Let consider a routine with nested loops 24b function

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Let consider a routine with nested loops [2.4.b]. ------------------------------------------------------------ FUNCTION SumProd (n: integer): integer; VAR result,k,i: integer; BEGIN result:= 0; FOR k:= 1 TO n [2.4.b] FOR i:= 1 TO k result:= result+k*i; SumProd:= result END ;{ SumProd } ------------------------------------------------------------ Entering procedure and initializing O (1) FOR loop ( n iterations) FOR loop ( n iterations) O (n*1) = O (n) O (n*n) =O (n 2 ) O( 1+n 2 ) =O (n 2 ) Inner O (1) Fig. 2.4.b. Temporal scheme of the algorithm [2.4.b] To analyze the time complexity of this algorithm, we make use of assumption (2), and assume that the inner loop is executed the maximum number of times possible. In reality this is not true because the inner loop iterates with limit k n (and not with limit n ) We can demonstrate that this simplifying assumption does not affect the final temporal estimation. In a more careful analysis, we observe that the first time the inner loop is encountered it iterates once, the second time it iterates twice, and so on. In that case, as k varies in the outer loop, it causes the inner loop to take time O(k) . Since k in the outer loop takes on the values 1 , 2, ... , n, the exact amount of time contributed by the outer and inner loops is [2.4.c]: ------------------------------------------------------------ nnOn () ()+=12 2 1 + 2 + 3 + ... + n = [2.4.c] ------------------------------------------------------------ As result, the exact number of iterations is presented in [2.4.c], which is O(n 2 ).
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In an analog manner, the time estimation of the sequence [2.4.d] is O (n n n) = O (n 3 ). ------------------------------------------------------------ FOR i:= 1 TO n DO FOR j:= 1 TO i DO FOR k:= 1 TO i DO [2.4.d] something that is {O(1)}; ------------------------------------------------------------ The exact number of iterations can be computed as in [2.4.e], observing that the two nested inner loops iterate i 2 times for each iteration of the outer FOR loop. ------------------------------------------------------------ 1 2 + 2 2 + 3 2 + ... + n 2 = = O (n 3 ) [2.4.e] 61)(21)(+ ⋅⋅ + nnn ------------------------------------------------------------ Example 2.4.c. Consider a program segment with a multiplicative loop , one that is nested within another loop, but whose iterations are not a linear function of the control variable of the outer loop [2.4.f]. ------------------------------------------------------------ m:= 1 FOR i:= 1 TO n DO BEGIN m:= m*2; FOR j:= 1 TO m DO [2.4.f] something that is {O(1)} END ; {FOR} ------------------------------------------------------------ In this case, as the outer loop control takes on the values 1, 2, 3, …, the inner loop iterates 2, 4, 8, …, times, giving a running time of order O ( 2 + 4 + 8 + ... + 2 n ) [2.4.g]. In fact we have here the sum of a geometric series in which each term is a constant multiple of the preceding one. ------------------------------------------------------------ 2 + 4 + 8 + ... + 2 n = [2.4.g] )()( nn 1 2O12122 + =−− ------------------------------------------------------------ This estimation is completely different from estimation O (n 2 ), which we are tempted to advance, because it is valuable only for non-multiplicative loops. 2.5 Profiling an Algorithm Suppose we assume that a program solving some problem has been devised, coded, proved correct, and debugged on a computer. The program was built by expressing a certain algorithm in terms of a programming language . We are interested in producing a performance profile , that is determining the precise amounts of time and storage this program will consume.
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In order to determine exact times, our computing system must be equipped with a clock whose time can be read using specific system functions .
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