The lead is wherever it is taken out

Charge flipping reconstructs a structure from diffraction amplitudes by a two-step loop: make a map from the current phases, reverse the sign of every point below a threshold, and keep the new phases while restoring the measured amplitudes. The heavy atom the flipping solver cannot see past found that one lead among nineteen carbons stops the loop entirely: the lead owns the map’s spread and its ripples are larger than any carbon. The lead taken out solves what no threshold could found the repair. Compute the lead’s own density from the same reflections, subtract it from the map before flipping, flip what is left, put the lead back. The lead cell went from nought to fifty-five per cent of its starts.

The repair used the lead’s exact position, and the essay ended on the obvious worry. In practice the lead’s position comes from a Patterson map and is known only approximately. Subtract it a fraction of a grid step from where it really is and the subtracted density no longer cancels the lead’s own; it leaves a positive lobe on one side and a negative one on the other. The essay estimated that for a weight of eighty-two that residual would pass a carbon’s height at a displacement of a few hundredths of the cell, and asked for two runs: the census with the lead displaced by known amounts, and a version that refines the lead’s position from the map as it goes.

This essay makes both. The residual is worse than the estimate. The success rate does not notice it. And the reason is not a subtlety of the flipping rule but a fact the earlier essay had already written down and not applied: a lead in a cell with no symmetry has no position to be wrong about.

The residual is larger than predicted #

The prediction can be checked without running the solver at all. The lead’s density is computed from the data’s own reflections at its true position and at a displaced one, and the largest value of the difference is compared with the peak a carbon makes on the same grid. A tenth of a grid step, three thousandths of the cell, already leaves 1.3 carbons. Half a step leaves six. One step leaves eleven, and by two steps the two copies of the lead have separated and the residual levels off near fourteen, a positive lead and a negative one side by side.

So the earlier estimate was optimistic by an order of magnitude. A lead at weight eighty-two is so much heavier than a carbon that a displacement far below the grid’s own spacing leaves an error larger than the atoms the solver is trying to find. If the subtraction worked by cancelling the lead, a position good to a few hundredths of the cell would not be nearly good enough; it would need to be good to a few thousandths.

The rate does not notice #

The figure at the head of this essay is the census. The same five arrangements and eight starts as the earlier one, the same threshold, the same two hundred and fifty cycles, and the subtracted lead displaced by 0, 0.1, 0.25, 0.5, 1, 2, 3, 4, 6, 8, 12 or 16 grid steps, each start in its own random direction. With the lead exactly in place the solver succeeds in 22 of 40 starts, the earlier essay’s fifty-five per cent. At a tenth of a step, where the residual already passes a carbon, it succeeds in 19. At three steps, eleven carbons’ worth of residual, in 18. At half the cell, where the subtracted lead is as far from the true one as the cell allows, in 9.

The numbers scatter, from 9 to 22, and the scatter has no trend. Forty starts at a success rate near forty per cent give a binomial standard deviation of about three starts, and every point lies within two and a half of them of the pooled rate, with no trend in either direction. Displacing the subtracted lead by any amount, to half the cell, leaves the success rate where it was, and far above the original solver’s, which with nothing subtracted solves none of the forty.

The second run asked whether the solver could correct a rough position by itself, by moving the subtracted lead every ten cycles to the nearest peak of the map, located to a fraction of a grid step. It does not help: at every displacement the refined runs succeed in 14 to 20 of 40, no better than the runs that held the position fixed. And it hardly moves the position. A refined run that started three steps off ends a median 3.1 steps off; one that started exactly in place drifts a fifth of a step away. The peak search, tested on the lead’s own density, moves a two-step guess onto the lead to within five hundredths of a step. In the solver’s map there is nothing to move it towards.

The structure is built where the lead is put #

The maps say why. In every solved run the reconstructed lead, the strongest peak in the final map, sits within half a grid step of the place the lead was subtracted, and the true lead’s place is empty. The carbons are at the right distances and directions from it. The solver has not found the structure at its true position with a defect; it has found the structure translated so that its lead lies under the subtracted one, and the success test, which compares the map with the true structure up to a shift and a choice of hand, recognises it. That shift is the whole of the phase problem’s freedom in this group, and the solver has spent it on the reference.

That is possible because nothing in the data fixes the origin. Three phases that do not move built the subject’s invariants on exactly this fact: shifting every atom by one vector changes every phase and no amplitude, so a cell with no symmetry can be described with its origin anywhere. The earlier essay noticed it in passing, that a solver told where one atom is has been told the origin. It follows that a single heavy atom in a cell with no symmetry has no position to get wrong. Wherever it is subtracted, there is a description of the structure in which it is exactly there, and the solver finds that description.

The residual is therefore real only at the start. In the first cycles the map, built from random phases, has no lead anywhere in particular, and the subtracted density is simply a fixed pattern the flipping must work around. As the phases improve the structure assembles round the reference, and once it has, the lead is where the reference is and the residual is gone. The order-of-magnitude error in the estimate never matters, because the estimate assumed the structure would stay put while the reference was wrong, and the structure moves instead.

Which half of the repair does the work #

The repair has two parts, and taking them apart says what kind of thing it is. The first part changes the threshold: subtracting the lead removes its peak from the map whose spread sets the flipping level, so the level drops. The second keeps the lead itself out of the flipping: it is put back after the flip, so its density is never reversed. Run separately, the first does nothing, nought of forty starts when the threshold is taken from the map without the lead and the whole map is then flipped against it. The second alone, flipping the remainder and not putting the lead back, solves one start in forty.

