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Discrete Mathematics 344 (2021) 112316
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Discrete Mathematics
journal homepage: www.elsevier.com/locate/disc
Leprechauns on the chessboard
Guillaume Escamocher, Barry O’Sullivan
Insight Centre for Data Analytics, School of Computer Science & Information Technology, University College Cork, Ireland
article info
Article history:
Received 14 August 2020
Received in revised form 10 December 2020
Accepted 15 January 2021
Available online xxxx
Keywords:
Constraint satisfaction
n-Queens Problem
Assignment diversity
abstract
We introduce in this paper leprechauns, fairy chess pieces that can move either like
the standard queen, or to any tile within kking moves. We then study the problem
of placing nleprechauns on an n×nchessboard. When k=1, this is equivalent to
the standard n-Queens Problem. We solve the problem for k=2, as well as for k>2
and n≤(k+1)2, giving in the process an upper bound on the lowest non-trivial value
of nsuch that the (k,n)-Leprechauns Problem is satisfiable. Solving this problem for
all kwould be equivalent to solving the diverse n-Queens Problem, the variant of the
n-Queens Problem where the distance between the two closest queens is maximized.
While diversity has been a popular topic in other constraint problems, this is not the
case for the n-Queens Problem, making our results the first major ones in the domain.
©2021 The Author(s). Published by Elsevier B.V. This is an open access article under the CC
BY license (http://creativecommons.org/licenses/by/4.0/).
1. Introduction
The n-Queens Problem asks how to place nchess pieces on an n×nboard such that no two pieces share the same
row, column or diagonal. Finding such an arrangement can be done in linear time for every napart from n=2 and n=3,
for which none exist [19,20]. Despite its relative easiness, the n-Queens Problem has proven to be very popular during
the last one and a half centuries, giving birth to several related combinatorial problems. Some of these variants remain
tractable, like the modular n-Queens Problem, where opposite sides of the board are merged to form a torus [22]. Some
of them on the other hand are NP-Complete, like the n-Queens Completion Problem which asks whether a given set of
queens that have already been placed on the board can be extended to a solution to the standard n-Queens Problem [9].
The complexity of others is still open, see for example the question of finding the lexicographically minimal solution to
the n-Queens Problem [8].
An intuitive n-Queens Problem variation that, to the best of our knowledge, has never been studied, is the Diverse n-
Queens Problem. In this problem, the goal is to find a solution that not only fulfills the original constraints, but also places
the queens as far away as possible from each other. The absence of work on the subject is regrettable, because diversity
is a popular topic in constraint programming [1,4,14,15,21], with numerous applications that include recommender
systems [27], exact satisfiability [5], car configuration [13,16], and architectural tests [17].
Finding a solution to the Diverse n-Queens Problem is equivalent to finding a solution to the standard n-Queens
Problem where all queens are more than kaway from each other, with kbeing a distance parameter corresponding to
the number of moves a king needs to move between two tiles. Indeed, in order to find a solution for the former, one can
simply solve the latter for kfrom 1 to n−1 and keep the largest kfor which a solution was found. In the other direction,
whether a solution to the Diverse n-Queens Problem has all queens more than kapart from each other determines if there
is a solution to the n-Queens problem with the additional distance constraint.
E-mail addresses: guillaume.escamocher@insight-centre.org (G. Escamocher), b.osullivan@cs.ucc.ie (B. O’Sullivan).
https://doi.org/10.1016/j.disc.2021.112316
0012-365X/©2021 The Author(s). Published by Elsevier B.V. This is an open access article under the CC BY license (http://creativecommons.org/
licenses/by/4.0/).
G. Escamocher and B. O’Sullivan Discrete Mathematics 344 (2021) 112316
To model the requirement that all queens be more than kaway from each other, we use a fairy1chess piece that we
name a leprechaun. Leprechauns can move either like a queen, or to any tile of the board that is at most krows and at
most kcolumns away from them, with kbeing their range. The problem of placing nqueens on an n×nboard such
that each one of them is (strictly) more than kaway from her nearest neighbor becomes then the problem of placing n
non-attacking range-kleprechauns on that same n×nboard. We call this problem the (k,n)-Leprechauns Problem. The
distance parameter kestablishes a hierarchy over (k,n)-Leprechauns Problems: the higher kis, the more tiles a range-k
leprechaun can potentially move to, and therefore the harder it is to find a configuration of nnon-attacking leprechauns.
In this paper, we take the first steps towards a characterization of the Leprechauns Problem depending on the
parameters kand n. As we mention among the definitions in the next section, a range-1 leprechaun is a queen, so the
(1,n)-Leprechauns Problem is already solved since it corresponds to the standard n-Queens problem. In Section 3, we
present the first of our two main results: a construction of a solution to the (2,n)-Leprechauns Problem for all n≥10.
While the (un-)satisfiability of this particular problem was already known for n<10 [10], our algorithm is the first, as
far as we know, to completely solve the (2,n)-Leprechauns Problem for any value of n.
For higher values of k, the (k,n)-Leprechauns Problem appears to be more challenging. We suspect, however, that the
frontier between unsatisfiability and satisfiability lies around the n=(k+1)2parabola. Indeed, as we show in Section 4,
the (k,n)-Leprechauns Problem is always unsatisfiable (except for the trivial case n=1) if n<(k+1)2, but can go either
way if n=(k+1)2, where the existence of a solution depends on the parity of k. Most of the section is devoted to the
proof of this result, while the remainder presents miscellaneous empirical observations. We conclude in Section 5.
2. Definitions
A (chess)board is a grid of dimensions n×ncontaining n2tiles. Each tile has coordinates (i,j), with ibetween 1 and n
indicating the column and jbetween 1 and nindicating the row. We use the standard convention of numbering the
columns from left to right and the rows from bottom to top. Diagonals can be labeled as well.
Definition 1 (Diagonals).The sum diagonal d+is the set of tiles {(i,j)|i+j=d+}. The difference diagonal d−is the set of
tiles {(i,j)|i−j=d−}.
To define our distance metric, we use the appropriately named chessboard distance:
Definition 2 (Chessboard Distance).Let t1and t2be two tiles of respective coordinates (i1,j1) and (i2,j2). We say that the
chessboard distance between t1and t2is the maximum between |i1−i2|and |j1−j2|.
The chessboard distance is sometimes called Chebyshev distance. Since this is the only distance metric we will be using,
we will from now on simply refer to it as the distance.
The game of chess uses figurines, or pieces, that can moved from one tile to another. The n-Queens Problem focuses
on the eponymous queen piece:
Definition 3 (Queen).Aqueen is a chess piece that can move from a tile (i,j) to any other tile (i′,j′) as long as one of the
following is true:
•i′=i
•j′=j
•i′+j′=i+j
•i′−j′=i−j
The first condition corresponds to a move on the same column, the second to a move on the same row. The last two
conditions correspond to a diagonal move. The queen is a part of the standard chess figurine set. For recreational purposes,
many non-standard chess pieces have been invented. We now introduce the leprechaun chess piece:
Definition 4 (Range-k Leprechaun).Arange-k leprechaun is a chess piece that can move from a tile t1to any other tile t2
as long as one of the two following conditions is true:
•t2can be reached from t1by a queen move.
•The distance between t2and t1is at most k.
A range-1 leprechaun is a queen. A range-2 leprechaun is an amazon, or superqueen, although it has been called many
names throughout history [3], like for example maharajah [7], and was even used as a substitute for the queen in ancient
Russian chess [12]. A range-3 leprechaun can be seen as the combination of a queen, a knight, a camel and a zebra, where
camels and zebras are fairy chess pieces that move to a tile (1,3) and (2,3) away respectively, leaping over intermediary
pieces.
