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Permutation

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EE364a Homework 3 solutions - Stanford Engineering …

see.stanford.edu

permutation P. Show that if f has a minimizer, then it has a minimizer of the form α1. (This means to minimize f over x ∈ Rn, we can just as well minimize f(t1) over t ∈ R.) Solution. (a) We first observe that when you multiply each Qi by some fixed Qj, you get a permutation of the Qi’s: QjQi = Qσ(i), i = 1,...,k, where σ is a ...

  Permutation

Heuristic (Informed) Search

ai.stanford.edu

3(N) = sum of permutation inversions = 4 + 6 + 3 + 1 + 0 + 2 + 0 + 0 = 16 is not admissible h(N) = number of misplaced tiles = 6 8-Puzzle Heuristics 4 1 7 5 2 3 6 8 STATE 4 6 7 1 5 2 8 3 18 h 1(N) = number of misplaced tiles = 6 is admissible h 2(N) = sum of the (Manhattan) distances of every tile to its goal position = 2 + 3 + 0 + 1 + 3 + 0 ...

  Permutation

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