The Friedman method, all permutations unknown

• Find message key
• Known plaintext, math attack.
• Known plaintext, tables attack.
• Dawson method, unknown rotor
• Dawson's method, unknown rotor and keyboard
• Konheim's method
• Hill Climbing, unknown rotor
• Hill Climbing, unknown rotor and keyboard
• Friedman's method v1, unknown rotor),
• Friedman's method v1, unknown Rotor and keyboard
• Friedman's method v1, unknown Rotor and lampboard
• Friedman's method v1, all permutations are unknown. This Page.
• Friedman's method v2, unknown Rotor
• Messages in-depth
• Remove lampboard permutation

Introduction

In the previous chapters, we studied the method of Friedman intervals v1. In all cases studied, part of the basic key was known (keyboard and/or lampboard).

In this chapter, we know nothing. In short, we do not know any permutation.

The method

The direct method

It is always possible to reconstruct some basic cipher-text sequences.

If we have enough sequences, we can attack the cryptogram like a gigantic simple substitution.

Reconstruct the Lampboard permutation

We can also reconstruct the Lampboard permutation. Then we removes it from the cryptogram. The resulting cryptogram is produced by two permutations: the rotor and the keyboard. We have already covered this problem in one of the previous chapters.

Probable word

There is a short-cut if you know the beginning of the text. We can easily associate a plain text letter with each of the basic cipher-text sequences found.

Example

We construct the cryptogram:

C:\H1_TOOLS> python hebern1_tui.py -R c5 -K LFS -L RFS 
	< MSGS\baskerville.pln > MSGS\basker5.cry

C:\H1_TOOLS> python groupe.py < MSGS\basker5.cry 
BLQFM XEQTM QRRXM AOORJ IUNKC XFSGT PONZQ RFRWN GHCGP OTLRG
YFPSG BWBOM MCADV PMBBS WTTAR HTUQX KPEKS MSOAG GFWDX AZACB
HPZXU XKLFI GTJDS HWTBX QZWFK CLXZX SSQPB STBHY JARWL MHGPV
CFZGT KSKZG UPPWS XQRPW LBUOO OKFYT MKBRF VDDOZ AMAGX URSPP
BEMJW ABKTX GKKSB MBXUR PRUET SMVXU YAGNT NFSPA GIHBK QAFDZ
KYBFE ROATQ EXBYP KINFR UYBOG DJWZK FNLMT XNZLY EYCZM MBXIN
UXVGE NPFYS YILTF QYPSB FCHRR SGDTC NKWFN WXLNE DLNMV MTTKX
MDPXY ZQMPZ YPVIB ZZLZJ UPCVJ QCEWR PNZPI JHDWL NGAWZ CEIAC
HQIFE SKPYU WFNQN EXEMH UPPWS WAOOS GRTCL MCUBZ IISQC LMFXL
JUGWO GZXVY BNFIQ RZXFV ZMBSL SOWOM YVLAU SIFFV IFEXE YLGVB
...

We analyze the cryptogram:

C:\H1_TOOLS> python friedman.py -I 5 < MSGS\basker5.cry
    A  B  C  D  E  F  G  H  I  J  K  L  M  N  O  P  Q  R  S  T  U  V  W  X  Y  Z
A:  0  2  0  2 10 12  0  0  0  1  3  1  1  5  6  1  0  2  8 10  5  8  2  6  0  2
B:  0  1  0  7 19  7  0  2  1  2  6  3  1 21  5  1  0  8 10 10  5 10  0  2  0  5
C:  0  0  1  1  3  1  0  3  0  1  0  1  0  4  1  0  0  1  4  1  4  2  0  3  0  1
D:  0  1  0  2  4  2  0  0  0  0  2  0  0  0  2  0  0  4  2  3  1  2  1  2  0  1
E:  0  0  0  0  2  0  1  0  0  0  0  0  0  1  0  0  0  0  0  0  0  1  0  1  0  0
F:  0  1  0  2  9  8  2  3  0  1  0  4  2  4  5  1  1  4  6  2  7  5  0  7  0  4
G:  0  0  0  0  1  1  1  0  0  0  1  0  1  1  1  0  0  0  1  1  1  5  0  2  0  1
H:  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0
I:  0  3  0  5 13  6  0  1  0  3  5  4  1  5  9  2  0  4  7  9  4  6  0  7  0  5
J:  0  3  0  5 10 17  2  6  0  3  1  5  1 12 14  1  2  0 10 13 11  9  1  9  0  8
K:  0  4  0  3 12  8  5  3  0  1  3  4  1  8 10  2  4  7 11 10  6  8  0  6  0  6
L:  0  6  0  7 11  3  1  2  0  2  1  5  1  8  6  4  4  3  8  6  5 11  0  6  0  1
M:  0  2  0  3  6  3  3  1  0  1  1  1  1  3  4  0  0  0  2  5  4  5  0  8  0  1
N:  0  2  0  4  3  1  1  2  0  0  1  2  3  1  1  1  0  1  1  2  2  4  0  1  0  2
O:  0  2  0  1  5  3  1  1  0  0  0  0  1  3  3  0  3  0  1  1  4  3  0  4  0  0
P:  0  1  0  1  4  2  0  2  0  2  1  1  1  1  3  1  0  1  2  6  3  5  1  5  0  2
Q:  0  0  0  0  4  6  0  0  0  2  0  0  0  3  0  0  0  1  3  2  1  5  0  2  0  1
R:  0  2  0  0  2  1  1  0  0  0  0  0  0  0  2  2  1  1  1  1  2  1  0  1  0  2
S:  0  6  0  1  3  3  2  1  0  0  0  0  1  7  5  0  2  2  6  6  6  7  0  5  1  4
T:  0  0  0  1  2  2  2  2  0  2  1  1  1  3  3  0  1  0  0  4  1  4  0  0  0  0
U:  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0
V:  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0
W:  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0
X:  0  3  0  5 22 12  3  7  0  4  5  3  4 12 15  2  2  7 18 13 13 12  1 14  0  7
Y:  0  3  0  8 13  4  1  4  0  1  0  1  1 10  4  2  1  5 13 10  8  7  1  9  0  1
Z:  0  2  0  1  6  0  1  0  0  0  0  0  0  6  4  0  0  1  1  4  2  4  0  1  0  1
    0 44  1 59165102 27 40  1 26 31 36 22118103 20 21 52115119 95124  7101  1 55

After a lot of work, we found the following basic cipher-text sequences (corresponding to the most frequent letters):

	a b c d e f g h i j k l m n o p q r s t u v w x y z
(1)	E S F H B X G M A D J R V W P D K C Z N Y U I Q T O
(2)	V E Q O K B T S C U K H A Y L X J G N M D F R W Z T
(3)	T K L P M J O B H X A W N S E R U Y I V C Q L Z F G
(4)	N X Z G P K U W Q O L Y I E M H D J R E F B C V A T

It is assumed that the message begins with the word "CHAPTER":

Let's analyze this beginning:

	Key	Crypto	Plain	Hypothesis
	a	B	C		
	b	L	H
	c	Q	A	Sequence 2 correspond to letter A	
	d	F	P
	e	M	T	Sequence 3 correspond to letter T
	f	X	E	Sequence 1 correspond to letter E
	g	E	R

Sequence 4 probably corresponds to either the letter O or the letter N, the two most frequent letters not yet associated with a sequence.

These deductions will help us enormously in finding the final solution.

If we then have a multitude of cribs, we can position them.