1Fairy chess pieces are chess pieces that are not part of the standard chess set.
2
G. Escamocher and B. O’Sullivan Discrete Mathematics 344 (2021) 112316
Fig. 1. A solution to the (2,14)-Leprechauns Problem.
Definition 5 (Attack).Let tbe a tile of the board. We say that a chess piece attacks t if it can move from its current tile
to tin one move. We say that a chess piece attacks another chess piece if the latter is placed on a tile attacked by the
former.
The n-Queens Problem can be generalized to numerous other (fairy) chess pieces, and the leprechaun is no exception.
Definition 6 ((k,n)-Leprechauns Problem).The (k,n)-Leprechauns Problem is to find a way to place nnon-attacking range-k
leprechauns on an n×nboard, or to prove that there is no such arrangement.
A solution to the (k,n)-Leprechauns Problem will be written as an ordered list of nrows, indexed by columns. In other
words, if the ith element of a solution is j, then this solution contains a leprechaun on tile (i,j).
The board pictured in Fig. 1 illustrates the notions defined in this section. Its dimensions are 14 ×14, and it
contains 14 leprechauns, each represented by a shamrock. The leprechaun in tile (11,6) is in Sum Diagonal 17, because
its coordinates add to 17, and in Difference Diagonal 5, because the row component of its coordinates subtracted
from the column component gives 5. It is the only leprechaun in these diagonals, which go from (3,14) to (14,3)
and from (6,1) to (14,9), respectively, and is also alone in Row 6 and Column 11. Its nearest neighbors are the
leprechauns in tiles (13,3) and (12,9), each at distance 3, so if it is a range-1 or range-2 leprechaun, then it does not
attack them. In fact, this board is a solution for the (1,n)- and (2,n)-Leprechauns Problems, which can be written as
(1,4,7,10,13,5,8,11,14,2,6,9,3,12). However this board is not a solution to the (k,n)-Leprechauns Problem for
k≥3, because several pairs of leprechauns are within distance 3 of each other.
To remain consistent with the labeling of the board axes, we will use ‘‘m ˙
odulo’’ and m˙
od instead of ‘‘modulo’’ and mod,
where a m˙
od b is equal to bif ais divisible by b, and to a mod b otherwise. So for example 28 m˙
od 14 =14.
3. The (2,n)-Leprechauns problem
The case of the range-1 leprechaun is equivalent to the n-Queens Problem, for which a linear-time algorithm to
find a solution for each n(apart from n=2 and n=3 for which there is no solution) has been known since the
nineteenth century [19,20]. In contrast, we are not aware of any algorithm, regardless of complexity, that can solve the
(2,n)-Leprechauns Problem for any n.
For the first few values of n, the number of solutions to the (2,n)-Leprechauns Problem is given at the On-line
Encyclopedia of Integer Sequences [25], where (2,n)-Leprechauns are called ‘‘superqueens’’. In 2004 a mention was made
on that web-site of code by Frank Schwellinger that could solve the problem in linear time, but the link is no longer active.
A cached version2reveals his algorithm, which appears to be incorrect when nis congruent to 2 or to 3 m ˙
odulo 6 and
n≥26. For n=50 for example, his construction places superqueens on tiles (11,44) and (36,19), which belong to the
same diagonal. A subsequent literature review of the problem [2] does not mention any algorithm that works for all n.
2https://web.archive.org/web/20040823045854/http://super.info.ms/.
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G. Escamocher and B. O’Sullivan Discrete Mathematics 344 (2021) 112316
The only results so far for the (2,n)-Leprechauns Problem are partial. If n≥10 is a prime or 1 less than a prime, then
there is a solution to the (2,k)-Leprechauns Problem [11]. If n<10, then there is a solution to the (2,n)-Leprechauns
Problem if and only if n=1 [10]. The problem has also been solved when the number of leprechauns to place is strictly
less than nand the board is a torus [18].
Many solutions to the n-Queens Problem are based on linear congruence algorithms, meaning that the queen in each
column is a constant number of rows (m ˙
odulo n) above the queen from the previous column. Because of arithmetical
properties, these methods have to vary the exact height of this step, and possibly make other adjustments to the
construction, depending on the value of nm˙
odulo 6. When a step of at least k+1 rows is allowed between queens
from consecutive columns, the additional distance constraint inherent to range-kleprechauns will be fulfilled and the
n-queens solutions can be used as is for the (k,n)-Leprechaun Problem. Unfortunately, in some cases (most often when
n≡2m˙
od 6 or n≡3m˙
od 6) strict conditions on the structure of n-Queens solution do not allow for the leprechaun
distance constraints to be satisfied without substantial alterations to the algorithm itself.
We are now going to present a linear-time algorithm that for any neither returns a solution to the (2,n)-Leprechauns
Problem, or indicates that none exists. Our algorithm also distinguishes cases depending on the congruence of nm˙
odulo
6, although the particular case of n≡2m˙
od 6 will be further partitioned, with its two main subcases being n≡2m˙
od 12
and n≡8m˙
od 12. For n≡1m˙
od 6, n≡3m˙
od 6, and n≡5m˙
od 6, we will use one of the n-Queens constructions
that allow for jumps of 3 rows from one column to the next [6]. We start with the cases 1 m˙
od 6 and 5 m˙
od 6, which are
treated in Algorithm 1.
Data: An integer nsuch that n≡1m˙
od 6 or n≡5m˙
od 6, and n=1 or n≥11.
Result: A solution to the (2,n)-Leprechauns Problem.
1Declare Sol, an array with ncells;
2Sol[1] ← 1;
3for c←2to ndo Sol[c] ← (Sol[c−1] + 3) m˙
od n;
4return Sol;
Algorithm 1: When n≡1m˙
od 6 or n≡5m˙
od 6.
Lemma 1. Let n be an integer such that n ≡1m˙
od 6or n ≡5m˙
od 6, and such that n =1or n ≥11. Then Algorithm 1
returns a solution to the (2,n)-Leprechauns Problem.
Proof. If n=1, then the solution consists of a single leprechaun and the satisfiability is trivial. Otherwise, we have
n≥11. Suppose that the leprechauns in tiles (i,j) and (i′,j′) attack each other. Assume without loss of generality that
i<i′. We know that the algorithm gives a solution to the n-Queens Problem [6], so the only way for these leprechauns
to attack each other is if they are within distance 2 of each other. So i′=i+1 or i′=i+2. So j′m˙
od n =j+3m˙
od n or
j′m˙
od n =j+6m˙
od n. The two rows above Row jare Rows j1and j2with j1m˙
od n =j+1m˙
od n and j2m˙
od n =j+2m˙
od n.
The two rows below Row jare Rows j−1and j−2with j−1m˙
od n =j+n−1m˙
od n and j−2m˙
od n =j+n−2m˙
od n.
Since n≥11, Row j′is neither one of the two rows above Row jnor one of the two rows below Row j. Since the queens
constraints are satisfied, we also know that Row j′is not Row j. So the leprechauns on tiles (i,j) and (i′,j′) cannot be
within distance 2 of each other. This proves the validity of the algorithm. □
Fig. 2 illustrates the solution output by Algorithm 1for n=13. Notice that the algorithm places a leprechaun on the
bottom left tile of the board (Line 2). This means that by cropping out the first row and the first column of the n×n
board, we get an arrangement of n−1 non-attacking leprechauns on an (n−1) ×(n−1) board, which is a solution to the
(2,n−1)-Leprechauns Problem. For example, the top right part of the board in Fig. 2 is a solution to the (2,12)-Leprechauns
Problem. Algorithm 2exploits the property to solve the cases n≡4m˙
od 6 and n≡6m˙
od 6.
