CSC152 2005F, Class 29: Algorithm Analysis (1)

Admin:
* Exams not graded yet.  Sorry.  Hopefully tomorrow.
* Challenge for tomorrow:
  Rewrite the recursive exponentiation with a while loop
  * Hint 1: Use the rule that x^(2k) = (x^2)^k
  * Hint 2: When doing tail recursion, we almost always introduce an accumulator.  You also often need one when converting recursive functions to while loops.
  * Hint 3: You may want to try tail recursion first
Overview:
* A motivating example: Exponentiation
  * Standard mechanism
  * Elegrant recursion
  * Pre-computer strategy
* Comparing algorithms
* Ignoring details

Problem: Compute x^n for non-negative integer n and BigDecimal x
  BigDecimal result = new BigDecimal(1.0);
  for (int i = 0; i < n ; i++) {
    result = result.multiply(x);
  }

Observation:
  x^(2k) = (x^k)^2

  public BigDecimal expt(BigDecimal x, int n)
  {
    // Base case: x^0 = 1
    if (n is 0) {
      return new BigDecimal(1.0);
    }
    // Recursive x^(2k) = (x^k)*(x^k) ; k = n/2
    else if (n is even) {
      BigDecimal tmp = expt(x, n/2);
      return tmp.multiply(tmp);
    }
    // Recursive: x^n = x^(n-1)*x
    else {
      return x.multiply(expt(x,n-1));
    }
  } // expt(BigDecimal, int)

Which of the two versions of expt do you prefer?  Why?
* First: 7
  * Conceptually simple
  * Short code
* Second: 1
  * Might be faster: Seems to do fewer computations
  * We certainly do fewer multiplications; does multiplication depend on the size of the number?

* Informal tests show that the second is correct, in spite of you naysayers
* Careful analysis should also suggest that is correct

Consider 3^32
For loop: 32 multiplications
Recursive: 
  * 1 division to get 16
    1 multiplication of 3^16 * 3^16
  * 1 division to get 8
    1 multiplication of 3^8 * 3^8 
  * 1 division to get 4
    1 multiplciation of 3^4 * 3^4
  * 1 division to get 2
    1 multiplciation of 3^2 * 3^2
  * 1 division to get 1
    1 multiplication of 3^1 * 3^1
  * 1 subtraction to get 0
    1 multiplication of 3^0 * 3
  * 6 multiplcations, 5 divisions; 1 subtraction
* If we double the exponent, we add only one mult. and division in the recursive case

Even though the second is more code, it is likely to be significantly faster

Experimentation

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How do you compute 5^20 without a computer?
* Run algorithm a or b by hand