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  "body": "\n## Statement\n\nLet $n \\ge 1$.\nCall a [[tiling|GKJK293DXS]] of $\\R^n$ a *cube tiling* if each tile is a [[translate|087VJM5Q77]] $[0, 1]^n + v$ of the unit cube, with $v \\in \\R^n$.\nThen the following are equivalent.\n\n1. Every cube tiling of $\\R^n$ contains two cubes that meet in a common $(n-1)$-dimensional face.\n2. $n \\le 7$.\n\n## Remarks\n\nOtt-Heinrich Keller conjectured in 1930 that clause 1 holds in every dimension.\nOskar Perron proved it for $n \\le 6$ in 1940.\nJeffrey Lagarias and Peter Shor showed in 1992 that it fails for $n \\ge 10$, and John Mackey showed in 2002 that it fails for $n = 8$, hence also for $n = 9$.\nThe case $n = 7$ was settled positively in 2020 by Joshua Brakensiek, Marijn Heule, John Mackey and David Narváez with a large computer search based on a SAT solver.\nThe counterexamples come from cliques in the *Keller graphs*, finite graphs whose large cliques correspond to cube tilings without face-to-face pairs.\nFor $n = 2$ the statement is easy to see.\nA tiling of the plane by unit squares splits into rows or into columns, and two neighbouring squares in a row or column share an edge.\n\n## Explanations\n\n### L1 human\n\nFill space with equal cubes, all facing the same way, with no gaps and no overlaps.\nKeller guessed that some two cubes must always touch along a whole face.\nThis is true in dimensions up to $7$ but false from dimension $8$ on.\n\n### L2 human\n\nIn the plane the conjecture is easy, and in three dimensions it can still be checked by hand.\nHajós turned the problem into a question about factorizations of finite abelian groups, and later the problem was reduced to finite graphs.\nThese reductions allowed computer searches, and the final case, dimension $7$, needed a computer-assisted proof.\nThe lattice version, where the translation vectors form a lattice, is Minkowski's conjecture, proved by Hajós in 1941 in all dimensions.\n",
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