Challenges from competitors

[Person314]

@TLW

Will reveal the answer to my challenge in a week (if nobody solves it). And if anyone wants clues I am happy to give them.

For your challenge, in order to not spoil the answer for people reading the forum I have multiplied my answer matrix by the matrix [1 2 3; 4 5 6; 7 8 9] and I got, [41 55 69; 41 49 57; 70 86 102]

[_straylight_]

Hello,

Here is another challenge, taken from the “Cold War Spy Pocket Manual”. In the appendix, the book contains a challenge a number of such challenges most of which I was able to solve somehow. This one, however, resisted up to now.

M@PIS M^QGQ T_^W_ NM@JR WO]]G KH[KM PI\BX [ATZU JMGG[ NIR^Q LGZXG IUCBU FEXGF IYOGW JG\@C M@SIU LKCAS G@S\Q FIGGW IFZ[V G^AJA SYP]Z GZ\\S KB\O\ C@Y\] ADT\P QCGIQ UEYB] CAXKX TEYBQ KZTG_ WGZZ[ EYGGP eMD[] LC}EZ M[[OG VC^W[ PCFK} EIZ\S GGZXU NFZFZ CBAF[ LUBOX IIGGX KMOLU XHTGS M^RAA XI[E[ JIGCU LBROF VVGOX RDXMS GDPKS GCGIQ @@TEQ

There is no further information about the cipher, but maybe someone around here has an idea what kind of encryption it might be.

[Madness]

Just to double-check: Are the “e” and “}” characters correct? They do not fit the pattern. Should they be “E” and “}”?

[Madness]

I meant “E” and “]”. Sorry.

[_straylight_]

Yes, it is indeed “e” and not “E”. But what’s the difference between “}” and “}”?

The complete challenge can be found here:
https://books.google.de/books?id=mDMZCwAAQBAJ&pg=PA190&lpg=PA190&dq=cold+war+spy+pocket+manual+m@pis&source=bl&ots=bbg2Ya9yiW&sig=ACfU3U1XlMjG2grEC_mIZesNDHIWszLF3Q&hl=de&sa=X&ved=2ahUKEwie8fSZuf7tAhULyqQKHUYDCXEQ6AEwDHoECBMQAQ#v=onepage&q=cold%20war%20spy%20pocket%20manual%20m%40pis&f=false

[Mattyrat2027]

Hi all,

A ciphering method I have come across, which I think I could share. Below is an encrypted plaintext, but the message has been encrypted by a series of moves rather than a key or a table. Hint: It is alike to an anagram (in the sense that it has the same letters in ciphertext to plaintext) but involving splitting into 2 letter groups. The plaintext below is the method used to get there.

ifsry tuoat ekoyr uemss gaaed npsil ititt ndogi arhpn stoxe acbnu eesad asifl lreot veneh tnemu eborl fteet srhtn eoysu iwcte hcalh teeti rtneh aprin itgeh snmil ppytu htmes easeg abkco teght renas dlpti htcep iehtr xeitt nfovi letee tgror puxs

If decrypted correctly it should come out as a message with no punctuation.

Best of luck, I’ll be interested as to who solves it. If hints ar needed, I’ll happily give a few.

[Madness]

Every now and then a really bad paper comes out about combining two classical ciphers and “proving” that the result is
secure and can be used in modern internet commerce and other super-secret stuff. Today I read a particularly bad one
(total crap) that combined a “triple-key transposition” that was really just a bunch of XORs with base64 encoding, with
the base64 done last. They claimed that the result looked so random that it must be secure. I laughed so much I fell off
my perch and banged my head.

To explain just how bad this is, imagine that you have a substitution cipher and that you encode the ciphertext with
base64 to get a final ciphertext. Base64 is an ENCODING, and is keyless. An eavesdropper can easily decode the base64,
and recover the intermediate ciphertext, then break that with some technique that you should know by now.

So I got to thinking, “wouldn’t it be more secure to encode with base 64 *before* the substitution cipher?”
I quickly disabused myself of this delusion (think about why) and decided to make a challenge based on this
mistaken idea. So… here is a ciphertext. The plaintext contains only capital letters and no spaces (to make it
easy on you). The plaintext was first encoded in base64, and then a monoalphabetic substitution (with 64-character
alphabet) was applied to the result. Your task is to break it.

Anyone who thinks this is impossible should go lie down for a while and watch cartoons. To be fair to the rest of you,
I tried it before posting and my program was successful. Have fun. If no one gets it, I can explain the approach, which
can also be applied to a lot of other ciphers. (That might be a hint to you cipherclockers out there.)

