ITEC 122 2008fall ibarland

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lect12a
program proof

Strong induction to prove the correctness of this code:

// Return x^n.
//
double pow( double x, int n ) {
  if (n==0) {
    return 1;
    }
  else {
    if (n%2==0) {
      return sqr( pow(x,n/2) );
      }
    else {
      return sqr( pow(x,n/2) ) * x;
      }
    }
  }

// Return x^2.
double sqr( double x ) { return x*x; }


(Note that this program can be written more concisely:)
// Return x^n.
//
double pow( double x, int n ) {
  return (n==0) ? 1 : sqr(pow(x,n/2)) * (n%2==0 ? 1.0 : x);
  }


Let P(n) be:
"For any x, pow(x,n) really does return xn."
Note how this bridges program-return-values with math (actual exponentiation).


  

    Base case: P(0) = "For any x, pow(x,0) returns xn".
    Well, by inspection of our code (and the meaning of Java's 'if'),
    our code returns 1.  Is that x0?
    (Hmm, it is except for when x=0…)
  
  

    Induction step: We must show that, ∀k≥0,
    P(0)…P(k) → P(k+1).
    That is, we want to conclude P(k+1) = "∀x, pow(x,k+1) 
    returns xk+1.
    

    We consider two cases:
    

      

        Case k+1 is even (that is, k+1=2m, for some m<k+1):
        Then, k+1 % 2 == 0 (since k+1 even),
        our code returns sqr(pow(x,(k+1)/2)).
        By our inductive hypothesis, P((k+1)/2), we know that
        pow(x,(k+1)/2) = x(k+1)/2,
        so sqr(pow(x,(k+1)/2)) = sqr(x(k+1)/2)
        =
        x(k+1)/2 * x(k+1)/2
        (by the meaning of calling sqr),
        = xk+1, as desired.
      
      

        Case k+1 is odd (that is, k+1=2m+1, for some m<k+1):
      
    
  

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