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  "body": "\n## Statement\n\nLet $P \\subset \\R^2$ be a simple polygon with $n \\ge 3$ vertices.\nSay that a point $g \\in P$ sees a point $x \\in P$ when the segment from $g$ to $x$ lies in $P$.\n\n1. There is a set of at most $\\floor{n/3}$ vertices of $P$ such that every point of $P$ is seen by at least one of them.\n2. For every $n \\ge 3$ there is a simple polygon $P$ with $n$ vertices that no set of fewer than $\\floor{n/3}$ points of $P$ sees entirely.\n\n## Remarks\n\nThe theorem is due to Václav Chvátal; the short proof below is due to Steve Fisk.\nTake a [[triangulation|Q6KKBP3RRP]] of $P$, and consider the graph whose edges are the sides of $P$ and the diagonals used.\nThis graph can be properly coloured with $3$ colours (its [[chromatic number|0VHJRXNAAS]] is $3$), by removing an ear triangle, colouring the rest, and giving the last vertex the missing colour.\nEach triangle then has one vertex of each colour.\nThe smallest colour class has at most $\\floor{n/3}$ vertices, and it contains a vertex of every triangle, which sees the whole triangle because triangles are convex.\n\nFor sharpness, a comb with $k$ narrow teeth has $n = 3k$ vertices, and no point sees into two different teeth, so $k = n/3$ guards are needed.\n\n## Explanations\n\n### L1 human\n\nPicture an art gallery whose floor plan is a polygon, and guards who stand still but can turn around.\nThe theorem says that one guard for every three corners is always enough.\nSome galleries, shaped like combs, really need that many.\n\n### L2 human\n\nThe proof is a model use of polygon triangulation: cut, colour, and count.\nThe bound $\\floor{n/3}$ is a worst case over all polygons; for a given polygon the minimum number of guards can be much smaller, and computing it is NP-hard.\nMany variants exist, such as orthogonal polygons, where $\\floor{n/4}$ guards suffice, and polygons with holes.\n",
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