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\title{Curriculum CoT for $9{\times}9$ Sudoku\\[2pt]
\large Rebuttal / Paper-Section Material}
\author{}
\date{Last updated: 2026--05--24}
\begin{document}
\maketitle
\noindent
This document is a comprehensive, paper-ready reference of (a) the data
pipeline, (b) the instruction-tuning prompt format, (c) the curriculum
and reward design, (d) the latent thought-token architecture, (e) the
multi-stage SFT-then-GRPO training recipe, and (f) the headline numerical
results --- so a rebuttal section can be assembled directly from this
document.
\bigskip
\hrule
\bigskip
\section{Task}
We use the model as a \textbf{per-cell value policy} for $9\times 9$
Sudoku. For a fixed target empty cell, the model emits a JSON set of
candidate digits that are ``i-consistent'' with the current grid
(definition in \S 4). We evaluate two metrics:
\begin{itemize}[leftmargin=*]
\item \textbf{per-cell exact set match} (\code{exact\_set\_match}) ---
predicted set equals the ground-truth i-consistent set;
\item \textbf{whole-puzzle solve rate} (\code{solve}) --- every empty
cell on a 20-empty puzzle produces an exact set match.
\end{itemize}
Because $\text{solve} = \prod \text{exact\_set\_match}$ across the $\sim 20$
empty cells of a puzzle, the two metrics are non-linearly coupled:
\[
\text{solve} \approx \text{exact\_set\_match}^{N_{\text{empty}}}
\]
so $0.95^{20} \approx 0.358$ and $0.97^{20} \approx 0.544$ --- every
percentage point of per-cell exact maps to a much larger swing in solve.
\section{Data pipeline}
\subsection{Puzzle generation}
Generated by \code{simple\_9x9\_curriculum/build\_dataset.py}:
\begin{itemize}[leftmargin=*]
\item Start from a base Latin-square grid; randomly relabel digits,
permute rows and columns within bands, and transpose.
\item Sample \code{empties=20} cell positions uniformly at random and
erase them.
\item Save 10\,000 train + 1\,000 eval puzzles (seed 0, seed 1).
\item Output JSONL files \code{data/sudoku\_t3\_20empty\_value\_qwen\_text\_stage1\_\{train,eval\}.jsonl}.
\end{itemize}
A single record contains:
\begin{lstlisting}
{
"prompt": "<full Qwen chat-templated prompt for one (puzzle, target_cell) pair>",
"completion": "[7,3,8,2,6,9,4,5,...]",
"metadata": {
"grid_size": 9, "box_size": 3, "empties": 20,
"empty_locs_1based": [[1,4],[1,9],...],
"target_triples_1based": [[1,4,7],[1,9,3],...]
}
}
\end{lstlisting}
The 20 \code{target\_triples} give the \textbf{solved} value at each of
the 20 empty positions, so per-cell training targets are always
available. At training time we expand each puzzle into 20 (puzzle,
target\_cell) examples.
\subsection{Cell-policy framing}
The model is never asked to solve a whole puzzle in one shot. Each
example is one (current\_grid, target\_cell) pair, and the supervised
target is the set of digits that are ``i-consistent'' with the current
grid (see \S 4). This turns Sudoku into a
\textbf{classification-into-a-set} problem and lets us share parameters
across cells, stages, and puzzle sizes.
\subsection{Multi-value oversampling (data-side trick)}
Implemented in \code{multi\_output\_cell\_policy/sft\_multi\_output\_train.py}
via \code{tokenizer.\_multi\_value\_oversample\_factor} and the CLI flags
\begin{lstlisting}
--multi_value_oversample_factor INT (default 1)
--train_target_size_min INT (default 0)
--train_target_size_max INT (default 0)
\end{lstlisting}
Inside the dataset builder, examples whose target set has more than one
digit are repeated \code{multi\_value\_oversample\_factor} times in the
training mix. This biases gradient steps toward exactly the cells the
model gets wrong (multi-value cells). Empirically, this is the single
biggest data-side lever --- see \S 10.
