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A Group Theoretic Solution to the Alien Tiles
Puzzle
Glenn C. Rhoads
3651 Karen Drive
Chesapeake Beach, MD 20732
email: rhoads@cs.rutgers.edu
1 Introduction
Alien Tiles is an on-line puzzle game invented by Dr. Cliff Pickover and is
located at http://www.alientiles.com/ You are given a square board of tiles,
each of which could be any one of the four colors which are ordered in the cycle
Red →Green →Blue →P urpl e →Red. Clicking on a tile rotates the color
of the selected tile and of every tile in the same row and column. The puzzle is
to find a sequence of clicks that transforms a starting state to a goal state. For
ease of description, I’ll assign the following integers to the colors; 0 for red, 1
for green, 2 for blue, and 3 for purple. With this assignment, the state of any
tile rotates through the sequence 0 →1→2→3→0. For uniformity with
how the game is played at the on-line site, I’ll assume that a goal position is
specified and that the starting position is chosen randomly.
2 Basic Facts
Clicking on a particular tile say xtimes will rotate the color of all tiles in the
same row and column xtimes regardless of how these xclicks are interspersed
with other clicks. Thus,
Fact 1 The order of clicks is irrelevant.
Hence, we need to be concerned only with how many times to click on each
tile. The number of clicks reduces to arithmetic modulo the number of colors,
four. Hence, to solve a puzzle, no tile ever needs to be clicked on more than
three times.
Definition 1 A Short Solution is a solution in which no tile is clicked on more
than 3 times.
Definition 2 Two solutions or patterns of clicks are Short Equivalent if for
each tile, the number of clicks is equivalent modulo four.
1
Short equivalence partitions the solutions into equivalence classes and each
class has at least one short solution. Intuitively, the short solutions are the ones
using a small, though not necessarily minimal, number of clicks and these are
the ones we want to find.
From now on, let nbe the dimension of our square board. A n×nboard
has n2tiles. Each tile is one of four possible colors and each tile is clicked on
0, 1, 2, or 3 times in a short solution. Hence, a n×nboard has 4n2starting
positions and 4n2possible short solutions. Thus,
Fact 2 For all board sizes, all starting positions are solvable if and only if each
position has a unique short solution.
Theorem 1 For odd n, some starting positions are not solvable.
Proof:
Case 1:n= 1 mod 4. Clicking once on every tile in the same row advances
every tile on the board forward by 1 mod 4.
Case 2:n= 3 mod 4. Clicking twice on every tile in the same row advances
every tile on the board forward by 2 mod 4.
Thus in both cases, some positions have more than one short solution which by
fact 2 implies the some positions do not have a solution.
3 Solving the Puzzle
Theorem 2 For even n, all positions are solvable.
Proof:
Case 1:n= 0 mod 4
You can independently cycle any tile forward by 1 mod 4 as follows. Click on the
particular tile three times, click on all other tiles in the same row and column
once, and click on all other tiles twice. By examining the cumulative effects of
these clicks on the various tiles, we see that the particular tile will cycle forward
by 1 mod 4 and that all other tiles will remain as they were.
Case 2:n= 2 mod 4
You can independently cycle any tile forward by 1 mod 4 as follows. Click on the
particular tile three times, click on all other tiles in the same row and column
three times, and click on all other tiles twice.
The sets of clicks described in the proof of Theorem 2 provide a method
for solving any even order puzzle. For any tile that is not equal to its goal
value, simply apply the described set of clicks the appropriate number of times.
Obviously, you can repeat this for each tile.
Corollary 1 For even n, all board positions have a unique short solution.
2
4 Case: n = 1 mod 4
Definition 3 A Quarter Identity (QI) is a pattern of clicks where each tile in
a row or column is clicked once.
Theorem 3 A pattern of clicks is an identity transformation if and only if it
is composed of X quarter identities where X = 0 mod 4.
Proof:
(⇐=) A QI advances each tile on the board once. Thus, any number of QIs, X,
where Xis a multiple of 4 is an identity transformation.
(=⇒) Let matrix Abe a pattern of clicks for the identity. Let Ai,j denote the
i,j’th element A. Pick a row and column of A, say row pand column q.
For i= 1,2,3, . . . , n, click on all entries in column i,Ap,i times.
For j= 1,2,3, . . . , n, click on all entries in row j, (Aj,q −Ap,q mod 4) times.
