This therefore leads me to conclude that for a 9 by 9 grid the formula for finding the T-Total is:
Part 2: Use grids of different sizes. Translate the T-Shape to different positions. Investigate relationships between the T-Total, the T-Numbers and the grid size.
To begin this part of the investigation I created several grid sizes ranging from 3 by 3 up to 10 by 10. The grids are shown below:
3 by 3 grid: T-Number is 8, and following the formula worked out in part 1:
(N-7) + (N-6) + (N-5) + (N-3) +N
= 5N-21
4 by 4 grid: T-Number is 15, and following the formula worked out in part 1:
(N-9) + (N-8) + (N-7) + (N-4) +N
= 5N-28
5 by 5 grid: T-Number is 22, and following the formula worked out in part 1:
(N-11) + (N-10) + (N-9) + (N-5) +N
= 5N-35
For the purposes of conserving space all grids are only 5 squares deep from now on. The end effect still remains the same.
6 by 6 grid: T-Number is 29, and following the formula worked out in part 1:
(N-13) + (N-12) + (N-11) + (N-6) +N
= 5N-42
7 by 7 grid: T-Number is 30, and following the formula worked out in part 1:
(N-15) + (N-14) + (N-13) + (N-7) +N
= 5N-49
8 by 8 grid: T-Number is 23, and following the formula worked out in part 1:
(N-17)+ (N-16) + (N-15) + (N-8) +N
= 5N-56
9 by 9 grid: T-Number is 38, and following the formula worked out in part 1:
(N-19)+ (N-18) + (N-17) + (N-9) +N
= 5N-63
10 by 10 grid: T-Number is 38, and following the formula worked out in part 1:
(N-21)+ (N-20) + (N-19) + (N-10) +N
= 5N-70
I noticed that every time the grid size is increased by 1 the variable at the end of the formula becomes another minus seven to what it already was before, e.g. 5N-21, 5N-28, 5N-35…… and so on. The constant value is always 5N, simply because there are always 5 numbers in the T-Shape. I also noticed a pattern between the grid size and the value to be subtracted at the end of the equation:
From this table I can work out a simple formula to find the T-Total of any sized grid. Using another T-Shape will help me prove the formula:
(Using a 3 by
3 grid as an
Example)
Where G is the grid size (in this example G = 3)
When worked out:
T-Total = (N-2G-1) + (N-2G) + (N-2G+1) + (N-G) + (N
= 5N-7G
Therefore the equation for finding the T-Total for any grid size is in the following formula:
I can now deduce that for a 1 by 1 size grid (although this is not possible as a 3 by 3 sized grid is necessary for a T-Shape) the equation would be 5N-7, for a 2 by 2 (again not possible to do) the equation would be 5N-14, for a 3 by 3 it would be 5N-21, and so forth up to huge numbers that would be very hard to work out using arithmetic – for example 231 by 231 would have a T-Total equation of 5N - (231*7) which works out as 5N-1617. Similarly a 45 by 45 grid size would work out as 5N- 315 and a 74 by 74 grid size would have an equation of 5N-518.
Part 3: Use grids of different sizes again. Try other transformations and combinations of transformations. Investigate relationships between the T-Total, the T-Numbers, the grid size and the transformations.
Part 3a: Rotations
I began this part by rethinking my initial formula. If I am to test the T-Shape in different rotations of 0˚, 90˚, 180˚ and 270˚ then the formula ‘5N-7G’ must be true for 0˚ as this is the angle the formula was based upon.
I next drew a T-Shape rotated 90˚ in the right on a 6 by 6 grid (I will test other grid sizes later) to give the following diagram:
The T-Number of this shape is 14 and the T-Total is 77.
From my previous findings on a 6 by 6 number grid the formula for finding a T-Total was 5N-42. If I attempt to use this formula I get 5(14)-42 which equals 28, so this is obviously not the formula I can use for 90˚ rotations. However, I simply used the same method I used to find out non-rotated T-Total formulas as I did with the rotated formulas. I drew another T-Shape and the differences between the T-Number and the rest of the numbers:
=
I then put these figures into another equation:
T-Total for 90˚ rotations on a 6 by 6 grid = (N) + (N+1) + (N-4) + (N+2) + (N+8)
= 5N + 7
I then tested this formula on another shape in the 6 by 6 table, using the figures 8, 9, 10, 4 and 16.
The calculated T-Total for this shape is 32, with the T-Number being 8. When putting 8 into the formula I’d just calculated I got 5(8) + 7 which equals 32 also.
