Extracurricular Challenges

[_madness_]

This is a familiar cipher, but there is something different about it.
Figure out what is different, and break the cipher.

RYJXH RYJPF HDRHA ADCHP JHALR RAJVJ BRFLR YYHLP VAHBI YJWFY LRJRO OFYLR
JXOPY LDDRL AAGOE BKIOE BRJBH BIJHB WLBYL DNHBB JPHBW HVOER YLDCJ PDOBR
YJDJW HRJHE DRJPL RGOXI HPPLH KJRYH RVJDC JHMDR YJCEP LRHBR YJWHE KYRJP
FHDCO DDLVA GDONJ RFJBR GXOEP OPRFJ BRGXL SJGJH PDOXH KJDYJ FHDSJ PGDAL
KYRJN HILHR JWYJP JTIJJ WLBKA GCHAJ IOEBR JBHBI JVJHP LBKHA HBKEL WDCLP
LRAJD DJTCP JDDLO BOBJO XRYOD JCJOC AJFYO NFJDO NJRLN JDJBI OEBRJ PHCCH
PJBRA GROOF JHMXO PRYJI HPJDH BWRHD MDOXA LXJRO OXJJV AJRON OSJOP WORYJ
RYLBK DRYHR FJNED RWOJS JPGWH GBJSJ PRYJA JDDRY JKLPA FHDCP JRRGF LRYRY
JJRYJ PJHAV JHERG OXHBH CCHPL RLOBL RFHDD YJEBW OEVRJ WAGFY OIHNJ XOPRY
JVJBJ XLROX RYJFH RJPD

[Astralica]

What’s different?
It seems like a regular substitution:
Hardest part was adding the spaces back.

Thought it best to remove the answer. I think the real challenge is not so much the decrypt as working out what the twist was! Harry

[Astralica]

In fact, here’s my own cipher that’s familiar, but has an interesting change.

The usual rules apply, if you want to post a puzzle or a challenge then please include the solution so I can see if it is suitable! Harry

It has been solved before, so it’s definitely possible.

[ByteInBits]

@-madness_
I do not see anything different about it.
Is there realy a ‘twist’ to it?

I hope Madness doesn’t mind me saying that the subtlety here is to understand how the substitution is designed. It is not arbitrary, though it looks like it might be. There is a small twist on a classic cipher, and he is interested in seeing if you can work out what type of cipher has been used/adjusted and what the adjustment was. Is that right Madness? Harry

[upsidedown]

I think this is a variation on the Caesar cipher, with:

• A 25-letter alphabet (Q removed)
• The cipher alphabet proceeds through all odd letters then all even letters
• A “shift” of 7 (ie. alphabet starts with H)

This is my substitution key:

ABCDEFGHIJKLMNOPRSTUVWXYZ (plain)
H I J K L M N O P R S T U (cipher, even)
 V W X Y Z A B C D E F G  (cipher, odd)

It is also similar to the Affine cipher, but with a & b coming from the plaintext letter rather than the key:

a = plain mod 2
b = plain div 2
cipher = a * 13 + b + 7    (mod 25)

This reminds me of one of the GCHQ puzzles (I think it was a Bletchley park interview test) where you had to work out the scheme through which a substitution key was generated.

[Astralica]

Sorry, Harry.
I initially added the plaintext solution yesterday, but it looks like I got flagged for sending too many messages, too quickly.

Here’s the plaintext:
THETHINGSWEDOOUTLASTOURMORTALITYTHETHINGSWEDOARELIKEMONUMENTSTHATPEOPLEBUILDTOHONORHEROESAFTERTHEYVEDIEDTHEYRELIKETHEPYRAMIDSTHATTHEEGYPTIANSBUILTTOHONORTHEPHARAOHSONLYINSTEADOFBEINGMADEOFSTONETHEYREMADEOUTOFTHEMEMORIESPEOPLEHAVEOFYOU

You can solve it by using a substitution with key TQWPKDVYBRNMZCJGSXLUEIHFOA followed by a progressive shift of -1. (Or you can do the progressive shift first.)

[victorlazlo]

if you finished 3.b we can meet for a coffee and make a team
ETHRWE HITSEP IDRSSE

[_madness_]

@upsidedown wins the prize. But… there is a way to define the affine key in the forward direction (you did it in the reverse direction).

[upsidedown]

Ah, I see it can be written more cleanly as a regular affine cipher with an alphabet length of 25:

a = 13
b = 7
encrypt letter (a, b) = (letter * a + b) % 25
decrypt letter (a, b) = ((letter - b) * mmi(a, 25)) % 25

[ByteInBits]

NCC 2024 BYTEINBITS FORUM CIPHER
================================

Hi codebreakers and cipher crackers,
Here is a cipher for you all to try and break.

Not sure which code will be easiest to copy from in the forum, if you use the one with
line breaks you may need to remove the breaks from your copy before decrypting.