Data: An integer n≥10 such that n≡4m˙
od 6 or n≡6m˙
od 6.
Result: A solution to the (2,n)-Leprechauns Problem.
1Declare Sol, an array with ncells;
2Sol[1] ← 3;
3for c←2to ndo Sol[c] ← (Sol[c−1] + 3) m˙
od (n+1);
4return Sol;
Algorithm 2: When n≡4m˙
od 6 or n≡6m˙
od 6.
Lemma 2. Let n ≥10 be an integer such that n ≡4m˙
od 6or n ≡6m˙
od 6. Then Algorithm 2returns a solution to the
(2,n)-Leprechauns Problem.
Proof. We just need to prove that Algorithm 2returns the same board as Algorithm 1, minus the bottom row rbottom and
the leftmost column cleft . Observe that the congruence is modulo n+1, which compensates for the lack of the bottom row
rbottom when Sol[c−1] + 3 is greater than n. So the only thing to check is that the result of the congruence m ˙
odulo n+1
is never n+1. We know that n+1 is congruent to either 1 or 5 m ˙
odulo 6, so it is prime with 3. Furthermore, the value
in the first cell of the solution is 3 (Line 2), and the height of the jump between consecutive columns is also 3 (Line 3),
4
G. Escamocher and B. O’Sullivan Discrete Mathematics 344 (2021) 112316
Fig. 2. A solution to the (2,13)-Leprechauns Problem.
so the first time that the result of the congruence will be n+1 is at the (n+1)th column, which is not used because of
the lack of the leftmost column cleft .□
For n≡3m˙
od 6, we use one of the rare constructions that allow a jump height of 3 for this particular case [6]. The
method is presented in Algorithm 3, and illustrated in Fig. 3.
Data: An integer n≥15 such that n≡3m˙
od 6.
Result: A solution to the (2,n)-Leprechauns Problem.
1Declare Sol, an array with ncells;
2Sol[1] ← 5; // Can actually be any value between 5 and n−9inclusive.
3for c←2to n
3do
4Sol[c] ← (Sol[c−1] + 3) m˙
od n;
5end
6Sol[n
3+1] ← Sol[n
3] + 7;
7for c←n
3+2to 2n
3do
8Sol[c] ← (Sol[c−1] + 3) m˙
od n;
9end
10 Sol[2n
3+1] ← Sol[2n
3] + 7;
11 for c←2n
3+2to ndo
12 Sol[c] ← (Sol[c−1] + 3) m˙
od n;
13 end
14 return Sol;
Algorithm 3: When n≡3m˙
od 6.
Lemma 3. Let n ≥15 be an integer such that n ≡3m˙
od 6. Then Algorithm 3returns a solution to the (2,n)-Leprechauns
Problem.
Proof. The reasoning is similar to the one used in the proof of Lemma 1, in that we already know that the algorithm
outputs a solution for the n-Queens Problem [6], so we only need to check that the distance constraints are satisfied.
Suppose that the leprechauns in tiles (i,j) and (i′,j′) attack each other. Assume without loss of generality that i<i′.
Since queens constraints are fulfilled [6], the only way for these leprechauns to attack each other is if they are within
distance 2 of each other. So i′=i+1 or i′=i+2. The height of the jump between successive columns is either 3 (Lines
4, 8, and 12) or 7 (Lines 6 and 10), and the latter does not occur consecutively (because it occurs at Columns n
3+1 and
2n
3+1, and since n≥15 we have n
3+1<2n
3+1−1), so j′can only be equal to j+3m˙
od n,j+7m˙
od n, or j+10 m˙
od n.
The two rows above Row jare Rows j1and j2with j1m˙
od n =j+1m˙
od n and j2m˙
od n =j+2m˙
od n. The two rows
below Row jare Rows j−1and j−2with j−1m˙
od n =j+n−1m˙
od n and j−2m˙
od n =j+n−2m˙
od n. Since n≥15,
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G. Escamocher and B. O’Sullivan Discrete Mathematics 344 (2021) 112316
Fig. 3. A solution to the (2,27)-Leprechauns Problem.
Row j′cannot be Row j, cannot be one of the two rows above Row j, and cannot be one of the two rows below Row j. So
the leprechauns on tiles (i,j) and (i′,j′) cannot be within 2 of each other. This proves the validity of the algorithm. □
Algorithm 3places the first column’s leprechaun in Row 5. Any row between Row 5 and Row n−9 inclusive would
have worked too [6]. For example, Fig. 3 illustrates for n=27 the output of Algorithm 3when placing the first column’s
leprechaun in Row 8 instead of Row 5. The result is still a solution to the (2,27)-Leprechauns Problem. Furthermore, we
never mention Line 2 in the proof of Lemma 3, so its correctness stands regardless of which row in this set is chosen.
On the other hand, placing the first column’s leprechaun in either one of the first four rows or one of the last nine ones
would lead to violated diagonal constraints if following the rest of Algorithm 3. In particular, we cannot use Algorithm 2’s
trick of placing a leprechaun in a corner of the board and then cropping one side row and one side column to get a solution
to the (2,n−1)-Leprechauns Problem.
When n≡2m˙
od 6, applying Algorithm 3to n+1 and removing the last column of the output gives an arrangement
of nnon-attacking leprechauns on an (n+1) ×nboard. Moving the last row’s leprechaun to the one row among the first
nthat is empty transforms this into an arrangement of nleprechauns on an n×nboard that respects row and column
constraints, however other constraints might not be satisfied. As an example, cropping out the last column in Fig. 3 and
moving the leprechaun in Column 15 from the last row to the now empty Row 13 places this leprechaun on the same
diagonal as the one on tile (24,4). To obtain a general solution for the case n≡2m˙
od 6 with this method, special care
has to be taken when choosing in which row to place the first column’s leprechaun. No row works for all n, but when
n≥20 it is possible for all but six small values of nto determine an acceptable row for the first leprechaun by looking at
the remainder of ndivided by 12, as is done in Algorithm 4. Even the other six values of ncan be treated with the same
algorithm, by using their own separate formula to find out in which row to place the first column’s leprechaun.
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G. Escamocher and B. O’Sullivan Discrete Mathematics 344 (2021) 112316
Data: An integer n≥20 such that n≡2m˙
od 6.
Result: A solution to the (2,n)-Leprechauns Problem.
1Declare Sol, an array with ncells;
2if n=20 or n=26 or n=32 or n=38 or n=44 or n=56 then
3Sol[1] ← n−10;
4end
5if n≥50 and n≡2m˙
od 12 then Sol[1] ← 8;
6if n≥68 and n≡8m˙
od 12 then Sol[1] ← 14;
7for c←2to n+1
3do
8Sol[c] ← (Sol[c−1] + 3) m˙
od (n+1);
9end
10 Sol[n+1
3+1] ← Sol[n+1
3] + 7;
11 for c←n+1
3+2to 2(n+1)
3do
12 Sol[c] ← (Sol[c−1] + 3) m˙
od (n+1);
13 end
14 Sol[2(n+1)
3+1] ← Sol[2(n+1)
3] + 7;
15 for c←2(n+1)
3+2to ndo
16 Sol[c] ← (Sol[c−1] + 3) m˙
od (n+1);
17 end
18 if n=20 or n=26 or n=32 or n=38 or n=44 or n=56 then
19 Sol[2(n+1)
3+2] ← n−5;
20 end
21 if n≥50 and n≡2m˙
od 12 then Sol[2(n+1)
3−3] ← 13;
22 if n≥68 and n≡8m˙
od 12 then Sol[2(n+1)
3−5] ← 19;
23 return Sol;
Algorithm 4: When n≡2m˙
od 6.
