General Big-Oh Rules

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Mathematical Expression

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T(N)= O ( F ( N ) )

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Relative Rates of Growth

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Growth of T ( N ) is Igrowth of F(N)

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TU@=OtF(N)I T(N)= @ ( F ( N ) )

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1 Growth of

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TIN) is 2 growth of F ( M

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Growth of T ( N ) is = growth of F(N)

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T Growth of

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T ( N ) is c growth of F ( N ) ,

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Meanings of the various growth functions

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of the loop condition is usually no more dominant than are the statements encompassing the body of the loop The running time of statements inside a group of nested loops is the running time of the statements (including tests in the innermost loop) multiplied by the sizes of all the loops The running time of a sequence of consecutive loops is the running time of the dominant loop The time difference between a nested loop in which both indices run from 1 to N and two consecutive loops that are not nested but run over the same indices is the same as the space difference between a two-dimensional array and two one-dimensional arrays The first case is quadratic The second case is linear because N + N is 2N, which is still O(N)Occasionally, this simple rule can overestimate the running time, but in most cases it does not Even if it does, Big-Oh does not guarantee an exact asymptotic answer-just an upper bound The analyses performed thus far involved use of a worst-case bound, which is a guarantee over all inputs of some size Another form of analysis is the average-case bound, in which the running time is measured as an average over all the possible inputs of size N The average might differ from the worst case if, for example, a conditional statement that depends on the particular input causes an early exit from a loop We discuss average-case bounds in more detail in Section 68 For now, simply note that, because one algorithm has a better worst-case bound than another algorithm, nothing is implied about their relative average-case bounds However, in many cases average-case and worst-case bounds are closely correlated When they are not, the bounds are treated separately The last Big-Oh item we examine is how the running time grows for each type of curve, as illustrated in Figures 61 and 62 We want a more quantitative answer to this question: If an algorithm takes T(N) time to solve a problem of size N, how long does it take to solve a larger problem For instance, how long does it take to solve a problem when there is 10 times as much input The answers are shown in Figure 610 However, we want to

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A worst-case bound a guaranteeOver all inputs of some size

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In an average-case bound, the running time is measured as an over all of the ~ossible inouts of size N

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Algorithm Analysis

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Observed running times (in seconds) for various maximum contiguous subsequence sum algorithms

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answer the question without running the program and hope that our analytical answers agree with the observed behavior We begin by examining the cubic algorithm We assume that the running time is reasonably approximated by T(N) = cN3 Consequently, T(1ON) = c( 1Om3Mathematical manipulation yields

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If the size of the input increases by a factor of f, the running time of a cubic program increases by a factor of roughly f 3

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If the size of the input increases by a factor of f, the running time of a quadratic program increases by a factor of roughly f 2

Thus the running time of a cubic program increases by a factor of 1000 (assuming that N is sufficiently large) when the amount of input is increased by a factor of 10 This relationship is roughly confirmed by the increase in running time from N = 100 to 1000 shown in Figure 610 Recall that we do not expect an exact answer-just a reasonable approximation We would also expect that for N = 10,000, the running time would increase another 1000fold The result would be that a cubic algorithm requires roughly 35 minutes of computation time In general, if the amount of input increases by a factor o f j the cubic algorithm's running time increases by a factor of f 3 We can perform similar calculations for quadratic and linear algorithms For the quadratic algorithm, we assume that T(N) = cN2 It follows that T(1ON) = c(lON)2 When we expand, we obtain

So when the input size increases by a factor of 10, the running time of a quadratic program increases by a factor of approximately 100 This relationship is also confirmed in Figure 610 In general, an ffold increase in input size yields an f 2-fold increase in running time for a quadratic algorithm