COSC 3100 Transform and Conquer Instructor: Tanvir

What is Transform and Conquer ? The 4th algorithm design technique we are going to study Three major variations Instance Simplification: Transform to a simpler or more convenient instance of the same problem Representation Change: Transform to a different representation of the same instance Problem Reduction: Transform to an instance of a different problem for which you know an efficient algorithm Two-step process: Step 1: Modify problem instance to something easier to solve Step 2: Conquer!

What is Transform and Conquer ? 1. Presorting Simpler instance Or Another representation Another problem’s instance Problem’s instance Solution 1. Computing Least Common Multiple 2. Counting Paths in a Graph 3. Reduction to Optimization Problems 4. Reduction to Graph Problems 1. Heapsort 2. Horner’s Rule 3. Binary Exponentiation

Trns. & Conq.: Presorting Many questions about a list are easier to answer if the list is sorted We shall see three examples Example 1: Checking element uniqueness in an array T(n) = (n-1)+(n-2)+…+1 = 𝒏−𝟏 𝒏 𝟐 є Θ(n2) E.g., 4 1 8 9 3 7 10 2 What is the brute-force idea ? Take say the first element and check whether it is in the rest of the array, take the second element and check if it is in the rest of the array, and so on…

Trns. & Conq.: Presorting Let us apply sorting first ALGORITHM PresortElementUniqueness(A[0..n-1]) sort the array A for i <- 0 to n-2 do if A[i] = A[i+1] return false return true 4 1 8 9 3 7 10 2 S O R T 1 2 3 4 7 8 9 10 T(n) = Tsort(n) + Tscan(n) є Θ(nlgn) + Θ(n) = Θ(nlgn)

Trns. & Conq.: Presorting Example 2: Computing a mode A mode is a value that occurs most often in a list E.g., in , mode is ? 5 1 7 6 5 Brute-force: scan the list and compute the frequencies of all distinct values, then find the value with the highest frequency How to implement brute-force ? 5 3 1 7 2 6 Value: Frequency: Store the values already encountered along with their frequencies, in a separate list. What is the worst-case input ? An array with no equal elements… C(n) = 0 + 1 + … + (n-1) = 𝒏−𝟏 𝒏 𝟐 є Θ(n2)

Trns. & Conq.: Presorting Example 2: Computing a mode (contd.) If we sort the array first, all equal values will be adjacent to each other. To compute the mode, all we need to know is to find the longest run of adjacent equal values in the sorted array

Trns. & Conq.: Presorting Example 2: Computing a mode (contd.) ALGORITHM PresortMode( A[0..n-1] ) sort the array A i <- 0 modefrequency <- 0 while i ≤ n-1 do runlength <- 1; runvalue <- A[i] while i+runlength ≤ n-1 and A[i+runlength] = runvalue runlength <- runlength+1 if runlength > modefrequency modefrequency <- runlength; modevalue <- runvalue i <- i+runlength return modevalue 1 2 3 7 8 9 10 while loop takes linear time, so the overall runtime is dominated by the time for sorting, Θ(nlgn)

Trns. & Conq.: Heapsort Another O(nlgn) sorting algorithm Uses a clever data structure called “Heap” It transforms an array into a Heap and then sorting becomes very easy Heap has other important uses, like in implementing “priority queue” Recall, priority queue is a multiset of items with an orderable characteristic called “priority”, with the following operations: Find an item with the highest priority Deleting an item with the highest priority Adding a new item to the multiset Let us study Heap first…

Trns. & Conq.: Heap A “heap” can be described as a binary tree, with one key per node, provided the following two conditions are met: Shape property: Binary tree is essentially complete, all levels are full except possibly the last level, where only some right most leaves may be missing Parental dominance (or heap property): key in each node is greater than or equal to the keys in its children (considered automatically satisfied for the leaves) This is actually the “max heap”, There is a corresponding “min heap”…

Trns. & Conq.: Heap Shape and Heap properties… Is it a Heap ? 5 7 2 1 10 5 7 2 1 10 5 2 4 7 1 Is it a Heap ? 10 5 7 2 6 1

