The cliff a cut invents

A numeral in the empty squares solved one Push run exactly. A run of $k$ coins with $g$ empty squares in front of it is worth

$$-\frac{1 - 2^{-kg}}{2^{k} - 1},$$

a binary expansion whose digits live in the gaps and whose limit is fixed by the colours. It also found that a strip of two runs is not the sum of them, on every split it tried, and closed by naming the object a repaired reading would need:

The rung above is the interaction. Two runs are not independent, and the dependence has a shape — pushing the front run gives the back run a gap — so the natural object is a correction term in the gap between them. Measuring it is a sweep over two-run strips with the gap varied, which is affordable.

The sweep is here. There is no correction term in the gap, and the reason is not about Push.

The sum, and how badly it fails #

Ninety-three strips of the form $.^{g_1} L^{k_1} R\, .^{g_2} L^{k_2} R$ fit inside eleven squares, and the value of the whole is the sum of its two runs on six of them. The disagreements are not small: ...LR..LLLR is worth about $-0.178$ and its two runs add to $-\tfrac{65}{64}$, a gap of over four fifths of a move on a strip whose total value is under a fifth.

Nor do they point one way. On some strips the whole is worth more to Right than the parts, on others considerably less, and the sign changes as the back gap grows. That already rules out the simplest shape a correction could have — a fixed charge for adjacency — before any of the structure below.

Widening the gap does not separate them #

The rung below’s proposal was that the interaction lives in the gap: two runs sitting next to each other interfere, two runs far apart interfere less, and a correction term in the separation would capture the difference. It is the natural guess and it is how nearly everything else on this site behaves — the sum is the object, and a sum is what a position becomes once its parts stop touching.

Push does not do that. Widen the gap and the strip’s value converges, and it converges to something that is not the sum. For a single coin behind one empty square against a two-coin back run the values run $-0.3125$, $-0.375$, $-0.390625$, $-0.394531$, $-0.395508$, and continue to $-0.3958$ — while the sum of the two runs at that separation is heading for $-\tfrac56$. The two sequences are not approaching each other at all.

So a correction term in the gap would have to tend to a non-zero constant as the gap grows without bound, which is another way of saying it is not a correction in the gap. Whatever is left over when two Push runs are put on one strip does not go away when they are moved apart.

The rate belongs to one run #

The convergence is geometric and the rate is exact: successive differences shrink by a factor of $2^{-k}$, where $k$ is the number of coins in the back run. Not the front run’s length, not the front run’s own gap, not the total length of the strip — the back run alone, on all twenty-six families.

That is the first quantity anywhere in this ladder that depends on one run rather than on the strip, and it has a clean reading. The gap is a store of moves for the back run: each extra empty square is one more push before the back run reaches anything, and the numeral already established that a run of $k$ coins resolves $k$ binary places per gap it spends. So an extra square of separation is worth $2^{-k}$ of whatever it was worth before, exactly as it is for a run standing alone against the cliff.

The gap therefore behaves like part of the back run’s own numeral rather than like a distance between two objects. The strip has one numeral in it, not two numerals and a correction.

What it converges to #

At infinite separation the two runs add exactly when the back run is one coin long. For $k = 1$ the limit is the front strip’s value less one — one being the back run’s own limiting value $-1/(2^1 - 1)$ — on all nine families, to the last place the extrapolation reaches. For $k = 2$ and $k = 3$ the limit is nowhere near the sum: $-0.396$ against $-0.833$, $-0.174$ against $-0.643$.

And the six strips that add exactly, at finite separation, are the same phenomenon caught early: all six have a single-coin back run, and all six have enough room in front of the front run that the back run’s coin never changes what the front run can do. The exception and the limit are one fact, met twice.

The failure is the cut #

There is an obvious objection to all of this, and following it up is what turns the measurement into an argument. Is a two-run strip a sum of anything?

