RESEARCH / Focused target
Optimality of the Ordered Zeckendorf Long Game Strategy
RELEASED FOUNDATIONS / PROPOSED CONNECTION
Research connections
All source anchors (8)
- W-Digit ConventionReleased anchor
- Raw W-Digit StringsReleased anchor
- Local Carry RulesReleased anchor
- Digit: NormalizeReleased anchor
- List Inversions and Local Replacement BoundsReleased anchor
- Carry Chains from a Run of Consecutive IndicesReleased anchor
- Newman ConfluenceReleased anchor
- Normal Form FunctionReleased anchor
Problem
The question
A state is an ordered list of Fibonacci numbers. The game starts with n copies
of F_1. Legal adjacent moves are (F_i, F_{i+1}) -> F_{i+2};
(F_1, F_1) -> F_2; (F_i, F_i) -> (F_{i-2}, F_{i+1}) for i > 2;
(F_2, F_2) -> (F_1, F_3); and switching an inversion (F_i, F_j) -> (F_j, F_i)
when i > j.
The Long Game Strategy uses the priority: all switch moves in any order, combine adjacent ones from the left, split from the right, then merge from the left.
Conjecture 1.7, quoted from arXiv:2508.20222v2:
“The LGS has the longest game length.”
Status statement, quoted from the same version:
“The longest-game strategy described in Conjecture 1.7 is supported by empirical simulations. However, a rigorous proof establishing its optimality remains an open problem.”
Formal statement, with the five moves represented exactly:
∀ n > 0, ∃ g, CompleteLGSRun n g
∀ n > 0, ∀ g, CompleteLGSRun n g ->
∀ h, LegalTerminalRun n h -> h.length ≤ g.length
Every permitted switch ordering is retained in the proved universal comparison. Priorities restart after every move, including a switch.
The paper says the conjecture is backed by exhaustive simulations rather than a proof. In its broader exact game-tree analysis it also states:
“Due to the combinatorial explosion in the number of game states and legal move sequences, a full resolution of the winner for n>25 was computationally infeasible.”
That second quote concerns winner computation, not directly Conjecture 1.7, but it identifies the same state-space obstacle. For length, the paper proves only upper and lower asymptotics and a structural lemma about repetitions under LGS.
Motivation
Our foothold
- Erasing order maps a game state to
RawDigits. Each merge and split move preserves Fibonacci value and is closely related to a frozen carry step; switch moves erase to the identity. - The frozen normalizer proves termination and unique canonical output for its own oriented carry system. This explains the shared terminal Zeckendorf state, but confluence intentionally forgets path lengths.
- The new content is an extremal refinement of rewriting: among ordered
value-preserving paths from
[F_1,...,F_1]to the sorted normal form, LGS maximizes length.
Gap
Missing bridges
D5/S1/Digit/Carry/OrderedGame/Completion.result : Conjecture17 proves the exact
source target with its definitions unchanged. For every positive n, including
n=1, a complete ordered LGS path exists; every permitted complete LGS path is
at least as long as every legal terminal competing path. The three new results
lgs_path_raw_erasure, complete_lgs_exists and result have exactly the
standard axiom closure propext, Classical.choice, Quot.sound.
Independent pre-Freeze source-fidelity and declaration-admission review passed.
The proof does not use the paper's at-most-one-repetition lemma.
