Friends, I couldn’t let it go.

Here’s the problem with what has been moderately agreed upon to be Swinburn’s double sestina: first, it isn’t widely agreed upon, and some say it’s random, and secondly (and perhaps most importantly), it doesn’t follow the logic of the sestina.

The sestina follows a couple of rules, or rather the same rule laid out differently, to make up it’s order. The basic explanation is retrogradatio cruciata, which may be rendered as “backward crossing”. The second stanza can be seen to have been formed from three sets of pairs (6–1, 5–2, 4–3), or two triads (1–2–3, 4–5–6). The 1–2–3 triad appears in its original order, but the 4–5–6 triad is reversed and superimposed upon it.. (swiped shamelessly from wikipedia as it said it so succinctly.) So, that translates to the 12 line stanza pretty simply:

12 1 11 2 10 3 9 4 8 5 7 6

So! The double sestina would actually be

A B C D E F G H I J K L
L A K B J C I D H E G F
F L G A E K H B D J I C
C F I L J G D A B E H K
K C H F E I B L A J D G
G K D C J H A F L E B I
I G B K E D L C F J A H
H I A G J B F K C E L D
D H L I E A C G K J F B
B D F H J L K I G E C A
A B C D E F G H I J K L
L A K B J C I D H E G F

So, unless I messed up somewhere along the way, it cycles back on line 11. And I haven’t figured out the last line yet. I’ll let you know when I do. And if you find a mess up, let me know in the comments.

Cheers!

Johnny