Lemma 4. Let n ≥20 be an integer such that n ≡2m˙
od 6. Then Algorithm 4returns a solution to the (2,n)-Leprechauns
Problem.
Proof. If n∈ {20,26,32,38,44,56}, the outputs are as follows:
n=20: (10,13,16,19,1,4,7,14,17,20,2,5,8,11,18,15,3,6,9,12)
n=26: (16,19,22,25,1,4,7,10,13,20,23,26,2,5,8,11,14,17,24,21,3,6,9,12,15,18)
n=32: (22,25,28,31,1,4,7,10,13,16,19,26,29,32,2,5,8,11,14,17,20,23,30,27,3,6,9,12,15,18,21,24)
n=38: (28,31,34,37,1,4,7,10,13,16,19,22,25,32,35,38,2,5,8,11,14,17,20,23,26,29,
36,33,3,6,9,12,15,18,21,24,27,30)
n=44: (34,37,40,43,1,4,7,10,13,16,19,22,25,28,31,38,41,44,2,5,8,11,14,17,20,23,26,
29,32,35,42,39,3,6,9,12,15,18,21,24,27,30,33,36)
n=56: (46,49,52,55,1,4,7,10,13,16,19,22,25,28,31,34,37,40,43,50,53,56,2,5,8,11,14,
17,20,23,26,29,32,35,38,41,44,47,54,51,3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48)
In each case, the bold value indicates the leprechaun that was moved in Lines 18–20. All of these outputs have been
confirmed to be solutions by a constraint solver, so we will now assume that n≥50 and n≡2m˙
od 12 (respectively
n≥68 and n≡8m˙
od 12).
Let us put aside for the moment the last three lines of the algorithm, and let us attempt to describe the state of the
board after Line 17. The board can be partitioned in three. The first part covers Columns 1 to n+1
3and is filled in Lines
5 to 9. The second part covers Columns n+1
3+1 to 2(n+1)
3and is filled in Lines 10 to 13. Finally the last part covers
Columns 2(n+1)
3+1 to nand is filled in Lines 14 to 17. In each part, the row of a leprechaun is determined by adding
3 m ˙
odulo n+1 to the row of the previous column’s leprechaun. Since n≡2m˙
od 6, n+1 is divisible by 3, so all
leprechauns in the same part are congruent to the same value m ˙
odulo 3. From Line 5 (respectively 6), we know that the
first leprechaun is placed in Row 8 (respectively 14), so the rows of the leprechauns in the leftmost part of the board
comprise all the numbers less than n+1 that are equal to 2 m ˙
odulo 3, in the following order: [8,11,...,n−3,n,2,5]
(respectively [14,17,...,n−3,n,2,5,8,11]). From that and Line 10, we know that the next leprechaun is placed in
Row 12 (respectively 18), so the rows of the leprechauns in the second part of the board comprise all the numbers less or
equal than n+1 that are equal to 3 m ˙
odulo 3, in the following order: [12,15,...,n−5,n−2,n+1,3,6,9](respectively
[18,21,...,n−5,n−2,n+1,3,6,9,12,15]). With Line 14 we now know that the next leprechaun is placed in Row
16 (respectively 22), so the rows of the leprechauns in the rightmost part of the board comprise all the numbers less
than n+1 that are equal to 1 m ˙
odulo 3, except the number that corresponds to the row of the leprechaun that would
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G. Escamocher and B. O’Sullivan Discrete Mathematics 344 (2021) 112316
Table 1
Leprechaun coordinates when n≥50 and n≡2m˙
od 12.
Column cRow Sum diagonal Difference diagonal
1 to n+1
3−2 8 +3(c−1) 4c+5−2c−5
n+1
3−1 to n+1
32, 5 n+1
3+(1,5) n+1
3−(3,5)
n+1
3+1 to 2(n+1)
3−3 12 +3(c−n+1
3−1) 4c+8−n−2c−8+n
2(n+1)
3−2 to 2(n+1)
33, 6, 9 2(n+1)
3+(1,5,9) 2(n+1)
3−(5,7,9)
2(n+1)
3+1 to n−4 16 +3(c−2(n+1)
3−1) 4c+11 −2n−2c−11 +2n
n−3 to n1,4,7,10 n+(−2,2,6,10) n−(4,6,8,10)
Table 2
Leprechaun coordinates when n≥68 and n≡8m˙
od 12.
Column cRow Sum diagonal Difference diagonal
1 to n+1
3−4 14 +3(c−1) 4c+11 −2c−11
n+1
3−3 to n+1
32, 5, 8, 11 n+1
3+(−1,3,7,11) n+1
3−(5,7,9,11)
n+1
3+1 to 2(n+1)
3−5 18 +3(c−n+1
3−1) 4c+14 −n−2c−14 +n
2(n+1)
3−4 to 2(n+1)
33, 6, 9, 12, 15 2(n+1)
3+(−1,3,7,11,15) 2(n+1)
3−(7,9,11,13,15)
2(n+1)
3+1 to n−6 22 +3(c−2(n+1)
3−1) 4c+17 −2n−2c−17 +2n
n−5 to n1,4,7,10,13,16 n+(−4,0,4,8,12,16) n−(6,8,10,12,14,16)
have been placed in Column n+1 if the loop had iterated once more. The order of the rows in the last part is as follows:
[16,19,...,n−4,n−1,1,4,7,10](respectively [22,25,...,n−4,n−1,1,4,7,10,13,16]), and all numbers less than
n+1 have been placed in the solution, with the exception of 13 (respectively 19), that would have been placed in cell
n+1. The formulas describing the coordinates of the leprechauns depending on which part of the board they belong are
detailed in Table 1 for n≥50 and n≡2m˙
od 12, and in Table 2 for n≥68 and n≡8m˙
od 12. The smallest size for which
these tables are relevant is n=50, which is too large to represent. However, while it is not actually a solution to the
(2,26)-Leprechauns Problem, Fig. 3 provides a convenient visual representation in a manageable size of the construction
when placing the first leprechaun in tile (1,8).
From Lemma 3 we know that Lines 5–17 produce an arrangement of nnon-attacking leprechauns on an (n+1) ×n
board. It remains to demonstrate that the operations done in Lines 21 and 22 are equivalent to moving the leprechaun
in the last row to an empty row, and that this move does not introduce a conflict.
From Table 1 (respectively 2), we know that the leprechaun in Column 2(n+1)
3−3 (respectively 2(n+1)
3−5) was initially
placed in Row 12 +3(2(n+1)
3−3−n+1
3−1) =12 +3( n+1
3−4) =n+1 (respectively 18 +3( 2(n+1)
3−5−n+1
3−1) =
18 +3( n+1
3−6) =n+1). We also know that no leprechaun was placed in Row 13 (respectively 19). Therefore Line
21 (respectively 22) describes moving the leprechaun in Row n+1 to an empty row, resulting in an n×nboard with
nleprechauns, and all row and column constraints satisfied. Moreover, the tables inform us that the leprechauns in the
previous two columns are in Rows n−5 and n−2, and that the leprechauns in the next two columns are in Rows 3 and
6. Since n≥50, this means that moving the leprechaun in tile ( 2(n+1)
3−3,n+1) (respectively ( 2(n+1)
3−5,n+1)) to tile
(2(n+1)
3−3,13) (respectively ( 2(n+1)
3−5,19)) does not put it within reach of another range-2 leprechaun. Therefore it only
remains to show that no other leprechaun shares a diagonal with the tile ( 2(n+1)
3−3,13) (respectively ( 2(n+1)
3−5,19)).