OrderedGame.greedy_attainmentconstructs the concrete full raw greedy continuation, recursively using the strict carry measure, and attains its rewardGat a binary nonadjacent endpoint. This is raw completion, not yet ordered completion or domination of all competitors.complete_greedy_rewardprovesw = G cfor every completeRawGreedyPath c e wwithCanonicalRaw e. Legal preferred successors are unique, so path induction agrees with the recursive concrete continuation.shared_input_merge_repairproves all five shared-input detours with unrestricted spectators, exact endpoints, and exact full reward gains.ones_terminal_promotionproves ones-first promotion against arbitrary terminal raw paths, including interleaved merges. The actual split-prefix cut and finite high cascade now discharge the singleton and binary exchange cases inOrderedGame/Optimality.raw_terminal_bound, proving the full boundweight ≤ Gfor every legal path to a canonical raw endpoint.terminal_raw_canonicalproves that actual ordered terminality implies that endpoint condition.Completion.lgs_path_raw_erasureproves the ordered/raw priority bridge without restricting switches.complete_lgs_existsminimizes a rank among actual legal moves and recurses on the carry measure paired with inversions to obtain a terminal continuation from every list.OrderedGamedefines ordered legality and the switch move, andpath_raw_erasureproves erasure into a labelled raw path with identical accumulated reward.OrderedGame/Attainment.lgs_move_potentialandlgs_path_potentialprove exact attainment of the inversion potential by actual ordered LGS moves and paths.Completion.resultcombines these with the raw weighted comparison to prove actual ordered length optimality.- Newman confluence and normal-form uniqueness say nothing about longest paths.
- LGS contains a tie phrase "switch moves (in any order)".
ASSUMED-UNVERIFIED: an uncommitted exhaustive search overn <= 16found equal minimum and maximum LGS lengths across all priority-one switch choices. If correct, that supports omitting a switch tie-breaker within that range; the unrestricted switch-phase independence is now proved byOrderedGame/Attainment.switch_normalization: all maximal legal zero-reward paths share their sorted endpoint and exact inversion length. The full all-switch comparison is now proved byCompletion.result. - The inversion count now exists here.
D5/S1/Digit/Carry/ListInversionsis frozen and suppliesinv : List Nat -> Nat, its append law, a three-block window decomposition, and four local replacement bounds, one per non-switch move, stated for arbitrary lists rather than for reachable states. Entries may be zero; nothing there assumes positivity. - The four moves themselves, and their termination, were already frozen under
a different vocabulary.
D5/S1/Digit/CarrydefinesCarrySteponRawDigitswith constructorsadjacent,double_zero,double_oneanddouble_succ, which are merge, merge-ones, split-twos and split in zero-based indexing, andD5/S1/Digit/NormalizeprovescarryStep_measure_decreasesfor every one of them and builds a terminating normalizer. A search by the problem's name reaches neither module; a search by the name of each operation reaches both. - The length side now has its first node, and it is a lower bound.
D5/S1/Digit/Carry/RunChainLowerBoundis frozen and states that for every startaand every lengthLthere exists aCarryStepschain of exactlyL * L / 4steps out of the raw digit vector carrying one token at each of theLconsecutive indices froma. Its proof is a construction rather than a pure existence argument: forLat least two, one merge at the bottom of the run followed by one split for each remaining duplicate carries that run to the run of lengthL - 2together with one isolated token inL - 1steps, and a strong induction in steps of two composes those blocks, with lengths zero and one as its bases. - That module supplies neither a matching upper bound nor a definition of
height, so maximality of the constructed chain is unproved and there is still no optimality statement. - The four
CarryStepconstructors of frozenD5/S1/Digit/Carrymatch the four moves of the unordered Zeckendorf game under the index shiftW_i = F_{i+1}:adjacentis merging consecutive Fibonacci numbers,double_zerois combining ones,double_oneis splitting twos anddouble_succis the general split. Applying arXiv:2009.09510 Theorem 1.2 from an arbitrary start rather than fromnones remains an external dependency, recorded below. Conditional on that transfer, the consecutive-run upper bound reduces to the run-length formula of one deterministic strategy, which is a formalizable target on its own.
Route
Proposed approach
The weighted split-only subproblem is now proved in
SplitStabilization.split_prefix_promotion, split_phase_promotion, and
greedy_split_optimality. Legal replay preserves the endpoint and does not
decrease full reward; first-overfire supplies the selected split's occurrence.
The comparison covers all complete split phases from arbitrary raw digits,
with ones priority and highest-duplicate priority restarted after every step.
The split-only theorem does not cover paths containing merges.
