# Mathematical term for strings with equivalent order of symbols?

What is the mathematical notation or terms for the following relation between two strings

Given a pattern and a string str, find if str follows the same pattern. Here follow means a full match, such that there is a bijection between a letter in pattern and a letter in str.

Example : "aab" follows "bbd". My question is how to say this with formal mathematical notation?

I assume you're working with finite strings and a finite number of possible characters, so I'm giving my answer on this assumption, but I believe all that follows can be extended to at least countably long strings and countably many characters with no problem.

Let $X$ be the set of all finite strings, which consist of characters from the finite set $S$. Let $\mathcal{F}$ be the set of all bijections $f:S \to S$. Define a relation on $X$ such that for two strings of the same length $(x_1x_2\ldots x_n)$ and $(y_1,y_2\ldots y_n)$ we have $(x_1x_2\ldots x_n) \sim (y_1y_2\ldots y_n)$ if and only if $(x_1x_2\ldots x_n) = (f(x_1),f(x_2)\ldots f(x_n))$ for some $f \in \mathcal{F}$.

Now, as a shorthand I will denote the sequence $(x_1x_2\ldots x_n) \in X$ by $x$. And I will denote the sequence $(f(x_1)f(x_2)\ldots f(x_n) \in X$ by $f(x)$.

You can check that the relation defined above on $X$ is an equivalence relation. It is clearly reflexive, which you can see by letting $f$ be the identity. It is symmetric, because if $x = f(y)$ for some $f \in \mathcal{F}$ then we have that $y = f^{-1}(x)$ with $f^{-1}$ also in $\mathcal{F}$. Finally, it is transitive, because if $x=f(y)$ and $y = g(z)$ for some $f,g \in \mathcal{F}$ then we have $x = (f\circ g)(z)$ with $f\circ g \in \mathcal{F}$.

As a concrete example, take $S$ to be the set of letters in the English alphabet. Then $X$ is the set of all finite words created from these characters, and $\mathcal{F}$ is the set of all permutations, or rearrangements of these letters. Then $aab \in X$ and $bbd \in X$. Now, take any permutation in $\mathcal{F}$ which sends $a \to b$ and $b \to d$. Call this permutation $f$. By the definition of the equivalence relation above we have that $aab \sim bbd$ since $aab = f(a)f(a)f(b)$.