4.6. Calculating Program Running Time¶

4.6.1. Calculating Program Running Time¶

This modules discusses the analysis for several simple code fragments. We will make use of the algorithm analysis simplifying rules:

If \(f(n)\) is in \(O(g(n))\) and \(g(n)\) is in \(O(h(n))\), then \(f(n)\) is in \(O(h(n))\).
If \(f(n)\) is in \(O(k g(n))\) for any constant \(k > 0\), then \(f(n)\) is in \(O(g(n))\).
If \(f_1(n)\) is in \(O(g_1(n))\) and \(f_2(n)\) is in \(O(g_2(n))\), then \(f_1(n) + f_2(n)\) is in \(O(\max(g_1(n), g_2(n)))\).
If \(f_1(n)\) is in \(O(g_1(n))\) and \(f_2(n)\) is in \(O(g_2(n))\), then \(f_1(n) f_2(n)\) is in \(O(g_1(n) g_2(n))\).

Example 4.6.3

We now analyze a code fragment with several for loops, some of which are nested.

Java
C++

sum = 0;
for (j=1; j<=n; j++) {     // First for loop
  for (i=1; i<=j; i++) {  //   is a double loop
    sum++;
  }
}
for (k=0; k<n; k++) {      // Second for loop
   A[k] = k;
}

sum = 0;
for (j=1; j<=n; j++)     // First for loop
   for (i=1; i<=j; i++)  //   is a double loop
      sum++;
for (k=0; k<n; k++)      // Second for loop
   A[k] = k;

This code fragment has three separate statements: the first assignment statement and the two for loops. Again the assignment statement takes constant time; call it \(c_1\). The second for loop is just like the one in Example 4.6.2 and takes \(c_2 n = \Theta(n)\) time.

The first for loop is a double loop and requires a special technique. We work from the inside of the loop outward. The expression sum++ requires constant time; call it \(c_3\). Because the inner for loop is executed \(j\) times, by simplifying rule (4) it has cost \(c_3j\). The outer for loop is executed \(n\) times, but each time the cost of the inner loop is different because it costs \(c_3j\) with \(j\) changing each time. You should see that for the first execution of the outer loop, \(j\) is 1. For the second execution of the outer loop, \(j\) is 2. Each time through the outer loop, \(j\) becomes one greater, until the last time through the loop when \(j = n\). Thus, the total cost of the loop is \(c_3\) times the sum of the integers 1 through \(n\). We know that

\[\sum_{i = 1}^{n} i = \frac{n (n+1)}{2},\]

which is \(\Theta(n^2)\). By simplifying rule (3), \(\Theta(c_1 + c_2 n + c_3 n^2)\) is simply \(\Theta(n^2)\).

Example 4.6.4

Compare the asymptotic analysis for the following two code fragments.

Java
C++

sum1 = 0;
for (i=1; i<=n; i++) {     // First double loop
  for (j=1; j<=n; j++) {  //   do n times
    sum1++;
  }
}

sum2 = 0;
for (i=1; i<=n; i++) {     // Second double loop
  for (j=1; j<=i; j++) {  //   do i times
    sum2++;
  }
}

sum1 = 0;
for (i=1; i<=n; i++)     // First double loop
   for (j=1; j<=n; j++)  //   do n times
      sum1++;

sum2 = 0;
for (i=1; i<=n; i++)     // Second double loop
   for (j=1; j<=i; j++)  //   do i times
      sum2++;

In the first double loop, the inner for loop always executes \(n\) times. Because the outer loop executes \(n\) times, it should be obvious that the statement sum1++ is executed precisely \(n^2\) times. The second loop is similar to the one analyzed in the previous example, with cost \(\sum_{j = 1}^{n} j\). This is approximately \({1 \over 2} n^2\). Thus, both double loops cost \(\Theta(n^2)\), though the second requires about half the time of the first.

Example 4.6.5

Not all doubly nested for loops are \(\Theta(n^2)\). The following pair of nested loops illustrates this fact.

Java
C++

sum1 = 0;
for (k=1; k<=n; k*=2) {    // Do log n times
  for (j=1; j<=n; j++) {  // Do n times
    sum1++;
  }
}

sum2 = 0;
for (k=1; k<=n; k*=2) {    // Do log n times
  for (j=1; j<=k; j++) {  // Do k times
    sum2++;
  }
}

sum1 = 0;
for (k=1; k<=n; k*=2)    // Do log n times
   for (j=1; j<=n; j++)  // Do n times
      sum1++;

sum2 = 0;
for (k=1; k<=n; k*=2)    // Do log n times
   for (j=1; j<=k; j++)  // Do k times
      sum2++;

When analyzing these two code fragments, we will assume that \(n\) is a power of two. The first code fragment has its outer for loop executed \(\log n+1\) times because on each iteration \(k\) is multiplied by two until it reaches \(n\). Because the inner loop always executes \(n\) times, the total cost for the first code fragment can be expressed as

\[\sum_{i=0}^{\log n} n = n \log n.\]

So the cost of this first double loop is \(\Theta(n \log n)\). Note that a variable substitution takes place here to create the summation, with \(k = 2^i\).

