Extracurricular Challenges

[_madness_]

Vigenère ciphers are all fine and good, but here is something more disturbing than burning witches. The following was encrypted with an interrupted Vigenère cipher, which (not witch) means that with every word the encryption starts over at the beginning of the key. Have fun.

ZSVVHHRIERIGSKGCBFRBWVRGXVCYSFQRQPOLLWFLLBRFQSKWVNHZSPWTVMQCGESLBGSKUCHMYULKOQIOYZXQ
LBGCPMVSESGHVHNHXRBRHJRBBQOZVRBZSKSNFWDOSGHXUGFBDIHLWVECNJPWVRVZVHLBWVRVOVRGCBFDBQIO
YZRBWVRJFBIGGLHFICEBCHKOARXOVOVRKSEPXUBJCVBZRIGDBQFCZWGJEOPYZWGVLHDBQLKCCSWVRFXLGQCG
VBQOWVRPOGHXUWVVGZOFLBDZYILLWFWCLHFKSNJREZNQDVSNZDBQECERXU

The MD5 of the plaintext, in all caps, with word spaces included but no punctuation, is 74c9865d328618150fb98c3021a52ac7.

[upsidedown]

Re: #98576

Tricky!

md5(“upsidedown” || uppercase plaintext) = fc975cb6448b9c96d87d8d8f4eff1567

[Zeta]

I have an explanation on how I solved this, although I don’t know how to mark text as spoilers. Therefore, I’ve Caesar shifted my solution in case other people want to have a go at it. If anybody knows how to spoiler mark on the forums, please let me know!

My solution:
OB WBHSFFIDHSR JWUSBSFS QWDVSF VOG HVS ORJOBHOUS HVOH MCI QOBBCH SOGWZM TWBR HVS CFWUWBOZ YSM ZSBUHV. VCKSJSF, WH BCK VOG OB SLHFSAS KSOYBSGG KVSFS HVS GOAS KCFRG KWZZ PS SBQFMDHSR WB HVS GOAS KOM. W GHOFHSR PM ZCCYWBU OH HVS ACGH TFSEISBH HFWUFOAG, KVWQV VODDSBSR HC PS “KJF”. QCADOFWBU HVWG KWHV HVS HFWUFOA “HVS”, WH GIUUSGHG HVOH HVS YSM PSUWBG KWHV “RCB”. W QCBHWBISR HC QVSQY TCF KCFRG W’R SLDSQH HC PSUWB KWHV “HVS”. HVS GSEISBQSG “KJFTL” OBR “KJFUL” GHCCR CIH HC AS PSQOIGS HVS TCIFHV ZSHHSF CBZM RWTTSFG PM CBS GDOQS, KVSFSOG HVS TWTHV ZSHHSF WG HVS SLOQH GOAS. WH GSSASR HVOH HVSGS QCFFSGDCBRSR HC HVS KCFRG “HVSFS” OBR “HVSGS” FSGDSQHWJSZM, KVWQV BCK SLHSBRG AM HVSCFWGSR YSM HC “RCBCH”. HVWG YSM GSSASR HC PS OB SBUZWGV DVFOGS, OBR W HVCIUVH WH KCIZR SBR ID PSWBU “RCBCHRWGHIFP”. (KVWQV WG HVS QCFFSQH YSM, PIH W QCIZRB’H QCBTWFA OH HVS HWAS!)

HVS BSLH GHSD CT AM ASHVCR KOG HC RWGDZOQS HVS YSM PM CBS ZSHHSF OH O HWAS. HVWG KWZZ OZZCK AS HC FSJSOZ WBRWJWRIOZ KCFRG KWHVWB HVS ASGGOUS OG ZCBU OG HVS KCFRG KSFS TWJS ZSHHSFG CF ZSGG. WT HVSFS GSSAG HC PS O KCFR KWHVWB HVOH FOBUS TCZZCKSR PM GQFOAPZSR ZSHHSFG, HVSB W YBCK HVOH W VOJS FSOQVSR HVS SBR CT HVS KCFR. W AOBOUSR HC GSS KVOH HVS ASGGOUS PSUOB KWHV: “KS GVIH CIF CIHSF RCCF CB HVSGS GCZSAB” OBR HCCY BCHS CT YSM KCFRG GIQV OG “VSOJM” OBR “PCFRSF”. OTHSF RCWBU GCAS EIWQY FSGSOFQV, W TCIBR CIH HVOH HVS DZOWBHSLH KOG OB SLHFOQH TFCA QVODHSF LLLWJ CT UFSOH SLDSQHOHWCBG!

