Alternating diagram of knots

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Is it true that every tame knot has at least an alternating diagram?

If yes, is it true that we can always obtain an alternating diagram by a finite number of Reidemeister moves from a diagram of a knot?

If yes, how can we do it?

I am reading GTM Introduction to Knot Theory and find they sort of assume this, which makes me think it should be evident but I cannot figure out.

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The former is true and is essentially a "checkerboard colouring" argument -- colour all the regions black and white, no two of the same colour sharing a border ( (ab)use Jordan curve theorem if you want to prove this ). Then as you travel towards a crossing, if there's black on your right as you approach, go under; otherwise, over.

The second is false according to mathworld, wolfram; this is a proof by counterexample (note that because that particular knot is non-alternating, you cannot make an alternating knot diagram by using Reidemeister moves as the positions you can get with Reidemeister moves are precisely the projections of the knot that you can get).

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A knot is called alternating if it has an alternating knot diagram. If there is a sequence of Reidemeister moves on a diagram for a knot that results in an alternating diagram, then the knot is an alternating knot. Because Reidemeister moves are a complete set of moves, given a diagram for an alternating knot, there is some sequence of Reidemeister moves that will give an alternating diagram.

Alternating knots are fairly special. If $K$ is a prime alternating knot that is not a torus knot, then $S^3-K$ can be given a complete hyperbolic metric of constant negative curvature. Using the geometrization theorem, it follows that any non-trivial satellite operation gives a non-alternating knot.

If the knot is prime and alternating, then every minimal-crossing-number diagram is an alternating diagram (see Alternating and Non-Altenating Knot projections with same crossing number?). It's not clear to me if there exists a bounded-time algorithm that can actually find such a sequence of Reidemeister moves, however! (There is always exhaustively trying all sequences of Reidemeister moves, but this is a priori an unbounded-time algorithm.)