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edited the code, so that the indentation pastes better into snappy.

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edited Mar 22, 2015 at 23:49

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Mark Bell has a wonderful answer for the knot tables. I thought I would provide an answer for the census data. Using SnapPy, the current version of the OrientableCuspedCensus, which is a combination of the Callahan-Hildebrand-Weeks census, Morwen Thistlethwaite's 8-tetrahedral census and Ben Burton's 9 tetrahedral census, appears to have 77 2-bridge knots according the following computation.

 CC = OrientableCuspedCensus(num_cusps=1)
 knotcount = 0
 for C in CC:
    if C.is_two_bridge():
            print C.identify(), C.is_two_bridge(), C.volume()
            knotcount = knotcount+1
  print knotcount

This function runs in the SnapPy application. Note: printing C.is_two_bridge() identifies the two bridge knot. Note that copying the code and then typing '%paste' on the command line is the easiest way to run the code yourself.

 CC = OrientableCuspedCensus(num_cusps=1)
 knotcount = 0
 for C in CC:
    if C.is_two_bridge():
            print C.identify(), C.is_two_bridge(), C.volume()
            knotcount = knotcount+1
  print knotcount

This function runs in the SnapPy application. Note: printing C.is_two_bridge() identifies the two bridge knot.

 CC = OrientableCuspedCensus(num_cusps=1)
 knotcount = 0
 for C in CC:
    if C.is_two_bridge():
            print C.identify(), C.is_two_bridge(), C.volume()
            knotcount = knotcount+1
 print knotcount

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created Mar 14, 2015 at 2:34

Neil Hoffman

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1
23
48

 CC = OrientableCuspedCensus(num_cusps=1)
 knotcount = 0
 for C in CC:
    if C.is_two_bridge():
            print C.identify(), C.is_two_bridge(), C.volume()
            knotcount = knotcount+1
  print knotcount

This function runs in the SnapPy application. Note: printing C.is_two_bridge() identifies the two bridge knot.