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Home Page Hebern's machines The 5 rotors machine home page
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IntroductionIn his secret report, Friedman had studied the possibility of reconstructing the wiring of the rotors in the case where the plain text of a message was known. Chance allowed him to demonstrate it. Brigman's featA cryptologist, Benjamin R. Brigman, was assigned to the army's signal service. Previously he had been in contact with Mr Hebern's company (Hebern Electric Code). This company sent a message to Brigman asking him to try to decipher it, probably to prove the hermiticity of the 5-rotor machine. The message was a poem presented with the separation of words. Here is this poem with the external key.
Key : GORDON-Z, write ELEANOR
Setting of wheels: 5-4-3-2-1, width #4 and #5 inverted.
KB BTR EKSMO DG TNS GDNX AAT XCN ICA
IDUSEA AJEF HI RGZ TKCD FP AQWDJ YD MON
ZK DA JGE ONW HXTCHQC WOSG WTMCP BN RF
GUUKHEJ II XHR WARHVH FQ QIKCW HGBQLY
PWVHHROT SMHLME PHGEEPNFY
Brigman, aided by the separation of words, searched a library for all the poems compatible with the structure of the cryptogram. Finally he found it:
KB BTR EKSMO DG TNS GDNX AAT XCN ICA
in the photo of her hero she can see
IDUSEA AJEF HI RGZ TKCD FP AQWDJ YD MON
things that do not show so clear to you
ZK DA JGE ONW HXTCHQC WOSG WTMCP BN RF
or me for her outlook ever seens to be
GUUKHEJ II XHR WARHVH FQ QIKCW HGBQLY
colored by her dreams in which golden
PWVHHROT SMHLME PHGEEPNFY
sunshine gleams endlessly
Note: While searching the web, I couldn't find the title of this poem. Some help would be appreciated! Friedman's featAfter Brigman joined the cipher department, he informed Friedman of his exploit. Friedman thus had an opportunity to prove his theories. At first, he rebuilds the basic cipher text sequences table. At the end of his analysis, Friedman was able to reconstruct the wiring of the rotors in position V. Finally, he found rotor I and the stator S equivalent to the action of the rotors in positions II, III and IV. Here is the solution he found: AL5: FIZDTMPGAVUBHQWLRCJXSOYKEN AL1: EHLNSVZCFKMQUYDJPRWAGIOTBX (234): YQWTBUMXKRDLIZGESFPVONHCAJ Note: Friedman did not use the key provided. He thought the key was generated by positioning the external key to GORDON-Z and entering the ELEANOR indicator. You can read Friedman's report for details of his rotor wiring recovery. IMPORTANT REMARK: Unfortunately the solution given by Friedman does not provide me with a correct decipherment. As a result I tried to reconstruct rotor wiring (I, II+III+IV, V). My reconstruction of the wiring of the rotor VIntroductionTo reconstruct the rotor V wiring, I used the established formulas to find the rotor wiring of a Hebern 1 rotor machine (plain known, using formula).
y = L[ R[ x + i ] - i ] The basic formula used here gives the ciphered letter (y) from the plain letter (x) undergoing the permutation of a rotor (R) and the Lampboard permutation (L). In our case, permutation L corresponds to pseudo stator equivalent to rotors I+II+III+IV. The permutation R corresponds to the inverse rotor V. Don’t forget we are in decipher mode. The value x corresponds to the decipherment of a letter of the cryptogram after elimination of the RFS permutation. The value y corresponds to the cipherment of the plain letter by the LFS permutation. There is a difficulty, we do not know the external key and therefore we do not know to which position the rotor I advances. This information is essential if we want to be able to consider the permutation L as a stator. A simple solution is to try the 26 possible positions of the RAW ratchet wheel. 1st step: Remove the influence of RFS and LFS permutationsNote: Friedman does not.C:\H1_TOOLS> type poem.cry KB BTR EKSMO DG TNS GDNX AAT XCN ICA IDUSEA AJEF HI RGZ TKCD FP AQWDJ YD MON ZK DA JGE ONW HXTCHQC WOSG WTMCP BN RF GUUKHEJ II XHR WARHVH FQ QIKCW HGBQLY PWVHHROT SMHLME PHGEEPNFY C:\H1_TOOLS> type poem.pln in the photo of her hero she can see things that do not show so clear to you or me for her outlook ever seens to be colored by her dreams in which golden sunshine gleams endlessly C:\H1_TOOLS> python striplamp.py TYOEUMXDFJQVKWBNSHCILRZAGP ^ poem.cry > poem_stri_L.cry C:\H1_TOOLS> type poem_stri_L.cry MOOAVDMQFCHYAPQYHPGXXAGSPTSXTHEQDXXJDIRTVYWAMSHIZXKNHJBHFCPWM HXJYDCPNRGASRKSNCQYNAFSZOPVIYEEMRDJTTGRVNXVRLRIKKTMSNRYOKUBZN LRRVCAQFRUFDZRYDDZPIB C:\H1_TOOLS> python hebern1_tui.py -R 0 -L LFS < poem.pln ^ > poem_stri_K.pln C:\H1_TOOLS> type poem_stri_K.pln YIFLSULOFOOZLSDLSDOBLSKXIBSSFLYIJBFLXFHOIOFBLOVBOKPSXDFOQOTOD MSZODLSDOTFPOOGSNSDBSSIBFOASKOPODSHAQLSDHDSXMBYIVLYKLJOPHSIBT IBLYISJPSXMBSIHPSBBPQ 2nd step: Try RAW at position NWe cut the messages into sections of 26 letters. We start by assuming that RAW is at position N. So the first 26 letters will be encrypted with rotor I stationary (and the following slices).
