When two thermographs can be added

Two hot fights on one board, and a player wants to know how hot the board is. The temperatures do not add — that was settled three rungs down, and how cold a sum of hot games can be measured where in the permitted range the answer actually lands. Both essays closed on the same suggestion:

The temperature of a sum is not determined by the components’ temperatures, and it may be determined by their thermographs — the whole diagrams rather than the single number each reduces to.

A thermograph is a pair of functions of temperature. Functions can be added. So the construction to test is obvious and there is nothing else it could be: add the two left walls pointwise, add the two right walls pointwise, and see whether the result is the sum’s own diagram.

Ninety-two of the 120 come out exactly right. The twenty-eight that do not are exactly the pairs in which both components are hot.

Two things that always hold #

Before the failure, the two successes, because they are what makes the construction worth trying at all.

The added walls always bound the true ones. At every temperature, the true left wall sits at or below the added left wall, and the true right wall at or above the added right wall. The constructed diagram is a container: the real one lives inside it, never outside. The two added walls meet at the larger of the two temperatures, since above it both parts have frozen, so the constructed diagram closes exactly at the bound and the true one closes no later. That is the thermographic statement of temperatures do not add.

The mast always comes out right. The mean of a sum is the sum of the means, on all 120 pairs, with no exception and no near miss. So the top of the diagram — the vertical line the two walls meet at, and the number the position is worth once the fighting is over — composes perfectly.

So what fails is never the top of the diagram and always the bottom. Every one of the twenty-eight failures has its first disagreement at $t = 0$ — at the feet of the walls, which is to say at the stops.

Why the top adds and the bottom does not #

The sweep reports the split — the mast exact on all 120, the feet wrong on twenty-eight — and does not say why the diagram should behave so differently at its two ends. The reason is in the definitions, and it is worth having, because it says the split is structural rather than a fact about this pool.

The mean is a limit, and limits survive bounded error. The mean of $G$ is the unique number $\mu$ such that $n \cdot G$ stays within a bounded distance of $n\mu$ however large $n$ gets — that is the definition, and the theorem is that such a number exists. Put two positions together and the bounds add: if $n \cdot G$ stays within $B_G$ of $n\mu_G$ and $n \cdot H$ within $B_H$ of $n\mu_H$, then $n(G+H)$ stays within $B_G + B_H$ of $n(\mu_G + \mu_H)$. A sum of two bounds is a bound, so the number characterised by the limit is $\mu_G + \mu_H$, and uniqueness closes it. Nothing in that argument cares what the components do to each other, because everything they do to each other is finite and the limit does not see finite.

The temperature is a threshold, and thresholds do not survive anything. It is the least tax at which no move is worth making, and worth making is a comparison that can flip on a single point. Two copies of $\{1 \mid -1\}$ are each worth a move at every tax below one; together they are worth a move at no tax at all, because each is the answer to the other. The quantity being asked for is the position of a boundary, and moving a boundary needs only an arbitrarily small change in what sits on either side of it.

That is the same distinction, in a different vocabulary, as the one between the score and the schedule. The score is what the board comes to in the long run and the schedule is a sequence of decisions, and a construction built by adding two finished diagrams has enough information for the first and none for the second.

It also predicts the shape of the failures rather than merely permitting it. If the discrepancy were an accumulation it would grow as the diagram is read upward, and it does the opposite. Above the height at which the added walls close — the larger of the two component temperatures — both diagrams are the same vertical mast at the same number, because the mean adds and the true diagram has frozen no later. Every disagreement therefore lives strictly below that height, in the band where a move is still worth making in at least one part, and every one of the twenty-eight is at its worst at the very foot. The error is a boundary being in the wrong place, not a quantity being slightly off, which is what a threshold argument predicts and an accumulating one does not.

The failure, drawn #

The clearest case is two copies of the same switch.

$\{1 \mid -1\} + \{1 \mid -1\} = 0$. Not approximately nought — exactly nought, a second-player win with no fight in it at all. Whoever moves first in one copy is answered in the other, and the two moves cancel. The added diagram says the board is worth between $-2$ and $2$ depending on who moves; the real board is worth nothing to anybody.

This is the mirroring argument in its plainest form, and it is why the failure is not a small numerical discrepancy that a correction term might absorb. Two copies of a fight are not a bigger fight. They are no fight, because each is the answer to the other.

The obvious next question is whether the construction is failing on identity rather than on hotness — whether it is confused by seeing the same diagram twice. It is not, and the cheapest way to show so is a pair of components that are not the same position at all.

