pset10_sol_04

pset10_sol_04 - Home Work 10 The problems in this problem...

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Unformatted text preview: Home Work 10 The problems in this problem set cover lectures C1 1 and C1 2 1. a. Define a recursive binary search algorithm. If lb > ub Return -1 else Mid := (lb+ub)/2 If Array(Mid) = element Return Mid Elsif Array(Mid) < Element Return Binary_Search(Array, mid+1, ub, Element) Else Return Binary_Search(Array, lb, mid-1, Element) End if End if b. Implement your algorithm as an Ada95 program. 46 . function Binary_Search (My_Search_Array : My_Array; Lb : Integer; Ub: Integer; Element : Integer) return Integer is 47 . mid : integer; 48 . begin 49 . if (Lb> Ub) then 50 . return - 1 ; 51 . else 52 . Mid := (Ub+Lb)/ 2 ; 53 . if My_Search_Array(Mid) = Element then 54 . return (Mid); 55 . elsif My_Search_Array(Mid) < Element then 56 . return (Binary_Search(My_Search_Array, Mid+ 1 , Ub, Element)); 57 . else 58 . return (Binary_Search(My_Search_Array, Lb, Mid- 1 , Element)); 59 . end if ; 60 . end if ; 61 . 62 . end Binary_Search; 63 . end Recursive_Binary_Search; c. What is the recurrence equation that represents the computation time of your algorithm? Recursive Binary Search Cost if (Lb> Ub) then c1 return - 1 ; c2 else c3 Mid := (Ub+Lb)/ 2 ; c4 if My_Search_Array(Mid) = Element then c5 return (Mid); c6 elsif My_Search_Array(Mid) < Element then c7 return (Binary_Search(My_Search_Array, Mid+ 1 , Ub, Element)); T(n/2) else c8 return (Binary_Search(My_Search_Array, Lb, Mid- 1 , Element)); T(n/2) end if ; c9 end if ; c10 In this case, only one of the recursive calls is made, hence only one of the T(n/2) terms is included in the final cost computation. Therefore T(n) = (c1+c2+c3+c4+c5+c6+c7+c8+c9+c10) + T(n/2) = T(n/2) + C d. What is the Big-O complexity of your algorithm? Show all the steps in the computation based on your algorithm. T(n) = T(n/2) + C T(n) = aT(n/b) + cn k , where a,c > 0 and b > 1 T(n) = log b a O ? ? a ? b k n k k O ? n log n ? a ? b b 1 = 2 , hence the second term is used, k ? ? k T(n) = 2. What is the Big-O complexity of : a. Heapify function A heap is an array that satisfies the heap properties i.e., A(i) A(2i) and A(i) A(2i+1). The heapify function at i makes A(i .. n) satisfy the heap property, under the assumption that the subtrees at A(2i) and A(2i+1) already satisfy the heap property. Heapify function Cost Lchild := Left(I); c1 Rchild := Right(I); c2 if (Lchild <= Heap_Size and Heap_Array(Lchild) > Heap_Array(I)) c3 Largest:= Lchild; c4 else c5 Largest := I; c6 if (Rchild <= Heap_Size) c7 if Heap_Array(Rchild) > Heap_Array(Largest) c8 Largest := Rchild; c9 if (Largest /= I) then c10 Swap(Heap_Array, I, Largest); c11 Heapify(Heap_Array, Largest); T(2n/3) T(n) = T(2n/3) + C = T(2n/3) + O(1) a = 1, b = 3/2, f(n) = 1, therefore by master theorem, log b a T(n) = O ( n log n ) = O ( n log 3/ 2 1 log n ) = O(1 * log n) = O(log n) The important point to note here is the T(2n/3) term, which arises in the worst case, when...
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pset10_sol_04 - Home Work 10 The problems in this problem...

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