The results, worked through: on these boards the five clue-shaped hints never help a backtracker, they range from a mild cost to a catastrophe, and the fill order decides how much damage they do; the scores are bimodal, not a smooth gradient; and the hint-count question is confounded by a free pinned-seam floor.
The study fixes five hints in the shape of Eternity II's own five clues, runs eight fill orders on the identical boards, and asks a simple question: what are those five hints worth? Every number below is over fifteen distinct generated instances, single core, eight seconds a run, re-scored out of 480 by one canonical scorer. The comparison is paired, every fill order sees the same boards, and the per-instance spread is shown rather than averaged away, because it turns out to matter more than any median.
Start with the distribution. Each dot is one board; the tick is the median.
One dot per board. The scores are bimodal — a board either nearly solves or the search stalls early, with little in between — so the spread, not a median, is the honest summary. The compact orders land high on most boards but are dragged down by the ones where a pinned piece walls off the sweep; the fragmenting orders sit low throughout.
The paired change from adding the five clue-shaped hints, same board with and without them. Every bar is at or below zero: the hints never help a backtracker here. On the compact orders they cost a little; on spiral-out and the hint-seeking order they are catastrophic, turning a board the solver nearly finishes without help into one it barely starts. A correct hint is still a hard constraint the fixed fill order must satisfy on arrival — and sometimes it cannot.
For contrast, the beam solver (not a backtracker) reaches a median of 453/480 on the same boards — the hints and board are far from unsolvable; the chronological backtracker simply cannot use them.
Each bar is the median score of a hint layout, split into the seams the pins complete for free (both endpoints pinned) and the seams the search actually earned. A clustered block banks a tall free floor — up to a quarter of the whole board — so its raw score flatters it. Compare the earned parts, not the totals.
Two things are visible at once. First, the scores are bimodal: on most orders a board either climbs into the 360s or stalls in the double digits, with little in between. A median drawn through that is a summary of a gap, not a centre, which is exactly why the dots are shown. Second, the orders split hard: the three compact sweeps (row-major, its bottom-up mirror, Verhaard's comb) sit high on most boards; the fragmenting orders sit low throughout.
But the compact-beats-fragmenting split is not the real finding, because it invites the wrong question, "which order best uses the hints?" The question that matters is whether the hints help at all. The second chart answers it, and the answer is no. It shows the paired change from adding the five hints: each order's score on a board minus its score on the same board with no hints.
Every bar is at or below zero. The five clue-shaped hints do not help a single fill
order. On the compact sweeps they cost only ten to twenty points. On spiral-out they
cost about ninety. And on the two hint-seeking orders they are ruinous: the
trace-hints order, which draws a skeleton between the clues before filling, loses
about three hundred and twenty-five, and the connect-hints-first flood loses
roughly three hundred and forty-five, taking an order that scores among the
best of all eight with no hints to the worst with them. The more deliberately an
order chases the hints, the more they cost it. Handing the solver five correct
pieces, in the puzzle's own clue geometry, made every version of it worse.
A pinned piece is not free information to a chronological backtracker; it is a hard constraint the fixed fill order must satisfy on arrival. When a compact sweep rolls down to a pinned interior cell, the piece is already there, and the row it just built has to match that piece's edges. Most of the time it can, at a small cost: the sweep steers around the constraint and loses a few tens of points. But on some boards the pinned piece contradicts what the frontier has committed to, and there is no local repair: the search hits a wall it cannot pass and thrashes below it. That is the stalled mode, and it is the hints that create it. On row-major, the boards that collapse into the double digits are precisely the ones where a clue-shape pin lands where the sweep cannot honour it; the same board with no pins climbs into the 360s and 370s.
This reframes the fill-order result. The order still matters (a compact sweep
survives the hint constraints with a scar of ten to twenty points while
connect-hints-first is destroyed by them), but what the order is buying is not
"using the hints well." It
is surviving them. The frontier is why: an order that keeps a single tight
frontier has room to route around a bad pin; an order that has already fragmented
into five open blobs has committed everywhere at once and cannot.
That frontier relationship, across the eight orders, is strikingly clean, and it is worth showing precisely because the frontier can be computed from an order's geometry with no solver at all, then checked against the measured scores:
The average open frontier an order holds predicts its median score closely across these eight orders. It is a strong descriptive relationship, not a law proved on eight points, and it stops settling the ranking among the fragmenting orders on the right (spiral-in holds a larger frontier than spiral-out yet scores higher). But the direction is exactly what the mechanism predicts: branching cost is multiplicative in the frontier, so an order that keeps the frontier small keeps the room to absorb a hostile pin.
For contrast, the beam solver, which is not a chronological backtracker and does not pay the frontier cost the same way, reaches a median in the 450s on these same hinted boards. The hints and boards are nowhere near unsolvable. It is specifically the chronological, fixed-order backtracker that cannot turn five correct pieces into progress.
The same question one level out: does adding more hints help? The answer is not a smooth "more is better", and the raw number hides which part is real.
The lower chart above splits each layout's score into two parts. The floor is the seams the pins complete for free, because both their endpoints are pinned to the true solution; the earned part is what the search actually found. A solid clustered block banks a tall floor (five 4×4 blocks pin a quarter of the whole board's seams before the search takes a single step), while a spread lattice, whose hints never touch, banks nothing. So a raw-score comparison hands clustered layouts a hundred-point head start that says nothing about whether the search made headway.
