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Independence and Domination Separation on
Chessboard Graphs
R. Douglas Chatham∗Maureen Doyle†Gerd H. Fricke∗
Jon Reitmann∗‡ R. Duane Skaggs∗§ Matthew Wolff¶
Abstract
Alegal placement of Queens is any placement of Queens on an order
Nchessboard in which any two attacking Queens can be separated by
a Pawn. The Queens independence separation number is the minimum
number of Pawns which can be placed on an N×Nboard to result in a
separated board on which a maximum of mindependent Queens can be
placed. We prove that N+kQueens can be separated by kPawns for
large enough Nand provide some results on the number of fundamental
solutions to this problem. We also introduce separation relative to other
domination-related parameters for Queens, Rooks, and Bishops.
1 Introduction
The well-known 8-Queens problem involves placing 8 Queens on an 8 ×8 chess-
board such that no two Queens attack each other. A collection of some refer-
ences to the N-Queens problem can be found in [15], [18] includes a variety of
other interesting chessboard problems, and a storage scheme for parallel memory
systems based on solutions to the N-Queens problem is given in [10]. Various
algorithms for solving the N-Queens problem are given in [1, 8, 9]. We primarily
use the notation and terminology of [12, 13].
In January 2004, the Chess Variant Pages [2] proposed a variation of the
traditional 8-Queens problem. The new problem is to place nine Queens on an
8×8 board by using Pawns to block all Queens that would otherwise attack
each other. An example in which three Pawns were needed to separate the nine
Queens was provided. A contest problem was to determine the least number of
∗Department of Mathematics and Computer Science, Morehead State University, More-
head, KY 40351 USA
†Department of Computer Science, Northern Kentucky University, Highland Heights, KY
41099 USA
‡Undergraduate student
§d.skaggs@moreheadstate.edu (Corresponding author)
¶Pyramid Controls, Inc., Cincinnati, OH 45203 USA
1
Pawns needed to separate these nine Queens. We say the chessboard (graph)
resulting from the addition of one or more Pawns is a separated board (graph).
Alegal placement of Queens is defined in [5] to be any placement in which any
two attacking Queens can be separated by a Pawn. The Queens independence
separation number,sQ(β, m, N ) is the minimum number of Pawns which can be
placed on an N×Nboard to result in a separated board on which a maximum
of mindependent Queens can be placed. More generally, for any set of natural
numbers Aand graph parameter πdefined on the Queens graph, we define
sQ(π, A, b) to be the minimum number of Pawns on a b×bboard which causes
the value of πto be an element of the set A. The upper separation number
relative to a given parameter, SQ(π, A, b), is the maximum cardinality of a
minimal set of Pawns that results in πtaking a value in A.
We consider sQ(π, A, b) for π=γ, β , i, where γis the domination number,
βis the independence number, and iis the independent domination number. If
A={a}and no confusion will arise, we denote sQ(π, A, b) by sQ(π, a, b) rather
than sQ(π, {a}, b). Similarly, sQ(π, [x, y], b) will be used when A={[x, y ] : x∈
N∪ {0}, y ∈N}.
It is proved in [5] that sQ(β, N + 1, N) = 1 for N > 5 and conjectured that
sQ(β, N +k, N) = kfor large enough N. We prove this conjecture, namely that
if N > max{87 + k, 25k}then N+kQueens can be safely placed on the board
by using kPawns to separate the Queens. We also consider these values for
other chess pieces such as Bishops (sB(π, A, b)) and Rooks (sR(π, A, b)).
2 Domination separation
A set Sof vertices in a graph is a dominating set if every vertex in the graph is
either in the set Sor is adjacent to a vertex in S. For a graph G, the minimum
number of vertices in a dominating set is denoted γ(G). A set of vertices S
is independent if no two vertices in Sare adjacent. The maximum cardinality
of a maximal independent set is denoted β(G) while the minimum cardinality
of a maximal independent set is denoted i(G). It is well known [12] that an
independent set is maximally independent if and only if it is dominating, so
i(G) is also called the independent domination number.
