CS Lunch Mary Allen Wilkes Wednesday 12:15 Kendade 307 2 Divide - - PowerPoint PPT Presentation

CS Lunch Mary Allen Wilkes Wednesday 12:15 Kendade 307

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Divide and Conquer

Divide-and-conquer. Break up problem into several parts. Solve each part recursively. Combine solutions to sub-problems into overall solution. Most common usage. Break up problem of size n into two equal parts of size n/2. Solve two parts recursively. Combine two solutions into overall solution in linear time.

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Mergesort

Complexity? Base case - O(1) Divide - O(1) Recursive cases ?? Merge - O(n)

mergesort(m, low, high) { if high == low { return m[low] } else if (high == low + 1) { return sort m[low] and m[high]; } else { middle = (low + high) / 2 left = mergesort(m, low, middle-1) right = mergesort(m, middle, high) return merge(left, right) } }

3 Slides13 - Recurrences, CountingInversions.key - March 18, 2019

mergesort Recurrence Relation

Note that if n > 2, merge sort of size n requires 2 merge sorts each of size n/2 followed by a merge of all n items: T(n) ≤ 2 ∙ T(n/2) + cn when n > 2 T(2) is O(1)

Problem: How do we solve T(n) for a O() value?

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Recursion Tree

T(n) T(n/2) T(n/2) T(n/4) T(n/4) T(n/4) T(n/4) T(2) T(2)

Solving recursively Solving base cases

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T(n) T(n/2) T(n/2) T(n/4) T(n/4) T(n/4) T(n/4) T(2) T(2)

1∙cn 2∙cn/2 4∙cn/ 4 n/2∙c # of problems at this level cost of a subproblem, excluding cost due to recursion

6 Slides13 - Recurrences, CountingInversions.key - March 18, 2019

Recursion Tree

T(n) T(n/2) T(n/2) T(n/4) T(n/4) T(n/4) T(n/4) T(2) T(2)

1∙cn 2∙cn/2 4∙cn/ 4 Cost at each recursion level n/2∙2c How many recursion levels?

7-1

Recursion Tree

T(n) T(n/2) T(n/2) T(n/4) T(n/4) T(n/4) T(n/4) T(2) T(2)

1∙cn 2∙cn/2 4∙cn/ 4 Cost at each recursion level n/2∙2c = cn = cn = cn = cn How many recursion levels?

7-2

Recursion Tree

T(n) T(n/2) T(n/2) T(n/4) T(n/4) T(n/4) T(n/4) T(2) T(2)

1∙cn 2∙cn/2 4∙cn/ 4 Cost at each recursion level n/2∙2c = cn = cn = cn = cn How many recursion levels? log2n

7-3 Slides13 - Recurrences, CountingInversions.key - March 18, 2019

Recursion Tree

T(n) T(n/2) T(n/2) T(n/4) T(n/4) T(n/4) T(n/4) T(2) T(2)

1∙cn 2∙cn/2 4∙cn/ 4 Cost at each recursion level n/2∙2c = cn = cn = cn = cn How many recursion levels? log2n Total = cn log2n

7-4

Generalizing the Recurrences

If recurrence involves dividing a problem into 2 pieces that are half the original size, and has linear cost

• utside of the recurrence => O(n log n)

Mergesort is just one example If recurrence involves solving 1 subproblem of half the size of the original, and has constant cost outside of the recurrence => O(log n) Binary search is just one example

8 Divide + Merge Subproblem size Num subproblems Total cost O(1) n/2 1 O(log n) O(n) n/2 1 O(n) O(n) n/2 2 O(n log n) O(n) n/2 q > 2 O(nlog q) O(n2) n/2 2 O(n2) 9 Slides13 - Recurrences, CountingInversions.key - March 18, 2019

Recommender Systems

Netflix tries to match your movie preferences with

• thers.

You rank n movies. Netflix consults database to find people with similar tastes. Netflix can recommend to you movies that they liked.

Doing this well was worth $1,000,000 to Netflix!!

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Counting Inversions

Similarity metric: number of inversions between two rankings. My rank: 1, 2, …, n. Your rank: a1, a2, …, an. Movies i and j are inverted if i < j, but ai > aj.

You Me 1 4 3 2 5 1 3 2 4 5 A B C D E Movies Inversions 3-2, 4-2

What is the brute force algorithm?

