Building linear sequences using geometric shapes.
So, a few days ago I was doing my math curriculum. There is a chapter named 'Patterns'. Basically, about sequences and stuff. One interesting part about that was it said to show 5x+7 as a geometric pattern using simple geometric shapes. Well, the book didn't define what a 'simple geometric' shape counts as but using a guide and examples from the book I figured it's mainly squares and triangles with the same side length (unit length). Using those 2 shapes we had to build sequences. This got me thinking is there a general construction method I can use to build the linear sequence.
So, I had to define somethings from the start. These are the rules I came up with,
- Each new shape must be adjacent to the previous shape.
- No loops with shapes (I will get to that)
And other rules I think (I forgot the rules at the time of writing if I might remember whilst writing). So, the first question is what do we count from the shapes? We count the sides/sticks needed to make the final shape. From the picture you can see we only count the total lines. For a square (starting) the number of lines is 4. And for a starting triangle it's 3. For adding another square to the existing square, the total lines become 7 since 2 squares next to each other share 1 line. So, it should have been 4+4 but 1 overlap so 4+4-1=7. Same goes with a triangle.
Now let's actually get into some equations. So first of let,
x = total amount of squares
y = total amount of triangles
z = overlaps between those squares and triangles
So, since I counted for overlaps, I can easily come up the total number of lines to be 4x+3y-z
N shapes arranged in a tree shape has N-1 adjacent sides. So, if we only arranged in tree shape meaning every new shape is places next to the previous shape, we can easily set z=x+y-1 since x+y is the total amount of shapes. So, plugging it back in we get,
4x+3y-(x+y-1)
= 3x+2y+1
So, the total number of lines effectively reduces to 3x+2y+1.
So now that we have established that let's get to a linear sequence of a+nd. From this we can see a starting block a then adding another block d. So, we can solve for a starting shape by setting 3x+2y+1=a. Now for the common difference d at first you might also just solve for 3x+2y+1=d but that would be wrong. Why? Because we saw from before a shape with d lines actually contributes d-1 lines to the total count since 1 side overlaps (I specifically made the construction that way!!). So, we actually solve for 3x+2y+1=d+1 or 3x+2y=d.
Now of course this has some limitations.
First of all, I removed loops. What is a loop? For example if you get 4 squares and arrange them such that it makes a bigger square it would be a loop because according to my assumption that z=x+y-1 in that case z=4+0-1 (No triangles in the shape) which gives us 3 but the actual overlap is 4 sides.
And there are some numbers which I can't represent using this construction method. Specifically, a=2 and d=1 any sequence having those cannot be represented. Why? Frobenius Coin Problem. Using 2 and 3 we can represent every integer greater than (2)(3)-2-3=1. Hence, we can't show those numbers.
Well, you might be also wondering wait is a=1 representable? Surprisingly yes! I was also shocked at this the first time. For example 5x+1 here a=1 so we solve 3x+2y=0 meaning x=0,y=0 and we solve 3x+2y=5 or x=1,y=1 the first term is 5(1)+1=6 if we add one triangle and one square in a house shape we get 3+4-1=6 lines. The next term is 5(2)+1=11 if we add another house shape that would be 5 lines (since 1 overlaps) so the next shape would have 11 lines perfectly consistent!
Now what about sequences like 4x-3? To solve this we use an algebra trick 4x-3=4x-4+1=4(x-1)+1 let y=x-1 so 4y+1 we use our method on that sequence but be careful the indexes shifted by 1 so don't freak out if it didn't match one to one you need to account for the shift in index.