Why one hot component is enough and two are too many #

The condition is exact — every pair with two hot components fails, every pair with fewer succeeds, and both halves are asserted in the code so that a change to the enumeration cannot quietly soften them. The mechanism is worth stating in terms of what a wall is.

A thermograph’s left wall at temperature $t$ is what Left gets by moving first when every move costs $t$. For a sum, Left moving first has a choice of which component to move in, and the answer to that move may come in the other component. Adding the walls assumes the two conversations are separate: that Left’s best in $G$ plus Left’s best in $H$ is Left’s best in $G + H$.

When only one component is hot, that assumption costs nothing: the cold component has no move worth making at any positive tax, so there is no cross-board answering to model and the walls of the cold part are a flat line the hot part’s walls are shifted along.

When both are hot, the assumption is exactly wrong. The correct play in a sum of two fights is not two independent fights; it is an alternation, with each player choosing which fight to enter and the opponent free to reply in the other. The added walls have no vocabulary for that choice, and the twenty-eight failures are twenty-eight instances of it. Which part to move in is the decision they leave out, and it is not recoverable from two finished diagrams.

Adding those same two fights is the third variety of failure, and it rules out the last easy explanation: that the construction only breaks where the sum collapses.

The twenty-eight, sorted #

The failures are not scattered across the pool. Seven of the fifteen day-two values that are not numbers are genuinely hot — the two half-unit switches $\{0 \mid -1\}$ and $\{1 \mid 0\}$, the wide switch $\{1 \mid -1\}$, and four positions carrying a star among their options. The other eight are cold: $1\ast$ and $-1\ast$, which are numbers with a star added, and the six all-small values $\ast$, $\uparrow$, $\downarrow$, $\uparrow\!\ast$, $\downarrow\!\ast$ and $\ast 2$.

Seven hot values give twenty-one unordered pairs of distinct ones plus seven pairs of a value with itself, which is twenty-eight. That is the entire failure set, exactly, with no member of it outside and no non-member inside.

The tidiness of that arithmetic is the strongest form the finding takes. It is not that the construction tends to fail on hot pairs; it is that hotness of both parts is the necessary and sufficient condition on this pool, so a reader can decide in advance, from two temperatures, whether the addition is going to be right — and the answer is that it is right only when it has nothing to do.

What is left standing #

The construction is not useless, and it is worth being precise about what survives, because two of the three results are usable and the third is not.

As a bound it is exact and free. The added walls contain the true diagram, so a solver holding the two parts’ diagrams can read off an interval for the sum’s stops and an upper bound for its temperature without touching the sum. That is a real saving on a board with many components, where computing the total’s diagram means canonicalising the total.

As an answer for the mast it is exact. The mean of the whole board is the total of the parts’ means, always, and the mean is what the board is worth once every fight is settled. So the score composes and the urgency does not, which is the same split the endgame, accounted for works with: a Go player’s count is an addition and their move order is not.

As a construction of the diagram it holds only where nothing is at stake. A pair with at most one hot component is a pair with a single fight in it, and the diagram of a single fight shifted by a cold thing is not a result anybody needed.

The other reading of the same pool is where a sum’s temperature actually lands inside the range the bound allows: most sums sit at the maximum, some collapse to the floor, and the few in between are the ones whose walls bend — which is where the added construction is furthest from the truth.

And the property that decides the answer is a single bit per component, which a reader can read off without any of this machinery: is the temperature above nought? Seven of the pool’s fifteen non-numbers clear it — $\{1 \mid -1\}$ at 1, the half-unit switches $\{0 \mid -1\}$ and $\{1 \mid 0\}$ at a half, and four positions carrying a star among their options, $\{1 \mid \ast\}$ among them. The other eight sit at nought and freeze under any tax at all. That bit is the test the construction turns out to be applying, and the only test it applies.

Why the bound is still the best available #

It is easy to read the last three sections as a wholly negative result, and one part of it is not.

A player or a solver holding two diagrams and wanting to know about their sum gets, for nothing, an interval containing the sum’s stops and a ceiling on the sum’s temperature. Those are not empty: the ceiling is what playing the hottest prices a move against, and the guarantee that rule comes with is stated in terms of exactly that number.

There is a second thing worth carrying, and it is the one a Go player already has. The mast is exact, so a board of any number of components has a count that is a plain addition — and that count is what the two players are arguing about once the fighting is over. What the construction fails to supply is the schedule: which fight to enter, and when. Splitting the answer that way is not a defeat, because it is the same split the practical theory has always made.

