T-Total the investigation relies on investigating relationships between our three variables; t-number, t-total, and grid size

T-Total

Introduction

The T-Total investigation involves looking at the relationship between what are known as the t-number and the t-total in an n grid. The t-number can be thought of as the number at the base of the t shape, and the t-total the sum of all the numbers in the T. Therefore taking the following grid:

The t-number is 31, grid size is 9 and t-total is 31+22+12+13+14=92

Therefore the investigation relies on investigating relationships between our three variables; t-number, t-total, and grid size. Further variable factors may be added along the way to increase the scope of the investigation.

Variable Declaration

Individual variables within the ‘T’

We shall first look at how we can derive t through looking at n and x.

Contents

Investigating t, n, and x

The first logical step to finding t is to look at the individual values x, a, b, c, and d. If we know these values, we can simply add them up to find t. As we already know x, it simply becomes a matter of finding the other values. There are specific rules governing the placement of each value in relation to the others- we can see that a is always directly above x, and that it is directly below c. These may seem obvious, but they are key to deriving solutions to the problem. Simplifying the problem even further, we can see that there are only two values we need to know to calculate the values of the other squares; the row and column of x.

To find the row and column of x is not particularly difficult, and brings us directly to a solution that will work on any sized grid. It is clear that the grid works on a cycle- resetting at the end of each row. The interval of that cycle- the number at which it resets, is simply the grid size. Therefore, the position in a row is simply the place of x within any phase of the cycle- it can be found irrelevant of its row. This is quite similar to modulo arithmetic, and we can see therefore that the place in the cycle can be found by the function (x mod n). This takes x/n, but instead of looking at the answer, it finds the remainder. Because only this is involved, the actual row does not affect the answer.

Because we are using modulo arithmetic, the actual row is ignored. For example, 8 and 13 are in the same column. 8 mod 5 gives 3, as does 13 mod 5.

However, we do also need to know the row that x lies in. This is slightly more difficult. The natural way to look at it would be to take x/n, but this need to be rounded down to achieve the correct function. Saying rounded down is not a useful term when working algebraically, so we need an operator to do the same effect. The perfect operator exists in the form of the step function. Therefore, we can now say that the row is equal to [x/n] We shall now move on to calculating a, b, c, and d, but to facilitate this we shall define short term variables for the row and column to be included in our solutions.

P = x mod n

L = [x/n]

We shall first look at a. This is one row above x, and thus will be in the same column. We know that the row is equal to l-1, and that it lies at position p within the row. By multiplying n and (l-1) we can find the number at the end of the last row, from which we can count on to reach a. Therefore, we see that a is equal to

N(l-1) + p. Extrapolating this:

Grid width * (number of rows to move down) + squares to count on to reach a

We can proceed to find similar rules for the rest of the numbers. B is two rows above x, giving n(l-2), and is in the proceeding column; therefore p-1. The whole thing is n(l-2)-(p-1). Knowing b, we can easily find c and d, simply adding 1 to p each time; n(l-2) + p, and n(l-2) + (p+1). We know have formulas to calculate each square within the T shape. However, what we really want is one formula to calculate t from n and x. We can easily do this by combining our existing formulae:

T= x + n(l-1) + p + n(l-2) +p – 1 + n(l-2) +p + n(l-2) + p + 1

T = 4nl – 7n + 4p + x

However, we still have a 2-stage formula (remember l and p need to be calculated first). To get a single stage, we substitute in l and p to show

T = 4n [x/n] - 7n + 4(x mod n) + x

Factorising:

T = x + 4(n[x/n] + (x mod n)) – 7n

We now have a formula for any grid size and any t-number. To make sure this works, we will briefly look at a few examples demonstrating the principle.

Verification

Take grid size 9, and the starting number 59. Using the above method, we see

that t = 59 + 4(9[59/9] + 59 mod 9) – 7*9

thus t = 59 + 4(9*6 + 5) – 63

thus t = 59 + 236 – 63

therefore t = 232

Looking at the grid we see that t = 59+50+40+41+42=232

Take grid size 13 and the starting number 72.