gCzNCweUCwsQWCmoOb8fCbVWgVhC9CnqVV88CbsUCweVCPeJNwmoNV8nVPeCOVhCOCz4WbO8WwBWNbhsWCVWVPhfVs8sCs8sCbVo
VP88Wwsg9COfVPsQW78sC7fsNbmJgCQfWPknNVQsCVVsWwQsWbnsgCBHVP88VP8nVPVWgCzfWwsWNbm8C7OggCzUOwegWCsCNmmC
CwsoCbkfVPmoObmCgCzCWbsgNCO89sQ89sOnOweVCm8sCs8sCbVoVPsCgCeoVPhsVPVgWC8fCbVWgVhC9CnqVV8qCwmUgCzfVPVW
Om8qWCsWCPV4gCn8Nb78WCV4WbzCOCzCVs8fWmQJOV8sWwQqOPmoObVfNbhoWbzJgCzfWP8fCbVWgVhC9CnqVV8PgCB8Vs8sCs8s
CbVoVsQs9Ps4VPknCbmuNwmCCbeJOPsCNVOHCwVsOCmUgsOrgVOr9VOsCbmsNVOqVPslWbnCVbVoVsmJW7VgNwmCCbQfWmOHOV8s
OwegOC8sCwVNCwVWOCzCOCOr9CnqVV8WgVhrgVOrNVQsCbmuVsmJW7VgOPmUgVOWNbeQWCeoVPefWPkrgCzfCmmCW7Os9sOsWwQq
OPmoO7Q4gPVQOVQrNVQsCbmuVsmJW7VggVQPOVQ8ObzsOsOqNbsgCmmPNVOfC7OqCwVPgCzrgCzfCmmJW78QNVOWNCQgW7QWNbhf
WwzsWsQCgPsCWbzl9V8sWPmfNwknC7VNCPegVsOs9sO4WbzCOCzCNwsWOVQ89sOnOweVCwmWCPsgVPm4VCkfCwknCs8sVwslOCzC
WbzCgPVKW78lOsB8OPVKOC8KgPVgOCmCC7VWOVQ8WwQlVCOsVPhsNC88WPmC9VOqOC7rOCO8WCsUOCn8WPVWW78qVPhsCw88Wwsg
9CsWCbVCCbmoCbmPOVOs9sOVNCkfC7QsVsQWVCQHNVQHVP7lNCzPN7QWOwmlOVQrNVQsCbmuVsmJW7VggVQfWsQqVbmPOCknVVQs
OPnqCmQsWwO8WwBsWCs8WPsCVPs4gP7sWmOWVPh8CbmWCwVOVCmgOCOrOCQfVVQsCb7CCPmogVOWW788ObmoNCkJW78QVbsWOPVW
gCBoOCOCW7OgNCzWCPegVsQsVwVoNwmCNVQ4gCm4gPsgNCQCOV8WWbzl9VOHgVQsWwQqOPmoObQfVVQsCbsoW7nsCwhsNCOqOmOH
gV8C9VQ89sfsCwQsWmN1

[Madness]

@_straylight_, in that case, I dunno. It seems odd to me that only three characters of about 300 would lie outside
the set of 32 characters @ through _, which form a contiguous set in ASCII and which vary only in their last 5 bits.
It seems so perfect that I am having trouble believing that “e” and “}” are not typos.

The link you posted is used up, georestricted, requires a thumbprint and DNA sample, or is broken. Can you furnish
a screen shot of the page? Thank you.

[_straylight_]

Hi Madness,
thank you very much for taking a look at the cipher. Here is (hopefully) a screenshot of the page displayed in google books:

[Person314]

For anyone who attempted my challenge from #52764 but was unable to get the answer, the cipher text used randomly generated letters, with one letter fixed as the key letter. In order to convert the plaintext into cipher text, each letter of the plain text is converted into a number and then that number of random letters are inserted into the cipher text (followed by the key letter).

As I mentioned earlier this year this is not secure at all, but perhaps it is more steganography than cryptography anyway. It should be noted that the limited alphabet was used because it would be incredibly obvious what was going on if any other number of letters was used as you would see a spike at the key letter on the frequency distribution.

[The-letter-wriggler]

@Person314 Your grid answer in post #52781 is referring to the @Kford-academy puzzle of 24th December #52753 who needs to confirm your answer.

[The-letter-wriggler]

Harry why has @Jbrintcrypt ‘s post not yet been put up?
I refer to 19th Dec. #52724 and 24th Dec. #52772
The person must be really disappointed at the neglect!

[It is a sad story of overworked elves, lost parcels and COVID-19. @Jbrintcrypt would be fully justified in feeling disappointed, and if they want to repost I promise to give it my attention. Harry]

[The-letter-wriggler]

@Mattyrat2027
Here is a message to you using Trigrams instead of Bigrams, the same coding applies:

htikniwtiluoahdbevneeromfoehcallagnefieuoydahtonvigtnehehtnitelehtlosrevifsidnuotofthtrsmevleasetdnnehvigiheistenfedexxd

Also you would benefit to learn to use the MD5 hash so we can answer without giving it away.
All plaintext is put in uppercase with no spaces and then the user-name put on the end, in uppercase also, then an md5 is generated from that.
In my case the MD5 of PLAINTEXT+TLW is: 7477126d4c9979c7ac55a8da19b27c77

This is a very simple cipher (no shame in that) and ideal for those relatively new to have a go at solving it.
Nice to see you participate, not many do. No one needs to be an expert, people of all levels should post their ciphers
but please be aware that you need to give Harry the plaintext and how it is enciphered before he will post.

[Madness]

@Person314,
It was easy after you gave the hint: WERE NO STRANGERS …
Thanks.

We should all remember Kerckhoff’s principle(s) and assume that the adversary already knows your algorithm.

[Madness]

@_straylight_, thanks, but now I’m even more suspicious: LBRØF ?

Thanks also to @TLW for finding an alternative link to the book.

[Author]

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