1. Design and Analysis of Algorithms Maximum-subarray problem and matrix multiplication
Haidong Xue
Summer 2012, at GSU

2. Maximum-subarray problem back ground
If you know the price of certain stock from day i to day j;
You can only buy and sell one share once
How to maximize your profit?

3. Maximum-subarray problem back ground
What is the brute-force solution?
max = -infinity;
for each day pair p {
if(p.priceDifference>max)
max=p.priceDifference; }
Time complexity?
𝑛 2 pairs, so O( 𝑛 2 )

4. Maximum-subarray problem back ground
If we know the price difference of each 2 contiguous days
The solution can be found from the maximum-subarray
Maximum-subarray of array A is:
A subarray of A
Nonempty
Contiguous
Whose values have the the largest sum

5. Maximum-subarray problem back ground
Day 1 2 3 4
Price 10 11 7 6
Difference -4
What is the solution?
Buy on day 2, sell on day 3
Can be solve it by the maximum-subarray of difference array?
Sub-array
1-1
1-2
1-3
1-4
2-2
2-3
2-4
3-3
3-4
4-4
Sum 1 -3 -4 -1 -5 3

6. Maximum-subarray problem – divide-and-conquer algorithm
How to divide?
Divide to 2 arrays
What is the base case?
How to combine the sub problem solutions to the current solution?
A fact:
when divide array A[i, …, j] into A[i, …, mid] and A[mid+1, …, j]
A sub array must be in one of them
A[i, …, mid] // the left array
A[mid+1, …, j] // the right array
A[ …, mid, mid+1….] // the array across the midpoint
The maximum subarray is the largest sub-array among maximum subarrays of those 3

7. Maximum-subarray problem – divide-and-conquer algorithm
Input: array A[i, …, j]
Ouput: sum of maximum-subarray, start point of maximum-subarray, end point of maximum-subarray
FindMaxSubarray:
if(j<=i) return (A[i], i, j);
mid = floor(i+j);
(sumCross, startCross, endCross) = FindMaxCrossingSubarray(A, i, j, mid);
(sumLeft, startLeft, endLeft) = FindMaxSubarray(A, i, mid);
(sumRight, startRight, endRight) = FindMaxSubarray(A, mid+1, j);
Return the largest one from those 3

8. Maximum-subarray problem – divide-and-conquer algorithm
FindMaxCrossingSubarray(A, i, j, mid)
Scan A[i, mid] once, find the largest A[left, mid]
Scan A[mid+1, j] once, find the largest A[mid+1, right]
Return (sum of A[left, mid] and A[mid+1, right], left, right)
Let’s try it in Java

9. 1 -4 3 2
Target array : 1 1
All the sub arrays: -4 -4 3 3 2 2 1 -4 -3 -4 3 -1 3 2
Max! 5 1 -4 3 -4 3 2 1 1 -4 3 2 2

10. 1 -4 3 2
Divide target array into 2 arrays
Target array : 1
We then have 3 types of subarrays: 1
All the sub arrays: -4 -4
The problem can be then solved by:
The ones belong to the left array 3 3
1. Find the max in left sub arrays
The ones belong to the right array 2 2 1 -4
2. Find the max in right sub arrays -3
The ones crossing the mid point -4 3 -1
3. Find the max in crossing sub arrays 3 2 5 1 -4 3 -4 3 2
4. Choose the largest one from those 3 as the final result 1 1 -4 3 2 2

11. 1 -4 3 2
FindMaxSub ( )
1. Find the max in left sub arrays 1 -4
FindMaxSub ( ) 3 2
2. Find the max in right sub arrays
FindMaxSub ( )
3. Find the max in crossing sub arrays 1 -4 3 2
Scan once, and scan once
4. Choose the largest one from those 3 as the final result

12. 3. Find the max in crossing sub arrays1 -4 3 2 -4
Sum=-4 1 -4
Sum=-3
largest 3
Sum=3 3 2
Sum=5
largest
The largest crossing subarray is :

13. Maximum-subarray problem – divide-and-conquer algorithm
What is the time complexity?
FindMaxSubarray:
if(j<=i) return (A[i], i, j);
mid = floor(i+j);
(sumCross, startCross, endCross) = FindMaxCrossingSubarray(A, i, j, mid);
(sumLeft, startLeft, endLeft) = FindMaxSubarray(A, i, mid);
(sumRight, startRight, endRight) = FindMaxSubarray(A, mid+1, j);
Return the largest one from those 3
𝑇 𝑛 =2𝑇 𝑛 2 +Θ(n)

14. Matrix multiplication
How to multiply two matrices?
Given matrix 𝐴 𝑛𝑛 𝑎𝑛𝑑 𝐵 𝑛𝑛 , 𝐶 𝑛𝑛 =𝐴𝐵
𝑐 𝑖𝑗 = 𝑘=1 𝑛 𝑎 𝑖𝑘 𝑏 𝑘𝑗
For each 𝑐 𝑖𝑗 , we need Θ(𝑛)
There are 𝑛 2 𝑐 𝑖𝑗 , so 𝑇 𝑛 = 𝑛 2 Θ(𝑛)=Θ( 𝑛 3 )
= −9 −3 −
Time Complexity?

15. Matrix multiplication divide-and-conquer algorithm
C=AB
𝐶 11 𝐶 12 𝐶 21 𝐶 22 = 𝐴 11 𝐴 12 𝐴 21 𝐴 22 × 𝐵 11 𝐵 12 𝐵 21 𝐵 22
𝐶 11 = 𝐴 11 × 𝐵 11 + 𝐴 12 × 𝐵 21
𝐶 12 = 𝐴 11 × 𝐵 12 + 𝐴 12 × 𝐵 22
𝐶 21 = 𝐴 21 × 𝐵 11 + 𝐴 22 × 𝐵 21
𝐶 22 = 𝐴 21 × 𝐵 12 + 𝐴 22 × 𝐵 22
Recurrence equation?
𝑇 𝑛 =8𝑇 𝑛 2 +Θ( 𝑛 2 )

16. Matrix multiplication divide-and-conquer algorithm
𝑇 𝑛 =8𝑇 𝑛 2 +Θ( 𝑛 2 )
What is the time complexity?
From Master method we know it is Θ( 𝑛 3 )

17. Matrix multiplication Strassen’s algorithm
𝑇 𝑛 =8𝑇 𝑛 2 +Θ( 𝑛 2 )
𝑇 𝑛 =7𝑇 𝑛 2 +Θ( 𝑛 2 )

18. Matrix multiplication Strassen’s algorithm
1. Perform 10 times matrix addition or subtraction to make 𝑆 1 𝑡𝑜 𝑆 10 from 𝐴 𝑖𝑗 𝑎𝑛𝑑 𝐵 𝑖𝑗
2. Perform 7 times matrix multiplication to make 𝑃 1 to 𝑃 7 from 𝐴 𝑖𝑗 , 𝐵 𝑖𝑗 𝑎𝑛𝑑 𝑆 𝑖
3. Perform matrix addition or matrix subtraction to obtain 𝐶 11 , 𝐶 12 , 𝐶 21 and 𝐶 22
𝑇 𝑛 =7𝑇 𝑛 2 +Θ( 𝑛 2 )= Θ( 𝑛 𝑙𝑜𝑔 2 7 )