, w->wdt, E, P);
k3_mmw(bt + (size_t)t * H, xt, w->b, w->wdt, E, H);
k3_mmw(fa, xt, w->f_a, w->wdt, E, D); /* one low-rank pair, all heads */
k3_mmw(z + (size_t)t * P, fa, w->f_b, w->wdt, D, P);
}
/* 2. ShortConv with fused SiLU, carrying state across calls */
float *cs = state ? state + (size_t)H * D * D : NULL;
k3_shortconv(q, q, w->q_conv, cs ? cs : NULL, P, K, T);
k3_shortconv(k, k, w->k_conv, cs ? cs + (size_t)P * hist : NULL, P, K, T);
k3_shortconv(v, v, w->v_conv, cs ? cs + (size_t)2 * P * hist : NULL, P, K, T);
/* 3. L2Norm on q and k ONLY, per head. v is deliberately left alone. */
for (int t = 0; t < T; t++)
for (int h = 0; h < H; h++) {
l2norm_(q + (size_t)t * P + (size_t)h * D, D, 1e-6f);
l2norm_(k + (size_t)t * P + (size_t)h * D, D, 1e-6f);
}
/* 4/5. beta and the decay chain */
for (int t = 0; t < T; t++) {
for (int h = 0; h < H; h++) bt[(size_t)t * H + h] = sigmoidf_(bt[(size_t)t * H + h]);
k3_kda_decay(z + (size_t)t * P, al + (size_t)t * P, z + (size_t)t * P,
w->A_log, w->dt_bias, H, D, c->gate_lb);
}
/* 6. recurrence, per head, with q pre-scaled by d_k^-0.5 */
float *S = state;
float *Sown = NULL;
if (!S) {
Sown = (float *)calloc((size_t)H * D * D, sizeof(float));
if (!Sown) k3_fatal_oom("KDA recurrent state", (size_t)H * D * D * sizeof(float));
S = Sown;
}
const float qscale = 1.0f / sqrtf((float)D);
for (int t = 0; t < T; t++)
for (int h = 0; h < H; h++) {
const size_t off = (size_t)t * P + (size_t)h * D;
for (int i = 0; i < D; i++) wr[i] = q[off + i] * qscale;
k3_kda_step(S + (size_t)h * D * D, o + off, wr, k + off, v + off,
al + off, bt[(size_t)t * H + h], D, D);
}
/* 7/8/9. head-wise RMSNorm, THEN the gate, THEN the output projection */
for (int t = 0; t < T; t++) {
const float *xt = x + (size_t)t * E;
float *ot = o + (size_t)t * P;
for (int h = 0; h < H; h++)
k3_rmsnorm(ot + (size_t)h * D, ot + (size_t)h * D, w->o_norm, D, c->rms_eps);
k3_mmw(gb, xt, w->g, w->wdt, E, P);
for (int i = 0; i < P; i++) ot[i] *= sigmoidf_(gb[i]);
k3_mmw(out + (size_t)t * E, ot, w->o, w->wdt, P, E);
}
free(Sown);
}
Nine numbered steps, and the numbering is not decoration. Step 3 normalises q and k
and deliberately leaves v alone. Step 6 pre-scales the query by one over the square root
of the head dimension before the recurrence rather than after it.
Steps 7, 8 and 9 are where invariants two and three live. The released model ends the
layer with a fused kernel from the fla library called FusedRMSNormGated, which takes
the raw gate and applies the sigmoid internally. The reference implementation instead does
an explicit RMSNorm followed by a multiply by the sigmoid of the gate. Those are the same
function only if the fused kernel is norm-first and gate-second, so
tools/verify_kda.py proves it rather than assuming it:
The plausible alternatives, gating before norming or norming the gate, both run cleanly and produce a different model.
No test catches that except comparing against the released code, which is exactly what that script does.
one KDA layer : 443740384 params (887.48 MB at bf16)
69 KDA layers : 61.24 GB at bf16 (KDA attention ONLY, not the whole trunk)
full trunk : 113.49 GB at bf16, 56,743,648,000 always-active params
one expert : 33030144 params (17.55 MB at MXFP4)
all 82432 experts: 1.45 TB at MXFP4 <- streamed from NVMe
per-sequence state, FIXED regardless of context:
KDA recurrent : 217.06 MB at bf16
ShortConv : 15.26 MB at bf16
MLA KV : 2.37 MB per position (24 MLA layers, EXPANDED k and v, fp32)
19.38 GB at 8192 context
310.04 GB at 131072 context
allocating and running ONE full-width KDA layer (fp32)...
weights: 1.77 GB
ran 4 tokens in 0.17 s (44 ms/token)
output all finite: YES, max |y| = 0.000003
state non-zero : yes
The two lines under "FIXED regardless of context" are the payoff. The KDA state is 217 MB and the convolution history is 15 MB, and neither number moves no matter how long the sequence gets, against 310 GB for the other attention mechanism at 131,072 positions.
That is reduction two: 69 of 93 layers have a memory cost completely independent of how much text you feed them.
7. Reduction three: MLA, one latent instead of ninety-six heads
The other 24 layers do global attention, because a purely recurrent stack cannot look back at an arbitrary earlier token with full precision. They use Gated Multi-head Latent Attention, and they are the ones that cost memory per position.
A normal attention layer with 96 heads would store, for every position, a key and a value for each head. MLA instead projects the token down into one small shared latent, caches only that, and rebuilds the per-head keys and values from it when they are needed.
The latent is 512 dimensions for the key and value content plus 64 more, and those 64 are where invariant four lives.
/* NoPE: the 64 rope dimensions are projected and cached, but never rotated. */
static void k3_mla_project(float *q, float *kv, const float *x, const K3MlaW *w,
const K3Cfg *c)
{
float qa[K3_MAX_QLORA]; /* down to 1536, norm, back up */
k3_mmw(qa, x, &w->q_a_proj, c->q_lora, c->hidden);
k3_rmsnorm(qa, qa, w->q_a_norm, c->q_lora, c->rms_eps);
k3_mmw(q, qa, &w->q_b_proj, c->n_heads * (c->qk_nope + c->qk_rope), c->q_lora);
/* one 576-wide projection: the entire per-position cache for this layer */
k3_mmw(kv, x, &w->kv_a_proj, c->kv_lora + c->qk_rope, c->hidden);
k3_rmsnorm(kv, kv, w->kv_a_norm, c->kv_lora, c->rms_eps);
/* the trailing qk_rope floats are left unnormalised and unrotated */
}
The head width is 192 (128 content dimensions plus those 64 carried-but-unrotated ones), so the softmax scale is the inverse square root of 192 rather than of 128. Getting that wrong is a change of about 22 percent in every attention score, and it produces perfectly readable output.
for (int t = 0; t < T; t++) {
const int p = cached + t;
for (int h = 0; h < H; h++) {
const float *qt = q + ((size_t)t * H + h) * qh;
float m = -INFINITY;
for (int s = 0; s <= p; s++) { /* causal: s <= p */
const float *ks = K3_KV_AT(s) + (size_t)h * kvd;
const float *kr = K3_ROPE_AT(s); /* shared slot */
double d = 0.0;
for (int i = 0; i < qn; i++) d += (double)qt[i] * (double)ks[i];
/* the rope slot is UNROTATED but still scored, and the SAME 64
* values serve every head */
for (int i = 0; i < qr; i++) d += (double)qt[qn + i] * (double)kr[i];
sc[s] = (float)d * scale;
if (sc[s] > m) m = sc[s];
}
double z = 0.0;
for (int s = 0; s <= p; s++) { sc[s] = expf(sc[s] - m); z += sc[s]; }
The score is two dot products added together. The first runs over the 128 content
dimensions, which are per head. The second runs over the 64 rope dimensions, which are
shared: K3_ROPE_AT(s) takes no head index, so all 96 heads score against the same 64
numbers. That is what makes the cache 576 wide instead of 96 × 320, and skipping the
second term is the quiet way to get a model that still speaks.
Ninety-six heads times 320 floats is 30,720 floats per position per layer. The latent is 576. That is 53 times smaller for mathematically identical output, and it is the difference between a context length you can use and one you cannot.