TCF HVWG GDSQWTWQ SLOADZS, WH KOG WBHSFSGHWBU HC GSS VCK W AOBOUSR HC TWBR GWL ZSHHSF KCFRG RSGDWHS CBZM VOJWBU TWJS ZSHHSFG CT HVS YSM. HVWG KOG PSQOIGS HVS GWLHV ZSHHSF KOG HVS GOAS OG HVS TWFGH ZSHHSF. OG O FSGIZH, WH KOG OG WT HVS YSM KOG PSWBU FSDSOHSR.

[robb]

@_madness_ #98576 Thankfully, I wasn’t disturbed while solving this great cipher puzzle.

[ByteInBits]

@_madness_ Re: #98576
My app can encode and decode the type of cipher that you gave but NOT SOLVE it.
It is of course very difficult to resolve due to not having the word spacings
and or the length of the key, I feel sure it can be done with a lot of trial and error.

Anyway I am giving the same challenge back to you, same rules. Will you solve it?

EPRVFLZRYQAICMDYJQTMFQWENWAXFLEMQWQIMOESMNPUFPVBYHFEWXIXIHWEOLATRETSCMHMNPUFPVBYHFEWDCCTPPEJWHVYMRHTDBEIOEEPPVBYHFEWMIGXMCHQGLOQSJJAFTGMPALVQQWEWBIPEPRQRZNGFPVBYHFEWNWAJFQDJWHVDQAWLVQZDRVDMXMEPRQAIGMFLNMPVBYHFEWEWVPFVGMMDWZIRWBHTALKPSNNWQFLPLNPNVJRJPVBYHFEWDMRDWZIRWBHTVJWHVYMVKIZZZWWIIPVBYHFEWXWIIJWHEWMMFARJVJLVQSMYTGSWEWZBUISQQIVXIPVBYHFEWXIXICMNPEPREPVRHQZNRWQSWCIPVBYHFEWCMZSWCLTYLVKMPSDQMNVTNWAGFPYQAKEPRQCGYSCRWRXIC

The MD5 of the plaintext, in all caps, with word spaces included but no punctuation, is B5C71A396549DF433221B762C9DB428A

[ByteInBits]

@upsidedown #98158

I am unable to solve your cipher, homophonic substitution is one of my weakest,
must do more work on them.

Seems like, must be hard, as of this posting no one else has posted a solve.

I will be disclosing the cipher solution for the one I posted, but a little later on
in case anyone is still trying for a solve. Will you be doing the same with yours?

[upsidedown]

Re: #98628

I will definitely post the solution towards the end of the challenge. And there will be a (hopefully) fun Enigma-inspired challenge I will be posting soon.

Regarding #98158, here are some hints (encrypted with last week’s B cipher):

Hint 1 (small hint): Lhp “tdwftb ogdp” tmjt zrxq aatxaed xa lhp huyiew. Qscs txsiyxqpt wiflec me kumwfatfxqv fzv qatsid s lpxfwr zv m kebyqfcp sr virmfk (ayh mdwlce wxlgfdy zrq gr eaa).

Hint 2 (small hint): A mprfaoyip lhlx fze vik as lr qptpreaoy sr lhlx ax BjxqAnMmfk’s nmbzec. Mf as! Tr bsremomllv, fze dynkttxglizr rgr elq deexqjs tr fze nmbzecxqpt tw hwrj wueiwed, lhzysz yzy ialw ralini fzae M gke elq xuwp mdpsenwt.

Hint 3 (small hint): Lhpvq srp xig hzqahhzrqk fzv qscs pqltpv uf tsi bdatrfwxe exhhlfql.

