Editing Eight queens puzzle (section)

==Counting solutions for other sizes ''n''==
===Exact enumeration===
There is no known formula for the exact number of solutions for placing ''n'' queens on an {{math|''n'' × ''n''}} board i.e. the number of [[independent set (graph theory)|independent set]]s of size ''n'' in an {{math|''n'' × ''n''}} [[queen's graph]]. The 27×27 board is the highest-order board that has been completely enumerated.<ref>{{Cite web|url=https://github.com/preusser/q27|title=The Q27 Project|via=GitHub}}</ref>  The following tables give the number of solutions to the ''n'' queens problem, both fundamental {{OEIS|id=A002562}} and all {{OEIS|id=A000170}}, for all known cases.

{| class="wikitable" style="text-align:right;"
!style="padding: 0em .5em;"|''n''
|fundamental
|all
|-
!style="padding: 0em .5em;"|1
|1
|1
|-
!style="padding: 0em .5em;"|2
|0
|0
|-
!style="padding: 0em .5em;"|3
|0
|0
|-
!style="padding: 0em .5em;"|4
|1
|2
|-
!style="padding: 0em .5em;"|5
|2
|10
|-
!style="padding: 0em .5em;"|6
|1
|4
|-
!style="padding: 0em .5em;"|7
|6
|40
|-
!style="padding: 0em .5em;"|8
|12
|92
|-
!style="padding: 0em .5em;"|9
|46
|352
|-
!style="padding: 0em .5em;"|10
|92
|724
|-
!style="padding: 0em .5em;"|11
|341
|2,680
|-
!style="padding: 0em .5em;"|12
|1,787
|14,200
|-
!style="padding: 0em .5em;"|13
|9,233
|73,712
|-
!style="padding: 0em .5em;"|14
|45,752
|365,596
|-
!style="padding: 0em .5em;"|15
|285,053
|2,279,184
|-
!style="padding: 0em .5em;"|16
|1,846,955
|14,772,512
|-
!style="padding: 0em .5em;"|17
|11,977,939
|95,815,104
|-
!style="padding: 0em .5em;"|18
|83,263,591
|666,090,624
|-
!style="padding: 0em .5em;"|19
|621,012,754
|4,968,057,848
|-
!style="padding: 0em .5em;"|20
|4,878,666,808
|39,029,188,884
|-
!style="padding: 0em .5em;"|21
|39,333,324,973
|314,666,222,712
|-
!style="padding: 0em .5em;"|22
|336,376,244,042
|2,691,008,701,644
|-
!style="padding: 0em .5em;"|23
|3,029,242,658,210
|24,233,937,684,440
|-
!style="padding: 0em .5em;"|24
|28,439,272,956,934
|227,514,171,973,736
|-
!style="padding: 0em .5em;"|25
|275,986,683,743,434
|2,207,893,435,808,352
|-
!style="padding: 0em .5em;"|26
|2,789,712,466,510,289
|22,317,699,616,364,044
|-
!style="padding: 0em .5em;"|27
|29,363,495,934,315,694
|234,907,967,154,122,528
|}
The number of placements in which furthermore no three queens lie on any straight line is known for <math>n \leq 25</math> {{OEIS|id=A365437}}.

=== Asymptotic enumeration ===
In 2021, Michael Simkin proved that for large numbers ''n'', the number of solutions of the ''n'' queens problem is approximately <math>(0.143n)^n</math>.<ref>{{Cite web|last=Sloman|first=Leila|date=2021-09-21|title=Mathematician Answers Chess Problem About Attacking Queens|url=https://www.quantamagazine.org/mathematician-answers-chess-problem-about-attacking-queens-20210921/|access-date=2021-09-22|website=Quanta Magazine|language=en}}</ref>  More precisely, the number <math>\mathcal{Q}(n)</math> of solutions has [[asymptotic growth]]
<math display="block">
\mathcal{Q}(n) = ((1 \pm o(1))ne^{-\alpha})^n
</math>
where <math>\alpha</math> is a constant that lies between 1.939 and 1.945.<ref>{{Cite arXiv|last=Simkin|first=Michael|author-link=Michael Simkin|date=2021-07-28|title=The number of $n$-queens configurations|class=math.CO|eprint=2107.13460v2|language=en}}</ref> (Here ''o''(1) represents [[little o notation]].)

If one instead considers a [[flat torus|toroidal]] chessboard (where diagonals "wrap around" from the top edge to the bottom and from the left edge to the right), it is only possible to place ''n'' queens on an <math>n \times n</math> board if <math>n \equiv 1,5 \mod 6.</math>  In this case, the asymptotic number of solutions is<ref>{{Cite arXiv|last=Luria|first=Zur|date=2017-05-15|title=New bounds on the number of n-queens configurations|class=math.CO|eprint=1705.05225v2|language=en}}</ref><ref>{{Cite arXiv|last1=Bowtell|first1=Candida|last2=Keevash|first2=Peter|authorlink2 = Peter Keevash|date=2021-09-16|title=The $n$-queens problem|class=math.CO|eprint=2109.08083v1|language=en}}</ref>
<math display="block">T(n) = ((1+o(1))ne^{-3})^n.</math>