A question I was given

$T(0) = 1,$ $T(1)=0,$ $T(n)= 2T(n-2)$

I think the possible solution is $T(n)=2^n$

Proof: by induction. Base Case:

$n=0$ $T(0)=2^0=1$

Inductive Hypothesis: Assume for some $n$. $T(n)=2^n$ Show for $n+1$

$T(n+1)= 2(T(n-1))$ $=2(2^{n-1})$ $=2^n$

which is what I proposed. Is this correct? I'm a little confused because it says that $T(1)=0$, so it wouldn't make sense for $t(n)=2^n$. But I don't know how else I could prove the recurrence relation given as stated above.


1 Answer 1


This can be verified using the Master Theorem for decreasing functions

The recurrence relation can be written as $T(n)=2T(n-2)+O(1)$.

Comparing with the general recurrence here $a=2$, $b=2, k=0$

Using Case 2 of the theorem we get $T(n)= O(2^{\frac{n}{2}}*n^0) = O(2^{\frac{n}{2}})$

Even if you don't know the theorem, for such standard recurrences, one way is to draw the recurrence tree for it, and you can quickly figure out the same.


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