Trns. & Conq.: Heap Note: Sequence of values on a path from the root to a leaf is nonincreasing There is no left-to-right order in key values, though 10 8 7 5 2 1 6 ALGORITHM Parent(i) return 𝑖 2 1 3 5 ALGORITHM Left(i) return 2i index: 1 2 3 4 5 6 7 8 9 10 ALGORITHM Right(i) return 2𝑖+1 value: 10 8 7 5 2 1 6 3 parents leaves

Trns. & Conq.: Heap Important properties of heaps: 8 2 5 7 1 6 3 H: 9 There exists one essentially complete binary tree with n nodes, its height is 𝑙𝑔𝑛 2. The root of a heap contains the largest element 3. A node with all its descendants is also a heap 4. Heap can be implemented as an array By recording its elements in the top-down, left-right fashion. H[0] is unused or contains a number bigger than all the rest. 1 2 3 4 5 6 7 8 9 10 H: 9 8 7 5 2 1 6 3 a. Parental node keys will be in first 𝑛 2 positions, leafs will Be in the last 𝑛 2 positions b. Children of key at index i (1 ≤ i ≤ 𝑛 2 ) will be in positions 2i and 2i+1. The parent of a key at index i (2 ≤ i ≤ n) will be in Position 𝑖 2 . Which gives: H[i] ≥ max{ H[2i], H[2i+1] } for i = 1, … , 𝒏 𝟐

Trns. & Conq.: Heap How can we transform an array into a heap? Two ways: 2 9 5 6 8 7 1. Start at the last parent at index 𝒏 𝟐 2 9 7 2. Check if parental dominance holds for this node’s key 2 6 5 8 9 8 3. If it does not, Exchange the node’s key K with the larger key of its children 6 5 7 n 1 2 3 4 5 6 9 4. Check if parental dominance holds for K in its new position and so on… Stop after doing for root. 2 9 7 6 5 8 6 2 8 6 2 5 7 This process is called “Heapifying” This method is called “bottom-up heap construction”

Trns. & Conq.: Heap 9 5 6 7 8 n 2 9 7 6 5 8 one parent Heapify ALGORITHM HeapBottomUp(H[1..n]) //Input: An array H[1..n] of orderable items //Output: A heap H[1..n] for i <- 𝑛 2 downto 1 do k <- i; v <- H[k] heap <- false while not heap and 2*k ≤ n do j <- 2*k if j < n // there are 2 children if H[j] < H[j+1] j <- j+1 if v ≥ H[j] // only left child heap <- true else H[k] <- H[j] k <- j H[k] <- v 2 9 5 6 7 8 n 1 2 3 4 5 6 2 9 7 6 5 8 one parent Heapify

Trns. & Conq.: Heap Let us analyze the HeapBottomUp(H[1..n])’s worst case Let’s take a heap whose binary tree is full Why ? 7 4 5 6 2 3 So, number of nodes, n = 2m-1 largest possible # of nodes at each level… n = 22-1 9 n = 23-1 So what is the height ? h = 𝒍𝒈𝒏 = 𝐥𝐠⁡(𝒏+𝟏) -1 = m-1 Let’s look at a worst-case… A key on level i must travel to the leaf-level h; Level i has (h-i) lower levels. Going down by 1 level requires 2 comparisons: One to determine the larger child, the other to determine if exchange is needed. Cworst(n) = 𝑖=0 ℎ−1 𝑗=1 2 𝑖 2(ℎ−𝑖) = 𝑖=0 ℎ−1 2(ℎ−𝑖) 2 𝑖 =2ℎ 𝑖=0 ℎ−1 2 𝑖 −2 𝑖=0 ℎ−1 𝑖 2 𝑖 So, fewer than 2n comparisons are required = 2h(2h-1) – 2(h-2)2h + 4 = h2h+1-2h-h2h+1+2*2h+1+4 = 2 ( n – lg(n+1) + 4 ) 𝒊=𝟏 𝒏 𝒊 𝟐 𝒊 =(𝒏−𝟏) 𝟐 𝒏+𝟏 +𝟐