Because the KV cache is the one thing that grows, the engine refuses to start rather than discovering the problem an hour in:
/* The context limit is the MLA KV cache, not any array size. */
if (incremental) {
const double kv_need = (double)(np + gen + 1) * K3_KV_BYTES_PER_POS;
const double avail = mem_available_bytes();
if (avail > 0.0 && kv_need > avail * 0.9) {
fprintf(stderr,
"\nREFUSING: the KV cache for %d positions needs %s but only %s is\n"
"available. This is a MEMORY limit, not an engine ceiling: MLA caches\n"
"expanded k and v in fp32 across 24 layers, so context costs ~2.37 MB per\n"
"position regardless of budget. Shorten the request, or use full\n"
"recompute (drop --incremental), which carries no KV cache at all.\n",
np + gen + 1, kb, ab);
return 2;
}
}
"This is a MEMORY limit, not an engine ceiling" saves somebody an afternoon of looking for a hardcoded constant that does not exist.
That is reduction three. Context costs 2.37 MB per position instead of 125.
8. Attention residuals: layers that look back
One more structural piece, unusual enough to be worth a section even though it costs no memory.
In a normal transformer each layer adds its output to a running residual stream. Kimi K3 does something different: each layer attends over the outputs of every preceding block, where a block is twelve layers, and learns how much of each to mix in.
void k3_attn_res(float *out, const float *src, const float *fold,
int nsrc, int n, float eps)
{
float *score = (float *)malloc((size_t)nsrc * sizeof(float));
if (!score) k3_fatal_oom("AttnRes scores", (size_t)nsrc * sizeof(float));
for (int s = 0; s < nsrc; s++) {
const float *v = src + (size_t)s * n;
double ss = 0.0;
for (int i = 0; i < n; i++) ss += (double)v[i] * (double)v[i];
const float inv = (float)(1.0 / sqrt(ss / (double)n + (double)eps));
double acc = 0.0; /* key: the NORMALISED source */
for (int i = 0; i < n; i++) acc += (double)(v[i] * inv) * (double)fold[i];
score[s] = (float)acc;
}
float m = score[0];
for (int s = 1; s < nsrc; s++) if (score[s] > m) m = score[s];
double z = 0.0;
for (int s = 0; s < nsrc; s++) { score[s] = expf(score[s] - m); z += score[s]; }
for (int i = 0; i < n; i++) out[i] = 0.0f;
for (int s = 0; s < nsrc; s++) {
const float p = (float)(score[s] / z);
const float *v = src + (size_t)s * n; /* the RAW source, not the key */
for (int i = 0; i < n; i++) out[i] += p * v[i];
}
free(score);
}
The keys are normalised before scoring and the values are the raw sources. Norming the values as well is the obvious-looking mistake, and it quietly rescales the residual stream.
void k3_decoder_layer_inc(float *h, float *block_residual, int *n_blocks,
const K3LayerW *w, const K3Cfg *c, int layer_idx,
int T, float *state, float *scratch,
float *kvc, float *ropec, int cached, int cap)
{
const int E = c->hidden;
const int maxb = c->n_layers / c->attn_res_block + 2;
float *pref = scratch; /* [T][E] the running residual */
float *tmp = pref + (size_t)T * E; /* [T][E] module output */
float *hin = tmp + (size_t)T * E; /* [T][E] normalised layer input */
float *foldA = hin + (size_t)T * E; /* [E] attention aggregator */
float *foldM = foldA + E; /* [E] mlp aggregator */
float *src = foldM + E; /* [maxb+1][E] source stack */
float *dgu = src + (size_t)(maxb) * E; /* [2*dense_inter] */
float *sub = dgu + (size_t)2 * c->dense_inter; /* scratch for the module */
/* norm gain and scoring projection collapse into one vector */
for (int i = 0; i < E; i++) {
foldA[i] = w->attn_res_norm[i] * w->attn_res_proj[i];
foldM[i] = w->mlp_res_norm[i] * w->mlp_res_proj[i];
}
memcpy(pref, h, (size_t)T * E * sizeof(float));
int have_prefix = 1; /* mirrors "prefix_sum is not None" */
/* aggregation before attention, only when snapshots already exist */
if (*n_blocks > 0) {
for (int t = 0; t < T; t++) {
for (int b = 0; b < *n_blocks; b++)
memcpy(src + (size_t)b * E,
block_residual + ((size_t)t * maxb + b) * E,
(size_t)E * sizeof(float));
memcpy(src + (size_t)(*n_blocks) * E, pref + (size_t)t * E,
(size_t)E * sizeof(float));
k3_attn_res(h + (size_t)t * E, src, foldA, *n_blocks + 1, E, c->rms_eps);
}
}
/* block boundary: snapshot the running residual, then CLEAR it */
if (layer_idx % c->attn_res_block == 0) {
for (int t = 0; t < T; t++)
memcpy(block_residual + ((size_t)t * maxb + *n_blocks) * E,
pref + (size_t)t * E, (size_t)E * sizeof(float));
(*n_blocks)++;
have_prefix = 0;
}
/* whichever attention was bound for this layer */
for (int t = 0; t < T; t++)
k3_rmsnorm(hin + (size_t)t * E, h + (size_t)t * E, w->in_norm, E, c->rms_eps);
if (w->kda) k3_kda_layer(tmp, hin, w->kda, c, T, state, sub);
else k3_mla_cached(tmp, hin, w->mla, c, T, sub, kvc, ropec, cached, cap);
if (have_prefix) for (size_t i = 0; i < (size_t)T * E; i++) pref[i] += tmp[i];
else { memcpy(pref, tmp, (size_t)T * E * sizeof(float)); have_prefix = 1; }
/* aggregation before the MLP, with no emptiness guard */
for (int t = 0; t < T; t++) {
for (int b = 0; b < *n_blocks; b++)
memcpy(src + (size_t)b * E,
block_residual + ((size_t)t * maxb + b) * E,
(size_t)E * sizeof(float));
memcpy(src + (size_t)(*n_blocks) * E, pref + (size_t)t * E,
(size_t)E * sizeof(float));
k3_attn_res(h + (size_t)t * E, src, foldM, *n_blocks + 1, E, c->rms_eps);
}
for (int t = 0; t < T; t++)
k3_rmsnorm(hin + (size_t)t * E, h + (size_t)t * E, w->post_norm, E, c->rms_eps);
if (w->moe) {
int idx[K3_MAX_TOPK]; float wt[K3_MAX_TOPK];
k3_moe(tmp, hin, w->moe, c, T, idx, wt, sub);
}
}
The layer never asks which kind it is; it checks whether w->kda was bound, so the layer
map from config.json is the only thing that decides. The two aggregations are not
symmetric: the one before attention is skipped when no snapshots exist yet, and the one
before the MLP has no such guard, because the reference has none either.
Does that do anything observable? The reference forward pass draws it. Printing the maximum absolute activation after every layer:
L45 KDA MoE 41.2 s |h| max 47.714829
L46 KDA MoE 38.9 s |h| max 62.183392
L47 MLA MoE 44.3 s |h| max 76.281532
L48 KDA MoE 36.1 s |h| max 2.902113
L49 KDA MoE 39.5 s |h| max 4.353188
Layer 47 to layer 48: the activation magnitude goes from 76.28 to 2.90, a factor of 26, in one layer. That is a block boundary: the running prefix was snapshotted and cleared, so the next layer starts from a fresh small residual.
Seven sawteeth, one per block, each climbing for twelve layers and then collapsing. Nobody drew that shape; the model did. The biggest peak is 136.7 at layer 59, dropping to 1.78 at layer 60. When a measured curve has exactly the period the code says it should, that is a decent sign the code matches the model.
9. Picking 16 experts of 896
The router scores every one of the 896 experts and picks sixteen. This is invariant five, the subtlest of the lot.
/* Score all n_experts, pick the top-k, weight them. The bias steers SELECTION only. */
void k3_router(int *idx, float *w, const float *x, const float *W, const float *bias,
int hidden, int n_experts, int topk, int renorm, float routed_scale)
{
float sc[K3_MAX_EXPERTS], ch[K3_MAX_EXPERTS];
#pragma omp parallel for schedule(static)
for (int e = 0; e < n_experts; e++) {
double acc = 0.0;
for (int i = 0; i < hidden; i++)
acc += (double)x[i] * (double)W[(size_t)e * hidden + i];
sc[e] = 1.0f / (1.0f + expf(-(float)acc)); /* the UNBIASED score */
ch[e] = sc[e] + (bias ? bias[e] : 0.0f); /* the SELECTION score */
}
char taken[K3_MAX_EXPERTS] = {0}; /* top-k by repeated max */
for (int j = 0; j < topk; j++) {
int best = -1;
for (int e = 0; e < n_experts; e++)
if (!taken[e] && (best < 0 || ch[e] > ch[best])) best = e;
taken[best] = 1;
idx[j] = best;
w[j] = sc[best]; /* the UNBIASED score again */
}
if (renorm) {
float s = 0.0f;
for (int j = 0; j < topk; j++) s += w[j];
if (s > 0.0f) for (int j = 0; j < topk; j++) w[j] /= s;
}
for (int j = 0; j < topk; j++) w[j] *= routed_scale;
}
sc[e] and ch[e] are both computed and used for different things. The biased score picks
the winners; the unbiased score weights them. Collapsing those two into one variable is a
two-character edit that changes the model.
