Divide And ConquerSmall And Large InstanceSolving A Small InstanceSolving A Large InstanceSort A Large ListFind The Min Of A Large ListRecursion In Divide And ConquerRecursive Find MinTiling A Defective ChessboardOur Definition Of A ChessboardPowerPoint PresentationA TriominoSlide 13Slide 14Slide 15Slide 16ComplexitySubstitution MethodMin And MaxMax ElementSlide 21Slide 22Large Instance Min And MaxMin And Max ExampleDividing Into Smaller InstancesSolve Small Instances And CombineTime ComplexityInterpretation Of Recursive VersionIterative VersionIterative Version ExampleComparison CountInitialize A HeapInitialize A Loser TreeDivide And Conquer•Distinguish between small and large instances.•Small instances solved differently from large ones.Small And Large Instance•Small instance.Sort a list that has n <= 10 elements. Find the minimum of n <= 2 elements.•Large instance.Sort a list that has n > 10 elements. Find the minimum of n > 2 elements.Solving A Small Instance•A small instance is solved using some direct/simple strategy.Sort a list that has n <= 10 elements.• Use count, insertion, bubble, or selection sort.Find the minimum of n <= 2 elements.•When n = 0, there is no minimum element.•When n = 1, the single element is the minimum.•When n = 2, compare the two elements and determine which is smaller.Solving A Large Instance•A large instance is solved as follows:Divide the large instance into k >= 2 smaller instances.Solve the smaller instances somehow.Combine the results of the smaller instances to obtain the result for the original large instance.Sort A Large List•Sort a list that has n > 10 elements.Sort 15 elements by dividing them into 2 smaller lists.One list has 7 elements and the other has 8.Sort these two lists using the method for small lists.Merge the two sorted lists into a single sorted list.Find The Min Of A Large List•Find the minimum of 20 elements.Divide into two groups of 10 elements each.Find the minimum element in each group somehow.Compare the minimums of each group to determine the overall minimum.Recursion In Divide And Conquer•Often the smaller instances that result from the divide step are instances of the original problem (true for our sort and min problems). In this case,If the new instance is a small instance, it is solved using the method for small instances.If the new instance is a large instance, it is solved using the divide-and-conquer method recursively.•Generally, performance is best when the smaller instances that result from the divide step are of approximately the same size.Recursive Find Min•Find the minimum of 20 elements.Divide into two groups of 10 elements each.Find the minimum element in each group recursively. The recursion terminates when the number of elements is <= 2. At this time the minimum is found using the method for small instances.Compare the minimums of the two groups to determine the overall minimum.Tiling A Defective ChessboardOur Definition Of A ChessboardA chessboard is an n x n grid, where n is a power of 2.1x1 2x2 4x4 8x8A defective chessboard is a chessboard that has one unavailable (defective) position. 1x1 2x2 4x4 8x8A TriominoA triomino is an L shaped object that can cover three squares of a chessboard. A triomino has four orientations.Tiling A Defective ChessboardPlace (n2 - 1)/3 triominoes on an n x n defective chessboard so that all n2 - 1 nondefective positions are covered.1x1 2x2 4x4 8x8Tiling A Defective ChessboardDivide into four smaller chessboards. 4 x 4One of these is a defective 4 x 4 chessboard.Tiling A Defective ChessboardMake the other three 4 x 4 chessboards defective by placing a triomino at their common corner.Recursively tile the four defective 4 x 4 chessboards.Tiling A Defective ChessboardComplexity•Let n = 2k.•Let t(k) be the time taken to tile a 2k x 2k defective chessboard.•t(0) = d, where d is a constant.•t(k) = 4t(k-1) + c, when k > 0. Here c is a constant.•Recurrence equation for t().Substitution Methodt(k) = 4t(k-1) + c = 4[4t(k-2) + c] + c = 42 t(k-2) + 4c + c = 42[4t(k-3) + c] + 4c + c = 43 t(k-3) + 42c + 4c + c = … = 4k t(0) + 4k-1c + 4k-2c + ... + 42c + 4c + c = 4k d + 4k-1c + 4k-2c + ... + 42c + 4c + c = Theta(4k) = Theta(number of triominoes placed)Min And MaxFind the lightest and heaviest of n elements using a balance that allows you to compare the weight of 2 elements.Minimize the number of comparisons.Max Element•Find element with max weight from w[0:n-1].maxElement = 0;for (int i = 1; i < n; i++)if (w[maxElement] < w[i]) maxElement = i; •Number of comparisons of w values is n-1.Min And Max•Find the max of n elements making n-1 comparisons.•Find the min of the remaining n-1 elements making n-2 comparisons.•Total number of comparisons is 2n-3.Divide And Conquer•Small instance.n <= 2.Find the min and max element making at most one comparison.Large Instance Min And Maxn > 2.Divide the n elements into 2 groups A and B with floor(n/2) and ceil(n/2) elements, respectively.Find the min and max of each group recursively.Overall min is min{min(A), min(B)}.Overall max is max{max(A), max(B)}.Min And Max Example•Find the min and max of {3,5,6,2,4,9,3,1}.•Large instance.•A = {3,5,6,2} and B = {4,9,3,1}.•min(A) = 2, min(B) = 1. •max(A) = 6, max(B) = 9.•min{min(A),min(B)} = 1.•max{max(A), max(B)} = 9.Dividing Into Smaller Instances{8,2,6,3,9,1,7,5,4,2,8}{8,2,6,3,9}{1,7,5,4,2,8}{8,2}{6,3,9}{6} {3,9}{1,7,5} {4,2,8}{1}{7,5} {4}{2,8}Solve Small Instances And Combine{8,2}{6} {3,9}{1}{7,5} {4}{2,8}{2,8}{6,6} {3,9}{3,9}{2,9}{1,1} {5,7}{1,7}{4,4}{2,8}{2,8}{1,8}{1,9}Time Complexity•Let c(n) be the number of comparisons made when finding the min and max of n elements.•c(0) = c(1) = 0.•c(2) = 1.•When n > 2, c(n) = c(floor(n/2)) + c(ceil(n/2)) + 2•To solve the recurrence, assume n is a power of 2 and use repeated substitution.•c(n) = ceil(3n/2) - 2.Interpretation Of Recursive Version•The working of a recursive divide-and-conquer algorithm can be described by a tree—recursion tree.•The algorithm moves down the recursion tree dividing large instances into smaller ones.•Leaves represent small instances.•The recursive algorithm moves back up the tree combining the results from the subtrees.•The combining finds the min of the mins computed at leaves and the max of the leaf