1. Topic: Divide and Conquer
General idea:
Divide a problem into subprograms of the same kind; solve subprograms using the same approach, and combine partial solution (if necessary).
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Topic: Divide and Conquer

2. Topic: Divide and Conquer
Find the maximum and minimum
The problem: Given a list of unordered n elements, find max and min
The straightforward algorithm:
max ← min ← A (1);
for i ← 2 to n do
if A (i) > max, max ← A (i);
if A (i) < min, min ← A (i);
Key comparisons: 2(n – 1)
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Topic: Divide and Conquer

3. Topic: Divide and Conquer
List List 1 List 2
n elements n/2 elements n/2 elements
min, max min1, max1 min2, max2
min = MIN ( min1, min2 )
max = MAX ( max1, max2)
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Topic: Divide and Conquer

4. Topic: Divide and Conquer
The Divide-and-Conquer algorithm:
procedure Rmaxmin (i, j, fmax, fmin); // i, j are index #, fmax, begin // fmin are output parameters
case:
i = j: fmax ← fmin ← A (i);
i = j –1: if A (i) < A (j) then fmax ← A (j);
fmin ← A (i);
else fmax ← A (i);
fmin ← A (j);
else: mid ← ;
call Rmaxmin (i, mid, gmax, gmin);
call Rmaxmin (mid+1, j, hmax, hmin);
fmax ← MAX (gmax, hmax);
fmin ← MIN (gmin, hmin);
end
end;
Spring-2005 Young
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Topic: Divide and Conquer

5. Topic: Divide and Conquer
Example: find max and min in the array:
22, 13, -5, -8, 15, 60, 17, 31, 47 ( n = 9 )
Index:
Array:
Rmaxmin(1, 9, 60, -8)
1, 5, 22, , 9, 60, 17
1, 3, 22, , 5, 15, ,7, 60, , 9, 47, 31
1,2, 22, , 3,-5, -5
(1)
(6)
(7)
(2)
(5)
(8)
(3)
(4)
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Topic: Divide and Conquer

6. Topic: Divide and Conquer
Analysis: For algorithm containing recursive calls, we can use recurrence relation to find its complexity
T(n) - # of comparisons needed for Rmaxmin
Recurrence relation:
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Topic: Divide and Conquer

7. Topic: Divide and Conquer
Assume n = 2k for some integer k
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Topic: Divide and Conquer

8. Topic: Divide and Conquer
Theorem:
Claim:
They don’t have to be the same constant.
We make them the same here for simplicity.
It will not affect the overall result.
where are a, b, c constants
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Topic: Divide and Conquer

9. Topic: Divide and Conquer
Proof: Assume n = ck
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Topic: Divide and Conquer

10. Topic: Divide and Conquer
If a < c, is a constant
T(n) = bn · a constant
 T(n) = O(n)
if a = c,
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Topic: Divide and Conquer

11. Topic: Divide and Conquer
If a > c,
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Topic: Divide and Conquer

12. Topic: Divide and Conquer
2. Matrix multiplication
The problem:
Multiply two matrices A and B, each of size
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Topic: Divide and Conquer

13. Topic: Divide and Conquer
The traditional way:
use three for-loop
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Topic: Divide and Conquer

14. Topic: Divide and Conquer
The Divide-and-Conquer way:
transform the problem of multiplying A and B, each of size [n×n] into 8 subproblems, each of size
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Topic: Divide and Conquer

15. Topic: Divide and Conquer
which an2 is for addition
so, it is no improvement compared with the traditional way
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Topic: Divide and Conquer

16. Topic: Divide and Conquer
Example:
use Divide-and-Conquer way to solve it as following:
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Topic: Divide and Conquer

17. Topic: Divide and Conquer
Strassen’s matrix multiplication:
· Discover a way to compute the Cij’s using 7 multiplications and 18 additions or subtractions
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Topic: Divide and Conquer

18. Topic: Divide and Conquer
Algorithm :
procedure Strassen (n, A, B, C) // n is size, A,B the input matrices, C output matrix
begin
if n = 2,
else
(cont.)
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Topic: Divide and Conquer

19. Topic: Divide and Conquer
else
Partition A into 4 submatrices: ;
Partition B into 4 submatrices: ;
call Strassen ( ;
call Strassen (
call Strassen ( ;
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Topic: Divide and Conquer

20. Topic: Divide and Conquer
call Strassen ( ;
end;
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Topic: Divide and Conquer

21. Topic: Divide and Conquer
Analysis:
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Topic: Divide and Conquer

22. Topic: Divide and Conquer
Assume n = 2k for some integer k
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Topic: Divide and Conquer

23. Topic: Divide and Conquer
3. Integer multiplication
The problem:
Multiply two large integers (n digits)
The traditional way:
Use two loops, it takes O(n2) operations
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Topic: Divide and Conquer

24. Topic: Divide and Conquer
The Divide-and-Conquer way:
Suppose x and y are large integers, divide x into two parts (a and b), and divide y into c and d. x: y: a b c d
So, transform the problem of multiplying two n-digit integers
into four subproblems of multiplying two -digit integers.
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Topic: Divide and Conquer

25. Topic: Divide and Conquer
Worst-Case is:
However, it is same as the traditional way.
Instead, we write:
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Topic: Divide and Conquer

26. Topic: Divide and Conquer
The algorithm by Divide-and-Conquer:
function multiplication (x,y)
begin
n = MAX ( # of digits in x, # of digits in y);
if (x = 0) or (y = 0), return 0;
else if (n = 1), return x*y in the usual way;
else
m = ;
a = x divide 10m;
b = x rem 10m;
c = y divide 10m;
d = y rem 10m;
p1= MULTIPLICATION (a, c);
p2= MULTIPLICATION (b, d);
p3 = MULTIPLICATION (a+b, c+d);
return ;
end;
Spring-2005 Young
CS 331 D&A of Algo.
Topic: Divide and Conquer