A Sticky Problem
The game presented by a Java applet below was published in 2002 the American Mathematical Monthly as problem #569 (Sung Soo Kim). A solution by Li Zhou appeared in v. 111, n. 4 (April, 2004), pp. 363-364.
A game starts with one stick of length 1 and four sticks of length 4. The two players move alternately. A move consists of breaking a stick of length at least two into two sticks of shorter lengths or removing n sticks of length n for some
The applet allows one to start with different initial configurations, but also with the standard one. At the outset, you can force the computer to make the first move by pressing the "Make move" button. Sticks are represented by a row of small squares that might remind of a chocolate brick. To break a piece click on a square to the right of the desired break line.
What if applet does not run? |
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Every position in the game is completely determined by a quadruple
Introduce
r = (k + m) mod 2 and s = (l + n) mod 3. |
We prove that the set
Let N be the set of positions not in P. It suffices to show that any move from a P-position leads to an N-position, and that for every N-position there is a move that leads to an P-position. (Note that
A move that removes i sticks of length i reduces coordinate i by i; such a move changes exactly one of r or s. There are four possible stick-breaking moves:
(1) | (k, l, m, n) (k+2, l-1, m, n), |
(2) | (k, l, m, n) (k+1, l+1, m-1, n), |
(3) | (k, l, m, n) (k, l+2, m, n-1), |
(4) | (k, l, m, n) (k+1, l, m+1, n-1). |
Each of these also changes exactly one of r or s. Hence every move from a P-position leads to an N-position. From an N-position, the moves that lead to a P-position are listed below. Keep in mind that r is either 0 or 1, while s is either 0, or, 1, or 2.
N-Position | Winning move | ||
---|---|---|---|
s - r = -1 | k(k - 1) if k > 0 (2) if m > 0 | ||
s - r = 1 | (1) if l > 0 (3) if n > 0 | ||
s - r = 2 | l(l - 2) if l > 1 (4) if n > 0 |
It remains to be shown that the specified moves are available. If
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Copyright © 1996-2017 Alexander Bogomolny
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