/* Stable LatentMoE for one token.
*
* Six steps, and step 4 is the one people get wrong: the RMSNorm is applied to the
* AGGREGATE of the weighted expert outputs, not to each expert individually. Norming
* per expert and then summing is a different function.
*
* The two shared experts run on the ORIGINAL full-width input, not the latent, and
* their output is added UNWEIGHTED. They are not part of the top-k sum. */
void k3_moe(float *out, const float *x, const K3MoeW *w, const K3Cfg *c,
int T, int *idx, float *wt, float *scratch)
{
const int E = c->hidden, L = c->latent, I = c->moe_inter;
float *z = scratch, *accL = z + L, *gu = accL + L, *act = gu + 2 * I;
float *edn = act + I;
for (int t = 0; t < T; t++) {
const float *xt = x + (size_t)t * E;
/* 1. route on the FULL width, not the latent */
k3_router(idx, wt, xt, w->gate, w->gate_bias, E,
c->n_experts, c->topk, c->moe_renorm, c->routed_scale);
/* 2. down-project into the expert latent */
k3_mmw(z, xt, w->down, w->wdt, E, L);
/* 3. run the chosen experts, accumulating in the latent */
for (int i = 0; i < L; i++) accL[i] = 0.0f;
if (w->src && w->src->getmany) w->src->getmany(w->src, w->layer, idx, c->topk);
for (int j = 0; j < c->topk; j++) {
K3ExpertQ q;
if (w->src->get(w->src, w->layer, idx[j], &q) != 0) {
k3_expert_drops++; /* counted, never silent */
continue;
}
k3_matmul_mxfp4(gu, z, q.p1, q.s1, L, I, K3_MXFP4_GROUP);
k3_matmul_mxfp4(gu + I, z, q.p3, q.s3, L, I, K3_MXFP4_GROUP);
k3_situ_glu(act, gu, I, c->situ_b1, c->situ_b2);
k3_matmul_mxfp4(edn, act, q.p2, q.s2, I, L, K3_MXFP4_GROUP);
for (int i = 0; i < L; i++) accL[i] += wt[j] * edn[i];
}
/* 4. norm the AGGREGATE, not each expert */
if (c->latent_norm) k3_rmsnorm(accL, accL, w->latent_norm, L, c->rms_eps);
/* 5. back up to full width */
float *ot = out + (size_t)t * E;
k3_mmw(ot, accL, w->up, w->wdt, L, E);
/* 6. shared experts, on the ORIGINAL input, added UNWEIGHTED */
const int SI = I * c->n_shared;
k3_mmw(gu, xt, w->sh1, w->wdt, E, SI);
k3_mmw(gu + SI, xt, w->sh3, w->wdt, E, SI);
k3_situ_glu(act, gu, SI, c->situ_b1, c->situ_b2);
k3_mmw(edn, act, w->sh2, w->wdt, SI, E);
for (int i = 0; i < E; i++) ot[i] += edn[i];
}
}
The activation inside each expert is SiTU-GLU, a gated unit with both halves passed through bounded tanh functions.
void k3_situ_glu(float *y, const float *x, int n, float b1, float b2)
{
const float *gate = x;
const float *up = x + n;
for (int i = 0; i < n; i++) {
const float g = gate[i];
/* the sigmoid takes the UNCAPPED gate */
const float a = b1 * tanhf(g / b1) * sigmoidf_(g);
const float u = b2 * tanhf(up[i] / b2);
y[i] = a * u;
}
}
The sigmoid reads the uncapped gate g, not the capped b1 * tanh(g / b1). Feeding it
the capped value gives a function that is still smooth, still bounded, still produces
fluent output, and is a different activation.
Because both factors are bounded, the product can never exceed 4 × 25 = 100, and the fixture drives it deliberately to that exact analytic cap:
PASS situ_glu n=48 worst=0.00x tol
bound check |out|=100.000 must be <= b1*b2=100.0 : ok
A fixture that only tested the near-linear region would pass an implementation with the caps left out entirely, because for small inputs the tanh is almost the identity.
Now the thing to remember for later. The Kimi K3 technical report describes a training technique called Quantile Balancing whose entire purpose is to flatten expert usage across the pool, so that no small set of experts dominates.
That is good for the model, and it is going to be very bad for the cache. Hold that thought.
One last thing about the continue in the expert loop. A dropped expert means one token
was computed with fifteen-sixteenths of its routed sum, and the run still finishes and
still prints a plausible token. So the engine counts them globally and refuses to exit
successfully:
/* Silent numerical corruption that exits 0 is indistinguishable from a good run. */
if (k3_expert_drops) {
fprintf(stderr,
"\nRUN INVALID: %ld routed expert load(s) failed and were dropped from\n"
"the MoE sum. The token ids above are CORRUPT. Re-run; if this repeats,\n"
"the shard set or the storage is at fault.\n", k3_expert_drops);
return 4;
}
return 0;
Note where that sits: after all the reporting and all the frees. A corrupt run still prints its full diagnostics and still cleans up. It simply does not exit zero.
10. Packing the trunk: 93 layers, one read each
The resident set is down to 113.49 GB. That is still far more than a consumer machine has, and it is the last big number in the ledger.
The trunk is spread across 96 shard files, interleaved with the experts. Reading one layer's worth means finding a few dozen tensors scattered through a 17 GB file. So before running anything, it is rewritten once into a layout that suits how it is actually read.
The key fact that makes this cheap is that each layer's trunk tensors already sit in one contiguous run inside its shard. So packing is 93 range copies, not a scatter-gather.
# 93 sequential range copies, one per layer, each verified contiguous first.
for L in range(n_layers):
names = [n for n in index if n.startswith(f"model.layers.{L}.") and not is_expert(n)]
shards = {index[n] for n in names}
if len(shards) != 1:
die(f"layer {L} spans {len(shards)} shards; refusing")
runs = sorted((offsets[n][0], offsets[n][1]) for n in names)
lo, hi = runs[0][0], runs[-1][1]
covered = sum(b - a for a, b in runs)
if covered != hi - lo:
# a gap would drag expert bytes along with it
die(f"layer {L} is not contiguous: {hi - lo - covered} bytes of gap")
out_off = align_up(out_off, ALIGN) # head aligned for O_DIRECT
copy_range(shard_path(shards.pop()), lo, hi, out_fh, CHUNK)
out_off += align_up(hi - lo, ALIGN) # and the tail, so reads never overrun
If a layer's tensors were not contiguous, copying the whole span would drag expert bytes along with it and bloat the trunk file. Rather than silently produce a 400 GB trunk, the packer stops.
packed 10/93 layers, 12.90 GB, 25 s (521 MB/s)
packed 20/93 layers, 24.31 GB, 47 s (520 MB/s)
packed 30/93 layers, 36.14 GB, 74 s (488 MB/s)
packed 40/93 layers, 47.55 GB, 99 s (480 MB/s)
packed 50/93 layers, 59.38 GB, 127 s (466 MB/s)
packed 60/93 layers, 70.79 GB, 153 s (462 MB/s)
packed 70/93 layers, 82.62 GB, 181 s (457 MB/s)
packed 80/93 layers, 94.02 GB, 208 s (451 MB/s)
packed 90/93 layers, 105.86 GB, 236 s (448 MB/s)
packed 93/93 layers, 108.81 GB, 244 s (447 MB/s)
wrote trunk.bin: 108.81 GB across 93 layers
largest layer run: 2.341 GB <- the streaming slot size
Four minutes, once, and layer L lives at a known offset and can be read in a single call. The last line sizes everything downstream: the largest layer run is 2.341 GB, so any buffer that has to hold one arbitrary layer must be at least that big.
11. Reduction four: streaming the trunk turns a floor into a dial
The trunk is 108.81 GB and every layer of it is used on every token. There is no sparsity to exploit and nothing to skip. So the question is not how to avoid reading it, it is where to keep it.
The design is a pinned prefix plus one rotating slot. Whatever fits in the budget gets pinned permanently, and the rest cycles through a single ring buffer. Critically, it is a prefix and not a cache.
The engine walks layers 0 through 92 in the same order on every token. That is a cyclic scan, and a cyclic scan is the pathological case for least-recently-used eviction: by the time layer 0 comes round again it is the least recently used thing in the cache, so it has always just been evicted.