In the second code fragment, the outer loop is also executed \(\log n+1\) times. The inner loop has cost \(k\), which doubles each time. The summation can be expressed as

\[\sum_{i=0}^{\log n} 2^i = \Theta(n)\]

where \(n\) is assumed to be a power of two and again \(k = 2^i\).

What about other control statements? While loops are analyzed in a manner similar to for loops. The cost of an if statement in the worst case is the greater of the costs for the then and else clauses. This is also true for the average case, assuming that the size of \(n\) does not affect the probability of executing one of the clauses (which is usually, but not necessarily, true). For switch statements, the worst-case cost is that of the most expensive branch. For subroutine calls, simply add the cost of executing the subroutine.

There are rare situations in which the probability for executing the various branches of an if or switch statement are functions of the input size. For example, for input of size \(n\), the then clause of an if statement might be executed with probability \(1/n\). An example would be an if statement that executes the then clause only for the smallest of \(n\) values. To perform an average-case analysis for such programs, we cannot simply count the cost of the if statement as being the cost of the more expensive branch. In such situations, the technique of amortized analysis can come to the rescue.

Determining the execution time of a recursive subroutine can be difficult. The running time for a recursive subroutine is typically best expressed by a recurrence relation. For example, the recursive factorial function calls itself with a value one less than its input value. The result of this recursive call is then multiplied by the input value, which takes constant time. Thus, the cost of the factorial function, if we wish to measure cost in terms of the number of multiplication operations, is one more than the number of multiplications made by the recursive call on the smaller input. Because the base case does no multiplications, its cost is zero. Thus, the running time for this function can be expressed as

\[T(n) = T(n-1) + 1 \ \mbox{for}\ n>1;\ \ T(1) = 0.\]

The closed-form solution for this recurrence relation is \(\Theta(n)\).

4.6.1.1. Case Study: Two Search Algorithms¶

The final example of algorithm analysis for this section will compare two algorithms for performing search in an array. Earlier, we determined that the running time for sequential search on an array where the search value \(K\) is equally likely to appear in any location is \(\Theta(n)\) in both the average and worst cases. We would like to compare this running time to that required to perform a binary search on an array whose values are stored in order from lowest to highest. Here is a visualization of the binary search method.

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4.6.1.2. Binary Search Practice Exercise¶

4.6.1.3. Analyzing Binary Search¶

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Function binarySearch is designed to find the (single) occurrence of \(K\) and return its position. A special value is returned if \(K\) does not appear in the array. This algorithm can be modified to implement variations such as returning the position of the first occurrence of \(K\) in the array if multiple occurrences are allowed, and returning the position of the greatest value less than \(K\) when \(K\) is not in the array.

Comparing sequential search to binary search, we see that as \(n\) grows, the \(\Theta(n)\) running time for sequential search in the average and worst cases quickly becomes much greater than the \(\Theta(\log n)\) running time for binary search. Taken in isolation, binary search appears to be much more efficient than sequential search. This is despite the fact that the constant factor for binary search is greater than that for sequential search, because the calculation for the next search position in binary search is more expensive than just incrementing the current position, as sequential search does.

Note however that the running time for sequential search will be roughly the same regardless of whether or not the array values are stored in order. In contrast, binary search requires that the array values be ordered from lowest to highest. Depending on the context in which binary search is to be used, this requirement for a sorted array could be detrimental to the running time of a complete program, because maintaining the values in sorted order requires a greater cost when inserting new elements into the array. This is an example of a tradeoff between the advantage of binary search during search and the disadvantage related to maintaining a sorted array. Only in the context of the complete problem to be solved can we know whether the advantage outweighs the disadvantage.

CS2 Software Design & Data Structures

Chapter 4 Algorithm Analysis

4.6. Calculating Program Running Time¶

4.6.1. Calculating Program Running Time¶

4.6.1.1. Case Study: Two Search Algorithms¶

4.6.1.2. Binary Search Practice Exercise¶

4.6.1.3. Analyzing Binary Search¶

4.6.2. Summary Exercise¶