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Tree search algorithm: how to quickly determine if A has a strategy with confidence in winning

The original question is: There are 99 stones, A and B play a game, each of which takes several stones in turn, and each move can take only 1, 2, 4 or 6 stones, one takes the last stone. If A is the first one to take the stones, how many stones will A take on the first turn?

This seems like a rather difficult tree search quiz, listing all branches, then process it from the bottom up: a sheet with A taking the last stone is marked as “victory”; for an intermediate node, that any B strategies could take, if A always has a way to reach the node marked as “win”, this node is also marked as “win”.

But this approach is quite laborious. Is there any smart algorithm to check if the strategy is “guaranteed to win” A?

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2 answers

O (n) solution

If we start with 1, 2, 4, or 6 stones, Ahe will always win, because he simply takes them all in the first turn.

If we start with 3, he Awill lose no matter what he does, because regardless of whether he takes 1 or 2, he Bwill take 2 or 1 next and win.

If we start with 5, he Awill win, taking 2 first, sending Bin the above case, where he will start with 3 stones.

7, A , 4, B 3.

8, A , : , B .

9, A 1 B 8, .

10, A 2 B 8 , .

, O(n): let win[i] = true if i stones are winnable for the first person to move

:

win[1] = win[2] = win[4] = win[5] = win[6] = true, win[3] = false
win[x > 6] = not (win[x - 6] and win[x - 4] and win[x - 2] and win[x - 1])

:

win[7] = not (win[1] and win[3] and win[5] and win[6])
       = not (true and false and true and true)
       = not false
       = true

, . .

O (1)

, : , A , B , , , k - 6, k - 4, k - 2, k - 1 .

win , :

win[k] = false if k = 3, 8, 11, 16, 19, 24, 27, 32...
=> win[k] = false iff k mod 8 == 3 or k mod 8 == 0

99, 99 mod 8 = 3, A .

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, , :

, 7, 7.

, n < 1000, , , 7, .

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56, O (1) 56 .

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