An LRU of 90 slots over a 93-layer cycle achieves a hit rate of exactly zero. Pinning the first N layers instead gives a deterministic hit rate of N/93, which for N = 90 is 96.8 percent. The obvious data structure is not merely suboptimal here; it is wrong in the worst possible direction, returning zero where the trivial approach returns almost one.
/* Pinned count and slot size are mutually dependent, so iterate to a fixed point. */
size_t slot = tr->max_run; /* start assuming nothing is pinned */
int npin = 0;
for (int pass = 0; pass < 4; pass++) {
size_t avail = budget > slot * (size_t)nring ? budget - slot * (size_t)nring : 0;
int n = 0;
size_t used = 0;
while (n < tr->n_layers && used + tr->lay[n].nbytes <= avail) {
used += tr->lay[n].nbytes;
n++;
}
size_t need = 0; /* largest run still not pinned */
for (int L = n; L < tr->n_layers; L++)
if (tr->lay[L].nbytes > need) need = tr->lay[L].nbytes;
if (need == 0) need = 4096; /* everything pinned: degenerate but valid */
if (n == npin && need == slot) break; /* converged */
npin = n;
slot = k3_align_up(need, K3_TRUNK_ALIGN);
}
There is a memory detail that matters a lot. O_DIRECT reads pin their destination pages,
and a 2.37 GB slot on 4 KB pages is about 578,000 pages, pinned and unpinned 93 times per
token, nearly 54 million page operations per token purely in bookkeeping. So the arenas
are allocated on 2 MB hugepages instead.
/* Add up EVERYTHING before allocating anything. */
const double need_b = w_trunk + w_model + w_cache + w_state + w_buf + w_kv;
const double have = mem_available_bytes(); /* MemAvailable, not MemFree */
if (need_b > have * 0.95) {
fprintf(stderr,
"\nREFUSING TO START: this needs %s and the machine has %s "
"available, a shortfall of %s.\n"
"Options: a larger box, a smaller --cache-gb, or fewer --layers.\n",
b6, b1, b2);
return 1;
}
That plan is explicitly a forecast and not a result. It omits the safetensors index, about 78 MB at full scale, reports requested budgets rather than actual reservations, and cannot see fragmentation. So the engine also measures the peak resident set afterwards and labels it, in the output, as the number to quote.
/* One full forward over T tokens, writing logits for the LAST position only. */
static int forward(Weights *w, const K3Cfg *c, K3Cache *cache, const int *ids, int T,
float *logits_last, float *scratch, float *h, float *br, float *kstate)
{
const int E = c->hidden;
const int maxb = c->n_layers / c->attn_res_block + 2;
const int P = c->kda_heads * c->kda_head_dim;
const size_t kper = (size_t)P * c->kda_head_dim + (size_t)3 * P * (c->conv_k - 1);
for (int t = 0; t < T; t++)
k3_embed_row(h + (size_t)t * E, w->mb.embed, w->mb.wdt, ids[t], E);
memset(br, 0, (size_t)T * maxb * E * sizeof(float));
/* incremental decode carries the KDA state forward; full recompute rebuilds it */
if (!w->kvc) memset(kstate, 0, kper * (size_t)w->n_bound * sizeof(float));
int nb = 0;
for (int L = 0; L < w->n_bound; L++) {
/* bring this layer in, and hint the next one so its read overlaps */
if (w->trunk) {
if (k3_trunk_bind(w->trunk, c, L, &w->lay[L]) != 0) {
fprintf(stderr, "trunk bind failed at layer %d\n", L);
return -1;
}
k3_trunk_prefetch(w->trunk, L + 1);
}
if (w->lay[L].lay.moe) {
w->lay[L].moe.src = &cache->src;
w->lay[L].moe.layer = L;
}
if (w->kvc && w->mla_slot[L] >= 0) {
const size_t kvper = (size_t)w->kv_cap * c->n_heads * (c->qk_nope + c->v_head);
const size_t rpper = (size_t)w->kv_cap * c->qk_rope;
const int mi = w->mla_slot[L];
k3_decoder_layer_inc(h, br, &nb, &w->lay[L].lay, c, L, T,
kstate + kper * (size_t)L, scratch,
w->kvc + kvper * (size_t)mi,
w->ropec + rpper * (size_t)mi,
w->cached, w->kv_cap);
} else {
k3_decoder_layer_inc(h, br, &nb, &w->lay[L].lay, c, L, T,
kstate + kper * (size_t)L, scratch,
NULL, NULL, 0, 0);
}
}
/* one model-level aggregator, beyond the two in every layer */
if (w->mb.out_res_norm && w->mb.out_res_proj) {
float *fold = scratch;
float *src = fold + E;
for (int i = 0; i < E; i++) fold[i] = w->mb.out_res_norm[i] * w->mb.out_res_proj[i];
for (int t = 0; t < T; t++) {
for (int b = 0; b < nb; b++)
memcpy(src + (size_t)b * E, br + ((size_t)t * maxb + b) * E,
(size_t)E * sizeof(float));
memcpy(src + (size_t)nb * E, h + (size_t)t * E, (size_t)E * sizeof(float));
k3_attn_res(h + (size_t)t * E, src, fold, nb + 1, E, c->rms_eps);
}
}
float *nrm = scratch;
k3_rmsnorm(nrm, h + (size_t)(T - 1) * E, w->mb.norm, E, c->rms_eps);
k3_mmw(logits_last, nrm, w->mb.lm_head, w->mb.wdt, E, c->vocab);
return 0;
}
That is the entire model in seventy lines. Embed the tokens, walk 93 layers binding each one as it arrives, apply one final aggregator across all the block snapshots, normalise the last position and project it to 163,840 logits.
The if (!w->kvc) line is the whole distinction between incremental decode and full
recompute in a single condition.
The first time the engine touched all 2.78 trillion parameters:
Kimi K3, pure C, released checkpoint
shards : 96
prompt : 5 tokens, generating 2
indexed 497220 tensors from 96 shards in 0.70 s
memory plan
trunk (STREAMED) 16.00 GB
embed + lm_head 4.70 GB
expert cache 6.00 GB
recurrent state 626.25 MB
buffers 6.68 MB
KV cache 0.00 B
TOTAL 27.33 GB
available 65.91 GB
trunk stream: 108.81 GB packed, 10/93 layers PINNED (13.16 GB), ring 1 x 2.37 GB
reads use O_DIRECT (page cache bypassed)
deterministic hit rate 10.8% (a cyclic scan defeats LRU, so a pinned
prefix is used instead)
peak RSS after loading weights: 4.78 GB (the plan above is a forecast, this is measured)
expert cache: 341 slots x 17.56 MB = 5.99 GB (0.41% of the 1.45 TB expert pool)
STEP TOKEN SECONDS CACHE HIT READ GB TOK/S
--------------------------------------------------------------------
0 2494 167.84 35.0 83.91 0.006
1 9 171.65 39.3 94.05 0.006
--------------------------------------------------------------------
2 tokens in 339.5 s, 169.75 s/token average
PEAK RSS for the whole run: 25.83 GB <- quote this, not the plan
It works, and it is 169.75 seconds per token, which is unusable. Almost everything in Part IV is about walking that number down to 10.66, and the thing that fixes it is not the obvious one.
Pushing the other way, with nothing pinned at all and an expert cache under two gigabytes:
trunk stream: 108.81 GB packed, 0/93 layers PINNED (0.00 GB), ring 2 x 2.37 GB
deterministic hit rate 0.0%
expert cache: 113 slots x 17.56 MB = 1.98 GB (0.14% of the 1.45 TB expert pool)
STEP TOKEN SECONDS CACHE HIT READ GB TOK/S
--------------------------------------------------------------------
0 17374 57.08 22.8 99.70 0.018
1 20829 27.72 0.0 25.83 0.036
2 10 26.95 0.0 25.83 0.037
3 427 27.25 0.0 25.83 0.037
--------------------------------------------------------------------
cache [final step]
requests : 1472 hits 0 (0.00%) misses 1472 evictions 1472
Zero layers pinned. An expert cache holding 0.14 percent of the pool. A cache hit rate
of exactly zero, with all 1,472 requests missing and all 1,472 evicting. And it still
emits 17374, 20829, 10, 427, the same correct answer every other configuration gives.
Holding the trunk resident instead spends 150 seconds just loading weights before the first token, asks for 315 GB, and needs a machine with 755 GB free to be allowed to start.
