poetic form · the Prosody Workshop

The Seventh Stanza Would Be the First

A sestina does not rhyme. It ends its lines on the same six words over and over, and shuffles them by one fixed rule between stanzas. Everyone teaches the rule. Almost nobody says why the poem stops at six stanzas, and the answer is not taste: apply the shuffle a seventh time and every word is standing exactly where it started. The form closes because the arithmetic underneath it closes. Below you can watch that happen, write one with the machinery keeping score, and turn a dial to the numbers where it does not work at all.

one engine, shared by page and verifier · nine poems, three languages, seven centuries · offline 147/147 · browser 57/57

The rule, and the thing the rule does

Six words. Stanza one ends its lines on them in order. To get the next stanza you read the previous one from the outside in, alternating bottom and top: last, first, second-last, second, third-last, third. 6 1 5 2 4 3. The troubadours called it retrogradatio cruciata, the crosswise backward step.

Here it is running on the six words Arnaut Daniel used in about 1200, in the poem that invented the form. Press step. Watch the seventh row.

The seventh stanza is the first stanza. Not approximately: the same word in every one of the six positions. A poet who wrote a seventh stanza would be writing the first one again, so the form stops at six and adds a three-line envoi instead. Six is not a choice about length. It is the order of a permutation.

And notice what the six stanzas together have done. Each of the six words has stood in each of the six positions exactly once, and no combination has happened twice. The grid is a Latin square. That is what the form is actually for, and the shuffle is a machine for producing it.

The studio

So write one. Put six words in the top row and the machine will tell you, line by line, which one you have to land on, and whether you landed on it. It keeps no other score: the metre, the sense and the argument are yours. Nothing is sent anywhere, and your draft survives a reload.

The check is exact-word and case-insensitive, and it ignores punctuation at the end of the line. It is deliberately strict, because the interesting thing about the form is where you have to fight it. Real poets bend the rule constantly, which is measured further down.

Why six, and what happens at the other numbers

Nothing above used the number six except by arriving at it. The shuffle is defined for any number of end-words, so ask the same question of all of them: start from 1, 2, 3 and so on, apply the shuffle over and over, and see whether every word visits every position before anything repeats.

Drag the dial. The wheel shows the positions, and an arc from each position to the one its word moves to next. When the form works, the arcs make a single closed thread through every point. When it fails, they fall into separate loops that never meet, and each loop is a set of words condemned to circulate among themselves.

separate loops
stanzas before a repeat
does the form close?

Every number from 2 to 60. Highlighted ones close.

Four end-words do not work, and the failure is easy to feel: the third word never moves. It ends the third line of every stanza forever. Seven does not work either, and eight splits the words into two families of four that never trade places. The numbers that do work start 1, 2, 3, 5, 6, 9, 11, 14, 18, 23, and they thin out. They are called Queneau numbers, after Raymond Queneau, who asked which lengths admit the form, and they are A054639 in the encyclopedia of integer sequences. Six is one of the small handful a poet could plausibly have stumbled on.

The shuffle is doubling

Here is the reason the good numbers look so arbitrary. Number the positions 1 to 6 and ask where the word in position k goes. It goes to 2k, unless 2k runs off the end, in which case it goes to 13 − 2k. One rule, two branches, and both branches are the same thing: multiply by two, modulo thirteen, and fold the answer back into 1 to 6 by ignoring the sign.

position k2k2k mod 13folded into 1..6the spiral says

Thirteen because six words means 2n + 1 = 13. So the question "does the sestina close?" is the question "does repeatedly doubling reach every number from 1 to 6 before returning?", which is a question about the multiplicative order of 2 modulo 13, and that is why the answer is a thin, irregular-looking set. The form is a fact about small primes wearing a Provençal costume.

It also explains the failures exactly. With four end-words the modulus is 9, which is not prime, and 3 divides it: doubling can never move 3 anywhere, so the third word is stuck. With eight the modulus is 17, doubling has order 8, but it passes through 16, which is −1, so the fold sends it home halfway and the words split into two orbits of four.

