The reading that survives too much

Push is Shove with the cliff replaced by a wall. A row of coloured coins sits on a strip; a player pushes one of their own coins one square to the left, shoving the contiguous run in front of it along with it, and the move is illegal if that run is against the wall. Every position is worth a number, which is the first thing the game has in common with Shove and nearly the last.

The obvious reading counts the room each coin still has: for each coin, the number of empty squares strictly in front of it that it could ever occupy, signed by colour and added up. The other way to move a row tested it, found it right rather more than half the time, and closed by describing the failures:

What the 282 failures have in common is the obvious thing to look at, and they are all positions with adjacent coins of opposite colours.

One square further out that is false, and the smallest witness is five squares long.

Sixty-six that were not supposed to exist #

Over strips of at most seven squares there are 2,186 positions. The reading is right on 1,114 of them and wrong on 1,072.

Of the 1,072 failures, 1,006 do have two coins of opposite colour side by side. Sixty-six do not.

The shortest is .LL.R: an empty square, two of Left’s coins, an empty square, one of Right’s. Nothing is adjacent to anything of the other colour. The reading counts one square of room for the Left pair and one for the Right coin, which cancel, and calls the position nought. The recursion makes it minus a half.

The description also fails in the other direction, and more loudly. Six hundred and four positions have two opposite coins adjacent and the reading gets them right anyway — so the property is neither necessary for a failure nor sufficient for one. It is a correlate, and a strong one, and the rung below reported it as though it were the mechanism.

That is worth naming as a failure mode rather than as a slip, because it is the ordinary way a description of a set of counterexamples goes wrong. Every failure at six squares does have an adjacent pair; the property was read off the failures that existed rather than tested against the ones that did not.

What actually decides it #

.LL.R says what the real condition is, once it is looked at as a game rather than as a picture.

Right’s only move is to push the R coin one square left. That gives .LLR, and now the colours are adjacent. The interaction the description was looking for has not happened yet on the board and it is one move away, and the value of the position is decided by what happens after that move as much as by what is on the strip now.

So the condition is not about this position. It is about every position reachable from it:

The reading is right whenever no position either player can force ever has two coins of opposite colour side by side.

Checked on all 2,186 strips, that is sound without exception. Two hundred and fifty-four positions satisfy it and the reading is right on all 254; not one position satisfies it and comes out wrong. The census asserts it, so a strip one square longer that broke it would stop the build.

And it is not necessary #

The other half of the finding is the more surprising one, and it is where the essay’s title comes from.

Eight hundred and sixty positions do mix — some line of play brings the colours together — and the reading gets them right anyway. So of the 1,114 positions the reading handles, fewer than a quarter are handled for the reason that the colours never meet.

The reading survives mixing three quarters of the time. That is not what a criterion with no exceptions on one side usually looks like: a condition that guarantees a rule and covers a fifth of the cases the rule works on is a condition with a lot of slack in it.

The slack has a plausible source and this page cannot confirm it. When the colours meet, the coin that gets blocked loses room and the coin that blocks it does not; if the two losses are equal and opposite the count is unchanged and the reading survives. That would make the failures the positions where the interaction is lopsided, which is a condition about how much room each side loses rather than about whether they touch — and it is what a quantitative version of the criterion would have to be about.

Sixty-six counterexamples out of 1,072 failures is a small minority, and it is fair to ask whether they matter when the description they refute is right on the other 1,006.

They matter for the reason a single counterexample always does: the sentence was they are all, and a description offered as the shape a repaired rule would take has to be exactly right or it repairs nothing. A rule built on the colours are adjacent would have declared .LL.R safe and returned nought for a position worth minus a half, and it would have done so on sixty-six strips at seven squares and on a growing share thereafter.

They also matter because of where they are. All sixty-six are five squares or longer, and the sweep the description came from stopped at six — where there are twelve of them, against fifty-two at seven. A description that is exactly right on every position up to a size and starts failing just past it is the most expensive kind to have written down, because nothing about the sweep that produced it looks thin.

