should equal:
T-Total = (n+19) + (n+18) + (n+17) + (n+9) + (n)
= 5n+63
Using our past knowledge we can see that the TT should =
TT = 5n + 63
= (5 x 2) + 63
= 73
Therefore we can now say that for an Upside down T on a 9 x 9 grid, 5n+63 is the common formula. Using previous knowledge I predict that this formula will only work on T’s on a 9 x 9 grid. I will now try to prove this by testing it on a 10 x 10 grid:
We can see that with the formula the TT would equal :
TT= 5n + 63
= 10 + 63
= 73
We can also see that if we used the previously known method of retrieving the TT then it would equal:
TT = 23 + 22 + 21 + 12 + 2 = 80
We can see that these two numbers are not the same and therefore I can draw the conclusion that this formula of 5n+63 only works on 9 x 9 grids.
Rotation of a T by 900 Anticlockwise
This is what a 9 x 9 grid T looks like when it is rotated 900 left. I will try to find an equation that shows the general principal of finding the T-Total.
From this we can see that:
A= n + 7 We can see that the T-Total should =
B= n-2
C= n-11 T-Total = 19 + 10 + 1 + 11 + 12 = 53
D= n-1
N= n From this we can see that algebraically 53
should equal =
TT= (n+7) + (n-2) + (n-11) + (n-1) + (n) =
= 5n - 7
Using this equation on the same T, from past knowledge I predict that:
5n – 7 = 53
I will now test this to see if it correct:
TT= 5n - 7
= 5 x 12 - 7
= 60 – 7
= 53
We can see that the two T-Totals are equal and therefore I can conclude that the equation for a 900 anti-clockwise rotated T that the equation is 5n - 7
Rotation of a T by 900 Clockwise
Using my past knowledge, I can predict that this T will be 5n + 7, just as 5n – 63 was countered by an upside down T with 5n + 63. I will now try to prove this theory.
This is what a T would look like if rotated 900 clockwise.
As you can see, the T-Total is
3 + 12 + 21 + 11 + 10 = 57
From this I will try to prove algebraically prove that this 57 can correctly prove my 5n + 7 theory.
We can see form the T above that:
A= n -7 We can see that the T-Total should =
B= n-2
C= n-11 T-Total = 19 + 10 + 1 + 11 + 12 = 53
D= n-1
N= n From this we can see that algebraically 53
should equal =
TT= (n-7) + (n-2) + (n-11) + (n-1) + (n) =
= 5n + 7
I will now test this theory:
TT= 5n + 7
= (5 x 10) + 7
= 57
We can see that this T-Total is equal to the predicted algebraic answer.
I have also noticed that the general formulas all run in a similar vein. I believe that the number that comes after the algebraic part of the equation is connected to the size of the grid that the T is on.
Grid Size and T-Totals
From my prediction, I will try to find an equation that means you will only need the grid size and the T-number to work out the T-Total.
For an original T, on a 9 x 9 grid, we can see that the general formula is 5n-63.
I will try to prove the –63 has a connection with 9, the grid size.
TT= 5n-63
= 5n – 9 x 7
We can see that –63 is –9 x 7. This shows that there is a possibility that the T-Total is connected to the grid size. I will now try and complete this prediction with an original T on a 7 x 7 grid and a 10 x 10 grid.
7 x 7 Grid
This is what a T would look like on a 7 x 7 grid. We can see that the T-Total would be:
16 + 9 + 1 + 2 + 3 = 31
I will now try to prove that 31 is equal to:
TT= 5n – (7 x g)
= 5 x 16 – (7 x 7)
= 80 – 49
= 31
From this, I am almost certain that if g is the grid size then the formula for any original T is 5n – (7 x g). However I will test one more T to prove my theory.
10 x 10 Grid
This is what a 10 x 10 T looks like. We can see that the T-Total would be:
1 + 2 + 3 + 12 + 22 = 40
To prove my prediction, I will try to prove that this T-Total of 40 is equal to:
TT= 5n – (7 x g)
= 5 x 22 – (7 x 10)
= 110 – 70
= 40
We can see that these 2 numbers are equal, and from this I can draw the conclusion that the general formula for an Original T is 5n – (7 x g).
Concerning Grid size and Upside Down T’s
From my last equation I can see that the grid size was part of the equation. As the equation for an upside down T is very similar to that of an original T (5n + 63), I can predict that the general formula for any Upside down T will be 5n + (7 x g). I will try to prove this again on a 9 x 9 grid, a 7 x 7 grid and a 10 x 10 grid.
9 x 9 Grid
This is what an upside down T looks like on a 9 x 9 grid. We can see that the T-Total is:
21 + 20 + 19 + 11 + 2 = 73
From my prediction, I will try to prove that 5n + (7 x g) can equal 73.
TT= 5n + (7 x g)
= 10 + 63
= 73
We can see that this does add up but I will try this formula again on a 10 x 10 grid and see if I can defiantly prove it.
10 x 10 Grids
This is what an upside down 10 x 10 grid T looks like. We can see that the T-Total should be:
TT = 2 + 12 + 23 + 22 + 21 = 80
I shall again test to see whether 5n + (7 x g) will equal 80.
