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### 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

- 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.
```
Associated Topics:
High School Discrete Mathematics

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