• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Investigate the number of winning lines in the game of connect 4.

Extracts from this document...

Introduction

Investigation

Connect 4

 X X X X

This is a winning line in the game of connect 4 on a 4x5 board. Winning lines can be horizontal, vertical and diagonal. Investigate the number of winning lines in the game of connect 4.

The task is asking me to find out how many winning lines (connects) when you are connecting 4 there are on any size board.

What am I going to do

I am going to find out how many connect 4 there are in a 4x5 board.

•I will change the size of the box, but keep one value the width constant. And I will find a pattern in the number of connects there are in the different size boxes.

•I will use algebra to find a general formula for a NxWidth (W) box.

•I will then increase the width (constant) by one and work out a formula for that box.

•I will then find a pattern in the formulas for the different size boxes, connecting 4, and I will make a formula for the formula.

•I will then change the number that I will connect. For example 2, 3 or 5.

Connect 4

Firstly I will do a box with the width constant as 5 and I will change the height.

Hx5 Box

Any Number=N

Connects=C

Height= H

Width =W

Hx5   1   2   3   4    5    6

Connects   2   4   6  17  28  39          first layer

11  11  11             second layer

The box height of 1 and 2 do follow the pattern so

they are excluded. The connects go up by 11 each

time.   There are only 2 layers so the equation we

use is this equation. C=aH+b (original equation)

Middle

As before the first 2 equations do not follow

the pattern so they are excluded. Also like before

the connects has 2 layers so we use this original

equation.

C=aH+b

We use the same method as before.

9=a3+b     (1)

24=a4+b     (2)

15=a           (2)-(1)

Substitute ‘a’ which is 15 back into (1)

9=15x3+b

9=45+b

b= -36

Substitute‘a’ and ‘b’ back into original equation

C=15H-36    that is the equation for the number

of connects in a Nx6 box. But since the

first 2 heights didn’t follow the

pattern we didn’t use them in the

equation so this equation doesn’t

work for them.

Connect 4

Hx4 Box

Hx4   1   2   3   4   5   6

Connects   1   2   3  10 17 24     first layer

7   7   7         second layer

Like before the first to equations do not follow the

pattern so they are excluded. Also like before the

connects has 2 layers so we use this original

equation.

C=aH+b

We use the same method as before.

3=a3+b     (1)

10=a4+b     (2)

7=a             (2)-(1)

Substitute ‘a’ back into (1)

3=7x3+b

3=21+b

b= -18

Substitute ‘a’ and ‘b’ back into original equation

C=7H-18

that is the equation for the number of connects in a Nx4

box. But since the first 2 heights didn’t follow the

pattern we didn’t use them in the equation so this

equation doesn’t work for them.

Formula For Connect 4

The formula for any box with a width of 4 is C=7H-18

The formula for any box with a width of 5 is C=11H-27

The formula for any box with a width of 6 is C=15H-36

Conclusion

(1)

18=a16+b4+z   (2)

32=a25+b5+z   (3)

14=a9+b        (3)-(2)       (4)

10=a7+b        (2)-(1)       (5)

4=2a            (4)-(5)

2=a

Substitute ‘a’ back into (4)

14=2x9+b

14=18+b

b=-4

Substitute ‘a’ and ‘b’ into (3)

32=2x25-4x5+z

32=50-20+z

32=30+z

z=2

So we now know what ‘a’, ‘b’ and ‘z’ are so we sub them back into the original equation which was Fo=aC²+bC+z

Fo=2C²-4C+2 is the fourth number equation

So the first number equation is 4

The second number equation is 3C-3

The third number equation is 3C-3

The fourth number equation is Fo=2C²-4C+2

So we now join them together.

In the connect number equations the first 2 numbers were in brackets and so were the second 2. So we have to group the first 2 equations and the second 2 in brackets. But each equation has to be in its own brackets so we need to use double brackets.

The formula for connect 3 was (4W-6)H-(6W-8) we will use it as a bass.

(first number equation xW-second number equation)H-(third number equation-Fourth number equation)

But each number equation needs to be surrounded by its own brackets.

((first number equation xW)-(second number equation))H-((third number equation)xW-(Fourth number equation))

((4W)-(3C-3))H-((3C-3)xW-(2C²-4C+2))

This formula finds out how many connects there are in any size box using any connecting number. E.g you could use a height of 5 and width of 4 and we are connecting 4. This would give you the answer of 17 which is correct.

But as before the formula does not work if the height is 2 or more numbers lower than the number you are connecting.

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related AS and A Level Core & Pure Mathematics essays

1. ## Numerical Method of Algebra.

I only got its range, -0.66514 < root of the equation used < -0.66513. So, I decide to get their error bound by: Average of the upper and lower error bound = = Upper Error Bound = Lower Error Bound = The Error Bound of the root of this equation

2. ## Functions Coursework - A2 Maths

values, would be, for integer values of x from -2 to 2. This would therefore represent what values for f(x) would be in the table displayed earlier for integer values of x from -2 to 2. As one can see, only the red line from x=2 goes up, implying that

1. ## Math Portfolio Type II - Applications of Sinusoidal Functions

1 305 6.87h December 1 335 7.50h Toronto Location: 44?N 79?W Date Day Time of Sunset January 1 1 16.80h February 1 32 17.43h March 1 60 18.07h April 1 91 18.72h May 1 121 19.30h June 1 152 19.85h July 1 182 20.05h August 1 213 19.68h September 1

2. ## Solutions of equations

0.665 -0.019 0.666 -0.008 0.667 0.004 0.668 0.016 Table 5 x F(x) 0.6665 -0.001997 0.6666 -0.0007995 0.6667 0.0004 0.6668 0.0016 Table 6 x F(x) 0.66665 -0.0002 0.66666 -0.00008 0.66667 0.00004 0.66668 0.00016 I know that root B of the equation y = 243x3-378x2+192x-32 lies somewhere between: x = 0.66666 and x = 0.66667.

1. ## The open box problem

X 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 V 12.5 13.824 14.812 15.488 15.876 16 15.884 15.552 15.028 14.336 13.5 After looking at this table I have concluded that the maximum volume for the open box dimensions of 6x6 is 16, and x (that makes the volume at it's maximum is)

2. ## Triminoes Investigation

19a + 5b + c = 10 - equation 6 - 7a + 3b + c = 6 - equation 7 12a + 2b = 4 - equation 9 Equation 8 - Equation 9 I am doing this to cancel out b and find the value of a.

1. ## Experimentally calculating the wavelength of an He-Ne laser by means of diffraction gratings

fringes for diffraction gratings Diffraction Grating Spacing Distance Between Central Beam and Fringes (� 0.002m) 1st order (� 0.002m) 2nd order (� 0.002m) 3rd order (� 0.002m) 600 lines/mm 2.080 N/A N/A 300 lines/mm 0.573 1.228 2.122 100 lines/mm 0.303 0.605 0.915 Note: the distance between the central beam and

2. ## Maths - Investigate how many people can be carried in each type of vessel.

We now take the newly formed equations (iv) and (v) and we will now eliminate one of the factors from them. 25y + 2z = 133 X2) y - z = 1 => 2y - 2z = 2 27y = 135 y = 5 Now that we know one of the factors, we will now start to utilise the technique of substitution in our problem. • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 