How to solve the recurrence equation
T(n) = T(n/2) + T(n/4)
For base case T(n)=1
I have already checked this and this for clue but it didn't help me solving it using iterative method
I just need to get to the general equation for this.
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What is unclear with those? You iterate and plug `n/2` and `n/4` for `n`. Such as `T(n) = [T(n/4) + T(n/8)]+ [T(n/8) + T(n/16)]`... Keep going... – OneCricketeer Oct 11 '16 at 21:16
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I already tried it, went upto T(n)=T(n/4)+2T(n/16)+3T(n/32)+T(n/64). But I could not get to the generic equation @cricket_007 – Muhammad Zohaib Nawaz Oct 11 '16 at 21:24
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I don't see how to use the "iterative method" to solve this.
But there's a similarity between the recurrence relation you have and the that for the fibonacci numbers, and that can be used to find a solution.
T(2^k) = T(2^(k-1)) + T(2^(k-2)). So assuming T(1) = T(2) = 1, T(2^k) = Fib(k). So for n a power of 2, T(n) = Fib(lg(n)). Since Fib(n) = Theta(phi^n), T(n) = Theta(phi^(lg n)) = Theta(n ^ lg(phi)) ~= n^0.7
Here Fib(n) is the n'th Fibonacci number, and phi = (1 + sqrt(5))/2.
Paul Hankin
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Appreciate it, but why did you take T(2^k) instead of T(n)? – Muhammad Zohaib Nawaz Oct 11 '16 at 21:30
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@infinityAssault because powers of two produce exact Fibonacci numbers. It's common in many recurrence relation solutions involving /2 (for example that for merge sort or binary search) since rounding effects make the solutions difficult if you don't. – Paul Hankin Oct 11 '16 at 21:36
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This is also given in the question "For solving Fibonacci the formula is Fn= 1/√5 (ϕn− (−ϕ)−n) and you can ignore the constants in this formula" but I did not get it how to use it – Muhammad Zohaib Nawaz Oct 11 '16 at 21:43
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formula for Fib(n) = Theta(phi^n). where did this Theta came from? Or how to go from Theta(n^log(phi)) = Theta ( n^0.2) to n^0.7 as you wrote in the answer above? @PaulHankin – Muhammad Zohaib Nawaz Oct 22 '16 at 03:29