CIPHERTEXT TO BREAK (no line breaks)
F 11 6 F H 11 13 E 11 3 G E 11 13 4 6 G 13 M H 11 9 H H 6 10 9 12 13 H 13 G L 4 11 H K 12 11 G D E 11 L H 6 C 11 D J 11 7 13 G 1 2 13 1 9 K 1 6 3 L 6 10 6 6 L F 11 6 F H 11 2 9 H H 13 12 12 3 4 11 1 6 3 6 K 4 11 H K 9 4 J 3 H D 11 E 9 6 E 7 6 D 9 C 11 4 L 6 10 6 6 L 13 G 1 2 13 1 9 K 1 6 3 13 E 11 4 3 12 12 11 4 4 K 3 H 1 6 3 2 9 H H 2 9 G K 13 H 4 11 K E 9 11 G L 13 G L D E 3 11 11 G 11 7 9 11 4 4 3 12 12 11 11 L 13 G 1 2 13 1 D J 11 10 6 6 L 1 6 3 L 6 D 6 L 13 1 2 9 H H M 11 K 6 E 10 6 D D 11 G D 6 7 6 E E 6 2 L 6 10 6 6 L 13 G 1 2 13 1 J 6 G 11 4 D 1 13 G L K E 13 G 8 G 11 4 4 7 13 8 11 4 1 6 3 C 3 H G 11 E 13 M H 11 M 11 J 6 G 11 4 D 13 G L K E 13 G 8 13 G 1 2 13 1 D J 11 M 9 10 10 11 4 D 7 11 G 2 9 D J D J 11 M 9 10 10 11 4 D 9 L 11 13 4 12 13 G M 11 4 J 6 D L 6 2 G M 1 D J 11 4 7 13 H H 11 4 D 7 11 G 2 9 D J D J 11 4 7 13 H H 11 4 D 7 9 G L 4 D J 9 G 8 M 9 10 13 G 1 2 13 1 F 11 6 F H 11 K 13 C 6 3 E 3 G L 11 E L 6 10 4 M 3 D K 6 H H 6 2 6 G H 1 D 6 F L 6 10 4 K 9 10 J D K 6 E 4 6 7 11 3 G L 11 E L 6 10 4 13 G 1 2 13 1 2 J 13 D 1 6 3 4 F 11 G L 1 11 13 E 4 M 3 9 H L 9 G 10 7 13 1 M 11 L 11 4 D E 6 1 11 L 6 C 11 E G 9 10 J D M 3 9 H L 13 G 1 2 13 1 F 11 6 F H 11 E 11 13 H H 1 G 11 11 L J 11 H F M 3 D 7 13 1 13 D D 13 12 8 1 6 3 9 K 1 6 3 J 11 H F D J 11 7 J 11 H F F 11 6 F H 11 13 G 1 2 13 1 10 9 C 11 D J 11 2 6 E H L D J 11 M 11 4 D 1 6 3 J 13 C 11 13 G L 1 6 3 2 9 H H 10 11 D 8 9 12 8 11 L 9 G D J 11 D 11 11 D J 10 9 C 11 D J 11 2 6 E H L D J 11 M 11 4 D 1 6 3 J 13 C 11 13 G 1 2 13 1

CIPHERTEXT TO BREAK: (wrapped characters to 40 per line)
F 11 6 F H 11 13 E 11 3 G E 11 13 4 6 G 13 M H 11 9 H H 6 10 9 12 13 H 13 G L 4 11 H K 12 11 G
D E 11 L H 6 C 11 D J 11 7 13 G 1 2 13 1 9 K 1 6 3 L 6 10 6 6 L F 11 6 F H 11 2 9 H H 13
12 12 3 4 11 1 6 3 6 K 4 11 H K 9 4 J 3 H D 11 E 9 6 E 7 6 D 9 C 11 4 L 6 10 6 6 L 13 G
1 2 13 1 9 K 1 6 3 13 E 11 4 3 12 12 11 4 4 K 3 H 1 6 3 2 9 H H 2 9 G K 13 H 4 11 K E 9
11 G L 13 G L D E 3 11 11 G 11 7 9 11 4 4 3 12 12 11 11 L 13 G 1 2 13 1 D J 11 10 6 6 L 1 6 3
L 6 D 6 L 13 1 2 9 H H M 11 K 6 E 10 6 D D 11 G D 6 7 6 E E 6 2 L 6 10 6 6 L 13 G 1 2
13 1 J 6 G 11 4 D 1 13 G L K E 13 G 8 G 11 4 4 7 13 8 11 4 1 6 3 C 3 H G 11 E 13 M H 11 M
11 J 6 G 11 4 D 13 G L K E 13 G 8 13 G 1 2 13 1 D J 11 M 9 10 10 11 4 D 7 11 G 2 9 D J D J
11 M 9 10 10 11 4 D 9 L 11 13 4 12 13 G M 11 4 J 6 D L 6 2 G M 1 D J 11 4 7 13 H H 11 4 D 7
11 G 2 9 D J D J 11 4 7 13 H H 11 4 D 7 9 G L 4 D J 9 G 8 M 9 10 13 G 1 2 13 1 F 11 6 F
H 11 K 13 C 6 3 E 3 G L 11 E L 6 10 4 M 3 D K 6 H H 6 2 6 G H 1 D 6 F L 6 10 4 K 9 10
J D K 6 E 4 6 7 11 3 G L 11 E L 6 10 4 13 G 1 2 13 1 2 J 13 D 1 6 3 4 F 11 G L 1 11 13 E
4 M 3 9 H L 9 G 10 7 13 1 M 11 L 11 4 D E 6 1 11 L 6 C 11 E G 9 10 J D M 3 9 H L 13 G 1
2 13 1 F 11 6 F H 11 E 11 13 H H 1 G 11 11 L J 11 H F M 3 D 7 13 1 13 D D 13 12 8 1 6 3 9 K
1 6 3 J 11 H F D J 11 7 J 11 H F F 11 6 F H 11 13 G 1 2 13 1 10 9 C 11 D J 11 2 6 E H L D
J 11 M 11 4 D 1 6 3 J 13 C 11 13 G L 1 6 3 2 9 H H 10 11 D 8 9 12 8 11 L 9 G D J 11 D 11 11
D J 10 9 C 11 D J 11 2 6 E H L D J 11 M 11 4 D 1 6 3 J 13 C 11 13 G 1 2 13 1

