Two to the power of the dyads

The algebra field counts solutions. A four-bar has two assemblies; a Gough platform has forty rather than the eighty a total-degree bound predicts; a five-position synthesis has a root count that comes out of a homotopy. In every case the count is obtained by solving something and looking at how many answers there are.

Here is one obtained without solving anything.

The prediction #

Hold a link, drive a neighbour, and the rest of a mechanism comes apart into the smallest sets that can be positioned one after another. When every one of those sets is a pair of links, the position problem is a sequence of small problems, each of which is two circles meeting.

Two circles meet in two points. So a pair contributes exactly two solutions, the choices are independent, and a chain that decomposes into $k$ pairs has

$$2^k$$

assemblies at a given input angle.

That is a statement about a count of configurations derived from a graph, with no dimensions anywhere in it. The four-bar is one pair, so two. Watt’s chain is two pairs, so four. Stephenson’s chain from a driving choice that decomposes is also two pairs, so four.

Checking it #

A prediction of this kind is worth exactly as much as the check, so the check is a count of distinct converged solutions from random seeds.

The system solved is the one this field’s mechanisms are drawn from: the pin coordinates are the unknowns, each link’s own pin-to-pin distances are the equations, the frame’s pins are held, and the driven link’s angle is prescribed. Seed it 240 times from random perturbations of a reference placement, keep the solutions that converge below $10^{-12}$, and count the distinct ones. Two solutions are the same when every pin agrees to a millionth, which is generous by six orders of magnitude against the residual and therefore cannot be merging genuinely different assemblies.

The four-bar returns two, which is the answer the field has carried since its first essay. Stephenson’s chain returns four. Watt’s chain returned eight, and eight is not a power of two the decomposition predicted — it is one power too many.

The four that were not poses #

Eight is not four, and the discrepancy is the useful part of this rung.

The equations are distances. Three distances fix a triangle up to reflection — the mirror image of a triangle has exactly the same three side lengths — so a link carrying three or more pins has a mirror image that satisfies every equation in the system exactly. Watt’s chain has two ternary links, one of which is the frame and therefore held; the other is free to be mirrored, and four of the eight solutions had it mirrored.

A reflected link is not a different pose of the mechanism. It is a different part — a plate machined as its own mirror image — and drawing one would be a figure showing a configuration the mechanism cannot reach, which is the one thing this site’s premise forbids.

The repair is a sign test. Take the first three pins of every link with three or more, compute the signed area of the triangle they make, and require it to match the reference placement. Continuation preserves the sign automatically; a cold solve does not, and a frame that has flipped is now reported as unconverged rather than drawn.

Watt’s four, drawn #

The four genuine assemblies of Watt’s chain are worth describing, because the multiplication is easy to state and hard to picture.

Hold the frame and set the driven link’s angle. The first pair now has both of its outer pins on placed links, so its free pin lies on two circles: two answers, one on each side of the line joining the placed pins. Call them up and down.

Choose one. Now the second pair has both of its outer pins placed — one on the frame, one on the first pair’s output — so it too has two answers, and the choice is independent of the first.

Four combinations: up-up, up-down, down-up, down-down. Each is a genuinely different mechanism configuration, each is reachable from a different starting assembly, and no continuous motion of the input takes one to another without passing through a limit position where a pair’s two circles become tangent.

That independence is the whole content of the exponent. If the second pair’s choice depended on the first, the count would not multiply, and the reason it does not depend is structural: the second pair is placed from links that are already fixed, so it does not know how they got there.

What a reflection looks like #

The mirrored solutions are worth describing too, because they are not obviously wrong when drawn and that is the danger.

The ternary link of Watt’s chain is a triangle with three pins. In a mirrored solution the three distances are exactly right, all three pins are exactly where two other links need them, and every equation in the system is satisfied to machine precision. What has happened is that the triangle has been turned over: its vertices go round the other way.

