Kryha Cryptanalyse - Konheim cryptogram

Home Page
Kryha Home Page
Kryha Cryptanalysis Home Page

Introduction

In 1982, Alan G. Konheim wrote an article on the cryptanalysis of Kryha cipher machine (see References).

He proposed a mathematical attack on a long cryptogram (over 2000 characters). The wheel used is assumed to be unknown. I confess I understood almost nothing of his mathematical proofs :-(

In the rest of this page, I found the plain text of the message, primarily by using Sacco's method and Friedman's method. I only partially utilize Konheim's method.

The Wheel Used

In fact, as shown in the solution, the offsets found correspond to the standard wheel but with a systematic offset of three notches instead of four. Is Konheim using an experimental model of the Kryha machine?

At the end of his article, Konheim thanks Dr. Kjell-Ove Widman of Crypto AG for his help regarding the historical aspects. Perhaps this person provided him with the description of the wheel?

The cryptogram

$ more MSGS/konheim.cry
CECZDHYCFWGNVRTUA MZEDCQAMRUDWGAKCQ GQUSIFIZBMXMDOGOY WTSZRJBDOLZESLLSO

VVFBPHTKYMFXDGNRN MIIPITFWNRRIZCRGN LMKNXXIDHAARQEGAH PLIYXSQYARQRQSSVX

BVLUIULXQGISVXIKU YNUIZQUQVIJFEEQNH YGKIXTNIKOOQYIBFB UUGPLRJOSIYBACPGV

QYFZOWOGKRLPHTKNE YLAPCOYZEXBJAYGWI EHNDHRHZKNVTVXJXS KCDNKNEKSLXMFEDGB

XSKSZSBRCKKFOJDCF EEYXAHWSYMFIOUDBC BYXPMHQSDGFZGHVYO MSYMUBTVNKXHMCECA

GNXAVGPCCVBBBIRFM UJQJGSKMHCGSHGXFR QMUUZOWWVWMABUAWT LNOUIGGLRHWTIKWBV

XARTAYYWYYCPWXNNZ BXBLYJJHUDEYTPLLE IJCGFKHACOIOCNIKL MFOINMZOYTHRMKAMC

RREFOSOPBPWRJOGVJ AZPZXAKUZIPUGVGZW WPFZKWWYITNAGLWSN ZBVKXRDVDQXFKJIWO

DSYLEKXANBMTKWVVT FNLNKDPJADRPFHPHD VTKVKBEZLNIPTGFSW LJEWNQUQQCESWKEUF

NWAGUHAVYMARMXKDI UAUCKXTSFKPTLENNU NEUUJDYFVRZMNEWJA YKLNCPDEROIWUXUWH

HZAISFAMOKSYIOKLG JYXZCIIDITHEPGDNO WYUXPWMDLPBGPKIAL VXFADSSXTXNCORDJG

NFQYOGOZZECVPQHJV FCEQPABYPFBJFZBAZ HYPQYPKICOPXKRKHR XAQJXTPSJUGKHDDTG

BDGGNQUQRCWBMFSSC PABNHPCCSHJPNKTNV CVLSJMOZFVITJZGGR JRGPLGSGVSMDOALQR

VUGPQKDCPKPOKHNPG DAHEMYFESLBPFZOMQ WEBOBDMGCEYDWCGXQ PJMZVWPPLXGWXCGHT

XWFCELCAPZOAISTKZ VPPPMVXZEQOVJZOIG DSYLEKBVGHSUATFZA XAQCBFJFARNFSSDQY

XOXUJMTXEUMQEYXVQ BWNOQLYCBFSXCDTWT IHLDNOYPKQOEKNVSQ HPWZTLZXZUYEFVKCS

AUJSLBRIHMMWLKFAV VEAPGOVZEKWOVKCTM OHLXODFIXZXPFVTML UHFFUIVOCLYLQVOEN

DFIHXAUWNNAUFTNAU CFNGCEMIJJWXLDWYD SMSTQBSHLWFSIDHJV MBGSEXSMGOMKROVKA

SIKHDODLJUQRTSZOZ SGVTVSISYFESAXOKF BWWXWRWNJSYNBETPC PICCAYMUKHIRNXWHI

YYWAVNVLZOZPGUJSL YDCKLBDDBRESGOJQU YKKGFOWSAIQETLNFC XRZPCUCOGXHOGAUMX

The key length

If we use the superposition method, we can calculate the length of the key and deduce the number of sectors of the wheel used. 

$ python3 superpose.py -c MSGS/konheim.cry -m 500 -t 110
   0 =>  2272, 1.000
  61 =>   115, 0.051
  82 =>   113, 0.050
  94 =>   113, 0.050
 116 =>   110, 0.048
 141 =>   113, 0.050
 157 =>   116, 0.051
 165 =>   117, 0.051
 221 =>   140, 0.062
 249 =>   116, 0.051
 320 =>   110, 0.048
 322 =>   120, 0.053
 327 =>   110, 0.048
 381 =>   120, 0.053
 442 =>   142, 0.062
 468 =>   111, 0.049
Two peaks are noticeable: 221 and 442 (17*26 et 17*13). Clearly, the wheel has 17 sectors.

Sacco's Method: Offsets between columns

We distribute the cryptogram letter by letter into the cells of a table where each cell contains letters belonging to the same alphabet (see sacco).

Each column in this table contains the same sequence of alphabets, but shifted from one another. We test all possible shifts between any two columns, whether adjacent or not. For each shift, we group the letters belonging to the cells assumed to belong to the same alphabet and calculate the index (IC). The maximum value of the IC indicates the right shift.

