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I asked a similar questionquestion a while back. In case H is solvable there is an algorithm (see http://arxiv.org/abs/math/0405122) but the complexity is not clear. If H is nipotent and S is a knot complement then M. Eisermann has shown that the $|Hom(\pi(S),H)|$ is constant (see http://www-fourier.ujf-grenoble.fr/~eiserm/Publications/twistseq.pdf). Agol has pointed out that is should be polynomial if H is dihedral in his answer to my question linked above.

I asked a similar question a while back. In case H is solvable there is an algorithm (see http://arxiv.org/abs/math/0405122) but the complexity is not clear. If H is nipotent and S is a knot complement then M. Eisermann has shown that the $|Hom(\pi(S),H)|$ is constant (see http://www-fourier.ujf-grenoble.fr/~eiserm/Publications/twistseq.pdf). Agol has pointed out that is should be polynomial if H is dihedral in his answer to my question linked above.

I asked a similar question a while back. In case H is solvable there is an algorithm (see http://arxiv.org/abs/math/0405122) but the complexity is not clear. If H is nipotent and S is a knot complement then M. Eisermann has shown that the $|Hom(\pi(S),H)|$ is constant (see http://www-fourier.ujf-grenoble.fr/~eiserm/Publications/twistseq.pdf). Agol has pointed out that is should be polynomial if H is dihedral in his answer to my question linked above.

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Eric Rowell
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I asked a similar question a while back. In case H is solvable there is an algorithm (see http://arxiv.org/abs/math/0405122) but the complexity is not clear. If H is nipotent and S is a knot complement then M. Eisermann has shown that the $|Hom(\pi(S),H)|$ is constant (see http://www-fourier.ujf-grenoble.fr/~eiserm/Publications/twistseq.pdf). Agol has pointed out that is should be polynomial if H is dihedral in his answer to my question linked above.