OrderedGame/Optimality.raw_terminal_bound now supplies full weighted raw-game
domination: every legal raw path to a canonical endpoint has reward at most the
concrete G, for arbitrary raw starts. Its proof handles interleaved merges via
the actual split-prefix cut, finite high cascades, and exact legal exchanges.
terminal_raw_canonical connects actual ordered terminal states to that endpoint
condition. Completion proves ordered/raw priority correspondence, finite
ordered LGS completion and the full Conjecture17. Exact ordered potential
attainment and arbitrary switch-phase completion are supplied by
OrderedGame/Attainment.
- Use one raw ordered
List Nat, decoded byNat.succ; multiplicities useMultiset.toFinsupp.Carry/OrderedGamedefines all five positional moves, their full carry rewards, legal paths, relational LGS andConjecture17. OrderedGame.path_potentialproves, for every legal finite path,length + inv(decode end) ≤ inv(decode start) + reward. It reuses the four frozen inversion bounds.path_raw_erasurepreserves the exact reward in the labelled raw carrier.Attainmentproves equality for actual LGS paths;Completion.lgs_path_raw_erasuresupplies the priority bridge. The raw weighted upper bound and the ordered-terminal-to-canonical bridge are now proved inOptimality.Carry/SplitStabilizationproves exact site balance, the first-overfire least-action bound, unique complete split counts and endpoint, and existence of complete split phases. Weighted preferred-split promotion and maximal reward among complete split-only phases are also proved.greedyDecision,G, andgreedy_attainmentnow give a concrete full raw greedy continuation by the strict carry measure.ones_terminal_promotionhandles arbitrary terminal competitors for the ones-first branch, andshared_input_merge_repairproves the five duplicated-input merge repairs.split_greedy_terminal_promotioncuts a genuine complete split prefix before the first merge.raw_terminal_boundthen completes the strict-successor induction using all split competitors before handling shared-input merges, and strong index descent in the binary branch.- Legal cascade replay is proved separately from recognition of a greedy prefix:
a split at j=a-1 changes the lower boundary, as does a merge at b=a+3.
high_cascadeandOptimality.singleton_merge_cascadepreserve the high inputs and reward coordinates under those lower-boundary changes. The higher-split, lower-split, predecessor and separated-merge exchanges are all Lean-checked. Replacement tails need only be legal and terminal. Completion.complete_lgs_existsconstructs finite completion using the combined carry/inversion measure, including n=1. Exact potential attainment and raw greedy comparison then yieldCompletion.resultfor every complete LGS run and every terminal competitor. No global maximum or unproved Bellman premise is required.
The following older maximum-value proof sketch is historical mathematical context; its claims remain
ASSUMED-UNVERIFIED. Nothing here is Lean-verified, the four move-specific
count calculations are not displayed, and it does not settle the conjecture. Write inv s for the inversion
count of a state and L u for the longest game length from a sorted state u.
- Decomposition.
longest s = inv s + L (sort s). The easy direction is thatinv sswitches bubblesintosort s. The other direction is induction over the move relation: the switch case is an equality because sorting is unchanged andinvdrops by one, and the merge/split case follows from the key lemma below together with the Bellman recursion atsort s. - Key lemma. Let a merge or split act on an adjacent pair of
swith values(a,b), givingt; lett'be the result of the same-valued move applied at the leftmost such pair ofsort s. Theninv t <= inv s + inv t'. Each of the four move kinds reduces to two monotonicity steps on the counts#{p in prefix : p > j}and#{q in suffix : q < j}plus one absorption using that prefix and suffix are disjoint parts of the same multiset. The lemma mentions no Fibonacci number and no reachability: it holds for arbitrary lists, exhaustively over every list of length at most six on values one to six. - Consequence. Every switch is an optimal move, with
longest s = longest t + 1. So LGS priority one is exactly optimal, and the conjecture reduces to the sorted case.
Position is no longer part of the problem. On a sorted state the prefix
lies at or below the smaller window value and the suffix at or above the larger
one, so only the new window entries can form inversions with the surroundings.
