Bored Guru Challenge

Solving the Bored Guru Cipher Challenge

Here is the ciphertext of the Bored Guru challenge with one minor typo in the original corrected: (The sequence CTKHUWIL should have had a space and been CTKH UWIL.)

WNXW, BCOIDV, PQOV EJSKII VEPEX HBMZF WWMVOZ UUBK SBYS. WBGGY EZTB EM CTKH UWIL, LSI YLLQ, ZBKQN IBLZ YXU EVNH QC PDFP LTYRPT OQAV YX NPX TT SOV CNDR XZ EVONV. NJDMM GMHLWZX KJ PKEK VAKO RLJR RMG OTUA KOD AZZ NCQ OBAOKO QP MWIGYLR JLR WOBWIZ IA NIULU MZ UGXGYL

This cipher does not yeild to the methods used to crack a simple substitution cipher so we will have to make some assumptions about what kind of code it is. We will assume that it is polyalphabetic, in that it uses a series of coding alphabets insetad of just one. We will also assume that the cycle of alphabets used repeats at least once in the message, because if it does not repeat at least once then the cipher would be of the type known as a one-time pad, and would not be crackable from a single sample of text such as this.

The first question, therefore, is how often does the cycle of alphabets repeat? Here's how we go about discovering this important piece of information. (See also attacking polyalphabetic ciphers with coincidence testing.)

The book Crytanalysis for Microcomputers by Caxton C. Foster has a number of handy appendices that list things of use for cryptogram solvers. Among the things included is a list of most common words by length. According to this list, there are 234 frequently occuring four-letter words in English text. The list begins with: THAT, WITH, THIS, FROM, HAVE, THEY, WERE, BEEN, WHEN, WILL, ...

We will begin by assuming that at least one of those words appears at least once somewhere in the message. Since it would be highly unusual, and grammatically questionable to begin a sentence with "THAT" followed immediately by a comma, we'll skip that first four letter word for now and move on to the next one: PQOV.

Now this is where the circular slide rule comes in handy. You can print out two copies of this image and cut the outer ring from one of them. Pin the two together in the center so the smaller ring spins freely inside the outer ring.

We are going to assume, for the moment, that PQOV translates to THAT and see if that assumption leads us anywhere. If it doesn't then we will try using the next four letter word, UUBK, for THAT and see if that cracks it open. Eventually, one of our four letter words from the list of common words will click and give us a crack to slip our fingers into.

We spin the dial on the circular slide rule so that the first plaintext letter T on the inner ring lines up with the first ciphertext letter P on the outer ring. Now holding the two sections of the circular slide rule tightly so they don't slip we look up each cipher text letter that comes after our chosen word on the outer ring and jot down the plaintext letter from the inner ring beneath it. The result looks like this:

WNXW, BCOIDV, PQOV EJSKII VEPEX HBMZF WWMVOZ UUBK SBYS. T INWOMM ZITIB LFQDJ AAQZSD YYFO WFCW WBGGY EZTB EM CTKH UWIL, LSI YLLQ, ZBKQN IBLZ YXU EVNH QC AFKKC IDXF IQ GXOL YAMP PWM CPPU DFOUR MFPD CBY IZRL UG PDFP LTYRPT OQAV YX NPX TT SOV CNDR XZ EVONV. NJDMM THJT PXCVTX SUEZ CB RTB XX WSZ GRHV BD IZSRZ RNHQQ GMHLWZX KJ PKEK VAKO RLJR RMG OTUA KOD AZZ NCQ OBAOKO KQLPADB ON TOIO ZEOS VPNV VQK SXYE OSH EDD RGU SFESOS QP MWIGYLR JLR WOBWIZ IA NIULU MZ UGXGYL. UT QAMKCPV NPV TYPQPX PS CAKKX TD DPFGTZ