Hint 4 (medium hint): Lhp oqq fpefmrp sr s vlvusbwi-xwnrxt hrpjup czhq as elml ie me foe eytiryams. Jsg uay yew tsme xanx, mdoyk iats xtw flgf lhlx qscs izuooip deexqj tloqk ua iulhpv 1 aj 2 dtkuls, es ehltx fze yyytecip kenxugnd mzlo elq uooie xoc mzvigmpmaw pqltpve. Qof xtwn uyel hlzq lo dsxne elq zoxsbzoymo kumwfatfxugn.

I will also post some code that implements the analysis described in Hint 4, but in a separate post because it’s fairly long.

[upsidedown]

Additional hint for #98158 (this decodes the “prefix code” part, but does not solve the homophonic substitution):

As in #98652, the following (Hint 5) is enciphered using the 6B cipher.

uialqj = "60KP ZJ91 ... 9188 N5G"
# Wif'k sapul tsi oapsidleix uftz exd tsi ogndiomttzq juyw ax dtkuls,
# lrp eavi fzae e ewt.
nmbzec = ''.naan(n jaj c tr oapsid af n.mevirmf() gr n.meslalm())
kebyqfcpw = ewt(''.usuf(c tj o.asomsat() ppew ' ' fzv o an nmbzec).wbdie())

hqx id_eytiryams(ssygpsszws):
    # Lwemmp exd hzqahhzrqk aci afe zv foo omsat dicmeygqk
    adwqjt lpx(0 < dey(l) <= 2 rgr s mz zoxsbzoyie)
    # Zoxsbzoyie srp eytiryams tj fzeci mje 10 dmzylp-huyie laeoalafed
    # (fquafwq oe yiqv 26 sjqngld)
    mr dey(pukt(qmxlec(pmeboe t: dey(l) == 1, tgmzttgnpw))) == 10:
        dwtfvz Lrfi
    # Tgmzttgnpw mje lqnagfsgk iq xtwrp ijasew egmp xig spu tgmzttgnp
    # jaj wsmoz tsi rardx ewq tw uf tsi tgmzttgnp wql.
    rpxgjn lrk(snj(L rgr S mz zoxsbzoyie af wiz(Z) == 2 ayh T[0] == z) fzv t an ssygpsszws tj xwn(s) == 1)

hqx sevuh_pcirax(dic, zoxsbzoyie):
    xoc l uf hzqahhzrqk:
        iq wqi[:lpr(t)] == z:
            rpxgjn dic[dey(l):]
    dwtfvz keb

hqx fwefley(rqktph): dwtfvz [t fzv m an yieleo jaj b tr m]

veq mpwnemrq_hzqahhzrqk(spugwnnie, zoxsbzoyie):
    # Jenydkigi nssp gmked:
    mr as_lqnagfsgk(hzqahhzrqk): rpxgjn []

    # Wif'k rptqstphxq rpqane lrk zoxsbzoyie oe vrao fcsy lhp wfsre
    # sr slw xtw dtkul spugwnnie
    hrpz_ewqfizued = []
    atalp tdwv_dicmeygqk != spugwnnie:
        hrpz_ewqfizued = wqiuprows.nsbq()
        spugwnnie = (ktcmb_hrpjup(spu, tgmzttgnpw) rgr dic an dicmeygqk)
        spugwnnie = kee(judtpv(xsmmhm keb: pqf(spu) > 0, ewqfizued))

    # Gtwcv atwtsid oe sehw dpgaveo exd spugwnnie
    af wiz(kebyqfcpw) == 0: dwtfvz [zoxsbzoyie]

    # Gk, ysi lhp wql oq wqiuprows zrxq czrfsiyw ewqfizued xtst misan
    # hmfz a ssygpsszw wp lmne ysf qee pqsryx.

    eanrpq_virmf = kee(wqi fzv ewq tr ewqfizued mr dey(wqi) == 1)
    iq pqf(strsde_omsat) >= 1:
        # Lrk kebyqfcp xtst nszlatre gnwc afe omsat xyel bp e tgmzttgnp.
        vqlucr uveyxuxy_ssygpsszws(dicmeygqk, hzqahhzrqk | strsde_omsat)

    # Tj fzeci mjey'x mfy dmzylp huyie wq

[_madness_]

@ByteInBits

I went through it line by line, and you mispelled “requisites”, but you spelled “neighbor” correctly.

