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I've added a twist to my normal cryptogram methods. I want to know if this twist makes the puzzle too difficult and, if so, how far back should I scale the efforts until the twist doesn't make it unsolvable. The cipher used is not that complicated but I don't know how much that complication is increased by the twist. Here is my method:

  1. Post the original cryptogram
  2. Wait $n$ days to see if someone finds the plain text ($n$ is currently unknown)
  3. If someone solves it, consider the latest cipher text posted to be acceptable and exit this loop.
  4. If the twist is the only step remaining between the plain text and the latest cipher text posted, consider the twist to be unacceptable for use with cryptograms and exit this loop.
  5. If nobody solves it, add a slightly simpler version of the cipher text to the post.
  6. Return to step 2 until the loop is exited.

WYYXMNDGDFWFDGVJTDYUTQDWYGXEYDQPEYEOMUCFAMEBQTUMQVLJEYMQOPNPMUVNGUVEGJJKUWNYFGNFYGGYMFGYPJFODZTDTALYCWMEPUMQQDDWEJGMYYVGJEJDQFEKYGQMUYNUPMVGZBUYTJJALYDFWWFYDTCBYHMJZHVVKYYWDCVUZFJDWUV

$n = 0.23$

The end of sentences has been included.

WYYXMNDGDFWFDGVJTDYUTQDWYGXEYDQPEYEOMUCFAMEBQTUMQVLJEYMQOPNPMUV. NGUVEGJJKUWNYFGNFYGGYMFGYPJFODZTDTALYCWMEPUMQQDDWEJ. GMYYVGJEJDQFEKYGQMUYNUPMVGZBUYTJJALYDFWWFYDTCBYHMJZHVVKYYWDCVUZFJDWUV.

$n = 0.00$

All punctuation and spacing has been included.

The cipher part is a simple substitution cipher.

WYYXM NDGDFW FDGV JTDY UT QDWY GXEY D QPEY E OMUCFAMEBQ TUMQ VLJEYMQOP NPMUV. NGU VEGJJ KUWNYFG NFYGGYM FGY PJFODZTDTA LY CWMEP UM QQDDWEJ. GMYY VGJEJ D QFEKYG QMUY NUPMV, GZBUYTJJA LYDFWWFYDTC BYHMJZHV VKYYWDC VUZFJDWUV.

$n = 8.69$

The twist is the only thing remaining between the plain text and the cipher. To clarify, to go from plain text to the last edit (above), the process is: Plain Text, Twist, Cipher. Per Step 4 of the method, the loop will now exit.

NEEVR WIHITN TIHS LFIE OF MINE HVAE I MDAE A CROGTYRAPM FORM SBLAERMCD WDROS.

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  • $\begingroup$ the million-dollar question. sometimes, you think a twist is simple, yet weeks go by with it unanswered, and other times, you think it's too difficult and someone solves it in less than an hour. there is definitely no hard-fast rule, but common sense is a little bit helpful, i think. $\endgroup$
    – JLee
    May 19, 2015 at 15:16
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    $\begingroup$ The main problem is that if we know 10 ciphers and invent 20 twists, we must attempt to decrypt it 200 times! And likely we won't make any progress... $\endgroup$
    – leoll2
    May 19, 2015 at 18:09
  • $\begingroup$ @leoll2 I've seen that comment elsewhere and it certainly makes sense. What about if I tell you it's a simple substitution cipher and give you the spacing and punctuation? $\endgroup$
    – Engineer Toast
    May 19, 2015 at 19:07
  • $\begingroup$ @EngineerToast I've already made that comment on another question, so probably you've seen it there. Knowing that it's a simple substitution cipher and the punctuation is a great hint! Thanks! $\endgroup$
    – leoll2
    May 19, 2015 at 19:08
  • $\begingroup$ I doubt this is a "simple" substitution cipher, I can't make any progress on it. I could be wrong though, curious what the answer is. $\endgroup$
    – Quark
    May 19, 2015 at 20:47

2 Answers 2

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The plain text is:

Never within this life of mine have I made a cryptogram from scrambled words.

The twist is:

Scrambling the interior letters of each word before applying the cipher

While it appears to make it robust against attack, it renders it useless for Puzzling.

