Minimality conjecture for 2-bridge knots with four boundary slopes

Let $K$ be a 2-bridge knot, and suppose that its boundary-slope set has exactly four distinct elements. A 2-bridge knot $K_1$ is minimal with respect to the Silver–Whitten partial ordering if $K_1\geq K_2$ implies that $K_1$ and $K_2$ represent the same knot in the ordering. Four-slope minimality conjecture. A 2-bridge knot with exactly four distinct boundary slopes is minimal with respect to the Silver–Whitten partial ordering. The paper explains that boundary-slope arguments eliminate most cases, but several cases require additional information about Alexander polynomials or character varieties and remain unresolved.