The road to wisdom? Well its plain and simple to express: Err and err and err again, but less and less and less.
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There are two steps to the method.
1. Guess the solution
2. Prove the solution by induction
For example, we try to get the upper bound of T(n) = T(n/2) + 1 (this is the recurrence relation for binary search)
My guess is o(lgn) so our guess should look like T(n) <= clgn for some c and n > N.
Then we prove the guess by induction.
We assume that T(k) <= clgk holds for k < n.
T(n) = T(n/2) + 1
<= clg(n/2) + 1 (note: we invoke the assumption here)
= clgn - clg2 + 1
= clgn - (clg2 - 1) (note: we rearrange them to the form desired - residual)
<= clgn when c >= 1/(lg2)
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