T = 72 + 4(13*[72/13] + 72 mod 13) – 7*13

Thus T = 72 + 4(65 + 7) – 91

Thus T = 72 + 288 – 91

Therefore T = 269

Looking at the grid we see ...

This is a preview of the whole essay

that t = 59 + 4(9[59/9] + 59 mod 9) – 7*9

thus t = 59 + 4(9*6 + 5) – 63

thus t = 59 + 236 – 63

therefore t = 232

Looking at the grid we see that t = 59+50+40+41+42=232

Take grid size 13 and the starting number 72.

T = 72 + 4(13*[72/13] + 72 mod 13) – 7*13

Thus T = 72 + 4(65 + 7) – 91

Thus T = 72 + 288 – 91

Therefore T = 269

Looking at the grid we see that t = 72+59+45+46+47 = 269

In both these cases, the value calculated by the formula is the same as the actual total, proving to all extensive purposes that the formula works.

Alternate formula

The above formula is perfectly capable of calculating the T-total for any values of x and n. However, it does not take long to notice that it has some drawbacks. Foremost, but not necessarily prohibitive, it is rather complicated to use and explain. The major problem, though, is that it is almost impossible to rewrite the formula to calculate x when given n and t, or n given x and t. The use of modulo and step functions means that when reversing the formula, certain numbers cannot be established or distinguished between (i.e., in mod 3 arithmetic, there is no difference between 11 and 17). I do not believe it is impossible to achieve, but it is convoluted enough that I will not attempt to do so in this piece of work. Instead, we shall take a step sideways to look at an alternate formula to solve this problem.

My original idea when first looking at the problem was that, by calculating the row and column of the t-number, I could work out similar values for the other numbers, and thus calculate their values. However, after I had worked out the above formulas, it became obvious that this step is in fact, unnecessary. To look at this next step in a ‘real world’ scenario, it is rather like trying to calculate your latitude and longitude in order to describe where the person standing next to you is. It is a lot more logical to say ‘half a meter to the left of me’. After all, if we are to believe Einstein, then everything only has a position in relation to something else. To return and apply this scenario to our problem, we can see that it is unnecessary to calculate the position of x in order to get the locations of the others when we can just say ‘one row above me’. Because rows loop every n (ie, each row is n long, and thus the distance between two points in the same column is n), we can just say that a (returning to our above notation) is equal to x-n. Doing this for all the numbers in the T, we get:

A=x-n c=x-2n b=x-2n-1 d=b-2n+1

Conglomerating these, and x, we can get a formula for the whole T:

T = x-n + x-2n + x-2n + x-2n +1-1 + x

T = 5x –7n

This formula is a lot simpler than the previous one, and also has the advantage of being easy to rewrite in terms of x and n:

X = (t+7n)/5

N = (5x - t)/7

Using the formula in the above tests, we find the following:

(5*59) – (7*9) = 295 – 63 = 232 (which is correct)

(5*72) – (7*13) = 360 – 91 = 269 (which is also correct)

To reverse these, and work out x or n:

X = (232+63)/5 = 59

N = (5*59 - 232)/7 = 9

X = (269+91)/5 = 72

N = (5*72-269)/7 = 13

All these numbers are correct.

Extension 1: Transformations and Vector Translations

Our transformations of the T shape are basically the rotations of the T, and we shall look at calculating t for all of these. We shall also look at relationships between each of the different rotations. We shall then look at vector translations of the T, both direct and through backwards calculation, as will be explained further when we reach it.

Transformations

Each of the four different orientations of the T can be easily expressed in terms of either of the methods we have described above. The process for working them out is exactly the same: deal with each square individually, and then combine them to reach a total for the entire shape. For purposes of description, we shall assign each of the four rotations a name; north, south, east, and west- with their direction corresponding to the way the T is pointing. For example, the T we have so far looked at is south pointing.