/* Offset and length are both 4096-aligned, so this is a plain pread with no fixup. */
static int load_run(K3Trunk *tr, int L, unsigned char *dst)
{
const K3Run *r = &tr->lay[L];
size_t got = 0;
while (got < r->nbytes) {
const ssize_t n = pread(tr->fd, dst + got, r->nbytes - got,
(off_t)(r->off + got));
if (n <= 0) return -1; /* a short read is a corrupt layer */
got += (size_t)n;
}
tr->bytes_read += got;
return 0;
}
12. An LRU cache for the experts
Now the other side of the read path: 1,472 expert fetches per token, each 17.56 MB, drawn from a 1.45 TB pool.
/* Three slot states, not two: a key, EMPTY, or INFLIGHT. */
static int pick_victim(K3Cache *c)
{
int best = -1;
uint64_t oldest = (uint64_t)-1;
for (int i = 0; i < c->nslot; i++) {
if (c->key_of[i] == K3_SLOT_INFLIGHT) continue; /* being read into RIGHT NOW */
if (c->key_of[i] == K3_SLOT_EMPTY) return i; /* free, take it */
if (c->pinned[i]) continue;
if (c->used_at[i] < oldest) { oldest = c->used_at[i]; best = i; }
}
return best;
}
Three details in twelve lines. INFLIGHT slots are skipped entirely rather than treated as
candidates, so a slot cannot be claimed twice. EMPTY returns immediately, because a free
slot is always better than evicting a live one. And pinned slots are skipped after the
empty test, so pinning never blocks the cheap path.
static int cache_getmany(K3ExpertSrc *self, int layer, const int *experts, int n)
{
K3Cache *c = (K3Cache *)self->ctx;
int slots[K3_MAX_TOPK];
/* phase 1: reserve serially, so no two experts take the same slot */
int nres = 0;
for (int j = 0; j < n; j++) {
int s = cache_lookup(c, layer, experts[j]);
if (s >= 0) { slots[j] = -1; continue; } /* already resident */
s = cache_pick_victim(c);
if (s < 0) { slots[j] = -1; continue; }
c->slot_id[s] = K3_SLOT_INFLIGHT;
slots[j] = s;
nres++;
}
/* phase 2: read in parallel, in disk-offset order */
int order[K3_MAX_TOPK];
cache_sort_by_offset(c, layer, experts, slots, n, order);
#pragma omp parallel for schedule(dynamic)
for (int k = 0; k < n; k++) {
const int j = order[k];
if (slots[j] < 0) continue;
if (!k3_expert_load_direct(c->st, layer, experts[j], c->arena + slot_off(c, slots[j])))
slots[j] = -2;
}
/* phase 3: publish only what arrived */
for (int j = 0; j < n; j++) {
if (slots[j] >= 0) c->slot_id[j] = expert_key(layer, experts[j]);
else if (slots[j] == -2) c->slot_id[j] = K3_SLOT_EMPTY;
}
return nres;
}
The slot sizing carries a similar guard. An expert is 17,547,264 bytes, which happens to be exactly 4,284 × 4,096, so the alignment the reads need holds on the released checkpoint by coincidence. Code that assumed the alignment rather than enforcing it would work on every shipped weight, which is why the cache fixture deliberately uses a non-conforming expert size. The released weights cannot exercise that path, so the test data was built to.
int64_t k3_expert_load(const K3St *s, const K3ExpertRef *r, unsigned char *buf)
{
if (r->contiguous) { /* one coalesced 17.55 MB read */
K3Tensor t;
memset(&t, 0, sizeof t);
t.name = (char *)"expert";
t.shard = r->shard;
t.off = r->off;
t.nbytes = r->nbytes;
t.dtype = K3_DT_U8;
t.ndim = 1;
t.shape[0] = r->nbytes;
return k3_st_read(s, &t, buf);
}
/* fallback: six separate reads, one per tensor */
static const char *W[3] = { "w1", "w2", "w3" };
char name[256];
int64_t got = 0;
for (int i = 0; i < 3; i++) {
snprintf(name, sizeof name, EXPERT_FMT, r->layer, r->expert, W[i], "weight_packed");
const K3Tensor *p = k3_st_find(s, name);
snprintf(name, sizeof name, EXPERT_FMT, r->layer, r->expert, W[i], "weight_scale");
const K3Tensor *c = k3_st_find(s, name);
if (!p || !c) return got;
got += k3_st_read(s, p, buf + (p->off - r->off));
got += k3_st_read(s, c, buf + (c->off - r->off));
}
return got;
}
That synthetic K3Tensor named "expert" does not correspond to anything in the
checkpoint. It exists purely so the six-tensor run can be handed to the same short-read
loop that reads any other tensor.
4. streaming expert cache
PASS prefetch_reads <= hits requests 24, hits 24, prefetch 24
PASS mixed batch and serial 0 of 24 wrong
CACHE TESTS PASSED
Random cold is the pattern that matters, because the router picks experts by relevance and the file lays them out by index, and those two orders have nothing to do with each other.
13. How big should that cache be? Ask the trace
Measuring cache size directly would mean running the model once per cache size on a machine
large enough to hold the whole expert pool. There is a shortcut, and it is a good one:
routing does not depend on the cache. The same prompt picks the same experts in the same
order no matter what the cache does, so one run can record every (layer, expert) request
and that single trace can be replayed at any capacity under any policy.
def belady(trace, cap):
"""Evict whatever is needed furthest in the future. A ceiling, not a policy."""
nxt = defaultdict(deque)
for i, k in enumerate(trace):
nxt[k].append(i)
resident, hits = set(), 0
for k in trace:
nxt[k].popleft()
if k in resident:
hits += 1
continue
if len(resident) >= cap:
victim = max(resident, key=lambda r: nxt[r][0] if nxt[r] else 1 << 60)
resident.discard(victim)
resident.add(k)
return hits / len(trace)
trace: 100096 requests, 10010 distinct experts, about 68 token(s)
distinct experts touched: 10010 of 82432 (12.14% of the pool)
if nothing were cached: 25.83 GB per token
total reuse: 90086 of 100096 requests are repeats (90.0%)
CACHE SLOTS LRU BELADY PIN+LRU GB READ/TOK SEC/TOK
----------------------------------------------------------------------------
8 GB 455 36.24% 39.42% 37.82% 16.47 13.35
16 GB 911 36.24% 42.61% 39.42% 16.47 13.35
32 GB 1823 36.24% 48.99% 42.57% 16.47 13.35
64 GB 3647 36.24% 61.74% 48.66% 16.47 13.35
128 GB 7294 49.19% 84.59% 62.86% 13.12 10.64
192 GB 10941 90.00% 90.00% 90.00% 2.58 2.09
1450 GB 82633 90.00% 90.00% 90.00% 2.58 2.09
compulsory misses: 10010 (every expert must be read at least once), a ceiling of
90.00% hit rate for ANY policy at ANY size on this trace.
Sixty-eight tokens touched 10,010 distinct experts, only 12.14 percent of the pool. Holding every expert this trace touched would need 175.65 GB, against 1,446 GB for the whole pool. Even a perfect cache for this workload is a fraction of the model.
Two things jump out. First, LRU is completely flat from 8 GB to 64 GB: an eightfold increase in capacity buys nothing at all. Second, Belady over that same range climbs from 39 percent to 62 percent, which means the flatness belongs to the policy and not to the workload.
At 64 GB there is a 25.5 point gap between what LRU achieves and what an optimal policy would achieve at exactly the same memory, which says the promising direction is a better replacement policy rather than more RAM.
CAVEAT, and it matters
This trace was recorded during a run that re-prefills the whole prefix every step, so
the same experts are legitimately touched ~68 times. Steady-state incremental decode
has far less reuse, and its hit rates will be LOWER than this curve suggests. Treat
these numbers as an upper bound on what caching can do, not a forecast.
Keep that table in mind. Part IV measures it directly, and the two do not agree.
Part III: Validation
The gate ladder
A tiny oracle first
The middle level is a tiny model with the same tensor graph: thirteen layers, hidden size 128, vocabulary 256.
Why thirteen and not five? Because attention residuals operate in blocks of twelve, and their failure modes cannot appear until two blocks are complete and a third is in progress. A five-layer model would never exercise the boundary logic at all.
3. full-model oracle gates on the 13-layer reference
checkpoint: 628 tensors loaded
layer map (0-based): KKKMKKKMKKKMM (M=MLA, K=KDA; dense layer = 0)
attn_res boundaries at: 0 3 6 9 12
prompt_ids 12, full_ids 32, tf_pred 32
all layer weights bound
GATE 1 teacher forcing : 32/32 positions match tf_pred
generated span : 20/20 <- must be exact
GATE 2 greedy decode : 20/20 generated tokens match full_ids
GATE 3 incremental : 20/20 generated tokens match full_ids <- KV cache + carried KDA state
VERDICT: ENGINE MATCHES THE REFERENCE EXACTLY
Three different execution paths giving identical token ids. All three are exact, because on a discrete argmax there is no such thing as a small error.
