# Noughts and Crosses Problem Statement:Find the winning lines of 3 in grids of n x n.

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Introduction

Noughts and Crosses

Problem Statement:

Find the winning lines of 3 in grids of n x n.

3 x 3 4 x 4 5 x 5 6 x 6

Possible Solution:

To work out how many winning lines of 3 there are on each grid I will draw in the lines (see appendix). Here are my results:

3 | 4 | 5 | 6 |

8 | 24 | 48 | 80 |

To find the rule I will use the difference method.

3 | 4 | 5 | 6 |

8 | 24 | 48 | 80 |

\ | / \ | / \ | / |

16 24 32

\ / \ /

- 8

Half of 8 is 4. So the rule will begin with 4n².

n | 3 | 4 | 5 | 6 |

a | 8 | 24 | 48 | 80 |

4n² | 36 | 64 | 100 | 144 |

B | -28 | -40 | -52 | -64 |

\ | / \ | / \ | / |

-12 -12 -12

a = answer, as in number of winning lines.

B = a - 4n²

This gives my rule so far as 4n²-12n.

(n² x 4)-(12 x n)

(9 x 4)-(12 x 3) = 36-36 = 0

0 + 8 = 8 +8

(16 x 4)-(12 x 4) = 64-48= 16

16+8= 24 +8

Therefore my rule is: 4n² - 12n + 8

To confirm my rule I will use an alternative method. I will break down “a” into the vertical, horizontal, and diagonal lines, find the rule for each then find the sum of the resulting rules.

Middle

This rule (the sum of the rules for vertical, horizontal and diagonal lines) is the same as the rule I found using the difference method, therefore it confirms that my rule is correct.

I am now going to look at grids of n x m.

In this case m = n+1

3 x 4 4 x 5 5 x 6 6 x 7

Possible solution:

To work out how many winning lines of 3 there are on each grid I will draw in the lines (see appendix). Here are my results:

3 | 4 | 5 | 6 |

14 | 34 | 62 | 98 |

To find the rule I will use the difference method.

14 34 62 98

\ / \ / \ /

20 28 36

\ / \ /

- 8

Half of 8 is 4. So the rule will begin with 4n².

n | 3 | 4 | 5 | 6 |

a | 14 | 34 | 62 | 98 |

4n² | 36 | 64 | 100 | 144 |

B | -22 | -30 | -38 | -46 |

\ | / \ | / \ | / |

-8 -8 -8

4n²- 8n

(n² x 4)-(8 x n)

(9 x 4)-(8 x 3)= 36-24= 12

12 + 2= 14 +2

(16 x 4)-(8 x 4)= 64-32= 32

32 + 2= 34 +2

Therefore the rule for n x m- where m= n+1 is: 4n²- 8n +2.

Again I am

Conclusion

Again I am going to confirm my rule by breaking down the winning lines into vertical, horizontal and diagonal, and then adding together the rules.

a | 3 | 4 | 5 | 6 | |

Vertical | 9 | 16 | 25 | 36 | n² * |

Horizontal | 5 | 12 | 21 | 32 | n²-4 * |

Diagonal | 6 | 16 | 30 | 48 | 2n²- 4n |

Total | 20 | 44 | 76 | 116 | 4n² - 4n -4 |

* by inspection I can see this rule.

.. I used the difference method to find this rule:

3 4 5 6

6 16 30 48

\ / \ / \ /

10 14 18

\ / \ /

- 4

This shows that the rule will begin with 2n².

n | 3 | 4 | 5 | 6 |

D | 6 | 16 | 30 | 48 |

2n² | 18 | 32 | 50 | 72 |

B | -12 | -16 | -20 | -24 |

\ | / \ | / \ | / |

-4 -4 -4

D = number of diagonal lines.

B= D-2n²

This means my rule so far is 2n²-4n.

(n² x 2)-(4 x 3)= 18-12= 6

(n² x 2)-(4 x 4)= 32-16= 16

Therefore the rule for diagonal lines in an n x n+2 grid is 2n²-4n.

4n² - 4n -4

This rule (the sum of the rules for vertical, horizontal and diagonal lines) is the same as the rule I found using the difference method, therefore it confirms that my rule is correct.

I am now going to find a rule for any n x m grid.

This student written piece of work is one of many that can be found in our GCSE T-Total section.

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