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(Just to be precise, in this question, the word "knot" means "ambient isotopy class of a (EDIT: smooth) knot in $S^3$.") A knot in $S^3$ is called prime if it is not the connected sum of two other non-trivial knots in $S^3$. Clearly any knot is a sum of prime knots, and it is a theorem that this decomposition is unique. One downside of this is that there are infinitely many prime knots (for example, all non-trivial torus knots are prime). Here's a vague version of the question - Is there a way to trade off the uniqueness result (and add finitely many operations) in exchange for starting with a finite list of knots? To try to make my question a little more precise, I'll define an "operation" as a function that takes a list of knots as input and outputs a finite list of knots. I'm not sure how to say this well, but I'd like to avoid very dumb operations like "fix an ordered list of all knots $K_1,K_2,\ldots$ and if $K_i$ is input, output $K_{i+1}$." However, I am interested in dumb answers, just not as dumb as that :-)
For example, the first paragraph, phrased in this language, is
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Is there a procedure for obtaining all knots in S^3?(Just to be precise, in this question, the word "knot" means "ambient isotopy class of a knot in $S^3$.") A knot in $S^3$ is called prime if it is not the connected sum of two other non-trivial knots in $S^3$. Clearly any knot is a sum of prime knots, and it is a theorem that this decomposition is unique. One downside of this is that there are infinitely many prime knots (for example, all non-trivial torus knots are prime). Here's a vague version of the question - Is there a way to trade off the uniqueness result (and add finitely many operations) in exchange for starting with a finite list of knots? To try to make my question a little more precise, I'll define an "operation" as a function that takes a list of knots as input and outputs a finite list of knots. I'm not sure how to say this well, but I'd like to avoid very dumb operations like "fix an ordered list of all knots $K_1,K_2,\ldots$ and if $K_i$ is input, output $K_{i+1}$." However, I am interested in dumb answers, just not as dumb as that :-)
For example, the first paragraph, phrased in this language, is
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