Before moving onto the next step I tested the formula on grids of different sizes. I chose a 4 by 4 grid to test the formula on:
The calculated T-Total for this shape is 32, and before even checking this into the formula it was plain to see that the T-Total was going to stay the same regardless of the grid size. To prove this I checked it into the formula and came out with 5(5) + 7 which equals 32. I can therefore prove that:
I then used simple logic to determine the formula for a 270˚ rotation. If 90˚ was 5N + 7 and the angle is simply 90˚ in the other direction (270˚ of a complete 360˚ rotation) then the formula must be 5N-7. I tested and tried this formula again on a 6 by 6 grid:
The T-Total for this shape is 42 when calculated. When put into the formula it becomes 5(7) + 7 which also equals 42. This proves my theory that:
As with the last formula, I used simple logic to solve the 180˚ rotation formula. If the formula for 0˚ is 5N + 7G then the logical formula to solve the 180˚ rotation would be 5N – 7G. I tried and tested the theory below:
The calculated T-Total for this T-Shape is 57 and the T-Total when put into the formula above also equals 57. This proves that
Part 3b: Translations:
For this part I will be translating the T-Shape ‘a’ numbers left (negative numbers) and right (positive numbers) and ‘b’ numbers up (positive numbers) and down (negative numbers) and then I will be able to rotate it from there. Translating across by ‘a’ units is shown in the diagram below (on a 9 by 9 grid):
Where:
----- and
----- are original T-Shapes
And
----- and
----- are the same T-Shapes
translated 4 places to the
right.
The T-Total for the RED T-Shape is 37 and its BLUE translation has a T-Total of 57. Likewise:
The T-Total for the GOLD T-Shape is 222 and its GREEN translation has a T-Total of 242.
I noticed immediately that both translated T-Shapes had T-Totals 20 values higher than their originals. The reason for this is that the shapes were translated 4 places to the right and each number in the shape is increased by one every time it is moved to the right. Multiply this by the 5 numbers in the T-Shape (5*4) and you end up with 20.
This is better explained below:
When translated 4 places to the right is actually:
(where ‘a’ is the value the T-Shape is being translated by).
I can now conclude a provisional formula for a 9 by 9 grid T-Total when using translations:
T-Total of translated T-Shape = (T-Total of original shape) + (Number of units ______________of translation across grid x 5)
Before I conclude on this I will test it on a different sized grid to check that this formula works for any sized grid:
Where
----- = Original T-Shape and
----- = Translated T-Shape 3 units across
The T-Total for the RED T-Shape is 101 and the T-Total for the BLUE is 116.
When putting this into my provisional formula this becomes:
T-Total of translated T-Shape = (101) + (3 x 5)
Which equals 116, proving my formula correct. I can now conclude that the formula for finding the T-Total for a shape translated across any sized grid is:
It is important to note that when translating to the left the units are negative and when translating to the right the units are positive.
I will now use the same method to find the formula needed to translate the T-Shape up and down:
Where
----- and
----- are original T-Shapes
And
----- and
----- are the same T-Shapes
translated 4 places down.
The T-Total for the RED T-Shape is 37 and its BLUE translation has a T-Total of 217
Likewise:
The T-Total for the GOLD T-Shape is 102 and its GREEN translation has a T-Total of 282.
Again I noticed a pattern immediately. Both translations were 180 values higher than their original T-Shapes. The reason for this is the difference between the numbers in the original T-Shapes and the ones in the translated T-Shapes are always 36 values higher respectively. For example ‘1’ in the original T-Shape becomes ‘37’ in the translated T-Shape. Multiply this 36 by the 5 numbers in the T-Shape and you end up with 180, which is the difference between the T-Totals of the untranslated and translated shapes. This is explained more clearly below:
When translated 4 places actually becomes:
(Where ‘b’ is the value the T-Shape is being translated by).
My provisional translation formula for translating T-Shapes up and down a grid is:
T-Total of translated shape = (T-Total of original T-Shape) + (Difference __________________________between T-Numbers of the two T-Shapes x 5)
To confirm my theory I will try it on a different grid size:
Where
----- = Original T-Shape and
----- = Translated T-Shape 3 units across
The T-Total for the RED T-Shape is 36 and the T-Total for the BLUE is 141.
When put into the provisional formula this equals:
T-Total of translated shape = (36) + (21 x 5)
which equals 141, proving my formula correct.
I can now conclude that the formula for finding a T-Total of a number translated up or down is:
It is important to note that when translating down the units are negative and when translating upwards the units are positive.
This concludes my T-Totals investigation.
By James Muir
