A cycle sums to a whole turn

Thirteen wrecked cuts in the census have no balanced pair of displaced chains, and they were set aside with a question: are three exceptional chains a three-cycle, with three chains rotating into one another’s places, or a pair with a stray attached?

The test proposed was that a three-cycle’s displacements sum to zero. It does not, and the census has exactly one clean three-cycle to prove it on.

The row #

g008 — the golden branch at a rise of 0.008 — cut at offset 6, keeping a lag of 8. Three of its eight chains are displaced and the other five sit at the common level.

The three are chains 6, 7 and 0, which are consecutive round the circle of eight, and their displacements are −89.4, −135.3 and −134.8 degrees. Every one of them goes the same way round.

Chain 0 is the chain the removed organ was on, which is the labelling the exchange uses and is the control’s indexing rather than the cut run’s.

What they sum to #

−359.6 degrees. One whole turn to four parts in ten thousand, on an azimuth grid whose step is a quarter of a degree.

That is not close to zero. It is the largest sum in the excluded set by a factor of two, and under the test as proposed it is the worst row in the table rather than the one clean answer in it.

Why a cycle sums to a turn #

Because a cycle is a rotation. Three chains rotating into one another’s places means chain 6 goes where chain 7 was, chain 7 where chain 0 was, and chain 0 where chain 6 was — and going round the circle once is one turn’s worth of displacement shared between three moves.

Zero is what a cancellation sums to, not what a rotation sums to. A balanced pair sums to zero because one chain goes forwards exactly as far as another goes back; three chains going the same way round add up rather than cancel.

The seventeen rows the exchange covers are all cancellations, and they close within 4.6 degrees — which is what makes them the control for this addition rather than a second population of it.

The three moves are not three equal moves #

A rotation of three chains through one another could have been symmetric — three displacements of a hundred and twenty degrees each, adding to a turn. It is not, and the departure has a shape worth reading.

Two of the three are −135.3° and −134.8°, which agree to half a degree — two steps of the azimuth grid, and closer to each other than either is to anything else in the row. The third is −89.4°, a third smaller.

The two equal ones are the interesting pair, because 135 degrees is not an arbitrary number on this stem. The exchanged pair on a lag-8 row measures 133° to 135° across the census, so the two equal moves here are one exchange step each, to within what the grid can distinguish.

Which recasts the cycle. It is not three chains rotating symmetrically; it is two chains that have each moved by one step in the same direction, and a third that has absorbed the remainder. Two one-step moves and a remainder is a chain of exchanges rather than a rotation, and the two descriptions are different objects even though they close the same way.

The remainder is checkable. Two steps of 135° is 270°, which folded into half a turn either way is 90° — and the third displacement’s magnitude is 89.4°, which is within a degree of it. So the third chain has moved by what the other two left over, and the number it moved by is fixed by the other two rather than free.

That is a stronger statement than “the three sum to a turn”, because a sum has one constraint in it and this has two: the two equal moves are each one exchange step, and the third is what closes the circle. A row with three displacements summing to a turn but with no pair of equal ones would satisfy the first and refute the second.

One row cannot establish it and one row is what the census has. What it does supply is a shape for the next three-cycle to be read against, which is more than a sum would have been.

So the test needed folding #

An angle folded into half a turn either way puts −359.6 degrees at +0.4 degrees, which is the reading that makes the row what it is.

Every sum in the reading is folded for that reason, with the raw value carried beside it so the distinction stays visible. On twelve of the thirteen rows the two agree; on this one they differ by a whole turn.

What folding costs elsewhere #

It is worth saying, because folding is used throughout this collection and its price is rarely stated. A displacement is folded because a lattice and its mirror are one lattice under this rule — nothing distinguishes a turn one way from the same turn the other.

The price is that two very different placements can fold to nearby numbers, and that a sequence crossing a half turn looks like a reversal. Every claim about a folded quantity’s size is safe; claims about its change need the unfolded values checked.

Here it is the other way round #

Which is the neat part. Usually folding hides a distinction and the unfolded values have to be checked; here the unfolded value hides the finding and the folded one shows it.

Both directions come from the same fact: a folded angle and its raw value differ by a whole turn and the reader has to know which question is being asked. How big is this displacement? wants one; do these displacements cancel? wants the other.

How close to a turn is close #

Four parts in ten thousand: −359.6 against −360. The azimuth grid is 1,536 samples of the circle, so one step is 0.234 degrees and the miss is under two steps.

Each of the three displacements is read from a mean over a residue class, so it is not a single grid value and can sit between steps. Three of them summing to within 0.4 degrees of a turn is not something the grid forces.

Is it a coincidence? #

The number to compare against is what three arbitrary displacements would give. The exceptional chains in this census run from 12.4 to 160.4 degrees from the common level, so three of them drawn at random would sum to somewhere in a range several hundred degrees wide.

Landing within 0.4 degrees of a whole turn out of that is about one chance in a thousand. That is a rough figure and it is the right order: this is not a sum that happens to be near a turn.

And the chains are consecutive #

The second piece of evidence. The three displaced chains are 6, 7 and 0 out of eight, which are adjacent round the circle — a rotation among three chains that are not neighbours would be a stranger object.

Adjacency is not proof: several of the other excluded rows have consecutive exceptional chains and do not close. What it does is make the reading coherent, which a sum near a turn among three scattered chains would not have been.

It also fits what the balanced rows do. The exchange runs one way round on every row it covers — the chain displaced forwards immediately precedes the chain displaced backwards — so adjacency is what a rearrangement in this census looks like.

What the other three-exception rows do #

Four rows in the census have exactly three exceptional chains. This one closes; the other three do not.

