To get a final formula for any e-number on a 10 X 8 grid, we need to convert the numbers in the E into algebra.
We need to write each square in terms of “e” to come up with an algebraic formula.
When adding up the squares we get the e-total so therefore the formula is:
E-total = e + (e + 1) + (e + 2) + (e + 8) + (e + 16) + (e + 17) + (e + 18) + (e + 24) + (e + 32) + (e + 33) + (e + 34)
E-total = 11e + 185
To prove this formula correct we need to try it on a random e-number, for example 9.
11e + 185 = e-total
(11x9) + 185 = e-total
99 + 185 = 284
9 + 10 + 11 + 17 + 25 + 26 +27 + 33 + 41 + 42 + 43 = 284
This proves that the formula is correct and we can now work out the e-total of any E on a 10 x 8 grid
I will now plot out an E on a 10 x 9 grid to see if I can find a formula for the e-total of and E on any grid size. The formula I already have (11e +185) is only correct when used on a 10 x 8 grid.
If we follow the example of the 10 x 8 grid, then we can work out the formula for this grid fairly quickly by converting the E into algebra and then checking it with a random number.
When converted to algebra the E is:
Therefore the formula should be:
e-total = e + (e + 1) + (e + 2) + (e + 9) + (e + 18) + (e + 19) + (e + 20) + (e + 27) + (e + 36) + (e + 37) + (e + 38)
e-total = 11e + 207
Now lets check with a random number, 10.
11e + 207 = e-total
(11x10) + 207 = e-total
110 + 207 = 317
10 + 11 + 12 + 19 + 28 + 29 + 30 + 37 + 46 + 47 + 48 = 317
This proves that the formula is correct and we can now work out the e-total of any E on a 10 x 9 grid
We have now found out that the formula for any E on a 10 x 9 grid. We must do the same with an E on a 10 x 10 grid. Then we will have enough information to get a formula for the e-total of any E on any grid size.
If we follow the example of the 10 x 8 and 10 x 9 grids, then we can work out the formula for this grid fairly quickly by converting the E into algebra and then checking it with a random number.
When converted to algebra the E is:
Therefore the formula should be:
e-total = e + (e + 1) + (e + 2) + (e + 10) + (e + 20) + (e + 21) + (e + 22) + (e + 30) + (e + 40) + (e + 41) + (e + 42)
e-total = 11e + 229
Now lets check with a random number, 11
11e + 229 = e-total
(11x11) + 229 = e-total
121 + 229 = 350
11 + 12 + 13 + 21 + 31 + 32 + 33 + 41 + 51 + 52 + 53 = 350
This proves that the formula is correct and we can now work out the e-total of any E on a 10 x 10 grid
With the information I have, I can now come of with a formula involving grid size which I will call “g”.
To come up with the formula we need to incorporate “g” into a drawn out E.
So the first e on a 10 x 8 grid would be:
When adding up the squares we get the e-total so therefore the formula is:
e-total = e + (e + 1) + (e + 2) + (e + g) + (e + 2g) + (e + 2g + 1) + (e + 2g + 2) + (e + 3g) + (e + 4g) + (e + 4g + 1) + (e + 4g + 2)
e-total = 11e + 22g + 9
Now we must check this formula with a random e number and random grid size.
Grid size = 6
E number = 2
11e + 22g + 9 = e-total
(11x2) + (22x6) + 9 = e-total
22 + 132 = 163
2+3+4+8+14+15+16+20+26+27+28 = 163
This proves the formula correct and I can now work out the e-total of any e-number on any grid size.
As each of the E’s so far has had an arm length of 3, I’m now going to try and work out the formula for any e number on any grid size with any arm length. (“x” will be used to represent arm length)
I will look at the top row first and put it in terms of x and e.
X = 5
The last square in the row will be e + (x-1) due the last square being e plus the width. The first square is e + 0 so 1 must be taken off of the total width.
I now need to find a formula for the top row using this information. To do this I need to find out the sum of the numbers added onto the each e along the top row. I will write this down twice, with one on top of the other. When I reverse the second one, and add it to the first I will get twice the total. Then no matter what the width of the arm, I will always get the total of all the numbers added on after the e in the top row.
R = row total (. . . means any arm length)
R = 1 + 2 + 3 . . . + (x – 2) + (x – 1)
R = (x – 1) + (x – 2) + (x – 3) . . . + 2 = 1
2R = x + x + x . . . + x + x
= x(x – 1)
R = ½ x(x – 1)
Now I need to e’s there are in the row, as these will be added onto this number. There is only 1 e in each square so therefore the number of e’s will be X x e
Therefore the formula for the top row is R = xe + ½ x (x-1)
To work out the middle row we do exactly the same except there is an extra 2g added to the e for each square.
Therefore the formula for the middle row is R = x(e + 2g) + ½ x(x – 1)
To work out the bottom row we do exactly the same except there is an extra 4g added to the e for each square.
Therefore the formula for the bottom row is R = x(e + 4g) + ½ x(x – 1)
The final E will look like this:
To get the final formula for arm length we need to add up the formulae from each of the rows and add up the 2 squares in between. (e +g and e +3g)
e-total = xe + ½ x(x – 1) + x(e + 2g) + ½ x(x – 1) + x(e + 4g) + ½ x (x – 1) + e + g + e + 3g
e-total = xn + x (n + 2g) + x (n + 4g) + 1.5x (x – 1) + 2n + 4g
e-total = xn + xn + 2gx + xn + 4gx + 2n + 4g + 1.5x (x – 1)
e-total = (3x + 2) (n + 2g) + 1.5x (x – 1)
Finally we must check that the formula works by choosing random numbers:
E number = 7
Arm length = 5
Grid size = 6
e-total = 7 + 8 + 9 + 10 + 11 + 13 + 19 + 20 + 21 + 22 + 23 + 25 + 31 + 32 + 33 + 34 + 35
e-total = 353
e-total = (3x + 2)(e + 2g) + 1.5x(x – 1)
e-total = (3 x 5 + 2)(7 + 2 + 6) + 1.5 x 5 x 4
e-total = 17 x 19 x 30
= 353
This proves the formula correct so finally we have a formula, which can find the e-total of any e on any grid size with any arm length.
