A pool built to have an answer

The worst value in its own interval swept the free coefficient in the rule score a component by its temperature less $\lambda$ times its hottest answer’s, and play where that is largest, and found that every $\lambda$ strictly between nought and one scores exactly the same. The reason was arithmetic: the rule reads only the ordering of $t - \lambda a$ across a board’s components, an ordering changes only where two components’ scores cross, and on that pool the crossings sit at $0{\cdot}2$, $0{\cdot}25$ and $1$. Three crossings, four cells, and one of them is nearly the whole interval.

It closed by naming the only route out:

A pool can be designed to have crossings wherever they are wanted, and a pool with twenty crossings spread across the unit interval would narrow the plateau to whichever cell of it is best. That is a construction rather than a sweep, which is a kind of work this ladder has not done, and it is the only route to a number rather than an interval.

The construction works. The plateau narrows to a cell of width one twelfth, all three board sizes agree on it, and it is between a quarter and a third.

Designing a pool #

Building a pool to make a measurement possible is a move this site has made before — a pool of deliberate traps was assembled to catch a rule out, and did not, which is the risk a designed pool runs. The design here is answerable to a weaker requirement: not that a particular rule lose, but that the rules be told apart at all.

The two quantities the rule reads are a component’s temperature and its hottest answer’s temperature, and both are computable before any game is played. So a pool can be chosen by its $(t, a)$ pairs, and the crossings — where $t_i - \lambda a_i$ meets $t_j - \lambda a_j$ — fall out of arithmetic on those pairs alone.

Positions of the form $\{T \mid \{A \mid -A\}\}$ make that easy. The outer switch has one hot option and the temperature and the answer come out of $T$ and $A$ directly, so a grid of nine values of $T$ against seven of $A$ gives forty-eight positions covering forty-eight distinct $(t, a)$ pairs.

From those, seven are chosen greedily: at each step, the candidate that adds the most distinct crossings inside the unit interval, with ties broken by the position’s own notation so that the pool is the same on every run and can be quoted.

The result has twelve crossings — $1/6$, $1/5$, $1/4$, $1/3$, $3/8$, $2/5$, $1/2$, $2/3$, $3/4$, $4/5$, $5/6$, $7/8$ — spread across the interval.

The unit interval is now thirteen cells, and inside each cell the rule is one rule. That is the whole of the construction: nothing about the choice of pool is about how the positions play, and everything about it is about where their scores cross.

One choice inside the greedy is worth flagging because it looks arbitrary and is not. The criterion is distinct crossings, counted as a set, rather than pairs that cross. Two pairs crossing at the same $\lambda$ add one cell between them, not two, so counting pairs would reward a pool that stacks its crossings on one point — which is the opposite of what a discriminating pool needs. The seven positions chosen include one with a temperature of nought and one whose answer is nought, both of which look like odd members of a pool of hot components; each is there because it crosses several others in places nothing else does.

The score is not flat #

Scoring the rule in each cell — at the cell’s midpoint, since inside a cell every $\lambda$ gives the same rule — the plateau is gone.

At three components, over all 84 boards the pool makes, the score runs $72, 73, 76, \mathbf{80}, 78, 73, 70, 66, 59, 53, 50, 44, 39$ across the thirteen cells. It rises to a peak and falls away on both sides, and the fall on the right is steep: playing at $\lambda$ near one gets 39 of 84 where the peak gets 80.

At two components two adjacent cells tie at a perfect 28 of 28. At three and at four a single cell wins outright. The cell $(1/4,\ 1/3)$ is the only one that is best at all three sizes.

That is the number the rung below could not have. Not an interval, not a plateau, but a cell of width one twelfth with three independent board sizes agreeing on it.

Where the proposed coefficients land #

Two numbers had been proposed for $\lambda$ before this page, and both come off badly.