North:

The north pointing T is very similar to the south facing one, and as such can be easily worked out: we simply have to add n in each case rather than subtracting it. We therefore see that:

A = n([x/n] + 1) + x mod n

Explaining this: A = length of row multiplied by number of rows add one (number at the end of x’s row) + places to move forward to reach A

B = n([x/n] + 2) + (x mod n + 1)

Explaining: B = row length + number of rows add two (we are now going down two rows) add the number of places to move forward to reach C, plus 1 to reach B

C = n([x/n] + 2) + (x mod n)

D = n([x/n] + 2) + (x mod n – 1)

Adding these together, along with x, we arrive at:

T = x + 4n[x/n] + 7n + 4(x mod n) Factorised as:

T = x + 4(n[x/n] + (x mod n)) + 7n

As we mentioned earlier, the case should be that we simply have to add n rather than subtracting it. With the final formulae for north and south, we see this is true; the only difference between them is the addition or subtraction of 7n at the end.

The alternate formula for north has a similar pattern, with only the sign of the 7n reversed:

5x + 7n

East and West:

For east and west, I have chosen to give the more complicated formulae as it is, and instead explain the alternate formulae, for in these it is easier to spot an interesting anomaly that deserves further enquiry. In this light the formulae are:

East = 3x + 2n[x/n] + 2(x mod n) – 7

West = 3x + 2n[x/n] + 2(x mod n) + 7

These formulae can be deconstructed to give individual values, and explained in terms such as we have used above; however, this would not add to this investigation in any way, and thus will not be included.

We move instead to the alternate formulae:

East = 5x – 7

West = 5x + 7

It is imminently visible that these equations are lacking one important factor, namely n. If this formula is correct (which it is), then it suggests that the grid length has no implications on the t-total. We shall first look at the maths behind this. Deconstructing the formula for a west facing T, we get the following formulae:

A = x – 1

B = x – n – 2

C = x – 2

D = x + n –2

These formula can be added up with x to get 5x –7 +n – n, which clearly explains why we have the above formulae, and demonstrates why the grid size has no effect on the value of t. I, however, personally find this answer to be of the kind that is irrefutable to deny, because of the mathematical proof, but which that part of the brain not working in algebra still finds rather circumspect. The best method I have of explaining this in non-mathematical terms is based along the following lines.

We can immediately discount the values x, a, and c on account of them all lying on the same line- obviously, grid size has no bearing on their value. We must therefore look at d and b, and their relation on the total.

D = x + n – 2

B = x – n – 2

The total value of these is 2x – 4. The contribution n plays has disappeared. This is because the steps further we must go back to reach d (and thus the smaller the number), the more steps we must take forward to reach b (and thus the larger the number). The differences in the combined total balance each other out, giving us a constant value for D+B irrelevant of grid size. Looking at a number of examples (pick a valid number from one of the above grids and test it on the other), we see this is true. Though this is not directly relevant to our investigation, it was an interesting anomaly and thus worth exploring.

Relationships between rotations

In the final segment of our look at transformations, we will look at direct relations between each of the rotations. For two pairs of these, we can see it is very easy. Transforming north to south, or east to west, necessitates only a flip in the sign of the final term. To transform, with whatever formula we have previously used, we need only, we need only add or subtract 14n (for north-south) or 14 (for east-west). Therefore:

N-S = -14n S-N = +14n

E-W= +14 W-E = -14

The more difficult transformation are from a NS T to an EW T. By this stage, it is fairly obvious that the alternate method is more suitable for expansion, both due to its lack of complexity and the fact that it will work backwards. This is therefore how we shall look at this problem. First, we recap the formulae for all four directions.

East = 5x – 7

West = 5x + 7

North = 5x + 7n

South = 5x – 7n

It is again obvious that once we can transform one, the others are easy. We shall therefore look at a NW transformation first, and then move on to the others using this result.

5x + 7n

5x + 7

Looking at this, it is not as difficult as it seems. The 5x can first be removed, equal as it is between equations:

7n + 0

0n + 7

Note that no other changes have been made; we have simply rearranged the formula so that the next step is easier to see. Written like this, we can simply subtract 7n and add 7 to breach the gap between the formulae. With this knowledge, we can easily rewrite for all the other transformations.