Here are the results (for adjacent columns and 5 and 2 columns apart): 

$     python3 chi2.py -c MSGS/konheim.cry -m 17 -t 1
       0     1     2     3     4     5     6     7     8     9    10    11    12    13    14    15    16
 0: 0.050 0.044 0.049 0.047 0.024 0.039 0.034 0.032 0.033 0.042 0.024 0.039 0.038 0.042 0.035 0.039 0.033
 1: 0.056 0.047 0.055 0.039 0.025 0.039 0.021 0.023 0.023 0.039 0.044 0.031 0.023 0.045 0.035 0.033 0.033
 2: 0.039 0.036 0.053 0.042 0.042 0.032 0.054 0.057 0.059 0.057 0.035 0.057 0.028 0.029 0.040 0.042 0.047
 3: 0.046 0.039 0.033 0.065 0.039 0.037 0.055 0.045 0.034 0.031 0.065 0.033 0.060 0.050 0.040 0.068 0.040
 4: 0.035 0.021 0.046 0.031 0.038 0.036 0.036 0.034 0.037 0.037 0.037 0.038 0.038 0.031 0.026 0.032 0.039
 5: 0.028 0.019 0.032 0.035 0.027 0.022 0.032 0.025 0.037 0.052 0.046 0.025 0.033 0.022 0.031 0.029 0.041
 6: 0.027 0.024 0.045 0.045 0.021 0.035 0.036 0.051 0.032 0.032 0.034 0.030 0.041 0.033 0.024 0.049 0.036
 7: 0.035 0.039 0.053 0.046 0.055 0.041 0.052 0.056 0.045 0.031 0.039 0.056 0.043 0.038 0.055 0.032 0.060
 8: 0.055 0.021 0.029 0.025 0.035 0.040 0.033 0.028 0.027 0.030 0.029 0.042 0.033 0.053 0.030 0.026 0.043
 9: 0.033 0.029 0.021 0.031 0.034 0.037 0.053 0.019 0.035 0.043 0.025 0.043 0.022 0.026 0.038 0.040 0.036
10: 0.039 0.033 0.040 0.024 0.039 0.040 0.023 0.062 0.041 0.046 0.029 0.042 0.027 0.025 0.032 0.042 0.038
11: 0.057 0.042 0.069 0.029 0.043 0.061 0.032 0.041 0.048 0.032 0.034 0.049 0.043 0.063 0.048 0.046 0.053
12: 0.024 0.051 0.040 0.045 0.050 0.045 0.060 0.034 0.061 0.033 0.036 0.046 0.038 0.041 0.033 0.037 0.040
13: 0.045 0.035 0.034 0.054 0.035 0.031 0.037 0.035 0.027 0.039 0.029 0.037 0.038 0.049 0.038 0.036 0.033
14: 0.036 0.039 0.039 0.028 0.028 0.043 0.015 0.035 0.016 0.045 0.034 0.022 0.038 0.046 0.034 0.026 0.035
15: 0.036 0.058 0.041 0.041 0.049 0.041 0.047 0.041 0.056 0.042 0.051 0.051 0.033 0.043 0.059 0.038 0.040
16: 0.052 0.058 0.029 0.059 0.048 0.029 0.051 0.041 0.033 0.029 0.072 0.030 0.055 0.028 0.051 0.072 0.033
17: 0.053 0.034 0.025 0.031 0.039 0.042 0.043 0.029 0.025 0.042 0.045 0.036 0.035 0.036 0.039 0.034 0.032
18: 0.025 0.030 0.027 0.031 0.043 0.034 0.028 0.028 0.031 0.035 0.042 0.028 0.039 0.016 0.032 0.036 0.026
19: 0.024 0.031 0.022 0.055 0.041 0.030 0.037 0.039 0.030 0.034 0.040 0.022 0.040 0.032 0.042 0.038 0.028
20: 0.029 0.043 0.044 0.039 0.027 0.043 0.057 0.052 0.045 0.026 0.045 0.034 0.029 0.042 0.045 0.038 0.055
21: 0.050 0.019 0.032 0.033 0.021 0.037 0.034 0.030 0.036 0.041 0.029 0.034 0.024 0.054 0.027 0.046 0.032
22: 0.025 0.029 0.045 0.030 0.033 0.035 0.045 0.028 0.041 0.031 0.028 0.035 0.025 0.029 0.034 0.031 0.033
23: 0.037 0.047 0.037 0.033 0.028 0.041 0.026 0.054 0.046 0.043 0.034 0.035 0.036 0.048 0.025 0.022 0.043
24: 0.058 0.034 0.065 0.034 0.041 0.065 0.039 0.035 0.045 0.025 0.040 0.046 0.030 0.074 0.048 0.032 0.042
25: 0.044 0.041 0.054 0.046 0.062 0.038 0.046 0.042 0.061 0.033 0.033 0.044 0.028 0.039 0.043 0.054 0.053
$     python3 chi2.py -c MSGS/konheim.cry -m 17 -t 5
       0     1     2     3     4     5     6     7     8     9    10    11    12    13    14    15    16
 0: 0.068 0.049 0.056 0.056 0.049 0.035 0.036 0.044 0.048 0.036 0.049 0.036 0.059 0.031 0.047 0.065 0.045
 1: 0.045 0.041 0.052 0.041 0.034 0.038 0.034 0.033 0.027 0.022 0.026 0.042 0.025 0.025 0.029 0.038 0.026
 2: 0.029 0.033 0.031 0.025 0.027 0.032 0.045 0.050 0.027 0.027 0.047 0.047 0.041 0.038 0.019 0.037 0.034
 3: 0.057 0.044 0.049 0.038 0.035 0.034 0.063 0.062 0.033 0.034 0.044 0.036 0.036 0.035 0.024 0.041 0.040
 4: 0.030 0.033 0.033 0.042 0.052 0.033 0.034 0.026 0.046 0.036 0.035 0.037 0.021 0.042 0.035 0.042 0.032
 5: 0.041 0.029 0.031 0.032 0.043 0.025 0.038 0.034 0.048 0.049 0.063 0.059 0.057 0.048 0.034 0.047 0.056
 6: 0.038 0.028 0.023 0.030 0.036 0.043 0.036 0.050 0.037 0.034 0.039 0.033 0.027 0.037 0.036 0.029 0.028
 7: 0.034 0.029 0.043 0.027 0.041 0.043 0.046 0.038 0.045 0.039 0.029 0.023 0.028 0.027 0.026 0.034 0.032
 8: 0.031 0.042 0.035 0.033 0.072 0.053 0.046 0.043 0.039 0.066 0.051 0.033 0.047 0.044 0.033 0.052 0.045
 9: 0.022 0.043 0.039 0.055 0.040 0.051 0.033 0.023 0.046 0.018 0.037 0.037 0.030 0.045 0.077 0.034 0.039
10: 0.032 0.047 0.048 0.029 0.042 0.046 0.034 0.028 0.036 0.046 0.043 0.034 0.043 0.046 0.033 0.037 0.033
11: 0.035 0.053 0.037 0.023 0.039 0.066 0.038 0.035 0.043 0.034 0.028 0.032 0.027 0.033 0.033 0.040 0.051