If you break the cipher without the proper algorithm being used then still WELL DONE
after all the breaking of the cipher is the whole point of being a spy!

But if you discovered and used the correct algorithm then encipher the following

IDECLARETHATIAMANELITECODECRAKER

and post the ciphertext to the forums.
The MD5 of the ciphertext should be E6A408E0D4BC8AC70181644B4E9AD166

[victorlazlo]

Henri René Albert Guy de Maupassant ~ From the Tomb

[robb]

@ByteInBits, I get the MD5 hash with ciphertext 9 L 11 12 H 13 E 11 D J 13 D 9 13 7 13 G 11 H 9 D 11 12 6 L 11 12 E 13 8 11 E

[upsidedown]

Re: #98114

md5(“upsidedown” || encrypt(“IDECLARETHATIAMANELITECODECRAKER”)) = cdb9dcfbc86baec0c794345b4149a4ca

Here’s a similar cipher which uses a prefix code and extends the pattern of the key to produce a homophonic substitution:

60KE VX91 9689 5382 92Z8 580H 0KES 6948 0Q84 8968 3J84 V898 491L 3DSS LOV8
8949 480Q J849 1963 8091 DJ82 FSY7 XD3S 9594 VQ96 89X3 3809 1WW5 OGH5 SU80
G81S C828 9T03 J52H 9682 E808 8184 3ZM8 9808 1JD8 3E80 3FK8 2978 92ZE NTS8
9G80 G818 491L 88JX 3915 1SK8 8J82 LZ09 7XKS E823 D80H 92H8 40LX 82F2 9059
1D94 X81E SPG9 6QJV 66XE VGWM 8980 W948 8HN8 0GD9 4X1V 91DE 03PG L5O9 1889
4X83 X918 8EX5 K978 4U82 8938 8E82 82DV 88J8 4969 782X 9190 H52W JX96 8091
D584 M895 OGYE 26M9 2097 8491 L97N 88JX D89Z 95KV 83M2 88Z8 8X80 88JX 893V
9682 84G9 65GD JX94 8284 FX96 2F82 96W8 2Q59 5889 482K S588 90Z8 8JQV D3D8
092H H840 ZQ1J V888 2ZQO J82G VD95 82H9 2889 4828 1898 2NFZ 8782 TXGD JZLM
X829 183H 8284 91X9 684G 963Y E849 082L 9728 81ZQ QD80 H92L 8491 WX89 S23H
NQH8 0F90 8289 N3SD 9484 88DJ XEX1 8289 XKXO X89F 8433 8291 WX89 3D94 8491
PQPZ HVGL 82XL KE5T D94X L80E 8283 8880 591S KDJX TX88 ESF5 9280 88ZG 388Z
D805 G915 S918 2OZQ 835T VGW3 8487 82DJ XQ80 9181 92XW 8291 D928 2684 G1J5
382X Y838 2918 889V 83YS GLPY D94Z L96E ZOG6 0848 888X 9188 V5G

md5(lowercase plaintext) = 07ccad93115a2015c6537bf75099f98f

[ByteInBits]

Well done @rob and @upsidedown both gave correct cipher answer.

May I say, along with a few others they are amongst the best solvers
over the forums, far better than I am!

[_madness_]

In 2005, information about the Russian/Soviet cipher machine, Fialka, was first released to the public. In response, the 2005 NCC final challenge was a simplified version of the Fialka.

There is a very good and highly detailed reverse engineering of the machine at
https://www.cryptomuseum.com/pub/files/Fialka_200.pdf

And it only took me 19 years to create three challenges based on the full machine. They are currently at MysteryTwister. They told me that I had to promote them on social media, lest I be punished. It is a very interesting machine, with some cool innovations. Some of you will enjoy emulating the machine and possibly solving the challenges, if Harry doesn’t bring the hammer down.

Do we count as social media? Harry

[Author]

Posts