i abcde fghij klmno pqrst uvwxy z
01234 56789 01234 56789 01234 5 (25)
line 0,y yifls ulofo ozlsd lsdob lskxi b
x MOOAV DMQFC HYAPQ YHPGX XAGSP T
line 1,y ssfly ijbfl xfhoi ofblo vbokp s
x SXTHE QDXXJ DIRTV YWAMS HIZXK N
line 2,y xdfoq otodm szodl sdotf poogs n
x HJBHF CPWMH XJYDC PNRGA SRKSN C
line 3,y sdbss ibfoa skopo dshaq lsdhd s
x QYNAF SZOPV IYEEM RDJTT GRVNX V
line 4,y xmbyi vlykl jophs ibtib lyisj p
x RLRIK KTMSN RYOKU BZNLR RVCAQ F
line 5,y sxmbs ihpsb bpq
x RUFDZ RYDDZ PIB
If the other rotors (1,2,3,4) do not move,
The following formula applies:
y = L[ R[ x + i ] - i ]If we are in the same line, the permutation L (corresponding to rotors 1+2+3+4) is identical, therefore:
y = R[ x + i ] - i ]
So, let's take line number 0:
If y == L, we have
i=D, R[A+3]-3 = R[D]-3
i=G, R[M+6]-6 = R[S]-6
i=M, R[A+12]-12 = R[M]-12
i=P, R[Y+15]-15 = R[N]-15
i=U, R[X+20]-20 = R[R]-20
So, let's take line number 1:
If y == S, we have
i=A, R[S+0]-0 = R[S]
i=B, R[X+1]-1 = R[Y]-1
i=Z, R[N+25]-25 = R[M]-25
If we compare the results:
Line 0: R[S] = R[M]-6
Line 1: R[S] = R[M]-25
We can see that the results are incompatible. Consequently, we can say that every
line of 26 letters there is rotor I moving forward. The RAW ratchet wheel
position isn't N.
Note: In our first test we were unlucky because the indices i tested are in
the range 0 to 25. On the other hand, if we notice an inconsistency with i
in a range of 10 to 15, we drastically reduce the number of tests to be
carried out. Likewise if we have confirmations.
2nd step: Try other RAW positionsNow we need to try a different RAW position. In total there are 26 possible values. Now let's test the value found by Friedman: the RAW G position.
i abcde fghij klmno pqrst uvwxy z
01234 56789 01234 56789 01234 5 (25)
line 0,y y iflsu l
x M OOAVD M
line 1,y ofooz lsdls dobls kxibs sflyi j
x QFCHY APQYH PGXXA GSPTS XTHEQ D
line 2,y bflxf hoiof blovb okpsx dfoqo t
x XXJDI RTVYW AMSHI ZXKNH JBHFC P
line 3,y odmsz odlsd otfpo ogsns dbssi b
x WMHXJ YDCPN RGASR KSNCQ YNAFS Z
line 4,y foask opods haqls dhdsx mbyiv l
x OPVIY EEMRD JTTGR VNXVR LRIKK T
line 5,y ykljo phsib tibly isjps xmbsi h
x MSNRY OKUBZ NLRRV CAQFR UFDZR Y
line 6,y psbbp q
x DDZPI B
So, let's take line number 3:
If y == S, we have
i=T, R[Q+19]-19 = R[J]-19
i=W, R[A+22]-22 = R[W]-22
Then
R[J] = R[W]-3
If y == D, we have
i=G, R[D+6]-6 = R[J]-6
i=J, R[N+9]-9 = R[W]-9
Then
R[J] = R[W]-3
Here, not only do we have no contradiction, but rather a confirmation.