12: 0.048 0.044 0.035 0.047 0.052 0.020 0.043 0.046 0.046 0.041 0.035 0.015 0.046 0.039 0.044 0.032 0.041
13: 0.057 0.076 0.067 0.051 0.032 0.044 0.035 0.039 0.033 0.025 0.030 0.039 0.041 0.036 0.033 0.034 0.046
14: 0.045 0.031 0.044 0.036 0.024 0.040 0.027 0.032 0.027 0.031 0.027 0.047 0.032 0.025 0.042 0.035 0.028
15: 0.038 0.040 0.035 0.030 0.035 0.028 0.051 0.039 0.045 0.024 0.053 0.048 0.049 0.047 0.039 0.037 0.032
16: 0.065 0.045 0.036 0.034 0.024 0.041 0.061 0.057 0.030 0.046 0.038 0.034 0.031 0.041 0.046 0.029 0.033
17: 0.037 0.039 0.054 0.050 0.046 0.042 0.042 0.033 0.043 0.039 0.029 0.019 0.037 0.032 0.023 0.035 0.050
18: 0.025 0.023 0.033 0.036 0.045 0.038 0.044 0.034 0.030 0.042 0.065 0.060 0.039 0.040 0.044 0.037 0.042
19: 0.027 0.025 0.033 0.033 0.032 0.043 0.041 0.043 0.035 0.033 0.034 0.049 0.036 0.030 0.041 0.016 0.028
20: 0.018 0.023 0.024 0.025 0.021 0.040 0.048 0.027 0.056 0.038 0.028 0.028 0.019 0.029 0.045 0.029 0.031
21: 0.038 0.037 0.039 0.049 0.074 0.030 0.036 0.034 0.046 0.078 0.058 0.037 0.041 0.037 0.034 0.048 0.047
22: 0.032 0.047 0.035 0.040 0.048 0.031 0.025 0.037 0.050 0.048 0.047 0.033 0.031 0.045 0.072 0.042 0.029
23: 0.048 0.042 0.040 0.027 0.037 0.047 0.027 0.027 0.029 0.042 0.032 0.027 0.051 0.035 0.030 0.033 0.030
24: 0.047 0.051 0.033 0.020 0.052 0.069 0.029 0.034 0.041 0.035 0.039 0.037 0.040 0.027 0.063 0.026 0.039
25: 0.030 0.048 0.025 0.053 0.036 0.039 0.040 0.028 0.059 0.044 0.020 0.026 0.045 0.033 0.040 0.021 0.029
$ python3 chi2.py -c MSGS/konheim.cry -m 17 -t 2
       0     1     2     3     4     5     6     7     8     9    10    11    12    13    14    15    16
 0: 0.040 0.058 0.047 0.035 0.052 0.055 0.034 0.046 0.024 0.043 0.048 0.017 0.032 0.041 0.040 0.047 0.030
 1: 0.055 0.044 0.057 0.043 0.054 0.046 0.034 0.042 0.044 0.025 0.044 0.045 0.077 0.055 0.039 0.036 0.040
 2: 0.049 0.027 0.040 0.044 0.032 0.036 0.027 0.031 0.036 0.042 0.034 0.051 0.055 0.029 0.057 0.033 0.042
 3: 0.025 0.034 0.044 0.033 0.038 0.019 0.025 0.019 0.033 0.044 0.033 0.018 0.039 0.039 0.042 0.033 0.033
 4: 0.036 0.016 0.036 0.025 0.033 0.035 0.031 0.036 0.070 0.026 0.033 0.033 0.046 0.040 0.041 0.031 0.035
 5: 0.051 0.042 0.038 0.053 0.045 0.046 0.048 0.061 0.021 0.046 0.054 0.039 0.043 0.048 0.046 0.039 0.081
 6: 0.047 0.033 0.030 0.038 0.040 0.029 0.055 0.041 0.032 0.029 0.040 0.047 0.037 0.026 0.056 0.051 0.038
 7: 0.052 0.021 0.049 0.021 0.021 0.030 0.032 0.025 0.060 0.048 0.033 0.017 0.029 0.046 0.034 0.050 0.040
 8: 0.040 0.036 0.031 0.036 0.025 0.021 0.031 0.044 0.021 0.056 0.027 0.025 0.030 0.026 0.037 0.041 0.049
 9: 0.037 0.039 0.033 0.036 0.038 0.033 0.061 0.071 0.036 0.027 0.028 0.036 0.026 0.037 0.020 0.034 0.044
10: 0.029 0.048 0.036 0.039 0.053 0.039 0.043 0.035 0.025 0.021 0.046 0.045 0.037 0.035 0.045 0.048 0.031
11: 0.042 0.022 0.039 0.018 0.032 0.038 0.041 0.032 0.034 0.039 0.040 0.040 0.039 0.037 0.032 0.038 0.026
12: 0.030 0.046 0.033 0.041 0.034 0.024 0.043 0.051 0.030 0.029 0.036 0.043 0.032 0.023 0.037 0.024 0.036
13: 0.047 0.043 0.045 0.031 0.060 0.050 0.035 0.024 0.026 0.045 0.028 0.038 0.036 0.046 0.034 0.053 0.028
14: 0.040 0.036 0.077 0.034 0.050 0.041 0.036 0.036 0.047 0.028 0.046 0.055 0.069 0.040 0.042 0.030 0.036
15: 0.021 0.041 0.039 0.046 0.031 0.048 0.021 0.037 0.032 0.033 0.053 0.055 0.049 0.044 0.053 0.049 0.040
16: 0.031 0.037 0.027 0.028 0.039 0.042 0.033 0.024 0.039 0.037 0.045 0.022 0.031 0.032 0.044 0.021 0.034
17: 0.044 0.027 0.036 0.021 0.034 0.020 0.035 0.046 0.066 0.030 0.033 0.025 0.032 0.030 0.036 0.039 0.027
18: 0.043 0.054 0.039 0.054 0.028 0.032 0.046 0.044 0.044 0.042 0.031 0.058 0.025 0.054 0.057 0.036 0.074
19: 0.041 0.037 0.028 0.021 0.040 0.032 0.047 0.031 0.026 0.024 0.046 0.043 0.045 0.027 0.039 0.040 0.042
20: 0.039 0.035 0.030 0.031 0.028 0.044 0.040 0.022 0.050 0.039 0.040 0.030 0.038 0.035 0.034 0.053 0.028
21: 0.028 0.042 0.031 0.033 0.026 0.041 0.033 0.027 0.028 0.057 0.042 0.034 0.026 0.035 0.029 0.041 0.046
22: 0.038 0.046 0.039 0.031 0.028 0.037 0.046 0.076 0.040 0.057 0.036 0.026 0.035 0.041 0.025 0.048 0.042
23: 0.045 0.051 0.037 0.039 0.042 0.044 0.048 0.038 0.017 0.024 0.056 0.041 0.043 0.048 0.048 0.057 0.034
24: 0.052 0.031 0.048 0.020 0.027 0.028 0.048 0.045 0.032 0.042 0.032 0.038 0.040 0.027 0.026 0.039 0.024
25: 0.032 0.045 0.047 0.031 0.041 0.022 0.035 0.045 0.039 0.032 0.033 0.039 0.040 0.027 0.041 0.033 0.021