Read the earned column across the spread lattices, whose floor is zero so earned is the score, and a threshold appears. A sparse spread, four to sixteen hints, earns almost nothing (in the twenties): the pins are just scattered constraints the sweep keeps tripping over, exactly the five-clue effect. But keep adding them and the picture flips. Twenty-five spread hints earn 127, and thirty-six, a six-per-line lattice, solve the board outright on most instances. Below the threshold the spread hints only get in the way; above it there are finally enough of them to carve the board into pieces small enough for the sweep to finish. It is not a gradient of help, but a wall the count has to clear.
The clustered blocks show the mirror image. Their raw score is mostly floor: five 4×4 blocks (eighty hints, a quarter of the board) do solve every instance, but they have pinned so much of the board that they have half-solved it by hand. Strip the floor away and the clustered layouts short of that extreme earn only double digits, coasting on the free seams. So the real count story is not "more hints help" or "more hints hurt", but: it takes a lot of correct hints, spread or clustered, to move a chronological backtracker at all, and until you reach that amount the extra pins are as likely to trip the search as to speed it.
The method page works the floor arithmetic through in full.
The community's hint-geometry write-up makes a sharper version of the placement claim: eighteen hints scattered on a lattice solve a 16×16 E2-like puzzle in minutes, while eighteen piled into contiguous top rows barely help, and you need eighty or more contiguous hints to match the scattered eighteen. That result was measured on one specific puzzle with one specific engine. We put its two exact layouts, the scattered rows-5 lattice and the eighteen-cell top block, on our own generated boards and solvers to see whether the direction holds.
The two 18-hint layouts from the hint-geometry write-up, measured here per fill path. Contrary to that page, on our boards the contiguous block scores higher than the scattered lattice for seven of the eight orders, often by a wide margin. The two studies measure different things: the community result is time to a full solution, where scattered hints reach the endgame; ours is matched-edge score at a short budget, where a contiguous top block gives a row-major sweep a large correct region to build against fast.
It does not. On our boards the contiguous eighteen score higher than the scattered eighteen for seven of the eight fill orders, and by a wide margin: a row-major sweep reaches a median of 377 with the contiguous block and only 138 with the scattered lattice. This looks like a flat contradiction, and it is worth being precise about why it is not quite one.
The two studies measure different things. The community result is about time to a full solution: scattered hints reach into the deep endgame where a backtracker spends almost all of its time, so they prune the expensive part, while a contiguous top block prunes only the cheap opening. Our number is matched-edge score at a short budget, and at eight seconds none of these strict backtrackers reaches the endgame at all. What a top-anchored contiguous block does buy, immediately, is a large correct region for the row-major sweep to build against, so the score climbs fast even though the hard part of the board is untouched. Scattered hints, by contrast, fragment the early fill exactly the way the five-clue shape did. So the two results are consistent once you separate "solves the whole board eventually" from "scores well in the first eight seconds": scattered placement helps the former and hurts the latter. The next section makes that split visible.
At 16×16 nothing solves in eight seconds, so the score is always a snapshot of a search still in its opening. To see the endgame effect the community reported, we need a board that fully solves. An 8×8 built to the same colour recipe does, in well under a second, so on it we can measure the quantity that actually matters: the number of search nodes a row-major backtracker needs to reach a complete solution. Fewer nodes means the hints did real pruning work. Below, a spread lattice and a matched contiguous block, at rising hint counts.
Each pair of dots is one matched hint count: a spread lattice and a contiguous block of the same size, on the same boards, over the seeds that solved. At four hints the scattered lattice barely solves. Around sixteen the picture inverts: the scattered lattice solves in a few thousand nodes while the matched contiguous block still needs nearly two million, a five-hundred-fold gap. Past that both become easy as the board fills up. This is the community hint-geometry result reproduced on our own boards, and it is the reason the 16×16 score comparison looked opposite: at 16×16 in eight seconds nobody reaches the endgame, so scattered hints never get to do the pruning that wins here.
This is the whole story in one chart, and it finally lines up with the community claim. At four hints the scattered lattice is worse than useless, solving only two of thirty boards, the same sparse-hints-trip-the-search effect from every other axis, while the matched contiguous block solves almost all of them. But cross a threshold around sixteen hints and the lines swap hard: a sixteen-hint scattered lattice solves in about four thousand nodes, while the matched contiguous block still grinds through nearly two million, a five-hundred-fold difference at the same hint count. Push further and both layouts become easy (by thirty-six hints a quarter of the board is pinned and either one solves in a few hundred nodes), so the scattered advantage is a window, widest where the count is enough to reach the deep search but not so large that the board is half-solved by hand. In that window, spread hints reach into the part of the search a backtracker actually struggles with and cut it off, exactly as Joe and Peter McGavin described; a contiguous block only ever prunes the easy opening. The 16×16 score comparison looked like it contradicted them only because at a short budget the search never lives long enough to reach the region where scattered placement pays off.
This is a result about strict, chronological depth-first backtrackers on generated 16×16 boards built to Eternity II's colour recipe, at a short (eight second) budget. Each of those scoping words earns its place. Chronological: the result is specifically about solvers that fill cells in a fixed order and must satisfy a pinned piece when they reach it, a beam solver, which does not, reaches the 450s on the same boards. Generated: the boards share E2's colour recipe and its five-clue shape, but they are not the official puzzle, and the study transfers no number to it. Short budget: none of these strict backtrackers solves within eight seconds, so this measures how far each reaches, not a race to 480; whether the "hints hurt" effect survives at much longer budgets is untested here.
Within that scope the lesson is sturdy and, we think, counterintuitive: for a chronological backtracker, five correct pieces placed in the puzzle's own clue geometry are not a gift but a constraint, and can cost far more than they give. What decides the damage is not the hints but the fill order that has to live with them and the order pays for a hint the way it pays for everything else, in the size of the frontier it keeps open. It is one more reading of why the puzzle resists: even correct information helps only a solver built to receive it.