Only the two together work. Flipping the remainder and putting the lead back amounts to reflecting the low parts of the map about the lead’s density rather than about zero: a point that would have been flipped to its negative is instead flipped to its mirror image about the reference. A reference that is never flipped is a fixed feature the loop can organise round, and that is what a single heavy atom gives it, whether or not it is in the right place.

The last four bars check one possible artefact. The map lives on a grid of thirty-two by thirty-two, and a lead subtracted on a grid point, half a step off, at the centre of a grid cell or a quarter step off is represented differently on that grid. The four placements succeed in 18, 15, 16 and 18 starts of forty, the same within the scatter. The rate is not set by where the lead falls relative to the grid either.

Where the earlier explanation stands #

The earlier essay’s account of why the subtraction works, that the lead’s ripples are larger than a carbon and removing the lead removes them, survives in a shifted form. It is true in the frame the solver ends up in, where the reference lies under the reconstructed lead and the ripples cancel. What the offset census adds is that the frame is chosen by the solver, not by the crystallographer. The ripples cancel because the structure moves to make them cancel, and a subtraction anywhere gives the structure a place to move to.

It also changes what the method needs. The earlier essay framed the repair as the heavy-atom method and charge flipping combined, the Patterson finding the lead and flipping finding the carbons. For one heavy atom in a cell with no symmetry, the Patterson’s contribution is nothing at all: any position will do, including one chosen without looking at the data. What the Patterson, which solving from the vector set showed can find heavy atoms from vectors alone, is needed for is everything else, the number of heavy atoms, their kind, and, as soon as there are two of them, the vector between them, which a translation cannot absorb.

How much freedom the origin has, group by group #

The freedom the solver exploited is a property of the group, and it shrinks quickly once the group has any operation besides translations. The same pattern, described twice found that the descriptions of one structure that differ only in where the origin sits are counted by the group’s normaliser, and in the group with no symmetry that normaliser contains every translation, a continuous family. That is why the census found no position worse than any other: every position is the origin of some correct description.

A centre of symmetry changes this completely. A structure with a centre is described with the origin on the centre, and there are only eight such places in a cell, the corners, the face centres and the body centre of the cell and their combinations at halves, the finite set two origins for one group and one crystal, and sixteen coordinate lists count. A heavy atom subtracted between two of them cannot be carried under itself by any translation the data allow, because a translation that is not by one of those halves would destroy the centre. In such a group a misplaced reference is a genuine error, and the residual figure, a carbon’s worth at a tenth of a grid step, would set how good its position has to be.

So the result here is specific in a way its numbers do not show. In the group with no symmetry, which is where charge flipping works, ignoring whatever symmetry the structure has, as the solver that knows no symmetry found, the reference costs nothing to place. Run the same repair with the symmetry imposed, and the Patterson’s estimate of the heavy atom’s position becomes a measurement that can be good or bad. The step from one regime to the other is the step from a continuous normaliser to a finite one.

That also explains why the refinement never moved the position. A peak search moves an estimate towards the nearest maximum of the map, and in a solver that relocates the structure to its reference, the nearest maximum is the reference itself, so the search confirms whatever it was given. In a symmetric group, where the structure cannot follow the reference off the permitted origins, the same search would have something to find.

What the census must refuse #

The claim refused is the one the earlier essay proposed to test: that the subtraction needs the lead’s position to within a few hundredths of the cell. A lead subtracted a quarter of the cell away still solves three of eight starts in the refusal’s small sample, and the census’s forty-start points say the same at every offset to half the cell.

What the pictures cannot show is the trajectory. The maps are the end of two hundred and fifty cycles, and the claim that the structure moves to meet the reference is read from where it ends, not watched as it happens. A movie of the cycles would show the lead appearing under the cross and the carbons gathering round it, and it is the one view of the process these figures do not give.

Where the census stops #

One heavy atom, and no symmetry. The whole result rests on the cell having no symmetry, so that every origin is equally good, and on there being one heavy atom, so that a single translation can put it anywhere. In a space group with symmetry the origin can sit only at a few places, those the group’s normaliser allows, and a lead subtracted between them could not be moved under by any translation the data permit. With two heavy atoms, one translation can place one of them under its reference but not, in general, both.

Small two-dimensional cells. Twenty atoms on a grid of thirty-two, one arrangement at a time, as in the earlier census. The argument about the origin does not depend on the size, but the rates do, and real structures have three dimensions, many more atoms and noisy data.

Forty starts per point. Each rate is known to about three starts either way, which is enough to say that no offset changes it by much and not enough to rule out a small dependence. A trend of a few per cent over the whole range would be invisible.

Who saw which part #

Charge flipping is Oszlányi and Sütő’s, from 2004. That the origin in the group with no symmetry is free, and that a single atom can therefore be placed anywhere, is as old as the heavy-atom method itself, which has always started by putting the heavy atom at a convenient place and building the rest round it. What the census adds is a measurement of how completely the flipping loop exploits that freedom: it does not merely tolerate a misplaced reference, it relocates the whole structure to match one.

Still open: two leads, and the vector between them #

With two heavy atoms the freedom runs out. A translation can put one of them under its subtracted reference, and then the other is wherever the true vector between them puts it. If the subtracted pair has the wrong vector, the second lead’s residual cannot be removed by moving the structure, and the earlier essay’s worry returns in full for that vector. The prediction is definite. The success rate should not depend on where the pair is subtracted, but it should fall with the error in the vector between them, on the scale the residual figure sets, which is a tenth of a grid step for a carbon’s worth of error. That is the census here run on a cell with two leads, with the subtracted pair’s separation perturbed rather than its position, and it would say how good a Patterson’s inter-lead vector has to be. It has not been run.