We will study all the combinations to be able to reconstruct the wiring
of the rotor V.
So, let's take line number 1: If y=O, then R[Q] = R[E]-2 = R[K]-3 = R[R]-11 So, let's take line number 2: If y=O, then R[Z]-6 = R[G]-8 = R[E]-12 = R[O]-15 = R[D]-22 = R[A]-24 If y=S, then R[B]-7 = R[Q]-16 = R[K]-19 = R[W]-23 So, let's take line number 3: If y=D, then R[N]-1 = R[J]-6 = R[W]-9 = R[S]-20 If y=O, then R[W] = R[D]-5 = R[B]-10 = R[F]-14 = R[Z]-15 If y=S, then R[A]-3 = R[X]-8 = R[E]-17 = R[J]-19 = R[W]-22 = R[C]-23 So, let's take line number 4: If y=O, then R[Q]-1 = R[J]-5 = R[T]-7We set R[A]=0 arbitrarily. Thanks to the previous deductions, we obtain the following wiring (partial wiring of the inverse of Rotor V):
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 D U Y O H K . . Q P . . . R . M X . . . . T F . I
We could continue (we didn't use lines 1 and 4 a lot and not lines 5 and
6 at all). If I can't find the complete wiring I can try all the remaining
possible combinations.
But I realized that my solution corresponds to
Friedman's solution to the key F and shifted:
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 D U Y O H K B V Q P W C L R G M X E S N J T F Z I
Here the cipher table of the Fridman's rotor V:
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 F I Z D T M P G A V U B H Q W L R C J X S O Y K E N B H Y C S L O F Z U T A G P V K Q B I W R N X J D M E C X B R K N E Y T S Z F O U J P A H V Q M W I C L D G D A Q J M D X S R Y E N T I O Z G U P L V H B K C F W E P I L C W R Q X D M S H N Y F T O K U G A J B E V Z F H K B V Q P W C L R G M X E S N J T F Z I A D U Y O G J A U P O V B K Q F L W D R M I S E Y H Z C T X N G H Z T O N U A J P E K V C Q L H R D X G Y B S W M F I I S N M T Z I O D J U B P K G Q C W F X A R V L E H Y J M L S Y H N C I T A O J F P B V E W Z Q U K D G X R K K R X G M B H S Z N I E O A U D V Y P T J C F W Q L L Q W F L A G R Y M H D N Z T C U X O S I B E V P K J M V E K Z F Q X L G C M Y S B T W N R H A D U O J I P N D J Y E P W K F B L X R A S V M Q G Z C T N I H O U O I X D O V J E A K W Q Z R U L P F Y B S M H G N T C P W C N U I D Z J V P Y Q T K O E X A R L G F M S B H Q B M T H C Y I U O X P S J N D W Z Q K F E L R A G V R L S G B X H T N W O R I M C V Y P J E D K Q Z F U A S R F A W G S M V N Q H L B U X O I D C J P Y E T Z K T E Z V F R L U M P G K A T W N H C B I O X D S Y J Q U Y U E Q K T L O F J Z S V M G B A H N W C R X I P D V T D P J S K N E I Y R U L F A Z G M V B Q W H O C X W C O I R J M D H X Q T K E Z Y F L U A P V G N B W S X N H Q I L C G W P S J D Y X E K T Z O U F M A V R B Y G P H K B F V O R I C X W D J S Y N T E L Z U Q A M Z O G J A E U N Q H B W V C I R X M S D K Y T P Z L FNote: We can notice that the wiring found by Friedman corresponds to the inverse of the Rotor V! Here the wiring of the Rotor V (not its inverse):
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 H M B S X P F Z V G N Q U E K J O T W C I L R D Y
We can verify our deductions, by removing the effect of the rotor V.
If our solution is correct we have a
series of simple transpositions.