The number of QIs performed is X= the sum of the p’th row of A, plus the
sum of q’th column of A, minus Ap,q .Xis also the number of clicks that effect
tile p, q which must be 0 mod 4 since Aproduces the identity. Hence, we have
performed X= 0 mod 4 QIs. Showing the following claim will complete the
proof.
Claim: We have constructed all of A. In other words, Abeing the identity
forces the remaining clicks uniquely.
We have clearly constructed row pand column qof A.
Let A′be the matrix of additional clicks on the remaining n−1×n−1 submatrix.
We need to show that A′
i,j = 0 mod 4 for all i, j. The sum of the additional
clicks in any row of A′and any column of A′must be 0 mod 4 in order to not
change the cell contents in row pand column q. More formally, for any A′
i,j :
n−1
X
k=1
A′
i,k = 0 mod 4 (1)
n−1
X
k=1
A′
k,j = 0 mod 4 (2)
Since Ais the identity; for any A′
i,j , the sum of the clicks in A′effecting
A′
i,j = 0 mod 4. More formally, for any A′
i,j ,
n−1
X
k=1
[A′
i,k] +
n−1
X
k=1
[A′
k,j ]−A′
i,j = 0 mod 4 (3)
Substituting for 1 and 2 into 3, we get A′
i,j = 0 mod 4.
3
Theorem 4 There are exactly 42n−2identity transformations.
Proof:
Since we need consider only short solutions, there are 42ndistinct ways of se-
lecting QIs. Only for 1/4 of them, do we have a multiple of 4.
But counting all these selections of QIs would count each identity transforma-
tion four times. To see this, suppose we have some selection of QIs producing
the identity. For each row, let rbe the number of corresponding QIs selected for
that row and for each column, let cbe the number of corresponding QIs. Now
for each row, select r+xmod 4 QIs and for each column, select c−xmod 4
QIs. We will get a distinct selection of QIs for each of the four values of xfrom
0 through 3. Clearly the number of clicks on each tile will be the same for each
such selection (since we add xclicks for the row it is in and subtract xclicks for
its column). I.e. All four of these distinct selections of QIs correspond to the
same pattern of clicks.
Furthermore, this is the only way to achieve a distinct selection of QIs cor-
responding to the same pattern of clicks. If we change the number of QIs
corresponding to any row say r(or equivalently any column c), then in order
to have the same number of clicks in each entry of row r(col c), we must must
make the opposite change to every col (row) which in turn implies that we must
make the same change to every row (col).
In summation, we divide out two separate factors of four from our total of 42n
ways to get 42n−2distinct identity transformations.
Corollary 2 There are exactly 4n2−2n+2 solvable positions.
Our solution method when n= 1 mod 4 is as follows. First consider the even-
order square subboard consisting of the first n−1 rows and the first n−1 columns.
We can fix this subboard any way we like by using the solution algorithm for
the 0 mod 4 case (with all clicks restricted to the subboard). Once we’ve fixed
this, we can then independently fix entry n, n. This fixes everything except
the remaining 2(n−1) entries in row nand column n. But by Theorem 4,
once we fix all but 2n−2 = 2(n−1) entries, these final entries will be in fact
completely determined and thus, we have solved the puzzle (or else the puzzle
was insolvable).
5 Case: n = 3 mod 4
Definition 4 A Half Identity (HI) is a pattern of clicks where each tile in a
row or column is clicked twice.
Lemma 1 If a pattern of clicks is composed of X = 0 mod 2 HIs, then it is an
identity transformation.
4
Proof:
A HI advances each tile on the board twice. Thus, any number of HIs, X, where
X is a multiple of 2 is an identity transformation.
Lemma 2 There are at least 4n−1identity transformations.
Proof:
Since we need consider only short solutions, there are 22ndistinct ways of se-
lecting HIs. Only for 1/2 of them, do we have a multiple of 2. But this counts
each such identity transformation two times. The proof of this is almost iden-
tical to the corresponding proof in Theorem 4 and is thus omitted for brevity.
Thus, we divide out two separate factors of two from the total of 22nways to
get 22n−2= (22)n−1= 4n−1distinct identity transformations.
Corollary 3 There are at most 4n2/4n−1= 4n2−n+1 solvable positions.
Theorem 5 When n = 3 mod 4, every tile can be independently cycled forward
by two colors.
