Editing 15 puzzle (section)

==Mathematics==
=== Solvability ===
[[File:15-Puzzle solved.png|thumb|A solved 15 Puzzle]]
{{harvtxt|Johnson|Story|1879}} used a [[parity (mathematics)|parity]] argument to show that half of the starting positions for the ''n'' puzzle are impossible to resolve, no matter how many moves are made. This is done by considering a binary function of the tile configuration that is [[invariant (mathematics)|invariant]] under any valid move and then using this to partition the space of all possible labelled states into two mutually inaccessible [[equivalence class]]es of the same size. This means that half of all positions are unsolvable, although it says nothing about the remaining half.

The invariant is the [[Parity of a permutation|parity of the permutation]] of all 16 squares plus the parity of the [[taxicab distance]] (number of rows plus number of columns) of the empty square from the lower right corner. This is an invariant because each move changes both the parity of the permutation and the parity of the taxicab distance. In particular, if the empty square is in the lower right corner, then the puzzle is solvable only if the permutation of the remaining pieces is even.

{{harvtxt|Johnson|Story|1879}} also showed that on boards of size ''m'' × ''n'', where ''m'' and ''n'' are both larger or equal to 2, all even permutations ''are'' solvable. It can be proven by induction on ''m'' and ''n'', starting with ''m'' = ''n'' = 2. This means that there are exactly two equivalency classes of mutually accessible arrangements, and that the parity described is the only non-trivial invariant, although equivalent descriptions exist.

{{harvtxt|Archer|1999}} gave another proof, based on defining [[equivalence class]]es via a [[Hamiltonian path]].

{{harvtxt|Wilson|1974}} studied the generalization of the 15 puzzle to arbitrary [[finite graph]]s, the original problem being the case of a 4×4 [[grid graph]]. The problem has some degenerate cases where the answer is either trivial or a simple combination of the answers to the same problem on some subgraphs. Namely, for [[path graph|paths]] and [[cycle graph|polygons]], the puzzle has no freedom; if the graph is [[disconnected graph|disconnected]], only the connected component of the vertex with the "empty space" is relevant; and if there is an [[articulation vertex]], the problem reduces to the same puzzle on each of the [[biconnected graph|biconnected]] components of that vertex. Excluding these cases, Wilson showed that other than one exceptional graph on 7 vertices, it is possible to obtain all permutations unless the graph is [[bipartite graph|bipartite]], in which case exactly the even permutations can be obtained. The exceptional graph is a regular [[hexagon]] with one diagonal and a vertex at the center added; only {{sfrac|1|6}} of its permutations can be attained, which gives an instance of the [[Automorphisms of the symmetric and alternating groups#Exotic map S5 → S6|exotic embedding of S<sub>5</sub> into S<sub>6</sub>]].

For larger versions of the ''n'' puzzle, finding a solution is easy. But, the problem of finding the ''shortest'' solution is [[NP-hard]]. It is also NP-hard to [[Approximation algorithm|approximate]] the fewest slides within an additive constant, but there is a polynomial-time constant-factor approximation.<ref name="Ratner1986">{{cite journal |last1=Ratner |first1=Daniel |last2=Warmuth |first2=Manfred |title=Finding a Shortest Solution for the N × N Extension of the 15-PUZZLE Is Intractable |journal=National Conference on Artificial Intelligence |date=1986 |url=https://www.aaai.org/Papers/AAAI/1986/AAAI86-027.pdf |url-status=live |archiveurl=https://web.archive.org/web/20120309151834/https://www.aaai.org/Papers/AAAI/1986/AAAI86-027.pdf |archivedate=2012-03-09}}</ref><ref name="ratner+warmuth">{{cite journal|last=Ratner|first=Daniel|author2=Warmuth, Manfred|title=The (n<sup>2</sup>−1)-puzzle and related relocation problems|journal=Journal of Symbolic Computation|year=1990|volume=10|issue=2|pages=111–137|doi=10.1016/S0747-7171(08)80001-6|doi-access=free}}</ref> For the 15 puzzle, lengths of optimal solutions range from 0 to 80 single-tile moves (there are 17 configurations requiring 80 moves)<ref>[[Richard E. Korf]], [https://dl.acm.org/citation.cfm?id=1455250 Linear-time disk-based implicit graph search], ''Journal of the ACM'' Volume 55 Issue 6 (December 2008), Article 26, pp. 29-30. "For the 4 × 4 Fifteen Puzzle, there are 17 different states at a depth of 80 moves from an initial state with the blank in the corner, while for the 2 × 8 Fifteen Puzzle there is a unique state at the maximum state of 140 moves from the initial state."</ref><ref>A. Brüngger, A. Marzetta, K. Fukuda and J. Nievergelt, [http://www.iro.umontreal.ca/~gendron/Pisa/References/BB/Brungger99.pdf The parallel search bench ZRAM and its applications], ''Annals of Operations Research'' '''90''' (1999), pp. 45–63.<br>:"Gasser found 9 positions, requiring 80 moves...We have now proved that the hardest 15-puzzle positions require, in fact, 80 moves. We have also discovered two previously unknown positions, requiring exactly 80 moves to be solved."</ref> or 43 multi-tile moves;<ref name="multi-tile">[http://forum.cubeman.org/?q=node/view/223 "The Fifteen Puzzle can be solved in 43 "moves""]. Domain of the Cube Forum</ref> the 8 Puzzle always can be solved in no more than 31 single-tile moves or 24 multi-tile moves (integer sequence [http://oeis.org/A087725 A087725]). The multi-tile metric counts subsequent moves of the empty tile in the same direction as one.<ref name="multi-tile" />