Let n≥50 such that n≡2m˙
od 12. The leprechaun lthat was moved to the tile ( 2(n+1)
3−3,13) is in the sum diagonal
d+=2(n+1)
3+10 and in the difference diagonal d−=2(n+1)
3−16. Since n+1 is divisible by 3, both d+and d−are even.
Let l′be a leprechaun such that land l′attack each other. Let c′be the column of l′, let d′
+be the sum diagonal of l′, and
let d′
−be the difference diagonal of l′. The possible values for d′
+and d′
−depending on c′are given in Table 1.
•If 1 ≤c′≤n+1
3−2: both d′
+and d′
−are odd, so there can be no conflict.
•If n+1
3−1≤c′≤n+1
3: since n≥50, we have d′
+≤n+1
3+5<2(n+1)
3+10 =d+, so d′
+̸= d+. Similarly, since n≥50
we have d′
−≤n+1
3−3<2(n+1)
3−16 =d−, so d′
−̸= d−.
•If n+1
3+1≤c′≤2(n+1)
3−3: in this case d′
+=4c′+8−nand d′
−= −2c′−8+n. Since n≡2m˙
od 12, and in particular
since n≡2m˙
od 4, we have d′
+≡2m˙
od 4, but d+=2(n+1)
3+10 ≡4m˙
od 4, so d′
+̸= d+. Since c′≥n+1
3+1, we
have d′
−≤ −2( n+1
3+1) −8+n=n+1
3−11, which is strictly smaller than 2(n+1)
3−16 =d−because n≥50, so
d′
−̸= d−.
•If 2(n+1)
3−2≤c′≤2(n+1)
3: both d′
+and d′
−are odd, so there can be no conflict.
•If 2(n+1)
3+1≤c′≤n−4: both d′
+and d′
−are odd, so there can be no conflict.
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G. Escamocher and B. O’Sullivan Discrete Mathematics 344 (2021) 112316
•If n−3≤c′≤n: since n≥50, we have d+=2(n+1)
3+10 <n−2≤d′
+, so d′
+̸= d+. Similarly, since n≥50 we
have d−=2(n+1)
3−16 <n−10 ≤d′
−, so d′
−̸= d−.
Therefore, if n≥50 and n≡2m˙
od 12, then the leprechaun ldoes not share a diagonal with another leprechaun.
If n≥68 and n≡8m˙
od 12, the proof is almost identical with the minor difference being that if n+1
3+1≤c′≤
2(n+1)
3−5, the congruences m ˙
odulo 4 of d′
+and d+are switched. In any case, the result is the same: moving the leprechaun
in Row n+1 to Row 19 in Line 22 does not violate any diagonal constraint. Since that was all that remained to show,
this concludes the proof. □
Now that we have treated each residue class m ˙
odulo 6, we collect them together in Algorithm 5, which solves the
(2,n)-Leprechauns Problem for all n.
Data: An integer n.
Result: Either a solution to the (2,n)-Leprechauns Problem or ‘‘No solution’’.
1if n≥2and n ≤9then return ‘‘No solution’’;
2if n=14 then return (1,4,7,10,13,5,8,11,14,2,6,9,3,12);
3if n≡1m˙
od 6or n≡5m˙
od 6then return Algorithm 1;
4if n≡4m˙
od 6or n≡6m˙
od 6then return Algorithm 2;
5if n≡3m˙
od 6then return Algorithm 3;
6if n≡2m˙
od 6then return Algorithm 4;
Algorithm 5: General case.
Theorem 1. Algorithm 5solves the (2,n)-Leprechauns Problem in linear time.
Proof. Each of Algorithms 1,2,3and 4loops through Columns 1 to nexactly once, so Algorithm 5’s time complexity is
clearly linear. As for its correctness:
Line 1: Already known result [10].
Line 2: Fig. 1 shows that this is indeed a solution.
Line 3: From Lemma 1.
Line 4: From Lemma 2.
Line 5: From Lemma 3.
Line 6: From Lemma 4.□
4. The (k≥3,n)-Leprechauns problem
4.1. When n is under (k+1)2
We know that the (1,n)-Leprechauns (n-Queens) Problem has a solution for n=1, has no solution for n=2 to n=3,
and has a solution for n≥4. We also now know that the (2,n)-Leprechauns Problem has a solution for n=1, has no
solution for n=2 to n=9, and has a solution for n≥10. This would seem to indicate a pattern in the behavior of the
(k,n)-Leprechauns Problem, where for n=1 there is a trivial solution, for nfrom 2 to some Nthe board is too small to
support a solution, and finally for sizes over Nthe board is large enough to accommodate leprechaun distance constraints.
We suspect that the transition between these two phases occurs around n=(k+1)2. To support this conjecture, we are
now going to present two formal results that demonstrate the importance of this particular function. First we show that
the whole area under this parabola is in the unsatisfiability region (apart from the trivial case n=1).
Proposition 1. Let k >0and n >1be two integers such that n <(k+1)2. Then there is no solution for the (k,n)-Leprechauns
Problem.
Proof. Let n′= ⌈ n
k+1⌉and for 1 ≤i,j≤n′, let the box Bi,jbe the set of tiles with coordinates (i′,j′) such that
(k+1) ×(i−1) +1≤i′≤min((k+1) ×(i−1) +k+1,n) and (k+1) ∗(j−1) +1≤j′≤min((k+1) ×(j−1) +k+1,n).
The dimensions of a box Bi,jare at most k+1×k+1, so any two tiles in Bi,jare at distance at most kfrom each other,
so at most one range-kleprechaun can be placed in a box Bi,j.
Consider the set of boxes Bthat contains the boxes Bi,j. If k+1 divides nthen all n′2boxes in Bwill be of dimension
(k+1) ×(k+1), otherwise Bwill contain (n′−1)2boxes of dimension (k+1) ×(k+1), n′−1 boxes of dimension
k′×(k+1), n′−1 boxes of dimension (k+1) ×k′and one box of dimension k′×k′for some 1 ≤k′<k+1. Moreover
Bforms a partition of the board, illustrated in Fig. 4.
If n′<k+1, then the number of boxes is fewer than the number of leprechauns to place, so we have n′=k+1.
Since at most one range-kleprechaun can fit in each box, and since the k+1 boxes B1,1,B1,2,...,B1,n′span exactly
k+1 columns of the board, we know that each of these boxes contains exactly one leprechaun. In particular the box
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G. Escamocher and B. O’Sullivan Discrete Mathematics 344 (2021) 112316
Fig. 4. Partition of the board into boxes.
B1,n′contains one leprechaun. With the same reasoning on the other vertical sets of boxes, we know that the other boxes
B2,n′,...,Bn′−1,n′at the top edge of the board (minus the last one which spans only k′columns) contain one range-k
leprechaun each. So these kboxes (n′=k+1 boxes at the top edge of the board, minus the last one) contain ktotal
range-kleprechauns. But they span k′rows, so they can contain at most k′total range-kleprechauns. So k′≥k. Since
n<(k+1)2we have k′<k+1 and therefore k′=k.