Proof:
Click on the particular tile twice, click on all other tiles in the same row and
column once, and click on all other entries twice. By examining the cumulative
effects of these clicks on the various tiles, we see that the particular tile will
cycle forward by 2 mod 4 and that all other tiles will remain uneffected.
Definition 5 Two positions are two-transformable if one position can be changed
into the other by advancing some subset of tiles by two.
Two positions are two-transformable if and only if they have the same pat-
tern of odds and evens. Now we only have to worry about achieving a desired
pattern of 0s and 1s. Any other attainable position P can be achieved by first
reaching the position of 0s and 1s that is two-transformable to P and then by
applying the necessary two transformations given in Theorem 5.
Our solution method when n= 3 mod 4 is as follows. First consider the
even-order square subboard consisting of the first n−1 rows and the first n−1
columns. We can fix this subboard into any pattern of 0s and 1s we like by
using the solution algorithm for the 2 mod 4 case (with all clicks restricted to
the subboard). Once we’ve fixed this, we can then independently fix entry n, n
to a 0 or a 1. We can then reach any two-transformable position, by cycling any
subset of the n2tiles forward by two.
How many positions can we solve by this method? We are first fixing (n−1)2
tiles in the subboard to one of two colors. We can reach 2((n−1)2)= 2n2−2n+1
positions this way. For each of them, there are two ways of fixing entry n, n
to a 0 or a 1. And for each way of doing both of the preceding, we can fix
5
each of the n2tiles in one of two ways (due to the two-transforms). There
are 2(n2)ways of doing this. Thus the total number of positions solvable using
this method is 2n2−2n+1 ·2·2(n2)= 22n2−2n+2 = (22)n2−n+1 = 4n2−n+1. By
corollary 3, this is the total number of solvable positions. Thus, this constitutes
a solution method. I.e., the remaining pattern of 0s and 1s that we did not
fixed is uniquely determined and hence must be set correctly as well, or else the
puzzle has no solution.
Corollary 4 There are exactly 4n2−n+1 solvable positions and exactly 4n−1
identity transformations.
Theorem 6 A pattern of clicks is an identity transformation if and only if it
is composed of X = 0 mod 2 HIs.
Proof: Since there are 4n−1identity transformations of this form and since
there are exactly 4n−1total identity transformations, there are no other identity
transformations.
For odd order boards, one way to determine if you can achieve the given goal
position is to fix every tile you can according to the solution method and check to
see if the remaining tiles are in fact set to the desired goal. The solution method
and the three basic transformations described in Theorem 2 and Theorem 5 do
imply a way to describe whether a given position is solvable but the resulting
condition is not particularly simple.
Let Abe the initial position, Bbe the goal position, and let C=B−A
where each entry is taken mod 4.
Theorem 7 Suppose n=1 mod 4. The puzzle is solvable if and only if the
following two conditions hold.
I For every i= 1,2,3,...,n−1,
Ci,n ≡
2
n−1
X
j=1
Ci,j
+
3
n−1
X
k=1,k6=i
n−1
X
j=1
Ck,j
+Cn,n
mod 4
II For every j= 1,2,3,...,n−1,
Cn,j ≡
2
n−1
X
i=1
Ci,j !+
3
n−1
X
i=1
n−1
X
k=1,k6=j
Ci,k
+Cn,n
mod 4
Theorem 8 Suppose n=3 mod 4. The puzzle is solvable if and only if the
following two conditions hold.
6
I For every i= 1,2,3,...,n−1,
Ci,n ≡
3·
n−1
X
k=1
n−1
X
j=1
Ck,j
+Cn,n
mod 2
II For every j= 1,2,3,...,n−1,
Cn,j ≡" 3·
n−1
X
i=1
n−1
X
k=1
Ci,k!+Cn,n #mod 2
6 Extension to Rectangular Boards
Let the dimension of a board be m×n. All of section 2 extends to the general
case except that there are now mn tiles on the board. Thus, all solvable positions
have a solution using at most 3mn clicks; the board has 4mn distinct positions;
and there are 4mn distinct short solutions.
6.1 Case: m and n are both even
The solutions for the even cases extend to handle all positions where both m
and nare 0 mod 4, and where both mand nare 2 mod 4.
Theorem 9 Suppose m = 0 mod 4 and n = 2 mod 4. Then any position is
achievable.
Proof: We can independently cycle the tile in row i, column jforward by one
mod 4 with the following pattern of clicks. Click three times on all tiles in
column jand click once on all tiles in row iexcept for the tile i, j .
