Nautile aka Charles Hamel's personal pages  A SHORTCUT TO WRITING THE COMPLEMENTARY / CYCLIC SEQUENCE
Please be aware that, despite the attention given to proofing and proofing again ,there may persist some
mistakes ( "may" that is  like when a bookseller state " books may have reminder marks"  : you can be
dead certain the book has them  ! )

Get the complementary

( ← )   means from RIGHT to LEFT
( → )   means from LEFT to RIGHT

0    2    4    1    3
#    #   #     #    #

Write the top line relating to the bottom rim--------------------------------------------
0        2        4        1        3        0        2       4       1       3        0       2       4       1       3

BOTTOM rim sequence written ( ← )  TO BE READ  ( → )
0        2        4        1        3        0        2      x
@      @      @      @      @      @      @      @

Reverse the top line to write the bottom line ( relating to top rim-------------------

BOTTOM rim sequence written ( ← )  TO BE READ  ( → )
0        2        4        1        3        0        2       x
@      @      @      @      @      @      @      @
x       2        0        3        1        4       2        0
TOP rim sequence written ( → )  TO BE READ  READ  (←)

BOTTOM rim sequence written ( ← )  TO BE READ  ( → )
0        2        4        1        3        0        2      x     this is relating to the ODD    B.R.T.L
@      /          \         /         \        /        \        @
x       2        0        3        1        4       2        0    this is relating to the EVEN  T.L.B.R
TOP rim sequence written ( → )  TO BE READ  READ  (←)

------------------------------------
BOTTOM rim sequence written ( ← )  TO BE READ  ( → )
U        O      U       O       U        O       U
0        2        4        1        3        0        2       x    this is relating to the ODD    B.R.T.L
@      /          \         /         \        /        \        @
x       2        0        3        1        4       2        0    this is relating to the EVEN T.L.B.R
U      O        U       O      U       O       U
TOP rim sequence written ( → )  TO BE READ  READ  (←)
------------------------------------

This will be reduced to

------------------------------------
BOTTOM rim sequence written ( ← )  TO BE READ  ( → )
0        1       2        3       4         5        6       7    number of CROSSING ROW on left
none  U      O        U       O        U       O   none
0        2        4        1        3        0        2     x      this is relating to the ODD    B.R.T.L
@        \         /         \        /        \        /      @
x       2        0        3        1        4       2        0    this is relating to the EVEN  T.L.B.R
none   O       U       O       U       O      U     none
7         6        5       4        3        2       1       0     number of CROSSING ROW on right
TOP rim sequence written ( → )  TO BE READ  READ  (←)
------------------------------------

the 2 extreme 0-@  pair being just there as place-holding items ( just as digit 0 is just a place
holding digit giving meaning to the other digits 1 to 9 in numbers ! ) A SHORTCUT TO WRITING THE COMPLEMENTARY / CYCLIC SEQUENCE

this will add a short cut to the first phase of above

We know how to compute the place away from 0 of a given number.
But that give them in disarray.
It would be nice to get the first after 0 then the second, then the third...

The value of the increment span  ( modular) between 2 successive numbers must be an integer.

This integer must be when multiplied by (-L) mod B  [or of (L)mod B in the other case] an
integer itself.

IS for increment span between 2 consecutive ( spatially ) numbers

IS = f(B)  / (-L)mod B   - for the complementary ( for the periodic = f(B)  / (+L)mod B )

or rather
IS =  f(B) / ((B-L ) modulus B)) for the complementary  ( = f(B) / (B - ((B-L ) modulus B))
for the periodic )

we must use (-L)mod B   or  ((B-L ) modulus B))  as "great modular step"

even / even leads to integer
odd / odd may lead to integer
even / odd may lead to integer

odd / even does not lead to integer so to cover that in case B is ODD by precaution we
add 1 and we will eventually "tweak" the formula for correction

so it become  IS = f(B) + 1 / (-L) mod B

so trying IS = (B +1)  / (-L) mod B  is not satisfying if  (B +1) < (-L) mod B

so to ensure that (B+1) will always be ( in the most economical fashion ) > (-L) mod B
we need a multiplying factor for B that must be kept at it barest minimum (Occam' razor )

IS = ((mf * B ) + 1 ) /  (-L)mod B
for the complementary

and for the periodic
IS = ((mf * B ) + 1 ) /  (+L)mod B

so to disambiguate

IScomp = ((mf * B ) + 1 ) /  (-L)mod B
ISper     = ((mf * B ) + 1 ) /  (+L)mod B

OR my preferred

IScomp =  ((mf * B ) + 1 ) /   ((B-L ) modulus B))
ISper     =  ((mf * B ) + 1 ) /  (B - ((B-L ) modulus B))
(mf) must be an integer, the smallest making IS an integer step leading to a closed
path , in other word a circuit.