\subsection{Where the bottleneck lives}
For 20-empty puzzles in stage 3, only $\sim 25\%$ of empty cells have a
multi-value target set (the rest collapse to one i-consistent value).
Yet those multi-value cells are responsible for the entire solve-rate
gap: they are the cells where the model under-predicts (returns a
singleton when the target is a 2- or 3-element set), and a single
failed cell kills the whole-puzzle solve. The reward shaping in \S 6
and the oversample in 2.3 both attack this single failure mode.
\section{Instruction format}
\subsection{System prompt}
(verbatim from \code{multi\_output\_cell\_policy/prompt\_builder.py})
\begin{lstlisting}
You are a Sudoku value policy.
This setup uses puzzles with about 20 empty cells.
You will be given one target empty cell.
Return ONLY one JSON object of the form {"values":[...]}.
The JSON object must contain exactly one key named "values".
The "values" field must be a JSON array of unique integers in [1,9].
You may return as many candidate values as you want, including one,
several, or many values.
Choose the number of returned values yourself based on which values seem
i-consistent.
The order of the values does not matter.
Do not output any explanation, markdown, punctuation outside JSON, or
extra text.
Current stage objective: i={i} consistency.
\end{lstlisting}
\subsection{User message}
\begin{lstlisting}
Sudoku grid (0 means empty):
<grid_to_text(grid)>
Empty cells in row-major order (20 total): (1,4), (1,9), (2,8), ...
Target cell to fill now: (R,C).
Turn: t/T.
Return only JSON with candidate values for this target cell: {"values":[...]}
\end{lstlisting}
We use the Qwen2.5-Instruct chat template
(\code{tokenizer.apply\_chat\_template}, \code{add\_generation\_prompt=True})
to wrap system + user into the actual prompt ids.
\code{max\_prompt\_length = 768}.
\subsection{Output format}
\begin{lstlisting}
{"values":[3,7]}
\end{lstlisting}
Strictly canonical JSON (single key \code{values}, sorted unique digit
list, no whitespace). Outputs are scored by \code{parse\_values\_json}
(\code{shared\_multi\_output\_policy.py}); any deviation collapses the
whole prediction to \code{parse\_ok=0} and a hard-coded malformed
penalty.
\code{max\_completion\_length = 24} tokens --- enough to emit any
9-digit set.
\section{Curriculum: stage-i consistency}
The curriculum lives in \code{\_stage\_i\_consistent\_values\_for\_grid}:
\begin{itemize}[leftmargin=*]
\item \textbf{Stage 1 --- $i=1$ (legal moves).} A value $v$ is $i=1$
consistent at cell $c$ iff placing $v$ at $c$ violates no Sudoku
constraint (row, column, $3\times 3$ box). This is just ``legal
candidates''.
\item \textbf{Stage 2 --- $i=2$.} $v$ is $i=2$ consistent at $c$ iff
(a) it is $i=1$ consistent AND (b) after placing $v$, every other
empty cell in the grid still has at least one $i=1$-consistent value
(i.e.\ placing $v$ does not immediately make the puzzle unsolvable
by 1-step propagation).
\item \textbf{Stage 3 --- $i=3$.} Same recursion one more level deep:
$v$ is $i=3$ consistent iff after placing $v$, every other empty cell
still has at least one $i=2$ consistent value.
\end{itemize}
This is bounded look-ahead constraint propagation. Stage-3 sets are
tighter than stage-2 sets which are tighter than stage-1 sets. The
curriculum goal at deployment time is stage-3.
In data, we use the same source records and just change \code{--stage\_i};
the target set is regenerated on the fly by
\code{stage\_i\_consistent\_values}.
\section{Latent thought-token architecture}
Base model: \textbf{Qwen/Qwen2.5-1.5B-Instruct} + LoRA
($r=32$, $\alpha=64$, dropout $=0.05$) on
\code{q,k,v,o,gate,up,down}. The latent variant adds \textbf{$k$
thought-token slots} between the prompt and the next-token logits.
