The set that can be reached is not the set that can reach

A distance is symmetric: the distance from A to B is the distance from B to A, so within three thousand kilometres of London names one set however the question is put. Every reach set on the three rungs below this one inherits that, silently, by being built out of a distance.

The previous rung replaces the distance with a travel time under a flow. The triangle inequality survives; the symmetry does not.

Two questions, and they are different questions #

Where can this vehicle get to in six hours? The cheapest route out, from here.

Where can a vehicle get here from in six hours? The cheapest route in, to here.

Under a distance these are the same set, and under any cost that is symmetric they are the same set. Under a flow they are not, and the reason is one line: a route that runs downwind on the way out runs upwind on the way back, so the two problems see the flow with opposite signs.

Computationally that is one flag: the same wavefront with the flow’s contribution negated. That is worth stating as an implementation detail because it is also the argument. Negating the flow is exactly what travelling the other way does to it, so the backward problem is the forward problem in a reversed field — and every property of the pair follows from that single substitution rather than from a second model. Everything else — the grid, the neighbours, the rim-finding — is identical, which is what makes the comparison a comparison rather than a comparison of two implementations.

How different they are #

The two areas agree to five significant figures, and that is not a coincidence: reversing a uniform flow is a reflection of the whole problem, so the backward set is the forward set turned through half a turn. Same shape, same area, opposite orientation.

Which is exactly what makes the overlap small. Two identical elongated sets pointing in opposite directions from a common point share only their middle, and the more elongated they are the less of a middle there is. So the same elongation that makes each set interesting makes the pair disagree, and the two findings of this ladder’s last two rungs are one finding read twice.

So the summary statistic that is easy to compute — the area — is the one that carries no information about the difference, and the statistic that carries all of it is a set comparison nobody performs.

How weak a flow it takes #

The first row is the control and the rest is the finding, and the shape of the column matters as much as its size.

The curve is steepest where the flow is weakest. A tenth of the vehicle’s speed is a light breeze or a slow current — the kind of thing a planner rounds away — and it already puts a fifth of the ground in one set and not the other.

That shape is worth pausing on, because it is the opposite of the usual reassurance. The usual reassurance is that a small perturbation has a small effect, and it is true of the area, which moves by one per cent at a flow of a tenth. It is not true of the set, which moves by twenty-two.

The control, which is the whole method #

With no flow the cost is symmetric and the two sets have to be the same set. They are, to 0.03 per cent of their union — which is the grid’s own resolution and not a physical difference.

That control is doing more work than it looks. The forward and backward computations run the same code with a sign flipped, so a bug in the sign, in the neighbour list or in the rim-finding would show up as an asymmetry with no flow. It does not appear, and every asymmetry above is therefore the flow’s.

The second control is the area. A uniform flow reversed is a reflection, so the two sets must have equal areas whatever their shapes, and they do to five figures. A measurement that produced two different areas would be reporting an artefact of the lattice’s own orientation.

Why the overlap is the right statistic #

Three ways of saying how far apart two sets are, and only one of them says anything here.

The difference of their areas is zero, exactly, and would license the conclusion that the two questions have the same answer. It is the statistic a reach-set summary actually reports.

The distance between their centroids is not zero and is not the point either: two sets can share a centroid and be disjoint, and these two do share one — the source is on the axis of both.

The overlap as a share of the union — the Jaccard index — is the one that carries the difference, because it is the only one of the three that is a statement about the sets rather than about a summary of them. It is 0.323 here, and it falls to 0.160 at a flow of three quarters.

That is a general caution about set comparisons and not a local one. A statistic computed from a set is not a comparison of sets, and the two coincide only when the sets are similar enough that the summary determines them — which is exactly the case a reach set under a symmetric cost is, and exactly the case this rung is not about.

Where this changes a practice #

Three places, and the first two are ordinary operations that reach sets are used for.

A coverage question is directional and is never asked that way, in the way a corridor’s width is not the width anybody meant. Which places can a rescue vessel reach in four hours and which places can reach the vessel in four hours are different sets, and a station sited to maximise the first is not sited to maximise the second. A prevailing current makes them differ by two thirds of their union, and the purpose has to be named before the answer is — which is this field’s standing rule, arriving in a question about sets rather than about projections.

A nearest-facility partition is not a partition. Nearest of many is a partition divides the sphere by which source is nearest, and under an asymmetric cost there are two divisions — by which facility can reach a place soonest, and which facility a traveller at that place can reach soonest — and they do not agree. Emergency response uses the first and the second is what a patient travelling under their own power experiences.

And an isochrone map has to say which way it was computed, in the same way every figure here says which projection it is drawn in. A drive-time isochrone from a shop is a different object from the isochrone to it, and a road network with one-way streets and asymmetric turn costs makes them differ without any flow at all.

The partition, which is where it gets expensive #

The sharpest consequence is not about one source. It is about several, and it is the one that breaks a construction this ladder already has.

Nearest of many is a partition divides the sphere by which of several sources is nearest, and the division is well defined because a distance is symmetric: nearest to and nearest from are the same relation. Under an asymmetric cost they are two relations, so there are two partitions, and a place can belong to one facility in the first and to another in the second.

The size of that disagreement follows from the numbers above: if a single source’s two sets differ by 68 per cent of their union at a moderate flow, the boundary between two sources’ catchments moves by a comparable fraction of the distance between them. A boundary is where two costs are equal, and both costs have moved.

