Section C, “From a game rule to a network”Worked two-frame construction: explicit weights and operations, not a trained network

Informal interpretation Motivation, interpretation or scope; not a formal claim.

1 Paper text

From a game rule to a network. The networks in this appendix are constructed explicitly from the game rules: their connections and weights are specified by an algorithm, rather than found by training. To see what the construction computes, revisit Figure 2. The preceding board $Q=(3,5,6)$ has nim-sum zero; after the opponent changes $5$ to $4$ , the current board is $Q'=(3,4,6)$ . Comparing the frames gives $\delta=5\oplus4=1$ . In parallel, the circuit proposes replacements $(3\oplus1,4\oplus1,6\oplus1)=(2,5,7)$ and tests which decrease their heaps. Only $2<3$ passes, so the output selects heap index $0$ with replacement value $2$ , giving $(2,4,6)$ with nim-sum zero. For $B=3$ and a two-bit index, the output fields are validity $1$ , index $00$ , and value $010$ .

Generated from paper_jmlr.tex, lines 1472–1474.

2 Explain this

The input already includes the missing clue: the preceding board had nim-sum zero. Comparing it with the new board reveals the opponent's change. For $(3,5,6)\to(3,4,6)$ the change is $1$ , so the replacements are $(2,5,7)$ . Only heap 0 can be reduced to its proposed value. With three-bit heaps and a two-bit index, the output is validity $1$ , index $00$ , value $010$ . These are fields, not a single integer.

Two-frame restoration, step by step

Exact construction trace Every value is computed from the construction’s definitions for the example shown.

This is the checked two-frame selector run on one example, gate group by gate group. It starts from the two frames of Figure 2 and ends with the binary output fields shown in Figure 7. Every value below is computed from the selector's definition for the boards shown. Choose another example to see several candidates at once, the fallback, or a terminal board. The two examples marked "outside the promise" do not start from a zero-nim-sum board after a legal move, so the guarantee does not apply to them.

Inside the promise. The earlier board has nim-sum 0 and the opponent decreased one heap, so the restoration theorem guarantees that the output restores nim-sum 0.

Earlier board Q (3, 5, 6)nim-sum 0

Current board Q′ (3, 4, 6)nim-sum 1; changed: heap 1

After the reply (2, 4, 6)nim-sum 0; reply in heap 0

Compare the frames bit by bit
Each heap is a 3-bit block. For every heap and every bit position, one two-input XOR compares Q with Q′, all in parallel.
Heap Q Q′ XOR Heap
0 3 011 3 011 000 same
1 5 101 4 100 001 changed
2 6 110 6 110 000 same
Extract the difference δ
Bit k of δ is the OR of column k of the XOR table: 001 = δ = 1.
At most one heap changed, so at most one row is nonzero and the OR equals the XOR of the rows, which is nim(Q) ⊕ nim(Q′).
Propose replacements and test them in parallel
For every heap at once: replace Q′_i by Q′_i ⊕ δ and test whether that is a decrease. The decisive bit is the highest set bit of δ (weight 1): a heap passes exactly when it has that bit set.
Heap Q′ Q′ ⊕ δ Test
0 3 011 2 010 passes: 2 < 3
1 4 100 5 101 fails: 5 > 4
2 6 110 7 111 fails: 7 > 6
Show the bit comparisons
The test x ⊕ δ < x is a fixed-depth circuit: an OR over the possible highest differing bit, requiring equality above it, a 1 in x and a 0 in the replacement at that bit.
- Heap 0: 011 vs 010 — the highest differing bit has weight 1; the heap has 1 there, so the replacement is smaller.
- Heap 1: 100 vs 101 — the highest differing bit has weight 1; the heap has 0 there, so the replacement is larger.
- Heap 2: 110 vs 111 — the highest differing bit has weight 1; the heap has 0 there, so the replacement is larger.
Candidates passing: 1.
Choose between cancelling and the fallback
cancelling move At least one replacement passes, so the flags are the passing tests and the fallback is masked off.
Heap passes nonzero fallback value flag s_i
0 1 1 3 − 1 = 2 1
1 0 1 4 − 1 = 3 0
2 0 1 6 − 1 = 5 0
Encode the first flagged heap as a binary index
For N = 3 heaps the construction uses r = ⌊log₂ N⌋ + 1 = 2 index bits (the encoder itself needs only N ≤ 2^r). The first flag is at heap 0, so the index bits are 00.
Show the priority encoder
f_i = s_i ∧ ¬s_j for all j < i keeps only the first flag; index bit k is the OR of f_i over the heaps i whose binary index has a 1 at bit k. This is a fixed-depth circuit, not a search.
Heap i i in binary s_i first f_i
0 00 1 1
1 01 0 0
2 10 0 0
Index bits (OR of the f_i in each 1-column): 00
Output the move
validity1index (2 bits)00value (3 bits)010
Decoded move: set heap 0 to 2. New board (2, 4, 6), nim-sum 0.