Four modes are implemented (\code{latent\_multi\_output\_cell\_policy/});
the winning mode for the final number is \textbf{\code{recurrent\_hidden}}:
\begin{quote}
\code{build\_recurrent\_hidden\_latent\_hidden(model, ids, mask, k)}
\begin{enumerate}[leftmargin=*,nosep]
\item Run the backbone once on the prompt. Keep
\code{base\_hidden = h[:,-1,:]}.
\item Set \code{latent\_token = base\_hidden}.
\item Repeat $k$ times: append \code{latent\_token} (as an embedding)
to the running sequence, run the backbone again on the extended
sequence, and replace \code{latent\_token} with the new last hidden
state.
\item After $k$ recursions, \code{latent\_hidden} is fed through the LM
head to produce the next-token distribution.
\end{enumerate}
\end{quote}
In equations, with $E$ the input embedding lookup, $f_\theta$ the
LoRA-decorated backbone, $U$ the LM head:
\begin{align*}
z_0 &= f_\theta\bigl(E([x_1,\dots,x_T])\bigr)_T \\
z_{j+1} &= f_\theta\bigl([E(x_1),\dots,E(x_T), z_0, z_1, \dots, z_j]\bigr)_{T+j+1},\quad j=0,\dots,k-1 \\
p(\cdot \mid x_{1:T}) &= \mathrm{softmax}(U z_k)
\end{align*}
The model can therefore ``iterate'' $k$ extra forward passes on the
same prompt before committing to a token, with the $k$ extra hidden
states carrying intermediate computation. Setting $k=0$ recovers the
vanilla baseline.
The other three latent modes are alternatives that we ablated:
\code{fixed\_slots} (concatenate $k$ trainable seed embeddings ---
Option-2), \code{latent\_seeds} (similar to \code{fixed\_slots}), and
\code{residual} (project $k$ extra hidden states back onto the base
hidden state via a learned residual). All modes share the SFT and GRPO
trainers; only the next-token logit function changes.
For the curriculum, we grow $k$ stage by stage:
\begin{center}
\begin{tabular}{ccl}
\toprule
\textbf{stage} & \textbf{num\_cot\_tokens} & \textbf{comment} \\
\midrule
1 & 1 & one extra recursion as soon as the model has the surface form \\
2 & 2 & two --- needed for 1-step propagation reasoning \\
3 & 3 & three --- needed for 2-step propagation reasoning \\
\bottomrule
\end{tabular}
\end{center}
\section{The reward function}
Defined in \code{multi\_output\_cell\_policy/rewards.py}.
Given target set $T$, predicted set $P$ (after JSON parse), let
\begin{itemize}[leftmargin=*,nosep]
\item \code{num\_good} $= |P \cap T|$
\item \code{num\_bad} $= |P \setminus T|$
\item \code{num\_missing} $= \max(0, |T| - \text{num\_good})$
\item \code{is\_exact} $= (P \neq \varnothing) \land (P = T)$
\item $\mathrm{tri}(n) = n(n+1)/2$ (rewards larger correct sets superlinearly)
\end{itemize}
Then
\begin{align*}
r &= \mathrm{tri}(\text{num\_good}) \cdot R_g \;-\; \text{num\_bad} \cdot P_b \\
&\quad - \indic[P=\varnothing]\, P_e \;-\; \indic[|P|=1, |T|>1, i<2]\, P_s \\
&\quad - \text{num\_missing}\cdot P_m \;+\; \indic[\text{is\_exact}]\, B_x \\
&\quad - \indic[|P|<|T|, |T|>1]\, P_c
\end{align*}
with parameters (this is the recipe that produced the 0.58/0.68 latent
solve):
\begin{center}
\begin{tabular}{cllr}
\toprule
\textbf{symbol} & \textbf{flag} & \textbf{role} & \textbf{value} \\
\midrule
$R_g$ & \code{--reward\_good\_value} & per-correct-value reward (triangular shape) & 1.25 \\
$P_b$ & \code{--penalty\_bad\_value} & per-extra-wrong-value penalty & 1.0 \\
$P_{\text{mal}}$ & \code{--penalty\_malformed} & flat penalty if JSON parse fails & 4.0 \\
$P_e$ & \code{--penalty\_empty} & flat penalty if predicted set is empty & 0.5 \\
$P_s$ & \code{--penalty\_singleton} & only at stage$<$2: punishes singleton on multi-value targets & 1.5 \\
$P_m$ & \code{--penalty\_missing} & per-missing-value (recall pressure) --- \textbf{NEW} & \textbf{0.75} \\
$B_x$ & \code{--exact\_match\_bonus} & only when $P = T$ --- \textbf{NEW} & \textbf{2.0} \\
$P_c$ & \code{--cardinality\_mismatch\_penalty} & when $|P| < |T|$ and $|T|>1$ --- \textbf{NEW} & \textbf{1.0} \\
\bottomrule
\end{tabular}
\end{center}
Parse failures short-circuit to $r = -P_{\text{mal}}$ and zero per-cell
metrics.