C:\H5_TOOLS> type ROTORS\poem_V.rot
AHMBSXPFZVGNQUEKJOTWCILRDY
C:\H5_TOOLS> python hebern5_tui.py -I 0:0:0:0:poem_V ^
-E AAAAASG -d < poem.cry
OSWADTAMGMMYXRIXRIMZXRPTUZRRJXKUWTIPDIOXVXITPXBTXWMLDZ
IXAXEUCOADUCGACUMYPUUTAWACBAAOBQHRUYHGHBUNRWTUBNBUJVDE
ZATXQZCYVKJNMWNMSXNJCVJATMJNKIVSSIT
My reconstruction of the wiring of the rotor IWe have already studied the method to find the wiring of the rotor I (link). The following example uses the same recipes but it is more realistic. C:\H5_TOOLS> python hebern5_tui.py -I 0:0:0:0:poem_R5jf ^ -E AAAAASG -d < poem.cry -D ... Key: AAAAASG 001: K ->MOOOOO-> O AAAAATH 002: B ->OBBBBB-> S AAAAAUI 003: B ->OVVVVV-> W AAAAAVJ 004: T ->AXXXXX-> A AAAAAWK 005: R ->VHHHHH-> D AAAAAXL 006: E ->DFFFFF-> T AAAAAYM 007: K ->MXXXXX-> A AAAAAZN 008: S ->QMMMMM-> M BBAAAAO 009: M ->FJJJJJ-> G BBAAABP 010: O ->CMMMMM-> M BBAAACQ 011: D ->HMMMMM-> M BBAAADR 012: G ->YQQQQQ-> Y BBAAAES ... RAW OPQRSTUVWXYZABCDEFGHIJKLMN Rotor V ABCDEDGHIJKLMNOPQRSTUVWXYZ (line 0) p intheph pK yiflsul c’ OBVXHFX (line 1) p otoofherheroshecanseething PK ofoozlsdlsdoblskxibssflyij c’ MJMMQCDYCDYMECDUFTEDDWCGTV (line 2) p sthatdonotshowsocleartoyou pK bflxfhoiofblovbokpsxdfoqot c’ FYUHYOCNCYFUCAFCVMPHEYCXCS (line 3) p ormeforheroutlookeverseens pK odmszodlsdotfpoogsnsdbssib c’ TKOXHTKJXKTMQUTTFXVXKAXXOA (line 4) p tobecoloredbyherdreamsinwh pK foaskopodshaqlsdhdsxmbyivl c’ RLDTQLJLATIDVFTAIATWNHSEXF (line 5) p ichgoldensunshinegleamsend pK ykljophsibtiblyisjpsxmbsih c’ CREKQNGWIMVIMBCIWKNWXFMWIG (line 6) p lessly pK psbbpq c’ YNBBYFIf we apply the following formula (cf. the concept of rotor) which corresponds to the value of the encrypted letter y(i) in relation to the plain letter x(i) and the offset D[m]: y(i) = x(i) + D[ x(i) + i ]We obtain the final formula (in the case where c' has the same value): c’ = pK + D[ pk + i ]1) In the case where c’ == M, we have: (i=1) M = O + D[ O + 1 ], 12 = 14 + D[P] => D[P] = 24 (i=2) M = P + D[ P + 2 ], 12 = 15 + D[R] => D[R] = 23 (i=3) M = T + D[ T + 3 ], 12 = 19 + D[W] => D[W] = 19 (i=5) M = B + D[ B + 5 ], 12 = 1 + D[G] => D[G] = 112) In the case where c' == J, we have: (i=1) J = F + D[ F + 1 ], 9 = 5 + D[G] => D[G] = 4 (i=4) J = P + D[ P + 4 ], 9 = 15 + D[T] => D[T] = 20We still don't know D[G], but we know two values of its wiring (relatively), if c' == M, then D[Q] = 11 but if c' == J, then D[G] = 4. Conclusion: if c' == J is the reference, then c’[M] = c’[J]-7. We can aggregate the two series by taking the offsets where c’= J is taken as a reference: D[G]=4, D[P]=17, D[R]=16, D[T]=20, D[W]=12 If we continue with other series where the value of c' is identical, we can reconstruct all the shifts... but in relative terms! We can also reconstruct S but it is simpler to do so after we have finished reconstituting the rotor R1. We obtain the following displacements for permutation R (that of Rotor I).
