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Analysis of algorithm

Analysis of algorithm - Introduction CS 344 1 Analysis of...

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Introduction CS 344 1 Analysis of Algorithms I NSERTION S ORT ( A ) for j 2 to n do key A [ j ] /* Insert A [ j ] into the already sorted A [1 ..j - 1] */ /* comparing sequentially from the end */ i j - 1 while i > 0 and A [ i ] > key do A [ i + 1] A [ i ] i i - 1 A [ i + 1] key Subhasish Mazumdar CS 344 Home Page Fall 2008

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Introduction CS 344 2 NOTATION indentation / whitespace while, for , comments var: local array length[A] complex/array objects: ptrs parameter passing: by value Subhasish Mazumdar CS 344 Home Page Fall 2008
Introduction CS 344 3 The Model 1-processor random-access machine Data types: integer and floating point word size fixed but big enough to hold a given input. Subhasish Mazumdar CS 344 Home Page Fall 2008

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Introduction CS 344 4 instruction set: assignment arithmetic ( + - * / ) compromise: 2 k takes constant time for small k . control (procedure call, return, if-then, if-then- else, while, for, repeat loops) Each line of pseudo-code takes a constant amount of time. Subhasish Mazumdar CS 344 Home Page Fall 2008
Introduction CS 344 5 Example : Input sequence < 89 , 12 , 3 , 21 , 15 > : 89 12 89 3 12 89 3 12 21 89 3 12 15 21 89 In-Place sort. 1 extra location. (A sorting algo- rithm is in-place when it needs only O (1) extra storage) Number of shifts (inner loop) = # of inversions (pairs (i,j) such that A [ i ] > A [ j ] ). Subhasish Mazumdar CS 344 Home Page Fall 2008

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Introduction CS 344 6 Cost #Times for j 2 to n c 1 n do key A [ j ] c 2 n - 1 /* Insert A [ j ] into the already... */ 0 n - 1 /* examining it sequentially..., */ 0 n - 1 i j - 1 c 4 n - 1 while i > 0 and A [ i ] > key c 5 X do A [ i + 1] A [ i ] c 6 Y i i - 1 c 7 Y A [ i + 1] key c 8 n - 1 Subhasish Mazumdar CS 344 Home Page Fall 2008
Introduction CS 344 7 where X = t 2 + t 3 + ... + t n = n j =2 t j Y = ( t 2 - 1) + ( t 3 - 1) + ... + ( t n - 1) = n j =2 ( t j - 1) Best Case occurs when t j = 1 for all j . T ( n ) = c 1 n + c 2 ( n - 1) + c 4 ( n - 1)+ c 5 ( n - 1) + c 8 ( n - 1) = ( c 1 + c 2 + c 4 + c 5 + c 8 ) n - ( c 2 + c 4 + c 5 + c 8 ) = Θ( n ) Subhasish Mazumdar CS 344 Home Page Fall 2008

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Introduction CS 344 8 Worst Case occurs when t j = j for all j . T ( n ) = c 1 n + c 2 ( n - 1) + c 4 ( n - 1) + c 5 ( S ( n ) - 1) + c 6 ( S ( n ) - n ) + c 7 ( S ( n ) - n ) + c 8 ( n - 1) = Fn 2 + Gn - H where F, G, H are ... = Θ( n 2 ) Average case : some probabilistic assumptions are necessary. Subhasish Mazumdar CS 344 Home Page Fall 2008
Introduction CS 344 9 Assume that half the time the shift is necessary. In that case, X = t 2 + t 3 + ... + t n = 2 + 0 + 4 + 0 + 6 + ... + n = 2 . (1 + 2 + 3 + ... + n 2 ) = ( n 2 )( n 2 + 1) Y = ... Clearly T ( n ) = Θ( n 2 ) Subhasish Mazumdar CS 344 Home Page Fall 2008

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Introduction CS 344 10 Alternatively, assume that half the array is shifted every time, i.e., that t j = j 2 Here too, T ( n ) = Θ( n 2 ) Thus, average case worst case.
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