\subsection{Why those three new terms exist (the breakthrough)}
Diagnosis: at the v3/v4 plateau, eval reported
\begin{lstlisting}
exact=0.95 precision=0.95 recall=0.95 solve=0.30 avg_set_size=1.000
\end{lstlisting}
across all checkpoints. Per-cell exact and precision/recall were all
near 0.95 but the model \textbf{always predicted a single digit}
(\code{avg\_set\_size=1.000}). On a multi-value target $T=\{8,9\}$,
predicting $\{8\}$ keeps precision $=1.0$, recall $=0.5$ and yet
\code{exact\_set\_match}$=0$. Solve $= \text{exact\_set\_match}^N$ is
catastrophic in $N$ ($=20$), so even a small fraction of multi-value
cells killed it.
Without any of the new terms the optimum of $r$ on a multi-value cell
is trivially ``predict the singleton you are most confident about'' ---
there is no upside to enumerate the second value. The three new terms
close exactly that hole:
\begin{itemize}[leftmargin=*,nosep]
\item $P_m$ (\code{penalty\_missing}) directly penalises recall;
\item $B_x$ (\code{exact\_match\_bonus}) makes $P=T$ strictly dominate any singleton;
\item $P_c$ (\code{cardinality\_mismatch\_penalty}) is a flat hammer whenever $|P|<|T|$.
\end{itemize}
After these terms were added, GRPO on the latent variant moved solve
from $\sim 0.30$ to $\sim 0.58$ (100-puzzle eval) over $\sim 200$
steps. The same fix is what we ported back into the baseline pipeline
this evening (see \S 10).
\section{Multi-stage warm-baseline pipeline (the recipe that worked)}
Master script:
\code{hard\_9x9\_stage1\_consistency\_queue/launch\_20empty\_warm\_baseline\_all\_latent\_modes\_stages123.sh}.
For each curriculum stage we run \textbf{three sub-phases in order}:
\begin{lstlisting}
[stage i]
(1) baseline warm SFT (no latent tokens, k=0, vanilla LM)
(2) latent SFT (k = i, latent mode = recurrent_hidden)
(3) latent GRPO (k = i)
\end{lstlisting}
\textbf{The warm baseline phase (1) is the trick that makes the
curriculum work.} At every stage transition the data distribution
changes ($i$ increases $\Rightarrow$ target sets shrink) and a new
latent slot appears. Doing a vanilla SFT on the new distribution first
lets the LM relearn the surface form on familiar parameters; THEN the
latent SFT adds the extra thought slot on top of an already-good policy.
When we tried to add a new latent slot directly on top of the previous
stage's GRPO checkpoint, training loss did NOT decrease.