We consider domination separation for Queens, Bishops, and Rooks. We
begin with a useful lower bound for the number of Pawns needed to reduce the
domination number of a given chessboard graph. For any chess piece Π, let ΠN
denote the order Ngraph associated with that piece.
Lemma 1 For any chess piece Πand γ(ΠN)> k,sΠ(γ, γ(ΠN)−k, N )≥k.
Proof. It is shown in [4] that removal of a single vertex from a graph cannot
decrease the domination number by more than 1, and removal of an edge alone
never decreases the domination number.
Since placing a Pawn has the same effect as removing one vertex and some
edges, the result follows inductively.
2
2.1 The Queens graph
The domination number γ(QN) is the minimum number of Queens needed to
ensure that every square of an N×Nboard is either occupied or under attack.
It can be shown that five Queens are necessary in order to dominate an 8 ×8
board. While the value of γ(QN) is unknown for all but small values of N, the
current best upper bound for the domination number of the Queens graph is
γ(QN)≤(101N/195) + O(1) and is due to Burger and Mynhardt [3].
A computer search of all possible cases for n≤9 gives the following propo-
sition.
Proposition 2 For N≤9, one Pawn never decreases γon a separated Queens
graph, and one Pawn increases γon a separated Queens graph only if N= 3 or
6.
It can be deduced from an example on page 15 of [18] that it is possible
to decrease the domination number of a separated Queens graph by using two
Pawns. This gives rise to the following two open questions.
Open Question 1 For which size boards will two Pawns decrease γ?
Open Question 2 Can two Pawns decrease γby two?
2.2 The Bishops graph
Values for independence in the Bishops graph were given in [5].
Proposition 3 ([5]) Let N > 0be given.
1. For N≥3,sB(β, 2N−1, N ) = 1.
2. For odd N≥3,sB(β , 2N, N ) = 1.
Since it is shown in [7] that γ(BN) = i(BN) = N, it follows that sB(γ, N, N ) =
sB(i, N, N ) = 0.
From [12], the open neighborhood N(v) of a vertex vis the set of all vertices
adjacent to vwhile the closed neighborhood of vis N[v] = N(v)∪ {v}. Let S
be a set of vertices with u∈S. A vertex vis said to be a private neighbor of u
with respect to Sif N[v]∩S={u}.
The proof of the following lemma is modeled after the proof of Theorem 1
in [7], which also appears as Theorem 6 in [6].
Lemma 4 Let Sbe a minimum (independent) dominating set of an even order
NBishops graph. If one Bishop attacks no other Bishop, then that Bishop has
a private neighbor other than the square it occupies.
Proof. We prove this by contradiction. Suppose we have a minimum (indepen-
dent) dominating set with a Bishop bthat attacks no other Bishop and has no
3
private neighbors other than its own square. Suppose, without loss of generality,
that bis on a white square.
Consider either of the diagonals occupied by b. Since bhas no private neigh-
bor other than its own square and since bdoes not attack other Bishops, every
square in the open neighborhood of bmust be attacked by another Bishop not
in the closed neighborhood of b. Such a Bishop that attacks one square of our
chosen diagonal cannot attack another square on that same diagonal. So the
white squares must contain at least as many Bishops as there are squares on
our diagonal.
If the diagonal has more than N/2 squares we have a contradiction, since in
a minimum (independent) dominating set on an even order Bishops graph N/2
are on black squares and N/2 are on white squares. (The Bishops graph has
domination number Nand is a disjoint union of two isomorphic halves, so each
half has domination number N/2.) So each diagonal occupied by bhas at most
N/2 squares. So the closed neighborhood of bhas at most 2(N/2) −1 = N−1
squares. But this contradicts the fact that every square on the Bishops graph
has at least Nsquares in its closed neigborhood.
Proposition 5 For even N,sB(γ, N −1, N )≥2and sB(i, N −1, N)≥2.
