Answer to Question 11E.4-2

  // countOrderedPairsBeginningWith(i,n) returns a count of
  // the ordered pairs of the form (i,j) where  1 <= j <= n and i > j.

  int countOrderedPairsBeginningWith(const int i, const int n)
  {
    int j, count = 0;

    j = 1;
    while(j <= n)
    {
      if(i > j)
      {
        count++;
      }
      j++;
    }
    return count;
  }

  // countOrderedPairs(n) returns a count of the number of
  // ordered pairs (i,j) where 1 <= i,j <= n and i > j.
  // For example, the ordered pairs containing integers from
  // 1 to 3 are: (1,1), (1,2), (1,3), (2,1), (2,2), (2,3),
  // (3,1), (3,2), (3,3).  In 3 of those ordered pairs, the first
  // number is larger than the second number.  So countOrderedPairs(3)
  // returns 3.

  int countOrderedPairs(const int n)
  {
    int i, count = 0;

    i = 1;
    while(i <= n)
    {
      count += countOrderedPairsBeginningWith(i,n);
      i++;
    }
    return count;
  }

Note. The definition of countOrderedPairsBeginningWith is strange because the loop keeps going until j > n. But values of j that are where j ≥ i are ignored. Changing the condition in the while-loop heading makes the inner if-statement unnecessary and makes the loop waste less time.

  int countOrderedPairsBeginningWith(const int i, const int n)
  {
    int j, count = 0;

    j = 1;
    while(j < i)
    {
      count++;
      j++;
    }
    return count;
  }

Note. Both of these functions can be simplified using for-loops.

  int countOrderedPairsBeginningWith(const int i, const int n)
  {
    int count = 0;

    for(int j = 1; j < i; j++)
    {
      count++;
    }
    return count;
  }

  int countOrderedPairs(const int n)
  {
    int count = 0;

    for(int i = 1; i <= n; i++)
    {
      count += countOrderedPairsBeginningWith(i,n);
    }
    return count;
  }