Linking the clues early does not help, it hurts

Joining clue pairs with early piece corridors feels like adding constraints. Counting says otherwise: a 1-wide corridor between even the two closest clues admits about two quadrillion distinct fillings, so it excludes almost nothing while spending pieces the endgame will need. In a controlled A/B, every corridor arm lost to its own control at every board size.

findingFindingdead endmeasuredrefutedWhy it resists Reduce the searchUpdated 2026-07-22

Reproducedeterministic — reproduces byte-for-bytereruns the search

The five clues are the only cells of the official puzzle whose contents you know for certain. Stare at the board for a minute and an idea offers itself: connect them. Lay a short chain of pieces from clue to clue, let each chain become a fixed spine, and hope the spines constrain everything that comes after. I liked the idea enough to measure it, and the measurement is a clean negative twice over. An exact count shows that a clue-to-clue corridor admits about two quadrillion distinct fillings, so committing to one constrains almost nothing; and in a controlled A/B, every arm that laid corridors early finished worse than the same solver without them, at every board size tested.

Five islands, and the bridges between them#

The clues sit at cells 34, 45, 135, 210 and 221, that is (x, y) = (2, 2), (13, 2), (7, 8), (2, 13) and (13, 13): four inset two cells from the corners, one just left of centre. Their pairwise Manhattan distances run from 10 to 22. No two clues are close to adjacent, so the shortest possible bridge between any pair is a chain of ten placements.

What a chain placement is worth depends on how many sides it has to match. With $n = 196$ interior pieces and $C = 22$ declared interior colours, a cell matching $k$ already-placed neighbours has an expected $b_k = 4n / C^k$ legal candidates: $b_1 \approx 35.6$ , $b_2 \approx 1.62$ , $b_3 \approx 0.074$ . A placement only prunes, in expectation, once $b_k$ drops below 1, which requires $k \ge 3$ matched sides. A 1-wide corridor is built entirely from $k = 1$ placements: each new piece touches only its predecessor, and inherits about 36 legal candidates at every step.

What one corridor is worth: two quadrillion fillings#

The branching table is a uniform model, so the checker also counts exactly on the real piece set. If anything the real set is looser: the number of (piece, rotation) pairs presenting a given interior colour on a given side averages 46.1 (range 43 to 49), against the model's 35.6. Build the transfer matrix $T[a][b]$ counting inner pieces that show colour $a$ on one side and colour $b$ on the opposite side, raise it to the corridor length, and the matrix powers give exact walk counts. For the length-10 corridor between the closest clue pair, with both endpoint colours fixed by the clues, that count is $2.59\times10^{15}$ walks (the mean over colour pairs). Walks may reuse a piece; applying the mean-field distinctness correction $\prod_{j=0}^{9}(1 - j/196) = 0.792$ leaves $2.05\times10^{15}$ fillings using distinct pieces. My first pass at this count, with a slightly tighter piece model, gave $1.9\times10^{15}$ ; the committed checker settles on $2.05\times10^{15}$ , and the conclusion does not move between the two.

A commitment that can be satisfied two quadrillion ways is not a constraint in any useful sense. Laying it excludes a vanishing fraction of the search space, while it withdraws ten pieces from the pool at $k = 1$ , exactly where the arithmetic says the search gets nothing back. Route two or three clue pairs and the bill is 10 to 20 pieces spent before a single genuinely constrained cell has been filled.

The A/B: every corridor arm lost#

Counting says the corridor buys nothing; it takes an experiment to show that it costs. The study's original run compared three arms that differ in nothing but the corridor phase: a control (clues pinned, then a max-contact greedy fill with restarts), a 1-wide corridor arm, and a 2-wide ribbon arm, each laying its clue-to-clue routes before the identical fill. It ran on scaled instances with faithful clue geometry, at board sizes N = 8 to 16, with 12 paired seeds per cell and 20 seconds per single-core run.

Board	Control mean	1-wide corridor	2-wide ribbon
8×8	92.50	-4.67	-3.00
10×10	144.17	-9.58	-5.50
12×12	209.83	-13.58	-6.75
14×14	285.67	-12.00	-7.83
16×16	377.17	-17.67	-8.08

Scores are matched edges, and each delta is an arm against its own control on paired seeds. Every corridor arm lost to its control at every size, with paired Wilcoxon $|z| \ge 2.80$ . The harm grows as the boards, and with them the corridors, get longer: a mean loss of 4.7 edges at N = 8 becomes 17.7 at N = 16, where the two score distributions separate outright (the control's worst seed scored 373; the corridor arm's best managed 365). And widening the corridor to a 2-cell ribbon, which lets its second rank land at 2 contacts instead of 1, roughly halves the damage at every size. That is exactly the width dependence the counting predicts, which is what ties the mechanism to the measurement.