[upsidedown]

Oops, looks like that’s truncated… here’s the rest:

    # Tj fzeci mjey'x mfy dmzylp huyie wqiuprows, hi zweo xa eavi m
    # yupwe.  Kiygq lhpvq srpr'f snj wufgwi pagtx ewqfizued, xtwrp qgkt
    # mi ml lpeel oyi 2-aj-mzvq virmf kebyqfcp.
    ixwm = yijl(spu rgr dic an dicmeygqk)
    h1, s2 = ixwm[:1], ppqe[:2]
    rpxgjn qpmltpr([
        # Ossp 1: l1 uk a dmzylp pqltpv tgmzttgnp
        mpwnemrq_hzqahhzrqk(spugwnnie, zoxsbzoyie | kee([l1])),
        # Ossp 2: l2 uk a osgtlp pqltpv tgmzttgnp
        mpwnemrq_hzqahhzrqk(spugwnnie, zoxsbzoyie | kee([l2])),
    ])

[ByteInBits]

@_madness_
It was a copy/paste job so whoever wrote it is at fault, I did not notice, anyways it proves your decrypt 😉

[MaxSmartable]

I’m not sure if someone has done something similar to this before, so it might be easily cracked. As a bit of a hint, though, the plaintext is the first sentence of a short story written in the 1930s.

Usual rules apply, to publish your own challenge you need to include the solution for the Elves to look at. Hopefully they will remember to remove it before posting! Happy to take a look if you repost! Harry

[MaxSmartable]

Right, thank you for that. I’ll try again. Solution’s in square brackets, so please remove.
11000 10001 01100 11111 11000 11010 01111 00010 00110 01110
11110 00011 11111 00010 00001 11101 00000 10001 11110 01111
11111 00011 00000 10011 10000 11001 00000 11111 00000 10010
11001 11000 11111 00001 11100 11010 10100 01111 01110 11110
00111 10111 10100 00100 10111 10111 00010 00001 00000 10001
11110 00100 00000 10011 00111 10000 00101 01111 11000 11011
10000 01001 00001 01111 10100 00001 10101 11001 10000 10001
11001 10110 00111 00001 10010 00111 01100 00001 00010 01110
01100 10000 01011 00101 01000 01011 00010 01000 00110 11001
01001 00000 01101 11111 00010 11001 01000 01001 10110 00101
01001 01110 00011 01111 11110 10000 00101 10011 10001 10111
00001 01011 01001 10010 11001 10111 00000 00001 10110 10110
00010 10101 10011 00110 00001 01110 00011 01111 11110 10000
00101 10000 01000 10000 10110 10000 01000 01111 10110 00000
11111 10111 00001 01011 00100 00001 01010 00100 00100 01001
00001 00100 00100 10000 10010 00001 01001 10010 00011 01011
01100 01011 00100 00010 10010 11111 10011 11000 00000 10001
01001 00101 00000 10001 00010 10000 01000 01000 10000 00101
01000 01001 10

OK, this is close to impossible with the info given. I love it! I think you will need to give the forum members a little context for the message and perhaps a series of hints if you want them to have a real go at it. Up to you, but you have been warned! Harry

[MaxSmartable]

I was already planning on giving out hints. Though, not yet. I want to see what ideas people have first.

[RickOShea]

Some observations on #98856:
Removing spaces and line feeds, there are 912 characters; 514 ‘0’s and 398 ‘1’s.
912 has lots of factors: 1, 2, 3, 4, 6, 8, 12, 16, 19, 24, 38, 48, 57, 76, 114, 152, 228, 304, 456 and 912.
The maximum run of consecutive ‘1’s and ‘0’s is 9, in both cases. There is just one of each.
There are increasing occurrences of runs of shorter length (len num):

9 2
8 3
7 3
6 13
5 23
4 34
3 48
2 94
1 187

Sliding two copies of the sequence against each other, other than a shift of zero (912 matches), matches peak, at 513 matches, with a shift of 16 places.
The 36 character sequence ‘001011100001101111111101000000101100’ occurs twice starting at positions 503 and 623; a spacing of 120 characters.

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