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Algorithm Efficiency Analysis: Insertion Sort and Binary Search

This text explores the efficiency of algorithms through analyzing segments, insertion sort, logarithmic orders, and the binary search algorithm, including their computational complexities. It includes examples, analyses, properties of logarithmic functions, and the efficiency of binary search.

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Algorithm Efficiency Analysis: Insertion Sort and Binary Search

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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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