Shotgun assembly of random jigsaw puzzles

Abstract

We consider the shotgun assembly problem for a random jigsaw puzzle, introduced by Mossel and Ross (2015). Their model consists of a puzzle-an n×n grid, where each vertex is viewed as a center of a piece. Each of the four edges adjacent to a vertex is assigned one of q colors (corresponding to "jigs," or cut shapes) uniformly at random. Unique assembly refers to there being only one puzzle (the original one) that is consistent with the collection of individual pieces. We show that for any > 0, if q ≥ n 1+ , then unique assembly holds with high probability. The proof uses an algorithm that assembles the puzzle in time n Θ(1∕ ) .

Reconstructing Random Jigsaws

“…The starting point of our approach to proving the 1-statement in Theorem 1.2 is the strategy adopted by Bordenave, Feige and Mossel [4] to show that J = J(n, q) is reconstructible with high probability when q ≥ n 1+ε for some fixed ε > 0. Given the deck D(J) of J, Bordenave, Feige and Mossel use the following procedure to identify the neighbours of a given tile J v with v ∈ [n] 2 .…”

“…We define γ(f ) to be the difference between the size of the vertex set of G f and the number of connected components of G f . We require the following observation due to Bordenave, Feige and Mossel [4]; we include the short proof for completeness.…”

“…For n, q ∈ N, an (n, q)-jigsaw is a q-coloured extended n × n grid, i.e., an extended n × n grid whose 2n(n + 1) edges are coloured using a set of q different colours which we take to be [q] for concreteness. The tile of an (n, q)-jigsaw corresponding to a vertex v ∈ [n] 2 is given by the colouring of the four edges incident to v; more precisely, writing e 1 = (0, 1), e 2 = (1, 0), e 3 = −e 1 and e 4 = −e 2 , if the edge between v and v + e i gets colour c i ∈ [q] for 1 ≤ i ≤ 4, then the tile corresponding to v is the tuple ( 4 . Finally, the deck of an (n, q)-jigsaw is the multiset of the tiles of the jigsaw, one for each vertex of [n] 2 .…”

“…From below, Mossel and Ross [9] showed that P(J(n, q) is reconstructible) → 0 when q = o(n 2/3 ) due to the presence of local obstacles to reconstruction: in this regime, a random (n, q)-jigsaw contains, with high probability, two configurations each consisting of two neighbouring vertices which may be 'exchanged' in the jigsaw, and this is easily seen to obstruct unique reconstruction; however, this argument does not extend to configurations involving single exchangeable vertices (and to a corresponding bound when q = o(n)) since the presence of two identical tiles in the deck does not necessarily prevent unique reconstruction. From above, Bordenave, Feige and Mossel [4] and Nenadov, Pfister and Steger [10] independently showed, for any fixed ε > 0, that P(J(n, q) is reconstructible) → 1 when q ≥ n 1+ε . Here, we improve on both of these bounds and prove the following nearly optimal result.…”

“…The two results contained in the statement of Theorem 1.2 are proved by very different methods: the '0-statement' follows from a double counting argument, while the proof of the '1-statement' is based on an isoperimetric argument which draws from (but is somewhat more involved than) the strategy used by Bordenave, Feige and Mossel [4] where one attempts to reconstruct a suitably large neighbourhood of a tile in order to identify its neighbours in the jigsaw.…”