Super-fun and often extra-hard ciphers from the crowd
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The русский царь cipher
The Russian alphabet has 33 letters:
АБВГДЕЁЖЗИЙКЛМНОПРСТУФХЦЧШЩЪЫЬЭЮЯ
абвгдеёжзийклмнопрстуфхцчшщъыьэюяBe careful, Ы looks like two letters, but is only one.
Here is a piece of typical Russian text:
Ниже приводятся задачи из задачника С. А. Рачинского «1001 задача для
умственного счета», которые могут быть использованы с учащимися на
уроках арифметики. Их изложение соответствует третьему (1899) изданию в
современном литературном и грамматическом оформлении. Устранены
замеченные опечатки, отдельные задачи подверглись редактированию.
Как и в задачнике С. А. Рачинского, словом «год» обозначается год
простой, т.е. в 365 дней, под словом «февраль» имеется ввиду этот месяц
в году не високосном. Иногда вместо слова «четверик» употребляется более
распространенное в крестьянстве в период выхода в свет задачника С. А.
Рачинского слове «мера». Предполагается, чте в стопе бумаги 480 листов.
Все задачи снабжены ответами — сразу после текста задачи.Now here is your ciphertext. Have fun!
ЩЗЪЗЩЁЙИЬЪКЪЛМЯЖСМИЫХЬЙЯКЬХДЖЯЛЩРВЪЙЁЪМГМЦЯЖНКНЫЁЦГЕЪБЮХДЖЯЛЩРЙКГЫЪЬЁЩ
МЦЕЯЭИБЪЁИЬЪЗГШЙИКНЫЁШЕКИЖЯМИЭИЩЮЪЬЪЁЯЖНЙИКНЫЁШЗЪСЪДЬЕЪБЮХДГВЮЬНЗЪЮЯЛЩ
МХПЙКЪВЮЗГЕИЬЪЗЪЙЪЛПНЙИЮЪКГЁЯЖНЮЯЛЩМЦКНЫЁЯДЛЕИЁЦЕИЙИЁНСГЁИЗЬЭИЮReply to #91416
%%%%%%%%%% Delete all below this line %%%%%%%%%%
Is the answer 178 roubles?
I think I understand that we are hiring a chef. That we pay one rouble in the first month and increase that by one rouble every month thereafter. I don’t quite have the arrangement for the holidays, but am guessing a payment of ten roubles on each occasion. What is the annual pay?…
Harry! You didn’t delete!
[Whoops, sorry. At this time of year the Elves are a bit distracted wrapping presents for the Santa run.]
KLDLW EFEOO WSDNR UWXWD NTYCE ZWUQC IJWSI CLQEZ WEYKP EWWED XNTIB LBIIW SBKXY HHJMU TGYDT WRKWL PBUQU TVTWI UIQCS UQFCU WJTGR KDPNQ NPUKN TAYIU QITMB QQKXP IAWKT ATBSJ PLGES QWCXT DJEED XYESA VUKZF ANNTD STRAN OFPDS TFEDI WWNIF QGXJC JSLCR WMECD STGEN BUHRM CKYKT DWVWK THECV RUTET MZWMC YNRJJ DQHLS MILGW JSAWP QCHNB QQPTR MJJSL ETRNP XDHRM CKYKT DWBQU DGPIA RUKIR MWWDE GTABQ QDHHS MVQKO OCQKY MWJNI WTBDA JXAQD PCENK XYCUN IRDUN MRKUQ SPLUG VUKLT IBKLQ CSXMF LFPCB ZRCWW MTWWR LJEBZ NYBWX BBUUQ QUSNI CCPCW INYKI WWNIQ CHFMS ESUDU BZNYB POKNI ULKTM RWDDP CNBRC MDDAN XOUIW SBKXU TLXCJ LBTPC IZUBI WWRRS BDDUJ GUTDE ZWJQP PLJJE UWYGW SKXLJ VZBKL DSTJQ ZUBLM MQZCL QDKSY UQHDJ CISUH MXZLP QCSSI PUBDC BZFEW SEJXK XUXMI QCDLO XARRC CIWWN ZUQDU QNOUE EMIBQ UPYDN RGLCH XTQNL BTOKM FRQZN WNKYC NFMNF ECIMM WYQOW DJRNL VIWWR NKDYN ZSXYD HMXBQ LCPSW KWBDP SVSCC QIOJA KDUGC WAKBY YNRWE VQCRN WZDXP CWRRW XIQWJ KXUGO WILML COSMJ MUGWS KJYKQ TMNFB DDEZW VCUWG WAWKT GTSVC DLLTQ UFPUY ONAJY LCPMG NKDPC KMRCU ATBSY BLHDB ZNWEW UJXJM QNTRN ETCIW SBWBU IMJCI PYCSA SJLEG DSMND LIWJA NLVIW WTNQC IDBZW DMPCN AQYKI TQQNS DHGSA KQKOR JJUQK OFIAC PMYEZ WKYSG TYSIT UOEZN JUQGE ZERDX PLDNF ECPJW AWKTH NJEUI QCSAC IUOAS MWADX PXMKU QKHPY SRKCI FASET UYCQG RKDST BENKD TTBZY UKIFM GYQPP EZWPB UYEVN JYOWF ANFKP PLBFB
Can you work out what the following says?
