•Remember that means for the first time around there would
be N-1 comparisons.
–The next time around there would be N-2 comparisons
(because we can exclude comparing the previously found largest! Its already in
the correct
spot!).
–The third time around there would be
N-3 comparisons.
–So...the number of comparisons would
be:
– (N-1)+(N-2)+...+
1 = N*(N-1)/2