Class Notes (835,600)
CS 240 (49)
Lecture

# Cheat Sheet #2 I tried to fit in as much details as I could in one page- its a useful for reviewing

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School
Department
Computer Science
Course
CS 240
Professor
Therese Biedl
Semester
Fall

Description
Bubble-up (Node i ):Time = O(logn) use 4 insert Heapsort /* Heapify: */Equal(n) QuickSort Running time details while (p(i ) exists) and (value(p(i)) < value(i)) { for i := n/2 downto 1 exchange values(i ) and value(p(i )); Bubble-down(i) i ← p(i );} /* ExtractMax one at a time */ while (n > 1) { Bubble-down (Node i ): Time:O(logn) use EMax Swap(A[n], A[1]); while (i has children) { n--; let j be the child of i with the largest value Bubble-down(1);} if value(j) > value(i ) then exchange values(i ) and value(j); QuickSortRecurseOnSmallerPart(int[] A, int left, int right): Rotations i ← j ; int myLeft := left, myRight := right; RandomPermute(A, n): otherwise for i ← 1 to n while (myLeft < myRight) do break;} exchange(A[i ], A[random(i , n)]) | int i := Partition(A, myLeft, myRight); | if (i-myLeft) < (myRight - i) then QuickSort(int[] A, int left, int right): | | QuickSortRecurseOnSmallerPart(A, myLeft, i-1); if left < right | | myLeft = i+1 | else int i := Partition(A, left, right); | | QuickSortRecurseOnSmallerPart(A, i+1, myRight); QuickSort(A, left, i-1); QuickSort(A, i+1, right); | | myRight = i-1 Partition(A, left, right): Theorem pivot := A[right]; The height of an AVL tree with n nodes is O(log n). i := left-1; Proof: What is the least number of nodes (nh) of an AVL-tree of height h? for j := left to right Recurrence relation: ni = ni−1 + ni−2 + 1. Predecessor: at most right in leftsubtree if A[j] <= pivot ni = Fi+3 – 1 = floor( feta^(i+3) / sqrt(5) ) + ½ - 1 Successor : at most left in rightsubtree i := i+1 Therefore, ni = equal( feta^I ) Flooring and ceiling swap(A[j], A[i]) Hence, n ≥ nh = equal(feta^I.) implies h = O(logn) 3/2 floor(1) ceiling (2) return i; 5/2 floor(2) ceiling (3) Running time of Partition: Extendable hashing: search 7/2 floor(3) ceiling (4) Equal(right − left + 2). Summation equations Given a key k, compute h(k) = a1a2 . . . aL. Auxiliary Space: Equal(1). ◮ Probe block B(a1 . . . ad ). ◮ Perform a binary search to find all items in the block. If the directly resides in internal memory, we have only a single memory Priority queue sort block transfer. for i ← 1 . . . n do PQ.insert(A[i]); for i ← n . . . 1 do Insert(x) is done as follows: ◮ Unsuccessful Search(x): A[i] = PQ.extractMax(); ◮ Search for x to find the proper block to insert ◮ Computing the hash value h(x) takes Equal(1). ◮ If the block has space then done ◮ A list is searched to the end of the list. ◮ If the block is full (i.e. has already S keys) and k < d, perform ablock split: ◮ By uniform hashi
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