Concrete LR schedule used for the champion run:
\begin{center}
\begin{tabular}{lllc}
\toprule
\textbf{phase} & \textbf{init from} & \textbf{LR} & \textbf{k} \\
\midrule
S1 baseline SFT & base Qwen & 2e-4 & 0 \\
S1 latent SFT & S1 baseline & 2e-4 & 1 \\
S1 latent GRPO & S1 latent SFT & 1e-6 & 1 \\
S2 baseline warm SFT & S1 GRPO & 5e-5 & 0 \\
S2 latent SFT & S2 baseline & 5e-5 & 2 \\
S2 latent GRPO & S2 latent SFT & 1e-6 & 2 \\
S3 baseline warm SFT & S2 GRPO & 5e-5 & 0 \\
S3 latent SFT & S3 baseline & 5e-5 $\rightarrow$ 1e-5 (champion) & 3 \\
S3 latent GRPO & S3 latent SFT & 5e-6 ($\beta=0$) & 3 \\
\bottomrule
\end{tabular}
\end{center}
Other shared knobs:
\begin{lstlisting}
LoRA: r=32 a=64 dropout=0.05 on q,k,v,o,gate,up,down
SFT: per_device_bs=8 grad_accum=2 nproc=8 -> eff_bs=128
GRPO: per_device_bs=4 grad_accum=2 nproc=8 -> eff_bs=64
num_generations=4 beta=0.0 max_prompt_length=1024
max_completion_length=24
multi_value_oversample_factor=5, exact_match_bonus=2.0,
penalty_missing=0.75, cardinality_mismatch_penalty=1.0
\end{lstlisting}
\section{GRPO settings that mattered}
\begin{itemize}[leftmargin=*]
\item \textbf{$\beta = 0$.} The KL anchor was harmful in every sweep
where we tried $\beta>0$. \code{s3\_grpo\_kl04} ($\beta=0.04$) peaked
at solve $=0.625$ (40p) at step 100 and regressed to $0.525$ by step
500.
\item \textbf{\code{num\_generations} $= 4$.} With \code{num\_generations}$=2$
we routinely saw \code{reward\_std}$=0$ (all sampled completions
identical $\Rightarrow$ no gradient). Bumping to 4 fixed it.
\item \textbf{Low LR.} \code{lr=5e-6} was the steadiest. \code{lr=1e-5}
peaked at step 200 (solve $0.65$) then collapsed back to $0.54$ ---
classic mode collapse.
\item \textbf{Effective bs $\geq 64$.} TRL's GRPOConfig requires
\code{eff\_bs * grad\_accum \% num\_generations == 0}; with 8 GPUs we
hit this trivially, but we caution single-GPU rerunners to set
\code{per\_device\_bs=4 grad\_accum=2 num\_generations=4}.
\item \textbf{\code{enable\_input\_require\_grads()} on the wrapped backbone.}
Required for TRL 0.15.x + PEFT LoRA + gradient checkpointing ---
otherwise the loss tensor produced by GRPOTrainer has
\code{requires\_grad=False} and \code{.backward()} raises. Also
\code{unwrapped.config.use\_cache = False}.
\end{itemize}
\section{Final hyperparameters table --- champion latent run}
\begin{center}
\begin{longtable}{lll}
\toprule
\textbf{group} & \textbf{hyperparameter} & \textbf{value} \\
\midrule
\endfirsthead
\toprule
\textbf{group} & \textbf{hyperparameter} & \textbf{value} \\
\midrule
\endhead
Backbone & model & Qwen/Qwen2.5-1.5B-Instruct \\
Backbone & dtype & bf16 \\
Backbone & LoRA target modules & q,k,v,o,gate,up,down \\
Backbone & LoRA $r$ / $\alpha$ / dropout & 32 / 64 / 0.05 \\
Latent & mode & \code{recurrent\_hidden} \\
Latent & \code{num\_cot\_tokens} (S1/S2/S3) & 1 / 2 / 3 \\
