A bend that never reaches the surface

The bend is in the stops built a value’s whole thermograph out of its options’ four stops, found it exact on all 1,459 non-number values born by day three, and found it failing one day later on exactly the values with a bent-walled option. It closed by naming the quantity that would turn one level deep from a description into a measurement:

The sharp version is a statement about how many levels a value needs, which is a number attached to a position and not a property of a day … the natural conjecture is that the number is the depth of the deepest bend in the position’s tree.

The number exists and is easy to define. It is small — nought, one or two on everything reachable here. And the conjecture about it is wrong, in a way that says something about what a thermograph is.

Defining the number #

The thermograph recursion is already a recursion, and truncating it is the whole construction.

At depth nought, replace every position by the switch on its two stops: straight walls falling at slope one from the left stop and rising from the right stop, meeting at the midpoint. That is precisely the assumption the rung below’s test makes about a value’s options — that their stops determine their walls — applied to the position itself. Above depth nought, build the envelopes out of the options’ truncated graphs in the ordinary way: a thermograph is built from its options, and the truncation changes what the options hand up rather than how they are combined.

Then $\mathrm{levels}(G)$ is the least depth at which the truncation reproduces $G$’s real thermograph — both walls, the temperature and the mean, exactly.

The scale reads off cleanly. $\mathrm{levels}(G) = 0$ says the two stops alone give the whole diagram, which is what a straight-walled value is. $\mathrm{levels}(G) = 1$ is the rung below’s test: the options’ stops suffice. Anything above one is a value that test cannot read.

On day three, 1,109 values need nought levels and 350 need one. Nothing needs two, which is the rung below’s exact on all 1,459 restated as a distribution rather than as a pass. On the pool built one day later — every $\{A \mid B\}$ over the 120 hottest day-three values, 944 distinct non-numbers — 407 need nought, 284 need one and 253 need two.

One thing about the definition is worth flagging before any of it is used. The number is the least depth at which the truncation is exact, and nothing here proves that every deeper truncation is exact too. It ought to be — a deeper truncation replaces a switch approximation by the thing it approximates — but ought to be is not a measurement, and on this site that distinction is the whole method. What is checked is the least depth, and the reader should read the number as the first depth that works rather than as the depth from which it works.

That is already worth having. The failures the rung below reported are not a fringe: a quarter of the deeper pool is out of reach of a one-level reading, and the reason a whole day was needed to see any of them is that day three has no value with a bent-walled option at all.

The conjecture, and the count against it #

The deepest bend in a position’s tree is the depth of the deepest subposition whose walls are not straight. It is a natural candidate and it has a natural argument behind it: a bend is what the reconstruction cannot see, so the reconstruction should have to descend to the deepest one.

It is an upper bound and it is not the answer. No value in either pool needs more levels than its deepest bend allows — that much is checked and it is what the argument really establishes. But on 124 of the 944 deeper values it is strictly too large, and on 122 of those it is too large by the whole distance: the value carries a bend one level down and needs nought levels.

Those are values whose own walls are straight. Their stops give their whole thermograph; a reader who never looked at an option at all would get the right diagram. And somewhere under one of those straight walls is an option with a corner in it.

What a bend has to do to be seen #

The correction is one sentence and it is the recursion again.

$\mathrm{levels}(G)$ is nought when $G$’s own walls are straight, and otherwise one more than the longest chain of bends below it. Equivalently: follow bends downwards from the top, stopping the moment a straight-walled position is reached, and count how far the chain goes.

That is right on every one of the 1,459 day-three values and on 941 of the 944 deeper ones. Two thousand four hundred of two thousand four hundred and three.

The reason it is the chain and not the depth is what a thermograph is made of. Each wall is an envelope — an upper envelope of the Left options’ contributions, a lower one of the Right’s — and an envelope keeps only the option that is on top at each height. Every other option, and everything underneath it, is discarded. So a bent option under a straight wall is a bend the diagram never consults, and the truncation is free to get it wrong. That is the same mechanism the dominated option has in the canonical form, arriving in a different currency: an option that is never chosen leaves no trace in what the position is worth.

Read that way, the two candidate readings are asking different questions. How deep is the deepest bend asks what is in the tree. How long is the chain of bends asks what the envelopes actually look at, which is the thing the reconstruction has to reproduce.

One value, watched arriving #

The worked value has a chain of two, so the stops-only reading fails and the one-level reading fails as well, and the way each fails is worth reading.

At depth nought — the position treated as a switch on its stops — it comes out with a temperature of a quarter. At depth one, the rung below’s test, the temperature comes out at five eighths. At depth two it is three eighths, which is the truth.

The one-level reading is not a near miss on the low side. It is too hot by a quarter, and the direction is the mechanism showing through. Treating a bent option as straight replaces its wall by the straight wall through its two stops, and a bent wall lies below the straight wall through its own stops — a bend is the value declining an option it would have to pay too much for. So a straight-walled reading of a bent option makes the option look better than it is, makes the position that owns it look hotter, and pushes the temperature up.

That also says why the errors do not quietly cancel. Every option treated as straight is over-valued in the same direction, so the errors accumulate on one side of the envelope rather than averaging out — which is why a value with a chain of two is not approximately readable at one level. It is wrong by a definite amount, computed here as a quarter of a move.

Which values need two #

The 253 values needing two levels are a quarter of the deeper pool, and what they have in common is easier to say from the construction than from the values.