7F, 4, 8, 10, 7F, 0, 7E, 9, 9, 9, 7E, 0, 1, 1, 7F, 1, 1, 0, 0, 0, 41, 7F, 41, 0, 3E, 41, 41, 41, 3E, 0, 7F, 4, 8, 10, 7F, 0, 7E, 9, 9, 9, 7E, 0, 7F, 40, 40, 40, 0, 0, 0, 0, 0, 0, 0, 0, 3E, 41, 41, 41, 22, 0, 0, 0, 41, 7F, 41, 0, 7F, 9, 9, 9, 6, 0, 7F, 8, 8, 8, 7F, 0, 7F, 49, 49, 49, 41, 0, 7F, 9, 19, 29, 46, 0, 0, 0, 0, 0, 0, 0, 3E, 41, 41, 41, 22, 0, 7F, 8, 8, 8, 7F, 0, 7E, 9, 9, 9, 7E, 0, 7F, 40, 40, 40, 0, 0, 7F, 40, 40, 40, 0, 0, 7F, 49, 49, 49, 41, 0, 7F, 4, 8, 10, 7F, 0, 3E, 41, 49, 49, 7A, 0, 7F, 49, 49, 49, 41, 0
The characters of the following text have been re-ordered using a sequence of pseudorandom numbers generated by a simple linear congruential generator. Can you recover the original text?
roemaoastolgtowfsthmcinssfcfgusf.ouirlecsoiateoruosnusinselhitfuiS
The form of the linear congruential generator is:
Xi = (aXi-1 + c) mod m with 0 <= xi <= m;
where the i values are sub-scripts.
a is the called the “multiplier”, c is called the “increment, and m is called the “modulus” of the generator. If c is set to zero, and, in this case, the generator is called a multiplicative generator.
In this case m is is the first prime number larger than the length of the text, c = 0. a and m are relatively prime. The linear congruential generator produces a repeating sequence of numbers. The period of a multiplicative congruential generator with multiplier a and modulus m depends on the smallest positive value of k for which
a^k = 1 mod m
In this case the period is 66.
[You told me “I think this could be a bit of a challenge” and I think that might be an understatement! I will be fascinated to see what people do with it! Harry]
Hi Harry, my post 9th November 2023 at 1:08 pm #92161
Is there some reason you have not put it up yet?
Maybe you do not like the plaintext – let me know, I will remake with another text.
(by the way what is the amount of time needed to submit again after making one post? I get Error – You move too fast. . . )
[I am enjoying it too much myself! I am delaying several hard challenges like this for after the competition when more people will be ready to tackle them. Hope that is OK! Harry]
Has anyone made any progress with madness’s Morse cipher (#91966)? It’s beginning to keep me awake at nights.
Harry, I hope you are okay to post the following partial solution to #91992.
Well done to anyone who managed to decrypt puzzle #91992. If you didn’t here are some pointers that should help. Feel free to post your own methods, observations and techniques.
Looking at the cipher.
Length (including spaces) = 1301 characters.
Length (excluding spaces) = 1086 characters.Letter distribution:
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
* * * * * * * * * * * * * * * * * * * * * * * * * *
* * * * * * * * * * * * * * * * * * * * * * * * * *
* * * * * * * * * * * * * * * * * * * * * * *
* * * * * * * * * * * * * * * * * *
o o o o o o o o o o o
* * * * * *
*
*Index of Coincidence = 0.0429
The letter distribution and IoC both indicate this is neither a mono-alphabetic substitution cipher, nor a transposition cipher.
A closer inspection reveals that the sequence HRMCKYKTDW occurs twice, once starting at position 213 and again at position 283. There are many other similar repetitions.
HRMCKYKTDW 213 283
MCKYKTDW 215 285
IWSBKX 59 479
NLVIWW 679 896Looking at the spacing between the repetitions and the factors of the spacing gives…
HRMCKYKTDW 70; factors: 2, 5, 7.
IWSBKX 420; factors: 2, 2, 3, 5, 7.
KYKTDW 70; factors: 2, 5, 7.
NLVIWW 217; factors: 7, 31.Seven is a common factor. Writing the sequence of characters into a block seven characters long gives:
KLDLWEF
…
BUHRMCK
YKTDWVW
…
XDHRMCK
YKTDWBQ
…We then have the common characters lined up above each other. Examining the characters in each columns gives a much more promising letter distribution and allows a decryption.
Soton?
[I think it means “why”, not where, but good effort! Harry]
Sorry, my letter distribution chart got mangled in post #92382, but I’m sure you call all work that out.
Answer to post #92229.
The original text is “Success consists of going from failure to failure without loss of enthusiasm”. A quote from Winston Churchill.
It isn’t so much finding the solution, as what you learn in trying.
The characteristic of the generator is:
a = 32, c = 0, m = 67, seed = 1
Period = 66
Sequence:
32, 19, 5, 26, 28, 25, 63, 6, 58, 47, 30, 22, 34, 16, 43, 36, 13, 14, 46, 65, 3, 29, 57, 15, 11, 17, 8, 55, 18, 40, 7, 23, 66, 35, 48, 62, 41, 39, 42, 4, 61, 9, 20, 37, 45, 33, 51, 24, 31, 54, 53, 21, 2, 64, 38, 10, 52, 56, 50, 59, 12, 49, 27, 60, 44, 1,
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