Choosing an Algorithm

















Can you have more than one algorithm that solves a given problem?

	Yes
	searching: sequential vs binary
	sorting: selection, insertion, merge, quick, heap


How do you choose which algorithm you will use?

	1. simplest
	2. smallest
	3. fastest


How do you know how much time an algorithm will use?

	1. measure the running time
	2. predict the running time through analysis

















Classwork
You may work with a partner.

Give a function T(n) for the number of time units used by the code.
(Assume each statement uses one time unit.)


	double sum = 0;
	for (int i = 0; i < n; i++)
	    sum += a[i];
	double mean = sum / n;
	return mean;

















Big Oh


Suppose you have T(n) = n^2 + 2n + 4
Does it make sense to define T(n) this precisely?

	NO

	1. when n is large, the high-order term dominates
		n^2 + 2n + 4 = 1002004 when n is 1000

	2. when n is small, time is not an issue

	3. constant factors depend on machine and compiler

	4. the rate of growth is what is important



How do you represent the growth rate of T(n)?

	use Big Oh notation
	(abstraction of time formula)

	1. keep the dominant term

	2. drop low-order terms

	3. drop constants



What's the definition of BigOh?

	T(n) is O(f(n))

	if there exist positive constants c and n0 such that

	T(n) <= c*f(n)

	for all n >= n0

















Finding The BigOh


How do you find the Big-Oh for an Algorithm?


What's the Running Time of a simple statement?

	small = 1;


What's the Running Time of a block of simple statements?

	temp = A[small];
	A[small] = A[i];
	A[i] = temp;


What's the Running Time of a Loop?

	1. find the BigOh for the body of the loop
	2. multiply by the number of times the loop repeats

	for (i = 0; i < n; i++)
	   A[i][i] = 1;

	for (i = 0; i < n; i++)
	  for (j = 0; j < n; j++)
	     A[i][j] = 0;


What's the Running Time of a combination of statements?

	for (i = 0; i < n; i++)
	  for (j = 0; j < n; j++)
	     A[i][j] = 0;
	for (i = 0; i < n; i++)
	   A[i][i] = 1;


What's the Running Time of a selection statement?

	1. find the BigOh for the if part
	2. find the BigOh for the else part
	3. use the larger of the two BigOh's

	if (x == 0)
	   for (i = 0; i < n; i++)
	      A[i][i] = 1;
	else
	   for (i = 0; i < n; i++)
	     for (j = 0; j < n; j++)
	        A[i][j] = 0;

















Classwork
You may work together with a partner.

Give the Big-Oh of the running time for each program fragment.

	for ( int i = 0; i < n; i++ )
	    sum++;

	for ( int i = 0; i < n; i += 2 )
	    sum++;

	for ( int i = 0; i < n; i++ )
	    for ( int j = 0; j < n; j++ )
	    	sum++;

	for ( int i = 0; i < n; i++ )
	    sum++;
	for ( int j = 0; j < n; j++ )
	    sum++;

	for ( int i = 0; i < n; i++ )
	    for ( int j = 0; j < n*n; j++ )
	    	sum++;

	for ( int i = 0; i < n; i++ )
	    for ( int j = 0; j < i; j++ )
	    	sum++;

	for ( int i = 0; i < n; i++ )
	    for ( int j = 0; j < n*n; j++ )
                for ( int k = 0; k < j; k++ )
		    sum++;

















The inside loop in the following code does not always repeat n times.
Does the code have a Big-Oh bound that is smaller than O(n^2)?


	for ( int i = 0; i < n; i++ )
	    for ( int j = 0; j < i; j++ )
	    	sum++;

















Classwork
You may work together with a partner.

What's the BigOh for this algorithm?

	for (i = 0; i < n-1; i++) {
	   small = 1;
	   for (j = i+1; j < n; j++)
	      if (A[j] < A[small])
	         small = j;
	   temp = A[small];
	   A[small] = A[i];
	   A[i] = temp;
	}

















What are some examples of bad Big-Oh style?

	O(2n)
	O(n + 2)
	O(n^2 + n)

















Classwork
You may work together with a partner.

What's the BigOh for this algorithm?

        int maxSum = 0;

        for( int i = 0; i < n; i++ )
            for( int j = i; j < n; j++ )
            {
                int thisSum = 0;

                for( int k = i; k <= j; k++ )
                    thisSum += a[ k ];

                if( thisSum > maxSum )
                {
                    maxSum   = thisSum;
                    seqStart = i;
                    seqEnd   = j;
                }
            }

        return maxSum;