The general lesson is the one this site keeps relearning. A property shared by every counterexample found is a property of the search, not of the game, until something has tried to break it — and the same shape of error is what a wider pool undid on the misère outcome table.

The decay, and what it is a fact about #

At one and two squares the reading is exact. At three it is right on sixteen of eighteen, at five on two thirds, at seven on 668 of 1,458 — a shade under half.

The trend is not evidence of anything subtle. A strip of n squares has 3ⁿ arrangements of Left coin, Right coin and gap, and the arrangements in which the two colours never meet are essentially the ones with all of one colour in front of all of the other, of which there are on the order of n². So the covered fraction falls like n²/3ⁿ, and the reading’s own success rate falls with it but far more slowly — which is the slack again, seen as a function of size.

Extrapolating gives the wrong impression in both directions. The criterion’s coverage goes to nothing, and the reading’s success rate is falling toward something that may well be positive. Nothing here says where, and a strip of eight squares is 6,560 positions and one rebuild away.

What the value actually is when they meet #

The failures are not near misses. The errors run from an eighth to more than six, they are dyadic, and their distribution is exactly symmetric — every error e occurs as often as −e, which the negation of the strip forces and which the census checks.

The dyadic denominators are the interesting part. .LR is worth −½, ..LR is −¾, ...LR is −⅞: a Left coin with m empty squares behind it and a Right coin immediately in front is worth −(1 − 2⁻ᵐ). Those are binary expansions, and a binary expansion read off a row of two colours is what a Hackenbush string is.

That is a strong hint and it is not a theorem. What it says is that the mixed positions are not unreadable; they are readable by a different reading, one that is multiplicative in the blocking rather than additive in the room. Finding it is a rung of its own and this page has one family of it.

What a stuck coin is worth #

The value of a stuck coin is nought and it is not nothing, which is worth a paragraph because it is the whole of the interference.

A coin pushed against the wall can never move again. It contributes no moves to its owner, so on the reading it contributes nought, and that part is right. What the reading misses is that it also occupies a square, and the square it occupies is a square some other coin might have wanted — so a stuck Left coin subtracts room from the Right coins behind it without adding anything to Left.

That is an asymmetry with a sign, and it is why the errors above are as large as they are. On .LL.R the Right coin’s single push jams it against the Left pair, and from then on nothing on the strip can move except into the one empty square at the front — so what looked like one move for each side is one move for Left and one wasted move for Right, and the position is worth minus a half rather than nought.

A reading that counts room has no way to charge for a square that is occupied uselessly, and that is the shape of every failure on this page. It is also why the repaired reading, if there is one, will be about positions rather than about coins: whether a square is uselessly occupied is a fact about the sequence of pushes and not about the piece standing on it. The board that falls apart is the standing example of a game where the pieces really are independent, and Push is the standing example of one where a reader would swear they are and they are not.

Why a wall and not a cliff #

It is worth saying what the wall is doing, because Shove and Push differ in one clause and the reading behaves completely differently on the two.

In Shove a coin pushed off the left end is gone. Every move therefore removes a coin, the game shortens monotonically, and the count of moves each player has is exactly the count of coins they have times the room in front — which is why the Shove reading is a theorem and this one is not.

In Push nothing is removed. A coin pushed against the wall is stuck, and a stuck coin is a piece that cannot participate — a zugzwang built into the board rather than arising in play. The count of stuck coins only goes up, which is the game’s termination proof, and it is also the reason the two colours interfere: a stuck Left coin blocks a Right coin that would otherwise have room, and the room it takes away is not room the Left coin gains.

So the difference between a theorem and a half-right heuristic is one clause of the rules, and the clause is whether a piece leaves the board.

What a reading has to survive to be a reading #

A rule that produces the right value most of the time is a different object from a rule that produces it, and this page’s whole difficulty is that the two are hard to tell apart from a hit rate.

