# Cyclic poetry

Certain fixed-verse poetic forms have a beautiful mathematical structure to them. This article is about the repetitive structure of the sestina, a thirty-nine line form in which certain words reappear throughout in different combinations. Consider this example by Elizabeth Bishop:

Sestina, Elizabeth Bishop.

September rain falls on the house.
In the failing light, the old grandmother
sits in the kitchen with the child
beside the Little Marvel Stove,
reading the jokes from the almanac,
laughing and talking to hide her tears.

She thinks that her equinoctial tears
and the rain that beats on the roof of the house
were both foretold by the almanac,
but only known to a grandmother.
The iron kettle sings on the stove.
She cuts some bread and says to the child,

It's time for tea now; but the child
is watching the teakettle's small hard tears
dance like mad on the hot black stove,
the way the rain must dance on the house.
Tidying up, the old grandmother
hangs up the clever almanac

on its string. Birdlike, the almanac
hovers half open above the child,
hovers above the old grandmother
and her teacup full of dark brown tears.
She shivers and says she thinks the house
feels chilly, and puts more wood in the stove.

It was to be, says the Marvel Stove.
I know what I know, says the almanac.
With crayons the child draws a rigid house
and a winding pathway. Then the child
puts in a man with buttons like tears
and shows it proudly to the grandmother.

But secretly, while the grandmother
the little moons fall down like tears
from between the pages of the almanac
into the flower bed the child
has carefully placed in the front of the house.

Time to plant tears, says the almanac.
The grandmother sings to the marvelous stove
and the child draws another inscrutable house.

## 1 The cyclic structure

As you can see, the form of the sestina relies on an inventory of six chosen words at the end of each line—in this poem, the words are

1. house
2. grandmother
3. child
4. stove
5. almanac
6. tears

A sestina contains six stanzas of six lines each. The six chosen words appear in a different order at the end of the lines in each stanza; they are permuted. In particular, the order is:

ABCDEF, FAEBDC, CFDABE, ECBFAD, DEACFB, BDFECA

As it turns out, the repeated reorderings form a particularly elegant mathematical pattern: not only are they permuations, but they arise from repeated applications of a single cyclic permutation.

A cyclic permutation can be described in several ways. One way is to describe the trajectory of all the parts;

$$1\rightarrow 2 \rightarrow 4 \rightarrow 5 \rightarrow 3 \rightarrow 6 \rightarrow 1$$

which here means that with each application of the cyclic permutation, the first word becomes the second, the second becomes the fourth, the fourth becomes the fifth, etc. To determine the order of the next stanza, simply apply this rearrangement rule to the words as they are ordered in your current stanza.

Another way is to describe the cyclic permutation as a permutation matrix $$P$$:

$$\, \begin{bmatrix} \cdot&\cdot&\cdot&\cdot&\cdot&1\\ 1&\cdot&\cdot&\cdot&\cdot&\cdot\\ \cdot&\cdot&\cdot&\cdot&1&\cdot\\ \cdot&1&\cdot&\cdot&\cdot&\cdot\\ \cdot&\cdot&\cdot&1&\cdot&\cdot\\ \cdot&\cdot&1&\cdot&\cdot&\cdot\\ \end{bmatrix}^k\begin{bmatrix} \text{house}\\ \text{grandmother}\\ \text{child}\\ \text{stove}\\ \text{almanac}\\ \text{tears}\\ \end{bmatrix}$$

Representing the transformation as the matrix $$P$$ has three nice properties: first, rows of the matrix (top to bottom) correspond to lines in the stanza, and columns (left to right) correspond to words. Hence, to know what word should be in line $$i$$ of the next stanza, you look at row $$i$$ and see which column $$j$$ has a 1 in it. Second, this row/column interpretation applies to the powers of this matrix, too; the successive powers of this matrix tell you where the original words go in subsequent stanzas. The first stanza is $$k=0$$ and so on for $$k=0,1,2,3,4,5$$. Third, if you multiply $$P^k$$ by the vector of words as shown here, you will get a vector with the words in the appropriate order for the $$k$$th stanza (again, where $$k$$ starts at $$k=0$$).

