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CS420 lecture five Priority Queues, Sorting

CS420 lecture five Priority Queues, Sorting. wim bohm cs csu. Heaps. Heap: array representation of a complete binary tree every level is completely filled except the bottom level: filled from left to right Can compute the index of parent and children parent(i ) = floor(i/2)

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CS420 lecture five Priority Queues, Sorting

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  1. CS420 lecture fivePriority Queues, Sorting wim bohm cs csu

  2. Heaps • Heap: array representation of a complete binary tree • every level is completely filled except the bottom level: filled from left to right • Can compute the index of parent and children • parent(i) = floor(i/2) left(i)= 2i index(i)=2i+1 (for 1 based arrays) • Heap property: A[parent(i)] >= A[i] 16 10 14 7 8 9 3 1 2 4 16 14 10 8 7 9 3 2 4 1

  3. Heapify To create a heap at i, assuming left(i) and right(i) are heaps, bubble A[i] down: swap with max child until heap property holds heapify(A,i){ L=left(i); R=right(i); if L<=N and A[L] > A[i] max=L else max = i; if R<=N and A[R]>A[max] max =R; if max != i { swap(A,i,max); heapify(A.max) } }

  4. Building a heap • heapify performs at most lgn swaps • building a heap out of an array: • The leaves are all heaps • heapify backwards starting at last internal node buildheap(A){ for i = floor(n/2) downto 1 heapify(A,i) }

  5. Complexity buildheap • Suggestions? ...

  6. Complexity buildheap • initial thought O(nlgn), but half of the heaps are height 1 quarter are height 2 only one is height log n It turns out that O(nlgn) is not tight!

  7. complexity buildheap height 0 1 2 3 max #swaps 0 = 21-2 1 = 22-3 2*1+2 = 4 = 23-4 2*4+3 = 11 = 24-5

  8. complexity buildheap Conjecture: height = h max #swaps = 2h+1-(h+2) Proof: induction height = (h+1) max #swaps: 2*(2h+1-(h+2))+(h+1) = 2h+2-2h-4+h+1 = 2h+2-(h+3) = 2(h+1)+1-((h+1)+2) n nodes O(n) swaps height 0 1 2 3 max #swaps 0 = 21-2 1 = 22-3 2*1+2 = 4 = 23-4 2*4+3 = 11 = 24-5

  9. Cormenet.al. complexity buildheap

  10. Differentiation trick (Cormenet.al. Appendix A)

  11. Heapsort heapsort(A){ buildheap(A); for i = ndownto 2{ swap(A,1,i); n=n-1; heapify(A,1); } }

  12. Complexity heapsort • buildheap: O(n) • swap/heapify loop: O(nlgn) • space: in place: n • less space than merge sort

  13. 14 8 7 7 8 7 4 1 4 2 1 4 2 14 1 2 14 8 4 2 1 2 1 1 4 2 4 7 8 14 7 8 14 7 8 14

  14. Priority Queues • heaps are used in priority queues • each value associated with a key • max priority queue S (as in heapsort) has operations that maintain the heap property of S • insert(S,x) • max(S) returning max element • Extract-max(S) extracting and returning max element • increase key(S,x,k) increasing the key value of x

  15. Extract max: O(logn) Extract-max(S){ // pre:N>0 max=S[1]; S[1]=S[N]; N=N-1; heapify(S) } O(log N)

  16. Increase key: O(logn) Increase-key(S,i,k){ //pre: k>S[i] A[i]=k; // bubble up while(i>1 and S[parent(i)]<S[i]){ swap(S,i,parent(i)); i = parent(i) } }

  17. Insert O(logn) • Insert(S,x) • put x at end of S • bubble x up like in Increase-key

  18. Decrease-key • How would decrease key work? • What would be its complexity?

  19. Quicksort • Quicksort has worst case complexity?

  20. Quicksort • Quicksort has worst case complexity O(n2) • So why do we care about Quicksort?

  21. Quicksort • Quicksort has worst case complexity O(n2) • So why do we care about Quicksort? • Because it is very fast in practice • Average complexity: O(nlgn) • Low multiplicative constants • Sequential array access • In place • Often faster than MergeSort and HeapSort

  22. Partition: O(n) Partition(A,p,r){ // partition A[p..r] in-place in two sub arrays: low and hi // all elements in low < all elements in hi // return index of last element of low x=A[p]; i=p-1; j=r+1; while true repeat j=j-1 until A[j]<=x repeat i=i+1 until A[i]>x if i<jswap(A,i,j) else return j }

  23. QuickSort Quicksort(A,p,r){ if p<r { q=Partition(A,p,r) Quicksort(A,p,q) Quicksort(A,q+1,r) } } Worst case complexity: when one partition is size 1: T(n)=T(n-1)+n=T(n-2)+(n-1)+n=T(n-3)+(n-2)+(n-1)+n= = O(n2)

  24. Quicksort average case complexity • Assume a uniform distribution: each partition index has equal probability 1/n. Thus

  25. Quicksort average case complexity We are summing all Ts up (T(q)) and down (T(N-q)), both are the same sums so

  26. Quicksort average case complexity • multiply both sides by n • substitute n-1 for n

  27. Quicksort average case complexity • subtract (2)-(3)

  28. Quicksort average case complexity • Which technique that we have learned can help us here?

  29. Quicksort average case complexity • repeated substitution!!

  30. Quicksort average case complexity

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