OC OY FO LB VN PI AS AK OP VY GE SK OV MU FG UW ML NO OE DR NC FO RS OC VM TU UT YE RP FO LB VN PI AS AK OP VI VK YE OC NK OC CA RI CV VL TS OC OY TR FD VC VO OU EG KP VO OY VK TH ZS CV MB TW TR HP NK LR CU EG MS LN VL ZS CA NS CK OP OR MZ CK IZ US LC CV FD LV OR TH ZS CL EG UX MI FO LB IM VI VK IU AY VU UF VW VC CB OV OV PF RH CA CS FG EO LC KM OC GE UM OH UE BR LX RH EM HP BM PL TV OE DR NC FO RS GI ST HO GI LC VA IO AM VZ IR RL NI IW US GE WS RH CA UG IM FO RS KV ZM GC LB CG DR NK CV CP YU XL OK FY FO LB VC CK DO KU UH AV OC OC LC IU SY CR GU FH BE VK RO IC SV PF TU QU MK IG PE CE MG CG PG GM OQ US YE FV GF HR AL AU QO LE VK RO EO KM UQ IR XC CB CV MA OD CL AN OY NK BM VS MV CN VR OE DR NC GE SK YS YS LU UX NK GE GM ZG RS ON LC VA GE BG LB IM OR DP RO CK IN AN KV CN FO LB CE UM NK PT VK TC GE FH OK PD UL XS UE OP CL AN OY NK VK BU OY OD OR SN XL CK MG LV CV GR MN OP OY OF OC VK OC VK VW OF CL AN YE FV UA VN RP NC WM IP OR DG LO SH IM OC NM LC CV GR MN OP OY HX AI FO OU EP
FO RS appear 3 times in the passage. It could be ST OP.
OR also appears many times. It could also be S TO P.
OC also appears many times but it appeared in doubles (OC OC) so it cannot be part of STOP.
R T O R O T
P F S OR P S F
If the first one is the real keyword, then the square must be:
The second one cannot be the keyword because when the two letters are on the same row, the one to the immediate right is taken.
So, the square is R__TO
<IP to DG>
<Is to DG>
So, now the square is R__TO or ____I
By trial and error, the second square is used.
<IP to DG>
<Is to pG>
So, the square can be G___I or ____I or ____I
R__TO R__TO R__TO
PD_FS G____ PD_FS
Trail and error shows that the second one would be the best.
<IM to DP>
<Is to ps>
This reveals the square:
<rs to MZ>
<rs to pZ>
There are two choices: Z___I or ____I
By trail and error, the second one is used.
<OD to SN>
<Os to pN>
p* = SN. In Playfair, the encryption of * is always above or below P except in a special case.
*=the letter encrypted as N
Therefore, the square is now N___I
Now, there are no more clues to solve the key square, so assumption is used. The Playfair square is filled after with the rest of the alphabet in alphabetical order, so the last two letters should be X and Y.
<OC> appears very often, so it could stand for <TH>, the most common used two-letter part in English.
T and O appear on the same row, so this is one of the special cases where both the plaintext letter and the ciphertext letter appear on the same row. If this happens, then the letter to the right of the ciphertext letter would be used to encrypt. From that, the square is now:
Again, there are no more clues, so we have to deduce plaintext letters by guessing.
<th is st LB VN> could be <th is st ag eN>
The only possible way is to have the two plaintext letters on one row, so the square becomes:
e*=VN In Playfair, the encryption of *, N, is always above or below e. But in this case, N is in the corner and the column under it is filled. So, this encryption must have been a special case, where the letter directly below the plaintext letter is used.
So, the square now looks like: N_EVI
<th is st ag ei sn AS>
According to the square, AS is an encryption of _M. If we fill it in, we get: this stage is n_m. It looks very much like: this stage is number six.
So, if the guess is correct, then the square is now:
<th is st ag ei sn um AK rs ix>=THIS STAGE IS NUMBER SIX
In Playfair, the encryption of K is always above or below A and the encryption of A is always above or below K. There is only 1 place in the square that satisfies the conditions.
The square is now: NKEVI
Now, only the letters Q and W are left.
<an di ti sa pl UW fa ir ci ph er>
It is probably: and it is a Playfair cipher.
Looking at the square, there is only one place where it can fit.
The square can now be completed:
th is st ag ei sn um be rs ix an di ti sa pl ay fa ir ci ph er st op th ef ol lo wi ng st ag ei sn um be rs ev en wi th in th ec on te xt of th is co mp et it io na nd it is en cr yp te da cx co rd in gt oa na df gv xt yp ec ip he rs to pw he ny ou at te mp tx to cr yp ta na ly se st ag es ev en yo uw il ls ex et ha ti ti sm or ec om pl ic at ed th an as tr ai gh tf or wa rd ad fg vx ci ph er st op un fo rt un at el yi ca nx no tg iv ey ou an ym or ec lu es st op ne wp ar ag ra ph in te rm so ft hi sx st ag et he sh ib bo le th th at yo us ho ul dt ak en ot eo fi sm ol yb de nu mn ew pa ra gr ap hy ou wi lx lp ro ba bl yh av en ot ic ed by no wt ha te ac hs ta ge is in ad if fe re nt ci ph er an di su su al ly in an ap pr op ri at el an gu ag es to ps ot he vi ge ne re st ag ew as in fr en ch an dt hi sp la yf ai rs ta ge is in en gl is hs to pi ft he pa tx te rn pe rs is ts th en th en ex ts ta ge wi lx lb ei ng er ma ns to pb ut do es th ep at te rn pe rs is tq ue st io nm ar kq
After removing ‘STOP', ‘NEW PARAGRAPH', extra letters and junk letters:
This stage is number six and it is a Playfair cipher. The following stage is number seven within the context of this competition and it is encrypted according to an adfgvx type cipher. When you attempt to cryptanalyse stage seven you will see that it is more complicated than a straight forward adfgvx cipher. Unfortunately I cannot give you any more clues.
In terms of this stage the shibboleth that you should take note of is Molybdenum
You will probably have noticed by now that each stage is in a different cipher and is usually in an appropriate language. So the Vigenère stage was in French and this Playfair stage is in English. If the pattern persists then the next stage will be in German. But does the pattern persist?