Except for the word SUEZ, which is random and meaningless (as will become clear in a moment) that doesn't look particularly promising, but then we didn't expect it to yet. We need to add some more possibilities from our assumed word "THAT. We do that by writing H, the next letter of "THAT" under PQOV. Then, just as we did before, we spin the circular slide rule until Q on the outer circle lines up with H on the inner circle. Once more we look up the ciphertext letters on the outer wheel and jot down the plaintext letters from the inner wheel in a second row. This is definitely the most boring and time-consuming part of the job. Once we get a finger hold into the key length it will go much faster. Anyway, the result of this second operation looks like this:

WNXW, BCOIDV, PQOV EJSKII VEPEX HBMZF WWMVOZ UUBK SBYS. T INWOMM ZITIB LFQDJ AAQZSD YYFO WFCW H VAJBZZ MVGVO YSDQW NNDMFQ LLSB JSPJ WBGGY EZTB EM CTKH UWIL, LSI YLLQ, ZBKQN IBLZ YXU EVNH QC AFKKC IDXF IQ GXOL YAMP PWM CPPU DFOUR MFPD CBY IZRL UG NSXXP VQKS VD TKBY LNZC CJZ PCCH QSBHE ZSCQ POL VMEY HT PDFP LTYRPT OQAV YX NPX TT SOV CNDR XZ EVONV. NJDMM THJT PXCVTX SUEZ CB RTB XX WSZ GRHV BD IZSRZ RNHQQ GUWG CKPIGK FHRM PO EGO KK JFM TEUI OQ VMFEM EAUDD GMHLWZX KJ PKEK VAKO RLJR RMG OTUA KOD AZZ NCQ OBAOKO KQLPADB ON TOIO ZEOS VPNV VQK SXYE OSH EDD RGU SFESOS XDYCNQO BA GBVB MRBF ICAI IDX FKLR BFU RQQ ETH FSRFBF QP MWIGYLR JLR WOBWIZ IA NIULU MZ UGXGYL. UT QAMKCPV NPV TYPQPX PS CAKKX TD DPFGTZ HG DNZXPCI ACI GLCDCK CF PNXXK GQ QCSTGM

Now what we are looking for is words, or pieces of words that read correctly diagonally down and to the right, just as our assumed word "THAT" reads diagonally down and to the right. Therefore we can eliminate those combinations that are obviously impossible to lighten our work load for the next round. For example, under the ciphretext word VEPEX in the first row, we have the two diagonal letters ZV (in red), the Z directly under the V of VEPEX, and the V on the second row down beneath the E of VEPEX. No word can possibly begin with ZV, so this is an impossible combination. In this next copy of the chart so far I will remove the impossible combinations to show what I mean.

WNXW, BCOIDV, PQOV EJSKII VEPEX HBMZF WWMVOZ UUBK SBYS. T I--O-M -I-IB -F--J AA---D ---O ---W H VA--Z- M-G-O Y-D-- NND--- L--- J--- WBGGY EZTB EM CTKH UWIL, LSI YLLQ, ZBKQN IBLZ YXU EVNH QC A---C I--F IQ --OL -A-P --M --PU --OUR ---D C-Y I-RL -- NS--- VQ-- VD T--Y L-Z- C-- P--H Q--HE Z--- PO- VM-Y H- PDFP LTYRPT OQAV YX NPX TT SOV CNDR XZ EVONV. NJDMM T--T --C--X SUEZ -B -TB -X --Z GRHV -D I-SRZ RN--Q -U-- C--I-- FHRM P- E-O K- J-- TEUI O- VM-EM EAU-- GMHLWZX KJ PKEK VAKO RLJR RMG OTUA KOD AZZ NCQ OBAOKO --LPADB -N -OIO -EOS -PNV --K S--E OSH E-D --U -FE-OS X--CNQO B- G-VB M-BF I-AI I-- FK-- BFU RQ- E-- F-RF-F QP MWIGYLR JLR WOBWIZ IA NIULU MZ UGXGYL. -T -A-KCPV -PV -----X -S -A--X -D --F--Z H- D-Z-PCI A-I G----- C- P-X-- G- Q--T--