Since k′=k, we have n=k(k+1) +k=(k+1)2−1, with the number of boxes in Bbeing (k+1)2. So every box in B
but one contains a range-kleprechaun. Since the k+1 boxes B1,n′,B2,n′,...,Bn′,n′at the top edge of the board only span
k′=krows of the board, it means that the empty box is among them. Similarly, since the k+1 boxes Bn′,1,Bn′,2,...,Bn′,n′
at the right edge of the board only span kcolumns of the board, it means that the empty box is also among them. So the
empty box is Bn′,n′and every other box in Bcontains exactly one range-kleprechaun.
Since the box B1,n′spans only krows of the board, the range-kleprechaun in that particular box attacks all tiles on
the top row of the box B1,n′−1, which is the intersection of B1,n′−1with the n−k′th row of the board. So there is no
range-kleprechaun on the top row of the box B1,n′−1. Similarly, there is no range-kleprechaun on the top row of the
boxes B2,n′−1,...,Bn′−1,n′−1. So the range-kleprechaun lin the n−k′th row of the board is in the top row of the box
Bn′,n′−1. Using a mirrored reasoning we also get that the range-kleprechaun l′in the n−k′th column of the board is in
the rightmost column of the box Bn′−1,n′. Since lis at most kcolumns to the right of l′, and since l′is at most krows
above l, the two range-kleprechauns attack each other. So there is no solution for the (k,n)-Leprechauns Problem when
n<(k+1)2.□
Note that at no point in the proof of Proposition 1 did we mention diagonal constraints. These only come in play for
n≥(k+1)2, which is where the board is large enough to contain an arrangement that fulfills all other constraints. As we
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G. Escamocher and B. O’Sullivan Discrete Mathematics 344 (2021) 112316
Fig. 5. Partition of the board into boxes when n=(k+1)2.
are now going to show, (k+1)2is the first size for which the (k,n)-Leprechauns Problem is neither always satisfiable nor
always unsatisfiable. Instead, whether there is a solution depends on the parity of k. We will also characterize solutions,
when they exist.
Proposition 2. Let k >0and n be two integers such that n =(k+1)2. Then there is a solution to the (k,n)-Leprechauns
Problem if and only if k is odd, in which case there are exactly two (symmetrical) solutions.
Proof. We define boxes the same way we did in the proof for Proposition 1: each box Bi,jis a square subset of the board
containing (k+1) ×(k+1) tiles, and the bottom left tile of Bi,jis the tile ((k+1) ×(i−1) +1,(k+1) ×(j−1) +1). Since
n=(k+1)2, there are nboxes, as illustrated in Fig. 5. Also, since each box can contain at most one leprechaun and since
nleprechauns need to be placed on the board, each box contains exactly one leprechaun.
We start by showing that within a column of boxes, all leprechauns must be located at the same row of their respective
boxes.
Lemma 5. Let i be such that 1≤i≤k+1. Then there exist some r such that for each 1≤j≤k+1, there is a leprechaun
in the rth row of Bi,j.
Proof. We know that there is a leprechaun on Row k+1. Let iAsuch that the box BiA,1is the bottom box that contains
a leprechaun in its k+1th row. Since leprechauns have a range of k, there cannot be a leprechaun in the first krows of
the box BiA,2. So BiA,2also contains a leprechaun in its k+1th row. Using the same argument for boxes BiA,3to BiA,k+1, we
can see that all boxes BiA,jcontain a leprechaun in their k+1th row.
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G. Escamocher and B. O’Sullivan Discrete Mathematics 344 (2021) 112316
We know that there is a leprechaun on Row k. Let iBsuch that iB̸= iAand the box BiB,1is the bottom box that contains
a leprechaun in its kth row. Since leprechauns have a range of k, there cannot be a leprechaun in the first k−1 rows of the
box BiB,2. So there is a leprechaun in either the kth or the k+1th row of BiB,2. But we know that BiA,2is the box (among
the ones on the second row of boxes) that contains a leprechaun in its k+1th row. Therefore BiB,2contains a leprechaun
in its kth row. Using the same argument for boxes BiB,3to BiB,k+1, we can see that all boxes BiB,jcontain a leprechaun in
their kth row.
Repeating the reasoning for Rows k−1 to 1 completes the proof of the Lemma. □
By switching row and column coordinates the same proof can be used to prove the corresponding property for
leprechauns within the same row of boxes:
Lemma 6. Let j be such that 1≤j≤k+1. Then there exist some c such that for each 1≤i≤k+1, there is a leprechaun
in the cth column of Bi,j.
We now show that within a column of boxes, the columns of the leprechauns either monotonically increase or
monotonically decrease.
Lemma 7. Let i be such that 1≤i≤k+1. Then one of the following is true:
•For all 1≤j≤k+1, there is a leprechaun in the jth column of Bi,j.
•For all 1≤j≤k+1, there is a leprechaun in the k +2−jth column of Bi,j.
Proof. From Lemma 5 we know that there exists some rsuch that all leprechauns in this column of boxes are in the rth
row of their columns.
Suppose that the order of the leprechauns’ columns is not monotonically decreasing. Therefore there exist j0,c1and c2
such that c1<c2, there is a leprechaun in the c1th column of Bi,j0, and there is a leprechaun in the c2th column of Bi,j0+1.
Suppose also that the order of the leprechauns’ rows in the row of boxes j0is not monotonically decreasing. Then there
exist i0,r1and r2such that r1<r2, there is a leprechaun in the r1th row of Bi0,j0, and there is a leprechaun in the r2th
row of Bi0+1,j0. Since there is a leprechaun in the c1th column of Bi,j0, we know from Lemma 6 that the leprechaun in the
r2th row of Bi0+1,j0is in the c1th column of this box. Since there is a leprechaun in the c2th column of Bi,j0+1and there is
a leprechaun in the r1th row of Bi0,j0, we also know from Lemmas 5 and 6that there is a leprechaun in the r1th row and
c2th column of Bi0,j0+1. Since c1<c2and r1<r2, this leprechaun is attacking the leprechaun located in the r2th row and
c1th column of Bi0+1,j0. So if the order of the leprechauns’ columns is not monotonically decreasing, then the order of the
leprechauns’ rows must be monotonically decreasing.
By replacing ‘‘decreasing’’ with ‘‘increasing’’ and ‘‘<’’ with ‘‘>’’ in the previous paragraph, we can also show that
if the order of the leprechauns’ columns is not monotonically increasing, then the order of the leprechauns’ rows
must be monotonically increasing. So if the order of the leprechauns’ columns is neither monotonically decreasing nor
monotonically increasing then we have a contradiction. So the order of the leprechauns’ columns is either monotonically
decreasing or monotonically increasing. □
As before, switching row and columns gives us a proof for the corresponding property on rows of boxes:
Lemma 8. Let j be such that 1≤j≤k+1. Then one of the following is true:
•For all 1≤i≤k+1, there is a leprechaun in the ith row of Bi,j.
•For all 1≤i≤k+1, there is a leprechaun in the k +2−ith row of Bi,j.
At this point, we have four remaining potential solutions:
1. The order of the leprechauns’ columns is monotonically increasing and the order of the leprechauns’ row is
monotonically decreasing: for all 1 ≤i,j≤k+1, there is a leprechaun at coordinates (j,k+2−i) in the box
Bi,j, corresponding to the board tile ((k+1) ×(i−1) +j,(k+1) ×(j−1) +k+2−i).