And the caveat that goes directly underneath it:
The strongest-sounding line in gates.txt is about a toy model. "VERDICT: ENGINE
MATCHES THE REFERENCE EXACTLY" refers to the 13-layer, hidden-128, vocab-256 oracle
- NOT the 2.8T checkpoint.
Every fixture was designed to fail a specific plausible wrong implementation. The router fixture reorders its top two experts on five of six rows, so an implementation that ignores the routing bias fails it. The SiTU-GLU fixture spans inputs from 0.1 to 1000, because in the near-linear region the bounded tanh is indistinguishable from the identity. A test that a plausible bug would survive is not a test.
Proving it on the full checkpoint
Two questions remain: is each of the 93 layers wired correctly, and does the whole stack produce the right numbers.
# A wrong binding does not miss by 1e-6, it misses by ~1.
tol = 1.2e-7 * math.sqrt(width) * 50
L87 KDA PASS 3 stages worst 0.00x budget 44s
L88 KDA PASS 3 stages worst 0.00x budget 46s
L89 KDA PASS 3 stages worst 0.00x budget 43s
L90 KDA PASS 3 stages worst 0.00x budget 45s
L91 MLA PASS 3 stages worst 0.00x budget 46s
L92 MLA PASS 3 stages worst 0.00x budget 49s
==============================================================================
LAYERS 0..92 93 passed, 0 failed (69 KDA, 24 MLA) 5639 s total
worst stage across all passing layers: 0.00x of its rounding budget
==============================================================================
VERDICT: ALL LAYERS CONFORM
Note the last two rows: L91 and L92 are both MLA, the architectural quirk the config announced at the very beginning, now visible in the verification output.
For the second question, a full forward pass from an independent implementation:
reference forward: 93 layers, 5 prompt ids, hidden 7168, vocab 163840
embedded 5 ids (BF16)
L0 KDA dense 20.7 s |h| max 0.145457
L1 KDA MoE 44.7 s |h| max 0.187096
L2 KDA MoE 41.8 s |h| max 0.463652
L3 MLA MoE 39.2 s |h| max 1.284419
...
L92 MLA MoE 36.7 s |h| max 31.064180
final position: argmax token 2494, logit 15.021948, mean -0.462536, max 15.021948
total 3608.5 s
Three thousand six hundred seconds. One hour for a single forward pass over five tokens in PyTorch, loading and freeing one layer at a time because the fp32 expansion would be about 227 GB. The C engine did the same forward in 169.73 seconds.
===== ELEMENTWISE LOGIT COMPARISON =====
prompt cross-check : both sides ran [3, 4, 5, 6, 7]
vocab : 163840
C argmax : 2494 (logit 15.021946)
reference argmax : 2494 (logit 15.021948)
top-10 overlap : 10/10
max |diff| : 7.867813e-06
relative to max|ref| : 5.237545e-07 (budget 5.1e-04)
mean |diff| : 1.231738e-06
correlation : 1.000000000
VERIFIED: the C engine's logits match the torch reference over the FULL 93-layer
stack on the released checkpoint, elementwise, and the argmax token agrees.
The largest disagreement anywhere in 163,840 values is 7.87 × 10⁻⁶, about a thousandth of the allowed budget, and the correlation prints as 1.000000000.
And the caveat, which the corpus states about itself:
The logit parity run used prompt ids 3,4,5,6,7, which decode to: $%&'(
That is synthetic junk, not text.
That same run reports `KV cache 0.00 B`, so the KV path is not exercised by the
parity check at all.
It is agreement on one position of a meaningless prompt: strong evidence about the arithmetic and no evidence at all about output quality.
The first tokens
prompt ids : 1008,10484,318,15383,387
prompt txt : The capital of France is
generated ids : 17374,20829,10,427,414,1008,606,142957
generated txt : Paris.",~+ "The Eiffel~
(~ marks a newline)
The first two tokens are 17374 = ' Paris' and 20829 = '.",' - the model answers
the question CORRECTLY.
Paris. The first token out of a 2.78 trillion parameter model running on a CPU in a few gigabytes of RAM is the right answer. The trailing quote and the "The Eiffel" continuation are not a defect: with no chat template and no instruction tuning in the loop, this is a base model continuing text. It has decided it is inside a JSON list of sentences about France, and it is continuing that list.
cap trunk cache s/token token ids decoded
96G 48G 40G 72.3430 17374,20829,10 Paris.",~+~
64G 32G 24G 67.8921 17374,20829,10 Paris.",~+~
48G 24G 16G 63.1822 17374,20829,10 Paris.",~+~
32G 12G 12G 63.4072 17374,20829,10 Paris.",~+~
24G 10G 7G 60.8162 17374,20829,10 Paris.",~+~
16G 6G 4G 68.4805 17374,20829,10 Paris.",~+~
12G 4G 2G 61.0702 17374,20829,10 Paris.",~+~
Every row that ran must show the SAME tokens. A differing row is a bug.
From 96 GB down to 12, the same three tokens. Not merely the same ids at every budget, but the same right answer.
Sustained generation: text in, text out
Four full generations, plain text in and plain text out, with the C tokenizer on both ends and no Python anywhere on the path.
All four run at --trunk-gb 110 --cache-gb 13 --incremental. That split gives almost
everything to the trunk and almost nothing to the expert cache;
Allocation beats capacity shows why it wins.
for (int g = 0; g < gen; g++) {
k3_cache_reset_stats(&cache);
const double ts = now_s();
int frc;
if (incremental) {
/* step 0 feeds the whole prompt; later steps feed only the new token */
const int base = w.cached;
const int nT = (g == 0) ? np : 1;
frc = forward(&w, &c, &cache, seq + base, nT, lg, sc, h, br, ks);
w.cached = base + nT;
} else {
frc = forward(&w, &c, &cache, seq, T, lg, sc, h, br, ks);
}
/* Abort the run rather than argmax a buffer the forward never wrote. */
if (frc != 0) {
fprintf(stderr, "forward pass failed at generation step %d; aborting.\n", g);
return 1;
}
const int nxt = argmax_(lg, c.vocab);
...
}
The sampler is one line, and it is the only one there is:
static int argmax_(const float *v, int n)
{ int b = 0; for (int i = 1; i < n; i++) if (v[i] > v[b]) b = i; return b; }
Greedy, with no temperature and no top-p. That is deliberate rather than unfinished, because greedy decoding is what makes the output identical at every memory budget, the property most of the testing here depends on.
trunk stream: 108.81 GB packed, 90/93 layers PINNED (108.19 GB), ring 1 x 1.29 GB
deterministic hit rate 96.8%
expert cache: 740 slots x 17.56 MB = 12.99 GB (0.90% of the 1.45 TB expert pool)
incremental decode: KV cache 70.96 MB for 24 MLA layers at 30 positions
STEP TOKEN SECONDS CACHE HIT READ GB TOK/S
--------------------------------------------------------------------
0 1040 53.04 100.0 89.21 0.019
1 149803 8.92 100.0 25.83 0.112
2 316 8.92 100.0 25.83 0.112
3 374 8.74 100.0 25.83 0.114
4 1491 8.85 100.0 25.83 0.113
5 261 8.91 100.0 25.83 0.112
--------------------------------------------------------------------
24 tokens in 255.8 s, 10.66 s/token average
--- generated text ---
Kelsey and I am a certified teacher. I have experience tutoring after school at the
middle school level. I have taught
----------------------
PEAK RSS for the whole run: 127.89 GB <- quote this, not the plan
The prompt was "Hello! My name is". That is fluent, grammatical, and it holds a persona across the whole span: it invented a name and stayed consistent with it for twenty-four tokens.
Look at the timing column, because that shape repeats in every run. Step 0 takes 53.04 seconds and reads 89.21 GB. Every step after it takes about 8.9 seconds and reads exactly 25.83 GB.
Given def fibonacci(n)::
--- generated text ---
if n <= 1:
return n
else:
return fibonacci(n-1) + fibonacci
----------------------
28 tokens in 299.3 s, 10.69 s/token average
A correct recursive Fibonacci. The base case if n <= 1: return n is right, the
indentation is right, and the recurrence is right, cut off only because the token budget
ran out mid-expression.