A half, which half of the smaller temperature found as a correction on the sente ladder and which the rung below noted as consistent with its plateau, scores 161 of 210 at four components against the best cell’s 190. It sits four cells above the best one. Nothing here refutes the sente ladder’s half — that is a different quantity in a different setting — but the hope that one coefficient would turn up on two ladders is not supported by the only pool able to test it.

One, the rung below’s own rule and the full discount, scores 72 of 210. It was already the worst value in its own interval on the natural pool; on this one it is worse by a wider margin, and for the same reason — it is a crossing, so two components tie and the rule falls through to list order.

And nought — playing in the hottest component, which is Hotstrat — scores 173. That is better than a half. On the rung below’s pool the comparison could not be made, because everything strictly inside the interval scored the same; here it can, and the answer is that a badly chosen $\lambda$ is worse than not discounting at all.

What a designed pool cannot say #

The construction produces a number and the number’s standing has to be stated carefully, because a designed pool is not a sample of anything.

The seven positions were chosen to make the rule’s ordering change as often as possible. That is exactly what a representative pool does not do: a pool of positions arising in play would have whatever crossings it happened to have, and there is no reason for them to be spread evenly. So the best $\lambda$ is between a quarter and a third is a statement about a discriminating instrument, in the same way that a hearing test measures a threshold on tones nobody listens to.

What the construction does establish, and what could not be established before:

• The plateau was a property of the pool. Two pools, the same rule, three cells and thirteen.
• The rule has a best coefficient at all. On the rung below’s pool the score was constant inside the interval, which is consistent with the coefficient not mattering. It matters: 80 against 39 on the same boards.
• And the coefficient is small. Every cell above $2/5$ scores worse than every cell below $3/8$, at all three sizes. Whatever the right number is, it is nearer a quarter than a half.

The honest form of the finding is conditional: if there is a pool-independent best $\lambda$, it is between a quarter and a third, because on the one pool able to tell cells apart, three sizes agree on that cell.

What the peak looks like #

The shape of the curve is worth a paragraph on its own, because a single peak is not the only thing the sweep could have produced.

At four components the thirteen cells give 173, 172, 185, 190, 181, 167, 161, 151, 124, 113, 107, 87, 72. That is one rise and one long fall, with no second peak and no oscillation — the score is unimodal in $\lambda$, on all three sizes.

Unimodality is not guaranteed by anything. A selection rule is a step function of its parameter and successive steps are different rules, so there is no reason in principle for the scores to be ordered at all: cells three and nine could easily have beaten cells four and five. That they do not is evidence that the parameter is measuring something real rather than that a particular cell happens to suit a particular board. It also means a coarser search would have found the same place — a bisection over cells would work — which matters if this construction is ever run on a pool large enough that scoring every cell is expensive.

The one place the curve is not smooth is at the top: 185, 190, 181 is a peak two cells wide at four components and one cell wide at three. The plateau has not been narrowed to nothing, and nothing here says it could be.

Why the rule is a step function at all #

It is worth stopping on the mechanism, because it is the reason this rung needed a construction rather than a sweep and it generalises past this rule.

A rule with a continuous parameter that selects rather than computes is a step function of its parameter. The rule here does not use $t - \lambda a$ as a number; it uses it to pick a component, and picking depends only on which score is largest. So the rule’s behaviour is constant between crossings and changes only at them, and the number of distinguishable rules is one more than the number of crossings the pool has.

That has a practical consequence for every fitted coefficient on this site. Fitting a selection rule’s parameter on a small pool fits a cell, not a number, and reporting the midpoint of the cell as the coefficient is reporting a number the measurement did not contain. A rule that beats the hottest reported $\lambda = 1$; the rung below found that to be the one value in the interval where the rule stops being a rule; and this page finds that the interval it sat in was four cells wide on that pool and thirteen on a pool built to look.

It also says what a pool needs in order to measure a parameter at all: crossings where the answer might be. That is a design criterion for a census, and it is not one this site has used before.

What the rule is actually saying #

Strip the coefficient out and the rule is a claim about tempo, and the coefficient is a claim about how much of it to believe.