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Divide and Conquer

4 8 10 2 1 5 12 11 3 7 6 9

Count inversions relative to a sorted list

12-1 Slides13 - Recurrences, CountingInversions.key - March 18, 2019

Divide and Conquer

4 8 10 2 1 5 12 11 3 7 6 9

Count inversions relative to a sorted list

4 8 10 2 1 5 12 11 3 7 6 9

Divide into 2 sublists of equal size

12-2

Divide and Conquer

4 8 10 2 1 5 12 11 3 7 6 9

Count inversions relative to a sorted list

4 8 10 2 1 5 12 11 3 7 6 9

Divide into 2 sublists of equal size

5 blue-blue inversions 8 green-green inversions 5-4, 5-2, 4-2, 8-2, 10-2

6-3, 9-3, 9-7, 12-3, 12-7, 12-11, 11-3, 11-7

Recursively count the inversions

12-3

Divide and Conquer

4 8 10 2 1 5 12 11 3 7 6 9

Count inversions relative to a sorted list

4 8 10 2 1 5 12 11 3 7 6 9

Divide into 2 sublists of equal size

5 blue-blue inversions 8 green-green inversions 5-4, 5-2, 4-2, 8-2, 10-2

6-3, 9-3, 9-7, 12-3, 12-7, 12-11, 11-3, 11-7

Recursively count the inversions

9 blue-green inversions 5-3, 4-3, 8-6, 8-3, 8-7, 10-6, 10-9, 10-3, 10-7 Total = 5 + 8 + 9 = 22. Combine by adding recursive counts and inversions across halves

12-4 Slides13 - Recurrences, CountingInversions.key - March 18, 2019

Divide and Conquer

4 8 10 2 1 5 12 11 3 7 6 9

Count inversions relative to a sorted list

4 8 10 2 1 5 12 11 3 7 6 9

Divide into 2 sublists of equal size

5 blue-blue inversions 8 green-green inversions

Recursively count the inversions

9 blue-green inversions Total = 5 + 8 + 9 = 22.

Combine by adding recursive counts and inversions across halves

Cost

13-1

Divide and Conquer

4 8 10 2 1 5 12 11 3 7 6 9

Count inversions relative to a sorted list

4 8 10 2 1 5 12 11 3 7 6 9

Divide into 2 sublists of equal size

5 blue-blue inversions 8 green-green inversions

Recursively count the inversions

9 blue-green inversions Total = 5 + 8 + 9 = 22.

Combine by adding recursive counts and inversions across halves

Cost O(1)

13-2

Divide and Conquer

4 8 10 2 1 5 12 11 3 7 6 9

Count inversions relative to a sorted list

4 8 10 2 1 5 12 11 3 7 6 9

Divide into 2 sublists of equal size

5 blue-blue inversions 8 green-green inversions

Recursively count the inversions

9 blue-green inversions Total = 5 + 8 + 9 = 22.

Combine by adding recursive counts and inversions across halves

Cost O(1) 2*T(n/2)

13-3 Slides13 - Recurrences, CountingInversions.key - March 18, 2019

Divide and Conquer

4 8 10 2 1 5 12 11 3 7 6 9

Count inversions relative to a sorted list

4 8 10 2 1 5 12 11 3 7 6 9

Divide into 2 sublists of equal size

5 blue-blue inversions 8 green-green inversions

Recursively count the inversions

9 blue-green inversions Total = 5 + 8 + 9 = 22.

Combine by adding recursive counts and inversions across halves

Cost O(1) 2*T(n/2) ???

13-4

Finding Inversions

Combine: count blue-green inversions Assume each half is sorted. Count inversions where ai and aj are in different halves. Merge two sorted halves into sorted whole.

Variation of mergesort

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10 14 18 19 3 7 16 17 23 25 2 11

Merge and Count

Merge and count step. Given two sorted halves, count number of inversions where ai and aj are in different halves. Combine two sorted halves into sorted whole.

numLeft = 6

Total:

15 Slides13 - Recurrences, CountingInversions.key - March 18, 2019

10 14 18 19 3 7 16 17 23 25 2 11

Merge and Count

Merge and count step. Given two sorted halves, count number of inversions where ai and aj are in different halves. Combine two sorted halves into sorted whole.

2 numLeft = 6

Total: 6

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10 14 18 19 3 7 16 17 23 25 2 11

Merge and Count

Merge and count step. Given two sorted halves, count number of inversions where ai and aj are in different halves. Combine two sorted halves into sorted whole.

