Line up full boards by score, counting matched internal joints out of 480
(the matched-edges convention used throughout this wiki), and the ladder
looks continuous: rung after rung, each occupied by some board somebody has
built. It has exactly one hole. No legal full placement of the 256 pieces
scores 479. That is a theorem about the published piece set, provable by
counting alone, and this page walks the whole argument: the evenness that
forces it, the defect floor of 2 that follows, and the census that bounds
how many one-move near misses can surround any solution. It is also one
entry in the theorem sweep, the survey of
what can be proven from the bag before any search runs.
256 tiles with four edges each carry 1024 half-edges. The gray frame color
accounts for 64 of them, exactly the number of outward-facing slots on the
rim (16 per side). The remaining 960 carry the 22 colors, and the census is
strikingly even:
| Color group | Half-edges |
|---|
| Gray frame edge | 64 |
| Five frame-join colors (1 to 5) | 24 each |
| Five interior colors (6 to 10) | 48 each |
| Twelve interior colors (11 to 22) | 50 each |
Every colored count is even. The colored total, 960, is exactly twice the
number of internal joints (2×16×15=480): in a frame-legal
full placement, where all 64 gray half-edges face outward, the colored
half-edges exactly fill the 960 internal joint slots with nothing left over.
None of this depends on where the pieces go. These are properties of the bag.
There is a pleasing shortcut hiding here: the evenness could have been
predicted without examining a single piece. The designers guarantee a
solution exists, and a solved board pairs every colored half-edge with a
same-colored partner, so each color's count is twice its joint count, which
is even by definition. The census merely confirms on the real set what
solvability already promised.
Fix any full placement with a legal frame (every rim-facing edge gray) and
fix a color c. Each of the 480 internal joints shows two half-edges. Some
joints show c on both sides, some on exactly one. Counting c's
half-edges:
h(c)=2⋅#{joints showing c twice}+#{joints showing c once}.
Since h(c) is even, the number of joints showing c exactly once is even
too. This holds for every color simultaneously, in every full placement,
solved or broken.
Now suppose a placement scores 479. It has exactly one broken joint, and a
broken joint shows two different colors, say a and b (if both sides
agreed it would be a match; gray cannot appear, since all 64 gray half-edges
face outward). Every other joint is matched, showing its color twice. So
exactly one joint shows a exactly once. One is odd. That contradicts the
evenness of h(a), and the placement cannot exist.
Write the defect of a board as 480 minus its score. Parity forbids defect
1 and forbids nothing else: at defect 2 the odd counts can absorb each other
in pairs, so the argument goes silent. Every legal full placement is
therefore either a solution or misses at least two joints. One more censused
fact closes the remaining loophole: no Eternity II tile is fixed by any
rotation (0 of 256), so a board cannot score 480 while differing from a
solution only by a tile spun in place. Score 480 means solution. Anything
else means 478 or below.
For today's leaderboards the missing rung is academic: the community's best
full boards stand at 470 under the center-clue-only convention and 464 with
all five clues placed (the records page keeps the full
ladder). But it changes what "almost solved" can ever mean. There is no 479
to pass through on the way to 480. The last step of the climb is 478 to
480, two joints at once, and any method that improves boards one joint at a
time is structurally unable to take it.
A solution exists; the puzzle was constructed from one. What sits right next
to it at 478? A board one move away from a solution must come from a move
that breaks exactly two joints, and only two kinds of single move can do
that. Both are countable in the bag:
- Twin swaps. Two interior tiles that agree, in some orientations, on
three of their four edges. Placed so the agreeing edges line up, exchanging
them disturbs one edge of each: two joints. The set contains exactly
50 such near-twin pairs.
- In-place turns. A tile whose repeated colors let a half-turn or a
quarter-turn keep two of its four edge colors in position, so turning it
where it stands breaks exactly the other two joints. The set contains
23 half-turn tiles and 3 quarter-turn tiles of this kind.
Adding them up, any solution has at most 50+23+3=76 defect-2
neighbours reachable by a single swap or rotation. This is an upper bound
from the bag: whether a given twin pair actually lines up inside a specific
solution depends on that solution, so the realized count for the designed
solution is open. The ceiling stands regardless. Around the very top of the
ladder, near misses are sparse, a few dozen boards at most, while rungs
further down are populated astronomically. It is the same scarcity that
forbidden patterns show at the 2×2
scale, read at the summit instead.
The census has a second story to tell. Colors 1 to 5 appear on zero interior
tiles; their 5×24=120 half-edges live entirely on the side edges
of the 60 rim tiles. The rim ring has exactly 60 joints, so in any solution
the five frame-join colors saturate them exactly, 12 joints per color, with
no slack at all. And because a rim tile's orientation is forced (gray
outward), each of the 56 edge tiles shows one fixed color to the interior
wherever it lands. Summed over the bag, the interior receives a fixed demand
vector over colors 6 to 22, namely (4, 5, 3, 3, 1, 1, 2, 3, 4, 6, 4, 2, 3,
6, 4, 3, 2), 56 inward edges in all. Whatever frame you build, the
interior's boundary bill is the same 17 numbers, known before any search
starts.
The evenness was spotted before the puzzle even shipped. In June 2007 David
Eddy opened a mailing-list thread titled Parity with the observation that
"the number of edges of each type is even", and drew the last-piece
consequence: if the final piece has four different edges, the final hole
must show the same four colors in some order, which he put at a 1 in 6
chance of matching (msg 332).
Christophe Weibel supplied the worked example showing the last piece
genuinely can fail to match its hole
(msg 335), and Brendan Owen,
who had studied the design, closed the thread with the practical verdict:
parity is correct for E2 by construction, and in a search it only ever helps
near the very end (msg 346).
All of that is true, and the thread stopped there. Pushed one step further,
the same observation contains the theorem above: the evenness does more than
make the last piece chancy, it deletes the 479 rung outright, sets the
defect floor at 2, and caps the one-move shell around every solution at 76.
Every number on this page reduces to a one-pass census of the official set:
the half-edge counts and their evenness, the 960 exact-fit identity, the
zero rotationally symmetric tiles, the 4/56/196 corner/edge/interior split,
the 50/23/3 defect-2 generators, the frame saturation, and the inward demand
vector. The checker recomputes each value beside its expected one and emits
a pass flag per claim; all 16 checks pass and the output is byte-stable
across runs. The
article, checker source and results are on GitHub,
and the reproduction block on this page points at the same topic. One
caveat on scope: the parity argument covers full 256-piece placements with a
legal gray frame. Partial boards, and boards that break the frame rule, are
outside it.