The transversal achievement game on a square grid

Abstract: In the transversal achievement game on the $n\times n$ board, two players alternately claim cells, and the first to own a transversal---a set of $n$ cells of which no two share a row or column---wins. Ranđelović showed that the first player wins for every $n\ge4$, while the game is a draw for $n=2,3$. We give an independent proof that the first player wins for $n\ge4$ that additionally establishes a bound on the length of the win: the given strategy forces a win by ply $2n+3$, i.e.\ on the first player's $(n+2)$-nd move, for every $n\ge4$. The proof yields a strategy that is fully determined by a fixed rule on the current position and can thus be implemented directly. We isolate the use of the hypothesis $n\ge4$ to two steps in the analysis, explaining why the argument fails at $n=3$. An exhaustive computational search implementing the strategy verifies it against every legal defense for $n=4,5,6$, confirming both the strategy's validity and that the $2n+3$ bound is attained in these cases. The main theorem has also been formalized and machine-checked in Lean 4.

• The paper proves that the first player wins for every grid size n≥4 using a fixed two-phase strategy that completes a transversal by ply 2n+3.
• Matching-theoretic lemmas identify threat cells as rectangles formed by exposed rows and columns, enabling the strategy to force blocks and create decisive double threats.
• Exhaustive verification for n=4, 5, and 6 and Lean 4 formalization corroborate the proof, while the optimality of the 2n+3 bound remains open.

Overview and context

The transversal achievement game, posed by Erickson as an open problem, is a positional game in the sense of Beck: two players alternately claim cells of an $n\times n$ grid, and the first to own a transversal—a set of $n$ cells with no two sharing a row or column—wins; if the board fills with neither player succeeding, the game is a draw (Ranđelović, 2024). Identifying the board with $K_{n,n}$ (cells are edges, rows and columns are the two vertex classes), a transversal is exactly a perfect matching, and a player's set $S$ wins when its matching number $\nu(S)=n$. Standard tools from positional game theory are useless here: the Erdős–Selfridge criterion requires $\sum_T 2^{-|T|} \le 1/2$, but the winning family of $n!$ transversals gives $n!\,2^{-n} \gg 1/2$, and the density of winning sets rules out pairing strategies.

Prior work established that the second player never wins (a strategy-stealing argument, since the winning family is monotone), that $n=1$ is a trivial win and $n=2,3$ are draws, and—via Ranđelović—that the first player X wins for every $n$0. The paper under review gives an independent proof of this result with a sharper conclusion: a fully explicit strategy, determined by a fixed rule on the current position, that forces a win by ply $n$1, i.e. with X's $n$2-nd stone, for every $n$3.

Structural lemmas

The proof rests on two lemmas about matchings. The first, on threat structure, characterizes the completing cells of a set $n$4 with $n$5: if $n$6 and $n$7 are the rows and columns exposed by maximum matchings of $n$8, then the set of cells $n$9 with $K_{n,n}$0 is exactly the combinatorial rectangle $K_{n,n}$1. The proof is a clean alternating-path argument: given a matching exposing row $K_{n,n}$2 and one exposing column $K_{n,n}$3, the symmetric difference component through $K_{n,n}$4 must be an even-length path terminating at the other exposed row, and toggling it yields a maximum matching exposing both $K_{n,n}$5 and $K_{n,n}$6. A corollary on tempo follows: a player with fewer than $K_{n,n}$7 stones has no threat at all, and a player holding an $K_{n,n}$8-matching missing exactly row $K_{n,n}$9 and column $S$0 has the unique completing cell $S$1.

The second lemma describes how the threat rectangle grows when stones are added to the missing row $S$2 or column $S$3. Adding $S$4 to an $S$5-matching $S$6 (with induced bijection $S$7) enlarges the exposed rows to $S$8, giving completing cells $S$9; adding stones to both the missing row and column produces a $\nu(S)=n$0 rectangle of completing cells. A remark notes that stones off row $\nu(S)=n$1 and column $\nu(S)=n$2 create no new completing cells, since they form an alternating path whose unique maximum matching is unchanged.

The strategy

X's strategy has two phases. In Phase 1 (moves $\nu(S)=n$3 through $\nu(S)=n$4), X builds an $\nu(S)=n$5-matching while maintaining the invariant that the open block $\nu(S)=n$6 of rows and columns untouched by X contains no O-stone. The rule is local: if O's most recent stone lies in $\nu(S)=n$7, X plays a free cell of $\nu(S)=n$8 in that stone's row, which evicts the stone from the shrinking block and restores the invariant; otherwise X plays any free cell of $\nu(S)=n$9. Feasibility holds because before each move the block has side at least $\sum_T 2^{-|T|} \le 1/2$0, so at least two choices exist. Throughout Phase 1, O holds at most $\sum_T 2^{-|T|} \le 1/2$1 stones and therefore cannot threaten or win.

At move $\sum_T 2^{-|T|} \le 1/2$2, with $\sum_T 2^{-|T|} \le 1/2$3, X uses a tie-break. A cell of $\sum_T 2^{-|T|} \le 1/2$4 is admissible if it and its opposite corner are free; playing it designates the remaining free corner $\sum_T 2^{-|T|} \le 1/2$5. The tie-break selects an admissible cell minimizing the parameter $\sum_T 2^{-|T|} \le 1/2$6, the number of O's stones lying in row $\sum_T 2^{-|T|} \le 1/2$7 or column $\sum_T 2^{-|T|} \le 1/2$8; a counting argument (an intersection of line-pair constraints) shows some admissible outcome has $\sum_T 2^{-|T|} \le 1/2$9. This is the first place $n!$0 is used, since the argument requires $n!$1.