Physically that is a different part. A plate with three bearings and a plate that is its mirror image are two components, and no motion of a mechanism turns one into the other — it would have to pass through the plane of the drawing.

The general statement is that the distance formulation is a relaxation. It describes rigidity correctly and orientation not at all, so its solution set is larger than the mechanism’s configuration space by a factor of two for every link that has an orientation to lose. A link with $k \geq 3$ pins has one; a binary link does not.

The alternative formulation — three coordinates per body rather than two per pin, with the pins expressed in each body’s own frame — has no such relaxation, because a body’s orientation is one of its unknowns. It is what the networks field uses and it is the right formulation for a rank computation. The distance form was chosen here for the solve because it has no body frames to choose, which is one fewer place for a sign to be wrong; the reflection is what that choice costs, and the sign test is the price paid for it.

Why the relaxed formulation is kept #

The obvious response to a formulation that admits four solutions which are not poses is to use the one that does not, and it is worth saying why this site does the opposite and pays a sign test instead.

The two formulations differ in what a link is. In the distance form a link is a set of mutual distances between pins, and the pins are the unknowns — two numbers each, seven pins, fourteen unknowns for a six-bar. In the body form a link is a rigid frame with an origin and an angle, and the pins are expressed in that frame — three numbers per link, eighteen for the same six-bar. The second is larger and it is exact: a body’s orientation is one of its unknowns, so a mirrored link is not a solution of anything, it is simply a body at a different angle, and the reflection cannot arise.

What the second formulation buys in exactness it spends somewhere else, and the place is not obvious until it is written. Every equation in it refers to a body frame, so every equation contains a choice — where the origin sits on the link, which direction counts as zero — and none of those choices is visible in the answer. A sign convention wrong on one link produces a mechanism that solves, closes, sweeps and is subtly the wrong shape. The distance form has nothing to get wrong in this way: a distance between two pins is a distance, with no frame in it and no convention attached, which is why it was chosen and why every figure in this field is drawn from it.

So the choice is not between a safe formulation and a dangerous one. It is between two failure modes, and they differ in the property that actually matters here: whether the failure is detectable by something cheap. A reflected link satisfies every equation to twelve figures and is caught by a single signed area, computed once per link with three pins and compared against the reference placement. A body frame with an inverted convention satisfies every equation too, and there is nothing comparable to test — the check would have to be a second implementation of the same geometry, which is not a check, it is a hope that two authors made different mistakes.

That is the general principle worth taking from this rung, and it is not the obvious one. A relaxation is safe to the extent that the things it lets in can be recognised. The distance form lets in exactly one family of impostors, that family has a closed-form signature, and the signature is a sign — so the relaxation costs a line of code and buys a formulation with no conventions in it. A tighter formulation whose own errors are invisible is the worse trade, however much better it sounds stated as this one cannot produce a wrong answer.

The bookkeeping is worth having explicitly, because it also says which chains are exposed. Only a link carrying three or more pins has a reflection at all, and only a link that is not the frame can take it, since the frame’s pins are held where they are. Watt’s chain has two ternary links, one of which is grounded in the arrangement solved here — one exposed link, one binary choice, and four genuine assemblies doubled to the eight the search returned. A four-bar has no exposed links and returns exactly its two. An eight-link chain with three free ternary links has more room for impostors than either, and the prediction the count makes about it is a thing to be tested rather than a thing already known.

That is the right way to leave it. The exponent this essay is named for predicts assemblies from the graph; the reflections are a second population, also predicted from the graph, and the total the solver returns is the sum. A search that comes back with a number matching neither is reporting something new, which is precisely what a prediction is for.

Why a four-bar would never have shown it #

This is the part worth carrying, because it is a general observation about where a defect hides.

Every link of a four-bar is binary. Two points have no handedness — there is no such thing as a mirrored bar — so the distance formulation is exact for a four-bar and there is nothing to catch.

