A Quick Mathematical Review

• Summations
  • We will use summations all of the time here.
  • Consider sort1

    for(i = 0 ; i < size-1; i++) {
       for (j = i+1; j < size ; j++) {
           if (a[i] > a[j]) {
              swap(a[i],a[j]);
           }
       }
    }
           

  • The outer loop is executed n (size) times.
  • The inner loop is executed n-1 times the first time through
  • The inner loop is executed n-2 times the second time through
  • The inner loop is executed n-k times the k^th time through
  • So in general it is (n-1) + (n-2) + (n-3) + ... + 3 + 2 + 1
  • or
  • (prove the final formula)
• He gives a list of rules for logs and exponents on 115, plus some more information in appnedix A.
• you should refer to these sections when you need them (I do)

Justification Techniques

• Our methods are from mathematics
• But we will not be quite as rigorous as a math professor would require.
• Methods
  • If you can, give a counter example. This is not useful for proving anything, but good for disproving.
  • Contradiction. Assume it is false, prove this assumption leads to a contradiction, therefore the original assumption was true.

    Assume:
       We can prove directly that a bound on sorting is O(n lg(n))
       We know that finding a convex hull is equivelent to sorting 
    
    To Prove The lower time bound on the convex hull problem is O(n lg(n))
        Assume this is false, the lower bound on the convex hull is less than
        O(n lg(n)), this means that we can find the couvex hull of a set of points
        in less than O(n lg(n)) time,  which implys that we can sort n integers in
        less than 0(n lg(n)) time, which is a contradiction, therefore the 
        lower bound on sorting is O(n lg(n)).
        

• Induction
  • We have just done strong induction
    • prove for n=1;
    • Assume for n=k;
    • Use this to prove for n=k+1
  • Weak induction
    • Prove for a base casee.
    • Assume true for all cases n<k
    • Prove true for n=k
    • Loop invariants
      • To prove a statement about a loop is correct.
        • Show the statement is true for the case before the loop begins
        • If the statement is true before the i^th iteration of the loop, show that it is true at the end of the i^th iteration
        • Showing the for the last iteration proves the statement.
      • For procedure IsInOrder in the example code, After iteration i, if inorder is true, the array up to position i is sorted.

        S0: Vacuosly, a single element array is in order.
        Si: We know that the array up to position i is in  order, and that
            the element at position i-1 is the largest.  If the element at
            position i is larger thant this, the array is still in order and
            the value of the variable is not changed, if it is smaller, the
            value of the variable is changed, and the the proposition no
            longer holds.