1) Investigate the relationship between the T-total and the T-number. 2) Use grid of different sizes. Translate the T-shape to different position. Investigate relationships between the T-total, the T-numbers and the grid size.

T-Totals

Introduction

If you look at the 9x9 grid with the T-shape, you can see that the total of the numbers added together is 37 because it is1+2+3+11+21 which equals 37.

This is what we call the T-total (37)

And T-number is the number at the bottom of the T-shape which in this case is 20

My tasks

The tasks I have been set are:

1) Investigate the relationship between the T-total and the T-number.

2) Use grid of different sizes. Translate the T-shape to different position. Investigate relationships between the T-total, the T-numbers and the grid size.

3) Use grids of different sizes again. Try other transformation and combinations of transformations. Investigate relationship between the T-total, the T-numbers, the grid size and the transformations.

Standard T-shapes

If I place the results from 7 T-Shapes into a table then it would look like this.

This table clearly shows that the numbers go up by 5 every time the T-number goes up by 1.

Now I can use trial and error to try to find the formula, I will then test what I believe to be the correct formula to see if it is.

Because there is a difference of 5 between the T-totals I think that it is a logical place to start the formula. I will believe that the T-total should be multiplied by this, then if the difference between the T-total and answer is the same. I will have found my formula for the upright T-shape on a 9 by 9 grid.

In this chart, T represents the T-number

I will now test my formula (5N-63) to check that it is correct

I will take T-number =68. If I use this with my formula I will get (still using a 9 by 9 grid) 5x68-63=277

It turns out that the T-total is 68+59+51+50+49 = 277. This means that my rule for normal upright T's on a 9 by 9 grid is correct. Now I shall move on to a standard upright T-shape on any sized grid.

Now I shall investigate with using various grid sizes with my T-shape. If ...

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In this chart, T represents the T-number

I will now test my formula (5N-63) to check that it is correct

I will take T-number =68. If I use this with my formula I will get (still using a 9 by 9 grid) 5x68-63=277

It turns out that the T-total is 68+59+51+50+49 = 277. This means that my rule for normal upright T's on a 9 by 9 grid is correct. Now I shall move on to a standard upright T-shape on any sized grid.

Now I shall investigate with using various grid sizes with my T-shape. If I now place the T-shape in the same place on an 11 by 11 grid as I did at the start of the investigation into the 9 by 9 grid, you can see that the T-number has risen by 4 and the T-total has risen by 6.

In this table I can see that the difference between the T-total is still 5 so the first bit of the formula is the same as in the 9 by 9 grid. To save time I will use a faster method to work out the total difference between the T-totals and all the other numbers inside the T- shape. In the last formula this number was 63.

24-1=23

24-2=22

24-3=21

24-13=11

TOTAL=77

I can now take an educated guess that this is the number that changes as the grid size alters. I now have to alter the formula to allow the size of the grid to be multiplied or added to another number to reach the difference of 77 I require. So I did this and I found the formula, with g representing the grid number.

5T-(7g)

I shall now test this formula on another grid size to make sure that it is correct.

With my formula I have calculated that the T-total for T-number 22 on a 6 by 6 grid will be 5x22-(7x6) =68

After studying the T-shapes for some time now and studying my formula, I have realize that A always equals T - 2 x The grid size (g) – 1, B equals T - 2g, C = T – 2g + 1, and D = T – g

As the T-Total is A+B+C+D+T I can conclude therefore that the

T-total = T – 2g – 1 + T – 2g + T – 2g + 1 + T – g + T

This can be simplified to 5T – 7g my prior formula.

I have now proven that my formula works on smaller sized grids. I have now worked out how to calculate any upright t on any sized grid just by using the grid size and the T-Number in the formula.

I will now rotate the T-shape so it is upside down and see what part of the formula changes. I will then try to work out a formula for all rotations.

I must conclude that my formula must be altered in some way. There is no change in numbers so I will try and change the minus sign to a plus and see if that formula works. The formula will be

5T+63 or 5T+ (7G)

I will now test this formula on a different sized grid to check if it works.

The T- number of this shape is 9 therefore

5x9+ (7x6)

T-total=87

9+15+20+21+22=87

Now my upside down T-shapes has been proven correct on different grid sizes. I will now try turning the T-shape anti clockwise on to its side and then the left. Then I will attempt to work out the formula.

As the shape has not rotated a full 180 degrees then the negative sign does not change to a positive sign. To work out what I need to minus from 5N, I will work out the difference. I am going to use the long method just in case the times by 7 rule for the grid number does not work.

12-11=1

12-10=2

12-19=-7

12-1=11

Total=7

Therefore

5x12-7=53

I will now check the formula to make sure that it works

T-number=12

12+11+10+19+1=53

This has proven that my formula works.

If I now rotate the shape a further 180 degrees. I predict that the formula will stay the same except that I now will have to change the minus sign in the formula to a plus.

I will now test this theory

So the formula I will use is

5x10+7=57

T-number =10

10+11+12+3+21=57

I have now proven this formula to be correct.

I will now try to work out a formula that I can use with vectors to work out the T-number and then use the formulas that I have already obtained to work out the rest.

This shape has been translated by a vector of +4.

-3

If my original T-number is 20 and I have moved it across right by 4, for every square across I have added to my original T-number. This also happens with the other vector but for every one square up or down the grid number is added or subtracted.

So in this case

X- Along in 1's

Y- up r down in g

Therefore

(T+X)+ (YG)

I will now test this

This T-shape has been moved by vector +3

-4

So if I use my formula

Red T-number =32

(32+3)+ (4x9) =71

Which is shown by the Blue T-number, now to work out the T-total all I have to do I use my formula from the first section.

Validation

I will now try and find a formula which can check if the T-Number is valid for that size of grid. As you can see, only some squares can be used as T-Numbers, the below diagram shows them: -

The grid is 9 by 9 but only the middle seven squares can be used as T-Numbers. This changes when the grid size changes because it’s always the grid size minus two. So I am going to divide the T-Number by the grid size.

When the T-Number is divided by the grid size we are left with a number and remainder is left over. The numbers without the remainder add one is the row that the T-Number is on and the remainder is the column number. With this information I am able to work out when the entire shape is on the grid. I will now try and prove this algebraically.

Number- the number add one is equal to the row so if the grid is 9-2= 7 and we know it is in the bottom seven then it is valid.

Remainder- this is the column of the T-number. We know this is correct because 9-2= 7 so the middle seven columns are valid and two is in the centre.

So the three steps to validating a T-number are:

1. T/S - The T-Number (T) is divided by the Grid Size (S) , the answer will be a number and a remainder (R=Row, C=Column).

T / S = R .C

2. Row - The row number +1 must be higher then 2 and lower then the grid size +1

(The inequality 2 > (R+1) < S+1)

3. Column - The column number must be lower then one and also lower then the grid size -1

(The inequality 1 > C < S-1)

If the two numbers fit the inequalities then the T-Number will fit inside the grid. I will now test the formula using random figures.

Conclusion

In conclusion I have worked out many different formulae, including how to work out the T-total from just the T-number, how to work out a rotated T’s T-total from the T-number, working the T-total out by using vectors, checking if all of a T-shape is in a grid, and checked everything I have done. I enjoyed this work as it used a lot of formulae and think I explained my method to the best of my abilities.

By Keir Wyndham-Ayres