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Think of a starting cube and the cube in the plane above where Bishop one-steps through an edge. Then think of the same starting cube and the cube in the plane above where Unicorn one-steps through a vertex. Choose the ones above so they have faces adjacent. The distance travelled center to center in the first case is root-2 and second case root-3. Now think of the position of the first destination cube sliding gradually to that of the adjacent second (Unicorn) cube, one cube sliding into the other of the same plane. Keep an imaginary line center to center from the below departure cube, and the line segment increases from position to position. Then every value for distance travelled between 1.414 and 1.732 is assumed along the way. Every possible value along the infinite Cantorian continuum, until we reach root-3. And at that position the cube is locked into where Einhorn one-steps.