14B. Recursive Scan Algorithms

We have seen how to write scan algorithms using loops. Anything you can do with a loop, you can also do with recursion. The preceding page included a definition of sum(A, n). Here are some additional examples.

Largest

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Let's define function largest(A, n), which returns the largest value among the first n values in array A, where n > 0.

Be sure to remember that the first n variables in array A are A[0], … A[n−1].

If n = 1, there is only one value, A[0], to consider. That one must be the largest.

If n > 1, compute m = largest(A, n−1), the largest of the first n−1 values in A. Then the largest of the first n values in A must either be m or A[n-1], whichever one is larger.

Here is a definition of largest(A, n) that uses those ideas.

  // largest(A,n) returns the largest of A[0],...,A[n-1].
  //
  // It is required that n > 0.

  int largest(const int A[], const int n)
  {
    if(n == 1)
    {
      return A[0];
    }
    else
    {
      int m = largest(A, n-1);
      return max(m, A[n-1]);
    }
  }

SumPrimes

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Here is an implementation of sumPrimes(n), which returns the sume of the prime numbers that are less than or equal to n. We implemented sumPrimes previously using a loop. Because of the filtering (considering only at prime numbers), there is one more case than in sum or largest.

Since the smallest prime number is 2, sumPrimes(n) = 0 when n < 2. Otherwise, compute m = sumPrimes(n−1). If n is not prime, then sumPrimes(n) = m. If n is prime, sumPrimes(n) = m + n.

  //==============================================
  // sumPrimes(n) yields the sum of the prime
  // numbers that are less than or equal to n.
  // For example, sumPrimes(5) = 2 + 3 + 5 = 10.
  //==============================================

  int sumPrimes(const int n)
  {
    if(n < 2)
    {
      return 0;
    }
    else 
    {
      const int m = sumPrimes(n-1);
      if(isPrime(n))
      {
        return m + n;
      }
      else
      {
        return m;
      }
    }
  }

  //===============================================
  // convertToInt2(s, n) yields the result of
  // converting the first n characters of s to an
  // integer. s must be a null-terminated string.
  //
  // For example, 
  //   convertToInt2("12345", 5) returns 12345,
  //   convertToInt2("12345", 4) returns 1234,
  //   convertToInt2("12345", 3) returns 123,
  //   convertToInt2("12345", 2) returns 12,
  //   convertToInt2("12345", 1) returns 1,
  //   convertToInt2("12345", 0) returns 0.
  //
  // Requirement: s must be a null-terminated string
  // that only contains decimal digits.
  //===============================================

  int convertToInt2(const char s[], const int n)
  {
    if(n == 0)
    {
      return 0;
    }
    else 
    {
      int lastDigit = s[n-1] - '0';
      return 10*convertToInt(s, n-1) + lastDigit;
    }
  }

  //===============================================
  // convertToInt(s) yields the integer described
  // by string s. For example, convertToInt("325")
  // = 325.
  //
  // Requirement: s must be a null-terminated string
  // that only contains decimal digits.
  //===============================================

  int convertToInt(const char s[])
  {
    return convertToInt2(s, strlen(s));
  }

Exercises

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1. Using recursion, write a definition of sumsq(n), which returns 1² + 2² + … n². Answer

2. Write a recursive definition of function numPrimes(n), which returns a count of how many of 1, …, n are prime. Answer