Proof. Suppose to the contrary that 1 Pawn is sufficient to reduce the domina-
tion number. That is, suppose there is a placement of N−1 Bishops that cover
all but one square on the board, which contains a Pawn. The square containing
the Pawn cannot be attacked by any of the Bishops, else removal of the Pawn
would yield a covering of the entire board with only N−1 Bishops.
Replacing the Pawn with a Bishop byields a minimum dominating set with
the property that no other Bishop attacks b. By Lemma 4, it must be the case
that bhas a private neighbor not attacked by any other Bishop. This contradicts
the assumption that N−1 Bishops can cover all but one square.
Proposition 6 For odd N≥3,sB(γ, N −1, N) = sB(i, N −1, N ) = 1.
Proof. First note from Lemma 1 that the domination number cannot decrease
by more than 1.
Label the columns from −kto kon a 2k+ 1 board, and label the rows from
0 to 2k+ 1. To reduce the domination number, place the Bishops on column
0 then replace the Bishop on row 0 with a Pawn. The resulting set of N−1
Bishops is a minimum independent dominating set.
Note that placing a single Pawn at the center of a 3 ×3 Bishops graph
increases the domination number by three to 6.
Proposition 7 For N > 3, the addition of a single Pawn increases the (in-
dependent) domination number on a separated Bishops graph by no more than
2.
Proof. We consider two cases, one in which the Pawn is not at the exact center
of the board, and one with the Pawn at the exact center. Note in the second
4
B
B
B
B
B
B
P0
1
˙
˙
˙
2k+ 1
−k· · · 0· · · k
Figure 1: For odd N≥3, sB(γ , N −1, N ) = 1.
case that the board must be of odd order.
Case I: Suppose the Pawn is not at the exact center of the board. If the Pawn
is not on the column closest to the center, place Bishops on that column. If
the Pawn is on the column closest to the center, then rotate the board prior to
placing the Bishops. Suppose, without loss of generality, that the Pawn is to the
left of the Bishops. The Bishops cover everything to their right while the Pawn
blocks at most two diagonals, both of which are to the left of the Pawn. Place
Bishops on those diagonals to the left of the Pawn. This yields an independent
dominating set with at most N+ 2 Bishops.
Case II: Suppose the Pawn is at the exact center of the board. Label this square
as the origin of a rectangular coordinate system. Place Bishops on squares
(0,±1), (−1,±1), and (1,±1). For 5 < j ≤N, place Bishops on (0,±(j−1)/2).
It is easy to see this yields an independent dominating set of order N+ 1.
Thus, addition of a single Pawn can never increase the (independent) domi-
nation number by more than 2.
Note that in certain cases the maximum increase of 2 is possible. Number the
rows and columns of the Bishops board with row 0 at the bottom and column
0 to the left. Computer searches for odd N≤11 show that the domination
number can be increased to N+ 2 by placing a Pawn in row b(N+ 1)/4cand
column b(N+ 4)/4c.
Conjecture 3 On an odd order Bishops board with N≥3, the domination
number is increased by 2if a single Pawn is placed in row b(N+ 1)/4cand
column b(N+ 4)/4c.
5
2.3 The Rooks graph
It is shown in [5] that for N≥k+2, kPawns suffice to increase the independence
number of the Rooks graph to N+k. We consider the number of Pawns needed
to change the domination number. Note that since NRooks are necessary to
dominate an N×Nboard [13], it follows that sR(γ, N , N) = 0.
Proposition 8 For N≥2,sR(γ, N −1, N ) = sR(i, N −1, N ) = 1.
Proof. Label the columns from 1 to Nand the rows from 1 to Nbeginning
in the upper left of the board. Suppose that a single Pawn is placed in column
r, where 1 ≤r≤N. Place Rooks in the remaining columns, with the row
determined by any permutation of the columns which contain Rooks.
Proposition 9 For N≥k,sR(γ, N −k, N ) = sR(i, N −k, N ) = k2.