Another representation is to use a stacked two-line representation, where you list the numbers in their original order and then write the destination of each number beneath it:

\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 2 & 4 & 6 & 5 & 3 & 1\\ \end{pmatrix}

Swapping the order of these top-bottom pairs, you can write an equivalent two-line representation of the permutation which more clearly shows the cyclic structure. (In this arrangement, the bottom of each entry matches the top of the next entry):

\begin{pmatrix} 1&2&4&\color{red} 5&3&6 \\ 2&4&\color{red}5&3&6&1 \\ \end{pmatrix}

## 2 Properties of cyclic permutations

Each of these representations has its own advantages. The trajectory notation helps you see that a word will return to its original position after exactly six movements—to see this, just trace what happens to any one entry as you move it successively. (Hence the reason why there are exactly six stanzas in a sestina; if we continued the pattern to a seventh stanza, we'd arrive at the original order which we've seen before.)

The permutation matrix $$P$$ is useful because it captures the spatial arrangement of poetic lines in order from top to bottom on paper, making it easier to quickly look up which word goes in each line in order. You can read the rows to see what word you should put in each line, or read the columns to see in which line each word goes. You can also clearly see that the reordering is a permutation, because there is exactly one 1 in each row and in each column. Moreover, you can see the "telescoping" structure of the rearrangement: the first two lines of the next stanza will consist of the outermost lines of the current stanza; the next two will be the almost-outermost lines; the final two will consist of the two innermost lines.

## 3 The half-stanza

After six different combinations expressed in six stanzas, the sestina's cycle is complete. Afterwards, the sestina concludes with a "half-stanza", consisting of three lines with two of the keywords per line.

As you can see with reference to the poem above, if the keywords were originally written in order 123456, in the final half-stanza, they appear in the order: 65/24/31.

We can write another matrix to express the order in which words occur in the stanza:

\begin{bmatrix} \cdot&\cdot&\cdot&\cdot&\cdot&1\\ \cdot&\cdot&\cdot&\cdot&1&\cdot\\ \cdot&1&\cdot&\cdot&\cdot&\cdot\\ \cdot&\cdot&\cdot&1&\cdot&\cdot\\ \cdot&\cdot&1&\cdot&\cdot&\cdot\\ 1&\cdot&\cdot&\cdot&\cdot&\cdot\\ \end{bmatrix}

Though as far as I have found, this order is arbitrary and has no special connection to the sestina permutation.

## 4 The sestina form is a uniquely special permutation

You can think of the sestina as one member of a family of different poetic forms, each based on a different cyclic permutation. For example, you could make a variant kind of sestina using the permutation $$1\rightarrow 2\rightarrow 3 \rightarrow 4 \rightarrow 5 \rightarrow 6 \rightarrow 1$$ instead.

This raises the question: how unique is the standard sestina form? Well, there are $$n!$$ choices of permutation for a list of $$n$$ words. As for specifically cyclic permutations, there are only $$(n-1)!$$ of them1.

For a sestina, this means that there are $$5! = 120$$ possible permutations, each one of which gives its own variant sestina form.

However, as the matrix representation reveals, the standard sestina form has a "telescoping" effect, where the outermost lines of the previous stanza become the first two lines of the next stanza, etc. This pattern, shown in the image below, is unique among all permutations, at least with respect to the following defining traits:

1. The permutation is a "spiral"2, as shown in the picture.
2. The permutation puts each word in a new place; no word keeps its old position. 3
3. The permutation moves from outside to inside. (Instead of 123456 becoming 435261).

Brackets or parentheses are another way of representing this spiral permutation process. If the six symbols below represent the six chosen keywords of your sestina, then the transformation looks visually like:

[({})] → ][   )(   }{

We can generalize this pattern to any number of keywords (corresponding to an equal number of stanzas). I like to call these the cyclic spiral fixed verse forms. Though I don't have a concise way to express this pattern mathematically, I can express it fairly succinctly as a LISP-like (Clojure) subroutine:

(defn next-stanza [wordlist]
(if (< (count wordlist) 2)
wordlist
(concat [(last wordlist) (first wordlist)]
(recur (butlast (rest wordlist))))))


In detail: to reorder the words for the next round, start with the list of words in their current order and an empty list of words for their new order. Until you run out of words in the old list, take off the last and first words. Write those words down on the new list.