We observe that columns 0 and 1 are shifted by 11 positions. Similarly, columns 2 and 3 are also shifted by 11 positions (IC=0.069). However, columns 3 and 4 are only shifted by 3 positions.

We thus measure all the shifts.

Finally, we notice a particular behavior: The maximum ICs occur twice per column and are always 13 positions apart. So, in the same column, we only have 13 alphabets repeated twice (which gives 26 lines). When we looked for the length of the key, the superposition method gave two peaks, one of which has a distance of 221=13*17. Now we understand why.

Furthermore, the columns are divided into two distinct families. Columns 0 and 1 belong to the same family and therefore share the same alphabets. However, columns 2, 3, and 4 belong to a different family of alphabets.

By studying all possible shifts, we obtain the following result:

We obtain two sets of values:

  • Columns 0, 1, 5, 6, 8, 10, 11, 15, and 16, with the following shifts :
    0, 11, -, -, -, 0, 11, -, 7, -, 11, 1, -, -, -, 3, 6
  • Columns 2, 3, 4, 7, 9, 12, 13, 14:
    -, -, 0, 11, 1, -, -, 0, -, 9, -, -, 3, 6, 4, -, -
Note: Shifts 1 and 14, or 11 and 24 are equivalent.

As a check of the previous results, we can calculate the IC for each of the 26 alphabets, divided (let's not forget) into two groups: 

$ python3 sacco_13.py -c MSGS/konheim.cry \
	-m 0,11,-1,-1,-1,0,11,-1,9,-1,11,1,-1,-1,-1,3,6
 0, 0.058  93 P:10,O:9,V:9,A:8,Y:8,
 1, 0.065  93 Y:12,M:10,D:9,R:9,O:8,
 2, 0.054  95 J:10,U:9,G:8,F:7,I:7,
 3, 0.066  95 A:12,H:11,Z:9,C:8,F:7,
 4, 0.059  94 V:10,E:9,H:9,I:9,B:6,
 5, 0.055  94 K:11,V:9,B:8,R:7,S:7,
 6, 0.050  91 B:9,F:9,R:7,W:7,E:6,
 7, 0.051  92 F:12,S:8,G:6,M:6,P:6,
 8, 0.073  91 S:17,B:12,C:6,M:6,X:6,
 9, 0.066  91 Q:14,N:10,B:8,R:7,A:6,
10, 0.057  91 T:11,P:10,F:7,G:7,A:6,
11, 0.062  91 M:12,U:9,W:8,A:7,C:7,
12, 0.065  92 N:12,J:11,G:9,O:9,A:6,
Debug: False

$ python3 sacco_13.py -c MSGS/konheim.cry \
	-m -1,-1,0,11,1,-1,-1,0,-1,9,-1,-1,3,6,4,-1,-1
 0, 0.073  83 C:16,Z:8,S:7,A:6,W:6,
 1, 0.065  83 O:13,H:10,I:7,E:6,L:6,
 2, 0.069  84 D:10,G:9,P:9,U:9,V:8,
 3, 0.072  83 D:12,C:8,I:8,P:8,R:7,
 4, 0.072  82 F:10,K:10,I:9,X:8,O:7,
 5, 0.058  82 U:9,E:8,S:7,X:7,Z:7,
 6, 0.058  81 B:8,M:8,Y:8,E:7,K:7,
 7, 0.067  82 K:11,P:10,N:9,I:7,X:7,
 8, 0.058  81 X:9,A:8,L:8,M:7,Z:7,
 9, 0.064  82 Q:12,N:10,G:7,U:7,C:5,
10, 0.073  82 T:10,V:10,J:9,K:8,A:7,
11, 0.066  82 L:12,G:9,R:6,S:6,W:6,
12, 0.068  82 G:11,F:9,C:8,D:7,H:7,
Debug: False
Note: since there are more than 80 letters in each alphabet, we could skip the following steps and try to decipher the cryptogram right away. If we have probable words, it's actually quite easy.

The Konheim's Method: The Value of Each Sector

Now, let's apply the Konheim method (the part that allows us to find the sectors) (see Konheim).

Two letters belong to the same alphabet if and only if: 

	s * L = L[c1] – L[c2]

  • L corresponds to the sum of the shifts of each sector. In the case of the standard wheel, L = 111.
  • s corresponds to the number of periods, i.e., the difference in position between the two letters divided by the number of sectors.
  • c1 corresponds to the index of the first letter modulo the number of sectors.
  • c2, likewise, but for the second letter.
  • L[x] = is the sum of the shifts of the sectors up to position x. = l[0] + l[1] + … + l[x]
  • l[i] corresponds to the shift caused by the i-th sector.

