Notice that $T$ is now done, and that we have hence increased the number of done stacks while returning to state 2.
We put these $d$ balls into an empty stack, which we now call storage stack and move to state 2 bellow. Otherwise there is some stack $s$, say, whose top $d$ balls are of the same colour, but whose next ball is of a different colour. If all stacks are done, we can move on to depth $d+1$. (state 1) Assume we are at depth $d$ with two empty stacks.Ĭall a stack done if its top $d+1$ balls are of the same colour.
BALLS PUZZLE GAMES PLUS
After performing this permutation, we will get all stacks having the top $d+1$ balls of the same colour plus two empty stacks and move to depth $d+1$ until we reach depth 4.
BALLS PUZZLE GAMES FULL
Moreover, the number $4$ can be replaced by any number.Īt each depth $d$, we will start with all stacks having the top $d$ balls of the same colour plus two empty stacks, and we will make a permutation of the top $d$ layers, each time moving the top $d$ balls of some full stack to a stack with at least $d$ empty slots. If we make this assumption about the starting configuration, this game is solvable (although quite boring) for any $n$, and only two empty stacks are needed. All we need is $n$ so large that no matter which $m$ top colors your choose to move first, the $4m$ balls your exposed at layer $3$ have very few common colors, and after making those moves, the newly exposed balls at layer $2$ are all new.ĪSSUMPTION : I assume that for each height $1\leqslant h \leqslant 4$, and for each colour $c$, there is a stack where there is a ball of colour $c$ at height $h$. some starting configurations will be unsolvable. If we moved all b's and c's, the only common color exposed is j, and moving either j will expose a new color ( e or f) and the game is at a dead-end.įurther thoughts: I wonder if, by similar logic, for any number of empty stacks $m$ (above discusses $m=2$), there exists large enough $n$ s.t. If we moved all a's and c's, the only common color exposed is d, and moving either d will expose a new color ( h or i) and the game is at a dead-end. This is the only legal move left, and whichever g you move, it will expose yet a new color ( k or l) and the game is at a dead-end. If we moved all a's and b's, the only common color exposed is g. So for the next four moves, we might as well move all four balls of another top color.
The same is true if the first four moves moved all four b's or all four c's. Moving all the a's exposed four new balls, but by construction, they are all different colors defg, and, if you move any of them into $E_2$, this exposes yet another new color i or k and the game is at a dead-end (thanks to for pointing this out). Therefore, the first four moves might as well move all four a's onto $E_1$. Moving another a to $E_1$ cannot possibly hurt, since nothing else can go onto $E_1$, and the a in $E_1$ cannot go anywhere else (except $E_2$, which doesn't improve the game state). The first move, without loss of generality, might be a to an empty stack $E_1$.Įvery future move involving an a must be either to $E_1$ or the other empty stack $E_2$, because every a started at the top and it is impossible to put another a on top of it. The key observation is that the first four moves might as well be all of the same (top) color: they can be filled arbitrarily with the remaining balls of colors d through l. This diagram uses $4$ balls of colors a, b, c, spread across the top layers, and $2$ or $3$ balls of colors d, e, f, g, h, i, j, k, l. Partial solution: Even for $n = 12$ and $2$ empty stacks, the game is not solvable for this initial configuration: a a a a b b b b c c c c
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