Trns. & Conq.: Heap Building a heap, Method 2 It’s called “top-down heap construction” Idea: Successively insert a new key into a previously constructed heap Attach a new node with key K after the last leaf of the existing heap Sift K up until the parental dominance is established 9 9 10 6 8 6 10 6 9 2 5 7 10 2 5 7 8 2 5 7 8 Insertion operation cannot take more than twice the height of the heap’s tree, so it is in O(lgn)

Trns. & Conq.: Heap Maximum key deletion from a heap Exchange root’s key with the last key K in the heap Decrease the heap’s size by 1 “Heapify” the smaller tree by sifting K down establishing “parental dominance” as long as it is needed 1 We want to delete the root 1 8 6 8 6 9 2 5 2 5 9 8 6 2 5 1 8 Efficiency is O(lgn) 8 1 6 5 6 2 5 2 1

Trns. & Conq.: Heapsort An interesting algorithm discovered by J. W. J. Williams in 1964 It has 2 stages Stage 1: Costruct a heap from a given array Stage 2: Apply the root-deletion operation n-1 times to the remaining heap

Trns. & Conq.: Heapsort Stage 1: Heap construction 2 9 7 2 9 7 6 5 8 6 1 2 3 4 5 6 9 7 2 9 7 6 5 8 6 5 8 1 2 3 4 5 6 2 9 8 6 5 7 9 2 6 8 1 2 3 4 5 6 9 8 2 9 8 6 5 7 2 5 7 6 5 7 1 2 3 4 5 6 9 2 8 6 5 7 9 2 1 2 3 4 5 6 2 8 9 6 8 2 5 7 9 8 6 5 7 6 5 7

Trns. & Conq.: Heapsort Stage 2: Maximum deletions 9 6 8 2 5 7 2 6 5 7 1 2 3 4 5 6 1 2 3 4 5 6 9 6 8 2 5 7 2 6 5 7 8 9 1 2 3 4 5 6 1 2 3 4 5 6 7 6 8 2 5 9 6 2 5 7 8 9 1 2 3 4 5 6 1 2 3 4 5 6 8 6 7 2 5 9 5 2 6 7 8 9 1 2 3 4 5 6 1 2 3 4 5 6 5 6 7 2 8 9 2 5 6 7 8 9 1 2 3 4 5 6 7 6 5 2 8 9

Trns. & Conq.: Heapsort What is Heapsort’s worst-case complexity ? Stage 1: Heap construction is in ? O(n) Stage 2: Maximum deletions is in ? Let’s see… Let C(n) be the # of comparisons needed in Stage 2 Recall, height of heap’s tree is 𝒍𝒈𝒏 C(n) ≤ 2 𝐥𝐠⁡(𝒏−𝟏) + 2 𝒍𝒈⁡(𝒏−𝟐) + … … + 2 𝒍𝒈⁡(𝟏) ≤ 2 𝒊=𝟏 𝒏−𝟏 𝐥𝐠⁡(𝒊) C(n) ≤ 2 𝒊=𝟏 𝒏−𝟏 𝒍𝒈⁡(𝒏−𝟏) = 2(n-1)lg(n-1) ≤ 2nlgn So, C(n) є O(nlgn) So, for two stages, O(n) + O(nlgn) = O(nlgn)

Trns. & Conq.: Heapsort More careful analysis shows that Heapsort’s worst and best cases are in Θ(nlgn) like Mergesort Additionally Heapsort is in-place Timing experiments on random files show that Heapsort is slower than Quicksort, but can be competitive with Mergesort What sorting algorithm is used by Collections.sort(List<T> list) of Java?

Trns. & Conq.: Problem Reduction Compute least common multiple of two nonnegative integers, lcm(m, n) Say m = 24, and n = 60 24 = 2*2*2*3 and 60 = 2*2*3*5 Take product of all common factors of m and n, and all factors of m not in n, and all factors of n not in m lcm(24, 60) = (2*2*3)*2*5 = 120 reduction Alg. A Solution to Problem 2 Problem 2 Problem 1 To be solved Solvable by Alg. A So, m*n = lcm(m, n) * gcd(m, n) lcm(m, n) = 𝒎 ∗ 𝒏 𝐠𝐜𝐝⁡(𝒎,𝒏) ! Is it a good algorithm? Notice, 24*60 = (2*2*2*3) * (2*2*3*5) = (2*2*3)2*2*5

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