The letters that are left are at least possible letter sequences, read diagonally down. Now we can repeat the process with the third letter of our assumed word "THAT", adding a third row of candidate plaintext letters. This time we only look up the ciphertext letters that have valid two-letter diagonals beneath them. We spin the inner dial so that O is over A, hold the wheel tightly to prevent slippage, and build the third row as before. The result looks like this:

WNXW, BCOIDV, PQOV EJSKII VEPEX HBMZF WWMVOZ UUBK SBYS. T I--O-M -I-IB -F--J AA---D ---O ---W H VA--Z- M-G-O Y-D-- NND--- L--- J--- A -VE--U -Y-Q- TN-L- -IYH-- -G-- -N-- WBGGY EZTB EM CTKH UWIL, LSI YLLQ, ZBKQN IBLZ YXU EVNH QC A---C I--F IQ --OL -A-P --M --PU --OUR ---D C-Y I-RL -- NS--- VQ-- VD T--Y L-Z- C-- P--H Q--HE Z--- PO- VM-Y H- -NS-- -LF- -Y OF-- G--X -E- -X-- LN--Z -N-- -JG -HZ- CO PDFP LTYRPT OQAV YX NPX TT SOV CNDR XZ EVONV. NJDMM T--T --C--X SUEZ -B -TB -X --Z GRHV -D I-SRZ RN--Q -U-- C--I-- FHRM P- E-O K- J-- TEUI O- VM-EM EAU-- --R- -F--B- -CMH KJ -B- FF -A- -ZPD JL -HA-H ZVPY- GMHLWZX KJ PKEK VAKO RLJR RMG OTUA KOD AZZ NCQ OBAOKO --LPADB -N -OIO -EOS -PNV --K S--E OSH E-D --U -FE-OS X--CNQO B- G-VB M-BF I-AI I-- FK-- BFU RQ- E-- F-RF-F -Y--ILJ WV -W-W HM-A DX-D DY- -FG- -AP MLL -O- -N-AW- QP MWIGYLR JLR WOBWIZ IA NIULU MZ UGXGYL. -T -A-KCPV -PV -----X -S -A--X -D --F--Z H- D-Z-PCI A-I G----- C- P-X-- G- Q--T-- CB -I-S-XD VX- IA---- -M -U-X- -L -S--K-

Once again we have created some impossible combinations. Under the cipher text ZBKQN in the second row we have a diagonal starting below the K that reads OHZ (in red). That is an impossible combination in English so we can strike it out. Likewise the TCF diagonal under the second P of PDFP in the third row (also in red) can be struck out. Notice under UUBK in the first row the K has an O under it, and it is possible for a word to end in O, but the next part of the diagonal starts a new word under SBYS, and the combination JN cannot possibly start a word, so the whole diagonal is struck out, including the O under UUBK.

Deleteing the impossible letter diagonals we end up with the chart looking like this:

WNXW, BCOIDV, PQOV EJSKII VEPEX HBMZF WWMVOZ UUBK SBYS. T -----M ---I- ----- A----- ---- ---- H ------ M---O ----- -N---- ---- ---- A ------ -Y--- T---- --Y--- ---- ---- WBGGY EZTB EM CTKH UWIL, LSI YLLQ, ZBKQN IBLZ YXU EVNH QC A---- ---- I- --OL ---P --- --P- ---U- ---- C-- ---- -- -S--- ---- -D ---Y L--- C-- ---H ----E ---- -O- ---- -- --S-- ---- -- O--- GI-- -E- ---- L---- ---- --G ---- -- PDFP LTYRPT OQAV YX NPX TT SOV CNDR XZ EVONV. NJDMM T--- --C--- --E- -B -T- -X --- GRHV -- I-SRZ R---- -U-- ---I-- ---M -- E-O -- J-- -EUI O- -M-EM EA--- --R- ----B- ---- K- -B- F- -A- --PD JL --A-H ZVP-- GMHLWZX KJ PKEK VAKO RLJR RMG OTUA KOD AZZ NCQ OBAOKO --L--D- -- --I- -EO- -PN- --- ---E -S- --D --- -F--O- ---C--O -- ---B --BF --AI --- ---- B-U --- E-- --R--F ----I-- W- ---- H--A D--D D-- ---- -A- M-- -O- ---A-- QP MWIGYLR JLR WOBWIZ IA NIULU MZ UGXGYL. -T -----PV --V ------ -S ----- -- ------ -- D-----I A-- G----- -- P---- -- ------ C- -I----- VX- -A---- -- -U--- -- ------

