152 : PROFESSOR TAIT ON KNOTS. — 
Now write the same equivalents for the same letters in the odd places, and ae 
scheme in its new lettering is 
AP CAREY en oak) sA 
or the following are equivalents in the even places 
DAFBCE 
DFEBAG, 
and each of these has, of course, five other equivalents found by the first of 
these two processes. 
But we may also start from the same intersection A by either of these paths, 
but in the reverse direction round the curve. To effect this we have only to 
read the scheme backwards, beginning at either A, and changing the lettering 
throughout in accordance with our plan. Thus, taking the last example, 
ADBACFDBECFE|A 
| es Dm ae Oa Se he 
we keep the terminal A unchanged, and write B, C, &c., for the 2d, 4th; &c., 
preceding letters. We have thus, as it were, the key for translating from the 
upper line to the lower. Apply this key to all the letters, and then write the’ 
result in the reverse order. Thus we get 
ACBECFDBEAFD|A 
This new scheme has for its even places | 
CHEE BAD 
which is equivalent (in this particular case) to ae second of the two direct 
schemes just given, viz.:— 
DFEBAC. 
Finally, if we read this reversed scheme from the A in the even places, its 
even letters become 
EAFCBD 
which (in this case) is the same as 
DAFBCE 
the even letters of the original scheme. 
The notation we shall employ is this—do, de, ro, re, signifying the even 
places of the four cases 
d o the direct scheme, read from A in the odd place 
d e the direct scheme, read from A in the even place 
r o the reversed scheme, read from A in the odd place 
r e the reversed scheme, read from A in the even place 

