How does this riddle from "Speak Memory" work?

April 3, 2004 2:29 PM

In Nabokov's autobiography, "Speak Memory," there is a puzzle of sorts. It goes like this (from pg. 70): "We subjected [Uncle Ruka] to a test one day, and in a twinkle he turned the sequence '5.13 24.11 13.16 9.13.5 5.13 24.11' into the opening words of a famous monologue in Shakespeare." I'm stuck, can anyone help?
posted by adrober to media & arts (16 comments total)

Are you looking for someone to tell you what the "opening words" in question are, or what the code is, or just a nudge in the right direction towards figuring it out for yourself?
posted by m-bandy at 2:40 PM on April 3, 2004

The biggest hint is the repetition of 5.13 24.11.
posted by eatitlive at 2:45 PM on April 3, 2004

Another hint:

It's the insanely obvious one.
posted by abcde at 2:49 PM on April 3, 2004

eatitlive is right...guessing the phrase isn't very hard, just by looking at the structure and symmetry of the thing (each number pretty clearly represents a letter).

The actual code is somewhat knottier.
posted by kickingtheground at 2:50 PM on April 3, 2004

You can just let the cat out of the bag and explain how it works. I'm not very good at codes. Thanks!
posted by adrober at 2:50 PM on April 3, 2004

Oh duh... "to be or not to be"... ok that was insanely easy. But how does the code work? That doesn't seem so easy.
posted by adrober at 2:51 PM on April 3, 2004

My original impression was it was the cryllic positions in the alphabet, but I dunno.
posted by abcde at 2:57 PM on April 3, 2004

It's a simple substitution cypher, but we only get 6 different letters with the phrase. I don't see an obvious pattern:

A-
B-24
C-
D-
E-11
F-
G-
H-
I-
J-
K-
L-
M-
N-9
O-13
P-
Q-
R-16
S-
T-5
U-
V-
W-
X-
Y-
Z-

Because there's so much repetition in the phrase, Nabokov probably just picked a few numbers at random. Or perhaps the numbers related to his synaesthetic perception of the letters...
posted by eatitlive at 3:06 PM on April 3, 2004

For what it's worth, the Russian version says on ochen' bystro obratil "5.13 24.11 13.16 9.13.5 5.13 24.11" v nachal'nye slova izvestnogo monologa Gamleta: 'he very quickly turned [code] into the opening words of Hamlet's famous monologue." I guess he decided to make it harder for the English-speaking audience. (I doubt there was any system to the substitution; it was just a matter of creating the necessary pattern. On the other hand, knowing Nabokov, it's possible he hid some arcane pattern in there, perhaps based on chess.)
posted by languagehat at 5:36 PM on April 3, 2004

eatitlive is correct, it's a cypher, not a classic sequence.

You can search for this sequence over at The On-Line Encyclopedia of Integer Sequences.

If sequences are of general interest to you, I suggest the links anchor in addition to this page.
posted by sequential at 5:51 PM on April 3, 2004

Well, i just took a crack at it.

Firstly, its clearly not a linear transformation (Ax+b) of a typical alphabet encoding. I also tried the sequence against letter frequency tables as an ordering scheme (englsh and russian) I tried a few pattern things and, well, got, nowhere, except for this which is intriguing:

Number the alphabet 1-26, A=1, Z=26

If you assume the first letter is T
then the next letter is T-5=O
the next is O+13=B
the next is B-24=E

Now, the next digit (13) is already "taken" so we place an O there instead of continuing with our +/- addition.

The two remaining letters are indexed from B
R=b+16
N=b+9

I stumbled upon this when I noticed that the number after every letter in eatitlive's table above always allowed one to reach another valid letter by counting that number of letters either up or down.
posted by vacapinta at 6:58 PM on April 3, 2004

Apparently we have some MEFites working late at the NSA...
[Who knew?]
posted by Fupped Duck at 8:11 PM on April 3, 2004

vacapinta, I can't seem to generalize your observations.

Starting with T, the next letter is five letters away. (Straight subtraction, 20 - 15 = 5)

From O, the next character is a space. O is 13 places away from B. (Straight subtraction, 15 - 2 = 13.)

From B, we have to get to E. Using straight subtraction, it's only three, which is wrong. Going the long route to E is 23. To make it equal to 24, you need to count from 1, starting at B. (Normally, you'd count from one starting at A.)

Oh, lookie, we've seen O. We know it's a 13 and we've broken the pattern. To this point, the number corresponding to each letter was a map of how to find the next letter.

New rule, how many places from B is R? 18 - 2 = 16. (Why B?)

From N, how many spaces is O? One. That didn't work. How many spaces from B? 13, which is also wrong. How many spaces from E? 9. (Why E?)

The last letters we have seen already, O, T, T, O, B and E.

Given we have three exceptions (R, N and the value for B) and a lot of repeated letters (T, O, B, E), I don't think we can encrypt the rest of the first sentence.

languagehat, I looked into it being chess related. Neither of the common notations fit into this pattern easily. Can't say for certain there isn't a connection. Having thought about it a bit more with the insights shared here, it does not look like a reasonable system can be derived from this example.
posted by sequential at 10:46 PM on April 3, 2004

sequential: I did state that I "got nowhere" :)

It did seem odd that every letter can be "indexed" from another and was worth sharing if it spurred someone else to a more valuable insight (except for that arithmetic mistake i made for B-E (oops) which now makes that, sadly, an irrelevant observation.)

I agree, it does look like its just a simple arbitrary substitution cipher but its still fun to see if Nabokov threw another trick in there. Cyrillic shouldn't play into it since this resolves to english text.
posted by vacapinta at 11:41 PM on April 3, 2004

Hmm, it might be helpful to have more context for the quote. Presumably if Nabokov wanted it broken, he would have made it a meaningful puzzle, rather than a random encryption technique.
posted by Hildago at 12:21 AM on April 4, 2004

It is a neat encryption idea, though, vacapinta, and something that might take a few minutes to crack for those used to straight substitution cyphers. The following is the same statement, encoded such that each number is the number of forward steps required to reach the next letter:

0.21 13.3 10.3 22.1.5 0.21. 13.3

Interestingly, any single letter's representation depends on the preceding letter. But in any repeating sequence of letters ('to be'), there is repetition after the second letter (or, in this case, beginning with the first, thanks to the double 't'.)

My own personal favourite substitution cypher is pretty arcane: I use a particular 5x5 magic square (as pictured here; the letters in the square, if converted to 'numeric equivalents' (a=1, b=2, etc., though q is removed so that p=16, r=17, etc.) form a 'true' magic square, which sums along diagonals as well as rows and columns), and then choose some path through the square that suits me on whatever day I happen to be writing. So sometimes it will be {1=A, 2=O, Y=3, ...} , but on others, {1=M, 2=W, 3=E, 4=P, ...}. Thus the particular cypher is very random-seeming, but always ties back to a single unguessable structure. All you need to do is guess the starting point and the path. Eventually, maybe I'll find an even more structured way to go about the process, relying on some personal piece of information to determine those two inputs. I love the idea of internal orderings creating external chaos.

Of course, cyphers are computational child's play, and practically meaningless as a way of really hiding information. All the more reason to apply an art to your obfuscations. I would expect nothing less from Nabokov.
posted by kaibutsu at 4:09 AM on April 4, 2004