The T-shape to the left is from a grid of width 8 (w=8). When I used the terms above, in the T-shape, and they all work. This is proof that the pattern I found works.
As in Part 1, if I add all the terms together, I will be able to obtain a suitable formula to work out the T-total of a T-shape on a grid of width (w).
(n) + (n-w) + (n-2w) + (n-2w-1) + (n-2w+1) = 5n – 7w
This calculation shows that the sum of the 5 terms within the T-shape is 5n – 7w, so I can make a formula which shows the relationship between the T-total, the T-number and the grid size:
T = 5n - 7w where T is the T-total, n is the T-number and w is the grid width
The table above shows the formula in use, by using the three examples at the beginning of Part 2. The formula has been shown to be correct.
These are the terms of the numbers in a T-shape when the width of the grid is (w); they are all in terms of (n).
If this T-shape (below left) is moved down by 1, and moved right by 1, and the width of the grid is 9 (w=9), I obtain a new T-shape. The old T-number is 20, and the T-total is 37, while in the new T-shape the T-number is 30 and the T-total is 87. If I rewrite these numbers in terms of
If the vertical translation is 1 (the T-shape is moved down by 1) then y = 1, while the horizontal translation is also 1 (the T-shape is moved right by 1) then x = 1. The diagram below left shows what
happens to the terms in the T-shape when the T-shape is moved down 1 and right 1.
The terms in the translated T-shape are quite similar to the original T-shape terms; (n), (n-w), (n-2w), (n-2w-1) and (n-2w+1). This means that the formula must have 5n and -7w in it, although they do not have to be together. Every term in the T-shape has had two things added to it; (x) which is horizontal translation, and (wy) which is grid width multiplied by the amount of vertical translation.
Each term has had (x) added to it because each square in the T-shape has been moved horizontally by (x). Because the numbers increase by 1 horizontally each time (e.g. 1 → 2 → 3) adding (x) would not make it move down (or up), just across i.e. it is horizontally translated.
Every term in the T-shape has also had (wy) added to it, and this is because they have been moved down by a number of rows i.e. a multiple of (w). This is vertical translation. When the vertical translation is 1, y = 1, and the grid width in this example is 9, so (wy) is 1 x 9 = 9. This means that the terms in the T-shape have been moved down by 9 squares, not 9 rows, and 9 squares is equal to 1 row. Therefore the T-shape has been moved down by 1 row.
When all of the terms in the translated (moved) T-shape are added together, I obtain this formula:
(n+x+wy) + (n+x+wy-w) + (n+x+wy-2w) + (n+x+wy-2w-1) +
(n+x+wy-2w+1) = 5n + 5x + 5wy – 7w and when this is factorised I get:
5(n + x + wy) - 7w
This formula has only been proven in one type of translation; down and right (1 type of diagonal translation, of which there are four types). This is an example of another type of diagonal translation.
T = 14 + 15 + 16 +25 + 35 = 105
T = (5 x 99) + (5 x -4) + (5 x (-6 x 10)) - (7 x 10) = 105
These two examples show that the formula works in at least half of all diagonal translations, so I will keep it:
T = 5(n + x + wy) - 7w where T is the T-total, n is the T-number, w is the grid width, x is the horizontal translation, and y is the vertical translation
NB: The T-shape must stay on the grid for the formula to remain valid; if one of the squares is not on the grid (i.e. an upside-down L shape), the formula will not work, as one of the terms is missing and the total (5n+5x+5wy-7w) is not reached.
Part 3
Use grids of different sizes again. Try other transformations and combinations of transformations. Investigate the relationships between the T-total, the T-number, the grid size, and the transformations.
There are 8 types of reflection overall in this project, but I will only investigate 4 of them.
Horizontal Reflection
When the line of reflection is set horizontally between the two rows beginning 41 and 51, I get a reflection- an inverted T-shape. In the original T-shape, quite simply, all the terms add up to (5n-7w).
To find the top square of the inverted T-shape, (n) had (2h) rows of (w), and another row added to it. This makes (n+(2h+1)w). The square below this one has had another row, (w) added to it, so it is (n+(2h+2)w). Again the square below this is (w) greater than this square, so it is (n+(2h+3)w). The square left of this one is the same minus 1, because it is 1 less than it: (n+(2h+3)w-1). The final square is (n+(2h+3)w+1) because it is one more than (n+(2h+3)w). All the terms contain (2h) because every square is a minimum of (2h) rows (2hw) away from each other.