Proof. Let m=sR(γ, N −k, N ). Suppose we have an order Nchessboard
with mPawns and N−kRooks such that the Rooks dominate those squares
not occupied by Pawns. First we note that no Rook attacks any of the Pawns
since otherwise we could remove the attacked Pawns, which contradicts the
minimality of m.
Next we note that if a row (column) contains p(>0) Pawns, then there
must be at least N−pRooks on the board, since we need a distinct Rook
to cover each of the empty squares in the row (column). Since we want the
(independent) domination number to be N−k, each nonempty row or column
must have at least kPawns. If k= 0, we need no Pawns. If k > 0, we need at
least one Pawn so at least one row has a Pawn. That row must have kPawns,
and each of those Pawns must occupy a column with at least kPawns. So the
board must have at least k2Pawns.
Finally, we note that k2Pawns suffice: Place the k2Pawns in the corner
formed by the first krows and kcolumns, and place the Rooks on the unoccupied
squares of the main diagonal. (See Figure 2).
Corollary 10 Placing two or three Pawns on an order Nboard reduces the
Rooks (independent) domination number by at most 1.
It is possible to increase the domination number on the Rooks graph by
adding Pawns. For example, by placing Pawns on all the squares of the second
and fourth columns and rows of a 5 ×5 board, we can increase the domination
number from 5 to 9.
Open Question 4 Under what conditions can adding Pawns increase the Rooks
domination number?
3 Queens independence separation
For an N×Nchessboard, we wish to place N+kindependent Queens.
6
P
P
P
P
P
P
P
P
P
R
R
R
R
R
R
Figure 2: For N= 9, sR(γ, N −3, N ) = 9.
Theorem 11 For each positive integer kand N > max{87+k, 25k},sQ(β , N +
k, N ) = k.
Proof Sketch. There are seven patterns to consider. Each begins with a known
solution [9] to a smaller N-Queens problem. The proof that each pattern holds
involves elementary but extremely tedious computations. Let n=N−k.
Pattern I. Let n≡0 mod 6, n≥30, and k < n/8.
Solution: Number the rows −k, · · · ,−1,0,· · · , n−1 and the columns 0,1,· · · , n−
1 + k. Place the Queens at (2i+ 1, i) for i= 0,· · · n/2−1, (2i−n, i) for i=
n/2,· · · , n−1, (−i, n−1−2i) for i= 1,· · · , k, and ((n−nmod 12)/4−3+2i, n+
i−1) for i= 1,· · · , k. Place the Pawns at ((n−nmod 12)/4−3+2i, n −1−2i)
for i= 1,· · · , k.
Pattern II. Let n≡4 mod 6, n≥22, and k < n/8.
Solution: Number the rows −k, · · · ,−1,0,· · · , n−1 and the columns 0,1,· · · , n−
1 + k. Place the Queens at (2i+ 1, i) for i= 0,· · · n/2−1, (2i−n, i) for
i=n/2,· · · , n−1, (−i, n −1−2i) for i= 1,· · · , k, and ((n−12 +nmod 12)/4−
2+2i, n +i−1) for i= 1,· · · , k . Place the Pawns at ((n−12 + nmod 12)/4−
2+2i, n −1−2i) for i= 1,· · · , k.
Pattern III. Let n≡1 mod 6, n≥23, and k < (n−1)/8.
Solution: Number the rows −k−1,· · · , n −2 and the columns −1,· · · , n −2+ k.
Place the Queens and Pawns as in Pattern I with n:= n−1. Place an extra
Queen at (−k−1, n −2 + k).
7
Pattern IV. Let n≡5 mod 6, n≥23, and k < (n−1)/8.
Solution: Number the rows −k−1,· · · , n −2 and the columns −1,· · · , n −2+ k.
Place the Queens and Pawns as in Pattern I with n:= n−1. Place an extra
Queen at (−k−1, n −2 + k).
Pattern V. Let n≡2 mod 6, n≥56, and k≤n/10.