Latent & \code{max\_latent\_slots} / seeds & 8 / 8 \\
Data & total empties & 20 \\
Data & train rows / eval rows & 10\,000 / 100 \\
Data & \code{multi\_value\_oversample\_factor} & 5 \\
Data & \code{mixed\_stage1\_ratio} (S1) & 1 \\
Data & \code{mixed\_stage2\_ratio} (S$\geq 2$) & 1 \\
SFT & per\_device\_bs / grad\_accum & 8 / 2 \\
SFT & \code{num\_epochs} (cap) & 64 \\
SFT & LR (S1 latent) & 2e-4 \\
SFT & LR (S2/S3 baseline warm + latent) & 5e-5 \\
SFT & LR (S3 latent champion \code{s3b\_lr1e5\_o5}) & 1e-5 \\
SFT & weight\_decay & 0.0 \\
SFT & gradient checkpointing & on \\
GRPO & per\_device\_bs / grad\_accum & 4 / 2 \\
GRPO & \code{num\_generations} & 4 \\
GRPO & LR & 5e-6 (S3); 1e-6 (S1, S2) \\
GRPO & $\beta$ (KL) & 0.0 \\
GRPO & \code{max\_prompt\_length} & 1024 \\
GRPO & \code{max\_completion\_length} & 24 \\
Reward & \code{reward\_good\_value} & 1.25 \\
Reward & \code{penalty\_bad\_value} & 1.0 \\
Reward & \code{penalty\_malformed} & 4.0 \\
Reward & \code{penalty\_empty} & 0.5 \\
Reward & \code{penalty\_singleton} & 1.5 \\
Reward & \code{penalty\_missing} & 0.75 \\
Reward & \code{exact\_match\_bonus} & 2.0 \\
Reward & \code{cardinality\_mismatch\_penalty} & 1.0 \\
Eval & early-stop on prec/recall & 0.98 \\
\bottomrule
\end{longtable}
\end{center}
\section{Headline results}
\subsection{Latent (with thought tokens, \code{recurrent\_hidden})}
\begin{center}
\begin{tabular}{llrrrrr}
\toprule
\textbf{eval} & \textbf{model / phase} & \textbf{step} & \textbf{exact} & \textbf{prec} & \textbf{recall} & \textbf{solve} \\
\midrule
\textbf{100p (auth.)} & \code{s3\_grpo\_baseline} (S3 GRPO, $\beta=0$, lr=5e-6) & 200 & 0.9665 & 0.9673 & 0.9680 & \textbf{0.580 (58/100)} \\
40p & \code{s3\_grpo\_sharp\_rwd} ($B_x{=}4$, $P_c{=}3$) & 300 & --- & --- & --- & \textbf{0.675 (27/40)} \\
40p & \code{s3\_grpo\_lr1e5} & 200 & 0.978 & 0.978 & 0.979 & 0.650 \\
40p & \code{s3b\_lr1e5\_o5} (S3 SFT champion) & 2400 & 0.974 & 0.974 & 0.975 & 0.600 \\
\bottomrule
\end{tabular}
\end{center}
\subsection{Vanilla baseline (no thought tokens, same Qwen2.5-1.5B + LoRA)}
\begin{center}
\begin{tabular}{llrrr}
\toprule
\textbf{sweep} & \textbf{best variant} & \textbf{best step} & \textbf{exact} & \textbf{solve (100p)} \\
\midrule
v3 (single-GPU, no oversample, no new reward) & \code{baseline\_3stage\_20260522} & --- & 0.730 & \textbf{0.000} \\
v4 (LR sweep, multi-GPU, original reward) & \code{pipe\_v\_sft\_extend} (S3 SFT extended) & 4000 & 0.948 & \textbf{0.400} \\
\textbf{v6 (this evening; ports latent reward + oversample)} & \code{v6\_i\_sft\_v\_oversample10} & running & 0.952$+$ & \textbf{0.440 (best so far)} \\
\bottomrule
\end{tabular}
\end{center}
The v6 sweep is still running --- \code{v6\_e/f/i} are in S3 SFT
continuation, GRPO follow-on phases queued. The \code{v6\_i} variant
has hit \textbf{solve $=0.44$} at SFT eval (new baseline best,
$+0.04$ over v4) and is still climbing.
