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?
The purpose of the simulation below is to help you come up with the right answer. Please try
it before you proceed to the solution. Click where you want to break them.
Answer
As many as there are small squares minus 1.
Proof (by induction)
- If there are just two squares we clearly need just one break.
- Assume that for numbers 2<m<N we have already shown
that it takes exactly m-1 breaks to split a bar consisting of m squares. Let there
be a bar of N squares. Split it into two with m1 and m2 squares,
respectively. Of course, m1 + m2 = N. By the induction
hypothesis it will take (m1-1) breaks to split the first bar and (m2-1)
to split the second one. The total will be 1 + (m1-1)+(m2-1) = N-1.
Proof #2
Let start counting how many pieces we have after a number of breaks.
The important observation is that every
time we break a piece the total number of pieces is increased by one. (For one bigger
piece have been replaced with two smaller ones.) When there is no pieces to break, each
piece is a small square. At the beginning (after 0 breaks) we had 1 piece. After 1 break
we got 2 pieces. As I said earlier, increasing the number of breaks by one increases
the number of pieces by 1. Therefore, the latter is always greater by one than the former.
Follow up
The basic fact explained above may appear in many different guises. For example, there
are quite edifying games based on the principle explained above (with every move a number
related to the game is increased by 1.) These games are not very challenging as such.
However, they furnish an edifying experience besides giving a chance for a knowledgeable
person to show off if he/she is the only one who knows the secret. Here are a few examples.
Problem #1
A fellow sawed 25 tree trunks into 75 logs. How many cuts did he perform? (Answer)
Problem #2
75 teams took part in a competition organized according to the olympic rules: teams met 1-on-1 with the defeated team getting dropped out of the competition. How many meets are needed to before one team is declared a winner? (Answer)
Game #1
Two players take turns breaking a bar. The last to break a piece wins the game.
An Aside
It's a great way to learn of odd and even numbers. Any one privy to the secret would know
what is preferable: to start the game or to be a second player - depending as whether the
total number of squares is even or odd.
Game #2
Marbles, checkers, or stones are arranged in several piles. A move consists in selecting a
pile and splitting it into two. The player to split the last pile is the winner. (Explanation:
it clearly does not matter how many piles one starts with. Imagine starting with a single
pile and then making a few moves "that do not count.")
Other simple games may be thought up to explain and reinforce the notion of parity,
i.e., the concepts that odd and even numbers are of different parities. For example,
Game #3
Write a sequence of numbers. For the entertainment sake, let one opponent write the
sequence and the other start the game. A move consists in writing a plus or a minus sign
between two adjacent terms. The first player wins if, with all signs inserted and computations
carried out, the result is odd. If the result is even, the second player wins. (Explanation:
The result does not depend on the particular distribution of signs at all. Adding or subtracting
an even (odd) number does not change (changes) the parity of the result. So the final result
will be odd iff the number of odd numbers in the sequence is odd.)
You may want to test your skills against your computer's.
Remark
Returning to the original problem of a chocolate bar, the number of moves needed to break it
into separate squares is invariant with regard to the actual sequence of moves. A less trivial
invariant may serve as a basis for a trick suitable for a magic show.
Yvan_Roux from Canada was inspired to make the following remark
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Congradulations!
We can use the same induction proof to prove that the result is true for a
puzzle or a 3D shape made of elementary pieces, as far as we do not break
the elementary pieces.
Yvan Roux
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References
- D. Fomin, S. Genkin, I. Itenberg, Mathematical Circles (Russian Experience), AMS, 1996
- P. Winkler, Mathematical Puzzles: A Connoisseur's Collection, A K Peters, 2004
Copyright © 1996-2009 Alexander Bogomolny