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alagarmobileSuivre

27 Mar 2023•0 j'aime•3 vues

27 Mar 2023•0 j'aime•3 vues

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Consider the following tiling problem. You aregiven a m X m board with m squares in each rowand m squares in each column where m is a power of 2. Onearbitrary square on the board is distinguished as special. You arealso given a supply of tiles, each of which looks like a 2×2board with one square removed (L shaped). The problem is to coverthe board with these tiles so that each square is covered exactlyonce, with the exception of the special square, which is notcovered at all. Such a covering is called a tiling. Show, usingPMI, that the tiling problem can always be solved. Solution The base case is tiling the 2x2 square with any special squaremissing: this is easy, because the L shaped tile will accomodateany of the four spaces being the open tile. Inductive step: assume boards of the size 2^n x 2^n can be coveredwith any special square- now show you can also do a 2^(n+1) x2^(n+1). This can be split into four 2^n x 2^n sized pieces,three of which will be entirely covered, and the other which has aspecial square. The one with the special square can becovered by induction. The three covered pieces can be coveredwith the one square closest to the center missing by the inductiveassumption, and these three central squares can be covered by anadditional L-shaped piece in the center..

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- 1. Consider the following tiling problem. You aregiven a m X m board with m squares in each rowand m squares in each column where m is a power of 2. Onearbitrary square on the board is distinguished as special. You arealso given a supply of tiles, each of which looks like a 2×2board with one square removed (L shaped). The problem is to coverthe board with these tiles so that each square is covered exactlyonce, with the exception of the special square, which is notcovered at all. Such a covering is called a tiling. Show, usingPMI, that the tiling problem can always be solved. Solution The base case is tiling the 2x2 square with any special squaremissing: this is easy, because the L shaped tile will accomodateany of the four spaces being the open tile. Inductive step: assume boards of the size 2^n x 2^n can be coveredwith any special square- now show you can also do a 2^(n+1) x2^(n+1). This can be split into four 2^n x 2^n sized pieces,three of which will be entirely covered, and the other which has aspecial square. The one with the special square can becovered by induction. The three covered pieces can be coveredwith the one square closest to the center missing by the inductiveassumption, and these three central squares can be covered by anadditional L-shaped piece in the center.