An error you can reproduce

Jacques Roubaud, the Oulipo mathematician and poet who did more than anyone to popularise this question, published 141 as a length that works. It does not. The shuffle on 141 end-words falls into three separate loops of 47, so a 141-word "sestina" would return to its opening arrangement after 47 stanzas with two thirds of the arrangements never reached. The encyclopedia records the correction; the dial above reproduces it if you have the patience to extend the range, and the offline verifier checks it directly.

What nine poets actually did

All of that is about the rule. Whether poems obey it is a different question, and an empirical one, so here are nine of them: the sestinas I could reach in a machine-readable text that is public domain worldwide, from the first one ever written to 1896. No claim that this is every such poem, only that these nine were chosen before any of them was measured. Their end-words were pulled out of the raw source files by a program, and the permutation between each pair of consecutive stanzas was measured.

poemlangdatewordsstanzasspirallatin square

Six of the nine use the spiral and nothing but the spiral, across seven hundred years and three languages, including both of the poems that double the length. That is a stronger result than I expected from a rule transmitted by imitation for that long. It also settles what a "double sestina" of the older kind is: Petrarch and Sidney both run twelve stanzas on six words with the ordinary spiral, and since the spiral closes at six, their stanza 7 reproduces their stanza 1 exactly, which the measurement confirms for both. The second half of a double sestina is the first half again, in a new set of sentences.

The three poems that depart are the interesting ones, and each departs differently.

Spenser rotates

Swinburne trades the spiral for rhyme, and can prove he had to

The twelve-word double sestina cannot be done, and is not

And they bend the words themselves

poemwroteforthe line

One question that is still open in the wrong place

A footnote for anyone who follows the sequence link. A054639 carries a question posted by David Wasserman on 30 August 2011 and never answered there: is the order of this permutation always equal to the length of the orbit of 1? He reported checking it up to 9450. It is a fair thing to ask, because it is false for permutations in general, and the encyclopedia's own entry says "of order n" where a commenter says "is an n-cycle", which are not the same claim.

The answer has existed since 1969 and is one click from the question, in a paper the entry already links. M. Bringer, in Mathématiques et sciences humaines, proves as Lemme 1 that the length of every cycle divides the length of the cycle containing the first element, and draws the corollary directly:

Corollaire. L’ordre de δn est égal à l’ordre de C1 ; en effet l’ordre de δn étant le ppcm des ordres de Ci, est égal à l’ordre de C1.

M. Bringer, "Sur un problème de R. Queneau", Mathématiques et sciences humaines 27 (1969), pp. 13–20, p. 16. numdam.org. Quoted from the Numdam scan, whose OCR renders δ as 8 and C1 as Cx; the symbols are restored here and the sentence is otherwise verbatim.

The argument is one line once you see the map as doubling. If the cycle through 1 has length p, then 2p ≡ ±1 modulo 2n+1, so for any position x, applying the shuffle p times sends x to ±x, and the fold into 1..n makes that x itself. Every cycle length divides p, so the least common multiple is p. Nothing is claimed as new here: this is Bringer's, and the point is only that the question and its answer have been sitting one hyperlink apart for fifteen years. What this page adds is a wider check. The scan below walks the actual cycle structure using no number theory at all, so that it cannot accidentally assume what it is testing.

published checkn ≤ 9,450
checked here
counterexamples

The check

Everything above is recomputed, not typed in. One module, research/sestina-studio/spiral.mjs, defines the permutation; this page loads that exact file and so does the verifier, which also asserts the two copies are byte-identical. verify-the-seventh-stanza.mjs runs 147 assertions offline: the spiral table against the canonical sestina order printed in the reference works, the arithmetic formula against the permutation for every n up to 2000, the Queneau numbers against the published terms of A054639, Asveld's three-case characterisation against the directly computed cycle structure, Roubaud's 141, and every number this page prints about the nine poems, recomputed from the raw source files whose hashes are in research/sestina-studio/sources/MANIFEST.md.

Where this ends, and what it does not know

Part of the Prosody Workshop. Texts from Wikisource and Project Gutenberg, all public domain worldwide; each file's URL, retrieval date and SHA-256 are in the source manifest. Sequence data cross-checked against the On-Line Encyclopedia of Integer Sequences, A054639.