NW = -7n + 7

NE = -7n - 7

SW = + 7n +7

SE = +7n - 7

To operate these in reverse, simply multiply by –1 (flipping all signs in the equation).

Vector Translations

As briefly mentioned above, we shall look at vector translations in two ways. The first involves reverse engineering the t-total, adding on the vector, and recalculating. The second involves a direct translation; we shall look at this in a minute. Surprisingly, the former is in principle easier to solve using the first method. Because we already have values for the row and the column, we need only to add in the values for the vector. We will look at the following vector:

α represents our movement along the row, and β our movement up the column. We simply have to add these into the correct parts of the formulae:

t = x + 4(n([x/n] - β) + (x + α) mod n) + 7n

Here, we have simply subtracted β from the number of rows down we are, and added α to the number of places across we are. However, we still face the problem of reverse engineering the total, something that is best done using the alternate formula, and this leaves us with a dual equation; not particularly elegant. We will therefore attempt to integrate this method into the alternate formula.

Using the alternate formula, we do not actually calculate the row and column that x lies in, and thus we must do more to the formula to add in the extra elements. Again looking at a south facing T, we start with this formula:

5x – 7n

This time, because we are looking at the alternate formula, we must individually work out how much α and β effect the position of x (using the original formula, we dealt with the row and position on their own, so we could simply insert the values and allow the pre-existing formula to calculate their effect.). We first remove the 7n part- this relates only to the grid size, and thus is irrelevant, and are left with:

We now need to modify x to include the vector effect. The effect of α is simple: it is simply the number of places we move forward to reach the new x. We simply add it on to get 5(x + α). The effect of β is slightly harder- each increase of 1 in β is a full cycle down the grid- i.e., we must add on 1n for each β. Therefore, our effect is βn. Given this, our final formula is

5(x + α + βn) + 7n

The last step is to integrate into this a formula for deconstructing the t-total to get x: in this way, we can take a single T-total, regress it to the value of x and n, and recalculate to the new T based on a vector, all in one step. Luckily, we have a suitable formula which we looked at earlier:

X = (t+7n)/5

Putting this in the other formula, we find:

T2 = 5(((t+7n)/5) + α + βn)

We have now moved from an indirect translation to an effectively direct one. Although we still process it in the same way, we can simply insert all the relevant values into the equation and come out with the answer, giving us an effectively all-encompassing formula for direct vector translation of a T. We could look at expanding this to all directions, it would not add anything to this investigation, and thus we will move directly to looking at our last sector for investigation; expanding the dimensions.

Variable dimension Ts

This investigation, as will be further discussed in the evaluation, is very difficult to expand in any meaningful direction; so much so that the expansion into further dimensions is one of the only ways to move further. This is said because our foray into higher dimensions necessitates the making of certain assumptions that, to be completely accurate, should not be made. The first of these, looking at a 3D grid, is the assumption of the way the T should be orientated. In a 3 dimensional grid, there are 12 different ways that the T can be orientated, rather than the 4 in a 2D grid (we shall look later at how to work out the number of possible rotations for an n-dimensional T). It is also more difficult to translate between the different rotations (although this is not something we will look at).

It is rather complicated to draw a realistic 3D grid, especially on a computer, without specific tools and expertise. We shall therefore attempt to describe the way in which we have drawn the grid. This is in no way an ideal description; as such there may be confusion in the mental images of the grid.

The 2D grid gives us our n (horizontal) dimension and our (unspecified) vertical dimension. We now envision a y dimension being extended back from this grid to create a 3D object. We now show the top plane of this grid facing along the y dimension:

As you can see, we are now looping first along the n dimension and then along the y dimension. Our second plane looks like this:

We have oriented the T as if we were ignoring the effect of the y dimension; i.e. the T lies along the

n-vertical plane.

