Analysis of Algorithms | Set 5 (Practice Problems)

3.6

We have discussed Asymptotic Analysis, Worst, Average and Best Cases , Asymptotic Notations and Analysis of loops in previous posts.

In this post, practice problems on analysis of algorithms are discussed.

Problem 1: Find the complexity of below recurrence:

         { 3T(n-1), if n>0,
T(n) =   { 1, otherwise

Solution:

Let us solve using substitution.
T(n) = 3T(n-1)
     = 3(3T(n-2)) 
     = 32T(n-2)
     = 33T(n-3)
       ...
       ...
     = 3nT(n-n)
     = 3nT(0) 
     = 3n
This clearly shows that the complexity 
of this function is O(3n).

 

Problem 2: Find the complexity of the recurrence:

        { 2T(n-1) - 1, if n>0,
T(n) =   { 1, otherwise

Solution:

 Let us try solving this function with substitution.
T(n) = 2T(n-1) - 1
     = 2(2T(n-2)-1)-1 
     = 22(T(n-2)) - 2 - 1
     = 22(2T(n-3)-1) - 2 - 1 
     = 23T(n-3) - 22 - 21 - 20
       .....
       .....
     = 2nT(n-n) - 2n-1 - 2n-2 - 2n-3
       ..... 22 - 21 - 20

     = 2n - 2n-1 - 2n-2 - 2n-3
       ..... 22 - 21 - 20
     = 2n - (2n-1) 
[Note: 2n-1 + 2n-2 + ...... +  20 = 2n ]
T(n) = 1
Time Complexity is O(1). Note that while 
the recurrence relation looks exponential
the solution to the recurrence relation 
here gives a different result.

 

Problem 3: Find the complexity of the below program:

function(int n)
{
    if (n==1)
       return;
    for (int i=1; i<=n; i++)
    {
        for (int j=1; j<=n; j++)
        {
            printf("*");
            break;
        }
    }
}

Solution: Consider the comments in the following function.

function(int n)
{
    if (n==1)
       return;
    for (int i=1; i<=n; i++)
    {
        // Inner loop executes only one
        // time due to break statement.
        for (int j=1; j<=n; j++)
        {
            printf("*");
            break;
        }
    }
}

Time Complexity of the above function O(n). Even though the inner loop is bounded by n, but due to break statement it is executing only once.

 

Problem 4: Find the complexity of the below program:

void function(int n)
{
    int count = 0;
    for (int i=n/2; i<=n; i++)
        for (int j=1; j<=n; j = 2 * j)
            for (int k=1; k<=n; k = k * 2)
                count++;
}

Solution: Consider the comments in the following function.

void function(int n)
{
    int count = 0;
    for (int i=n/2; i<=n; i++)

        // Executes O(Log n) times
        for (int j=1; j<=n; j = 2 * j)

            // Executes O(Log n) times
            for (int k=1; k<=n; k = k * 2)
                count++;
}

Time Complexity of the above function O(n log2n).

 

Problem 5: Find the complexity of the below program:

void function(int n)
{
    int count = 0;
    for (int i=n/2; i<=n; i++)
        for (int j=1; j+n/2<=n; j = j++)
            for (int k=1; k<=n; k = k * 2)
                count++;
}

Solution: Consider the comments in the following function.

void function(int n)
{
    int count = 0;

    // outer loop executes n/2 times
    for (int i=n/2; i<=n; i++)

        // middle loop executes  n/2 times
        for (int j=1; j+n/2<=n; j = j++)

            // inner loop executes logn times
            for (int k=1; k<=n; k = k * 2)
                count++;
}

Time Complexity of the above function O(n2logn).

 

Problem 6: Find the complexity of the below program:

void function(int n)
{
    int i = 1, s =1;
    while (s <= n)
    {
        i++;
        s += i;
        printf("*");
    }
}

Solution: We can define the terms ‘s’ according to relation si = si-1 + i. The value of ‘i’ increases by one for each iteration. The value contained in ‘s’ at the ith iteration is the sum of the first ‘i’ positive integers. If k is total number of iterations taken by the program, then while loop terminates if: 1 + 2 + 3 ….+ k = [k(k+1)/2] > n So k = O(√n).

Time Complexity of the above function O(√n).

 

Problem 7: Find a tight upper bound on complexity of the below program:

void function(int n)
{
    int count = 0;
    for (int i=0; i<n; i++)
        for (int j=i; j< i*i; j++)
            if (j%i == 0)
            {
                for (int k=0; k<j; k++)
                    printf("*");
            }
}

Solution:Consider the comments in the following function.

void function(int n)
{
    int count = 0;

    // executes n times
    for (int i=0; i<n; i++)

        // executes O(n*n) times.
        for (int j=i; j< i*i; j++)
            if (j%i == 0)
            {
                // executes j times = O(n*n) times
                for (int k=0; k<j; k++)
                    printf("*");
            }
}

Time Complexity of the above function O(n5).

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