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# Binary Puzzle is NP–complete

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We prove that the generalization of the puzzle game Binary Puzzle, also known as Binary Sudoku, is NP-complete.
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Binary Puzzle is NP–complete
Marzio De Biasi
marziodebiasi [at] gmail [dot] com
July 2012
Version 1.01: second version with ﬁgures and more details
http://www.fractalmuse.org
Abstract
We prove that the puzzle game Binary Puzzle, also known as Binary
Sudoku, is NP-complete.
1 Introduction
Binary Puzzle (also known as Binary Sudoku) is an addictive puzzle played on
a n × n grid; intially some of the cells contain a zero or a one (ﬁxed cells); the
aim of the game is to ﬁll the remaining empty cells according to the following
rules:
Each cell should contain a zero or a one.
No more than two similar numbers next to or below each other are allowed.
Each row and each column should contain an equal number of zeros and
ones.
Each row is unique and each column is unique.
An example of a solved game is shown in Figure 1.
Figure 1: A solved Binary Puzzle game.
1
In line with the recent focus on the complexity of puzzle games [2] [1], we
study how hard it can be to solve a Binary Puzzle game; in particular we prove
that given a partially ﬁlled n × n grid, it is NP-complete to decide if the game
has a valid solution.
In Section 2 we formally deﬁne the decision problem associated to the puzzle
game; in Section 3 we brieﬂy describe the planar 3CNF NP–complete problem;
in Section 4 we describe the gadgets that can be used to reduce a planar 3CNF
problem to a Binary Puzzle game and in Section 5 we give the details of the
reduction to prove the NP–completeness of the game.
2 Deﬁnitions
Deﬁnition 2.1. decision problem BINARY PUZZLE :
Instance: A Binary Puzzle game, i.e. a partially ﬁlled n ×n grid speciﬁed
with a string g {⊥, 0, 1}
n
2
( is used to represent an empty cell).
Question: Does a valid solution for the game exist? I.e. can we ﬁll the
empty cells with a zero or a one following three rules:
R1. no more than two similar numbers next to or below each other are allowed;
R2. each row and each column should contain an equal number of zeros and
ones;
R3. each row is unique and each column is unique.
We also deﬁne a variant of the game, that is equal to Binary puzzle but rules
R2 and R3 are ignored.
Deﬁnition 2.2. decision problem BARE BINARY PUZZLE :
Instance: An n × n Binary Puzzle game.
Question: Does a valid solution for the game exist? I.e. can we ﬁll the
empty cells with a zero or a one in a way that:
R1. no more than two similar numbers next to or below each other are allowed.
3 Planar 3CNF
A CNF formula is planar if the bipartire graph between clauses and literals
plus all edges (x
i
, ¯x
i
) forms a planar graph.
For example the planar graph corresponding to the planar CNF formula:
(x
1
¯x
2
x
3
) (x
2
¯x
3
x
4
) (x
3
x
4
¯x
5
)
is shown in Figure 2.
The problem of deciding whether a planar CNF formula is satisﬁable or
not is NP–complete [3] and it remains NP–complete even with the additional
constraint that each clause must contain exactly three literals (planar 3CNF ).
2
Figure 2: A planar 3CNF formula (x
1
¯x
2
x
3
) (x
2
¯x
3
x
4
) (x
3
x
4
¯x
5
).
We prove the NP–hardness of BARE BINARY PUZZLE using the following idea:
given a planar 3CNF formula, we can reduce the maximum degree of its graph
to degree 3 replacing the edges from each variable to the clauses with a cascade
of degree 3 “Y” nodes. Then we can build its orthogonal representation in
polynomial time [4]. From the orthogonal representation we build an equivalent
Binary Puzzle grid replacing the literal pairs (x
i
, ¯x
i
), the clauses c
i
and the
(orthogonal) edges with a corresponding gadget.
Every gadget is a closed 21 × 21 portion of the grid with a border of ﬁxed
cells; some of the inner cells will be ﬁxed, and some will be empty. A gadget
can interact with the adjacent gadgets only through the middle empty cells; we
will call them interface cells (see Figure 3). The construction of each gadget
is such that not all combination of values 0, 1 can be assigned to the interface
cells: some combinations lead to an invalid conﬁguration i.e. the empty cells of
the inner part of the gadget cannot be ﬁlled without breaking rule R1 of the
game. We will also logically distinguish between input interface cells and output
interface cells: a particular value placed on an input cell of a gadget will force
a corresponding value on the output cell(s).