2. The order of the leprechauns’ columns is monotonically decreasing and the order of the leprechauns’ rows is
monotonically increasing: for all 1 ≤i,j≤k+1, there is a leprechaun at coordinates (k+2−j,i) in the box
Bi,j, corresponding to the board tile ((k+1) ×(i−1) +k+2−j,(k+1) ×(j−1) +i).
3. The order of the leprechauns’ columns is monotonically increasing and the order of the leprechauns’ rows is
monotonically increasing: for all 1 ≤i,j≤k+1, there is a leprechaun at coordinates (j,i) in the box Bi,j,
corresponding to the board tile ((k+1) ×(i−1) +j,(k+1) ×(j−1) +i).
4. The order of the leprechauns’ columns is monotonically decreasing and the order of the leprechauns’ rows is
monotonically decreasing: for all 1 ≤i,j≤k+1, there is a leprechaun at coordinates (k+2−j,k+2−i) in
the box Bi,j, corresponding to the board tile ((k+1) ×(i−1) +k+2−j,(k+1) ×(j−1) +k+2−i).
In the third potential solution, the leprechaun in the box B2,1is on the board tile (k+2,2) and the leprechaun in the
box B1,2is on the board tile (2,k+2). These two leprechauns are attacking each other, so the third potential solution is
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G. Escamocher and B. O’Sullivan Discrete Mathematics 344 (2021) 112316
Fig. 6. Potential solution for k=2 and n=(k+1)2=9.
not a solution. In the fourth potential solution, the leprechaun in the box B1,1is on the board tile (k+1,k+1) and the
leprechaun in the box B2,2is on the board tile (2k+1,2k+1). These two leprechauns are attacking each other, so the
fourth potential solution is not a solution.
Now only two potential solutions remain. In the first one, the leprechaun in the box B1,1is on the board tile (1,k+1).
In the second one, the leprechaun in the box B1,1is on the board tile (k+1,1). Therefore the two potential solutions are
distinct. We now show that they mirror each other through the main diagonal.
Lemma 9. Let i0and j0be two integers such that 1≤i0,j0≤n. Then there is a leprechaun on tile (i0,j0)in the first potential
solution if and only if there is a leprechaun on tile (j0,i0)in the second potential solution.
Proof. ⇒Suppose that there is a leprechaun on tile (i0,j0) in the first potential solution. Then there are 1 ≤i,j≤k+1
such that i0=(k+1) ×(i−1) +jand j0=(k+1) ×(j−1) +k+2−i. Let i′and j′be such that i′=jand j′=i. We know
that there is a leprechaun on tile ((k+1) ×(i′−1) +k+2−j′,(k+1) ×(j′−1) +i′) in the second potential solution.
Since i′=jand j′=i, we have (k+1) ×(i′−1) +k+2−j′=j0and (k+1) ×(j′−1) +i′=i0. So there is a leprechaun
on tile (j0,i0) in the second potential solution.
⇐Since each potential solution has the same number of leprechauns, we only needed to prove one direction. □
Since both remaining potential solutions are symmetrical, we will from now on only consider the first one. For ease of
future reference, we give in Table 3 the coordinates in this potential solution of the leprechaun in Bi,jand of its nearest
neighbors. We also illustrate the potential solution for an even value of kin Fig. 6 and for an odd value in Fig. 7. In
the latter case, no two leprechauns are attacking each other, and we therefore have an actual solution for k=3 and
n=(k+1)2=16. In the former case however, there are several cases of leprechauns attacking each other, for example
the two leprechauns in (1,3) and (6,8). This proves that there is no solution to the (2,9)-Leprechauns Problem. More
generally:
Lemma 10. Let k >0and n be two integers such that k is even and n =(k+1)2. Then in the first potential solution to the
(k,n)-Leprechauns Problem, the leprechauns in boxes B1,k/2and Bk/2+1,k+1attack each other.
Proof. Let l1be the leprechaun in the box B1,k/2and let l2be the leprechaun in the box Bk/2+1,k+1. We know from Table 3
that the coordinates of l1are (i1,j1) with i1=(k+1) ×(1 −1) +k/2 and j1=(k+1) ×(k/2−1) +k+2−1, while the
coordinates of l2are (i2,j2) with i2=(k+1) ×(k/2+1−1) +k+1 and j2=(k+1) ×(k+1−1) +k+2−(k/2+1).
Therefore
j1−i1=(k+1)×(k/2−1) +k+2−1−(k+1) ×(1 −1)−k/2
=(k+1) ×k/2−(k+1) +k+1−k/2
=(k+1) ×k/2−k/2
=k×k/2
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G. Escamocher and B. O’Sullivan Discrete Mathematics 344 (2021) 112316
Table 3
Coordinates in the first potential solution of the leprechaun lin the box Bi,jand of the leprechauns in the surrounding
boxes.
Box Coordinates Difference with l’s
coordinates
Distance to l
Bi,j((k+1) ×(i−1) +j,(k+1) ×(j−1) +k+2−i) (0,0) 0
Bi+1,j((k+1) ×i+j,(k+1) ×(j−1) +k+1−i) (k+1,−1) k+1
Bi+1,j+1((k+1) ×i+j+1,(k+1) ×j+k+1−i) (k+2,k) k+2
Bi,j+1((k+1) ×(i−1) +j+1,(k+1) ×j+k+2−i) (1,k+1) k+1
Bi−1,j+1((k+1) ×(i−2) +j+1,(k+1) ×j+k+3−i) (−k,k+2) k+2
Bi−1,j((k+1) ×(i−2) +j,(k+1) ×(j−1) +k+3−i) (−k−1,1) k+1
Bi−1,j−1((k+1) ×(i−2) +j−1,(k+1) ×(j−2) +k+3−i) (−k−2,−k) k+2
Bi,j−1((k+1) ×(i−1) +j−1,(k+1) ×(j−2) +k+2−i) (−1,−k−1) k+1
Bi+1,j−1((k+1) ×i+j−1,(k+1) ×(j−2) +k+1−i) (k,−k−2) k+2
Fig. 7. Potential solution for k=3 and n=(k+1)2=16.
and
j2−i2=(k+1) ×(k+1−1) +k+2−(k/2+1) −(k+
1) ×(k/2+1−1) −k−1
=(k+1) ×k+k+2−k/2−1−(k+1) ×(k/2) −k−1
=(k+1) ×k−k/2−(k+1) ×(k/2)
=(k+1) ×(k/2) −k/2
=k×k/2
So j1−i1=j2−i2. So l1and l2are on the same diagonal. So l1and l2are attacking each other. □
So if kis even, then the only potential solution (modulo reflection) to the (k,n)-Leprechauns Problem is not an actual
solution. This completes the proof of the Proposition for the case when kis even. Consequently we now assume that kis
odd.
14
G. Escamocher and B. O’Sullivan Discrete Mathematics 344 (2021) 112316
One of the consequences of Lemma 8 is that there is a leprechaun in each row of the board, so the row constraints of
the problem are satisfied. Similarly, from Lemma 7 we know that there is a leprechaun in each column of the board and
therefore that the column constraints of the problem are satisfied.
As can be seen in Table 3, for all 1 ≤i,j≤k+1, the range-kleprechaun in the box Bi,jis out of reach of the range-k
leprechauns in the surrounding boxes. Therefore the proximity constraints are also satisfied.
We are now going to show that the last remaining constraints, the diagonal ones, are satisfied by the first potential
solution. Let lbe the leprechaun in the box Bi,jwith 1 ≤i,j≤k+1 and let l′be the leprechaun in the box Bi′,j′with
1≤i′,j′≤k+1 such that either i′̸= ior j′̸= j. Let d+(respectively d−) be the sum (respectively difference) of l’s
coordinates, and let d′
+(respectively d′
−) be the sum (respectively difference) of l′’s coordinates.