Their folded sums are −16.9, +41.6 and −173.6 degrees, which are not near zero and not near a turn. So three exceptions is not a category with one answer, and the question as posed — are three exceptions a cycle or a pair plus a stray? — has the answer sometimes one and sometimes neither.

Which is a real answer #

Not a satisfying one, and it is the shape the census supports. Of four rows with three exceptions, one is a clean three-cycle with each chain moving about a third of a turn, and three are rearrangements whose displacements do not cancel and do not close.

The three that do not are not a pair plus a stray either. A pair plus a stray would have two displacements equal and opposite with a third left over, and none of the three has that.

What closes besides this one #

Six more rows close, and none of them is a three-cycle: they carry four, five and six exceptional chains, and their folded sums are −0.2, +2.6, +3.0, −6.2, −8.0 and −10.5 degrees.

Those are cancellations rather than rotations — several displacements in both directions adding to nothing. So closed covers two different objects and only one row in the census is the rotating kind.

Why only one #

Unknown, and the honest answer is that thirteen rows on seven lattices is not a population. What can be said is that the cycle sits at a lag of 8 with three of its eight chains moving, which is a small fraction of the profile.

The other closed rows have most of their chains moving. So the difference between a rotation and a cancellation may be about how much of the profile is involved, and one instance of the first cannot support that.

What the row is worth #

It is one row. What it is worth is that it exists at all: the question the excluded set was set aside with was whether cycles happen, and one clean instance answers it.

It also fixes the test. Anybody adding up a set of chain displacements to look for a rearrangement now knows to fold, and knows that the sum to look for is a whole turn as readily as nothing.

A test that could have been checked first #

The correction cost nothing and could have been made before any arithmetic ran. Three chains rotating into one another’s places sum to a whole turn is a sentence about rotations rather than about this census.

That it was not is the ordinary way of these things: a test gets written down at the end of one piece of work as a suggestion for the next, and a suggestion is not checked the way a claim is. The habit worth taking from it is to check the arithmetic of a proposed test before running it.

The general form #

A quantity defined modulo something, tested against a value that is only correct in one representative. Angles are the obvious case and this collection is full of them.

Two other instances are on record. A settled divergence at 220.9 degrees is the same arrangement as one at 139.1, so a mean over both representations belongs to neither; and a displacement near a half turn folds to a large negative, which is why a transition’s sign has to be checked unfolded.

What the row looks like in the profile #

Worth describing in the raw, since the circle picture is already an interpretation. The profile is how far each organ above the hole sits from where the control put it, and it is read over the last hundred and twenty organs of the run.

Folded onto a lag of eight it gives eight classes. Five of them sit within a few degrees of one another — that is the common level — and three sit 89 to 135 degrees below it.

Nothing in that description says rotation. The rotation is the reading that the three chains below the level are the same three chains, moved into one another’s places, and the sum is what supports it.

What would distinguish a rotation from three coincidences #

A stronger test than the sum, and it is available. If chains 6, 7 and 0 rotate into one another’s places, then the organs of chain 6 should sit where the control put chain 7’s, and so on round.

That is checkable against the control’s own positions rather than against a sum, and it has not been done. It would turn a suggestive arithmetic into a direct observation, and it is the obvious next thing this row deserves.

Why the sum was the test proposed #

Because it is one line of arithmetic on numbers already computed, and the direct check needs the control’s positions organ by organ. That is the ordinary trade and it is usually the right one: run the cheap test, and run the expensive one only if the cheap one says something.

The cheap test said something. It also had the wrong number in it, which is the argument for checking a proposed test’s arithmetic before running it rather than after.

The lattice it sits on #

g008 is the golden branch at a rise of 0.008, which is one of the census’s ten lattices and sits at 84 per cent of the golden 5/8 rung. It wrecks at four offsets — 6, 7 and 8 keeping a lag of 8, and 4 keeping a lag of 5.

Three of its four rows are balanced pairs and go to the exchange. Only the cut at offset 6 has three exceptions, and it is the one that turns out to be a cycle.

So the census’s one rotation sits on a lattice whose other cuts are entirely ordinary, which means nothing about the lattice explains it. Whatever makes offset 6 different is about the offset.

And the rise it sits at #

0.008 is the rise at which the second wall of a slot goes free on this rung — or rather, it is well below the rise where that happens, since the transition is at 0.00997 and the free stretch runs down from there.

That is almost certainly a coincidence and it is noted because both readings are about the same lattice and a reader meeting it twice deserves to know they are unrelated. One is about removing two organs and one is about removing one.

What a reader should carry #

That the census contains one clean three-cycle, at g008 cut at offset 6, whose three consecutive chains each move about a third of a turn the same way and sum to −359.6 degrees.

And that the test it was set — sum to zero — would have called it the worst anomaly in the table. A rotation sums to a turn; a cancellation sums to nothing; and both read as zero once folded, which is why every sum here is.

What the picture at the top shows #

Eight marks round a circle, one per chain of the lag this cut kept, with chain 0 at the top — the chain the removed organ was on.

Three of the eight are filled and carry their displacements: −89.4, −135.3 and −134.8 degrees. They are consecutive round the circle and all three point the same way. The number in the middle is their sum and the words beside it say what it is.

The one line #

g008 cut at offset 6 displaces three consecutive chains by −89.4, −135.3 and −134.8 degrees, summing to −359.6 — one whole turn to four parts in ten thousand, on a grid whose step is a quarter of a degree.

That is the census’s one clean three-cycle, and the test it was set would have rejected it, because a rotation sums to a turn rather than to nothing. Two of its three moves are one exchange step each and the third is what closes the circle, which is a second constraint the sum alone does not carry.