Ordering: Difference between revisions
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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>. | ||
Thus the number of elements in columns <math>0, ..., x-1</math> is <math>(x)(x+1)/2</math>. | Thus the number of elements in columns <math>0, ..., x-1</math> is <math>(x)(x+1)/2</math>. | ||
Thus our formula is <math>z = \frac{x(x+1)}{2} + y</math> | Thus our formula is | ||
<math display="block">z = \frac{x(x+1)}{2} + y</math> | |||
To calculate the inverse formula: | To calculate the inverse formula: |
Revision as of 18:26, 26 May 2020
Related to counting, countabiliy
Diagonal
The goal is to get formulas for the following:
Figure 1: \(\displaystyle \begin{bmatrix} 0 & 2& 5& 9&\\ 1 & 4& 8& &\\ 3 & 7& & &\\ 6 & & & &\\ \end{bmatrix} \)
This is a bijection to Figure 2 where y is shifted by x and the matrix is flipped upside down:
Figure 2: \(\displaystyle \begin{bmatrix} & & & &\\ & & 5& &\\ & 2& 4& &\\ 0 & 1 & 3& ... &\\ \end{bmatrix} \)
The is a 1-1 mapping \(\displaystyle \mathbb{Z}^2 \to \mathbb{Z}\).
We first derive the function from \(\displaystyle \mathbb{Z}^2\) to \(\displaystyle \mathbb{z}\) as shown in Figure 2.
Let \(\displaystyle (x,y)\) be the coordinates in \(\displaystyle \{(0,0), (1, 0), (1,1), ...\}\) which will map to \(\displaystyle \({0, 1, 2, ...\}\).
First note that \(\displaystyle \sum_{0}^{k}i = \frac{(k)(k+1)}{2}\).
Thus the number of elements in columns \(\displaystyle 0, ..., x-1\) is \(\displaystyle (x)(x+1)/2\).
Thus our formula is
\[z = \frac{x(x+1)}{2} + y\]
To calculate the inverse formula:
Given an integer \(\displaystyle z\), we want to find \(\displaystyle (x, y)\)
The formula is figure 1 is as follows: