Connect Four: Grids
Date: 02/03/2003 at 12:57:46 From: Robert Subject: High School Math Recently I encountered a problem I cant seem to solve. It is related to the grids in the game in Connect 4. I need to investigate different ways of putting 4 counters in a straight line in different size grids (squares and rectangles). For example in a 4 by 4 grid, I can find 10 combinations of putting four counters in a straight line, and in a 5 by 5 grid 28. I've done several of these but I can't find a pattern or rule. What I want to find is a formula that allows me to find the combination of a rectangle or square of any size.
Date: 02/03/2003 at 17:09:26 From: Doctor Greenie Subject: Re: High School Math Hi, Robert - The rows of 4 consecutive markers can be in either (a) horizontal rows, or (b) vertical rows, or (c) diagonal rows from lower left to upper right, or (d) diagonal rows from upper left to lower right. For each of these types of 4 markers in a row, analyze the number of ways you can make them on an n-by-n board. For vertical rows, there are n columns, and the number of rows of 4 consecutive markers you can make in each column is (n-3), so the number of vertical rows of 4 consecutive markers on an n-by-n board is n(n-3). The analysis is exactly equivalent for the horizontal rows; the number of horizontal rows of 4 consecutive markers on an n-by-n board is also n(n-3). For the diagonal rows from lower left to upper right, the number of rows of 4 of consecutive markers you can make is (n-3) on the main diagonal, 1 less than that on each of the diagonals one square away from the main diagonal, 2 less than that on each diagonal two squares away from the main diagonal, and so on, until there is just room for one such diagonal row. The total number of such diagonals is then 1 + 2 + 3 + ... + (n-4) + (n-3) + (n-4) + ... + 3 + 2 + 1 A little algebra shows this sum to be equal to (n-3)(n-3). And again the analysis for the diagonals in the other direction is the same as for these diagonals. So the total number of ways of making a row of 4 consecutive markers horizontally, vertically, or diagonally on an n-by-n board is n(n-3) + n(n-3) + (n-3)(n-3) + (n-3)(n-3) = (n+n+(n-3)+(n-3))(n-3) = (4n-6)(n-3) = 2(n-3)(2n-3) Note that this result agrees with the numbers you found for n=4 and n=5: n=4: 2(4-3)(8-3) = 2(1)(5) = 10 n=5: 2(5-3)(10-3) = 2(2)(7) = 28 I hope this helps. Please write back if you have any further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/
Date: 02/05/2003 at 11:25:41 From: Robert Subject: Thank you (High School Math) Thank you very much. This seems reasonable. I appreciate your help.
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum