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chapter25

2. Backtracking and time complexity
Backtracking is used to get the schedule.
Time complexity
O(n) if the jobs are sorted.
Total time: O(n log n) including sorting.
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3. Weighted Interval Scheduling: Bottom-Up
Input: n, s1, s2, …, sn, f1, f2, …, fn, v1, v2, …, vn
Sort jobs by finish times so that f1f2 … fn.
Compute p(1), p(2) , …, p(n)
M[0]=0;
for j = 1 to n do
M[j] = max { vj+m[p(j)], m[j-1]}
if (M[j] == m[j-1]) then B[j]=j-1 else B[j]=p(j) /*for backtracking
m=n;
while ( m ≠0)
{ print “job m”; m=B[m]; }
2019/4/10
chapter25

4. Weighted Interval Scheduling: Bottom-Up
Input: n, s1, s2, …, sn, f1, f2, …, fn, v1, v2, …, vn
Sort jobs by finish times so that f1f2 … fn.
Compute p(1), p(2) , …, p(n)
M[0]=0;
for j = 1 to n do
M[j] = max { vj+m[p(j)], m[j-1]}
if (M[j] == m[j-1]) then B[j]=j-1 else B[j]=p(j) /*for backtracking
m=n;
while ( m ≠0)
{ print “job m”; m=B[m]; }
Solving recurrence equation is not required.
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5. Interval with Maximum Sum.
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7. Algorithm:
O(n)
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8. Int maxSum (int A[], int i, int j){ if (i==j){
maxSum(int A[], int i, int j) finds the interval [p, q] in array A[i…j] with max sum and return the interval [p, q].
Int maxSum (int A[], int i, int j){
if (i==j){
return [i, i] as the interval and A[i] as the sum }
m=(i+j)/2;
([p1,q1], s1) = maxSum(A[], i, m);
([p2, q2], s2) = maxSum(A[], m+1, j);
([p3,q3], s3) = sumBothSides(A[], i, j, m)
([p, q], s) = the max of the three
return [p, q} and s.
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9. f(n)=2f(n/2)+O(n). So, f(n)=O(nlog n).
Running time:
f(n)=2f(n/2)+O(n). So, f(n)=O(nlog n).
Open Problem: Can we design a faster algorithm? (come back later after learning other techniques)
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10. Part-D1 Binary Search Trees
Dictionaries
二○一九年四月十日
Part-D1 Binary Search Trees
< 6 2
> 9 1 4 = 8
Binary Search Trees

11. Binary Search Trees (§ 10.1)
A binary search tree is a binary tree storing keys (or key-value entries) at its internal nodes and satisfying the following property:
Let u, v, and w be three nodes such that u is in the left subtree of v and w is in the right subtree of v. We have key(u)  key(v)  key(w)
External nodes do not store items
An inorder traversal of a binary search trees visits the keys in increasing order 6 9 2 4 1 8
Binary Search Trees

12. Cost of the tree: f6*1+f2*2+f9*2+f1*3+f4*3+f8*3
< 6 2
> 9 1 4 = 8
Cost of the tree: f6*1+f2*2+f9*2+f1*3+f4*3+f8*3
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15. Assume that e1<e2<…en
Algorithm:
Assume that e1<e2<…en
Calculate the sum of fi
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16. Running time: A bit disappointing???
Using a technique called memoization, we can have a polynomial time algorithm.
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