Now we've really narrowed down the possible locations of the assumed repeat of the key. The remaining diagonals are at least sensible letter combinations in English. So now we can add the final letter of our assumed word "THAT" and see what turns up on the other diagonals. That "ANY" just below WWMVOZ on the first row is looking pretty promising, as does the complete word "COG" under YXU on the second row. We may be on to something! The process is the same as before, lining up cipher text V above plaintext T on the wheel and completing the diagonals we already have.

WNXW, BCOIDV, PQOV EJSKII VEPEX HBMZF WWMVOZ UUBK SBYS. T -----M ---I- ----- A----- ---- ---- H ------ M---O ----- -N---- ---- ---- A ------ -Y--- T---- --Y--- ---- ---- T ------ --N-- -Z--- ---T-- ---- ---- WBGGY EZTB EM CTKH UWIL, LSI YLLQ, ZBKQN IBLZ YXU EVNH QC A---- ---- I- --OL ---P --- --P- ---U- ---- C-- ---- -- -S--- ---- -D ---Y L--- C-- ---H ----E ---- -O- ---- -- --S-- ---- -- O--- GI-- -E- ---- L---- ---- --G ---- -- ---E- ---- -- -R-- -UG- --G ---- -Z--- G--- --- C--- -- PDFP LTYRPT OQAV YX NPX TT SOV CNDR XZ EVONV. NJDMM T--- --C--- --E- -B -T- -X --- GRHV -- I-SRZ R---- -U-- ---I-- ---M -- E-O -- J-- -EUI O- -M-EM EA--- --R- ----B- ---- K- -B- F- -A- --PD JL --A-H ZVP-- ---N -----R ---- -V --V -R --T ---P VX C--L- LHBK- GMHLWZX KJ PKEK VAKO RLJR RMG OTUA KOD AZZ NCQ OBAOKO --L--D- -- --I- -EO- -PN- --- ---E -S- --D --- -F--O- ---C--O -- ---B --BF --AI --- ---- B-U --- E-- --R--F ----I-- W- ---- H--A D--D D-- ---- -A- M-- -O- ---A-- -----X- -H ---- -Y-- PJ-- PK- ---- --B -X- --O ----I- QP MWIGYLR JLR WOBWIZ IA NIULU MZ UGXGYL. -T -----PV --V ------ -S ----- -- ------ -- D-----I A-- G----- -- P---- -- ------ C- -I----- VX- -A---- -- -U--- -- ------ -N --G---- -JP --Z--- -- --S-- -- ------

Our diagonal ANY got clobbered since there is no word ANYT?? in English. COG still looks good, and we picked up a little surprise. The first word in the third row reads TURN which is a perfectly good word. The test of whether it's real or not is to count how far it is from our assumed word THAT, and then count that many letters after TURN to see if there is a second repeat of the cycle. If there is we can be pretty sure we've discovered the key length. Let's try it:

Counting 78 letters from THAT to TURN, we count another 78 and find "-FRAI-", which looks suspiciously like "AFRAID". I'd say we found the key length of 78 letters. COG doesn't fit that key length, but maybe it fits half the key length, giving an actual key of length 39. A quick check shows that to be a washout since at 39 letters past TURN there is nothing valid.