Shove is Push with one clause changed: a coin slides alone rather than shoving the run in front of it. Nothing worth fighting over established that Shove’s values are read straight off the board by counting free squares, exactly, everywhere, and that reading adds across a genuine disjunctive sum because it is a count of independent resources.

It fails at this cut too, on 89 of the same 93 strips.

That settles what is broken. Cutting a strip at a gap does not produce two independent games, because the back half of a cut strip has no cliff of its own. A Push or Shove strip is a line with a wall at one end; every coin’s moves are counted against the distance to that wall. Cut the strip and the front half keeps the real cliff, while the back half is handed a brand new one — and the back half’s real barrier was never a wall at all but the front half’s coins, which move.

So the sum the rung below could not find was never a sum of games. It was a comparison between one position and two different positions that happen to look like its halves, and the quantity being measured — the difference between them — is the cost of the cliff the cut invented.

A cut that invents a boundary condition is not a decomposition #

The finding generalises past Push, and it is worth stating in the form that would have prevented the search this ladder spent two rungs on.

A decomposition is a claim that the position is a sum: that its moves partition into moves of the parts, and that no move of one part changes what is legal or valuable in another. The usual way to check it is to look for interference between the parts. What this page shows is a second way for the claim to fail that no amount of looking for interference will catch — the parts, as written down, are not positions of the same game.

A Push strip is a line of coins with a wall at one end. That wall is not a piece on the board and it is not drawn; it is the boundary condition every value on the strip is computed against. Cut the strip in the middle and the front piece keeps the wall. The back piece does not have one, so writing it down as a strip means giving it a new wall where the cut was — and a wall is exactly what the front piece’s coins were not, because coins move and a wall does not.

So the cut does not divide the position into two; it produces one genuine subposition and one new game. The difference between the strip and the pair is not a coupling to be measured, it is the price of that substitution, and no correction term expressed in the two halves’ values can name it because one of the halves is not a half.

The test that catches this is the one run above and it is worth having as a habit. Change one clause of the rules and see whether the same cut still fails. Shove’s values are a plain count of free squares and add across any genuine sum, and Shove’s cut fails on 89 of the same 93 strips — so what failed was the cut, not the arithmetic, not Push’s values, and not anything about how the runs interact.

That is a different failure from the two the rest of this site records. A global move rule breaks a sum by letting one region’s legality depend on another’s. A component that carries state breaks it by needing to remember something. Here nothing is remembered and nothing reaches across: the geometry simply does not have two pieces in it to begin with, and the boundary a reader draws is the whole of the discrepancy.

What a store of moves is worth #

The rate is the part of this worth turning over slowly, because it says what a gap actually is in this game.

Consider one coin standing behind $g$ empty squares with a cliff beyond them. Each push spends one square, so the position is a chain of positions differing only in a counter, and its value walks through a binary expansion $-\tfrac12, -\tfrac34, -\tfrac78, \dots$ — that is the rung below’s numeral, and it is why the digits live in the empty squares.

Now put a run in front of it. The back run’s pushes still spend squares one at a time, and what has changed is only where the spending stops: instead of running out at the cliff, it runs out when the back run meets the front run’s coins. The count of squares before that happens is the separation plus whatever the front run itself leaves, and the value is the same expansion truncated at a different place.

So the separation is not a distance between two games. It is a prefix of one game’s numeral, and that is exactly why the rate is $2^{-k}$ with $k$ from the back run and nothing else: the rate is the base of the expansion, and the expansion belongs to whichever run is doing the spending.

Which also explains the single-coin exception without appealing to the census. A back run of one coin has nothing behind it to shove, so when it arrives at the front run it simply stops — it never merges with it, and the two runs never become one longer run. A back run of two or more coins does merge, and a merged run is a different run with a different numeral, which is the interaction that no distance removes.

What a reading would have to be #

Three rungs of this ladder have been looking for a way to read a Push strip without evaluating it, and the shape of what is left is now fairly clear.

