Let $K$ be a knot diagram in the plane. Let $C$ be a coloring for $K$ by a quandle $X$. Recall that the relation $x*y =z$ holds at each crossing where $x,z$ are colors of the under arcs , $y$ is the color of the over arc with orientation normal directed tawards the under arc colored by $z$. My question is : Suppose a knot $K$ can be colored by a quandle $X$. Suppose I start from any crossing and color the over arc by an arbitrary color in $X$ and the under arc by also an arbitrary color, and then apply the relation to get the color of the other under arc at the chosen crossing, then we trace the knot diagram and at each crossing we apply the quandle relation, if I do this can I guarantee that i’ll obtain a valid coloring by $X$. If so then we can obtain all possible colorings of the knot by a given quandle by this way, right?
2026-02-22 21:31:44.1771795904
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How to obtain all possible colorings of a knot diagram by a given quandle
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@KyleMiller's answer points out that choosing the color of two strands at a crossing might over-determine the coloring. Conversely, it can happen that such a choice under-determines the coloring. That is to say, there are knots such as $9_{35}$ (not sure that this is the example with fewest crossings) with the property that choosing colors for any two strands will not uniquely determine the colors of all the other strands. This is just to say that there are knots whose fundamental quandle cannot be generated by just two elements.
No. A simple counterexample is a crossing from a Reidemeister I twist. If you color the overstrand by anything other than the current color of the understrand, you will immediately get a contradictory coloring. This contradiction can be deferred arbitrarily far out in time by using Reidemeister moves to put arbitrarily many crossings on the twist.
You can reduce the number of choices by putting a knot into bridge position, coloring the bottom set of bridges arbitrarily, then checking that the top bridges are colored consistently. For a dihedral coloring, this is a system of $b$ unknowns and $b$ equations (with $b$ the bridge number). There is no reason to believe that this system of equations will always have either rank $0$ or rank $b$, which is what your conjecture would imply.