What this rules out, and what it leaves open#

The negative is precise: laying 1-wide (or 2-wide) clue-to-clue corridors as an early commitment hurts, and more corridor hurts more. The same arithmetic that kills them also points at where constraint is real: cells placed with 2 or more contacts, since only $k \ge 3$ prunes outright and $k = 2$ comes close. Growing compact clue-anchored regions, where most cells arrive with several contacts, is a different move and is untouched by this experiment.

The result also fits what the hint studies keep finding. On a related 16×16 puzzle, hint position beats hint count because a hint's value lies in reaching the part of the board the search struggles with; and in the lab hint study, the five official clues alone never helped a chronological backtracker on the boards tested. A corridor is the extreme way to misspend that gift: it cashes the five fixed cells immediately, in the cheapest region of the search, and pays for it with the piece pool. Like the negatives in the theorem sweep, this one comes priced: for clue linking to pay, its cells would have to arrive with three contacts, and a 1-wide path never does.

The counting side reproduces exactly: the checker in the topic directory recomputes the clue geometry, the branching table, the colour supply and the transfer-matrix corridor counts from the official piece set in about a second, and its output is committed as results/corridor_counts.json. The A/B table is the study's original measurement: the packaged rerun of the three arms is specified in the topic's reproduction plan and its tables are not yet committed, so read those deltas as the record of one run, with the fresh-run confirmation still pending.

Five islands, and the bridges between them#

What a chain placement is worth depends on how many sides it has to match. With

n = 196

interior pieces and

C = 22

declared interior colours, a cell matching

k

already-placed neighbours has an expected

b_k = 4n / C^k

legal candidates:

b_1 \approx 35.6

b_2 \approx 1.62

b_3 \approx 0.074

. A placement only prunes, in expectation, once

b_k

drops below 1, which requires

k \ge 3

matched sides. A 1-wide corridor is built entirely from

k = 1

placements: each new piece touches only its predecessor, and inherits about 36 legal candidates at every step.

What one corridor is worth: two quadrillion fillings#

T[a][b]

counting inner pieces that show colour

a

on one side and colour

b

on the opposite side, raise it to the corridor length, and the matrix powers give exact walk counts. For the length-10 corridor between the closest clue pair, with both endpoint colours fixed by the clues, that count is

2.59\times10^{15}

walks (the mean over colour pairs). Walks may reuse a piece; applying the mean-field distinctness correction

\prod_{j=0}^{9}(1 - j/196) = 0.792

leaves

2.05\times10^{15}

fillings using distinct pieces. My first pass at this count, with a slightly tighter piece model, gave

1.9\times10^{15}

; the committed checker settles on

2.05\times10^{15}

, and the conclusion does not move between the two.

k = 1

, exactly where the arithmetic says the search gets nothing back. Route two or three clue pairs and the bill is 10 to 20 pieces spent before a single genuinely constrained cell has been filled.

The A/B: every corridor arm lost#

Board	Control mean	1-wide corridor	2-wide ribbon
8×8	92.50	-4.67	-3.00
10×10	144.17	-9.58	-5.50
12×12	209.83	-13.58	-6.75
14×14	285.67	-12.00	-7.83
16×16	377.17	-17.67	-8.08

Scores are matched edges, and each delta is an arm against its own control on paired seeds. Every corridor arm lost to its control at every size, with paired Wilcoxon

|z| \ge 2.80

. The harm grows as the boards, and with them the corridors, get longer: a mean loss of 4.7 edges at N = 8 becomes 17.7 at N = 16, where the two score distributions separate outright (the control's worst seed scored 373; the corridor arm's best managed 365). And widening the corridor to a 2-cell ribbon, which lets its second rank land at 2 contacts instead of 1, roughly halves the damage at every size. That is exactly the width dependence the counting predicts, which is what ties the mechanism to the measurement.

What this rules out, and what it leaves open#

k \ge 3

prunes outright and

k = 2

comes close. Growing compact clue-anchored regions, where most cells arrive with several contacts, is a different move and is untouched by this experiment.

Five islands, and the bridges between them#

What one corridor is worth: two quadrillion fillings#

The A/B: every corridor arm lost#

What this rules out, and what it leaves open#

Keep exploring

Five islands, and the bridges between them#

What one corridor is worth: two quadrillion fillings#

The A/B: every corridor arm lost#

What this rules out, and what it leaves open#