Heap	Q	Q′	XOR	Heap
0	3 011	3 011	000	same
1	5 101	4 100	001	changed
2	6 110	6 110	000	same

Heap	Q′	Q′ ⊕ δ	Test
0	3 011	2 010	passes: 2 < 3
1	4 100	5 101	fails: 5 > 4
2	6 110	7 111	fails: 7 > 6

Heap	passes	nonzero	fallback value	flag s_i
0	1	1	3 − 1 = 2	1
1	0	1	4 − 1 = 3	0
2	0	1	6 − 1 = 5	0

Heap i	i in binary	s_i	first f_i
0	00	1	1
1	01	0	0
2	10	0	0

How this relates to the Lean proof. The selector is Family.restS (B, N): the local difference circuits LocalRestore.dC, the candidate tests of SelectCircuit.varSel and the priority encoder PriorityEncoder.encode, read on the two-frame input LocalRestore.tf. Family.rest_semantics proves that, on the promised domain, these formulas and the compiled network output a legal move that restores nim-sum 0. Outside the promise the circuit still computes the values shown, and no theorem says that the result restores 0.

What this does not show. The selector does not find a first winning move from an arbitrary board. Given a zero-nim-sum predecessor it maintains the invariant (Theorem 5.12(iii)). The boxes of Figure 7 group operations; the stages here group gates, not network layers.

Depicted in Lean: Family.rest_semantics, LocalRestore.restore_equiv, LocalRestore.localDelta_eq, LocalRestore.restC_eval, SelectCircuit.varSel_correct, AttainCircuit.cancel_iff_bit, PriorityEncoder.encode_correct, Family.rest_ac0, Family.rest_net. Explains the paper’s Figure 2, Figure 7.

Computed in your browser by a short re-implementation of the definitions above (tools/visuals/nimvis-core.js). It is not itself checked by Lean. At build time it agreed with Lean's own evaluation of those definitions on all 12,299 reference cases. Lean's evaluation fixes the output wires, the difference δ and the decoded move; the intermediate tables (flags, first-flag bits) are re-derived from the same definitions.

3 How this is formalised

An explanatory numerical instance of the checked construction, not a separately added Lean theorem. LocalRestore.dC_eval identifies the difference bits; ThresholdInst.restNet_correct proves the full network returns a legal move restoring zero under the preceding-zero and legal-transition hypotheses. The binary interface is PriorityEncoder.restBin.

Coverage record
Paper anchor	`par:construction-example` (manuscript label)
Displayed as	Section C, “From a game rule to a network”
Claim	Worked two-frame construction: explicit weights and operations, not a trained network
Coverage category	Informal interpretation
Statement source	paper_jmlr.tex lines 1472–1474; SHA-256 of the normalised block source `c1a74bbbea9021c2…`
Depends on	Appendix C, “Local differences and candidate moves”, Appendix C, “A binary index is part of the circuit”
Explanations	`explain/par:construction-example`, `formal/par:construction-example`
Build evidence	fresh isolated rebuild, audit report

4 Lean statement and source

No Lean statement corresponds to this claim. This is interpretation or motivation, not a formal mathematical claim.

Build evidence: fresh isolated rebuild of all 38 bundled modules with Lean (version 4.26.0, started 2026-10-10T15:21:56Z, finished 2026-10-10T15:27:00Z: 38/38 modules compiled, 1588 axiom reports, no errors. See Build for the commands and the independent audit of 2026-10-10 (dated evidence).

Section C, “From a game rule to a network”Worked two-frame construction: explicit weights and operations, not a trained network

1 Paper text

2 Explain this

Two-frame restoration, step by step

Compare the frames bit by bit

Extract the difference δ

Propose replacements and test them in parallel

Choose between cancelling and the fallback

Encode the first flagged heap as a binary index

Output the move

3 How this is formalised

4 Lean statement and source