Foundations of Algorithms, Fourth EditionRichard Neapolitan, Kumarss Naimipour

Chapter 2Divide-and-Conquer

Divide and Conquer

• In this approach a problem is divided into sub-problems and the same algorithm is applied to every subproblem ( often this is done recursively)

• Examples– Binary Search (review algorithm in book)– Mergesort (review algorithm in book)– Quicksort

Figure 2.1 : The steps down by a human when searching with Binary Search. (Note: x = 18)

Complexity of Binary Search

Since this and many other divide and conquer algorithms are recursive you will recall that we can determine their complexity using recurrence relations.

For Binary Search we have

T(n) = T(n/2) + 1

=[T(n/4)+1]+1 = T(n/22) + 2

=[T(n/8+1]+ 2 = T(n/23)+ 3

… =T(n/2k)+k

What is T(n/2k)+k

We if we let k get larger until n=2k then

we see that k = log2n. Why?

Consequently the relation becomes

T(n) = T(1) + log2n

T(n) = log2n

Since n/2k is 1 if they are equal and T(1) =1

MergeSort

Recall in this algorithm we divide the array into two equal parts and sort each half prior to merging. The recurrence relation is clearly

T(n) = 2T(n/2) + n

Recall that Merging is O(n) right?

Figure 2.2: The steps done by a human when sorting with Mergesort.

O(n)

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

T(n) = 2T(n/2) + n

T(n) = 2T(n/2) + n = 2[ 2T(n/22) + n/2] + n

= 22T(n/22) + 2n = 2[2T(n/23) + n/22] +2n

=23T(n/23) + 3n

…

=2kT(n/2k) + kn

If n=2k then we have

T(n) = nT(1) + (log2n)n = n+ nlog2n

= O(nlog2n)

QuickSortWorks in situ!

Void quicksort(int low, int high)

{

int pivot;

if (high > low){

partition(low, high, pivot);

quicksort(low, pivot-1);

quicksort(pivot+1,high);

}

Figure 2.3: The steps done by a human when sorting with Quicksort. The subarrays are enclosed in rectangles whereas the pivot points are free.

PartitionStudy this carefully

void partition (int low, int high, int&pivot)

{

int I,j, pivotitem;

pivotitem = S[low]; // select left item (hmmm)

j=low;

for (i=low+1; i<=high; i++) There are many ways

if (S[i] < pivotitem){ to write this function!

j++; All have a complexity

swap S[i] and S[j]; of O(n).

}

pivot= j;

swap S[low] and S[pivot];

}

Complexity of Quicksort

The complexity of this algorithm depends on how good the pivot value selection is . If the value is always in the middle of array then the best case complexity is

T(n) = n + 2T(n/2)

Which we already have determined is

T(n) = n log2 n

Worst case for Quicksort

This clearly will occur if each pivot value is less than (or greater) all the elements of the array. IE the array is split into 1 and n-1 size pieces. This gives a recurrence relation of T(n) = T(1) + T(n-1) + n-1

Time to sort left array right array partition

Worst Case analysisT(n) = T(1) + T(n-1) + n-1 = T(n-1) + n

Assume the answer is n(n-1)/2

check it out !

n(n-1)/2 = 0 + (n-1)(n-2)/2 + n-1

= (n-1)(n-2)/2 + 2(n-1)/2

=((n-1)(n-2)+ 2(n-1))/2

= (n-1)(n-2+2)/2 = n(n-1)/2 ☺

Quick Sort AnalysisQuicksort’s worst case is θ(n2)

Does this mean that quick sort is just as bad as say selection sort, insertion sort and/or bubble sort. No!

Its all about average case performance. The average case performance for these three is θ(n2) as well.What is the average case complexity for QS?

Average Case Analysisassume prob. pivotpoint is p

See HW 22 p 86 for above conversion

Average case continued

Multiplying by n

Subtracting these equations we have

Average case QS continued

Assume we get

,

Applying some simple math we have

Which give

)

Matrix Multiplication (Strassen)

Lets look at the product of two 2 by 2’s

Clearly after you do the homework #26m1=(a11+a22)(b11+b22) m5=(a11+a12)b22

m2=(a11+a22)b11 m6=(a21+a11)(b11+b12)

m3=a11(b12+b22) m7=(a12+a22)(b21+b22)

m4=a22(b21+b11) will give the following!

And the answer is

Original method

8 mult, four add/sub

Strassen’s method

7 mult, and 18 add/sub

Hmmmm! So what’s the big deal?

Big Matrices 2n by 2n

Where

C11 is the upper left hand corner of the matrix of size n/2 by n/2. The others are similarly defined. Now

m1=(A11+A22)(B11+B22)

Is the sum and product of matrices

Our function is then

void Strassen(int n, A,B, C)// these are nxn mats

{

if (n<= threshold) computer C=AxB normally

Partition A and B into eight submatrices

strassen(n/2, A11+A22, B11+B22, M1);

strassen(n/2, A21+A22, B11, M2)

etc // making 7 recursive calls

}

NOT EIGHT!

Complexity

T(n) = 7T(n/2) + cn2

Which is

T(n) = θ(nlg7) = O(n2.81)Using the general theorem.

The best know is

Coppersmith and Winograd with a time complexity of

O(n2.376)

Why am I using big O here?

Recalling General TheoremSee page 588

Assume for n>1 and n a power of b, T(1)=d

Just a side note

Suppose we has 8 recursive calls instead of 7 in the above case. Then the recurrence relation would be

T(n) = 8T(n/2) + cn2

This has a complexity of what?

When not to use divide and conquer

• An instance of size n is divided into two or more instances each almost of size n.

• An instance of size n is divided into almost

n instances of size n/c, where c is a constant