Now that we know the key length we can toss out all the leftover garbage and write our solution so far under the cipher text as follows:

WNXW, BCOIDV, PQOV EJSKII VEPEX HBMZF WWMVOZ UUBK SBYS. ---- ------ THAT ------ ----- ----- ------ ---- ---- WBGGY EZTB EM CTKH UWIL, LSI YLLQ, ZBKQN IBLZ YXU EVNH QC ----- ---- -- ---- ---- --- ---- ----- ---- --- ---- -- PDFP LTYRPT OQAV YX NPX TT SOV CNDR XZ EVONV. NJDMM TURN ----- ---- -- --- -- --- ---- -- ----- ----- GMHLWZX KJ PKEK VAKO RLJR RMG OTUA KOD AZZ NCQ OBAOKO ------- -- ---- ---- ---- --- ---- --- --- --- -FRAI- QP MWIGYLR JLR WOBWIZ IA NIULU MZ UGXGYL. -- ------- --- ------ -- ----- -- ------

I've marked the first letter of a new key cycle in red so I don't loose track of where it is as more of the message gets filled in.

First, we can go ahead and fill out the word AFRAID, since that's the only possible English word with those letters. After doing so we will set the slide rule with O of the cipher text over our new letter A in AFRAID. Then, holding the wheel stationary again, we will look up the plaintext letters for the letter to the left of each of our marked red letters and fill those in. Then we will repeat the process for the final D of AFRAID and add one more letter after THAT and TURN by using the wheel as before.

WNXW, BCOIDV, PQOV EJSKII VEPEX HBMZF WWMVOZ UUBK SBYS. ---- -----H THAT T----- ----- ----- ------ ---- ---- WBGGY EZTB EM CTKH UWIL, LSI YLLQ, ZBKQN IBLZ YXU EVNH QC ----- ---- -- ---- ---- --- ---- ----- ---- --- ---- -O PDFP LTYRPT OQAV YX NPX TT SOV CNDR XZ EVONV. NJDMM TURN A---- ---- -- --- -- --- ---- -- ----- ----- GMHLWZX KJ PKEK VAKO RLJR RMG OTUA KOD AZZ NCQ OBAOKO ------- -- ---- ---- ---- --- ---- --- --- --- AFRAID QP MWIGYLR JLR WOBWIZ IA NIULU MZ UGXGYL. -- ------- --- ------ -- ----- -- ------

With our two new letters in AFRAID we can put O over A and get O for C in QC just before TURN and H for V in BCOIDV just before THAT. The two letter word before TURN is probably TO, but might also be SO, DO, or GO. Let's try TO first. Putting Q over T on the wheel it looks like the D in BCOIDV will have to be G. A word that ends in GH is reasonable, so we try the word before AFRAID. That Q turns into T, according to the wheel, so NCQ looks like --T. Now the chart looks like this:

WNXW, BCOIDV, PQOV EJSKII VEPEX HBMZF WWMVOZ UUBK SBYS. ---- ----GH THAT T----- ----- ----- ------ ---- ---- WBGGY EZTB EM CTKH UWIL, LSI YLLQ, ZBKQN IBLZ YXU EVNH QC ----- ---- -- ---- ---- --- ---- ----- ---- --- ---- TO PDFP LTYRPT OQAV YX NPX TT SOV CNDR XZ EVONV. NJDMM TURN A---- ---- -- --- -- --- ---- -- ----- ----- GMHLWZX KJ PKEK VAKO RLJR RMG OTUA KOD AZZ NCQ OBAOKO ------- -- ---- ---- ---- --- ---- --- --- --T AFRAID QP MWIGYLR JLR WOBWIZ IA NIULU MZ UGXGYL. -- ------- --- ------ -- ----- -- ------

Consulting the list of three-letter words that end in T, it looks like we have our choice of "HIT AFRAID", "ART AFRAID", "YET AFRAID", "LET AFRAID", "ACT AFRAID", "CUT AFRAID", "NOT AFRAID", "SET AFRAID". "PUT AFRAID", "GOT AFRAID" and a few others that pretty far-fetched. I'm going to place my bet on "NOT AFRAID" and "ACT AFRAID" as the two most likely with "YET AFRAID" in reserve in case those both fail. So lets try the first one. Once more, we use the assumed letters NO in NOT to turn the wheel and add a few more letters in front of each of the secgtions we have done so far.