TT = 5n + (7 x g)
= 5 x 2 + (7 x 10)
= 10 + 70
= 80
We can see that these two T-Totals add up, but to definitely prove my theory, I will try it once more on a 7 x 7 grid.
7 x 7 Grid
This is an Upside Down T on a 7 x 7 grid. We can see that the T-Total is:
TT= 17 + 16 + 15 + 9 + 2 = 59
To certify my theory I will try to prove that 5n + (7 x g) is equal to 59.
TT= 5n + (7 x g)
= 5 x 2 + (7 x 7)
= 10 + 49
= 59
We can see that these two T-Totals are yet again the same and from this I can draw that conclusion that the general formula for any upside down T is 5n + (7 x g).
I will not try to find any more equations for rotated 900 T’s as I know that the formula stays constant on any grid.
Vectors
When we know the T-Total of a T, it is a lot easier to have a formula which we can use to allow us to move the T to anywhere on the grid and still know the formula.
We can see that if we want to find out the T-Total of the original T, we can use the original formula. However, if we want to find out the 2nd T-Total, we can use the formula for the original T and still work out the T-Total for the second T.
We know that the formula for an ordinary T is 5n – 7g, or 5n – 63 on a regular 9 x 9 grid. We can therefore incorporate this formula into a new formula. We know that if we move the T one space across then each of the numbers will get bigger by one. As there are 5 “squares” on the T, we can see that the T-Total will increase by 5, meaning that the formula for moving a T one space across is (if we call the number of squares X):
5n – 7g + X
From this we can also see that if we move it back from the second T to the original one that the formula will be:
5n – 7g – X
We now know how to move a T across but we also need to work out the formula for moving it down. We can again use the original formula, 5n – 7g and incorporate this into our new formula.
We can see that if we try to move the T down by one square that each of the numbers increases by 9, which is also the size of the grid. From this we can see that if we multiply the grid size by the number of squares (X) and add this to the original formula then we will find the formula. I predict that this formula will be:
5n – 7g + (X x g)
I will now test this formula on another two T’s, on a 10 x 10 grid.
We can see that using the normal way, the two T-Totals would be:
1 + 2 + 3 + 12 + 22 = 40
11 + 12 + 13 + 22 + 32 = 90
Hopefully, when I apply the formula to the first T, I will get 90 as my answer.
TT= 5n – 7g + (X x g)
= 5 x 22 – 7 x 10 + (5 x 10)
= 110 – 70 + 50
= 40 + 50
= 90
We can see that the prediction works, so I believe that the formula is correct. We can use this to work out that the formula for moving a T back to the original position would be:
5n – 7g – (X x g)
This is Vectors completed. I can now move a T anywhere on a any grid and work out its T-Total from the original T.
Stretching T’s
This is what a T would look (on a 9 x 9 grid) like if you stretched it to 5 long and 5 wide, instead of the regular 3 x 3:
We can see that the numbers if see algebraically would equal:
We can see that if we put all of these together, we would get:
9n – 26g
We can use this formula to interpret a formula that will satisfy any T of any size. Hopefully I will be able to do this so that you can work out the T-Total from the T size (i.e. 3 x 3 or bigger), the grid size and the actual T-number. Looking at the table below we can see that patterns emerge from the equations for stretched T’s and hopefully this will take us a step closer to finding a general formula.
From these results, we can see that the number of n’s is always two times the size of the t, -1. From this I can draw the conclusion that:
N = 2T – 1
However, there is no clear pattern for the relationship between the t size and the number of g’s. For this I will use an Nth number table and hope that I can find a constant.
0 7 26 57 100
7 19 31 43
12 12 12
From this we can see that the constant number comes on the 2nd line of the table that shows us that the formula we have to use is:
Ax2 + bx + c
We know that 12 is 2a so A is 6. We also know that a + b is equal to the first number in the G sequence and the previous number. This means that a + b is 7 which means that b is 1. We know that c will equal the number previous to the first G number, which is 0, so our final formula is:
6x2 + x
We know that the first formula (ie for a 3 x 3 T) that the number of g’s is 7. We can collate all of this into a table and see what wee can do.
We can see that because y is equal to (x-1)/2, we can find a formula with x in it. We can then use this to find the general formula.
………………………………
= 6/4 x (x2 – 2x + 1) + ((x-1)/2)
= 6x2 – 12x +6 + 2n – 2
4 4
= 6x2 - 10x + 4
4
= 3 x2 – 5 + 1
- 2
I believe this to be the general formula and will now test it. X is the size of the T.
On a 3 x 3 T:
3 x 9 x 5 x 3 + 1
- 2
= 27 - 15 + 2
2 2 2
= 14
2
= 7
We can see that this is the number of g’s in the equation, which shows that the formula is correct. I will test it one more time and then will be able to draw to a conclusion.
On a 5 x 5 T:
3 x 15 x 5 x 5 + 1
- 2
= 45 - 25 + 2
2 2 2
= 18
2
= 9
We can see that this is definitely correct and as such I can come to a conclusion that the general formula for any T shape is:
3 x2 – 5 + 1
2 2
I have found many equations in my coursework and all of them have been important in helping me find this equation.
