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There is a non-empty finite set $\ K,\ $ say, of plates. Initially, there are $\ p_0(k)\ $ stones on the $k$-th plate, where $\ p_0(k)\in\mathbb Z_{_{\ge0}}\ $ for each $\ k\in K.$

So far, it is like a NIM game, and even the moves look similar. Namely, a move amounts to removing a positive number of stones from an arbitrary single plate, i.e. the $\ n$-th move creates a function $\ p_n:K\to\mathbb Z_{_{\ge0}}\ $ such that $\ p_n(k)=p_{n-1}(k)\ $ for every plate $\ k\in K$ but for one plate $\ \kappa_n\in K\ $ for which $\,\ 0\le p_n(\kappa_n)<p_{n-1}(\kappa_n).$

But now, we diverge from NIM games. The game is finished the moment $\ p_n\ $ is a constant function; since then, there are no more legal moves.

In JUSTICE game, the winner is the one who played the last move.

In INJUSTICE, the winner is the first player who cannot make a legal move.

REMARK  If $\ p_0\ $ is the constant $0$-function then the first payer won INJUSTICE while the second player won JUSTICE.


Question Who wins which game (as a function of $\ |K|\ $ and $\ p_0)?$




After I created the games JUSTICE & INJUSTICE, I posted them on day 2007-06-14, on alt.pl.matematyka:

https://alt.pl.matematyka.narkive.com/Nzi0PiPA/justice-and-injustice-new-games-created-by-wlod-wh

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Here is a complete winning strategy for the Justice game.

One wins the Justice game simply by following the usual Nim strategy, with all the same winning positions and moves (except if the position is already constant and with an odd number of piles).

The usual winning Nim strategy is to create a Nim-balanced position, balanced in the sense that if each pile is represented as a sum of distinct powers of two, then each power of two occurs an even number of times overall for the position. The main Nim observation is that if a position is unbalanced, then there is a balancing move, and if a position is balanced, every move is unbalancing.

The key observation for the Justice game is that one can never move from a balanced Nim position to a winning Justice position.

If the number of plates is even, this is clear, since any constant position with an even number of plates is also Nim balanced, but every move on a balanced Nim position will unbalance it.

If the number of plates is odd, then to move to a constant position, one must have reduced a tall pile to match the constant height. Since the rest of the position would have an even number of same-height piles, that part would be balanced by itself, and so if the whole position had been balanced before, the tall pile must have had height 0, impossible. (Thanks to a comment of Edward Lockhart on Twitter.)

So by playing the Nim balancing moves, one will win Justice.

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    $\begingroup$ Very nice. Thank you. #### A time ago I introduced and solved a game that simultaneously generalizes a NIM generalization and games such as reaching a number, say 100, when increasing it from 0 but not more than by a constant. It was still by using powers of 2. But this time, for J. somehow this didn't occur to me -- I guess I wanted this game to be genuinely different from NIM but no such luck in the even case, as you have proven indeed. $\endgroup$
    – Wlod AA
    Aug 3 at 12:00
  • $\begingroup$ Somehow, despite my earlier enthusiasm, now I have some doubts. When all plates store the same number of stones by one that has a higher number of them, then the balancing doesn't matter -- the player on the move wins regardless of the balance situation. $\endgroup$
    – Wlod AA
    Aug 3 at 21:43
  • $\begingroup$ In the odd case, that situation will not arise, as I argue in my answer, since the position you describe from which the win would be obtained would not have been balanced. And in the even case, that situation does not arise, since the opponent would have achieved a balanced position, which is impossible for them. $\endgroup$ Aug 3 at 23:02

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