The situation where m= 2 mod 4 and n= 0 mod 4 is symmetric to the
above and thus, every position is solvable when both dimensions are even.
6.2 Case: m and m are both odd
The solutions to the odd cases extend to handle all positions where both m
and nare 1 mod 4, and where both are 3 mod 4. So consider the case where
m= 3 mod 4 and n= 1 mod 4 (the case m= 1 mod 4 and n= 3 mod 4 is
obtained by symmetry).
Lemma 3 If a pattern of clicks is composed of Xcolumn HIs and Yrow QIs
such that either
(X = 0 mod 2 and Y = 0 mod 4) or
(X = 1 mod 2 and Y = 2 mod 4),
then it is an identity transformation.
7
Proof: A column HI advances each tile on the board twice and a row QI ad-
vances each tile on the board once. Thus, any number of column HIs, X, and
row QIs, Y, where (X= 0 mod 2 and Y= 0 mod 4) or (X= 1 mod 2 and
Y= 2 mod 4) is an identity transformation.
Lemma 4 There are at least 4m−12n−1identity transformations.
Proof:
There are 2ndistinct ways of selecting column HIs and 4mdistinct ways of
selecting row QIs. But only 1/4 of them satisfy the conditions of Lemma 3.
But this counts each identity transformation twice. To see this, suppose we
have some selection of row QIs and column HIs producing the identity. For
each row, let rbe the number of corresponding QIs selected for that row and
for each column, let cbe the number of corresponding HIs. Now for each row,
select r+ 2xmod 4 QIs and for each column, select c−xmod 2 HIs. We will
get a distinct selection of QIs and HIs for x= 0 and x= 1. Now for each tile,
we add 2xclicks for the row it is in and subtract 2xclicks for its column (since
there are two clicks for each column HI). Thus, both these distinct patterns of
QIs and HIs correspond to the same pattern of clicks.
Furthermore, this is the only way to achieve a distinct selection of row QIs
and column HIs corresponding to the same pattern of clicks. If we change the
number of HIs corresponding to any column say c, then in order to have the same
number of clicks in each entry of column c, we must must make the opposite
change (but twice as large) to every row which in turn implies that we must
make the same change to every column.
In summation, we divide out factors of four and two from the total of 4m2n
ways to get 4m−12n−1distinct identity transformations.
Corollary 5 There are at most 4mn /(4m−12n−1) = 22mn−2m−n+3 solvable po-
sitions when m=1 mod 4 and n=3 mod 4.
Theorem 10 For any pair of tiles in the same row, we can independently cycle
both of them forward by two.
Proof: Click once on the each tile in the pair, twice on all other tiles in the
same row, and once on all other tiles in the same column as one of the pair.
Our solution method when m= 3 mod 4 and n= 1 mod 4, is as follows.
First solve the upper-left m−1×n−1 subboard by applying the solution for when
m= 2 mod 4 and n= 0 mod 4. (restricting all clicks to the sub-board). Then
for each tile in row mexcept the one in column n, we can independently cycle
that tile plus the tile in column nforward by 2. Then we can set the tile in row m
8
column n. Thus, we can solve 4(m−1)(n−1) ·2n−1·4 = 4mn−m−n+1 ·2n−1·22=
22mn−2m−n+3 positions. By corollary 5, this is the total number of solvable
positions and hence, the remainder is forced.
Corollary 6 There are exactly 22mn−2m−n+3 solvable positions and exactly
4m−12n−1identity transformations.
Theorem 11 A pattern of clicks is an identity transformation if and only if it
is composed of Xcolumn HIs and Yrow QIs such that either
(X = 0 mod 2 and Y = 0 mod 4) or
(X = 1 mod 2 and Y = 2 mod 4),
Proof: There are 4m−12n−1identity transformations of this form and there are
exactly 4m−12n−1identity transformations in total, hence these are all of the
identity transformations.
The only remaining situation is when mis odd and nis even (meven and
nodd is symmetric). This breaks up into three distinct cases.
6.3 Case: m = 1 mod 4, n is even
Lemma 5 If a pattern of clicks is composed of X column QIs where X = 0 mod
4, then it is an identity transformation.
Proof:
A column QI advances each tile on the board once. Thus, any number of column
QIs, X, where Xis a multiple of 4 is an identity transformation.
