5 Finish Heaps Building a heap from an array, in O(n) time Idea: The leaves are already heaps.Joining two adjacent (sub) heaps with a common root, it suffices to heapify (trickle down from the root). It takes O(h) time (h = distance from the local root to the leaves), for at most n/2h+1 nodes on that level.Total time: S n h/2h+1 = O(n), because S h/2h <2
6 Recap Linear Time Algorithms: O(n log n) Time Algorithms for sorting: Compute Sum, Product of n numbersFind Min/Max of n numbersMerge 2 arrays of n elements (total)Partition an array into 2 around a pivotO(n log n) Time Algorithms for sorting:Merge SortHeap SortQuick Sort (on average)O(log n) Time algorithms:Binary search
7 Lower Bounds Can we do better? Why not? Lower bounds prove that we cannot hope for a better algorithm, no matter how smart we are.Only very few lower bound proofs are knownMost notorious open problems in Theoretical Computer Science are related to proving lower bounds for very important problemsReading: Ch. 13 textbook
8 Input Lower BoundCompute the sum of n numbers: all numbers must be looked at, otherwise the answer might not be correctAdversary argument: assume there is a smart algorithm which computes the sum without looking at all the n inputs. An adversary goes and modifies the input not looked at, then run the algorithm again. It should give the same answer (because it didn’t look at the modified input data), but this is not the correct answer.
9 Adversary Arguments Hmmm… Yes! Is a7 < a9? Mr. Algorithm: thinks he has a fast way of solving the problemMs. Adversary: forces algorithm to work hard by given the worst possible answer
10 Adversary Arguments The answer is 3 This was the input Now try againThe answer is 3
11 Adversary ArgumentsWrong! The input was this time!The answer is 3If some questions were not asked, the Adversary tricks the poor Algorithm to try again, on a different input data, with the same answers to the same questions but with a different correct final answer….
12 Adversary Argument for FindMin: need n-1 questions No.Is a1 < a2?a3=7a4=6No.Is a3 < a4?Is a4 > a5?a5=5Yes.Minimum is a5!Wrong! It is a2!
13 If less than n-1 questions, the graph of comparisons is disconnected The adversary can re-arrange the data so that the answer is different
14 Binary Decision Trees ? Yes No Model algorithms based on successive answers to yes/no questions
16 Worst case time: depth of tree A binary tree of depth h has < 2h leavesA binary tree with N leaves must have depth at least log2NThis gives a lower bound on the worst case time to find an answerIf the number of possible answers is N, then the algorithm MUST ask at leastlog N questions
17 Lower Bound for Sorting Number of possible sorted orders = number of all possible permutations of n elements = n!Hence any comparison-based algorithm for sorting must take at least log n! = O(n log n) time
18 Lower Bound for Searching In-class exerciseHow many possible answers for the searching question?What is the log of that?What is the lower bound for searching?