A Pentomino Exclusion Problem
Kurt M. Anstreicher
Dept. of Management Sciences
University of Iowa
Iowa City, IA 52242
Introduction A pentomino is a generalization of a domino that covers 5 squares.
Many interesting problems involving pentominoes, and their further generalization
polyominoes, are described in Solomon Golomb's classic book [2]. One class of
pentomino problems considers the possibility, or impossibility, of using pentominoes
to construct a given gure in the plane. For example, it is known [2, p.8] that if
any 2 2 region is removed from an 8 8 checkerboard, then the remaining squares
can be covered using each of the twelve distinct pentominoes exactly once. Such a
construction, with the center 2 2 square removed, is illustrated below.
Another class of problems, described in Chapter 3 of [2], are pentomino exclusion
problems. Typical problems of this type start with an 8 8 board, and then consider
covering squares so as to make it impossible to place certain pentominoes on the
board. For example, it is known [2, p.27] that 12 dominoes can be placed so as to
simultaneously exclude all pentominoes from the 8 8 board, for example using the
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following placement:
(1)
A monomino covers one square. The placement of dominoes in (1) implies that
24 monominoes suce to exclude pentominoes from the 8 8 board. In [2, p.27]
it is conjectured that 24 monominoes are in fact required to do so, but this has
apparently never been proved. Recently Robert Bosch [1] described a discrete optimization formulation of this pentomino exclusion problem, which can be solved
using linear programming subproblems in a branch-and-bound framework to prove
that 24 monominoes are required to exclude pentominoes from an 8 8 board. In
this note we give a direct proof.
The 4 4 board Our analysis of the 8 8 board is based on the minimal
arrangement of monominoes that excludes pentominoes on a 4 4 board. It is
certainly possible to exclude pentominoes on a 4 4 board using 5 monominoes, as
shown by the following pattern:
(2)
It turns out that the arrangement of (2) is the essentially unique minimal arrangement that excludes pentominoes on a 4 4 board. We prove this in the next lemma.
In the next section we will use Lemma 1 to \build up" to a result for the 8 8
board.
Lemma 1. A 4 4 board requires at least 5 monominoes to exclude pentominoes.
Moreover, the only such placement of 5 monominoes, up to rotation, is given by (2).
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Proof. First consider a 4 2 board, with the squares labeled:
A
C
E
G
B
D
F
H
It is clear that to exclude pentominoes this board requires at least 2 monominoes.
Moreover, for 2 to suce they must be placed on one of the following pairs: CD, EF ,
CF , or ED. Since a 4 4 board consists of 2 of these 4 2 boards, if pentominoes
can be excluded with less than 6 monominoes, then one of the 4 2 boards must
have exactly 2 monominoes placed on it, and they must use one of the 4 possible
combinations of squares listed above. We consider the possible cases in turn.
Assume that the left half of the 4 4 board has only two monominoes, and they
are placed on the squares CD. The board now looks as follows:
?
?
?
?
?
?
I
K
M
O
J
L
N
P
where a \?" means that no monomino is allowed. In order to exclude pentominoes
we immediately need monominoes on M and O. After that, it is easy to see that if
only one more monomino is put down, there is always an opening for a pentomino.
It follows that in this case 6 monominoes are required to exclude pentominoes. The
situation is similar if the rst 2 monominoes are placed on EF .
Next assume that the left half of the board has only two monominoes, and they
are placed on squares CF . The board now looks as follows:
?
?
?
?
?
?
I
K
M
O
J
L
N
P
It is clear that at least 2 monominoes must be placed in the region IJKL. Moreover
if 3 were placed there, we would still need at least 1 in MNOP . Therefore if
pentominoes can be excluded using a total of less than 6 monominoes there must be
exactly 2 in IJKL. By inspection these two must be placed on either IK , IL, or
JK ; the other possibilities (IJ , KL, and JL) would allow a pentomino on the upper
half of the 4 4 board. In the rst two cases (IK and IL), if only one monomino
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is placed on MNOP , a pentomino is allowed somewhere on the board. However in
the case JK , placing a monomino on O excludes pentominoes, and this is the only
choice of a single square in MNOP that does so. We thus obtain the placement of
5 monominoes given in (2). If the two monominoes on the left half of the board are
place on DE instead of CF , the same argument results in a 90 clockwise rotation
of this pattern.
The same analysis can be repeated assuming that the right half of the board
contains only two monominoes. The CD and EF placements again require a total
of 6 monominoes to exclude pentominoes. The DE and CF placements result in
unique placements of 5 monominoes that exclude pentominoes that correpsond to
90 counterclockwise, and 180 degree rotations of the pattern in (2).
The 8 8 board We will now use Lemma 1 to obtain a result for the 8 8 board.
As an intermediate step we rst characterize the minimal placement of mononimoes
that excludes pentominoes from a 4 8 board.
Lemma 2. A 4 8 board requires at least 11 monominoes to exclude pentominoes.
Moreover, any placement of 11 monominoes that excludes pentominoes contains the
following con guration, possibly after rotation and/or re ection:
?
?
?
?
?
?
?
?
?
?
?
?
? ? ?
Proof. Clearly a 4 8 board requires at least 10 monominoes to exclude pentominoes, from Lemma 1. Moreover, if pentominoes can be excluded with 10 or 11
monominoes then one half of the 4 8 board must contain the con guration from
(2), possibly rotated.
