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  "body": "\n## Definition\n\nLet $x_1, \\ldots, x_n \\in \\R^d$ and let $1 \\le k \\le n$.\nFor a [[partition|08QMKTWF9T]] of $\\set{1, \\ldots, n}$ into non-empty sets $S_1, \\ldots, S_k$, let $m_j = \\frac{1}{\\abs{S_j}} \\sum_{i \\in S_j} x_i$ be the mean of cluster $j$.\nThe *$k$-means problem* is to find a partition that minimises the within-cluster sum of squares\n$$W(S_1, \\ldots, S_k) = \\sum_{j=1}^k \\sum_{i \\in S_j} \\abs{x_i - m_j}^2,$$\nwhere $\\abs{\\cdot}$ is the [[Euclidean norm|0NPPGHQNXQ]].\n*$k$-means clustering* is the [[clustering|D9Y3A9RH6J]] given by a solution, or by an approximate solution, of this problem.\n\n## Remarks\n\nLloyd's algorithm starts from $k$ centres and repeats two steps: assign each $x_i$ to a nearest centre, then replace each centre by the mean of its assigned points.\nNeither step increases $W$, and there are finitely many partitions, so the algorithm stops after finitely many steps, at a local minimum that need not be global.\nExample: for $0, 1, 10, 11 \\in \\R$ and $k = 2$, the partition $\\set{0, 1}, \\set{10, 11}$ has means $0.5$ and $10.5$ and $W = 4 \\cdot 0.25 = 1$.\nNon-example: the partition $\\set{0}, \\set{1, 10, 11}$ has means $0$ and $22/3$ and $W = 546/9 \\approx 60.7$, so it is not a solution.\nThe mean is used because, for fixed clusters, $m_j$ is the point $c$ that minimises $\\sum_{i \\in S_j} \\abs{x_i - c}^2$.\n",
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