Every value in that pool is $\{A \mid B\}$ for two hot day-three values, so its options are exactly $A$ and $B$ and its chain is one more than the longer of their chains. A day-three value has a chain of nought or one — nought when its walls are straight, one when they bend — so a pool value needs two levels exactly when at least one of its two options is one of the 350 day-three values that bend, and its own walls bend as well.

That second clause is the whole content of the correction, and it is why the count is 253 rather than the 122 + 253 that the deepest-bend reading would give. Building $\{A \mid B\}$ over a bent $A$ does not oblige the result to bend: the envelope may be governed by $B$ throughout, or the bent stretch of $A$’s wall may sit above where the mast cuts in, and in either case the corner is built into the position and never appears in its diagram.

So the population divides three ways rather than two. There are values with no bend anywhere, which the stops read. There are values with a bend that surfaces, which cost a level each time the chain steps down. And there are values carrying a bend that is structurally present and diagrammatically absent — 122 of them here — which the stops read as easily as the first group. Only the middle class is what the bend is the condition was about, and separating the third from it is what this rung adds.

The three it misses #

Three values in 2,403 have a chain of two and need one level, and they are not three separate accidents. All three carry the same Right option, $\{-1/2, \{1 \mid -1\} \mid -2\}$, whose wall bends; and on all three the bend is dominated at the heights where it would have made a difference, so the one-level reading gets the envelope right anyway.

That is the same phenomenon that breaks the deepest-bend conjecture, one level further down. The chain reading asks whether each position on the chain bends; it does not ask whether the bend itself is the part of the option the envelope uses. A bend that lies in a stretch of the wall the envelope never reaches is invisible for exactly the reason a bent option under a straight wall is.

So the chain is a bound too, and a much tighter one: too large on three values in 2,403 where the deepest bend is too large on 124. What would make it exact is a reading that asks not does this position bend but does the bend govern — and that is a question about which option is on top at which height, which is a thermograph, which is the thing being avoided. The pattern is familiar from half a follow-up out: each successive refinement of a cheap reading costs a little more of the machinery it was invented to avoid.

Whose diagram this is #

The thermograph is Berlekamp’s, from the Go endgame work of the 1970s, and it reached the general theory through Winning Ways — a picture of a position’s value as a function of a tax on moving, with the two walls as the two players’ scores and the height where they meet as what is at stake. It was designed to be drawn, and much of what makes it useful is that a reader can see a temperature rather than compute one.

The question this ladder has been asking for four rungs is what happens when the drawing is taken away — when the same information has to come out of numbers attached to positions rather than out of a picture. That is not a hostile question to ask of a diagram. It is the way to find out what the diagram is for: a quantity that survives the translation was never really about the picture, and a quantity that does not is where the picture is doing work no list of numbers does.

The level count is an answer of the second kind. It says exactly how much of the drawing a stops-only reading has to reconstruct before it stops being an approximation, and it says that the amount is not a property of how deep the tree goes. Reading a thermograph is where the picture is set out and where a reader learns to take a temperature off it by eye; this page is the account of what that eye is doing that a table of stops cannot.

What this does not establish #

The pools are shallow and the number may not be. Nothing here needs more than two levels, and that is a fact about how far the construction reaches rather than about the quantity. A value with a chain of five would need five, if one could be built; the $\{A \mid B\}$ pool over day three has chains of two and no more, and the pool one layer further up is beyond what the canonical-form machinery will do here in reasonable time. So the number is small is not a claim this page makes — what it claims is that the number is well defined and that the chain reads it.

The comparison is of diagrams, not of temperatures. A truncation is counted right only when its walls agree with the real ones everywhere, along with the temperature and the mean. A weaker test — say, getting the temperature right — would give a smaller number and a different answer, and would be a different question: how deep does a reading have to go to get this quantity right, which is worth asking separately and is not asked here.

And the level count is not a cost. The bend is in the stops established that the stops recursion reaches every subposition a thermograph reaches, so a two-level reading is not twice the work of a one-level reading in any useful sense. The number measures how much of the recursion’s structure the answer depends on, not how long it takes. A cheap reading that needs two levels is still cheap; it is just not a reading in terms of stops any more.

The three exceptions are three and not a class. Naming a shared Right option and a plausible mechanism for it is a description of three cases, and three cases will fit almost any description offered. The honest statement is that the chain reading is wrong on three values, that all three share an option, and that the account given above of why has not been tested against a fourth because there is no fourth in this population.

Normal play, and short games throughout. Every value here is born in finitely many moves and every thermograph is the ordinary one. A loopy game has a thermograph too and it is a different object, with walls that need not meet, and nothing on this page is a statement about one.

Where the ladder goes next #

The switches anchor has nine rungs: the switches and their mean values, what a number does to a fight, the fight with no midpoint, the bend as the condition, half a follow-up out, the second level of stops, the bend from the stops themselves, and now how deep a reading has to go.

The rung above is the governing bend. The chain reading is wrong on three values because it counts bends rather than bends that matter, and the fix is a predicate — does this option’s bend fall in the stretch of wall the envelope uses — which is computable from the same objects the chain reading already has. Three exceptions is a small target and that is the point: a refinement that fixes three values in 2,403 and breaks none would be the first exact reading on this ladder, and one that fixes two of the three would say the residue is something else again. The pool is built, the exceptions are named, and the predicate is one function.

Two neighbours are worth the trip. A second level of stops is where the reading’s error was first measured and given a size, and reading it beside this page is the clearest statement of the difference between how wrong and how deep. And the two numbers at the top is the other place on this site where a quantity turned out to depend on a bounded amount of what is beneath it, and the two together are the site’s account of how shallow a hot position’s arithmetic usually is.