#awk - big This didn't turn out as short as I had hoped, but I took a slightly different approach to dealing with the graph, so I'm posting the ungolfed version anyway.

I took advantage of the fact that an icosahedron (20 sided polyhedron) under rotations preserving orientation is isomorphic to the alternating group of degree 5 (5 element permutations having an even number of even length cycles). I then choose two permutations with cycle length 5 as "left" and "right", and I choose one permutation with cycle length 2 as "back". Using these, I build up the graph from one room by walking the Hamiltonian path (2xRRRLLLRLRL, using 3xRB in each room to capture the 3 possible directions).

function meta(z,a,b,c,d) { if(z==COMPOSE) { split(a,c,""); split(b,d,""); return c[d[1]]c[d[2]]c[d[3]]c[d[4]]c[d[5]]; } if(z==WALK) { split(R" "R" "R" "L" "L" "L" "R" "L" "R" "L,c); for(b = 1; b <= 20; b++) { map[a] = b; a = meta(COMPOSE,meta(COMPOSE,a,R),B); map[a] = b; a = meta(COMPOSE,meta(COMPOSE,a,R),B); map[a] = b; a = meta(COMPOSE,meta(COMPOSE,a,R),B); a = meta(COMPOSE, a, c[b % 10 + 1]); } } if(z==TEST) { a = map[meta(COMPOSE,U,L)]; b = map[meta(COMPOSE,U,R)]; c = map[meta(COMPOSE,U,B)]; if(a==W||b==W||c==W) print "You smell the wumpus"; if(a==P||b==P||c==P) print "You feel a breeze"; if(map[U]==W) { print "You have been eaten by the wumpus"; exit; } if(map[U]==P) { print "You have fallen into a bottomless pit"; exit; } } if(z==ARROWTEST) { if(A==W) { print "You have slain the wumpus!"; exit; } else { for(a in p) if(map[a]==W) break; W=map[meta(COMPOSE,a,v[int(rand()*3)+1])]; } } } BEGIN { COMPOSE = 0; WALK = 1; TEST = 2; ARROWTEST = 3; L = 35214; R = 35421; B = 35142; split(R" "L" "B,V); meta(WALK,L); W = int(rand()*19)+2; P = int(rand()*19)+2; U = L; meta(TEST); } { d=int($0/10); m=$0%10; if(m==5) U = meta(COMPOSE,U,V[d]); else if(m==4) { A = map[meta(COMPOSE,U,V[d])]; meta(ARROWTEST); } meta(TEST); }