The criterion that cannot exist ruled out every reading built from lengths, colours and run structure taken as separate quantities. The numeral rung solved the one-run case exactly and found the errors depending on the order of the colours inside a run. This page says the strip is one expansion rather than several, with the rate belonging to the rearmost run.

Put together, those say a reading has to be sequential: it starts at the back of the strip and works forwards, each run contributing digits at its own base and each gap advancing the expansion, with the whole thing terminating at the cliff. That is a very different object from a sum of per-run values, and it is much closer to what a Hackenbush string is — a numeral read off in one direction, where cutting the string at a point does not give two numerals that add.

The Hackenbush comparison is worth pushing, because it is the one place on this site where the analogous cut is legitimate. A Hackenbush stalk decomposes by the colon principle rather than by addition — what the colon respects sets out exactly which operation the cut corresponds to, and it is the ordinal sum, not the disjunctive one. If a Push strip has a decomposition at all, that is the shape to look for: an operation under which the front of the strip dominates the back completely, which is what the cliff does to everything above it.

What this does not say #

Push may still decompose somewhere. Nothing here rules out a genuine decomposition of a Push strip along some other line — a decomposition into a sum whose parts are not sub-strips, say, or a strip whose two halves have been given the right boundary conditions. What is ruled out is the split the rung below proposed and the correction it hoped for.

The limits are extrapolated. No strip of infinite length is evaluated. Each limit is read off the last difference at the rate the previous figure establishes, and the rate itself is a measurement over three or more terms rather than a theorem. A family whose ratio drifted at the fiftieth term would move its limit and the figure would not know.

The sweep is small in one direction. Runs of one, two and three coins with one to three squares in front, capped at thirteen squares for the separation sweep and eleven for the sums. A four-coin back run would converge at a rate of $2^{-4}$ and would need more length than the evaluator has.

And the single-coin rule is a finding, not a proof. It holds on all nine families with $k = 1$ and on none of the seventeen with $k > 1$, which is exactly the shape a real rule has and exactly the shape a coincidence over a small pool also has. The argument for it — that a single coin has nothing behind it to shove — is a sentence and not an induction.

The convention, named #

Normal play throughout, and every value computed by the game recursion and reduced to canonical form.

Push is played on a strip with a cliff at the left end. A player moves one of their own coins one square towards the cliff, and the whole run of coins in front of it moves with it; the move is legal when there is an empty square in front of that run. Shove is the same rule with the run left standing: a coin slides one square and pushes nothing.

A run here is a maximal block of coins, and the families swept are $.^{g} L^{k} R$ — $g$ empty squares, $k$ blue coins, one red — which is the family the rung below solved exactly.

The sum of two runs is the value of one strip plus the value of the other, each evaluated on a strip of its own with its own cliff. That is the object the page is about, and the last section is the argument that it was the wrong object.

The limit is an extrapolation: the last value computed plus the last difference multiplied by $r/(1-r)$, where $r$ is $2^{-k}$ for the back run. It is a reading of a sequence and never the value of a position, and every figure that prints one says so.

Where the ladder goes next #

The push anchor reaches five rungs to here, and the rung above answers the question this page’s last section asks — what a third run does to the rate — with a result that is more definite than the question expected.

Read from the back forwards finds the rate to be the rearmost run’s, for every gap and at every distance. Widen the front gap of a three-run strip, two whole runs away from the back, and the value still dies at the last run’s rate. The rates do not compound and the intervening runs contribute nothing to the exponent at all.

That is a strong statement about where the information in a Push strip lives, and the control it comes with is what makes it one. Shove — the game one clause away — does compound, on the same strips and by the same measurement. So the non-compounding is a fact about Push’s move rule rather than about strips, about geometry, or about the way the sweep was taken, and the one clause separating the two games is where it comes from.

Read back to this page, that says the cliff a cut invents is not a local feature that later structure can soften. Whatever sits in front of the rearmost run, however wide the gaps between, the rate is fixed at the back — which is why a cut made anywhere ahead of it changes the value by a bounded amount and never changes how fast the value dies.