28 September 2006

9/28 cryptogram solution

Here is yesterday's cryptogram:
Frequency analysis shows us that E is the most common letter. Assuming that E cannot represent itself, we have to move on to the next most common letter. In this case, it is a three-way tie between H, S, and Z, all of which appear 9 times. All other letters appear 5 times or less. So we can begin with a safe assumption that H, S or Z represents e.

We have several two-letter words: HI (x2), OZ, SW, and IQ. We also have one three-letter word: HMZ. We already know the is the most common three letter word, and we just discovered that H, S or Z probably represents e. So it looks as if we have ourselves a start: H represents t, M represents h, and Z represents e:
We also see now that our two-letter word HI is now tI. It is difficult to imagine this being anything other than to, so I represents o. There are also three instances of EE, and one of those is in a four-letter word QSEE. The most common double-letters in order of frequency are SS, EE, TT, FF, LL, MM, OO. The only common four-letter words with a double-letter are will and well. Thus E represents l (here is a case where frequency analysis fails us: l is the most common letter when we would expect it to be closer to 11th most common):
to Oe NOle to QSll leSWVTe SBtellSPeBtlU SW the lNWt FToLVAt oQ ASCSlSYNtSoB. – OeTtTNBL TVWWell
Notice that while I thought QSEE might be will or well (and I said E represents l), I didn't make Q represent w. Why? Because of the two-letter word IQ (which we see already as oQ). ow just doesn't make sense, but of makes sense, and so does fill or fell instead of will or well. So Q represents f. Further, we can guess at the second word Oe in context with the first and second. Oe is either be, me, or we, but the third word NOle answers it for us: neither to me Nmle or to we Nwle make sense, but to be Nble could easily be to be able and makes sense. Thus O represents b, and N represents a:
to be able to fSll leSWVTe SBtellSPeBtlU SW the laWt FToLVAt of ASCSlSYatSoB. – beTtTaBL TVWWell
W appears 5 times but we haven't figured it out yet. Here's our chance: our four-letter word laWt is undoubtedly last, so W represents s. This also turns our two-letter word SW into Ss, meaning S represents i as we already found a. This also shows us that fill was the four-letter word we stumbled over earlier:
to be able to fill leisVTe iBtelliPeBtlU is the last FToLVAt of AiCiliYatioB. – beTtTaBL TVssell
The remainder of the words begin to appear: leisVTe is leisure, this V represents u and T represents r. This also gives us enough to realize that bertraBL russell is bertrand russell: B represents n and L represents d:
to be able to fill leisure intelliPentlU is the last FroduAt of AiCiliYation. – bertrand russell
Finally we have revealed enough to find our last few words: P represents g, U represents y. F represents p, A represents c, C represents v, and finally Y represents z. Our solution (with capitilization added):
To be able to fill leisure intelligently is the last product of civilization. – Bertrand Russell
Here is the frequency analysis matchup:


Not as straightforward as before, thus the reason this cryptogram was a little bit more difficult to get started (at least I thought so). We needed a wide variety of methods to attack this one: frequency analysis, common two and three-letter words, common double letters, and even common double letters within a four-letter word. This I think is a very good example of the strength of using two methods in combination with each other.
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