A component with temperature $t$ is worth $t$ to move in, and the opponent’s best answer inside that component is worth $a$ to them. Playing there therefore starts an exchange rather than banking a move, and the exchange returns some of what it takes. The rule’s $t - \lambda a$ says: value the component at what moving there gains, minus a fraction of what the answer gives back.

$\lambda = 1$ says the answer is fully paid for — the exchange is a wash except for the difference. $\lambda = 0$ says the answer costs nothing, which is playing the hottest and is the standing rule this ladder is trying to beat. A small $\lambda$ therefore says something specific and slightly surprising: the answer matters, and it matters much less than it costs.

The mechanism is that the answer need not be played immediately. A component’s hottest answer sits on the board as an option for the opponent, and by the time they take it the board has cooled, so its full temperature over-states what they will actually get from it. A coefficient near a quarter is a way of saying that about three quarters of the answer’s nominal value evaporates before it is collected. Whether that is the right account is not something this page can test — but it is the kind of statement a coefficient of a quarter suggests and a coefficient of one does not.

What this does not settle #

The pool is seven positions and the boards are sums of them. 28, 84 and 210 boards at two, three and four components, all built from the same seven. A larger designed pool would give more cells and a narrower answer, and the greedy construction extends without change; seven was chosen because twelve crossings is already enough to break the plateau and because the boards grow as the fourth power.

The components are not positions from a game. They are brace forms with a chosen temperature and answer, which is what makes the design possible and what makes the pool artificial. Whether a pool of real Domineering or Go regions has crossings anywhere useful is a separate measurement, and it is the one that would say whether the number means anything to a player.

The greedy is greedy. It maximises distinct crossings one position at a time and there is no claim that seven positions cannot do better. An exhaustive search over seven-subsets of forty-eight is twelve million pools and was not run; a randomised search over four thousand draws found six-position pools with eight or nine crossings where the greedy’s first six have more, which is weak evidence that the greedy is not badly off and is not a guarantee.

The optimum is measured against exact play. A board is played exactly when the rule’s score equals what optimal play scores, computed by the ordinary recursion over the sum. Nothing here is about a bound; it is about how often a cheap rule is right.

And the worst case does not track the score. The best cell loses three points on its worst board at four components, and the cell below it loses three as well; the cell at $\lambda$ near nought loses only one. So a reader who cared about the worst case rather than the average would choose differently, and this page reports the count of exact boards because that is what the rung below reported.

And the figures show numbers rather than play. Every table here is a score, and the thing a reader might want — a board on which the best cell plays one component and $\lambda = 1$ plays another, with the two lines of play beside each other — is not drawn. It would be drawable and it would be one board, and one board would not distinguish the rules: the difference between cell four and cell seven is nineteen boards in eighty-four, so any single example is either the majority case or a cherry. A rule that beats the hottest draws the lines of play for the rule at $\lambda = 1$ and is the right page for that picture.

Normal play throughout, and Left plays the rule while Right plays optimally, as on every rung of this anchor.

Where the ladder goes next #

The coupons anchor has eight rungs: the environment, when to leave it, two games in one environment, how big the answer is, the quantity that does not order a board, the rule that beats the hottest, the rate it charges, and now the pool that can measure the rate.

The rung above is a second pool. One designed pool gives one answer, and the way to find out whether the answer is about the rule or about the pool is to design another — a different grid of $(t, a)$ pairs, chosen greedily by the same criterion, with crossings in different places. If the best cell of the second pool overlaps $(1/4, 1/3)$ the coefficient is a property of the rule; if it does not, then the rule’s parameter is a per-pool quantity and there is no number to find, which would be a cleaner negative result than any amount of further sweeping.

Two neighbours are worth the trip. The quantity that does not order a board is where discounting by the answer’s temperature was first tried and found to be a disaster in the other direction, and it is the reason the interesting range is the unit interval at all. And how big the answer is is where the answer’s temperature became a quantity worth reading, and it is the measurement everything on this ladder is a coefficient on.