Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the following paper, just in the first paragraph the author defines what $l$-equivalence for two m-tuples $\in [0,l]^m$ means. Can somebody please give me a more precise definition of what he means? I am not even sure what $[0,l]^m$ stands for, although I expect its just $\{0,1,\dots,l\}^m$.


enter image description here

share|cite|improve this question
I think it just means that (a,b,x) and (a,b,y) are 2-equivalent. I'vve not been able to understand the proof yet. – user50336 Nov 27 '12 at 18:12

According to the first sentence of the paper, $[0,\ell]$ is indeed $\{0,1,\dots,\ell\}$, and it’s clear that $[0,\ell]^m$ is, as usual, the set of $m$-tuples of elements of $[0,\ell]$. Suppose that $\langle x_1,\dots,x_m\rangle$ and $\langle x_1',\dots,x_m'\rangle$ are $m$-tuples in $[0,\ell]^m$. Suppose first that $\ell$ occurs at least once in each of these $m$-tuples. Let $k$ be the largest index such that $x_k=\ell$; $x_k$ is the last occurrence of $\ell$ in $\langle x_1,\dots,x_m\rangle$. Similarly, let $x_{k'}'$ be the last occurrence of $\ell$ in $\langle x_1',\dots,x_m'\rangle$. Then $\langle x_1,\dots,x_m\rangle$ and $\langle x_1',\dots,x_m'\rangle$ are $\ell$-equivalent iff $k=k'$, and $x_i=x_i'$ for $i=1,\dots,k$. If neither $\langle x_1,\dots,x_m\rangle$ nor $\langle x_1',\dots,x_m'\rangle$ contains an $\ell$, the agreement requirement is vacuous, so they are automatically $\ell$-equivalent.

For example, the sequences $\langle 2,1,2,0,1\rangle$ and $\langle 2,1,2,1,0\rangle$ in $[0,2]^5$ are $2$-equivalent, and the $2$-equivalence class of these two sequences also contains $\langle 2,1,2,0,0\rangle$ and $\langle 2,1,2,1,1\rangle$; it is in fact

$$\Big\{\langle 2,1,2,a,b\rangle:\langle a,b\rangle\in[0,1]^2\Big\}\;.$$

The $2$-equivalence class of $\langle 1,1,0,1,0\rangle\in[0,2]^5$, which doesn’t contain $2$ at all, is $[0,1]^5$: the $5$-tuples that don’t contain a $2$ are the ones that agree with $\langle 1,1,0,1,0\rangle$ up through the last occurrence of $2$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.