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  "body": "\n## Statement\n\nLet $k \\ge 2$ and $t \\ge 1$ be integers.\nLet a necklace be a row of beads of $t$ colours.\nSuppose colour $i$ occurs exactly $k a_i$ times, for some integer $a_i \\ge 0$.\nThen the necklace can be cut in at most $(k - 1) t$ places so that the pieces can be divided among $k$ people fairly.\nFairly means that each person receives exactly $a_i$ beads of colour $i$, for every $i$.\n\n## Remarks\n\nFor $k = 2$ the result follows from the [[Borsuk–Ulam theorem|QHJ0GBQE2E]].\nModel the necklace as the interval $[0, 1]$, with each bead a short subinterval.\nA point $x$ of the sphere $S^t \\subset \\R^{t+1}$ encodes a cut into $t+1$ consecutive pieces of lengths $x_1^2, \\dots, x_{t+1}^2$.\nThe sign of $x_j$ tells which of the two people gets piece $j$.\nFor each colour, take the amount the first person gets minus half the total; this defines an odd map $S^t \\to \\R^t$.\nBy Borsuk–Ulam it has a zero, which is a fair split with at most $t$ cuts.\nA final step moves the cuts to bead boundaries.\nThe general case is due to Noga Alon, by induction on the prime factors of $k$ using a generalisation of Borsuk–Ulam.\n\nThe bound is sharp: if the beads of each colour lie in one contiguous block, each block needs $k - 1$ cuts.\n\n## Explanations\n\n### L1 human\n\nThieves steal a necklace with several kinds of jewels and want to share each kind equally, using as few cuts as possible.\nWith two thieves and $t$ kinds of jewels, $t$ cuts always suffice, wherever the jewels sit.\nThe surprising proof uses a theorem from topology about maps on spheres.\n\n### L2 human\n\nThis is a standard showcase of topological methods in combinatorics, next to Lovász's proof of the Kneser conjecture.\nThe continuous version, where colours are replaced by measures on an interval, is the form to which Borsuk–Ulam applies directly, much as in the [[ham sandwich theorem|GDXX60SNG5]].\nThe bound $(k - 1) t$ cannot be lowered in general, as the example of contiguous colour blocks shows.\n",
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