Since Z occurs significantly more often than any other ciphertext character, we might conjecture that d_K(Z) = e. The remaining ciphertext characters that occur at least ten times (each) are C, D, F, J, M, R, Y. We might expect that these letters are encryptions of (a subset of) t, a, o, i, n, s, h, r, but the frequencies really do not vary enough to tell us what the correspondence might be.

At this stage we might look at digrams, especially those of the form -Z or Z-since we conjecture that Z decrypts to e. We find that the most common digrams of this type are DZ and ZW (four times each); NZ and ZU (three times each); and RZ, HZ, YZ, FZ, ZR, ZV, ZC, ZD, and ZJ (twice each). Since ZW occurs four times and WZ not at all, and W occurs less often than many other characters, we might guess that d_K(W) = d. Since DZ occurs four times and ZD occurs twice, we would think that D_K(D) ∈ {r, s, t}, but it is not clear which of the three possibilities is the correct one.

If we proceed on the assumption that d_K(Z) = e and d_K(W) = d, we might look back at the ciphertext and notice that we have ZRW and RZW both occurring near the beginning of the ciphertext, and RW occurs again later on. Since R occurs frequently in the ciphertext and nd is a common digram, we might try d_K(R) = n as the most likely possibility.

At this point, we have the following:

              ------end---------e----ned---e------------
              YIFQFMZRWQFYVECFMDZPCVMRZWNMDZVEJBTXCDDUMJ

              --------e----e---------n--d---en----e----e
              NDIFEFMDZCDMQZKCEYFCJMYRNCWJCSZREXCHZUNMXZ

              -e---n------n------ed---e---e--ne-nd-e-e--
              NZUCDRJXYYSMRTMEYIFZWDYVZVYFZUMRZCRWNZDZJJ

              -ed-----n-----------e----ed-------d---e--n
              XZWGCHSMRNMDHNCMFQCHZJMXJZWIEJYUCFWDJNZDIR

Our next step might be to try d_K(N) = h, since NZ is a common digram and ZN is not. If this is correct, then the segment of plaintext ne - ndhe suggests that d_K(C) = a. Incorporating these guesses, we have:

              ------end-----a---e-a--nedh--e------a-----
              YIFQFMZRWQFYVECFMDZPCVMRZWNMDZVEJBTXCDDUMJ

              h-------ea---e-a---a---nhad-a-en--a-e-h--e
              NDIFEFMDZCDMQZKCEYFCJMYRNCWJCSZREXCHZUNMXZ

              he-a-n------n------ed---e---e--neandhe-e--
              NZUCDRJXYYSMRTMEYIFZWDYVZVYFZUMRZCRWNZDZJJ

              -ed-a---nh---ha---a-e----ed-----a-d--he--n
              XZWGCHSMRNMDHNCMFQCHZJMXJZWIEJYUCFWDJNZDIR

Now, we might consider M, the second most common ciphertext character. The ciphertext segment RNM, which we believe decrypts to nh-, suggests that h- begins a word, so M probably represents a vowel. We have already accounted for a and e, so we expect that d_K(M) = i or o. Since ai is a much more likely digram than ao, the ciphertext digram CM suggests that we try d_K(M) = i first. Then we have:

              -----iend-----a-i-e-a-inedhi-e------a---i-
              YIFQFMZRWQFYVECFMDZPCVMRZWNMDZVEJBTXCDDUMJ

              h-----i-ea-i-e-a---a-i-nhad-a-en--a-e-hi-e
              NDIFEFMDZCDMQZKCEYFCJMYRNCWJCSZREXCHZUNMXZ

              he-a-n-----in-i----ed---e---e-ineandhe-e--
              NZUCDRJXYYSMRTMEYIFZWDYVZVYFZUMRZCRWNZDZJJ

              -ed-a--inhi--hai--a-e-i--ed-----a-d--he--n
              XZWGCHSMRNMDHNCMFQCHZJMXJZWIEJYUCFWDJNZDIR

Next, we might try to determine which letter is encrypted to o. Since o is a common letter, we guess that the corresponding ciphertext letter is one of D, F, J, Y. Y seem to be the most likely possibility, otherwise, we would get long strings of vowels, namely aoi from CFM or CJM. Hence, let’s suppose d_E(Y) = o.

The three most frequent remaining ciphertext letters are D, F, J, which we conjecture could decrypt to r, s, t in some order. Two occurrences of the trigram NMD suggest that d_E(D) = s, giving the trigram his in the plaintext (this is consistent with our earlier hypothesis that d_E(D) ∈ {r, s, t}). The segment HNCMF could be an encryption of chair, which would give d_E(F) = r (and d_E(H) = c) and so we would then have d_E(J) = t by process of elimination. Now, we have:

              o-r-riend-ro--arise-a-inedhise--t---ass-it
              YIFQFMZRWQFYVECFMDZPCVMRZWNMDZVEJBTXCDDUMJ

              hs-r-riseasi-e-a-orationhadta-en--ace-hi-e
              NDIFEFMDZCDMQZKCEYFCJMYRNCWJCSZREXCHZUNMXZ

              he-asnt-oo-in-i-o-redso-e-ore-ineandhesett
              NZUCDRJXYYSMRTMEYIFZWDYVZVYFZUMRZCRWNZDZJJ

              -ed-ac-inhischair-aceti-ted--to-ardsthes-n
              XZWGCHSMRNMDHNCMFQCHZJMXJZWIEJYUCFWDJNZDIR

It is now very easy to determine the plaintext and the key for Example 1.10. The complete decryption is the following:

Our friend from Paris examined his empty glass with surprise, as if evaporation had taken place while he wasn’t looking. I poured some more wine and he settled back in his chair, face tilted up towards the sun.¹

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¹P. Mayle, A Year in Provence. A. Knopf, Inc., 1989.

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