Solution: Number the rows −k, · · · ,−1,0,· · · , n−1 and the columns 0,1,· · · , n−
1 + k. Place the Queens at ((n−6)/4 + ((n−2) mod 12)/4+2i, n −1 + i)
for i= 1,· · · , k, (−i, n −8−2i) for i= 1,· · · , k , ((2i+ 4) mod n, i) for
i= 0,· · · , n/2−1, and ((2i+ 7) mod n, i) for i=n/2,· · · , n −1. Place
the Pawns at ((n−6)/4 + ((n−2) mod 12)/4 + 2i, n −8−2i) for i= 1,· · · , k.
Pattern VI. Let n≡3 mod 12, n≥87, and k≤n/24.
Solution: Number the rows −k, · · · ,−1,0,· · · , n−1 and the columns 0,1,· · · , n−
1 + k. Place the Queens at (−i, n −11 −3i) and (n/3−3 + 3i+ ((n−3)
mod 12)/6, n−1+i) for i= 1,· · · , k, ((3i+5) mod n, i) for i= 0,1,· · · , n/3−1,
((3i+ 9) mod n, i) for i=n/3,· · · ,2n/3−1, and ((3i+ 13) mod n, i) for i=
2n/3,· · · , n−1. Place the Pawns at (n/3−3+3i+((n−3) mod 12)/6, n−11−3i).
Pattern VII. Let n≡9 mod 12, n≥87, and k≤n/24.
Solution: Number the rows −k, · · · ,−1,0,· · · , n −1 and the columns 0,1,
· · · , n−1+ k. Place the Queens at (−i, n −11−3i) for i= 1,· · · , k, (n/3+1, n),
(n/3+3i−1, n−1+i) for i= 2,· · · , k, ((3i+5) mod n, i) for i= 0,1,· · · , n/3−1,
((3i+ 9) mod n, i) for i=n/3,· · · ,2n/3−1, and ((3i+ 13) mod n, i) for i=
2n/3,· · · , n−1. Place the Pawns at (n/3+1, n−14) and (n/3+3i−1, n−11−3i)
for i= 2,· · · , k.
The bounds on Ngiven in Theorem 11 are not best possible for all k. By
computer searches with N < 91, we have been able to show that N≥7 and
N≥8 are sufficiently large for k= 2 and k= 3 respectively. From Theorem
11, we see that N > 25kis large enough for k≥4. However, this bound can
most likely be decreased as well.
The following proposition provides some examples for which more than k
additional Pawns are needed to increase the independence number by k.
Proposition 12 Some values of sQ(β , m, N).
1. (Zhao [19]) sQ(β, 6,5) = 3.
2. sQ(β, 8,6) = 3.
3. sQ(β, 9,5) = 16.
4 An inverse problem
Given a board of order N, consider the maximum number of independent Queens
that can be placed on the board if kPawns have already been placed. Note that
8
the maximum number of Queens that can be placed is no more than N+k. In
many cases, the maximum may be considerably less.
The following proposition answers a question of Hammons [11] and describes
possible locations of Pawns if N+kindependent Queens and kPawns are to
be placed.
Proposition 13 If N+kQueens and kPawns are placed on an N×Nboard
so that no two Queens attack each other, then no Pawn can be on the first or
last row, first or last column, or any square adjacent to a corner.
Proof. First we note that a Pawn divides a row or column into at most two
independent parts, so pPawns divide a row or column into at most p+ 1 parts.
Suppose there is a Pawn on the first row. If there are pPawns on that row,
there are at most pQueens in the first row. With k−pPawns in other rows,
we can place at most k−p+N−1 Queens in those rows. So we can place at
most p+k−p+N−1 = N+k−1 independent Queens, which contradicts the
fact that there are N+kindependent Queens on the board. So no Pawn can
be on the first row. Symmetric arguments show there are no Pawns on the last
row, or the first or last column.
Now consider the squares diagonally adjacent to a corner. Without loss of
generality, suppose a Pawn is in the second row and second column. There
must be a Queen in the first column of the second row, else there are at most
N+k−1 parts of rows to place Queens. Similarly, there must be a Queen in the
first row of the second column. However, these two Queens attack each other.
