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CSC 236            Sample Solutions for Week  3 Tutorial          Fall 2007
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A.  Make a conjecture about the values of n for which n^2 <= 2^n.
    Prove your conjecture.

    Explore:  Compute table of values for n^2 and 2^n for small n.

    Conjecture:  n^2 <= 2^n for all n >= 4.

    0.  Define P(n):  n^2 <= 2^n.

    1.  Base Case:  Prove P(4): 4^2 <= 2^4.
            4^2 = 16 = 2^4
        Hence, P(4).

    2.  Ind. Hyp.:  Let n >= 4 and suppose P(n): n^2 <= 2^n.

    3.  Ind. Step:  Prove P(n+1): (n+1)^2 <= 2^{n+1}.

            (n+1)^2 = n^2 + 2n + 1 <= 2^n + 2n + 1  (by IH)

        Q:  What next?  What would help us conclude what we want?
        A:  If we knew 2n + 1 <= 2^n.
        Q:  How could we prove this?
        A:  Idea: maybe show 2n + 1 <= n^2 and use IH.
        Q:  What do we know about n that could help?
        A:  n >= 4 (from IH).

            2n + 1 <= 2n + n = 3n <= n*n = n^2, because n >= 4, so
            continuing with our reasoning:
                (n+1)^2 <= 2^n + 2n + 1
                        <= 2^n + n^2  (by the argument just given)
                        <= 2^n + 2^n = 2^{n+1}  (by IH).
        Hence, P(n+1).

    4.  Conclusion:  By induction, \-/ n >= 4, P(n), i.e., n^2 <= 2^n for
        all n >= 4.

B.  When computing the product of n distinct real numbers a_1 a_2 ... a_n,
    there are many ways to order the multiplication operations that have to
    be performed.  For example, the product of a_1 a_2 ... a_5 could be
    computed in the order (a_1 * a_2) * ((a_3 * (a_4 * a_5)), or in the
    order a_1 * ((a_2 * (a_3 * a_4)) * a_5), or in many other ways.

    Make a conjecture about the number of multiplication operations
    required to compute the product a_1 a_2 ... a_n of n distinct real
    numbers, for any n -- in particular, does this number depend on the
    order of the multiplications?
    Prove your conjecture.

    Conjecture:  For all n >= 1, it takes exactly n-1 operations to compute
    the product of any n numbers, no matter how it is parenthesized.

    0.  Define P(n):  It takes exactly n-1 operations to compute the
        product of any n numbers, no matter how it is parenthesized.

    Q:  What form of induction should we use?
    A:  Either would work!  Let's practice complete induction -- simple
        induction will be mentioned at the end.

    Complete induction version

    Q:  How do we know whether we should start with base cases?
    A:  Are there any values of n for which we cannot conclude P(n) from
        the fact that P(k) for all k < n?

    1.  Base Case:  Prove P(1): it takes 0 operation to compute the product
        of any one number.  This is trivially true, so P(1).

    2.  Ind. Hyp.:  Let n > 1 {since we've already proved P(1) and we're
        about to try to prove P(n)} and suppose \-/ k in {1,...,n-1}, P(k),
        i.e., for 1 <= k < n, it takes exactly k-1 operations to compute
        the product of any k numbers, no matter how it is parenthesized.

    3.  Ind. Step:  Prove P(n): it takes exactly n-1 operations to compute
        the product of any n numbers, no matter how it is parenthesized.

        Since n > 1 (by IH), we have to perform at least one operation to
        compute the product of a_1 ... a_n.  Consider the *last* operation
        performed: (a_1 ... a_i) (a_{i+1} ... a_n), for some 1 <= i < n.
        By the IH, it takes i-1 operations to compute the product of
        a_1 ... a_i, and it takes n-i-1 operations to compute the product
        of a_{i+1} ... a_n (since 1 <= i < n -> n-1 >= n-i > 0 so the IH
        applies to a_{i+1} ... a_n).  Hence, it takes 1 + i-1 + n-i-1 = n-1
        operations to compute the product a_1 ...  a_n.  Since we
        considered arbitrary ways to compute the product a_1 ... a_n, P(n)
        holds.

    4.  Conclusion:  For all n >= 1, P(n), i.e., it takes exactly n-1
        operations to compute the product of n numbers, no matter his it is
        parenthesized.

    Simple induction version

     -  Main idea of inductive step:  Consider the product a_1 ... a_{n+1}.
        For some pair of adjacent numbers a_i a_{i+1}, 1 <= i <= n, perform
        an operation to multiply them: a_i * a_{i+1} = a'_i.  The rest of
        the product looks like a_1 ... a_{i-1} a'_i a_{i+2} ... a_{n+1},
        and it consists of n numbers so by the IH, it will take n-1
        operations to compute it.  Together with the first operation, this
        adds up to n operations, as required.

     -  Easy exercise: write this up formally -- ask during office hours if
        you want to double-check your answer.