By applying this formula to cells belonging to the same alphabet (derived from the shifts found by Sacco's method) we establish a whole set of equations: 

  L[1]-L[8]=9L		L[5]-L[15]=3L		
  L[3]-L[7]=2L		L[4]-L[9]=8L
  L[11]-L[0]=12L	L[7]-L[9]=9L
  L[1]-L[15]=5L		L[10]-L[0]=2L
  L[9]-L[14]=8L		L[7]-L[13]=6L
  L[2]-L[14]=4L		L[11]-L[1]=10L
  L[1]-L[16]=8L		L[12]-L[14]=L
  L[6]-L[0]=2L

We can make the following deductions 

  L[0]=7L		L[5]=7L		L[10]=9L	L[15]=4L
  L[1]=3L		L[6]=9L		L[11]=6L	L[16]=L[1]-8L
  L[2]=L[1]+l[2]	L[7]=L[2]	L[12]=L[2]+10L
  L[3]=L[2]+2L		L[8]=0		L[13]=L[2]-6L
  L[4]=L[2]+12L		L[9]=L[2]+4L	L[14]=L[2]+9L
There are several ways to solve these equations. Personally, I first calculated L. To do this, I tested the following equations for all possible even values of L from 2 to 26. L must be even because the same letter appears twice in each column. 

	L[0] - L[1] = L * 11
	L[10] - L[11] = L * 3
	L[16] - L[0] = L * 7 => L[0] - L[16] = -L * 7
The only value of L that gives plausible values for sectors is 16. We also obtain l[0] = 8, l[1] = 6, and l[11] = 4.

Using these values, we can deduce some of the sectors and their cumulative sum. 

	l[3]=6, l[4]=4, l[6]=6, l[13]=4, l[14]=6, l[16]=4.
	L[0]=l[0]=7L=8, L[1]=9L=14, L[5]=7L=8, L[6]=9L=14,
	L[8]=0, L[10]=9L=14, L[11]=6L=18, L[15]=4L=12,
	L[16]=L=L[1]-8L=16 (verification)
However, for the moment, we are unaware of L[2], on which all the other sectors depend.

We can restrict the possible values of these sectors: 

	L[8]=0 = L[7]+l[8] = L[5]+l[6]+l[7]+l[8] = 8+6+l[7]+l[8]
	=> l[7]+l[8] = -8-6 = -14 = 12 (modulo 26)

	L[5]=8 = L[4]+l[5] = L[2]+12L+l[5] = L[1]+l[2]+10+l[5]=8
	=> l[2]+l[5]=10

	L[15]=12 = L[11]+l[12]+l[13]+l[14]+l[15] = 19+l[12]+4+6+l[15]
	=> l[12]+l[15] = 10
	
	L[10]=14 = L[8]+l[9]+l[10]= 0 + l[9]+l[10]
	=> l[9]+l[10] = 14

 	L[2] = L[1]+l[2] and L[7]=L[6]+l[7] and L[1]=L[6]=14
	=> l[2]=l[7]

	L[12]=L[2]+10L = 14+l[2]+10L=18+l[12]
	=> l[2]=l[12]
Similarly, we deduce that l[5] = l[15] and finally l[2] = l[9].

What is the value of l[2]? I tried several values until all the equations involving L[2] were consistent. I found the value to be 5. From there, all the other sectors can be deduced.

In short, here are the shifts: 

	0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16
	8  6  5  6  4  5  6  5  7  5  9  4  5  4  6  5  4

Note: Determining the value of the sectors was complicated because we had two families of alphabets. The number of teeth required to move from one family to the other was not easily deducible.

Friedman's Method

Now that we know the wheel, we can apply Friedman's method to find the alphabets (see Friedman).

First attempt

$ python3 friedman_end.py -c MSGS/konheim.cry -w 8,6,5,6,4,5,6,5,7,5,9,4,5,4,6,5,4 
...
Interior Alphabet:  KP.U.BZ.C.N.FRJV..OQ.SIWGM , Collision:  True
The method fails, why?

Second attempt

Perhaps we need to move the wheel forward before we start calculating?

In conclusion, we find that the fixed (outer) alphabet corresponds to the direct alphabet (ABC...Z) but shifted. If we shift the inner alphabet by the same amount, we obtain the normal direct alphabet. 

$ python3 friedman_end.py -c MSGS/konheim.cry -w 6,5,6,4,5,6,5,7,5,9,4,5,4,6,5,4,8
    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
 0:13  1  5  0  0  0  0  5  4  3  3  0  0  1  9 10  0  4  1  5  4  7  4  0  8  6
 1: 2  4  1  1  6  0  1 10  7  0  5  6  3  1 13  1  3  2  5  0  4  2  0  0  2  4
 2: 1  9  1  0  2  2  3  5  0  2 12  2  5  1  0  4  2  6  7  6  4 11  0  0  1  7
 3: 5  8  3  4  7  0  4  0  0  0  7  3  8  0  1  5  1  1  0  5  0  5  3  2  8  1
 4: 5  1  0  0  6  6  5  1  6  3  3  3  4  1  5 10  0  0  0 12  6  4  4  0  6  0
 5: 0  0  0  5  2  2  9  0  5  5  0 12  1  0  5  0  0  6  6  0  4  3  6  4  6  1
 6: 0  3  0  5  6  6  8  1  7 11  5  1  0  3  0  1  3  6  1  1  9  5  6  2  5  0
 7: 2  0  8 12  4  4  0  0  8  0  6  0  6  0  0  8  3  7  4  0  7  0  2  1  0  1
 8: 1  4  2  4  0 13  8  6  2  2  7  2  6  1  4  6  5  3  9  0  0  0  0  5  2  0
 9: 8  0  3  1  0  2  5  0  0  4  3  8  7  4  2  0  4  1  4  2  0  2  5  9  0  7
10: 7  6  3  1  2  1 11  5  3  9  1  0  3 15  8  1  3  1  0  0  0  3  4  3  2  0
11: 6  1 16  5  3  5  3  3  2  0  1  1  2  2  5  2  0  0  7  4  0  0  6  1  0  8
12: 0  6  5  0  6  7  0  9  9  0  0  1  2  0  2  2  6  1  4  1  0 10  5  4  8  5
13: 3  3  1  0  8  0  0  4  2  1  1  3  0  4  1  6  6  0  7  4  9  4  1  7  0  7
14: 7  8  2  3  3  0  0  0  1  5  1  7  0 12  4  6 14  8  2  2  0  0  0  1  4  2
15: 7  3  5  0  0  3  0  3  1  9  8  7  0  0  6  2  0  0  0 10  5 10  0  0  2  1
16: 1  2  0  9  6  3  0  6  0  9  2  3 10  0  9  1  2  7  2  0  2  1  0  3 12  3
17: 0  2  0 10  3  3  9  0  2  2  5  0  0  2  2  9  0  2  0  0  9  8  0  6  4  6
18: 3  7  4  2  9 10  1  0  1  0  2  4  4  5  1  7  0  9  2  4  3  2  7  4  0  0
19: 0  0  2  0  2  3  3  1  7  4 11  5  3  9  1 10  2  2  0  3  0  2  0  7  5  0
20: 2  0  5  6  2  0  1  2  0  4  0  4 15  7  0  2  0  1  4  0  9  1  8  7  7  4
21: 0  1  8  7  0  9 11  7  3  5  2  2  2  6  0  2  6  1  0  0  6  0  2  0  0  2
22:12  3  8  7  0  7  0 11  1  1  5  1  1  0  0  1  0  4  5  3  0  2  7  6  1  9
23: 2  5  0  1  0 10  0  0  9  0 10  0  2  6  7  0  2  1  5  3  0  0  1  8  3  7
24: 4 15  8  0  0  1  4  1  2  0  0  1  5  4  0  0  5  0 20  5  3  2  1  6  1  4
25: 3  2  5  0  5  0  7  4  4  0  0  1  0 10  3  0 12  2  2  3  7  2  5  2  3  0