2 3 numLeft = 5

Total: 6

17

10 14 18 19 3 7 16 17 23 25 2 11

Merge and Count

Merge and count step. Given two sorted halves, count number of inversions where ai and aj are in different halves. Combine two sorted halves into sorted whole.

7 2 3 numLeft = 4

Total: 6

18 Slides13 - Recurrences, CountingInversions.key - March 18, 2019

10 14 18 19 3 7 16 17 23 25 2 11

Merge and Count

Merge and count step. Given two sorted halves, count number of inversions where ai and aj are in different halves. Combine two sorted halves into sorted whole.

7 10 2 3 numLeft = 3

Total: 6

19

10 14 18 19 3 7 16 17 23 25 2 11

Merge and Count

Merge and count step. Given two sorted halves, count number of inversions where ai and aj are in different halves. Combine two sorted halves into sorted whole.

7 10 11 2 3 numLeft = 3

Total: 6 + 3

20

10 14 18 19 3 7 16 17 23 25 2 11

Merge and Count

Merge and count step. Given two sorted halves, count number of inversions where ai and aj are in different halves. Combine two sorted halves into sorted whole.

7 10 11 14 2 3 numLeft = 2

Total: 6 + 3

21 Slides13 - Recurrences, CountingInversions.key - March 18, 2019

10 14 18 19 3 7 16 17 23 25 2 11

Merge and Count

Merge and count step. Given two sorted halves, count number of inversions where ai and aj are in different halves. Combine two sorted halves into sorted whole.

7 10 11 14 2 3 16 numLeft = 2

Total: 6 + 3 + 2

22

10 14 18 19 3 7 16 17 23 25 2 11

Merge and Count

Merge and count step. Given two sorted halves, count number of inversions where ai and aj are in different halves. Combine two sorted halves into sorted whole.

7 10 11 14 2 3 16 17 numLeft = 2

Total: 6 + 3 + 2 + 2

23

10 14 18 19 3 7 16 17 23 25 2 11

Merge and Count

Merge and count step. Given two sorted halves, count number of inversions where ai and aj are in different halves. Combine two sorted halves into sorted whole.

7 10 11 14 2 3 18 16 17 numLeft = 1

Total: 6 + 3 + 2 + 2

24 Slides13 - Recurrences, CountingInversions.key - March 18, 2019

10 14 18 19 3 7 16 17 23 25 2 11

Merge and Count

Merge and count step. Given two sorted halves, count number of inversions where ai and aj are in different halves. Combine two sorted halves into sorted whole.

7 10 11 14 2 3 18 19 16 17 numLeft = 0

Total: 6 + 3 + 2 + 2

25

10 14 18 19 3 7 16 17 23 25 2 11

Merge and Count

Merge and count step. Given two sorted halves, count number of inversions where ai and aj are in different halves. Combine two sorted halves into sorted whole.

7 10 11 14 2 3 18 19 23 16 17 numLeft = 0

Total: 6 + 3 + 2 + 2

26

10 14 18 19 3 7 16 17 23 25 2 11

Merge and Count

Merge and count step. Given two sorted halves, count number of inversions where ai and aj are in different halves. Combine two sorted halves into sorted whole.

7 10 11 14 2 3 18 19 23 25 16 17 numLeft = 0

Total: 6 + 3 + 2 + 2 = 13

27 Slides13 - Recurrences, CountingInversions.key - March 18, 2019

Counting Inversions: Implementation

Sort-and-Count(L) { if list L has one element return (0, L) Divide the list into two halves A and B (rA, A) ← Sort-and-Count(A) (rB, B) ← Sort-and-Count(B) (rC, L) ← Merge-and-Count(A, B) r = rA + rB + rC return (r, L) }

28 Counting Inversions: Implementation

Merge-and-Count (A, B) { curA = 0; curB = 0; count = 0; mergedList = empty list while (not at end of A && not at end of B) { a = A[curA]; b = B[curB]; if (a < b) { append a to mergedList; curA++; else { append b to mergedList; curB++; count = count + num elements left in A } } if (at end of A) append rest of B to mergedList; else append rest of A to mergedList; return (count, mergedList); }

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Cost of Sort-and-Count?

Sort-and-Count(L) { if list L has one element return (0, L) Divide the list into two halves A and B (rA, A) ← Sort-and-Count(A) (rB, B) ← Sort-and-Count(B) (rC, L) ← Merge-and-Count(A, B) r = rA + rB + rC return (r, L) }

30 Slides13 - Recurrences, CountingInversions.key - March 18, 2019