So the added diagram is the right object to carry; it is simply not the sum’s diagram. Calling it the sum’s diagram would be the error, and it is an easy error to fall into because the construction is exact on the ninety-two pairs a reader is most likely to try first — a fight and a star, a fight and an up, a fight and a number with a star on it. Every one of those works. It is only when a second real fight arrives that the picture stops describing the board.

The hypothesis a sum theorem would need #

The measurement says what a real theorem here has to look like, and it is not a repair of the addition.

A correction term is the first thing a reader reaches for — some quantity depending on both parts, subtracted from the added walls to give the true ones. The two-copies case rules that out in one line. The correction there would have to be the whole of both walls: added, the diagram runs from $-2$ to $2$ at the foot; truly, it is a point. A term large enough to fix that is a term carrying all the information, and a correction that carries all the information is not a correction.

Every failure is at the foot of the diagram, and the foot is the stops. The stops of a sum are not the sums of the stops — that is the same statement in the language where the fight stops uses — so a construction that gets the mast right and the stops wrong is a construction with the correct asymptote and the wrong boundary condition.

A theorem that produced the sum’s diagram would have to build it upward from the two option sets rather than sideways from the two finished diagrams, which is exactly what a thermograph is built from its options’ describes for a single position. Running that recursion on a sum means enumerating the sum’s options, and the sum’s options are moves in either part — at which point the parts’ diagrams have been discarded and the whole saving with them.

That is a real obstruction rather than a gap in the sweep. The information the construction needs is which component the winner moves in at each temperature, and neither diagram carries it, because each was computed without the other on the board.

The same failure taken further makes the point harder still. Take eight copies of $\{1 \mid -1\}$ rather than two: the pile’s value alternates between nothing and one fight’s worth as the count goes odd and even, and no reading of the single diagram predicts that. The collapse is a fact about the pairing rather than about the arithmetic, and a construction that reads one diagram at a time has no way to know how many copies are on the board.

What the sweep cannot say #

The pool is the fifteen values born by day two that are not numbers, of which seven are hot. That is a small pool of hot components and their temperatures take only two values, $\tfrac12$ and $1$. A pool with a wider range of temperatures, or with components whose walls bend more than once, might exhibit a case that the at most one hot component condition does not describe — and the sweep would have to be re-run to find out rather than reasoned about.

The second limit is that the construction tested is the pointwise sum and only that. A weighted construction, or one that shifted the walls before adding them, is a different proposal; nothing here rules out a cleverer combination rule, and the essay’s negative result is about the obvious rule rather than about all rules.

There is also a class of position this pool does not contain at all: a game with a number in the middle of it, so that one player’s best move settles part of the board and leaves the rest hot. Those are the ordinary positions of a real endgame, and every value here is either a fight or an infinitesimal with nothing in between.

The third is that the failures are counted and not sized except at their worst. The widest disagreement in the sweep is 2, on the two-copies pair; whether the typical failure is close or far is a distribution this page reports only at its extreme.

The convention, named #

Normal play, disjunctive sum: a move in exactly one component, and the player unable to move anywhere loses. Every wall above was computed as an exact piecewise-linear function by the recursion in the site’s evaluator rather than sampled, so a temperature is a number the figure derived and not a place two curves appear to meet.

The tax is charged per move on both sides, which is the standard convention and the one that makes cooling and the thermograph the same object read two ways. Under a convention that taxed only one player the walls would not be comparable and the addition would not even be a candidate.

Where the ladder goes next #

The thermograph anchor reaches seven rungs: how to read one, what an endgame account does with it, that temperatures do not add, how the diagram is built from the options, what a second bend means, where in the permitted range a sum lands, and now that the diagrams do not add either.

The rung above is the loopy case, and it is the one direction where a construction from two diagrams is genuinely the state of the art. Positions that can return to themselves have no bottom for the recursion to reach, so generalised thermography answers with a pair of diagrams — an optimistic wall and a pessimistic one — and the question of what a bend means when there are two of them and they disagree is a rung with its own machinery.

Two neighbours are worth the trip. How cold a sum of hot games can be is the rung below, where the same 120 pairs are measured for where their temperature lands rather than for whether their diagram is constructible. And the same fight, eight times over is the special case where the two components are identical, which is the worst case for this construction and the best case for the mean.