After the tie-break, X owns an $n!$2-matching $n!$3 missing row $n!$4 and column $n!$5 with $n!$6 free, so by the tempo corollary X threatens $n!$7 while O, with $n!$8 stones, has nothing; O is forced to block at $n!$9 at ply $n!\,2^{-n} \gg 1/2$0. Phase 2 then executes one of two mirror-image plans. A row $n!\,2^{-n} \gg 1/2$1 is live if both $n!\,2^{-n} \gg 1/2$2 and $n!\,2^{-n} \gg 1/2$3 are free; since only the $n!\,2^{-n} \gg 1/2$4 O-stones in the critical line pair can kill rows, at least $n!\,2^{-n} \gg 1/2$5 live rows exist. Plan (i) plays $n!\,2^{-n} \gg 1/2$6 then $n!\,2^{-n} \gg 1/2$7 for distinct live rows $n!\,2^{-n} \gg 1/2$8 with a free cross cell: the first move creates a single threat on $n!\,2^{-n} \gg 1/2$9, forcing O's block, and the second creates a double threat on $n=1$0 and $n=1$1. O can block only one, and X completes at ply $n=1$2. The existence of a valid pair follows from a counting argument: the $n=1$3 ordered pairs of live rows exceed the $n=1$4 obstructing O-stones. In the exceptional case $n=1$5 with $n=1$6, where O's stones already form a perfect matching of the complement block, a structural lemma shows the cross cell is automatically free when $n=1$7 is chosen as X's last Phase-1 row.

Why the defense fails

A lemma establishes that O never wins or acquires a useful threat during the endgame. The argument splits on the parameter $n=1$8 into three cases. When $n=1$9 meets column $n=2,3$0 (plan (i)) or row $n=2,3$1 (plan (ii)), O's forced blocks all lie on one line $n=2,3$2, and the bound $n=2,3$3 gives $n=2,3$4 at every ply where X must move freely. The subtle sub-case is $n=2,3$5 with $n=2,3$6: here the crude additive bound fails since O already holds an $n=2,3$7-matching. But O's maximum matching is then unique, and its unique completing cell $n=2,3$8 is occupied by X's last Phase-1 stone, so O has no threat at all. The paper is careful to note that a stronger claim is false: in a worked $n=2,3$9 example, O does threaten a free cell at ply $n$00, harmlessly, since X completes first.

The hypothesis $n$01 is isolated to exactly two steps: the tie-break lemma needs $n$02, and the structural lemma needs the step $n$03. Everything else survives at $n$04; the counting fails at the tie-break precisely, and the paper observes that O survives $n$05 by exactly one tempo.

Computational and formal verification

The strategy, being a fixed rule, was implemented and verified exhaustively against every legal O defense for $n$06:

$n$07	Terminal lines	Nodes explored	Maximum win ply	Runtime
4	4,875	6,075	11	0.3 s
5	485,760	550,224	13	23.2 s
6	75,799,185	82,103,245	15	5,725 s

Every terminal line ends in an X win, and in each case some branch attains ply $n$08, so the bound is tight for this strategy at all three sizes. Winning occurs only at plies $n$09, $n$10, or $n$11, with the slowest outcome comprising roughly 28–32% of lines. All three proof cases occur in practice; the exceptional sub-case with $n$12 is rare (about 0.05% of lines at $n$13) and becomes rarer as $n$14 grows. The verifier also checks the proof's structural invariants at every position, not merely the final win.

Independently, the main theorem, the $n$15 draw, and the supporting lemmas were formalized in Lean 4 and machine-checked by the Lean kernel, with the formalization generated with assistance from the Aristotle system (Achim et al., 1 Oct 2025). The combination of an exhaustive search, invariant checking, and kernel-checked formalization makes the result unusually well corroborated.

Limitations and open questions

The paper is explicit about what it does not settle. The bound $n$16 is an upper bound achieved by one particular strategy; the search confirms only that this strategy admits a defense surviving to ply $n$17, not that no faster X strategy exists. The paper conjectures that $n$18 is the exact game length under optimal play for every $n$19, but the general lower bound is open. Nor does the work address the extremal questions raised by Ranđelović concerning the minimum size $n$20 of a winning family of transversals and the threshold $n$21 above which every family is winning. The verification is exhaustive only for $n$22; for larger boards, correctness rests on the proof and formalization alone.

Conclusion

The paper reproves that the first player wins the transversal achievement game for all $n$23, replacing the normal-form case analysis of prior work with a fixed invariant (an O-free open block), two matching-theoretic lemmas, and a single case split on the parameter $n$24. The resulting strategy is fully explicit, wins by ply $n$25, and is corroborated by exhaustive search at $n$26—where the bound is attained—and by a Lean 4 formalization. The analysis also pinpoints why the game turns at $n$27: the tie-break counting requires two O-stones, so the second player survives at $n$28 by exactly one tempo.