Engaging math activities for the summer break - Day 5
Breaking chocolate bars is one of my most favorite activities.
Assume you have a chocolate bar consisting, as usual, of a number of squares arranged in a rectangular pattern. Your task is to split the bar into small squares (always breaking along the lines between the squares) with a minimum number of breaks. How many will it take?
This is how it may look after 4 "steps" when you start with a 5×6 bar.
What is the setup? There is a benefit from breaking the same size bar repeatedly. So it is probably wise to draw and then copy several square grids of the same size. These will be cut with scissors.
What's to observe? Curiously, however you go about breaking a bar, the number of breaks it takes to break it entirely is always the same; for a 5×6 bar the count is always 29! There is no possible optimization. The request to minimize the number of breaks was what I would call a red herring.
What about a proof? There are several proofs of this claim. The one I prefer uncovers another red herring concealed in the formulation of the problem. It is not important that you break chocolate bars. What you do is actually counting the number of squares in a bar - and this certainly does not depend on the sequence of breaks.
To see why, generalize! Imagine that now you have to count beans in a pile. You proceed in an unconventional way. What it takes is shifting the point of view. Instead of counting the beans let's count the piles. So, right now in front of you there is a single pile. Count it. Say "One". Now split the pile anyhow and count the number of piles again. There are two, so say "Two". Keep on splitting the piles into two. Every time you do, the number of piles grows by 1. You'll have to stop when each pile contains exactly 1 bean, meaning that the last pile count gives you the number of beans in the original pile. The number of "breaks" is 1 less than the number of beans.
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