This is a \"pilesplitting problem\". Here is the situation: you start with n obj

Question

This is a "pilesplitting problem". Here is the situation: you start with n objects in a pile, and split it into two smaller piles. Then, for each new pile, you continue the splitting process until (finally) there are n piles of size one. At each splitting operation, you compute the product of the size of the two smaller piles. Once there are n piles, sum up all the products computed. The result will be a function of n, and will not (as it turns out) depend on how each pile is split along the way. Conjecture what this function is, and prove that your conjecture is correct. (You wil

l need to use "strong induction", as discussed in class when we proved the Fundamental Theorem

of Arithmetic.)

Here's an example, for n= 5.

Split the pile into 3 and 2. (The product of this split is 6.)

Split the 3 pile into 2 and 1. (The product is 2.)

Split the first 2

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pile into 1 and 1. (The product is 1.)

Split the second 2-pile into 1 and 1. (The product is 1.)

Now, the total is 6 + 2 + 1+1=10

I do not get it, please explain.

Explanation / Answer

Solution:

Will be splitting a pile at a time into two piles, multiply the numbers of chips in the two new piles and keep adding the results. The process stops when there is no pile with more than 1 chip.

For example, let start with 9 chips:

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somehow know that the result is 36! The secret is that the result is always the same: it does not depend on how the piles are split; but only on the initial size of the very first pile. Obviously, if you want to keep your audience in the dark, do not perform the trick with the same starting number of chips more than once.There is a very simple proof of this fact.

Here is the proof that uses mathematical induction.

Assume we start with N = 2 chips. The only way to split such a pile is to halve it into two piles of 1 chip each. The computed number is just 1. Of course it's independent of how you split the pile; for there is just one way to perform this feat. Note that starting with N = 1 leads to the number 0. We split nothing. One can argue it's the same as having a pile with 0 chips which contributes 0 to the total regardless of the number of chips in other piles.

Now, assume that the result has been established for all numbers less than N and let there be N > 2 chips in the original pile. Split it into two with n and m chips, respectively. We have n, m > 0 and n + m = N. By the inductive assumption, proceeding with the first pile we'll get the number n(n - 1)/2 regardless of how we actually proceed. For the second pile, we'll get m(m - 1)/2. The total is mn + n(n - 1)/2 + m(m - 1)/2 which, after a series of simplifications, yields (m + n)(m + n - 1)/2 = N(N - 1)/2 which

           i)   is dependent on neither original nor consecutive splits

           ii) is exactly the number we expected.

Piles Which is broken What's added Total 9 9 3*6 18 3,6 3 1*2 20 1,2,6 6 3*3 29 1,2,3,3 3 1*2 31 1,2,1,2,3 2 1*1 32 1,1,1,1,2,3 2 1*1 33 1,1,1,1,1,1,3 3 1*2 35 1,1,1,1,1,1,1,2 2 1*1 36 1,1,1,1,1,1,1,1,1 - -

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