I'm confused as to the solution for the snowball question. To start with, I have two specific questions:

(1) Each pair $a_i,b_j$ will account for one term (and why ONE term)? What is meant by term here? The coefficient, $c_k$ of the polynomial C? Or maybe the $x$ value at the $kth$ spot?

(2) Why is $c_k$ the number of such pairs?

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    $\begingroup$ Could you copy the relevant information into the question (with attribution of course!) to make the question as self-contained as possible? $\endgroup$ Commented Nov 16, 2013 at 14:08

1 Answer 1


Here is the idea of the proof. Let $a_i,b_i$ be the distance thrown by the $i$th male/female. Using FFT, we calculate $$ \left(\sum_i x^{a_i}\right) \left(\sum_j x^{b_j}\right) = \sum_{i,j} x^{a_i+b_j}. $$ Each pair $i,j$ satisfying $a_i + b_j = k$ contributes one term $x^k$ to the polynomial on the right. Hence the coefficient of $x^k$ is the number of pairs $i,j$ such that $a_i + b_j = k$.

If things still aren't clear, I suggest you try a few examples.

  • $\begingroup$ First you didn't even try to answer my specific questions. And now we have 3 summation signs, where there were none in the solution. So I am now more confused. $\endgroup$
    – lars
    Commented Oct 19, 2013 at 23:21
  • $\begingroup$ Moreover, what is $x^k$? Is it a coefficient? $\endgroup$
    – lars
    Commented Oct 19, 2013 at 23:23
  • $\begingroup$ Finally, you need an argument to go from Each pair i,j satisfying $a_i+b_j=k$ contributes one term $x^k$ to the polynomial on the right. To the claim that the coefficient of $x^k$ = number of pairs that equal $k$ $\endgroup$
    – lars
    Commented Oct 19, 2013 at 23:29
  • $\begingroup$ Here $x^k$ is a monomial ($x$ is known as a "formal variable"). You can also think of it as a polynomial having a single monomial. I suggest you go over the FFT algorithm again - they must have covered polynomials when they discussed the algorithm, especially given the application to multiplying univariate polynomials. $\endgroup$ Commented Oct 20, 2013 at 0:20
  • $\begingroup$ Re your other comment, I agree that an argument is required, but only if you want to be very formal. Otherwise, I suggest you try a few examples and see how it works. $\endgroup$ Commented Oct 20, 2013 at 0:20

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