Suppose that d′
+=d+. From Table 3 we have:
(k+1) ×(i′−1) +j′+(k+1) ×(j′−1) +k+2−i′=(k+1) ×(i−1) +j+(k+1) ×(j−1) +k+2−i
⇓
(k+1) ×i′+j′+(k+1) ×j′−i′=(k+1) ×i+j+(k+1) ×j−i
⇓
k×i′+(k+2) ×j′=k×i+(k+2) ×j
⇓
k×(i′+j′)+2j′=k×(i+j)+2j
So (2j′m˙
od k)=(2j m˙
od k). So either (j′m˙
od k)=(j m˙
od k) or (j′m˙
od k)=((j+k/2) m˙
od k). Since kis odd, we have
(j′m˙
od k)=(j m˙
od k). Moreover j′̸= j(because otherwise we would also have i′=iand we assumed either i′̸= ior
j′̸= j). We have 1 ≤j,j′≤k+1, so either j=1 and j′=k+1 or the other way around. Without loss of generality,
assume the former. The equality above becomes then:
k×(i′+k+1) +2(k+1) =k×(i+1) +2
⇓
k×(i′+k)+2k=k×i
Dividing by kwe get i=i′+k+2. Since 1 ≤i,i′≤k+1, we have a contradiction. Therefore d′
+cannot be equal
to d+.
Suppose now that d′
−=d−. From Table 3 we have:
(k+1) ×(i′−1) +j′−(k+1) ×(j′−1) −k−2+i′=(k+1) ×(i−1) +j−(k+1) ×(j−1) −k−2+i
⇓
(k+1) ×i′+j′−(k+1) ×j′+i′=(k+1) ×i+j−(k+1) ×j+i
⇓
(k+2) ×i′−k×j′=(k+2) ×i−k×j
⇓
k×(i′−j′)+2i′=k×(i−j)+2i
So (2i′m˙
od k)=(2i m˙
od k). So either (i′m˙
od k)=(i m˙
od k) or (i′m˙
od k)=((i+k/2) m˙
od k). Since kis odd, we have
(i′m˙
od k)=(i m˙
od k). Moreover i′̸= i(because otherwise we would also have j′=jand we assumed either i′̸= ior
j′̸= j). We have 1 ≤i,i′≤k+1, so either i=1 and i′=k+1 or the other way around. Without loss of generality,
assume the former. The equality above becomes then:
k×(k+1−j′)+2(k+1) =k×(1 −j)+2
⇓
k×(k−j′)+2k=k×(−j)
Dividing by kwe get k−j′+2= −j, or equivalently j=j′−k−2. Since 1 ≤j,j′≤k+1, we have a contradiction.
Therefore d′
−cannot be equal to d−. Therefore the diagonal constraints are satisfied, and the first potential solution
is an actual solution to the (k,n)-Leprechauns Problem when kis odd and n=(k+1)2. Since the second potential
solution is a reflective image of the first potential solution, it is an actual solution as well. This completes the proof of the
Proposition. □
We can combine the last two propositions to completely solve the (k,n)-Leprechauns Problem for all configurations
of kand nsuch that nis at or below the parabola corresponding to the function (k+1)2.
Theorem 2. Let k >0and n >0be such that n ≤(k+1)2. Then the (k,n)-Leprechauns Problem is satisfiable if and only if
k is odd and n =(k+1)2, in which case there are exactly two symmetrical solutions, or if n =1, in which case there is exactly
one trivial solution.
15
G. Escamocher and B. O’Sullivan Discrete Mathematics 344 (2021) 112316
Table 4
Smallest non-trivial board for which there is a solution to the
(k,n)-Leprechauns Problem.
k(k+1)2noffset
1 4 4 0
2 9 10 1
3 16 16 0
4 25 28 3
5 36 36 0
6 49 52 3
7 64 64 0
8 81 82 1
9 100 100 0
Table 5
Number pof solutions to the (k,n)-Leprechauns Problem for some combinations of kand
n. For each k, the first value of nshown is the first n>1 for which there exists at least
one solution.
k=3k=4k=5k=6
n p n p n p n p
16 2 28 10 36 2 52 6
17 34 29 286 37 74
18 4 30 696 38 0
19 112 31 10016 39 48
20 516 32 201332 40 68
21 7312 41 808
22 81324
23 1056560
24 13443944
25 171919446
We have fully treated the case when n≤(k+1)2, and proven that it covers the board sizes where the additional
distance constraints introduced by leprechauns cannot be satisfied.
4.2. When n is over (k+1)2
It is natural to wonder what is for each kthe first non-trivial (that is, not 1) nfor which there exists a solution to the
(k,n)-Leprechauns Problem. From Theorem 2, we know that for odd kthe answer is exactly (k+1)2. Since a solution to
the (k+1,n)-Leprechauns Problem also satisfies all constraints from the (k,n)-Leprechauns Problem, this implies that
the demarcation lies between (k+1)2and (k+2)2for all k.
For the first few values of k, we collect in Table 4 the smallest size n>1 for which a board is large enough to admit
a solution to the (k,n)-Leprechauns Problem. The entries for odd kwere filled by using the result from Theorem 2. For
even k, at least up until 8, the smallest value of n(>1) for which the (k,n)-Leprechauns Problem is satisfiable appears
to be equal to the smallest prime number larger than (k+1)2, minus 1. That this value is an upper bound comes from a
construction for the n-Queens Problem when nis either prime or one less than a prime [24]. We can, in a manner similar
to the proofs of Lemmas 1 and 3, generalize this construction to the (k,n)-Leprechauns Problem when nis big enough. To
verify that this is indeed the smallest eligible n, we used exhaustive search to look for a solution to the (k,n)-Leprechauns
for each nbetween the known bounds, and did not find any.
Interestingly, while for k=1 and k=2 there is only one continuous unsatisfiability phase (from 2 to 3 and from
2 to 9 respectively), this is not true for all k. For the (5,n)-Leprechauns Problem in particular, there is no solution for
2≤n≤35, there is a solution for n=36 and for n=37, but there is no solution for n=38.
We give in Table 5 the number of solutions to the (k,n)-Leprechauns Problem for the first few values of k≥3 and
n≥(k+1)2. These numbers were provided by an anonymous reviewer, to whom we are grateful. Similar results exist
for the (1,n)- (queens [26]) and (2,n)- (superqueens [25]) Leprechauns Problems.
5. Conclusion
We have made the first strides towards solving the Diverse n-Queens Problem by studying the equivalent (k,n)-
Leprechauns Problems. We have given a vertical result, in the form of an algorithm that can solve the (2,n)-Leprechauns
Problem, as well as a horizontal one, in our characterization of the Problem for n≤(k+1)2. The latter provides strong
theoretical evidence that the phase transition from unsatisfiability to satisfiability in the (k,n)-Leprechauns Problem
occurs around n=(k+1)2.
In addition of diversity, potential future work on the (k,n)-Leprechauns Problem could try to apply it on representation,
which is a related but different notion that has been studied in other constraint fields [23].
16
G. Escamocher and B. O’Sullivan Discrete Mathematics 344 (2021) 112316
Declaration of competing interest
The authors declare that they have no known competing financial interests or personal relationships that could have
appeared to influence the work reported in this paper.
Acknowledgments
This material is based upon works supported by the Science Foundation Ireland under Grant No. 12/RC/2289-P2 which
is co-funded under the European Regional Development Fund.
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