\subsection{Stage-by-stage trajectory (latent, 40-puzzle eval)}
\begin{lstlisting}
S1 SFT : exact ~ 0.85, solve ~ 0.20
S1 GRPO : exact ~ 0.90, solve ~ 0.20
S2 SFT (no oversample) : exact ~ 0.94, solve ~ 0.20-0.25 <- the wall
S2 SFT + multi_value_oversample=5 : exact ~ 0.96, solve ~ 0.30-0.35
S2 GRPO + new reward terms : exact ~ 0.96, solve ~ 0.35-0.40
S3 SFT (s3b_lr1e5_o5 step 2400) : exact 0.974, solve 0.600 <- SFT champion
S3 GRPO (s3_grpo_baseline step 200,100p): exact 0.967, solve 0.580 <- 100p champion
S3 GRPO (s3_grpo_sharp_rwd step 300,40p): solve 0.675 <- 40p peak
\end{lstlisting}
\subsection{Latent vs baseline gap (head-to-head, same 100p eval, same prompts)}
\begin{center}
\begin{tabular}{lrrrrr}
\toprule
\textbf{model} & \textbf{exact} & \textbf{prec} & \textbf{recall} & \textbf{solve} & \textbf{solved/100} \\
\midrule
Latent \code{recurrent\_hidden}, S3 GRPO & 0.9665 & 0.9673 & 0.9680 & \textbf{0.580} & 58 \\
Vanilla baseline, \code{v6\_i} (best at time of writing) & 0.952 & 0.952 & 0.952 & \textbf{0.440} & 44 \\
\bottomrule
\end{tabular}
\end{center}
Gap on 100-puzzle solve: $\approx$ \textbf{$+0.14$ absolute / $+32\%$
relative} for latent over the strongest baseline we have.
\section{Why the latent works (interpretation hypotheses)}
These are the working hypotheses the experiments are consistent with;
none is fully proven and ablations are still WIP.
\begin{enumerate}[leftmargin=*]
\item \textbf{Constraint-propagation depth.} Stage-3 i-consistency is
essentially 2-ply lookahead. With $k=3$ recurrent hidden tokens the
model gets exactly three extra forward passes between prompt and
output --- one for the legality check, one for 1-step propagation,
one for the second step of propagation. Empirically the gap to the
no-thought-token baseline appears at stages where multi-step
propagation matters (stage 2 onward; stage 1 numbers are essentially
identical).
\item \textbf{Multi-value cells require enumeration, which a singleton
softmax can't do in one forward pass.} A vanilla LM at 1.5B
parameters predicts essentially deterministically once temperature is
low; for a target set $\{8, 9\}$ the LM picks one of the two and
stops. The latent model can use one of the recurrent hidden steps to
``consider'' each option without committing yet, which is exactly
the failure mode in the data (\code{avg\_set\_size} $= 1.000$ for the
baseline, $\approx 1.05$ for the latent S3 model on the same eval).
\item \textbf{Stable curriculum capacity growth.} Adding a new latent
slot at every stage gives the model a ``fresh slate'' of
representational capacity at the exact transition where the task
gets harder. The warm-baseline SFT between stages prevents the new
slot from corrupting the previously learned policy. Without warm
baseline, training loss did not decrease at all (we observed this
directly when we tried to skip the warm baseline).
\item \textbf{GRPO without latent slots is starved of variance.} With
\code{max\_completion\_length} 24 and the model essentially
deterministic, GRPO's 4 sampled completions per prompt collapse to a
single answer --- \code{reward\_std}$=0$, no gradient. With latent
recurrence + the new \code{exact\_match\_bonus} reward, the model
occasionally samples a 2-element set, gets a much higher reward, and
that prompt gets a real gradient signal.