And given "Kimi K3 is a mixture-of-experts language model. It works by":
--- generated text ---
routing each token through a small subset of its total parameters, which keeps
inference fast and cheap relative to its size. The model is trained on a large corpus of
----------------------
32 tokens in 361.6 s, 11.30 s/token average
"routing each token through a small subset of its total parameters" is exactly what Part II implements. That run also shows the prefill cost scaling: its prompt is 17 tokens rather than 5, and step 0 took 97.98 seconds and read 200.67 GB, roughly double the five-token prompts, while every subsequent step still read exactly 25.83 GB.
Four prompts is a demonstration and not a benchmark. But the shape is consistent: about 10.7 to 11.8 seconds per token in sustained decode, at 127.9 GB of peak memory, producing text that is grammatical, factual and in one case correct Python.
Part IV: Measurements
The memory ladder: 8 GB to 224 GB
The memory limit has to be enforced. Telling the engine to use 8 GB on a 228 GB machine measures nothing, because nothing stops it from using more.
# MemorySwapMax=0 matters as much as MemoryMax: without it an over-budget rung
# swaps instead of dying, and its s/token measures swap bandwidth.
systemd-run --scope --user -q \
-p MemoryMax=${TOT}G -p MemorySwapMax=0 \
./bin/k3 "$MODEL" --ids "$IDS" --gen "$GEN" \
--trunk "$TRUNK" --trunk-gb "$TR" --cache-gb "$CA" --incremental \
--out "$OUT/$tag.json" > "$OUT/$tag.log" 2>&1
=== LADDER COMPLETE ===
total_gb pin_layers cache_gb s_per_tok trunk_hit gb_read peak_rss_gb
8 0 0.49 32.69 0.0 25.83 8.24
12 0 2.79 31.41 0.0 25.83 10.53
16 3 4.39 32.21 2.8 25.83 16.00
24 7 7.58 31.85 6.6 25.83 23.95
32 11 10.80 31.44 10.3 25.83 31.90
48 19 17.19 29.76 17.9 25.83 47.80
64 27 23.59 28.60 25.4 25.83 63.71
96 43 36.39 24.40 40.5 18.11 95.51
128 60 49.19 29.40 56.5 17.51 128.18
160 76 61.99 26.31 71.5 17.28 159.98
192 90 77.00 21.32 84.7 16.65 191.83
224 90 108.98 19.21 84.7 14.53 223.82
ids, every row: 17374,20829,10,427,414,1008,606,142957
Read that last line first. Every rung produced byte-identical output. Twelve budgets spanning a factor of 28, and the token ids are the same in every one.
Raw data: docs/data/memory-ladder.tsv.
Going from 8 GB to 224 GB takes 32.69 s/token down to 19.21. That is 28 times the memory for 1.70 times the speed. If you are choosing hardware, the jump from 8 GB to 64 GB buys 14 percent. The memory is not where the speed is.
Ask for 8 GB and it uses 8.24. Ask for 224 and it uses 223.82. Every rung lands on its own budget.
At the top rungs 90 of 93 layers are pinned, so the steady-state trunk hit rate is 96.8 percent. The table says 84.7 percent because these are eight-token runs and token zero pays the full pinning cost: seven tokens each hit 90 layers out of 744 total binds, which is exactly 84.7 percent.
The cache that was not participating
Look at the gb_read column again:
8 GB budget -> 25.83 GB read per token
12 GB budget -> 25.83 GB read per token
16 GB budget -> 25.83 GB read per token
24 GB budget -> 25.83 GB read per token
32 GB budget -> 25.83 GB read per token
48 GB budget -> 25.83 GB read per token
64 GB budget -> 25.83 GB read per token
Seven consecutive budgets read exactly the same number of bytes. Over that range the expert cache grows from 28 slots to 1,344, a factor of 48, and the bytes moved do not change by a single decimal.
A cache that grows 48 times and changes the bytes moved by zero is not working inefficiently. It is not participating at all.
The simulation predicted flat 36.24 percent with a knee at 128 GB; the reality is flat 0.0 percent with a knee at about 36 GB of arena. Its own caveat had warned us: the trace came from full-recompute decode, which touches each expert about 68 times and manufactures reuse that steady-state decode does not have.
Why is the cache so useless? The answer is in the model, not the engine. Quantile Balancing flattens expert usage across the pool by design, and flat usage is precisely what defeats a least-recently-used cache. With no hot subset, a few gigabytes of arena retain nothing worth keeping. The weak caching is a property of Kimi K3, not a defect in this implementation.
That leaves a measurement problem:
cache [final step]
slots : 740 of 17.56 MB = 12.99 GB arena (740 resident, 0 pinned)
requests : 1472 hits 1472 (100.00%) misses 0 evictions 1032
of those hits : 38940 came from the batch prefetch, i.e. read from disk
this token; TRUE resident hit rate 0.00%
read from disk: 25.83 GB in 5.06 s (5104 MB/s while loading)
A hit rate of 100.00 percent sitting two lines above a resident hit rate of 0.00 percent, for the same 1,472 requests.
The first counts a request satisfied from the arena, but the batch prefetch put the expert there microseconds earlier by reading it off disk, so it reads about 100 percent at every cache size and tells you nothing about avoided I/O. The second counts only experts already resident before the step began, which is the quantity we care about, but it is measured against a window that resets every step.
An expert that had to be evicted is one that was not retained. At the 36.39 GB rung that
gives 1 - 1032/1472 = 29.89% retained, against 18.11/25.83 = 70.1% of the bytes still
being read. Those two agree to the decimal, computed from completely different counters.
At the 8 GB budget every step reads 25.83 GB, flat forever. At 96 GB the reads fall from 99.70 to 23.58 to 19.90 to 14.07 across the first four steps as the arena fills with experts that get used again. That declining curve is what a working cache looks like, and it does not appear anywhere below about 36 GB of arena.
Allocation beats capacity
If the expert cache does nothing below 36 GB, where should the memory go instead?
The arithmetic is not close. The trunk is 108.81 GB re-read in full on every token. The experts are 25.83 GB per token. That is 4.2 times the bytes and 2.9 times the time.
A gigabyte given to the trunk pins roughly one more layer and removes about 1.17 GB per token of traffic that would otherwise happen with certainty. A gigabyte given to the expert cache removes, below the knee, nothing measurable.
So it was tested directly. Fix the total memory, vary only the split:
# Fractions of the budget given to the trunk, cache-heavy first so the sweep runs
# AGAINST the hypothesis rather than with it.
FRACS="0.10 0.25 0.60 0.86"
total_gb trunk_gb cache_gb s_per_tok hit_pct gb_read peak_rss
128 12.3 110.7 28.38 44.02 14.46 127.89
128 30.8 92.2 25.20 42.93 14.74 127.53
128 49.2 73.8 25.69 33.97 17.06 128.14
128 73.8 49.2 18.37 32.20 17.51 128.19
128 98.4 24.6 19.46 0.0 25.83 127.36
128 110.0 13.0 16.80 0.0 25.83 127.85
From 28.38 s/token down to 16.80, at identical total memory: 1.69 times faster from allocation alone.
Now look again, because that table contains a result that runs the wrong way. The fastest configuration reads 25.83 GB per token with a 0.0 percent cache hit rate. The slowest reads only 14.46 GB with a 44.02 percent hit rate.
The winner moves 79 percent more expert bytes than the loser and beats it by 69 percent anyway. Optimising the number everyone would naturally optimise, the cache hit rate, actively takes you to the slower configuration.
So the rule is simple: give the trunk everything until it is fully pinned, and only then feed the expert cache.
On how strong this is: the trend is not perfectly monotonic. There are two inversions inside the 128 GB series and one at 32 GB, all well inside the noise. The claim rests on the rank correlation across twelve points at two different totals, Spearman ρ of −0.886 and −0.714 plus a mechanism that predicts it. The direction is solid; the figure of 1.69 has not been replicated three times, so treat the magnitude as approximate.
Raw data: docs/data/trunk-cache-split.tsv.
Measuring the measurement
Before quoting any timing, how repeatable is a timing on this machine? Three runs of one identical configuration, back to back:
[09:36:39] --- replicate 1/3 (trunk 110 / cache 13, gen 8, identical every time) ---
8 tokens in 118.2 s, 14.78 s/token average
[09:39:14] --- replicate 2/3 ---
8 tokens in 117.3 s, 14.67 s/token average
[09:41:22] --- replicate 3/3 ---
8 tokens in 161.1 s, 20.14 s/token average
Mean 16.53 s/token, standard deviation 3.13, spread 33.1 percent. A third of the measurement is noise.
