5,337
edits
Ordering: Difference between revisions
→Diagonal
| Line 28: | Line 28: | ||
The is a 1-1 mapping <math>\mathbb{Z}^2 \to \mathbb{Z}</math>. | The is a 1-1 mapping <math>\mathbb{Z}^2 \to \mathbb{Z}</math>. | ||
We first derive the function from <math>\mathbb{Z}^2</math> to <math>\mathbb{ | We first derive the function from <math>(x,y) \in \mathbb{Z}^2</math> to <math>z \in \mathbb{Z}</math> as shown in Figure 2. | ||
Let <math>(x,y)</math> be the coordinates in <math>\{(0,0), (1, 0), (1,1), ...\}</math> which will map to <math>\({0, 1, 2, ...\}</math>. | Let <math>(x,y)</math> be the coordinates in <math>\{(0,0), (1, 0), (1,1), ...\}</math> which will map to <math>\({0, 1, 2, ...\}</math>. | ||
First note that <math>\sum_{0}^{k}i = \frac{(k)(k+1)}{2}</math>. | First note that <math>\sum_{0}^{k}i = \frac{(k)(k+1)}{2}</math>. | ||
| Line 40: | Line 40: | ||
To calculate the inverse formula: | To calculate the inverse formula: | ||
Given an integer <math>z</math>, we want to find <math>(x, y)</math> | Given an integer <math>z</math>, we want to find <math>(x, y)</math>. | ||
Inverting <math>\frac{x(x+1)}{2}</math>, we get: | |||
<math display="block"> | |||
\begin{align} | |||
x &= \left\lfloor \frac{-1 + \sqrt{1+8z}}{2} \right\rfloor\\ | |||
y &= z - \frac{x(x+1)}{2} | |||
\end{align} | |||
</math> | |||
The formula is figure 1 is as follows: | The formula is figure 1 is as follows: | ||