L = 5  B = 4 as in HALL p 18

IScomp = ((mf * 4 ) + 1 ) /   ((4-5 ) modulus 4)) =  ((mf * 5) + 1 ) / 3  = 2
ISper     =  ((mf * 4 ) + 1 ) /  (4 - ((4-5 ) modulus 4 ))  = ((mf * 5) + 1 )/ 1 = 3

IScomp = (mf * 4) + 1 / (-5) mod 4 = (mf * 4) + 1 / 3 = 2
ISper = (mf * 4) + 1 / (+5) mod  = (mf * 4) + 1 / 2 = 3

mf= 1    mf=1
IScomp = 2
ISper = 3

L = 7  B = 5

IScomp = ((mf * 5 ) + 1 ) /   ((5-7 ) modulus 5)) =  ((mf * 5) + 1 ) / 3  = 2
ISper     =  ((mf * 5 ) + 1 ) /  (5 - ((5-7 ) modulus 5))  = ((mf * 5) + 1 )/ 2 = 3

IScomp = (mf * 5) + 1 / (-7) mod 5 = (mf * 5) + 1 / 3 = 2
ISper = (mf * 5) + 1 / (+7) mod 5 = (mf * 5) + 1 / 2 = 3
this is nice as 5 + 1 = a multiple of 3  so mf = 1
mf= 1    mf=1
IScomp = 2
ISper = 3

now for L= 17  B = 13

IScomp = ((mf * 13 ) + 1 ) /   ((13-17 ) modulus 13)) =  ((mf * 13) + 1 ) / 9  = 3
ISper     =  ((mf * 13 ) + 1 ) /  (13 - ((13-17 ) modulus 13))  = ((mf * 13) + 1 )/ 4 = 10

IScomp = ((mf * 13 ) + 1 ) /  (-17) mod 13 = ((mf * 13 ) + 1 ) / 9
ISper     = ((mf * 13 ) + 1 ) /  (+17) mod 13  = ((mf * 13 ) + 1 ) / 4

mf = 2    mf = 3

IScomp =  ((2 * 13 ) + 1 ) / 9   =     26+1/9        =      27 / 9       = 3
ISper    = ((3 * 13 ) + 1 ) / 4     =    39 + 1 / 4       =      40 / 4    = 10

Apply modulus B of course

0     3     6     9     12      2     5    8     11   1     4    7   10   [13 == 0)]

(-17) mod 13 = 9
0     3      x      x      x      2      x     x      x    1      4    x     x

this works
You get the complementary at a fast pace and the just reverse it ! to get the periodic
0    10   7    4    1    11    8    5    2    12    9    6    3
the other way is

DW ( Digit Written)   then its place after 0 is  DW * (L) modulus B
(too late to change but instead of Digit Written it should have been Number Written)
(1 * (-17)) mod 13    =   -1 * 9     ==     -17 mod 13   ===      9th place
(2 * (-17)) mod 13    =     2 * 9    ==      -34 mod 13   ===      5th place
(3 * (-17)) mod 13    =     3 * 9    ==      -51 mod 13   ===      1rst place
(4 * (-17)) mod 13    =     4 * 9    ==      -68 mod 13   ===    10th place
...
(12 * (-17)) mod 13   =  12 * 9    ==    -204 mod 13   ===      4th place

now for L=14  B =5
mf = 1    mf = 3
Scomp =  ((1 * 5 ) + 1 ) / 1   =     6/1       =      6
Sper    = ((3 * 5 ) + 1 ) / 4     =    15 + 1 / 4       =      16 / 4    = 4

Now for L=15  B = 7
mf = 5    mf = 1
Scomp =  ((5 * 7 ) + 1 ) / 6   =     35+1/6        =      36 / 6       = 6
Sper    = ((1 * 7) + 1 ) / 1     =    7 + 1 / 4       =      8 / 4    = 2

Now for L=15  B = 8
mf = 1    mf = 6
Scomp =  ((1 * 8 ) + 1 ) / 9   =     8+1/9        =      9 / 9       = 1
Sper    = ((6 * 8) + 1 ) / 7     =    48 + 1 / 7       =      49 / 7    = 7 Copyright 2005 Sept - Charles Hamel / Nautile -
Overall rewriting in August 2006 . Copyright renewed. 2007-2012 -(each year of existence)

Url : http://charles.hamel.free.fr/knots-and-cordages/