Writing c_v for the multiplicity of v, the inversions created by each move
are then exact counts: c_{i-1} + c_i - 2 for a split at i > 2, c_2 - 2
for a split of twos, c_{a+1} - 1 for a merge at a, and c_1 - 2 - k for
the merge of ones whose window starts after k other ones. Only the last
depends on position, and it strictly decreases in k, so the leftmost
merge-ones is the unique best merge-ones, strictly so whenever more than one is
available. The other three do not depend on position at all. Conditional on the
proposed decomposition, these inversion formulas would settle the positional
choices, and they give "combine adjacent ones from the left" a mechanism: the
leftmost choice keeps the most inversions.
The remaining problem is a position-free recurrence. Substituting those four counts and eliminating the merge-ones position parameter turns the sorted-state longest length into a recurrence on multisets alone:
L(M) = 0 when no branch applies, otherwise 1 + max over applicable branches of
merge-ones (c_1 >= 2) : (c_1 - 2) + L(M - 2*{1} + {2})
split-twos (c_2 >= 2) : (c_2 - 2) + L(M - 2*{2} + {1,3})
split i (i > 2, c_i >= 2) : (c_{i-1} + c_i - 2) + L(M - 2*{i} + {i-2,i+1})
merge a (c_a, c_{a+1} >= 1) : (c_{a+1} - 1) + L(M - {a,a+1} + {a+2})
Conditional on the same decomposition, the following four guarded Bellman equalities would suffice for Conjecture 1.7, one per region of the LGS priority, the four regions being disjoint. Necessity over all multisets is not established:
c_1 >= 2implies the merge-ones branch attains the maximum;c_1 <= 1and somei > 2withc_i >= 2implies the branch at the largest suchiattains it;c_1 <= 1,c_2 >= 2and no suchiimplies the split-twos branch attains it;- all multiplicities at most one and non-terminal implies the merge at the
smallest available
aattains it.
Two formulations recorded here earlier were stronger than necessary, and both were false as stated. Branch values below are quoted as inner maxima, excluding the move itself; a total continuation length is one larger.
- "Any split with index above two is optimal" is false. On the sorted state
(3,3,4,4)the split at 3 has inner maximum 5 while the split at 4 has 7; the same failure occurs on(1,3,3,4,4)and on(1,1,3,3,4,4). The correct rule selects the largest splittable index, which on a sorted state is exactly the rightmost splittable pair, so the paper's "splits, starting from the right" is a genuine rule rather than an artefact of presentation. - "Split-twos dominates every merge" is false without the guard. The state
{2,2,3,3,7,8}, which hasn = 2F_2 + 2F_3 + F_7 + F_8 = 65and is reachable from 65 ones by building each block separately, has total continuation length 7 through split-twos and 8 through a merge. Its certificate isL(2,5) = 0,L(1,1,5) = 1,L(4,4) = 1,L(2,3,4) = 2,L(1,1,3,4) = 3,L(1,2,2,4) = 4,L(1,3,3,3) = 5,L(2,2,3,3) = 7. Only the guarded form, in which split-twos is taken when no higher split exists, survives. The mechanism is that an independent merge elsewhere can still be followed by the optimal higher split, whereas taking split-twos first has already given up a move; local independence preserves that loss rather than repairing it.
Falsifier
What would falsify this route
A complete certificate consists of a smallest n, an exact legal terminal LGS
run g, and another exact legal terminal run h with length h > length g. If
switch ordering changes LGS length, two LGS runs with unequal lengths falsify
the universal proposed formalization even if one remains globally maximal.
Evidence
Evidence to collect
Construct a memoized exact DAG for n <= 30:
- canonicalize states only by the ordered Fibonacci-index tuple; do not quotient by multiplicity;
- compute exact
height(S)and retain a maximizing successor witness; - enumerate every LGS tie choice and compare its length to
height(start_n); - emit, for each state reached by LGS, whether its chosen move is height-maximizing;
- when a priority exchange first fails locally, save both suffix paths even if the global conjecture still holds.
This extends the paper's simulation into proof-certificate-shaped data and directly probes the formal ambiguity.