 A:13  2  1  5  5  0  0  2  1  8  7  6  0  3  7  7  1  0  3  0  2  0 12  2  4  3 : 94
 B: 1  4  9  8  1  0  3  0  4  0  6  1  6  3  8  3  2  2  7  0  0  1  3  5 15  2 : 94
 C: 5  1  1  3  0  0  0  8  2  3  3 16  5  1  2  5  0  0  4  2  5  8  8  0  8  5 : 95
 D: 0  1  0  4  0  5  5 12  4  1  1  5  0  0  3  0  9 10  2  0  6  7  7  1  0  0 : 83
 E: 0  6  2  7  6  2  6  4  0  0  2  3  6  8  3  0  6  3  9  2  2  0  0  0  0  5 : 82
 F: 0  0  2  0  6  2  6  4 13  2  1  5  7  0  0  3  3  3 10  3  0  9  7 10  1  0 : 97
 G: 0  1  3  4  5  9  8  0  8  5 11  3  0  0  0  0  0  9  1  3  1 11  0  0  4  7 : 93
 H: 5 10  5  0  1  0  1  0  6  0  5  3  9  4  0  3  6  0  0  1  2  7 11  0  1  4 : 84
 I: 4  7  0  0  6  5  7  8  2  0  3  2  9  2  1  1  0  2  1  7  0  3  1  9  2  4 : 86
 J: 3  0  2  0  3  5 11  0  2  4  9  0  0  1  5  9  9  2  0  4  4  5  1  0  0  0 : 79
 K: 3  5 12  7  3  0  5  6  7  3  1  1  0  1  1  8  2  5  2 11  0  2  5 10  0  0 :100
 L: 0  6  2  3  3 12  1  0  2  8  0  1  1  3  7  7  3  0  4  5  4  2  1  0  1  1 : 77
 M: 0  3  5  8  4  1  0  6  6  7  3  2  2  0  0  0 10  0  4  3 15  2  1  2  5  0 : 89
 N: 1  1  1  0  1  0  3  0  1  4 15  2  0  4 12  0  0  2  5  9  7  6  0  6  4 10 : 94
 O: 9 13  0  1  5  5  0  0  4  2  8  5  2  1  4  6  9  2  1  1  0  0  0  7  0  3 : 88
 P:10  1  4  5 10  0  1  8  6  0  1  2  2  6  6  2  1  9  7 10  2  2  1  0  0  0 : 96
 Q: 0  3  2  1  0  0  3  3  5  4  3  0  6  6 14  0  2  0  0  2  0  6  0  2  5 12 : 79
 R: 4  2  6  1  0  6  6  7  3  1  1  0  1  0  8  0  7  2  9  2  1  1  4  1  0  2 : 75
 S: 1  5  7  0  0  6  1  4  9  4  0  7  4  7  2  0  2  0  2  0  4  0  5  5 20  2 : 97
 T: 5  0  6  5 12  0  1  0  0  2  0  4  1  4  2 10  0  0  4  3  0  0  3  3  5  3 : 73
 U: 4  4  4  0  6  4  9  7  0  0  0  0  0  9  0  5  2  9  3  0  9  6  0  0  3  7 : 91
 V: 7  2 11  5  4  3  5  0  0  2  3  0 10  4  0 10  1  8  2  2  1  0  2  0  2  2 : 86
 W: 4  0  0  3  4  6  6  2  0  5  4  6  5  1  0  0  0  0  7  0  8  2  7  1  1  5 : 77
 X: 0  0  0  2  0  4  2  1  5  9  3  1  4  7  1  0  3  6  4  7  7  0  6  8  6  2 : 88
 Y: 8  2  1  8  6  6  5  0  2  0  2  0  8  0  4  2 12  4  0  5  7  0  1  3  1  3 : 90
 Z: 6  4  7  1  0  1  0  1  0  7  0  8  5  7  2  1  3  6  0  0  4  2  9  7  4  0 : 85

10
[[10, 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, 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]]
[[10, 0], [18, 5], [5, 15], [15, 11], [2, 18], [13, 17], [1, 21], [0, 7], [6, 2], 
[20, 24], [24, 23], [12, 1], [23, 16], [14, 8], [21, 9], [8, 4], [25, 20], [7, 19], 
[3, 14], [4, 10], [16, 6], [9, 13], [22, 3], [11, 12], [17, 25], [19, 22]]

