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The recurrence relation is defined as follows:

$$f(0) = 1$$ $$ f(1) = 1$$ $$ f(2n) = f(n)$$ $$ f(2n+1) = f(n) + f(n-1)$$

I am supposed to calculate $f(10)$, however I have a hard time wrapping my head around this; I can't see any way the relation terminates to the base case.

For instance letting $n = 2$, a simple case supposedly, We get the relation: $$ f(4) = f(2), f(5) = f(2) + 1 $$ If we roll it even further the recurrence just gets larger and larger, not smaller. I don't see a way to solve this using a classical divide and conquer approach.

I am wondering if this is some sort of mistake made by the person that created this question? If not, what am I overlooking?

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  • $\begingroup$ The way the recurrence works is that in order to evaluate $f(k)$, if $k = 2n$ is even then you use the third line, and if $k = 2n+1$ is odd then you use the fourth line. $\endgroup$ – Qiaochu Yuan Aug 7 '16 at 2:37
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$f(10)=f(5)=f(2)+f(1)=f(1)+f(1)=1+1=2.$

The recurrence is well-defined because the relations defining $f(n)$ always replace the argument n of the function with smaller and smaller values of n. Eventually these values must reach 0 or 1, at which point the base cases take care of the situation.

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  • $\begingroup$ Thanks! I was thinking about it the wrong way, I understand it now. I think I'll be able to solve the rest of this problem set. $\endgroup$ – Liz Aug 7 '16 at 3:07

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