Measurements taken so far are recorded with their sampling domains, because two of the claims corrected above were first believed on the strength of a domain too small to contain their counterexamples.
Over every reachable sorted state with 2 <= n <= 20: the position-free
recurrence agrees with the original definition at all 797 states; the four exact
inversion counts hold in 5064 of 5064 move windows; the leftmost merge-ones is
optimal among merge-ones in 550 of 550 states, strictly so in 451 of them; and
the merge at the smallest index is optimal among merges in 585 of 585 states, of
which 276 have merges at two or more distinct indices.
Reachable states are still the wrong domain for the domination claims, since
the recurrence is defined on every multiset. The n <= 20 cutoff misses the
n = 65 split-twos counterexample entirely; the split-at-3 counterexample
(3,3,4,4) has n = 16 and does lie inside that range, but only above
n <= 13, which is where these claims were first believed. Over 14978 arbitrary
multisets of length at most eleven with entries at most thirteen: merge-ones
dominates merge in 1576 of 1576 instances, dominates split-twos in 188 of 188,
and equals the best split in 1294 of 1294; the best split dominates split-twos
in 1377 of 1377 and merge in 10454 of 10454. On that same domain the four
guarded equalities hold in 1776, 10356, 429 and 1957 instances respectively.
Unguarded split-twos against merge fails in 56 of 1749 instances, which is how
the {2,2,3,3,7,8} family was found independently of the analysis predicting
it.
Write U(M) for the length of the longest chain of carry steps out of the raw
digit multiset M, and run a L for the multiset carrying one token at each of
the L consecutive indices from a. The readings below are enumeration on the
domains stated with them.
Over a in 1 to 6 and L in 1 to 13, U(run a L) equals floor(L^2 / 4) in
78 of 78 instances, with no observed dependence on a. For the proposed
equality over all consecutive runs, the lower bound is the frozen module cited
in the gap above and the upper bound remains open.
All union readings below place the first run at start a = 1 in zero-based raw
indices, and a gap counts the empty indices between consecutive runs.
Over 684 unions of runs whose gaps are all at least two, being 108 two-run
states with lengths in 1 to 6 and gaps in 2 to 4 together with 576 three-run
states with lengths in 1 to 4 and both gaps in 2 to 4, U of the union equals
the sum of floor(L_i^2 / 4) in 684 of 684 instances. Over the 100 states of
the same shape with every gap equal to one, that sum is exact in 10; among the
36 two-run members of that sample the measured U is above the sum in 30
instances, equal in 6 and below in none.
Within the 81-pair gap-one sample with L_1 and L_2 in 1 to 9, the measured
excess over the additive sum is zero when L_1 is one and ceil(L_2 / 2) when
L_1 is at least two, in 81 of 81 instances. Applying
that correction once per gap is not sufficient: over the 64 three-run states
with lengths in 1 to 4 and both gaps one it is exact in 52, and each of the 12
failures has middle length exactly one and first length at least two, with the
measured U above the prediction. Those samples suggest that a run block emits
a token into the empty index above the run, adjacent to a length-one middle run,
so that the two can merge and reach the next gap; that is a reading of the
sample, not a classification of U.
Two routes to the upper bound are closed by counterexample, so that the next attempt need not repeat them.
A potential of the form sum_i v(m_i) + sum_{i<j} w(|m_i - m_j|) is excluded by
the following algebra, which is written out here and is not kernel-verified.
Requiring it to equal floor(L^2/4) on every run determines it: independence of
the start forces v constant, length one forces v = 0, and the lengths then
force w(d) = 1 for odd d >= 1 and 0 for even d >= 2, leaving w(0)
free.
That is the statement that floor(L^2/4) counts the pairs of run indices at odd
distance. Every member of that calibrated family rises on the step
(2,4,6,6) -> (2,4,4,7): both sides carry exactly one pair at distance zero, so
the w(0) term cancels, while the pairs at odd distance go from none to three.
The rise is therefore 3 whatever w(0) is, and no member of the family
decreases on every step. For w(0) in {0,1} the same family is not even an
upper bound: it gives at most 1 on (0,0,2), whose longest chain has length 2.
