Lab Module C -- Writing tree and linear recursive procedures

1. Build experience writing tree recursive and linearly recursive procedures.
2. Be able to translate back and forth between C++ functions using loop constructs and recursively expressed Scheme procedures.

Reading: SICP (Structure and Interpretation of Computer Programs), finish Section 1.2

Homework exercises (Due at start of lab 26 Feb 01):

1. SICP Exercise 1.12 (Pascal's triangle).

2. The following C++ procedure computes the alternating sum of the squares of the positive integers up to n. Thus alternating_sum_squares(4) returns -10. Translate alternating_sum_squares into Scheme.

   int alternating_sum_squares(int n) // assume n > 0.
   { int a = 1;
     int sign = -1;
     for(int i = 2; i <= n; ++i)
     { a += sign*i*i;
       sign = -sign;
     }
     return a;
   }
   

   Remark: As an alternative to directly translating this method, you may if you wish work out an alternate method for (alternating-sum-squares n).

3. Many interesting number theory problems concern numbers with a geometric property. Most basically, the number n = k*k is a "square" because n dots could be arranged in a k by k square. The number n = k(k+1)/2 is a triangular number for a similar reason. We won't explore triangular numbers here, but rather will consider numbers which can be written as a sum of squares (you can think of stacking the squares of dots like a child playing with blocks of various sizes). Some famous open questions in number theory concern how each positive integer may be written as a sum of squares.

   It can be challenging to find the fewest squares necessary to make a given number. As a start in that direction, the following C++ procedure determines if a positive integer can be written as the sum of two squares. It assumes the procedure isqrt(n) returns the integer square root, i.e., the greatest integer r such that r*r <= n < (r+1)*(r+1). Translate it into Scheme. You may use your integer square root procedure from the module B topping off exercise.

   int isqrt(int n); // defined elsewhere
   
   bool sum2squaresp(int n)
   { int a = 1; 
     int b = isqrt(n);
     while (a <= b)
     { while (a*a + b*b < n) a = a + 1;
       if (a*a + b*b == n) 
         return true;
       else 
       { // try a lower top.
         b = b - 1; 
         a = a - 1;
       }
     }
     // all possibilities failed, so...
     return false;
   }
   

   Remark: Some C/C++ programmers use a p at the end of a function name to indicate that it is a predicate (a bool valued function). But name your Scheme predicate sum2squares? . The question mark available in Scheme names is an even better convention.

   ---

   Flourishes (these are optional problems):

4. SICP Exercise 1.11 (both parts).
5. Define a predicate (sum-of-k-squares? n k) which returns true just in case the n can be written as the sum of exactly k squares. You may assume n and k are both positive. Hint: this problem has something in common with the count-change example in the text. You might start like this:

   (define (sum-of-k-squares? n k) (limited-sum-of-k-squares? n k (isqrt n)) )
   

   where you define

   (limited-sum-of-k-squares? n k b) 
   

   to return true if n is the sum of the squares of k numbers, each of which is no larger than b. Thus (limited-sum-of-k-squares 6 3 2) is true, but (limited-sum-of-k-squares 6 3 1) is false,