Two leads need their vector to half a step

Charge flipping rebuilds a structure from its diffraction amplitudes by a loop of two steps: make a map from the current phases, reverse the sign of every point below a threshold, then keep the new phases and restore the measured amplitudes. A heavy atom stops the loop, because its ripples are larger than any light atom, and the lead taken out solves what no threshold could found the repair: compute the heavy atom’s own density, subtract it before each flip, put it back after. The lead is wherever it is taken out then found that, with a single lead in a cell with no symmetry, the subtracted position does not matter. Subtracted a tenth of a grid step away or half the cell away, the lead let the solver succeed at the same rate, because the solver built the whole structure round whatever place it was given. A lead in such a cell has no position to be wrong about: every place is the origin of some correct description.

That essay ended on the case where the freedom should run out. With two heavy atoms a translation can carry one of them under its subtracted reference, and then the other sits wherever the true vector between them puts it. If the subtracted pair has the wrong vector, the second lead’s residual is stuck in the map, and the worry the first lead escaped returns for the vector. The prediction was definite in shape and in scale: the success rate should not depend on where the pair is put, should fall with the error in the vector between the two, and should fall on the scale of a tenth of a grid step, where a misplaced lead’s residual first exceeds a carbon.

The census here makes the measurement. The shape is right. The scale is not: the solver tolerates five times the predicted error.

A cell with two leads #

The cell is the earlier lead cell with one carbon exchanged for a second lead: eighteen carbons and two leads, at the same five random arrangements on a grid of thirty-two by thirty-two, with the amplitudes the same Gaussian atoms produce. In the five arrangements the two leads lie 12.0, 2.5, 8.9, 13.1 and 4.2 grid steps apart, so the pair is sometimes close enough for the two densities to overlap and sometimes nearly half the cell apart.

One thing had to change, and it is reported because it changes the comparison. At the settings that served the one-lead cell, a flip threshold of 0.6 standard deviations and 250 cycles, the solver with both leads subtracted exactly solves only three of forty starts. A small sweep of the threshold and the length of the run found 0.8 standard deviations and 500 cycles best for this cell, and every rate below uses them, with sixteen starts on each arrangement, eighty runs a point. The two-lead cell is a harder problem than the one-lead cell, and the rates here are not comparable in level with the earlier ones, only in shape.

Both leads, or nothing #

The controls settle what the method needs before any error is introduced. With nothing subtracted the solver solves none of eighty starts, as it solved none of the one-lead cell’s forty. With one of the two leads subtracted it still solves none. The repair is not a matter of removing most of the heavy density; a single lead left in the map is enough to stop the loop, exactly as a single lead did in the cell that had only one. With both subtracted where they are, it solves twenty-one.

Two freedoms survive from the one-lead case. Moving the pair together by six grid steps, which keeps the vector and changes only the origin, solves twelve of eighty. That is fewer than twenty-one, by more than the noise in either number would make likely but not by enough to be sure: a binomial count near twenty out of eighty is uncertain by about four either way, and the difference is about one and a half combined standard deviations. Whether a rigid translation of the pair costs anything is not settled by eighty starts. Subtracting the second lead at the inverted vector, on the other side of the first, solves eighteen, and that freedom is exact rather than approximate: a pair of points is centrosymmetric, so the inverted pair is the original pair turned over, and the structure turned over has the same amplitudes in a cell with no symmetry. The Patterson’s sign ambiguity costs nothing, because it is the enantiomorph ambiguity every structure in this group already has.

The vector #

The census proper moves the second lead and leaves the first alone. Each of the eighty runs at an error $e$ displaces the second lead’s subtracted density by $e$ grid steps in a direction drawn afresh for the run, and the first lead is subtracted exactly. The figure at the head of this essay is the result. At errors of 0, 0.1 and 0.25 grid steps the solver succeeds 21, 19 and 18 times in eighty, the same rate within the noise. At half a step it succeeds 11 times. At one step, two steps and four steps it succeeds never: nought in eighty, three times over.