K:  3  5 12  7  3  0  5  6  7  3  1  1  0  1  1  8  2  5  2 11  0  2  5 10  0  0 , 0
S:  6  1  4  9  4  0  7  4  7  2  0  2  0  2  0  4  0  5  5 20  2  1  5  7  0  0 , 5
F:  3  3  3 10  3  0  9  7 10  1  0  0  0  2  0  6  2  6  4 13  2  1  5  7  0  0 ,15
P:  2  2  6  6  2  1  9  7 10  2  2  1  0  0  0 10  1  4  5 10  0  1  8  6  0  1 ,11
C:  4  2  5  8  8  0  8  5  5  1  1  3  0  0  0  8  2  3  3 16  5  1  2  5  0  0 ,18
N:  2  5  9  7  6  0  6  4 10  1  1  1  0  1  0  3  0  1  4 15  2  0  4 12  0  0 ,17
B:  1  3  5 15  2  1  4  9  8  1  0  3  0  4  0  6  1  6  3  8  3  2  2  7  0  0 ,21
A:  2  1  8  7  6  0  3  7  7  1  0  3  0  2  0 12  2  4  3 13  2  1  5  5  0  0 , 7
G:  3  4  5  9  8  0  8  5 11  3  0  0  0  0  0  9  1  3  1 11  0  0  4  7  0  1 , 2
U:  3  7  4  4  4  0  6  4  9  7  0  0  0  0  0  9  0  5  2  9  3  0  9  6  0  0 ,24
Y:  3  1  3  8  2  1  8  6  6  5  0  2  0  2  0  8  0  4  2 12  4  0  5  7  0  1 ,23
M:  3  5  8  4  1  0  6  6  7  3  2  2  0  0  0 10  0  4  3 15  2  1  2  5  0  0 , 1
X:  3  6  4  7  7  0  6  8  6  2  0  0  0  2  0  4  2  1  5  9  3  1  4  7  1  0 ,16
O:  4  2  8  5  2  1  4  6  9  2  1  1  0  0  0  7  0  3  9 13  0  1  5  5  0  0 , 8
V:  2  3  0 10  4  0 10  1  8  2  2  1  0  2  0  2  2  7  2 11  5  4  3  5  0  0 , 9
I:  6  5  7  8  2  0  3  2  9  2  1  1  0  2  1  7  0  3  1  9  2  4  4  7  0  0 , 4
Z:  4  2  9  7  4  0  6  4  7  1  0  1  0  1  0  7  0  8  5  7  2  1  3  6  0  0 ,20
H:  1  2  7 11  0  1  4  5 10  5  0  1  0  1  0  6  0  5  3  9  4  0  3  6  0  0 ,19
D:  3  0  9 10  2  0  6  7  7  1  0  0  0  1  0  4  0  5  5 12  4  1  1  5  0  0 ,14
E:  2  3  6  8  3  0  6  3  9  2  2  0  0  0  0  5  0  6  2  7  6  2  6  4  0  0 ,10
Q:  3  3  5  4  3  0  6  6 14  0  2  0  0  2  0  6  0  2  5 12  0  3  2  1  0  0 , 6
J:  1  5  9  9  2  0  4  4  5  1  0  0  0  3  0  2  0  3  5 11  0  2  4  9  0  0 ,13
W:  3  4  6  6  2  0  5  4  6  5  1  0  0  0  0  7  0  8  2  7  1  1  5  4  0  0 , 3
L:  1  3  7  7  3  0  4  5  4  2  1  0  1  1  0  6  2  3  3 12  1  0  2  8  0  1 ,12
R:  2  4  2  6  1  0  6  6  7  3  1  1  0  1  0  8  0  7  2  9  2  1  1  4  1  0 ,25
T:  3  3  5  3  5  0  6  5 12  0  1  0  0  2  0  4  1  4  2 10  0  0  4  3  0  0 ,22
Interior Alphabet:  KRUYTBZHCNXFDJLPEVOAQSIWGM , Collision:  False

$ python3 kryha_tui.py -o -i KRUYTBZHCNXFDJLPEVOAQSIWGM -d -m \
	-s 8,6,5,6,4,5,6,5,7,5,9,4,5,4,6,5,4 < MSGS/konheim.cry \
	>| /tmp/konheim_pseudo.txt

$ python3 ic.py /tmp/konheim_pseudo.txt
QEBFPPRBLCMBOCLOJXKZBBSXIRXQFLKXKAMOBAFZQFLKEXPZLKZBOKBARPBO
...
0.06708669010983559

$ pypy3 hc_substi.py -c /tmp/konheim_pseudo.txt
0 17170.500085762113 THEISSUEOFPERFORMANCEEVALUATIONANDPREDICTIONHASCON
1 17105.547620270587 THEISSUEOFPERFORMANCEEKALUATIONANDPREDICTIONHASCON
2 17170.500085762113 THEISSUEOFPERFORMANCEEVALUATIONANDPREDICTIONHASCON
...
9 17170.500085762113 THEISSUEOFPERFORMANCEEVALUATIONANDPREDICTIONHASCON

Key:  XYZABCDEFGHIJKLMNOPQRSTUVW  Score:  17170.500085762113
THEISSUEOFPERFORMANCEEVALUATIONANDPREDICTIONHASCONCERNEDUSERSTHROUGHOUT
THEHISTORYOFCOMPUTEREVOLUTIONPERIODINFACTCOMMAASINANYOTHERTECHNOLOGICAL
DEVELOPMENTCOMMATHEISSUEISMOSTACUTEWHENTHET...

The key (Solution)

  • Internal Alphabet: WGMKRUYTBZHCNXFDJLPEVOAQSI
  • External Alphabet: ABCDEFGHIJKLMNOPQRSTUVWXYZ
  • Sectors: 8,6,5,6,4,5,6,5,7,5,9,4,5,4,6,5,4
  • Move the Internal disk before ciphering
Note: The machine does not appear to be a standard machine (see previous paragraphs). 

Standard Wheel: 3, 2, 3, 1, 2, 3, 2, 4, 2, 6, 1, 2, 1, 3, 2, 1, 5
Konheim Wheel : 8, 6, 5, 6, 4, 5, 6, 5, 7, 5, 9, 4, 5, 4, 6, 5, 4
-3            : 5, 3, 2, 3, 1, 2, 3, 2, 4, 1, 6, 1, 2, 1, 3, 2, 1
If we subtract 3 from each sector and shift the values to the left, we obtain the values of the sectors of the standard wheel!

