Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 9 pdf
... + E) to compute G . • O(VE) to run BELLMAN-FORD. • (E) to compute w. • O(V 2 lg V +VE) to run Dijkstra’s algorithm | V | times (using Fibonacci heap). • (V 2 ) to compute D matrix. Total: ... t (k−1) kj . T RANSITIVE-CLOSURE(E, n) for i ← 1 to n do for j ← 1 to n do if i = j or (i, j) ∈ E[G] then t (0) ij ← 1 else t (0) ij ← 0 for k ← 1 to n do for i ← 1 to n do for j...
Ngày tải lên: 13/08/2014, 18:20
... Data Structures Lecture Notes 14-1 Solutions 14 -9 Preface This document is an instructor’s manual to accompany Introduction to Algorithms, Second Edition, by Thomas H. Cormen, Charles E. Leiserson, ... invariants is like mathematical induction: Instructor’s Manual by Thomas H. Cormen Clara Lee Erica Lin to Accompany Introduction to Algorithms Second Edition by...
Ngày tải lên: 13/08/2014, 18:20
... 5. • MAX-HEAPIFY is applied to subtrees rooted at nodes (in order): 16, 2, 3, 1, 4. 1 23 4567 891 0 1 23 4567 891 0 4 13 291 0 14 8 7 16 41 23 16 9 10 14 8 7 16 14 10 893 241 7 A i 234567 891 01 Correctness Loop ... E 2 [ X j ] = 0 2 · 99 100 + 100 2 · 1 100 − 1 2 = 100 − 1 = 99 . Summing up the variances of the X j gives Var [ V n ] = 99 n. 5-12 Solutions for Chapter 5...
Ngày tải lên: 13/08/2014, 18:20
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 3 docx
... 7-3 81640 395 p,j ri 81640 395 prj 18640 395 p,i rj 18640 395 p,i rj 1864 0 395 prji 18640 395 prji 13640 895 prji 13640 895 pri 16540 893 pri i A[r]: pivot A[j r–1]: not yet examined A[i+1 j–1]: known to ... overview [The treatment in the second edition differs from that of the Þrst edition. We use a different partitioning method—known as “Lomuto partitioning”—in the second...
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Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 4 pot
... pointer) to the new slot, and updating the pointer in the slot that pointed to j to point to the new slot. Then insert the new element in the now-empty slot as usual. To update the pointer to j , ... lists. If we want to delete elements, it’s better to use doubly linked lists.] • Slot j contains a pointer to the head of the list of all stored elements that hash to j [or t...
Ngày tải lên: 13/08/2014, 18:20
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 5 potx
... O(h) time, analogous to the changes we made for persistence in insertion. But to do so without using parent pointers we need to walk down the tree to the node to be deleted, to build up a stack ... keys are not distinct, because in order to Þnd the path to the node to delete—a particular node with a given key—we have to make some changes to how we store things in the tre...
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Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 6 pps
... row-major order, i.e., row-by-row from top to bottom, and left to right within each row. Column- major order (column-by-column from left to right, and top to bottom within each column) would also ... long it takes to get through S 1,1 . • If j ≥ 2, have two choices of how to get to S 1, j : • Through S 1, j −1 , then directly to S 1, j . • Through S 2, j −1 , then transfer ove...
Ngày tải lên: 13/08/2014, 18:20
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 7 ppsx
... to go to get to 1 or 1/4, starting from 1/2, rate of increase of differs. • For α to go from 1/ 2to1 ,num increases from size /2tosize, for a total increase of size /2. increases from 0 to size. ... sorting A and B into monotoni- cally increasing order works as well. Lecture Notes for Chapter 16: Greedy Algorithms 16-3 Solution to S ij is (solution to S ik ) ∪ { a k } ∪ (s...
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Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 8 potx
... edge it visits to the cycle C, which is then added to the Euler tour T , when we return from a call to VISIT(u), all edges entering or leaving vertex u have been added to the tour. When we advance ... B to the end of A, instead splice B into A right after the Þrst element of A. We have to traverse B to update representative pointers anyway, so we can just make the last element o...
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Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 10 potx
... the input to a half-cleaner is a bitonic 0-1 sequence, then for the output: • both the top and bottom half are bitonic, • every element in the top half is ≤ every element in the bottom half, ... 27-5 Bitonic sorter: ≤ bitonic sorter half-cleaner n n/2 n/2 n/2 n/2 sorted bitonic bitonic sorted sorted bitonic sorter 0 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 0 0 0 1 1 1 bitonic ... a...
Ngày tải lên: 13/08/2014, 18:20