The fall is sharp. Between a quarter of a step and one step, three thousandths and three hundredths of the cell, the rate goes from its full value to nothing, and nothing in the one-lead census looked like it: there, sixteen grid steps of error cost nothing. The prediction’s shape holds exactly. A wrong vector is an error no translation can absorb, and the solver cannot work round it.

The maps show what a failure looks like. With the vector exact the carbons appear round both leads. With the vector two steps wrong the strongest features of the map do sit at the two subtracted places, within a quarter of a grid step of each in the three arrangements checked, because the solver puts back the subtracted densities after every flip. But the carbons never organise round them. The leads are where they were put, and there is no structure for which that pair of places is right.

The residual it tolerates #

The prediction’s scale came from the residual: the difference between a lead’s density at its true place and at a displaced one, which the one-lead essay computed and which is the same here for the second lead. It exceeds a carbon’s peak, 1.3 carbons, at a tenth of a grid step, and the earlier essay took a carbon’s worth of spurious density as the threshold beyond which the solver would be misled. The figure puts the census beside it. At a tenth of a step, 1.3 carbons, nothing happens. At a quarter of a step the residual is 3.1 carbons and still nothing happens. At half a step it is 6.1 and the rate halves. At one step it is 10.9 carbons and the solver never succeeds.

The solver absorbs three carbons’ worth of stray density and not eleven, so the scale the prediction used was too strict by about a factor of three in density and five in distance. The reason is plausibly the flipping rule itself. The residual is a positive lobe and a negative one side by side, and the flip reverses the negative lobe along with every other low point of the map, which turns a pair of opposite lobes into a pair of same-signed ones of a carbon’s height or two, a feature the loop can carry for a while as a spurious atom. A lobe of six carbons or more is too large to pass for anything the map contains, and it stays. That reading is offered as an explanation of the numbers, not derived from them.

It is the vector, not the position #

The census displaced only the second lead, so a reader could ask whether the solver minds the second lead’s position rather than the vector. The one-lead result says it should not mind a position, and a direct test agrees. Split the same error between the two leads, half on each, in opposite directions, so that the vector between them is wrong by $e$ and neither lead is where it belongs. At half a step that gives 6 successes in eighty against 11 with the error on one lead, the same order given the noise, and at one and two steps both give none. The solver can move the structure to put one lead under its reference, and then the other is off by the vector error, whichever lead the error was put on. The quantity the solver needs is the vector between the leads, and it needs it to about half a grid step.

Arrangement by arrangement #

The pooled rates hide a spread that the five arrangements make visible, and it qualifies the half-step figure. Broken down by arrangement, sixteen starts each, the counts with the vector exact are 4, 3, 8, 0 and 6; at a quarter of a step they are 3, 1, 10, 2 and 2; at half a step, 1, 0, 10, 0 and 0. One arrangement, the third, with its leads 8.8 grid steps apart, carries almost the whole rate at half a step and does not notice the error at all. The other four lose everything they had by half a step, and two of them have mostly lost it by a quarter. The arrangement with its leads closest, 2.5 grid steps apart so that their densities overlap, falls from three to one to none.

So the honest statement is a range rather than a number. For most arrangements the vector must be right to about a quarter of a grid step; for some, half a step is no trouble at all. What decides which is not the separation alone, since the most tolerant arrangement has its leads at a middling distance and the least tolerant ones span the whole range of separations. It is presumably how the carbons sit relative to the two lobes a misplaced lead leaves, whether the spurious features the flip makes of them fall where carbons are or where nothing is, and five arrangements are too few to say. This is the same lesson the phase problem taught in its general form: what a solver can recover depends on the structure as well as on the data, and a rate pooled over a handful of structures is a statement about those structures.

It also changes what the pooled zero at one step means. It is not that the average arrangement fails there; every one of the five fails there, the tolerant one included, in all sixteen starts. One step is beyond every arrangement’s tolerance, and a quarter of a step is within most.