m 0 1 2 3 4 5 6 7 8 9 10 11 12
A B C D E F G H I J K L M
R 9 13 21 1 12 19 4 10 3 22 15 7 14
m 13 14 15 16 17 18 19 20 21 22 23 24 25
N O P Q R S T U V W X Y Z
R 6 24 17 11 16 2 20 8 0 12 18 5 23
From the formula PI[x] = x + D[x] (cf. The concept of rotor), we can reconstruct the wiring for the rotor I:
J O X E Q Y K R L F Z S A T M G B H U N C V I P D W
My reconstruction of the wiring of the stator SWe have already studied the method to find the wiring of the rotor S, equivalent to rotors II+III+IV (link). The following example uses the same recipes. First, we apply the action of Rotor I to the plain text after the action of permutation K (Keyboard). C:\H1_TOOLS> python hebern5_tui.py -I poem_R1jf:0:0:0:0 ^ -E AAAAASG < poem.pln -D ... Key: AAAAASG 001: I ->YDDDDD-> E AAAAATH 002: N ->ILLLLL-> V AAAAAUI 003: T ->FYYYYY-> G AAAAAVJ 004: H ->LSSSSS-> C AAAAAWK 005: E ->SUUUUU-> L AAAAAXL 006: P ->UCCCCC-> O AAAAAYM 007: H ->LSSSSS-> C AAAAAZN 008: O ->OFFFFF-> M BBAAAAO 009: T ->FJJJJJ-> J BBAAABP (line 0) p intheph pK yiflsul pKR(0) DLYSUCS c’ OBVXHFX (line 1) p otoofherheroshecanseething PK ofoozlsdlsdoblskxibssflyij pKR(1) FJFFIZMPZMPFWZMRCEWMMJZVEY c’ MJMMQCDYCDYMECDUFTEDDWCGTV (line 2) p sthatdonotshowsocleartoyou pK bflxfhoiofblovbokpsxdfoqot Pkr(2) CPRUPDZXZPCRZNCZYFAUWPZSZT c’ FYUHYOCNCYFUCAFCVMPHEYCXCS (line 3) p ormeforheroutlookeverseens pK odmszodlsdotfpoogsnsdbssib pKR(3) EHDSUEHJSHEFIREECSYSHNSSPN c’ TKOXHTKJXKTMQUTTFXVXKAXXOA (line 4) p tobecoloredbyherdreamsinwh pK foaskopodshaqlsdhdsxmbyivl pKR(4) BQMEIQJQNEOMYCENONEKXUTWSC c’ RLDTQLJLATIDVFTAIATWNHSEXF (line 5) p ichgoldensunshinegleamsend pK ykljophsibtiblyisjpsxmbsih pKR(5) ZBWHIXVKOFYOFWZOKHXKSCFKOV c’ CREKQNGWIMVIMBCIWKNWXFMWIG (line 6) p lessly pK psbbpq pKR(6) PXLLPC c’ YNBBYF The permutation S is easily deduced: Since S is the same for every 26 letters, in the following expression, i can have any value. (pKR(i))S = c’Thus the following correspondences are valid whatever the value of i: d.S = o, l.S = b, y.S = v, ...According to the formula “y = PI[x]”, we therefore have: o = S[d], b = S[l], v = S[y], x = S[s], h = S[u], f = S[c], m = S[f], j = S[j], ...Ultimately, the permutation S is as follows: 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 P R F O T M Z K Q J W B D A I Y L U X S H G E N V C My solution (my feat ;-)Here are the rotor wirings that I reconstructed:
Rotor V : AHMBSXPFZVGNQUEKJOTWCILRDY (poem_R5jf)
Rotor I : JOXEQYKRLFZSATMGBHUNCVIPDW (poem_R1jf)
Rotor II+III+IV: PRFOTMZKQJWBDAIYLUXSHGENVC (poem_234jf)
Thanks to them, I can decipher the cryptogram:
C:\H5_TOOLS> python hebern5_tui.py ^ -I poem_R1jf:0:poem_234jf:0:poem_R5jf ^ -E AAAAASG -d < poem.cry INTHEPHOTOOFHERHEROSHECANSEEDHINGSTHATDONOTSHOWSOCLEARTOYOU ORMEFORHEROUTLOOKEVERSEEMSTOBECOLOREDBYHERDREAMSINWHICHGOLD ENSUNSFINEGLEAMSENDLESSLYNote: We notice that there are some errors. Where do they come from?
Another cryptogram encrypted with the same keyIn the NSA archives, we find a cryptogram undoubtedly created by Friedman (see the Web links section bellow). This cryptogram was encrypted with the same key as the poem deciphered by Brigman. This cryptogram is undoubtedly given as an exercise given to cryptology students.
