Spring 2001  CS100M  Solutions to Bonus Exercise E4

1. (last student is from M) = (m is odd).

note: You might be more used to seeing "if and only if" instead of "=".

2. inv: The parity (remainder mod 2) of the number N of M students is constant
   Proof by case analysis on the loop body:
   + if both students are from M, then N <-- N-2, so parity is unchanged.
   + else (either or both are from J) N is unchanged, so parity is unchanged.

Remark: The loop invariant ("storyline") is a powerful tool that clearly
        and succinctly explains what is happening!