Solving this system of linear equation.

In summary, the problem is with aligning the tile numbers with the white lines in the isometric game. The goal is to determine the column and row of a given tile based on its index, and vice versa. The formula for getting the column and row of a tile index is x = Index%width and y = floor(Index/width), while the formula for getting the tile index from a column and row is x+(height/2-((height&1)^1))-(y/2) for the column and x+(y+1)/2 for the row. However, these formulas do not work and further calculations on Wolfram Alpha are needed to get the correct result.
  • #1
slicer4ever
5
0
hey good people, I'm having a problem with how to solve a problem I'm having.

first of all, I am working on an isometric game, and align my tile's are layed out like so:

image.png


so, here's the problem:
the numbers are arranged in how the tile's are layed out in memory, however, i want to be able to get a patch of tile's relative to how the white lines are layed out.

for example: column 1 is 6, 9, and 13. 2 is 3, 7, and 10, and 3 is 1, 4, and 8.
so my input would be something like this: the bottom tile's index in the world(in this case that is tile #6), and then i'd specify which tile i was last on, and finally i'd pass the column and row size to get the chunk of.

the goal is to quickly determine which column/row the last tile is on, increment first by column, then by row. so i'd iterate over tiles 6, 9, 13, 3, 7, 10, 1, 4, and 8.

so, I've figured out how to specify a tile index, and get it's column/row with the following formula:

Code:
//above example, the width and height can be taken as 3, and 5.
//note all divisions are integers, and round down.
void Get(int Index){
  int x = Index%width;
  int y = floor(Index/width);
  int column = x+floor(height/2-((height&1)^1))-floor(y/2);
  int row = x+floor((y+1)/2);
}

however, i also need to be able to input the column/row and get the tile index(or at least the x/y).

this is the math I've tried to do, but it's obviously not correct:

Code:
int x = Index%width;
int y = Index/width;
int column = x+(height/2-((height&1)^1))-(y/2);
int row = x+(y+1)/2;


width = 3, height = 5 = (5/2 = 2 as int)
Index(x, y) = column, row

Index: 0 (0, 0) = 2, 0
c = x+2-y/2;
 2 = x+2-y/2
-2 =  -2
 0 = x-y/2
+y/2 =+y/2
y/2 = x
*2   = *2
 y = 2x 

r = x+(y+1)/2
0 = x+(2x+1)/2
0  = x+x ?//not sure if my math here is correct, as it's integer math(or you can think of it as all divisions are rounded down), the 1/2 would round out right?
0 = 2x
/2  = /2
0 = x
y = 2x = 2(0) = 0

Index: 3(0, 1) = 2, 1
c = x+2-y/2
 2 = x+2-y/2
-2 =  -2
 0 = x-y/2
+y/2 = +y/2
y/2 = x
*2  = *2
y = 2x

r = x+(y+1)/2
1 = x+(2x+1)/2
1 = x+x ?
1 = 2x
/2 = /2
0 = x
y = 2x = 2(0) = 0 != 1

as you can see that doesn't work. however, after inputing the equations into wolfam:
http://www.wolframalpha.com/input/?i=%28x%2Bfloor%28h%2F2%29-floor%28y%2F2%29+%3D+c%2C+x%2Bfloor%28%28y%2B1%29%2F2%29+%3D+r%29+where+x%3D+0+and+y+%3D+2+and+h%3D5

i take the results of that, and plug in c and r into the equation to get the x/y:

http://www.wolframalpha.com/input/?i=%28x%2Bfloor%28h%2F2%29-floor%28y%2F2%29+%3D+c%2C+x%2Bfloor%28%28y%2B1%29%2F2%29+%3D+r%29+where+c%3D+1+and+r+%3D+1+and+h%3D5

wolfam is able to get the correct result in the integer solution panel, but i don't know how it got that result.
 
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  • #2
i dislike having to bump, but i really could use some help with this problem.
 

1. How do I solve a system of linear equations?

To solve a system of linear equations, you need to use algebraic techniques such as substitution, elimination, or graphing. First, identify the variables in the equations and the number of equations in the system. Then, choose a method to solve the system based on the given information.

2. What is the purpose of solving a system of linear equations?

Solving a system of linear equations allows you to find the values of the variables that satisfy all of the equations in the system. This can help you find the intersection point of two lines or the common solution to multiple equations.

3. Can I use a calculator to solve a system of linear equations?

Yes, most scientific or graphing calculators have built-in functions to solve systems of linear equations. However, it is important to understand the algebraic concepts behind the solution method used by the calculator.

4. What are some real-life applications of solving systems of linear equations?

Solving systems of linear equations can be used in various fields such as engineering, economics, and physics. For example, it can help determine the optimal production levels in a factory or the best route for a delivery truck.

5. Are there any limitations to solving systems of linear equations?

Yes, there are some cases where a system of linear equations may not have a unique solution or may not have a solution at all. This can occur when there are too many or too few equations, or when the equations are not independent of each other.

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