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Computing an Order of an Algorithm Segment

Efficiency of Algorithms: Analyzing algorithm segments, Insertion sort algorithm Logarithmic Orders Binary Search Algorithm. Computing an Order of an Algorithm Segment. Consider the following algorithm segment : max:=a[1]; min:=b[1] for i:=2 to n if max < a[i] then max:=a[i] ;

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Computing an Order of an Algorithm Segment

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  1. Efficiency of Algorithms:Analyzing algorithm segments,Insertion sort algorithmLogarithmic OrdersBinary Search Algorithm

  2. Computing an Order of an Algorithm Segment • Consider the following algorithm segment: max:=a[1]; min:=b[1] for i:=2 to n if max < a[i] then max:=a[i] ; if min > b[i] then min:=b[i] next i if min ≥ max then med:=(min+max)/2 • Analysis: Number of loop iterations: n-1 Comparisons in each iteration: 2 Thus, elementary operations in the loop: 2(n-1) Operations after the loop: 3 Total number of operations in the segment: 2(n-1)+3 = 2n+1 • Order of the segment: O(n)

  3. The Insertion Sort Algorithm • Insertion sort arranges the elements of one-dimensional array a[1], …, a[n] into increasing order. • for i:=2 to n (insert a[i] into the sorted sequence a[1],…,a[i-1]) key:=a[i] for j:=1 to i-1 if key<a[j] then goto (*) next j (*)(shift the values of a[j],…,a[i-1] to a[j+1],…,a[i]) for k:=i to j+1 a[k]:=a[k-1] next k a[j]:=key next i

  4. The Insertion Sort Algorithm:Example • Arrange array <8, 2, 7, 9, 1, 4> into increasing order. 8 2 7 9 1 4 2 8 7 9 1 4 2 7 8 9 1 4 2 7 8 9 1 4 1 2 7 8 9 4 1 2 4 7 8 9

  5. The Insertion Sort Algorithm:Analysis • For each i (outer loop), maximum number of elementary operations is i – 1 . • Thus, the total number of elementary operations: 1+2+…+(n-1) = n(n-1)/2 = .5n2 - .5n • The insertion sort algorithm is O(n2) .

  6. Logarithmic function • Definition: The logarithmic function with base b (b>0, b1) is the following function from R+ to R: logb(x) = the exponent to which b must raised to obtain x . Symbolically, logbx = y  by = x . • Property: If the base b>1, then the logarithmic function is increasing: if x1<x2 , then logb(x1) < logb(x2) . Note: Logarithmic function grows very slowly, e.g., log2(1,024)=10, log2(1,048,576)=20 .

  7. A property of logarithmic function • Proposition 1:Ifk is an integer and x is a real number with 2k x < 2k+1, thenlog2x = k . • Proof: 2k x < 2k+1 log2(2k)  log2(x) < log2(2k+1) (since log2x increasing) k log2(x) < k+1 (by definition of log2x) log2x = k (by definition of floor function) ■

  8. An application of logarithms • Question: Given a positive integer n, how many binary digits are needed to represent n? • Solution: Binary representation of n: 1ck-1ck-2…c2c1c0 which corresponds to n = 2k + ck-1∙2k-1 + … + c2∙22 + c1∙2 + c0 Since ci 1, n = 2k + ck-1∙2k-1 + … + c2∙22 + c1∙2 + c0  2k + 2k-1 + … + 22 + 2 + 1 = (2k+1-1) / (2-1) (as a sum of geometric sequence) = 2k+1-1 <2k+1(1) On the other hand, n = 2k + ck-1∙2k-1 + … + c2∙22 + c1∙2 + c0 ≥ 2k(2) Combining (1) and (2):2k n < 2k+1(3) Based on (3) and Proposition 1, k = log2n and the number of binary digits is log2n + 1.

  9. Exponential and Logarithmic Orders • For all real numbers b and r with b>1 and r>0 andfor all sufficiently largevalues of x,  logbx  xr; ( which implies that logbx is O(xr))  xr bx ( which implies that xr is O(bx) ) • For all real numbers b with b>1 andfor all sufficiently largevalues of x, x  x logbx  x2 (which implies that xis O(x logbx) and x logbxis O(x2) )

  10. Binary Search Algorithm • The algorithm searches for an element x in an ascending array of elements a[1],…,a[n] . • Algorithm body: index:=0, bot:=1, top:=n while (top ≥ botandindex=0) mid :=  (bot+top) / 2 if a[mid] = xthen index := mid if a[mid] > x thentop := mid-1 elsebot := mid+1 end while Output: index (Ifindex has the value 0 thenx is not in the array; otherwise, index gives the index of the array where x is located.)

  11. Binary Search Algorithm: Example • Suppose a[1]=Amy, a[2]=Bob, a[3]=Dave, a[4]=Erin, a[5]=Jeff, a[6]=John, a[7]=Larry, a[8]=Mary, a[9]=Mike, a[10]=Sam, a[11]=Steve, a[12]=Tom. (sorted in alphabetical order) • Search for x=Erin. • The table tracing the binary search algorithm:

  12. The Efficiency of the Binary Search Algorithm • At each iteration, the length of the new subarray to be searched is approximately half of the previous one. • If n = 2k+m, where 0  m < 2k, then n can be split approximately in half k times. • Since 2k  n < 2k+1, then k = log2n(by proposition 1) • Thus, the number of iterations of the while loop in a worst-case execution of the algorithm is log2n+1. • The number of operations in each loop is constant (doesn’t increase with n). • Thus, the binary search algorithm is O(log2n) .

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