We describe each gadget using the following convention in the ﬁgures: gray
cells represent ﬁxed border cells, blue cells represent the ﬁxed cells that “imple-
ments” the logic of the gadget, orange cells represent the interface cells.
We used a simple constraint solver program to test every combination of 0, 1
values in the interface cells of each gadget in order to check if a corresponding
valid conﬁguration of the inner empty cells exist.
The Variable gadget has two output interface cells T, F , and is used to simulate
an assignment of truth value to a variable. As shown in Figure 4 it is not
possible to correctly ﬁll the Variable gadget if we place a 1 in both the interface
cells.
3
Figure 3: The 21 × 21 border of ﬁxed cells common to all gadgets, and the
interface (empty) cells.
T F Valid
0 0 YES
0 1 YES
1 0 YES
1 1 NO
4
I1 O1 O2 Valid
0 0 0 YES
0 0 1 NO
0 1 0 NO
0 1 1 NO
1 0 0 YES
1 0 1 YES
1 1 0 YES
1 1 1 YES
The Split gadget is used to duplicate an edge and has one input interface cell
I1 and two output interface cell O1, O2. As shown in Figure 5 the two output
cells can contain 1 only if I1 contains 1.
The edge gadgets are used to connect the other gadgets and simulate the edges
of the planar 3CNF formula. They are of two types: Line gadget and Turn
gadgets. As shown in Figure ??, Both have an input interface cell I1 and an
output interface cell O1: if cell O1 contains a 1 then cell I1 must contain a 1,
too.
The OR gadget is used like a logical OR gate: it has two input interface cells
I1, I1 and an output interface cell O1. As shown in Figure 7, in order to ﬁll cell
O1 with a 1 at least one of the two inputs must be 1.
gadgets has its output interface cell O
1
forced to 1. As shown in Figure ??, it
has three input interface cells are I
1
, I
2
, I
3
and at least one of them must be
ﬁlled with a 1.
We can extend the set of gadgets rotating their inner logic by 90, 180 and
270 degrees (the ﬁxed border blocks remain unchanged).
5
I1 O1 Valid
0 0 YES
0 1 NO
1 0 YES
1 1 YES
I1 I2 O1 Valid
0 0 0 YES
0 0 1 NO
0 1 0 YES
0 1 1 YES
1 0 0 YES
1 0 1 YES
1 1 0 YES
1 1 1 YES
6
I1 I2 I3 Valid
0 0 0 NO
0 0 1 YES
0 1 0 YES
0 1 1 YES
1 0 0 YES
1 0 1 YES
1 1 0 YES
1 1 1 YES
7
5 Reduction
Given a planar 3CNF formula F with n variables x
1
, x
2
, ..., x
n
and m clauses
c
1
, c
2
, ..., c
m
; we ﬁrst reduce the max degree of the corresponding planar graph
to degree 3 adding some split nodes between the variables and the clauses:
if the positive literal x
i
is connected to clauses c
i
1
, c
i
2
..., c
i
k
k 1 split nodes y
i
1
, ..., y
i
k1
and replace the k edges (x
i
, c
i
j
) with new edges
(x
i
, y
i
1
), (y
i
1
, y
i
2
), ..., (y
i
k2
, y
i
k1
) and (y
i
1
, c
i
1
), (y
i
2
, c
i
2
), ...,(y
i
k1
, c
i
k1
), (y
i
k1
, c
i
k
).
We do the same for the negative literals ¯x
i
.
Then we build an orthogonal representation of the resulting planar graph
that can be embedded in a square grid of size N = (n + m)
2
) (see an example
in Figure 9).
Figure 9: Orthogonal representation of the planar 3CNF formula of Figure 2;
y
3
and y
4
are the split nodes added to reduce the maximum degree to degree 3.
