Onye's Algorithm for determining the mod z colorability of a pretzel knot!!!!!!!!!!!!

[onye]

This is an algorithm that I discovered myself. I don't know if
somebody else had discovered this previously. But the algorithm holds
true generally.



For any pretzel knot (p,q,r), alternating or non alternating, such
that the relationship pq+qr+pr=k holds, the pretzel knot (p,q,r) can
be said to be mod z colorable iff k is a multiple of z.



For example, the pretzel knots (8,-7,5) and (7, 7, -2) are both tri-
colorable. This is because ((-7*8)+(-7*5)+(8*5)=-51) and ((-2*7)+
(-2*7)+(7*7)=21). Since -51 and 21 are both multiples of 3, we may
conclude that the two pretzel knots are tri-colorable. By the same
reasoning, the first knot is mod 17 colorable while the second pretzel
is not. On the other hand, the second pretzel knot is mod 7 colorable
while the first pretzel knot is not.



Watch out for the proof in my next posting.

[marg...@math.utah.edu]

Hi Onye,

Are you sure about the (8,-7,5) example? How can we check this is
right?

Dan