Define an algebraic structure $A_{n}$ by letting $$A_{n}=(\{1,\dots,2^{n}-1,2^{n}\},*_{n})$$ where $*_{n}$ is the unique operation such that $x*_{n}1=x+1\mod 2^{n}$ for $$x\in\{1,\dots,2^{n}-1,2^{n}\}$$ and where $$x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)$$ for all $x,y,z$. Then the algebra $A_{n}$ is called the $n$-th classical Laver table (or simply called a Laver table as they were called back in the early 90's.).
Let $\mathcal{A}$ be the collection of all isomorphism classes of quotients of subalgebras of the classical Laver tables $(A_{n})_{n\in\omega}$.
If $X\in\mathcal{A}$, then let $I_{X}$ denote the unique element such that $I_{X}*x=x,x*I_{X}=I_{X}$ for $x\in X$. We shall call an algebra $X\in\mathcal{A}$ critically simple if whenever $\simeq$ is a non-identity congruence on $X$ there is some $x\in X\setminus\{1_{X}\}$ with $x\simeq 1_{X}$.
Suppose $\lambda$ is an ordinal and $\phi:\mathcal{A}\rightarrow\lambda$ is a surjective function. Then we shall call $\phi$ a strength calibration function if it satisfies the following properties for $X,Y\in\mathcal{A}$:
If $X$ is a subalgebra or a quotient of $Y$, then $\phi(X)\leq\phi(Y)$.
If $f:X\rightarrow Y$ is a homomorphism such that if $f(x)=1_{Y}\rightarrow x=1_{X}$, then $\phi(X)\leq\phi(Y)$.
If the function $\phi$ is a strength calibration function, we shall call $\phi$ efficient if the problem of determining whether $\phi(X)<\phi(Y)$ from the multiplication tables of $X$ and $Y$ is solvable in polynomial time (I mean polynomial in terms of $|X|+|Y|$). For which ordinals $\lambda$ does there exist an efficient strength calibration function $\phi:\mathcal{A}\rightarrow\lambda?$