Three tests separate them and they get progressively harder to pass.

Agreeing on a sample is the weakest. A reading that is right three quarters of the time is right three quarters of the time, and nothing follows: there is no theorem it can be plugged into, no sum it can be used inside, and no position for which it can be trusted without checking.

Agreeing on a described class is stronger, and it is what this page goes looking for. If the failures had a description — a condition on the strip that a reader could check — the reading would be exact on the complement and would be usable there. That is what a quantitative criterion would have supplied, and it is what the rung above shows cannot exist.

Adding across a sum is the strongest and is what makes a reading worth having at all in this subject. A value is only useful because it composes; a reading that gives the right number for a component and the wrong one for a board of components has bought nothing, because a board is what a player is looking at.

Push’s reading fails the second and third together, and it fails them for one reason: the strip is not a sum of its runs, so there is nothing for a per-run reading to compose over. Which is why the ladder above spends its remaining rungs on what a run’s numeral is rather than on patching the hit rate — the hit rate was never the thing that was wrong.

What the census does not say #

Three limits.

The criterion is one-way and is stated that way. It says when the reading is safe and says nothing when it fails. A solver could use it as a shortcut — check reachability, and if the colours never meet, count instead of searching — and the check is itself a search over the reachable positions, so whether it pays depends on how much cheaper reachability is than evaluation. It is cheaper, and by how much is not measured here.

The errors are described and not explained. Symmetric, dyadic, and up to six and a sixth: three facts about a distribution, none of which predicts an individual value. The .LR family is a formula for one shape and there are many shapes.

Nothing here is a proof that the criterion is sound. It has been checked on 2,186 strips and no exception exists among them; the argument for it is a sentence — if the colours never meet, no coin ever loses room to the other side, so the count of room is a count of moves — and turning that sentence into an induction over the reachable set is a paragraph nobody has written.

And seven squares is where the sweep stops. Every count on this page is over strips of at most seven, and the two headline numbers — the sixty-six and the 254 — are counts rather than proportions of anything that continues.

The convention, named #

Normal play. A push moves one coin and the contiguous run in front of it one square left; it is illegal when that run reaches the wall. Every value here is computed by the recursion, reduced to canonical form, and checked to be a number — a Push position that came out anything else would stop the build, since the whole page assumes there is a count to compare against.

The reading is the analogue of the Shove reading, stated for coins rather than moves: each coin contributes the number of empty squares strictly in front of it that no earlier coin of either colour occupies, signed by colour.

Where the ladder goes next #

The push anchor has two rungs to here, and the four above take this page’s request for a quantitative criterion and refuse it, replace it, and then find the limits of the replacement.

The criterion that cannot exist settles the request with three strips of four squares. .LLR, .LRL and .RLL have the same length, the same reading, the same coins and the same single run — and their readings are wrong by $1\tfrac14$, $\tfrac14$ and $\tfrac12$. The error is a fact about the order of the colours, so no statistic over lengths, counts and runs can express it, and 207 of 805 statistical classes carry more than one of the three.

A numeral in the empty squares then supplies what does work, and the two ingredients turn out to be the other way round from the guess. The colours pick a fraction — $-1$, $-\tfrac13$, $-\tfrac17$, $-\tfrac1{15}$ — and the empty squares supply the binary precision, so a run of $k$ coins before one of the other colour, with $g$ gaps, is worth exactly $(1 - 2^{-kg})/(2^k - 1)$.

And it does not compose. The cliff a cut invents finds why: a strip cut at a gap is not two positions, because a Push strip is a line with a wall at one end and cutting hands the back half a wall it never had. Widen the gap and the value converges geometrically to a limit that is not the sum, and Shove — whose reading is exact everywhere and adds across genuine sums — fails at the same cut on 89 of 93 strips.

Read from the back forwards closes the anchor with where the information lives. The convergence rate is the rearmost run’s, at every gap and every distance, and it does not compound across runs — while Shove, one clause away, does.