By now you should know how to use the wheel so I won't bother with all the minute details.

WNXW, BCOIDV, PQOV EJSKII VEPEX HBMZF WWMVOZ UUBK SBYS. ---- --OUGH THAT T----- ----- ----- ------ ---- ---- WBGGY EZTB EM CTKH UWIL, LSI YLLQ, ZBKQN IBLZ YXU EVNH QC ----- ---- -- ---- ---- --- ---- ----- ---- --- --NT TO PDFP LTYRPT OQAV YX NPX TT SOV CNDR XZ EVONV. NJDMM TURN A---- ---- -- --- -- --- ---- -- ----- ----- GMHLWZX KJ PKEK VAKO RLJR RMG OTUA KOD AZZ NCQ OBAOKO ------- -- ---- ---- ---- --- ---- --- --- NOT AFRAID QP MWIGYLR JLR WOBWIZ IA NIULU MZ UGXGYL. -- ------- --- ------ -- ----- -- ------

That --OUGH word has to be THOUGH, and I'm betting the word after TURN is AROUND. It fits, so I'm going to try it and see what happens.

WNXW, BCOIDV, PQOV EJSKII VEPEX HBMZF WWMVOZ UUBK SBYS. ---- THOUGH THAT THINGS ----- ----- ------ ---- ---- WBGGY EZTB EM CTKH UWIL, LSI YLLQ, ZBKQN IBLZ YXU EVNH QC ----- ---- -- ---- ---- --- ---- ----- ---- --- WANT TO PDFP LTYRPT OQAV YX NPX TT SOV CNDR XZ EVONV. NJDMM TURN AROUND ---- -- --- -- --- ---- -- ----- ----- GMHLWZX KJ PKEK VAKO RLJR RMG OTUA KOD AZZ NCQ OBAOKO ------- -- ---- ---- ---- --- ---- --- -RE NOT AFRAID QP MWIGYLR JLR WOBWIZ IA NIULU MZ UGXGYL. OF PUS---- --- ------ -- ----- -- ------

By now it should be obvious that the rest of the solution is just a matter of playing the three repetitions of the key back and forth against each other to get enough letters to guess a new word, and using the letters from that new word to fill in a few more letters in the other two repetitions of the key cycle. A quick glance at the dictionary shows that PUS---- has to be PUSHING which gives us four more letters following THINGS (WORT-) and AROUND (PACK), and bit by bit all the missing parts are filled in. -RE has to be ARE, so that gives us another letter on the left hand side of the key group, and so it goes, chipping away one batch of letters at a time.

WNXW, BCOIDV, PQOV EJSKII VEPEX HBMZF WWMVOZ UUBK SBYS. ---W, THOUGH, THAT THINGS WORT- ----- ------ ---- ---- WBGGY EZTB EM CTKH UWIL, LSI YLLQ, ZBKQN IBLZ YXU EVNH QC ----- ---- -- ---- ---- --- ---- ----- ---- --U WANT TO PDFP LTYRPT OQAV YX NPX TT SOV CNDR XZ EVONV. NJDMM TURN AROUND PACK -- --- -- --- ---- -- ----- ----- GMHLWZX KJ PKEK VAKO RLJR RMG OTUA KOD AZZ NCQ OBAOKO ------- -- ---- ---- ---- --- ---- --- ARE NOT AFRAID QP MWIGYLR JLR WOBWIZ IA NIULU MZ UGXGYL. OF PUSHING --- ------ -- ----- -- ------

By now you should understand the process, so the rest of the solution is left up to you.

By the way...

The only reason this method worked is that the alphabet used to create the message was in correct alphabetical order. If the alphabet had been scrambled then we could only use this method if our wheel matched the scrambled alphabet, and since there are precisely 403,291,461,126,605,635,584,000,000 ways to scramble the alphabet it's not likely we could have guessed the right one!

With a scrambled alphabet we would have to apply more powerful techniques that are best carried out by computer.

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