If $X$ is a self-distributive algebra, then define $x^{[n]}$ for all $n\geq 1$ by letting $x^{[1]}=x$ and $x^{[n+1]}=x\*x^{[n]}$. The motivation for this question comes from the following fact about self-distributivity on one generator.



> 
> **Theorem:** Suppose that $X$ is a self-distributive algebra generated by one element. Then for all $x,y\in X$, there are $m,n$
>  with $x^{[m]}=y^{[n]}$.
> 
> 
> 


Define the Fibonacci terms by letting $t\_{1}(x,y)=y,t\_{2}(x,y)=x,t\_{n+2}(x,y)=t\_{n+1}(x,y)\*t\_{n}(x,y)$.


An algebra $(X,\*,1)$ is said to be a reduced permutative algebra if $x\*(y\*z)=(x\*y)\*(x\*z),x\*1=1,1\*x=x$ for all $x,y,z\in X$ and where for all $x,y\in X$, there is some $n$ where $t\_{n}(x,y)=1$. We say that a reduced permutative algebra $(X,\*,1)$ is critically simple if whenever $\simeq$ is a congruence, either $\simeq$ is the trivial congruence or $x\simeq 1$ for some $x\neq 1$.


Suppose that $(X,\*,1)$ is a finite reduced permutative algebra and $x,y\in X$ and suppose that $p\geq 0$. Then does there a reduced permutative algebra $(Y,\*,1)$, natural numbers $m,n$ and a homomorphism $\phi:Y\rightarrow X$ and $x',y'\in Y$ where $\phi(y')=y,\phi(x')=x$ and $x'^{[m]}=y'^{[n]}$ and where
$x'^{[m+p]}\neq 1$? What if $(X,\*,1)$ is critically simple?



Attraction in Laver tables
I have a method of generating counterexamples even if $(X,\*,1)$ is critically simple. In this post, all algebras will be assumed to be finite reduced permutative
self-distributive algebras. To do this, we provide an example of an algebra algebra $X$ generated by $x,y$ where there does not exist a algebra $Y$ generated by $x\_1,y\_1$ and a surjective homomorphism $\phi:Y\rightarrow X$ with $\phi(x\_1)=x,\phi(y\_1)=y$ and where $Y$ has more critical points than $X$.


The examples of the algebras $X$ are quite rare, and I currently only know of one technique for producing such an algebra. Let $X\_1$ be the algebra with only one element which we shall denote by $x\_1$ and also by $y\_1$. We then by induction build a sequence of critically simple algebra $(X\_1,\dots,X\_n)$ where $|\text{crit}[X\_j]|=j$ for $1\leq j\leq n$ and where $X\_j$ is generated by 2 elements $x\_j,y\_j$ where $y\_j\*y\_j=1$ and where there is a surjective homomorphism $\phi\_j:X\_j\rightarrow X\_{j-1}$ with $\phi(x\_j)=x\_{j-1},\phi(y\_j)=y\_{j-1}$, and we quite the induction process when we cannot extend the sequence $(X\_1,\dots,X\_n)$ any further in the sense that there does not exist an algebra $X\_{n+1}$ with $\text{crit}[X\_{n+1}]=n+1$ and a surjective homomorphism $\phi\_{n+1}:X\_{n+1}\rightarrow X\_n$ along with generators $x\_{n+1},y\_{n+1}$ such that $\phi\_{n+1}(x\_{n+1})=x\_n,\phi\_{n+1}(y\_{n+1})=y\_n$ and $y\_{n+1}\*y\_{n+1}=1$.
After we recursively build our sequence of critically simple algebras $(X\_1,\dots,X\_n)$, we extend the sequence some more to obtain more critically simple algebras $(X\_n,\dots,X\_m)$ and homomorphisms
$\phi\_j:X\_j\rightarrow X\_{j-1}$ where $|\text{crit}[X\_j]|=j$ and $\phi\_j(x\_j)=x\_{j-1},\phi\_j(y\_j)=y\_j$ and where $X\_j$ is generated by $x\_j,y\_j$ until we cannot extend $X\_m$ any more to an algebra $X\_{m+1}$ generated by $x\_{m+1},y\_{m+1}$ with $|\text{crit}[X\_{m+1}]|=m+1$ and $\phi\_{m+1}(x\_{m+1})=x\_m,\phi\_{m+1}(y\_{m+1})=y\_m$.


In this case, $Y\_m$ with the generators $x\_m,y\_m$ is our counterexample.


To go from $X\_j$ to $X\_{j+1}$, we use my (possibly inefficient) backtracking algorithm that takes a critically simple algebra $X$ and generating set $(x\_a)\_{a\in A}$ and returns all possible (up-to generator preserving isomorphism) critically simple algebras $Y$ with generating sets $(y\_a)\_{a\in A}$ and homomorphisms $\phi:Y\rightarrow X$ with
$\phi(y\_a)=x\_a$ for $a\in A$ and where $|\text{crit}[Y]|=|\text{crit}[X]|+1$.


For critically simple algebras $X$ with multiple generators $(x\_a)\_{a\in A}$, I have no algorithm or theory other than my exhaustive search that finds all critically simple algebras $Y$ generated by $(y\_a)\_{a\in A}$ along with a homomorphism
$\phi:Y\rightarrow X$ with $\phi(y\_a)=x\_a$ for $a\in A$, and I have no way of knowing what critically simple algebras $Y$ might look like, so I have no way of obtaining a different kind of counterexample.