The number of possible positions of the 24 puzzle is {{sfrac|25!|2}} ≈ {{val|7.76e24}}, which is too many to calculate [[God's number]] feasibly using brute-force methods. In 2011, lower bounds of 152 single-tile moves or 41 multi-tile moves had been established, as well as upper bounds of 208 single-tile moves or 109 multi-tile moves.<ref>[http://forum.cubeman.org/?q=node/view/238/2557#comment-2557 "24 puzzle new lower bound: 152"]. Domain of the Cube Forum</ref><ref>[https://web.archive.org/web/20160307005428/http://juropollo.xe0.ru/stp_results_mxn_en.htm "m × n puzzle (current state of the art)"]. Sliding Tile Puzzle Corner.</ref><ref>[http://forum.cubeman.org/?q=node/view/241 "208s for 5x5"]. Domain of the Cube Forum.</ref><ref>[http://forum.cubeman.org/?q=node/view/241/2566#comment-2566 "5x5 can be solved in 109 MTM"]. Domain of the Cube Forum.</ref> In 2016, the upper bound was improved to 205 single-tile moves.<ref>[http://forum.cubeman.org/?q=node/view/559 "5x5 sliding puzzle can be solved in 205 moves"]. Domain of the Cube Forum.</ref>

===Group theory===
The transformations of the 15 puzzle form a [[groupoid]] (not a group, as not all moves can be composed);<ref>Jim Belk (2008) [https://cornellmath.wordpress.com/2008/01/27/puzzles-groups-and-groupoids/ Puzzles, Groups, and Groupoids], The Everything Seminar</ref><ref>[http://www.neverendingbooks.org/the-15-puzzle-groupoid-1 The 15-puzzle groupoid (1)], Never Ending Books</ref><ref>[http://www.neverendingbooks.org/the-15-puzzle-groupoid-2 The 15-puzzle groupoid (2)], Never Ending Books</ref> this [[Group action (mathematics)#Generalizations|groupoid acts]] on configurations.

Because the combinations of the 15 puzzle can be generated by [[Permutation#Definition|3-cycles]], it can be proved that the 15 puzzle can be represented by the [[alternating group]] <math>A_{15}</math>.<ref>{{cite web | access-date=26 December 2020 | first1=Robert | last1=Beeler | url=https://faculty.etsu.edu/beelerr/fifteen-supp.pdf | title=The Fifteen Puzzle: A Motivating Example for the Alternating Group | publisher=East Tennessee State University | website=faculty.etsu.edu/ | archive-date=7 January 2021 | archive-url=https://web.archive.org/web/20210107214840/https://faculty.etsu.edu/beelerr/fifteen-supp.pdf | url-status=dead }}</ref> In fact, any <math>2 k - 1</math> sliding puzzle with square tiles of equal size can be represented by <math>A_{2 k - 1}</math>.