Thus, there can be no Pawn in the second row and second column. Symmetric
arguments complete the proof.
Based on computer searches, we conjecture that for N≥10 a Pawn can be
placed in any square not included in the above proposition in order to allow
placement of N+ 1 independent Queens on an N×Nboard.
5 Counting fundamental solutions
Our primary focus so far has been to determine whether even a single solution
exists for a given Nand k. A more interesting and difficult problem arises when
we consider the total number of solutions to separation problems. A fundamental
solution of a chessboard problem is a class of solutions such that all the members
of the class are rotations or reflections of one another. Knuth [16] mentions the
existence of a method for counting the number of solutions to the N-Queens
problem without actually generating the solutions. Note, however, this method
is still exponential. We discuss an implementation of a method for generating
fundamental solutions of the Queens independence separation problem which is
an improvement of traditional backtracking techniques.
9
5.1 Exact covers and dancing links
Given a set, S, an exact cover of Sis a set {S1, S 2, . . . , Sk}of subsets of S
where all elements of Sare in one and only one subset Sj. A typical problem
is to find all exact covers for a given set Sunder some additional constraints
for the subsets. Solutions to this problem are found using backtracking and
recursion. Backtracking and recursion require potential solutions to have ele-
ments ‘appended’ and ‘removed’ when either a solution is found or a constraint
violated.
The Dancing Links (DLX) algorithm was first presented in 1979 by Hitotu-
matu and Noshita [14]; over twenty years later, Knuth [16] named and publicized
the technique as an efficient method for implementing backtracking algorithms.
In particular, [16] provides applications of DLX for solving various forms of
exact cover problems. DLX is a more efficient way to implement backtracking
algorithms due to a data structure which, by design, provides faster ‘append’
(cover) and ‘removal’ (uncover). The primary goal in using this method is to
save computation time rather than space.
The generalized cover problem does not require that all elements of Sbe in
the set of subsets {S1, S2, ..., S k}. For example, consider the N-Queens problem.
The rows of the chessboard require an exact cover with one Queen in each row,
as do the columns. The diagonals, however, can have at most one Queen to
have a valid solution. Therefore, the N-Queens problem is a generalized exact
cover problem.
As defined in [16], a universe is a multi-dimensional data structure composed
of intersecting double, circularly linked lists. The components of the universe
are either a header node, Column Header, or Column Object. The universe is
an expanded representation of the N×Nchessboard.
Knuth defines each row of the chessboard as a rank R, the column of the
chessboard as a file F, negative diagonal from lower right to upper left as A, and
positive diagonal from lower left to upper right as B. As a result, each block
of the chessboard is defined by a combination of specific, horizontally linked,
Column Objects R,F,A, and B. The rank and file columns are the primary
columns and must be covered. The two diagonals are secondary columns and
may each contain zero or one Queen.
The header node links together the Column Headers of the universe. There
is one Column for each R,F,A, and B. The Column Headers contain the name
of the universe Column (e.g., R1) and link vertically the number of chessblocks
Column Objects.
The Column Objects link horizontally to the other three Columns to define
the chessboard block.
5.1.1 N-Queens universe
The N-Queens universe consists of a header node, Column Headers, and Column
Objects. The header node is created and linked to itself in the constructor of
its class. Then, 2NColumn Headers are added to represent the Nranks and N
10
Figure 3: Chessboard and Corresponding Universe
columns. Then 2(2N−1) Column Headers are added for the negative diagonal
Aand positive diagonal B.
The Column Objects are the final object types added. Each rank and file
have NColumn Objects. The number of Column Objects for Aand Bis de-
termined by the number of chessblocks contained in the diagonal. For example,
if A0is defined to be the main negative diagonal, it has NColumn Objects. If
ANis the corner diagonal, it contains 1 Column Object.
