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back to boardFor prove it just use some maths See on one iteration of "for": let be x and y on start:x1 and y1. then after each change it'll be for x1 x2...x3; for y1 y2...y3; so you have to find out x3 and y3 through x and y on start(x1 and y1); Something like that x3=y1. Good luck and sorry if my English is not enough good for explaining it) |
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