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    $\begingroup$ *facepalm I wonder why nobody got it. What line of thinking could possibly allow somebody to solve this? Now if you said you scrambled the words in advance, MAYBE this could be solvable. $\endgroup$
    – Quark
    May 28, 2015 at 16:57
  • $\begingroup$ That's an interesting point. I'm not sure how to make a puzzle out of that without simply stating that the words were scrambled and I'm trying to get better at making puzzles instead of just "I did a thing now you undo the thing" $\endgroup$
    – Engineer Toast
    May 28, 2015 at 19:24
  • $\begingroup$ Rectification of the last line: The hmaun eye is vrey good at fdinnig flwas in rdeabale txet and tehn coverntnig it to rdaaelbe txet on the fly, so to spaek. Tihs has a bnuch of erorrs, and you may not eevn konw, prodvied taht the frist and lsat lteters are the smae, and all the ltteers are the smae in the "smacbreld" and "unscrambled" vraiant, and you are not pyanig any anetttoin to the txet. Tihs, of crouse, maens taht all 1-, 2- and 3- lttres wrods are not scamlrbed. It is byeond me how it wroks, but ask the noolieugstrs and ohsohitaollgptms who konw mroe tahn me. $\endgroup$
    – garr890354839
    Oct 11, 2016 at 13:07
  • $\begingroup$ So, the answer is not this plaintext, but a plaintext followed by CIPHER $\endgroup$
    – garr890354839
    Oct 13, 2016 at 5:16
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What is the key to this nonsense now?

WYYXM NDGDFW FDGV JTDY UT QDWY GXEY D QPEY E OMUCFAMEBQ TUMQ VLJEYMQOP NPMUV.

NEEVR WIHITN TIHS LFIE OF MINE HVAE I MDAE A CROGTYRAPM FORM SBLAERMCD WDROS.

Knowing even part of the plaintext also helps massively, as seen here.

The partial resultant key from the known plaintext is

W=n, Y=e, X=v, M=r, D=i, G=h, F=t, N=w, V=s, J=l, T=f, U=o, Q=m, E=a, P=d, O=c, C=g, A=y, B=p, L=b

That key get us:

NEEVR WIHITN TIHS LFIE OF MINE HVAE I MDAE A CROGTYRAPM FORM SBLAERMCD WDROS. WHO SAHLL kONWETH WTEHHER THE DLTCIzFIFY BE GNRAD OR MMIINAL. HREE SHLAL I MTAkEH MROE WODRS, HzPOEFLLY BEITNNTEIFG PEhRLzhS SkEENIG SOzTLINOS

Unscrambling the words gets us:

Never within this life of mine have I made a cryptogram from scrambled words. Who shall kONWETH whether the DLTCIzFIFY be grand or minimal. Here shall I MTAkEH more words, HzPOEFLLY benefitting PEhRLzhS seeking SOzTLINOS.

Using inference, two of the three missing keys are:

Since the missing letter in hopefully is a u, then Z=u and since the missing letter in Maketh is K, then k is itself.

Then the string is now:

Never within this life of mine have I made a cryptogram from scrambled words. Who shall knoweth whether the difficulty be grand or minimal. here shall I maketh more words, hopefully benefiting PEhRLUhS seeking solutions

The missing answer is this "H" character. I know that the letters "I", "R", and "S" are NOT used in the cipher. Since there are only 4 unused characters, it is easy to go through all 4 in turn.

The used letters are: ypgiath..lkbrwcdm..fosnveu, in alphabetical order they are: abcdefghi.klmnop.rstuvw.y., so the missing letters are j, q, x, and z.

PEJRLUJS if H=j

The problem with q's in english: YOU NEED 1 U FOR 1 Q, so therefore H is not q

PEXRLUXS if H=x

PEZRULZS if H=z

Since one of the 3 is actually a proper English word, then I have found a key!!!

The key is:

W=n, Y=e, X=v, M=r, D=i, G=h, F=t, N=w, V=s, J=l, T=f, U=o, Q=m, E=a, P=d, O=c, C=g, A=y, B=p, L=b, Z=u, K=k, H=z, I=j, R=x, S=q

And, by the way, your cryptogram is secure, PROVIDED THAT you don't give part of the plaintext, or part of the key.

Now, the only thing left is to check it.

And it works!!!

NEEVRWIHITNTIHSLFIEOFMINEHVAEIMDAEACROGTYRAPMFORMSBLAERMCDWDROSWHOSAHLLKONWETHWTEHHERTHEDLTCIUFIFYBEGNRADORMMIINALHREESHLALIMTAKEHMROEWODRSHUPOEFLLYBEITNNTEIFGPEZRLUZSSKEENIGSOUTLINOS

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