Figure 3 shows a mapping of a 4×4 chessboard to the Dancing Links universe
for solving the N-Queens problem.
0000
1P2 2
3333
4444
Figure 4: Recomputed Rows with 1 Pawn
11
5.1.2 N+k Queens Universe
When one Pawn is placed on the chess board, it divides the chessboard blocks for
its row, column, negative, and positive diagonal. In the Dancing Links universe,
this added Pawn splits the corresponding four Columns (R,F,A,B), resulting
in four additional and unique Columns to be covered.
To analyze N+kwe solve the exact cover problem used in the N-Queens
problem with the addition of a preprocessor to determine all valid Pawn place-
ments, iterate through the Pawn placements, and determine unique row and
column ids.
Given a valid Pawn placement, the total rows and columns of the chessboard
are recomputed. If a Pawn splits a row, k, then up to Pawn is row k, and after
the Pawn is row k+ 1. The next column row is then row k+ 2. Figure 4 shows
the row identifiers when a Pawn is placed in location (1,1).
6 Conclusions and Future Work
Computer counts of fundamental solutions to the independence separation prob-
lem have provided some results for small Nas shown in Tables 1, 2, and 3. An
improvement of the algorithm used in [5] for N+ 2 Queens resulted in the dis-
covery of additional fundamental solutions for the cases N= 9,10,and 11 that
were not included in [5].
Using C++, we implemented sequential and parallel algorithms with back-
tracking as well as the previously described parallel algorithm with Dancing
Links on the Midas cluster located in the Department of Mathematics and
Computer Science at Morehead State University. The Midas cluster consists of
a head node and eight compute nodes running OSCAR [17] under the Fedora
Core 2 Linux distribution. The compute nodes are recycled Gateway E-4200
computers with a combined 3072MB RAM, while the head node is an IBM
NetVista with 512MB RAM. The nodes communicate through a switched Gi-
gabit Ethernet private network.
The times given in Table 4 are the average elapsed seconds over five to
ten trials for each value of Nfor the N+ 1 independence separation problem.
Our current version counts the total number of solutions and at its conclusion,
compresses the solutions into a set of fundamental solutions. It is obviously
faster to derive an algorithm that computes the fundamental solutions first and
determines the total solutions later.
Currently, the N+k-Queens universe is prepended to the N-Queens universe
solver, which then uses DLX. In the future we hope to reduce runtimes even
more by utilizing DLX to add and remove Pawns from the solution set.
Acknowledgements This research was funded in part by NASA KY-EPSCoR
Grant NCC5-571 and Morehead State University Faculty Research Grant 225229.
The authors thank Drew Henderson, Mark Jarvis, and Owen Wagoner of the
MSU Office of Information Technology for invaluable assistance.
12
NSolutions Fundamental solutions
5 0 0
6 16 2
7 20 3
8 128 16
9 396 52
10 2288 286
11 11152 1403
12 65172 8214
13 437848 54756
14 3118664 389833
15 23387448 2923757
16 183463680 22932960
Table 1: N+ 1 Queens and 1 Pawn on N×Nchessboard
NSolutions Fundamental solutions
6 0 0
7 4 1
8 44 6
9 280 37
10 1304 164
11 12452 1572
12 105012 13133
Table 2: N+ 2 Queens and 2 Pawns on N×Nchessboard
NSolutions Fundamental solutions
7 0 0
8 8 1
9 44 6
10 528 66
11 5976 751
Table 3: N+ 3 Queens and 3 Pawns on N×Nchessboard
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13
NSequential Parallel with backtracking Parallel with DLX
6 0.002 0.002 0.0065
7 0.014 0.005 0.0198
8 0.064 0.020 0.0366
9 0.572 0.142 0.1011
10 3.051 0.635 0.2213
11 28.555 4.521 1.4471
12 173.007 27.018 6.0479
13 1667.834 261.106 42.3711
14 1812.864 256.9734
15 17353.400 1987.509
16 14536.260
Table 4: Elapsed time in seconds of sequential versus parallel implementations
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