We build an equivalent BARE BINARY PUZZLE game P of size 21N × 21N
replacing:
nodes x
i
, ¯x
i
and edge (x
i
, ¯x
i
) with a Variable gadget X
i
;
each split node y
i
j
i
j
(with the input interface cell
I
1
directed toward the corresponding source variable);
each clause c
j
j
;
the (orthogonal) edges with the corresponding Edge gadgets.
5.1 BARE BINARY PUZZLE is NP–hard
Theorem 5.1. The decision problem BARE BINARY PUZZLE is NP–hard.
Proof. First, we notice that each step of the construction of P (degree reduction,
orthogonal representation and equivalent game construction) can be performed
in polynomial time and space.
We must prove that the resulting game P has a solution if and only if the
original planar 3CNF formula F has a solution.
8
() Suppose that the game P has a solution, then every Clause gadget has
at least one input interface cell containing 1, which is also the output cell of
an Edge gadget; but if an output cell of an Edge gadget is 1 then its input cell
must be a 1, too. If we follow the Edge gadgets we can reach one (or more)
Split gadget, but, again, if an output cell of a Split gadget is 1 its input cell
must be 1, too. At the end we reach one of the two output interface cells of a
i
and it must contain 1. We set the associated variable x
i
to
the corresponding truth value. Each Clauses can generate only one assignment
(T
i
= 1 (exclusive) or F
i
= 1); and in this way we can ﬁnd an assignment that
satisﬁes all the clauses of the formula F . If a Clause gadget is not reached by
the above procedure, the corresponding variables can be set arbitrarily.
() In a similar manner if there is a truth assignment that satisfy the
formula F , we can set the output interface cell of Variable gadget X
i
in this
way: if x
i
= true then T
i
= 1, F
i
= 0 else T
i
= 0, F
i
= 1, and there is a valid
conﬁguration for their inner empty cells. The output cells of the Split gadgets
can be set according to the value of their input cells (if I1 = 1 then O1 = O2 = 1
else O1 = O2 = 0, and this guarantees that there is a valid conﬁguration for
their inner empty cells. The same can be applied to Edge gadgets. Finally every
clause in the formula is made true by the assignment, and this guarantees that
at least one of the input cells of each Clause gadget is set to 1 and all of them
have a valid conﬁguration.
5.2 BINARY PUZZLE is NP-complete
Theorem 5.2. The decision problem BINARY PUZZLE is NP–hard.
Proof. We must prove that adding the rules:
R2. each row and each column should contain an equal number of zeros and
ones;
R3. each row is unique and each column is unique.
don’t change the diﬃculty of the problem (in particular don’t make it easier).
We can use a direct reduction from BARE BINARY PUZZLE using two tricks
that transform a BARE BINARY PUZZLE instance B into a BINARY PUZZLE in-
stance P .
R2. Equal number of zeros and ones in each row and column.
If B is the given BARE BINARY PUZZLE n ×n starting grid, let B be the same
grid with the ﬁxed 0s and 1s inverted. We can build a new 2(n+2)×2(n+2) grid
P placing a copy of B on the top–left (B
1
) and bottom–right (B
4
of P , and a copy of B on the top–right (B
2
) and bottom–left (B
3
of P . Between the four grids there is a frame of empty cells of width 4; we call
F
1
, F
2
, F
C
, F
3
, F
4
the ﬁve parts of the frame (see Figure 10).
9
B
1
F
1
B
2
F
2
F
C
F
3
B
3
F
4
B
4
Figure 10: The grid P obtained by cloning the original starting grid B.
It is easy to see that if P has a valid solution, then B has a valid solution:
just keep the top–left n × n portion (B
1
) of the solution of P .
Vice versa if B has a valid solution S, then we can build a valid solution for
P that doesn’t violate rule R2 in the following way:
put the values of S on top–left, bottom–right quadrants (S
1
, S
4
);
put the inverted values of S on top–right, bottom–left quadrants (
¯
S
2
,
¯
S
3
),
extend S
1
to the right (over frame F
1
) placing an inverted copy and a copy
of its last column (x
1
, ..., x
n
):
x
1
¯x
1
x
1
y
1
x
2
¯x
2
x
2
y
2
... ... ... ...