What a Patterson map has to deliver #

A heavy atom’s position usually comes from a Patterson map, the map of all vectors between atoms, where vectors between heavy atoms stand out because they are products of two large scattering factors. Solving from the vector set found heavy-atom positions from those vectors alone. With one heavy atom the earlier essay found that the Patterson’s estimate was not needed at all, since any position would do, the continuous freedom of origin that the same pattern, described twice counts by the group’s normaliser. With two, it is needed, and it is exactly the right thing: the Patterson measures vectors, and the vector between the two leads is what the solver needs. It does not need the leads’ positions, which the Patterson does not give without an origin choice, and it does not need the vector’s sign, which the Patterson cannot give.

The census says how good that vector has to be: half a grid step of the map the solver works on, and comfortably good at a quarter. On this grid a step is a thirty-second of the cell, so the requirement is a few thousandths of the cell, which a Patterson peak between two heavy atoms ordinarily delivers, since the peak’s width is set by the resolution and its centre can be located to a fraction of that. Two heavy atoms are a measurement problem; one was not. And three phases that do not move is the reminder that the one quantity the data fix absolutely in this group is a combination that no choice of origin changes, which is what the vector between two atoms is in real space.

Where the census stops #

Eighty starts a point. The rates are uncertain by about four either way near twenty, which is why the cost of moving the pair together and the difference between one-sided and split errors at half a step are reported as unresolved rather than as effects. The fall from a quarter of a step to one step, from eighteen to nought, is far outside that noise.

One cell, two dimensions, no symmetry. The leads are two of twenty atoms on a grid of thirty-two in a plane, in the group with no symmetry, where the origin is free and where charge flipping is usually run even on symmetric structures, as the solver that knows no symmetry found. In a group with a centre of symmetry the origin is not free, it may sit only at the eight places two origins for one group counts, and, as the earlier essay argued, even one lead’s position becomes a measurement; the census here is the first step out of the free case, not the general one.

Settings tuned to the cell. The threshold and run length were chosen by a sweep on this cell with both leads exact, and the vector census used them throughout. A different choice would change the level of every rate and could, in principle, change the tolerance. The sweep found the rate at the one-lead settings low enough, three in forty, that a census at those settings could not have resolved a fall.

No refinement of the vector. Every run here holds the subtracted pair fixed for all five hundred cycles. With one lead, refining its position from the map every ten cycles did nothing, because the structure moved to the reference and the nearest peak was always the reference itself. With two leads the refinement could in principle do something, since the second lead’s true density sits a fraction of a step from its reference and a peak search might walk towards it. The census does not test that, and the one-lead experience, where the reconstructed leads sit at their subtracted places even in failed runs, suggests that a peak search on the solver’s own map would mostly confirm the error it was given rather than correct it.

What the pictures cannot show is the loop. The maps are the end of five hundred cycles, and the account of why a residual of three carbons is absorbed and one of eleven is not is read from end states and from the flip’s arithmetic, not from watching the residual’s lobes through the cycles.

Who saw which part #

Charge flipping is Oszlányi and Sütő’s, from 2004, and the heavy-atom method, with its use of the Patterson map to find heavy atoms from their vectors, goes back to Patterson in 1934 and the structure work that followed. That a pair of heavy atoms fixes the origin up to the pair’s own symmetry, where one heavy atom fixes nothing, is how the heavy-atom method has always worked, and the heavy atom the flipping solver cannot see past is where heavy atoms first stopped the loop here. What the census adds is the measured tolerance of the flipping loop to an error in the pair’s vector: a quarter of a grid step free, half a step halving the rate, one step fatal.

What the census must refuse #

The refused claim is the one the previous result invites: that with two leads, as with one, the place where the pair is subtracted does not matter. It matters through the vector. A second lead four grid steps off solves none of the refusal’s small sample where the exact pair solves several, and the census’s eighty-start points say the same from one step on.

Still open: three leads and a hand #

Two points are always centrosymmetric as a pair, which is why the inverted vector cost nothing here. Three points in general position are not: three heavy atoms fix the hand as well as the origin, and their mirror image is a different set of vectors. A cell with three leads subtracted with the wrong hand would test whether the solver, which in this group never had to choose between a structure and its enantiomorph, is forced to the hand the subtracted triangle dictates, and whether it fails, as it failed here for a wrong vector, when that hand is wrong. The census here would run unchanged with a third lead and the triangle reflected. It has not been run.