To add these terms together, I must multiply out the brackets:
(n+(2h+1)w) + (n+(2h+2)w) + (n+(2h+3)w) + (n+(2h+3)w-1) + (n++(2h+3)w+1) =
(n+2hw+w) + (n+2hw+2w) + (n+2hw+3w) + (n+2hw+3w-1) + (n+2hw+3w+1) = 5n + 10hw + 12w and when this is factorised I get:
5(n + 2hw) + 12w
In the example above, the T-total for the reflected T-shape is 430 (when the 5 numbers are added together). When (n) is 22, (w) is 10 and (h) is 2, the formula tells me that the T-total is also 430. This proves that it works.
T = 5(n + 2hw) + 12w where T is the T-total, n is the T-number, w is the grid width, and h is the distance from the horizontal line of reflection
NB: The other type of horizontal reflection puts the original T-shape underneath the line of reflection. The formula is: T = 5n – (10hw + 18w).
Vertical Reflection
When the line of reflection is set between the columns beginning with 5 and 6, a new T-shape is formed, to the right of the grid. This is the second yellow T-shape.
The reflected T-shape’s T-number equivalent would be (n+2h+3), and the (n-w) equivalent would be (n-w+2h+3). The (n-2w) equivalent is (n-2w+2h +3). The (n-2w-1) equivalent would be (n-2w+2h+3-1), and the (n-2w+1) equivalent is (n-2w+2h+3+1). Effectively, all the terms in the original T-shape have had (2h+3) added to them, so that they move across by this amount. There is no vertical movement, which is why there are no (hw) in the new terms.
By adding the new terms together I can get a formula for vertical translation:
(n+2h+3) + (n+2h+3-w) + (n+2h+3-2w) + (n+2h+3-2w-1) +
(n+2h+3-2w+1) = 5n + 10h +15 – 7w and when this is factorised I get:
5(n + 2h + 3) - 7w
In the example above, when the 5 numbers of the reflected T-shape are added together, the T-total is 75. Using the formula, when (n) is 22, (w) is 10 and (h) is 2, the T-total is 75, which means the formula works.
T = 5(n + 2h + 3) - 7w where T is the T-total, n is the T-number, w is the grid width, and h is the distance from the horizontal line of reflection
NB: There is another kind vertical translation, where the original T-shape is on the right of the grid, and the reflection is on the left. The formula I found for this was T = 5n – 10h – 15 – 7w or T = 5(n – 2h – 3) – 7w.
Diagonal Reflection (1)
The T-shape in the top-left corner is the original, while the one in the bottom-right corner is the reflection.
The reflection’s T-number equivalent is (n+(2h+1)w+2h+1) because it has been moved down by ((2h+1)w), or (2hw+w) i.e. it has been moved down by (2h+1) rows. It has also been moved right by 2h+1 (if (h) is 2, then it has been moved across by 5). The term to the right of the T-number equivalent is almost the same, but it has been moved across by 1 more, so it must be (n+(2h+1)w+2h+2). The term to the right of this is 1 more again, so it has to be (n+(2h+1)w+2h+3). The term above this, is 1 row (1w) less than this term, but it has not horizontally, so it should be (n+2hw+2h+3). The last term, two rows below the previous one, must contain (2h+2)w in it, because the previous term contained (2h)w, and was 2 rows (2w) above it. It has not moved horizontally either, so it must be (n+(2h+2)w+2h+3).
This is what I get when I add all the terms together:
(n+(2h+1)w+2h+1) + (n+(2h+1)w+2h+2) + (n+(2h+1)w+2h+3) + (n+2hw+2h+3) + (n+(2h+2)w+2h+3) = 5n + 10hw + 10h +5w + 12 and when this is factorised I get:
5(n + 2hw + 2h + w) + 12
In the example, the T-total is 447 (worked out by adding up the numbers). Using the formula, when (n) is 33, (w) is 10, and (h) is 2, the calculated T-total is 447, which means that the formula works.
T = 5(n + 2hw + 2h + w) + 12 where T is the T-total, n is the T-number, w is the grid width, and h is the distance from the horizontal line of reflection
There are three other types of diagonal reflection; using this 45° line, the original T-shape can be put in the bottom-right hand corner, and the formula for this is T = 5n – 10hw – 20w – 10h – 13 (or T = 5(n – 2hw – 4w - 2h) – 13), or I can put the original T-shape in either of the two remaining corners when I use a 45° line that is perpendicular to the one used above (2 diagrams shown in rectangular box); the formula for the first of the diagonal reflections in the rectangular box is T = 5n + 10hw + 5w – 10h – 12 or T = 5(n + 2hw + w – 2h) – 12. The formula for the last of the 8 cases of reflection is T = 5n + 10hw + 20w + 10h + 15 or T = 5(n + 2hw + 4w + 2h + 3).
NB: The (n) (T-number) denotes where the original T-shape is.