Assume that the left half contains 5 monominoes, and assume that they are
placed exactly as in (2). The 4 8 board is then
?
?
?
?
?
?
?
?
?
?
?
A
E
I
M
B
F
J
N
C
G
K
O
D
H
L
P
In order to exclude pentominoes, monominoes are now immediately required on E ,
I , and M . Moreover, each of the 2 3 regions BCDFGH and JKLNOP requires
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2 monominoes, for a total of 7. Therefore a total of at least 12 monominoes will
be required in this case. If the left half is rotated 90 clockwise the same argument
applies.
Next assume that the left half has the con guration obtained by rotating the
pattern from (2) 90 counterclockwise. The situation is then
?
?
?
?
?
?
?
?
?
?
?
A
E
I
M
B
F
J
N
C
G
K
O
D
H
L
P
To exclude pentominoes a monomino is certainly required on A, and as described
above the 2 3 regions BCDFGH and JKLNOP each require 2 monominoes. As
a result, if a total of less than 7 monominoes are to be placed on the right half,
only one can be used in the region EIM . This monomino must be placed on I ,
since either E or M would immediately permit a pentomino. A monomino is then
required on N , and since the 3 3 region BCDFGHJKL requires 3 monominoes,
none are available for O or P . This results in the con guration of the lemma. Note
that it is certainly possible to place 3 monominoes in BCDFGHJKL so as to
exclude pentominoes, for example by using F , G, and L.
If the left half is obtained by rotating the pattern from Lemma 1 180 then
the above argument results in a (vertically) re ected version of the same pattern.
The entire argument, repeated with the right half containing only 5 monominoes,
produces rotations and/or re ections of the same pattern.
Theorem 3. There is no placement of 23 monominoes on an 8 8 board that
excludes pentominoes.
Proof. The 8 8 board contains 4 disjoint 4 4 boards. We will refer to the
upper-left 4 4 board as UL, and the others similarly as UR, LL, and LR. Clearly
if 23 monominoes are placed on the 8 8 board then one of these 4 4 boards has
no more that 5 monominoes, and if the placement excludes pentominoes then none
can have fewer than 5, by Lemma 1. Thus one of the 4 4 boards has exactly 5
monominoes. Assume without loss of generality that UL has 5 monominoes. By
Lemma 2, if the placement excludes pentominoes then UR and LL each must have
at least 6 monominoes. Since LR must have at least 5, at most 1 of LL and UR can
have more than 6, and neither can have more than 7. It follows that the number of
monominoes in the di erent 4 4 boards must correspond to one of the following
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three possibilities:
5 6
6 6
5 6
7 5
5 7
6 5
Since the third can be obtained from the second by a diagonal re ection it suces
to consider the rst two cases.
In the rst case, Lemma 2 can be applied to the upper and left halves of the
8 8 board. The result is that if pentominoes are excluded, the 8 8 board must
have the following pattern:
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ?
However, this arrangement obviously permits a pentomino at the center of the board,
and therefore cannot correspond to a placement that excludes pentominoes.
The other possibility is that UL and LR both have 5 monominoes, UR has 6,
and LL has 7. In this case Lemma 2 can be applied to the upper and right halves
of the 8 8 board. There are 2 possible rotations of the pattern in (2) for each of
UL and LR, resulting in a total of 4 cases. The rst possibility is:
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
(3)
where the con guration in UR is a consequence of Lemma 2. This pattern obviously
permits pentominoes. Another case, which is equivalent, is obtained by re ecting the
pattern in (3) about the \back diagonal" that goes from the lower-left to the upperright corner. A third case corresponds to rotating LR in (3) 90 counterclockwise,
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and this pattern also admits pentominoes. The nal possibility is:
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
A
C
?
?
?
?
B
D
?
?
?
?
?
?
?
?
?
where the pattern in UR is obtained by applying Lemma 2 to both the upper and
right halves of the 8 8 board. Since UR is assumed to have only 6 monominoes
only one can be placed in ABCD, and it is clear that no such placement can exclude
pentominoes. This completes the proof.
Andy Liu (private communication) has suggested the following interesting generalization of Theorem 3. Note that the pattern in (1) is periodic; one can think of
taking the upper-left 3 3 square and repeating it. It follows that this pattern can
be used to exclude pentominoes from an in nite planar checkerboard using monominoes on 4=9 of the squares. The fraction of the squares on an in nite planar board
that must be covered in order to exclude individual pentominoes varies from 1=5 to
1=3; see [2, p.28] and [3]. However, the fraction of squares that must be covered to
simultaneously exclude all pentominoes from an in nite planar board appears to be
an open problem. Based on Theorem 3 and (1) we propose the following conjecture.
Conjecture. On an in nite planar checkerboard, at least 4=9 of the squares must
be covered with monominoes in order to exclude pentominoes.
References
[1] R.A. Bosch, A pentomino exclusion problem, OPTIMA 60 (1998), 3.
[2] S.W. Golomb, Polyominoes: Puzzles, Patterns, Problems, and Packings (Second
edition), Princeton University Press, Princeton, NJ, 1994.
[3] M.S. Klamkin and A.C.F. Liu, Polyominoes on the in nite checkerboard, J.
Combin. Theory Series A 28 (1980), 7-16.