Recursion overblow

Computer Science Level 3 

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def fun(p,q):
    if p==0:
        return q
    else:
        return fun(p-1, p+q)

Given the above function what is the value of fun(10000, 9950) ?


The answer is 50014950.

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Beakal Tiliksew by Beakal Tiliksew , 24, Ethiopia

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2 solutions

Claim: f ( p , q ) = q + 1 2 p ( p + 1 ) . f(p,q) = q + \tfrac12 p (p+1). Proof by recursion on p p :

If p = 0 p = 0 then f ( 0 , q ) = q f(0, q) = q is correct.

Assume that the equation is true for some value p p . Then f ( p + 1 , q ) = f ( p , p + 1 + q ) = ( p + 1 + q ) + 1 2 p ( p + 1 ) = q + ( p + 1 ) + 1 2 p ( p + 1 ) = q + 1 2 ( p + 2 ) ( p + 1 ) , f(p+1, q) = f(p, p+1+q) = (p+1+q) + \tfrac12 p(p+1) \\ = q + (p+1) + \tfrac12 p(p+1) = q + \tfrac12(p+2)(p+1), showing that the equation is then also true for p + 1 p + 1 .

Therefore the equation holds for all integer values p ≥ 0 p \geq 0 .

The answer, then, is 9950 + 1 2 ⋅ 10 000 ⋅ 10 001 = 50 014 950 . 9950 + \tfrac12\cdot 10\:000\cdot 10\:001 = \boxed{50\:014\:950}.

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fun ( p , q ) = q + ∑ i = 1 p i = q + p ( p + 1 ) 2 \text{fun}(p,q)=q+\sum_{i=1}^{p}i=q+\frac{p(p+1)}{2}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

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