Splitting U over components of the support joined at a fixed distance also
fails. On (0,0,1,1,6,6) enumeration gives U = 6 against a component sum of
5 at each of the thresholds 1, 2, 3 and 4, its support splitting the same way
at all four. Over random multisets of length 2 to 7 with
entries at most 12 that have at least two components, exact splitting holds in
2634 of 3750 at distance 1, 2976 of 3338 at distance 2, 2594 of 2658 at distance
3 and 1832 of 1835 at distance 4; Whether some larger distance would restore exact
splitting was not tested, and the reason separated runs are additive above is
not established here.
The published strategy, in contrast, attains U on every state sampled: over
the runs with start in 0 to 3 and length in 1 to 11 it is exact in 44 of 44, and
over 3000 random multisets of length 2 to 7 with entries at most 9 it is exact
in 3000 of 3000. On runs with start 1 and length L in 2 to 14 the
recorded lengths satisfied greedy(L) = (L - 1) + greedy(L - 2), with lengths
zero and one as bases. That this recursion sums to floor(L^2/4) is the parity
argument already carried by the frozen module's own proof. Whether the recursion
itself holds for every run is an unverified proof obligation.
None of the enumeration above is a proof. The kernel-verified
material cited in this section comprises the chain-length lower bound and its
recurrence arithmetic in D5/S1/Digit/Carry/RunChainLowerBound.
Triage
Scope assessment
theorem. The full Conjecture17 now has a kernel-checked proof, including
ordered/raw priority correspondence, complete ordered LGS existence and the
universal comparison with all permitted switch choices retained. Independent
pre-Freeze source-fidelity and declaration-admission review passed. Related suppliers were contributed
in merged PRs #7495, #7575, #7643 and #7651; this work does not replace them.
ASSUMED-UNVERIFIED
Unverified assumptions
- Bounded checks of the arXiv history, exact-title and ordered-longest searches, and OpenAlex W4414447583 verified no later exact resolution in that scope. These checks do not establish worldwide priority or novelty of the exchange-lemma route.
- The stronger raw terminal upper bound is kernel-checked in
OrderedGame/Optimality.raw_terminal_bound, and the ordered priority and existence bridges are kernel-checked inOrderedGame/Completion. U(run a L) = floor(L^2 / 4)is an enumeration reading onain 1 to 6 andLin 1 to 13, and the union-of-runs readings hold only on the domains stated with them. Independence froma, the gap-one closed form and the cascade explanation are all unproved outside those samples. Of the run-length claims only the lower bound is kernel-verified.- The calibration of the pairwise potential, the identity between
floor(L^2/4)and the count of odd-distance run pairs, and the exclusion it yields are written-out algebra with no frozen GID behind them. - Cusenza et al.'s Lemmas 2.2 and 2.3 explicitly allow arbitrary starting game states. Their unordered theorem alone does not establish ordered, state-dependent reward comparison. The direct weighted proof above establishes that comparison without assuming a transfer from their theorem.
- The alternative route through the unordered smallest-merge theorem is not
used by
Completion.result. The unordered game's longest length is treated in arXiv:2009.09510, whose Theorem 1.2 reads "The longest game on anynis achieved by applying split moves or combine 1's (in any order) whenever possible, and, if there is no split or combine 1 move available, combine consecutive indices from smallest to largest", and whose Lemmas 2.1 to 2.3 are each stated "starting from any game state". Two features of that theorem match measurements above that were taken independently of it: that split and combine-ones may be applied in any order matches the observed equality between the merge-ones branch and the best split branch, and "smallest to largest" matches the smallest-merge rule. The theorem concerns the unordered game only, and transferring it to the ordered length requires identifying the two on states whose multiplicities are all at most one, which is not established here. - The bibliography of arXiv:2508.20222v2 cites a different paper by an overlapping author group, "Winning strategy for multiplayer and multialliance Zeckendorf games". Citation by author surname alone is ambiguous between the two; the identifier above is the one meant here.