C:\H5_TOOLS> type poem_msg2.cry QAHOX OSXAO WYKVN XDKJG TTIKN TIHKB ZDFQK FYCBA DDXXI WNIHY YLZTV PVWAG WHWEQ KRNZY MTFHZ OCHPO YXKXB CXYVW PPXUA GYVRQ QYHEW UPMNS TOGHB AEKLF UYYTR LGWKQ OC C:\H5_TOOLS> python hebern5_tui.py ^ -I poem_R1jf:0:poem_234jf:0:poem_R5jf ^ -E AAAAASG -d poem_msg2.cry POEONCEWROTEQUOTETHEREISNOTHINGTHEINGENUITYOFMANCANINVENTTHAT THEINGENUITYOFMANCANNOTUNRAVELUNQUOTEDOESNTTHISHELPTOPROVEHIS CONTENTION Back to Friedman rotorsI was very intrigued by the fact that I could not use Friedman's rotors to decipher the poem ciphered. Were they false? I manually tried to find a transformation (inversion, shift, mirror) that would allow me to move from Friedman's wiring to those I discovered. I did not succeed. I then (for other reasons) developed a signature program which allows me to identify a rotor independently of whether it has undergone a rotation or a twist effect. This program is described in my page rotor. It followed the same approach as my manual tests. I tested my program on Friedman rotors. I was happily surprised that they had the same signatures as mine. I incidentally concluded that my manual tests were marred by errors. So the twist effect was the main reason they differed. Here are the signatures of rotors I, V and II+III+IV in mode direct and in reverse mode. Rotor I, with my wiring (JOXE) and Friedman's (EHLN). C:\H1_TOOLS> python signature.py -R ^ =JOXEQYKRLFZSATMGBHUNCVIPDW -r -S -p PI direct PI : JOXEQYKRLFZSATMGBHUNCVIPDW Signature: [4, 8, 6, 11, 7, 11, 6, 19, 19, 19] PI reverse PI : MQUYDJPRWAGIOTBXEHLNSVZCFK Signature: [1, 3, 3, 3, 4, 5, 5, 1, 4, 3] C:\H1_TOOLS> python signature.py -R =EHLNSVZCFKMQUYDJPRWAGIOTBX -S -p PI : EHLNSVZCFKMQUYDJPRWAGIOTBX Signature: [1, 3, 3, 3, 4, 5, 5, 1, 4, 3]Rotor V, with my wiring (AHMB) and Friedman’s (FIZDT) C:\H1_TOOLS> python signature.py ^ -R =AHMBSXPFZVGNQUEKJOTWCILRDY -r -S -p PI direct PI : AHMBSXPFZVGNQUEKJOTWCILRDY Signature: [1, 6, 4, 14, 16, 4, 17, 15, 19, 21] PI reverse PI : ADUYOHKBVQPWCLRGMXESNJTFZI Signature: [2, 16, 3, 15, 18, 2, 16, 19, 20, 24] C:\H1_TOOLS> python signature.py ^ -R =FIZDTMPGAVUBHQWLRCJXSOYKEN -S -p PI : FIZDTMPGAVUBHQWLRCJXSOYKEN Signature: [2, 16, 3, 15, 18, 2, 16, 19, 20, 24]Rotor II+III+IV, with my wiring (PRFO) and Friedman’s (YQWT) C:\H1_TOOLS> python signature.py ^ -R =PRFOTMZKQJWBDAIYLUXSHGENVC -r -S -p PI direct PI : PRFOTMZKQJWBDAIYLUXSHGENVC Signature: [1, 13, 8, 4, 18, 12, 10, 5, 18, 12] PI reverse PI : NLZMWCVUOJHQFXDAIBTERYKSPG Signature: [5, 18, 24, 19, 20, 23, 8, 14, 17, 5] C:\H1_TOOLS> python signature.py ^ -R =YQWTBUMXKRDLIZGESFPVONHCAJ -S -p PI : YQWTBUMXKRDLIZGESFPVONHCAJ Signature: [5, 18, 24, 19, 20, 23, 8, 14, 17, 5]Conclusion : The Friedman’s rotors had the same signatures as mine but they are reversed compared to mine. How do I get from Friedman's wiring to mine? All I had to do was inspect the cipher tables for each rotor. Here is the answer :
Conclusion: Friedman has found the wiring of the rotors. They differ from mine because of the twist effect. We note that they are all in reverse mode. On the other hand, I still don't understand how Friedman could decipher a message using his rotors. References
Web Links
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