NON-ALTERNATE =b KNOTS. 772 



both even, the form is a link. Hence to have a non-alternate knot, A (or B) must 

 furnish at least five crossings and the remaining three connections at least one each. 

 The knot is at least an eightfold. 



Second, let all six connections exist. If each is a single crossing the form is a link 

 of three threads. If one connection has two crossings, and the other five one each, the 

 form has two threads. In all other cases in this class the form is at least an 

 eightfold. 



Class V. and higher classes do not exist in orders lower than Order Eight. This 

 completes the proof of the theorem that there exists no non-alternate ± knot with 

 fewer than eight crossings. 



7. No reduced, non-alternate knot with three consecutive overs has feiver than eleven 

 crossings. 



Since in classes II., III., and IV., order n, it is always possible to go from any part 

 of the leading partition to any other part of the same partition with not more than two 

 crossings, no sequence of three (three consecutive overs) can exist in these classes. 

 Hence no reduced eightfold knot with a sequence of three can exist. 



All other ninefolds of class V. can be derived from the ninefold 35, below, by 

 erasing crossings at some of the existing connections of the leading partition and 

 adding the crossings erased at other connections. Although it requires 

 three crossings by every path to go from A to E, nevertheless the form as 

 it stands cannot have three consecutive overs, for this would require that 

 we should have three consecutive overs on a closed circuit of five crossings ; 

 but this is impossible, since, in order to have in a reduced knot three con- 

 secutive overs on a closed circuit, the circuit must have at least seven crossings. If 

 (BC), (CD), or (DB) were erased it would be possible to go from A to E with two 

 crossings. If an A (or E) crossing be erased, say (AB), we have a link, and for knot 

 must add the crossing erased to (BC), (EC), (BD), or (ED). But this will leave D and 

 C connected by the 2-gon A and the crossing (DC). By the transposition of these it 

 becomes possible to go from A to E with two crossings. Hence no reduced ninefold 

 knot can have a sequence of three. 



All class V. tenfolds are to be had from the ninefold 35 by the process above 

 described, except that the number of crossings added must exceed by one the number 

 erased. If none be erased and one added, we have either a link or the tenfold 87, 

 which can have no sequence of three, from A to E, except upon a five-crossing circuit. 

 As above, we may not erase (BC), (CD), or (DB). If (BA) be erased and two crossing 

 added in any way to (BC), (EC), (BD), (ED), we again have, after transposition of 

 2-gon A and the crossing (CD), only two portions of the thread between A and E. If 

 one crossing be added to any one of the four mentioned connections and the other to 

 (DA), (CA), or (DC), A can still be brought into the same position with respect to E. 

 If the second crossing be added to (CE) or (DE) there will be two threads. Hence no 

 class V. tenfold can have a three sequence. 