That is the bar every timing claim has to clear, and it rules a line through several of the smaller ladder steps. The 12 GB rung sitting four percent under the 8 GB rung is not an effect. The 11 percent trunk-first improvement at a 32 GB total is not an effect. The dip at 128 GB is not an outlier, it is dispersion.
Two results clear the bar comfortably: the ladder spans 70 percent end to end, and trunk-first at 128 GB is 69 percent. Both are more than twice the floor.
Every non-timing result is untouched, because counts and byte totals are not stopwatch readings. The byte-identical output across twelve budgets, the flat 25.83 GB across seven rungs, the 0.0 percent retention from 28 slots to 1,344, the trunk hit rate tracking the pinned fraction. No amount of scheduling jitter moves any of those.
So where does a third of a measurement go? Two runs of the same configuration, taken from two different sweeps:
run A (from the ladder) run B (from the split sweep)
trunk (STREAMED) 73.80 GB trunk (STREAMED) 73.80 GB
expert cache 49.20 GB expert cache 49.20 GB
60/93 layers PINNED (72.34 GB) 60/93 layers PINNED (72.34 GB)
2802 slots 2802 slots
requests 1472 evictions 998 requests 1472 evictions 998
binds 744, hits 420 (56.5%) binds 744, hits 420 (56.5%)
read 374.99 GB in 138.40 s read 374.99 GB in 63.84 s
(2709 MB/s) (5874 MB/s)
29.40 s/token 18.37 s/token
They pinned the same layers, allocated the same slots, made the same number of requests, evicted the same number of experts and read exactly the same 374.99 GB. Every count matches. The only difference is that one got 2,709 MB/s out of the disk and the other got 5,874, a factor of 2.17 on identical work.
The variance is not scheduling jitter or NUMA placement. It is the device.
Raw data: docs/data/replication.tsv.
Storage is the whole game
Between 40.9 and 60.6 percent of wall clock is spent waiting on the disk, at every rung. So the same binary on two different devices is two different engines.
A run at 31.71 s/token on local NVMe and a run at 70.62 on a network volume differ by 2.2 times, and none of that is attributable to anything in the source. Any comparison between two configurations has to hold the device fixed, or it is measuring the device.
It also means the tool that measures storage cannot use dd:
# dd is one sequential stream at queue depth 1. The expert path is random
# 17.55 MB O_DIRECT reads, up to 16 outstanding. On network storage they diverge.
SZ = 17547264 # one Kimi K3 routed expert, to the byte
N = 40 # reads per measurement
for qd in (1, 4, 16):
measure(path, qd)
One optimisation result. Switching the arenas from 4 KB pages to 2 MB hugepages, measured as a clean A/B on one binary using an environment variable rather than two builds:
########## A: 4 KB pages (previous behaviour)
4 tokens in 45.5 s, 11.37 s/token average
read 153.02 GB in 24.36 s (6282 MB/s)
########## B: 2 MB hugepages (new)
4 tokens in 43.0 s, 10.76 s/token average
read 153.02 GB in 22.64 s (6758 MB/s)
=== output equality (a speedup that changes tokens is worthless) ===
A: [161427, 11294, 58776, 123595]
B: [161427, 11294, 58776, 123595]
IDENTICAL
The tokens are identical, which is the first thing to check about any optimisation. The timing improved by 5.4 percent, which is comfortably inside the 33 percent noise floor, so it is not claimed as a win. The mechanism is sound and the measurement does not establish the size.
/* An A/B between two builds compares two binaries. An A/B on one binary
* compares one decision. */
Why the trunk is not quantised
The trunk is 108.81 GB of bfloat16. Quantising it to int8 would halve that, and to int4 would quarter it. Every other engine of this shape offers a bit-width knob. This one has exactly two weight types and no knob at all:
enum { K3_WF32 = 0, K3_WBF16 = 1 };
The reason is that the cost was measured. The study sampled 31 attention tensors from the released checkpoint over HTTP range reads, 384 rows each, quantised with symmetric per-row scaling.
type tensor int8 mean int4 mean ratio
------------------------------------------------------------------
KDA L13.o_proj 0.01188 0.21122 17.8x
MLA L3.kv_b_proj 0.00736 0.13355 18.2x
MLA L11.o_proj 0.01399 0.24612 17.6x
------------------------------------------------------------------
MEAN over KDA-layer tensors 0.01046 0.18746 17.9x (n=2)
MEAN over MLA-layer tensors 0.00948 0.17154 18.1x (n=18)
MEAN over ALL sampled tensors 0.00961 0.17383 18.1x (n=31)
Int8 costs about one percent and int4 about seventeen, a ratio of 18 that holds across every tensor sampled. And the mean understates the damage, because the tail is much worse.
The worst individual rows at int4 reach 45 percent, 56 percent and 65 percent relative error. Those are not rounding artefacts.
There is also a strong hint from the model authors. The technical report says the experts are MXFP4 with quantisation-aware training, "while all non-expert components remain in higher precision". That list of non-expert components is exactly this trunk. It was deliberately not quantised, and it was never trained to tolerate four bits.
A lossless stream costs seconds per token, and those seconds are recoverable by giving the engine more RAM, which is exactly what the memory ladder demonstrated. Accuracy lost to four-bit rounding is not recoverable at any budget.
Four caveats, all of which the study states about itself: this is weight reconstruction error and not output quality; it sampled 384 rows per tensor rather than whole tensors; it covered attention projections only, with no MoE, no embeddings and no output head; and no downstream logit or token comparison was run at int4, so the quality cost is bounded rather than measured.
Raw data: docs/data/trunk-quantisation.txt.
Part V: Reference
Scope
- No chat template. Base model continuations, not replies.
- Greedy decoding only. No temperature, no top-p, no top-k. Greedy is what keeps the output identical across budgets.
- No chunked prefill. The ceiling is 32,768 tokens, but a 21,000 token prompt is one quadratic pass.
- Context is bounded by memory, not by the engine.
- No vision. MoonViT-V2 is fully specified in
config.jsonat 27 layers, and has zero code here. That absent encoder is 0.057 percent of the checkpoint and downloadable on its own, which makes implementing it a 0.9 GB job rather than a 1.56 TB one. - No SIMD in the KDA recurrence. The matmuls have AVX2 paths; the recurrence is still scalar C.
- No quality benchmark. No perplexity, no task eval. At 11 seconds per token that is days of compute, and it would measure Kimi K3 rather than this engine.
What comes next, in priority order: ROADMAP.md.
Closing the ledger
Every parameter at bfloat16 is 5.56 TB. The experts already ship at 0.53125 bytes per weight, which takes it to 1.56 TB on disk. Routing means only 16 of 896 experts fire per layer, so 1.45 TB of that never needs to be in memory at all, leaving 113.49 GB. Streaming the trunk a layer at a time turns that last floor into a dial, and the dial goes down to a measured 8.24 GB.
The point was never the speed. At 8 GB it takes about half a minute per token, and pretending otherwise would be silly. The point is that the model fits, that it produces the same tokens at 8 GB as it does at 224, and that the gap between "you need a datacentre" and "you need a desktop" was four decisions about where bytes live rather than any change to the model itself.
Documentation
QUICKSTART.md | the setup above, condensed to commands |
ARCHITECTURE.md | how the model maps onto the code |
PERFORMANCE.md | the memory ladder, measured, with its noise floor |
TUNING.md | picking a budget and a split |
BENCHMARKING.md | measuring without fooling yourself |
TESTING.md | what each gate establishes |
API.md | the C interface, for embedding the engine |
ROADMAP.md | scope, and what comes next |
docs/data/ | measurement output; every figure above is transcribed from it |
docs/images/ | every diagram and equation, with the mermaid and Python sources that generate them |
kimi-k3-tech-report.pdf | the model's technical report |
Development
make test # the gate that stays green, no weights, no network, no Python
make help # the documented targets
make asan # AddressSanitizer and UBSan
make portable # generic AVX2, without -march=native
CI runs the same gates: a GCC and Clang matrix under -Werror, sanitizers over the parsers
and the cache, the full weightless suite, and ruff and shellcheck as blocking checks.
Fixtures are generated from the PyTorch reference and committed;
tests/fixtures/README.md records what makes each one
adversarial and how to regenerate it. CONTRIBUTING.md is the place to
start.
License
Apache 2.0, see LICENSE. Vendored third-party components, and the
modifications made to them, are declared in NOTICE.
Kimi K3 is created and released by Moonshot AI under its own license. This repository contains no model weights and grants no rights to them; the technical report is included for reference and remains the property of its authors.
Built for the proposition that a trillion-parameter model should not require a trillion-dollar rack.