The site’s whole linkage repertoire up to the six-bar is binary. So a formulation with this defect would have passed every test the collection could have offered it, right up to the moment a ternary link appeared — and the ternary link appeared in a field about graphs, where the mechanisms are generated rather than chosen.

That is the same shape as the degenerate-chain failure of the mobility count: a rule tested only on inputs already known to be well-behaved, in a collection where all the inputs were well-behaved, until a field arrived that generated its inputs.

And how the discrepancy was noticed #

By a count disagreeing with a count, which is the field’s standing method.

Nothing was wrong with the solve. Every one of the eight solutions had a residual below $10^{-12}$; every one was a genuine root of the system as written; a plot of any of them looks like a mechanism. What said something was wrong was that the graph predicted four.

Without the prediction there would have been nothing to notice. Eight assemblies is a perfectly plausible number for a six-bar, the extra four are geometrically sensible-looking configurations, and no residual, no rank and no convergence test distinguishes them. The only instrument that fired was an integer computed from a graph.

What the prediction does not cover #

Three limits, and they are the boundary of the whole rung.

It applies only to all-dyadic decompositions. A chain whose groups include a set of four links has no such formula: the group’s position problem is a polynomial system that does not factor into quadratics, and its root count is a genuine algebraic question rather than a power of two. Four of the sixteen eight-link chains have no all-dyadic driving choice at all, so the prediction does not apply to any way of driving them.

It counts solutions over the reals and over the complexes alike, and only some of them assemble. Two circles meet in two real points, in one, or in none, and the last case is a real mechanism failing to close. So $2^k$ is the count where every intersection is real, and at a configuration where a pair has separated, the count drops. That is exactly the limit position of the pair, and it is where the two things called jamming live.

And it says nothing about which branch a mechanism is on. A count of four is not a way of choosing among four, and following one continuously through a motion is the continuation problem rather than a counting one. Every sweep in the topology field is a continuation for exactly that reason.

Where this sits in the field #

The algebra field’s other counts are hard-won. Twenty-eight rather than forty took a monodromy computation; the Gough platform’s forty took a homotopy over eighty paths; the five-position generator count took a total-degree solve and a check that every root reached its points.

This one is a graph and an exponent. It covers a much narrower class — all-dyadic mechanisms, which is most of the classical repertoire and a minority of the ten-link census — and within that class it is free, exact, and available before any dimension is chosen.

It is also the only count in this field that is a property of the topology. Every other one depends on the dimensions somewhere — the number of real solutions varies with the parameters even when the number of complex ones does not, which the field measures directly. Two to the power of the dyads does not vary at all, because the two circles meet in two points whatever their radii, and a change of dimensions can only take intersections away rather than add them.

What would make it a harder theorem #

It is worth saying what the honest generalisation would look like, because the exponent is a special case of something the field could ask properly.

For a mechanism whose decomposition contains a group of four links, the position problem is a polynomial system and its root count is a genuine algebraic question — one that would be answered the way the field answers all of them, by a total-degree bound and a homotopy that reports how many of the paths arrive. That number would then be the factor the group contributes, in place of the two a pair contributes, and the assembly count would be a product of per-group root counts rather than a power of two.

That computation is not built. What it would produce is a count for the ninety ten-link chains that have no all-dyadic driving choice, and a check on the general claim that the assembly count factors over the decomposition at all — which is obvious for pairs and is not obvious in general.

What is established here is narrower and exact: for every chain and every driving choice whose decomposition is all pairs, the assembly count is two to the power of the number of them, and the count from a solve agrees.

It is recorded as owed rather than as a limitation, because the machinery for it exists in this field already and the missing piece is only the work of applying it to a group of four links rather than to a platform. What that would settle is whether the assembly count factors over the decomposition is a theorem or a coincidence of the dyadic case — and the honest position today is that it is stated for pairs, checked for pairs, and unknown beyond them.