Question

T(n) = T(n-2) + 1 for all n > 2. where T(1) = T(2) = 1. Find the even case and odd case.

Recursion relation

Answer 1

Given that T(n) = T(n - 2) + 1

T(n) = (T(n - 4) + 1) + 1 (By putting T(n - 2) = T(n - 4) + 1 while assuming n-2 > 2)

T(n) = T(n - 4) + 2

Now put value of T(n - 4) = T(n - 6) + 1 in above equation assuming that n - 4 > 2

T(n) = T(n - 6) + 3

From this, we make a hypothesis that T(n) = T(n - 2*k) + k (where k is any integer)

To prove the above hypothesis, we apply mathematical induction on k

Base case : k = 1

Then above becomes T(n) = T(n - 2) + 1 (This is true by given recurrence in question)

Hypothesis step : we assume that T(n) = T(n - 2*k) + k is true for k.

Inductive step : We will prove it for k + 1 given that it is true for k

To prove : T(n) = T(n - 2*(k + 1)) + (k+1)

T(n) = T(n - 2*k - 2) + (k + 1)

T(n) = T((n - 2*k) - 2) + (k + 1) [Equation 1]

Now T(n - 2*k) = T((n - 2*k) - 2) + 1 [From recurrence relation given in question]

So putting this value in Equation 1, we get

T(n) = T(n - 2*k) + k

and we know that above equation is true from hypothesis step,

So our hypothesis is true that T(n) = T(n - 2*k) + k from some integer k

Now,

Odd case : n is odd

we know that n is odd and 2*k must be even (for any integer k), So n - 2*k is odd.

so we will find value of k for which n - 2*k = 1

k = (n - 1)/2 (Note that n is odd so n - 1 is even there (n - 1)/2 must be integer)

now put value of k = (n - 1)/2 in equation T(n) = T(n - 2*k) + k

T(n) = T(n - 2*(n - 1)/2) + (n - 1)/2

T(n) = T(n - (n - 1)) + (n - 1)/2

T(n) = T(1) +  (n - 1)/2

T(n) = 1 + (n - 1)/2 (Given T(1) = 1)

T(n) = (n + 1)/2  where n is odd

Even Case : n is even

We know

T(n) = T(n - 2*k) + k from some integer k

n is even and 2*k must also be even (for any integer k), So n - 2*k must also be even

We will find value of k for which n - 2*k = 2

k = (n - 2)/2

put value of k in equation T(n) = T(n - 2*k) + k

T(n) = T(n - 2*(n - 2)/2) + (n - 2)/2

T(n) = T(n - (n - 2)) + (n - 2)/2

T(n) = T(2) + (n - 2)/2

T(n) = 1 + (n - 2)/2 (T(2) = 1)

T(n) = n/2  where n is even

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