Short strategies in the even game of Barmpalias, Zhang and Zhan

Interval characterisation of an immediate Player-1 win in the k-even game, for every k >= 2: she can end the game on her current move iff some interval [lo,hi] has both endpoints unused by her, every even number of [lo,hi] already used by Player 2, and at least k of its numbers unused by her

Explicit Player-1 winning strategy in the 2-even game, winning in two moves: play 2,4; answer Player 2's 2 with 1,3 and his 4 with 3,5. Every legal reply is covered and the certificate is re-checked from the raw rules

Explicit Player-1 winning strategy in the 3-even game that wins within FOUR of her moves: a 76-node tree with a child for every legal Player-2 reply, kernel-checked from the raw rules

The k = 3 / k >= 4 forcing dichotomy: for k = 3 the move 2n-1, 2n, 2n+1 leaves Player 2 the single reply 2n, which may be brand new; for k >= 4 every Player-1 move that leaves exactly one reply e forces e to sit two away from an even number Player 2 has ALREADY used – so at k >= 4 her first move can never be a forcing one

For every k >= 4, Player 1 has NO winning strategy of length at most three in the k-even game (hence none of length at most two); and for every k >= 2 she never wins on her first move

Forcing never helps Player 1, at any k >= 1: if her move leaves Player 2 exactly one legal reply e, then the invariant Safe k S1 S2 ('no interval all of whose even numbers Player 2 has used holds k numbers Player 1 has not used', i.e. she cannot finish this move) is preserved by that reply

Isolated replies are safe: for every k ≥ 4, if Player 1 cannot finish and Player 2 answers with an even number e such that neither e−2 nor e+2 is one he has used, then she cannot finish afterwards either — whatever k numbers she just played. The covered interval containing e is trapped in [e−1, e+1], three numbers

The crowding bound: for Player 1 to leave Player 2 no isolated reply, every even of her span must lie in the 2-neighbourhood S2 ∪ (S2+2) ∪ (S2−2), so ⌊k/2⌋ ≤ 3·|S2|. Equivalently, an explicit Player-2 rule — while 3·|S2| < ⌊k/2⌋ he has, against every legal k-set she can play and with no case analysis, a reply after which she still cannot finish

For every k ≥ 4, Player 1 has NO winning strategy of length d in the k-even game whenever 6·d ≤ k; equivalently any Player-1 win of length d has k < 6·d. Combined with EG5: none of length ≤ max(3, ⌊k/6⌋). The first bound on the open case whose strength grows with k, and it also holds from mid-game positions (3·(|S2|+d) ≤ ⌊k/2⌋)

For every k ≥ 4, Player 1 has NO winning strategy of length four in the k-even game — not even from a position where she has already played and Player 2 has not; and none of length three once he has answered once. With EG5 (length ≤ 3) the no-go ladder now reaches four for the whole open range. Proved by an explicit three-round Player-2 rule whose middle step is "answer as far from your first number as her span allows"

The counting no-go: a covered interval is at most 2·|S2|+1 numbers long, so Player 1 has no winning strategy of length d in the k-even game whenever 2·(|S2|+d) ≤ k — from the empty position, whenever 2·d ≤ k. The shortest Player-1 win, if one exists, is longer than k/2, and the bound is exactly sharp at k = 2. Supersedes EG9 (6·d ≤ k)