Because it is very difficult to describe how we have drawn the grid, it is equally difficult to describe how we have calculated the formula. Because of the extra degree of complexity involved, we have abandoned the original formula and are now looking at the alternate formula. While the original method would have proved quite interesting (at first looks, I believe that it would involve a double modulo calculation, to calculate the position in the ny plane and then isolate each of the individual dimensional positions), it is far more than is needed to solve this, and too complicated to be worth it.

So, we start again with our formula for the 2D grid:

5x – 7y,

and again we break our triangle down into it’s component numbers:

a = x – ny (the value of x, minus the loop of a whole plane, rather than just a line)

b = x – 2ny – 1 (this time, we are going up two planes, and moving back one place)

c = x – 2ny

d = x – 2ny + 1

We can therefore add this up, along with x, to reach our solution:

T = 5x – 7ny

This is quite an interesting, although predictable, result. We see that the only effect the other dimension has is to increase the size of the loop, by a factor of it’s own length. We can therefore, rather tentatively, suggest a method for calculating the t-total in a truly n dimensional grid:

T = 5x – 7(G1G2G3…Gη)

Where G is each direction, η is the number of dimensions, excluding our vertical dimension, which is of course irrelevant.

The shift into truly n-dimensional grids raises some difficult questions, such as how exactly we are to model a higher dimensional grid, whether it is possible to meaningfully insert a 2D T in to it, how to do so, and whether the maths is sound in these higher dimensions. These questions shall be, mercifully, completely ignored for the purposes of this investigation.

As was mentioned earlier, we shall now take a brief look at how we are to calculate the number of possible rotations of a T in an n dimensional grid. This number will naturally increase quite dramatically as we increase the number of possible dimensions. Our first approach is to think of it almost in a probabilistic sense, in that each dimension adds another multiplier to the number of possibilities. However, we see also that the number of dimensions that the shape we are rotating occupies contributes as a factor to this, which complicates the equation. Because we only need a formula for a two dimensional T shape, we see it is easier instead to visualise the T and see how the possibilities work. Firstly, we see that for each dimension, there are two ways in which the T can be oriented- our idea or north and south, east and west, for example. We therefore get the original equation 2η. However, each of these different Ts can be further flipped between lying along any of the other dimensions. Therefore, we must multiply our 2η by one less than the number of dimensions, or

η – 1. Our complete equation is therefore 2η(η – 1).

Conclusion and Evaluation

We have now looked at a wide variety of equations addressing a number of ways of looking at and expanding the original problem. We have attempted to make logical progression between them, and hopefully have discussed each problem fully and accurately. In cases, as mentioned, this has not been done, because in most cases it would have been close repetition of already covered topics. Even as we have looked at the problem, I believe we have strayed very close to the line of irrelevant data.

I maintain, as was my original belief, that this investigation was poorly expandable, with little extra to look at beyond the obvious that doesn’t just recover the same topics. The fact that we have had to stray into n-dimensions, something that is clearly not totally applicable to this, is testimony to the fact. This investigation was, to be blunt, far too simple to provide any truly interesting results, or to be extended.

While I say this, there are many ways we could have extended this investigation, but few of them moved further from the original problem. For example, we could have looked at every single rotation of the T in 3 dimensions, but this would have not moved the investigation any further beyond the one rotation we looked at. We would simply have been faced with pages of tedious, repetitive algebra; something which is of no help to anyone. Looking at the original formula in 3 dimensions would have been interesting, but would have been completely anomalous to the investigation; it would not have shown us anything further than the alternate formula, and would have been unnecessarily complicated. There is no point to such an expansion. Had I not discovered it first, the inclusion of the original formula in this investigation would have been purely superfluous.

To surmise, I believe this investigation was covered as thoroughly as it could be, and that the results produced are valid and justified. I do believe that a lot of the material is tedious and repetitive; I also feel this was unavoidable due to the nature of the problem set. I further believe that there is nothing further I could do to extend this investigation without either moving completely out of the purview of the original problem, or covering the same territory already covered.

Nicholas Clarke 11AH

Clevedon Community School