\end{enumerate}
\section{Reproducibility}
\noindent
Code repository: \url{https://github.com/Avra98/curriculum_cot} \\
Latent checkpoints: \url{https://huggingface.co/Avra98/sudoku-latent-recurrent-hidden-20empty-stages} \\
Baseline checkpoints: \url{https://huggingface.co/Avra98/sudoku-9x9-20empty-baseline-1p5b-sweep}
Key scripts:
\begin{itemize}[leftmargin=*,nosep]
\item Master orchestrator (latent, 9-phase warm-baseline pipeline):
\code{hard\_9x9\_stage1\_consistency\_queue/launch\_20empty\_warm\_baseline\_all\_latent\_modes\_stages123.sh}
\item Vanilla baseline pipeline:
\code{\_runs/baseline\_1p5b\_pipeline\_v4.sh} (with v6 launchers
\code{\_runs/launch\_baseline\_push\_v6.sh})
\item SFT trainer (vanilla):
\code{multi\_output\_cell\_policy/sft\_multi\_output\_train.py}
\item GRPO trainer (vanilla):
\code{multi\_output\_cell\_policy/grpo\_multi\_output\_train.py}
\item SFT trainer (latent):
\code{latent\_multi\_output\_cell\_policy/sft\_latent\_multi\_output\_train.py}
\item GRPO trainer (latent):
\code{latent\_multi\_output\_cell\_policy/grpo\_residual\_projector\_latent\_train.py}
\item Reward function: \code{multi\_output\_cell\_policy/rewards.py}
\item Prompt builder: \code{multi\_output\_cell\_policy/prompt\_builder.py}
\item Stage-i consistency:
\code{multi\_output\_cell\_policy/shared\_multi\_output\_policy.py}
\item 100-puzzle evaluator: \code{analysis/eval\_stage2\_checkpoint.py}
\end{itemize}
To reproduce the latent champion (1.5B, 9-phase, $\sim 16$ GPU$\cdot$h
on $8\times$H100 80GB):
\begin{lstlisting}
export STAGE1_BASELINE_ADAPTER_DIR=/path/to/stage1_baseline_seed_adapter
bash hard_9x9_stage1_consistency_queue/launch_20empty_warm_baseline_all_latent_modes_stages123.sh
\end{lstlisting}
To reproduce the v6 baseline push (single-GPU per variant, $\sim 6$
GPU$\cdot$h):
\begin{lstlisting}
bash _runs/launch_baseline_push_v6.sh
\end{lstlisting}
\appendix
\section{The reward fix as a one-line patch}
The single most consequential code change in this whole project, as a
self-contained patch on \code{multi\_output\_cell\_policy/rewards.py}:
\begin{lstlisting}[language=Python]
# new args (default 0 preserves legacy behaviour)
penalty_missing: float = 0.0
exact_match_bonus: float = 0.0
cardinality_mismatch_penalty: float = 0.0
num_missing = max(0, len(target_set) - num_good)
is_exact = bool(predicted_values) and (set(predicted_values) == target_set)
# ... base reward (triangular_number(num_good)*reward_good_value
# - num_bad*penalty_bad_value)
if num_missing > 0:
reward -= num_missing * penalty_missing
if is_exact:
reward += exact_match_bonus
if len(predicted_values) < len(target_values) and len(target_values) > 1:
reward -= cardinality_mismatch_penalty
\end{lstlisting}
Defaults are zero so old runs are unaffected; the recipe sets
$(P_m, B_x, P_c) = (0.75, 2.0, 1.0)$ for the vanilla recipe and
$(1.0, 4.0, 3.0)$ for the ``sharp\_rwd'' variant.
\section{The warm-baseline trick as a sequence diagram}
\begin{lstlisting}
Stage 1 Stage 2 Stage 3
--------- --------- ---------
[base Qwen] | |
| | |
v v v
S1 baseline SFT -> S2 baseline SFT -> S3 baseline SFT
(no latent, k=0) (no latent, k=0) (no latent, k=0)
| | |
v v v
S1 latent SFT -> S2 latent SFT -> S3 latent SFT
(k=1) (k=2) (k=3)
| | |
v v v
S1 latent GRPO -> S2 latent GRPO -> S3 latent GRPO
(k=1, b=0, lr 1e-6) (k=2, b=0) (k=3, b=0, lr 5e-6)
| | |
[final policy]
\end{lstlisting}
Every arrow is \code{init\_adapter\_dir = <previous output>}. Each row
is a ``slot in the curriculum''; the column adds reasoning capacity
($k\mathrel{+}=1$) and moves to a harder target distribution
($i\mathrel{+}=1$). The diagonal across the diagram is the actual
training trajectory.
\bigskip
\noindent\emph{End of report.}
\end{document}