S
1
... F
1
S
2
... ... ... ...
x
n
¯x
n
x
n
y
n
extend S
2
to the left (over frame F
1
) placing an inverted copy and a copy
of its ﬁrst column (y
1
, ..., y
n
):
x
1
¯x
1
x
1
y
1
¯y
1
y
1
x
2
¯x
2
x
2
y
2
¯y
2
y
2
... ... ... ... ... ...
S
1
... F
1
... ... S
2
... ... ... ... ... ...
x
n
¯x
n
x
n
y
n
¯y
n
y
n
ﬁll frames F
2
, F
3
, F
4
in a similar manner;
ﬁll the central 4 × 4 frame F
C
inverting its borders (see Figure 11).
The resulting grid is a valid solution for P because it generates no more
than two consecutive numbers next to or below each other; and by construction
it contains an equal number of 0s and 1s in each row and column. See ﬁgure
Figure 12 for a full example of the construction.
10
x ¯x x y ¯y y
¯x x ¯x ¯y y ¯y
x ¯x x y ¯y y
z ¯z z w ¯w w
¯z z ¯z ¯w w ¯y
z ¯z z w ¯w w
Figure 11: Filling the central frame F
C
: the conﬁguration is valid for all values
of x, y, z, w {0, 1}.
R3. Uniqueness of each row and column.
In order to complete the reduction from BARE BINARY PUZZLE to BINARY
PUZZLE we trasform P to an equivalent grid P
0
with distinct rows and columns.
In order to do this we ﬁrst extend the 2(n + 2) × 2(n + 2) grid P to its right
and to its bottom adding two more empty columns and two more empty rows.
Then we add to its right and to its bottom dlog
n+3
2
e unique sequences of binary
strings, where each digit is represented with two ﬁxed 2 × 2 digit blocks (see
Figure 13 and Figure 14). Each 2 × 2 block contain exactly two 1s and two 0s,
so they keep an equal number of 0s and 1s (don’t break rule R2).
Again, if P
0
has a solution, then it can be easily transformed to a valid
solution for the original BARE BINARY PUZZLE game B (just keep the upper–left
Vice versa if B has a solution, it can be transformed to a valid solution S
0
for
P
0
: use the same construction used for P and extend it to the new two columns
and the new two rows with the same technique used for extending S
1
: make
a double inversion of the last column and the last row. This guarantees that
there are no conﬂicts with the adjacent numbers in the ﬁxed digits blocks (no
more than two consecutive 0s or 1s). By construction the two new columns and
rows have an equal number of 0s and 1s and the same applies to the sequences
of ﬁxed digits; so S
0
satisﬁes rule R2. Furthermore the sequences of ﬁxed digits
make each row and column unique and thus S
0
satisﬁes rule R3, too. So S
0
is
a valid solution for the BINARY PUZZLE problem P
0
(see ﬁgure Figure 12 for a
full example of the construction of the solution of P
0
).
Corollary 5.3. The decision problem BINARY PUZZLE is NP-complete.
Proof. The input of the problem is a string of length n
2
; a solution of the game
can also be represented with a string of the same length and its validity can
be checked in linear time O(n
2
), so BINARY PUZZLE is contained in NP and
NP–hard, thus it is NP–complete .
6 Conclusion
Is this proof correct? .... 010110010011001101010011 !?!
11
Figure 12: The construction of a Binary 12 × 12 puzzle board solution with
equal number of 0s and 1s starting from a 4 × 4 Bare Binary puzzle solution
(upper–left box).
0 1
1 0
1 0
0 1
Figure 13: 2 × 2 ﬁxed blocks representing digit 0 and digit 1.
D0 ... D0 D0
D0 ... D0 D1
P D0 ... D1 D0
... ... ...
... ...
... D1 ... D1 D1
D0 D1 ... ... D1 D1 D0 ... D0
D0 D0 ... ... D1 D0 D1 ... D0
... ... ... ... ...
D0 D0 ... ... D1 D0 D0 ... D1
Figure 14: The grid P
0
built from P adding ﬁxed blocks (D0, D1) to make each
row and column unique.
12
Figure 15: The ﬁnal expansion of the solution of a BARE BINARY PUZZLE prob-
lem with a sequence of 2 × 2 ﬁxed “digits” that make each column and row
unique (rule R3) and thus make it a valid solution for the reduced BINARY
PUZZLE problem, too.
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13
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