The plain text

$ python3 kryha_tui.py -o \
	-i WGMKRUYTBZHCNXFDJLPEVOAQSI \
	-e ABCDEFGHIJKLMNOPQRSTUVWXYZ \
	-s 8,6,5,6,4,5,6,5,7,5,9,4,5,4,6,5,4  -d -m \
	< MSGS/konheim.cry | fold -w 70
THEISSUEOFPERFORMANCEEVALUATIONANDPREDICTIONHASCONCERNEDUSERSTHROUGHOU
TTHEHISTORYOFCOMPUTEREVOLUTIONPERIODINFACTCOMMAASINANYOTHERTECHNOLOGIC
ALDEVELOPMENTCOMMATHEISSUEISMOSTACUTEWHENTHETECHNOLOGYISYOUNGTHEPERSIS
TENTPURSUITOFPRODUCTSWITHIMPROVEDCOSTHYPHENPERFORMANCECHARACTERISTICST
HENCONSTANTLYLEADSTODESIGNSWITHUNTRIEDANDUNCERTAINFEATURESPERIODFROMTH
EINITIALCONCEPTIONOFASYSTEMSARCHITECTURALDESIGNTOITSDAILYOPERATIONAFTE
RINSTALLATIONPERIODINTHEEARLYPLANNINGPHASEOFANEWCOMPUTERSYSTEMPRODUCTC
OMMATHEMANUFACTURERMUSTUSUALLYMAKETWOTYPESOFPREDICTIONPERIODTHEFIRSTTY
PEISTOFORECASTTHENATUREOFAPPLICATIONSANDTHELEVELSOFSYSTEMWORKLOADSOFTH
ESEAPPLICATIONSPERIODHERETHETERMWORKLOADMEANSCOMMAINFORMALLYCOMMATHEAM
OUNTOFSERVICEREQUIREMENTSPLACEDONTHESYSTEMPERIODWESHALLELABORATEMOREON
WORKLOADCHARACTERIZATIONINSECTIONONEPOINTFOURPERIODTHESECONDTYPEOFPRED
ICTIONISCONCERNEDWITHTHECHOICEBETWEENARCHITECTURALDESIGNALTERNATIVESCO
MMABASEDONHARDWAREANDSOFTWARETECHNOLOGIESTHATWILLBEAVAILABLEINTHEDEVEL
OPMENTPERIODOFTHEPLANNEDSYSTEMPERIODHERETHECRITERIONOFSELECTIONISWHATW
ECALLCOSTHYPHGNPERFORMANCETRADEOFFPERIODTHEACCURACYOFSUCHPREDICTIONRES
TSTOACONSIDERABLEEXTENTONOURCAPABILITYOFMAPPINGTHEPERFORMANCECHARACTER
ISTICSOFTHESYSTEMCOMPONENTSINTOTHEOVERALLSYSTEMHYPHENLEVELPERFORMANCEC
HARACTERISTICSPERIODSUCHTRANSLATIONPROCEDURESAREBYNOMEANSSTRAIGHTFORWA
RDORWELLESTABLISHEDPERIODONCETHEARCHITECTURALDECISIONSHAVEBEENMADEANDT
HESYSTEMDESIGNANDIMPLEMENTATIONSTARTEDCOMMATHESCOPEOFPERFORMANCEPREDIC
TIONANDEVALUATIONBECOMESMORESPECIFICPERIODWHATISTHEBESTCHOICEOFMACHINE
ORGANIZATIONQUESTIONWHATISTHEOPERATINGSYSTEMTOSUPPORTANDWHATARETHEFUNC
TIONSITSHOULDPROVIDEQUESTIONTHEINTERACTIONSAMONGTHEOPERATINGSYSTEMCOMP
ONENTSHYPHENALGORITHMSFORJOBSCHEDULINGCOMMAPROCESSORSCHEDULINGCOMMAAND
STORAGEMANAGEMENTHYPHENMUSTBEUNDERSTOODCOMMAANDTHEIREFFECTSONTHEPERFOR
MANCEMUSTBEPREDICTEDPERIODTHETECHNIQUESUSEDFORPERFORMANCEEVALUATIONAND
PREDICTIONDURINGTHEDESIGNANDIMPLEMENTATIONPHASESRANGEFROMSIMPLEHANDCAL
CULATIONTOQUITEELABORATESIMULATIONPERIODCOMPARINGTHEPREDICTEDPERFORMAN
CEWITHTHEACTUALACHIEVEDPERFORMANCEOFTENREVEALSMAJORDEFECTSINTHEDESIGNO
RERRORSINTHESYSTEMPROGRAMMINGPERIODITISNOWAWIDELYACCEPTEDBETIEFTHATTHE
PERFORMANCEPREDICTIONANDEVALUATIONPROCESSSHOULDBEANINTEGRALPARTOFTHEDE
VELOPMENTEFFORTSTHROUGHOUTPERIOD

$ python3 kryha_tui.py -o -i WGMKRUYTBZHCNXFDJLPEVOAQSI \
	-w konheim  -d -m < MSGS/konheim.cry | fold -w 70 | head -5
THEISSUEOFPERFORMANCEEVALUATIONANDPREDICTIONHASCONCERNEDUSERSTHROUGHOU
TTHEHISTORYOFCOMPUTEREVOLUTIONPERIODINFACTCOMMAASINANYOTHERTECHNOLOGIC
ALDEVELOPMENTCOMMATHEISSUEISMOSTACUTEWHENTHETECHNOLOGYISYOUNGTHEPERSIS
TENTPURSUITOFPRODUCTSWITHIMPROVEDCOSTHYPHENPERFORMANCECHARACTERISTICST
HENCONSTANTLYLEADSTODESIGNSWITHUNTRIEDANDUNCERTAINFEATURESPERIODFROMTH

$ cat keys/konheim.whl
4, 2, 1, 2, 0, 1, 2, 1, 3, 1, 5, 0, 1, 0, 2, 1, 0

Reference

  • Cryptanlysis of a Kryha Machine, by A.G. Konheim. Article of a book: Lecture Notes in Computer Science, Edited by G. Goos and J. Hartmanis, 159, Cryptography, Proceeding, Burg Feuertein, (1982), Edited by Thomas Beth, Springer-Verlag, Berlin Heidelberg New York.