What makes it expensive rather than merely interesting is that neither partition is wrong. A response service wants the inward one — which station can get here soonest — and a patient travelling under their own power experiences the outward one, and a single map cannot show both.

This collection’s own machinery inherits it. The misassignment a page partition causes is measured against a ground partition that is assumed unique, and under an asymmetric cost there is no such thing to measure against.

What a symmetric cost would have to be #

It is worth asking what it would take for the two questions to have one answer, because the condition is stricter than it looks.

A cost is symmetric when the charge for a step is the same in both directions. Under a flow that requires the flow’s component along every step to be zero — which means no flow at all, not merely a weak one. There is no partial version: any flow whatsoever, at any strength, makes the two sets differ, and the table above is the size of the difference rather than a threshold for its existence.

That is worth setting beside the isotropy question of the previous rung, because the two conditions are independent. A cost can be symmetric and not isotropic: a terrain where going uphill and downhill cost the same but east–west is cheaper than north–south, which gives one reach set that is not round. It can be isotropic and not symmetric: a uniform vertical drift, which gives a round set displaced from its source. And a distance is both, which is why three rungs of this ladder could treat a reach set as a ball without ever saying so.

The catalogue of what breaks is therefore two-dimensional, and the reach sets a chart draws sit somewhere in it. Naming which of the two properties a cost has is one sentence and it decides what a reader may do with the picture.

The operational form is a pair of names rather than one. A chart that says reachable has answered one of the two questions and not said which, and a reader planning a return leg needs the other. Two overlays, drawn from the same solver with the flow reversed, cost one extra solve and remove the ambiguity entirely.

Where the model stops #

The flow is uniform for the ladder above, and uniformity is what makes the two areas exactly equal. A structured flow — the jet from the previous rung — breaks that: the reflection is no longer a symmetry of the field, so the forward and backward sets have different areas as well as different shapes, and the area then stops being the reassuring statistic it is here.

The cost keeps the triangle inequality, which every result here relies on: Dijkstra is only correct on non-negative costs, and a flow faster than the vehicle would produce places from which there is no return at all. Keeping the flow under the vehicle’s own speed is why the ladder stops at 0.75.

And the two sets are compared by area, not by shape. The Jaccard overlap says how much ground differs and not where, and the figure says where. A summary that carried both would be a shape statistic, and there is no standard one.

The generalisation #

A cost that is not symmetric is not a distance, and every construction that assumed a distance has silently become two constructions.

That is the whole rung and it is not about flows. A quasi-metric — one that keeps the triangle inequality and drops the symmetry — has forward balls and backward balls, forward Voronoi cells and backward Voronoi cells, and a forward and a backward version of every derived object. The literature on those objects is much smaller than the literature on the symmetric ones, and the reason is that the symmetric case is where the intuition is.

The cartographic form of the warning is short. Whenever a map shows a set defined by a cost, ask which direction the cost was measured in, and if the answer is it does not matter, ask whether the cost is symmetric — because if it is not, the map has answered a question nobody asked and has not said which.

Who found it, and when #

Asymmetric costs are old in operations research and are the reason a routing algorithm distinguishes a graph from a digraph. The specific consequence — that a reachability set has a forward and a backward version — is standard in that literature and is stated as a definition rather than as a discovery.

What is not standard is drawing them. Isochrone maps are a cartographic product of the nineteenth century — Francis Galton’s isochronic chart of 1881 shows travel times from London — and the whole tradition draws the outward set, because that is the question a traveller leaving asks. A chart of the inward set, for the same place and the same time, would look different by two thirds of its area, and this collection can find no published pair.

Which of the two each application actually wants #

The tradition draws the outward set because that is the traveller’s question, and a good many of the uses an isochrone is now put to are not the traveller’s. Sorting them is worth doing because the answer is not the same one twice.

Outward from the marked place. An ambulance or a fire appliance leaving a station: the vehicle travels from the station to the incident, so the set that matters is where the station can reach. A delivery round from a depot. A field team dispatched from a base. In all of these the marked point is the origin of the journey and the classical isochrone is the right picture.

Inward to the marked place. A hospital’s catchment: patients travel to the hospital, so the question is which places can reach it. A shop’s trade area, a school’s intake, a station’s commuter shed, a polling place’s accessibility. In every one of these the marked point is the destination and the classical isochrone answers the wrong question.

And the second list is longer than the first, which is the awkward part. Most of what isochrones are used for in planning, retail and public-service analysis is a catchment — a set of people who can get to a facility — and the tool draws the set the facility can get to.

The two coincide only if the cost is symmetric, and the essay’s measurement is that a very weak flow is enough to separate them substantially. Terrain, one-way systems, tidal currents, prevailing winds, asymmetric traffic and gradient all do it, and none of them is exotic.

There is a second-order consequence for anybody comparing facilities. Two hospitals compared by their outward isochrones are being ranked on a quantity neither of them uses, and the ranking can invert against the inward sets — a site that is easy to leave and hard to reach is exactly the kind of place a one-way system or a valley produces, and it scores well on the wrong map.

The practical instruction is one line. Compute the set in the direction the journey is actually made, which for a catchment means reversing the cost — running the search on the reversed graph, or the flow field negated — and label the map accordingly. Every routing engine can do this and almost no analysis asks it to, because the default is the direction that has been drawn since Galton.

Where the ladder goes next #

Five rungs build sets from a distance and then from a cost, and every one of them is about a single source. What none of them asks is what happens when the sources compete: a partition under an asymmetric cost has two versions, they disagree, and which of them a boundary belongs to is a question the ground cannot settle.