Chapter 7: Analysis of the six-wheel Hagelin cipher machine 
where indicators are unciphered

--- Explanations:

Problem 1:
==========

1) First step: find the absolute positions of each message:

Indicator: YQPFI => Message Key: YYQPFI: 
indx = 24, 23, 16, 15, 5, 8

total    =  89705175 * (24+1) + 56787276 * (23+1) + 
            92587950 * (16+1) + 82090450 * (15+1)  +
            42697200 * (5+1)  + 41755350 * (8+1)
position = (total % 101405850) – 1  
         = (7124947699 % 101405850) – 1 = 26,538,198 
 
We calculate the positions for each message:
YYQPFI: 26,538,198
DDVUKN: 26,538,203
HHCDOA: 26,538,207
RRMNFK: 26,538,217
VVQRJO: 26,538,220

2) Now, we can put the messages "in depth" and search for 
probable word (example: we try the word "British" at the beginning 
of the second message)

 198 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9
(1)A G J V R F L L Z J D C F I A O N R T E J M Q A Z F L D D B S B ...
                     a n d z a p r i
(2)                  I Z X L A X Y W L E I X N D N Y K M T Q X E O ...
                     b r i t i s h z
          
(3)                                      N Q U D A F M K Q H Y R E ... 
          

(4)          D G K G K B F J E M G C V O H B Y E B T R T T R Y N S ...
                     z p a v e d z t

(5)                                            S L G J Y A L W T N ...

3) The following steps are known (rebuild the plain text of message
in depth, then calculate key and find lugs and pins setting).

Annexe: Algorithm (in Python) for find the absolute position 
from message Key
#===========================================================
alpha_roues = [
  "ABCDEFGHIJKLMNOPQRSTUVWXYZ",
  "ABCDEFGHIJKLMNOPQRSTUVXYZ",
  "ABCDEFGHIJKLMNOPQRSTUVX",
  "ABCDEFGHIJKLMNOPQRSTU",
  "ABCDEFGHIJKLMNOPQRS",
  "ABCDEFGHIJKLMNOPQ"
]
constC38 = [
	89705175,
	56787276,
	92587950,
	82090450,
	42697200,
	41755350
]
def calpos( cle ):
	total = 0
	for i in range( len(cle) ):
		indx = alpha_roues[i].find( cle[i] )
		v = constC38[i] * ( indx + 1 )
		total += v
	return (total % 101405850) – 1
#===========================================================


Problem 2:
==========
We don’t use new methods.