## 5 Any-length sestinas and automata

I note in passing a connection to automata theory. I had thought that the parenthesis-like matching of this spiral pattern would make it possible to use a kind of pushdown transducer to read in a list of words and print them out in the proper sestina order. However, I don't believe that this is possible for pushdown automata. Instead, here is a different result: considering your sestina keywords to be part of an alphabet $$\Sigma$$, there is a linear bounded transducer 4 which reads the word and prints out the sestina rearrangement:

Linear bounded transducer algorithm for performing sestina permutations

First, let's assume a convention about the input format and starting state. The input is written onto a tape, with out-of-bounds symbols ■ in the two cells on either side of the input. The tape head begins at the first input symbol.

2. If ■ is not in the current cell, go to 1.
4. If ■ is in the current cell, halt.
5. Print the contents of the current cell.
6. Overwrite the current cell with ■.
8. If ■ is not in the current cell, go to 7.
10. If ■ is in the current cell, halt.
11. Print the contents of the current cell.
12. Overwrite the current cell with ■.
13. Go to 1.

To confirm that this algorithm works, I have provided Python code for simulating linear bounded transducers (including this one as an example).

stop = None # the "out of bounds" symbol

verbose_output = False

### DEFINING LINEAR BOUNDED AUTOMATA
### (they are like Turing machines with finite tape)

def write_cell(state, symbol) :

def move_left(state) :

def move_right(state) :

def writeout(state, symbol) :
state["print"] += [symbol]

def writeout_current_cell(state) :

def halt(state) :
return -1

def LBA_initial_state(word) :
return {"tape_contents" :
[stop]*2 + list(word) + [stop]*2,
"print" : []}

def lba_process_instructions(state, instructions) :
## an "instruction" is a function that takes in the LBA's current
## next. If no number is provided, proceeds to the next
## instruction.

instruction_ptr = 0
while 1 :
if verbose_output :
print "ln "+str(instruction_ptr)+".\t current state:\t", state["tape_contents"]
ret = instructions[instruction_ptr](state)
if ret is None :
instruction_ptr += 1
elif ret == -1 :
return state["print"]
else :
instruction_ptr = ret

if instruction_ptr >= len(instructions) or instruction_ptr < 0 :
return "OUT OF BOUNDS EXCEPTION"

def lba_sestina(word) :

state = LBA_initial_state(word)
instructions = [
move_right,
lambda s : 0 if read_cell(s) != stop else None,
move_left,
lambda s : halt(s) if read_cell(s) == stop else None,
lambda s: write_cell(s, stop),
move_left,
lambda s : 6 if read_cell(s) != stop else None,
move_right,
lambda s : halt(s) if read_cell(s) == stop else None,
lambda s: write_cell(s, stop),
lambda s: 0
]

return lba_process_instructions(state, instructions)

bishop_words = ['house','grandmother','child','stove','almanac','tears']

w = bishop_words  # alternatively, uncomment the next line
# w = "123456"
for i in range(6) :
print list(w)
w = lba_sestina(w)


## Footnotes:

1

You can see this in at least two ways. First, imagine taking an arbitrary permutation and putting arrows in between the elements as we did with the trajectory representation, for example: $$2\rightarrow 4\rightarrow 1 \rightarrow 3\rightarrow 2$$. This makes any permutation into a cyclic permutation. But because in a cyclic representation, it doesn't matter which element you list first, the permutations 2413, 4132, 1324, and 3241 all refer to the same cyclic permutation. There are in general $$n$$ starting points for a cycle of $$n$$ elements, all of which result in the same cyclic permutation. So we divide out the duplicates by dividing by $$n$$. Another strategy is to argue recursively: if you have a cycle of $$n$$ elements, you can make a cycle of $$n+1$$ elements by inserting $$n+1$$ anywhere into the cycle. There are $$n$$ choices of insertion point and each results in a different $$n+1$$ cycle. The base case is that for two elements, there is only one cyclic permutation.

2

One way of defining this is that the line numbers of the words that appear next to each other after the permutation all add up to the same sum, 7. For example, after one reordering, 123456 becomes 615243. 6+1 = 5+2 = 4+3.)

3

This rules out other spirals such as 123456 becoming 16-25-34.

4

A linear bounded automaton (LBA) is a machine slightly more powerful than a pushdown automaton and slightly less powerful than a Turing machine: rather than an infinite amount of writeable tape, an LBA has only an amount of tape proportional to the size of the input. Also, a general note: any kind of automaton (Turing machine, pushdown automaton, finite state machine) can be made into a transducer by equipping it with the ability to print symbols to a second, write-only tape.