Lemma 6 There are at least 4n−1identity transformations.
Proof:
There are 4ndistinct ways of selecting column QIs. Only for 1/4 of them, do
we have a multiple of 4 and furthermore, every such selection corresponds to a
distinct identity.
Corollary 7 There are at most 4mn−n+1 solvable positions.
We can solve the puzzle, when m= 1 mod 4 as follows. First solve the upper
m−1×nsubboard using the method from the corresponding both even case
(again with the clicks restricted to the sub-board). Then we can set the tile in
row mcolumn 1 to one of four distinct values. Doing this will cycle all the tiles
in column one forward by an equal amount. But re-solving the upper m−1 rows
from such a position will have no net effect on the entire bottom row! Thus, the
number of positions we can solve using this method is 4(m−1)n·4 = 4mn−n+1
which by Corollary 7 is the total number of solvable positions and hence, the
remainder is forced.
9
Corollary 8 There are exactly 4mn−n+1 solvable positions and exactly 4n−1
identity transformations.
Theorem 12 A pattern of clicks is an identity transformation if and only if it
is composed of X column QIs where X = 0 mod 4.
Proof: There are 4n−1identity transformations of this form and there are ex-
actly 4n−1identity transformations in total, hence these are all of the identity
transformations.
6.4 Case: m = 3 mod 4, n = 0 mod 4
Lemma 7 If a pattern of clicks is composed from either X column HIs where
X = 0 mod 2, or a row QI for every row plus Y column HIs where Y = 1 mod
2, then it is an identity transformation.
Proof:
A column HI advances every tile on the board twice and thus, Xcolumn HIs
where X= 0 mod 2 is an identity transformation. A row QI for every row
advances each tile on the board twice. Thus, a row QI for every row plus Y
column HIs where Y= 1 mod 2 is an identity transformation.
Lemma 8 There are at least 2nidentity transformations.
Proof:
There are 2ndistinct ways to select column HIs. For half of them we have
a multiple of two. Thus, there are 2n−1ways of selecting an even number of
column HIs and 2n−1ways of selecting an odd number of column HIs. There
is only one way to select a row QI for every row. Thus, there are at least
2n−1+ 2n−1= 2nidentity transformations.
Corollary 9 There are at most 4mn/2n= 22mn−nsolvable positions.
Theorem 13 We can cycle any two distinct tiles in the same row forward by
two.
Proof:
Click zero times on both of the particular tiles, click twice on all other tiles in
the same column as one of the pair, and click once on all the remaining tiles
(including those tiles in the same row as the pair). The cumulative effect of
these clicks is to cycle both tiles of the pair forward by two while all other tiles
remain unchanged.
To solve the puzzle when m= 3 mod 4 and n= 0 mod 4, first solve the top
m−1 rows using the solution method from the m= 2 mod 4, n= 0 mod 4 case
10
(as usual, all clicks are restricted to the subboard). For each tile in the bottom
row except for tile m, n, we can advance it and tile m, n forward by two mod
four. Then we have two choices for tile m, n, we can leave it as it is or we can
click on it twice. If we click on it twice, this will advance all tiles in row mand
column nby two. Now re-solving the first m−1 rows will also restore the bottom
row except for tile m, n which will be uneffected. Thus using this method we
can solve 4(m−1)n·2n−1·2 = 4mn−n·2n= 22mn−npositions. Corollary 9 implies
that this is the total number of solvable positions and hence, the remainder is
forced.
Corollary 10 There are exactly 22mn−nsolvable positions and exactly 2niden-
tity transformations.
Theorem 14 A pattern of clicks is an identity transformation if and only if it
is composed from either X column HIs where X = 0 mod 2, or a row QI for
every row plus Y column HIs where Y = 1 mod 2.
Proof: There are 2nidentity transformations of this form and there are ex-
actly 2nidentity transformations in total, hence these are all of the identity
transformations.
6.5 Case: m = 3 mod 4, n = 2 mod 4
Lemma 9 If a pattern of clicks is composed from either X column HIs where
X = 0 mod 2, or a row QI for every row plus X column HIs where X = 0 mod
2, then it is an identity transformation.
Proof:
A column HI advances every tile on the board twice and thus, Xcolumn HIs
where X= 0 mod 2 is an identity transformation. A row QI for every row is
an identity transformation. Thus, a row QI for every row plus Xcolumn HIs
where X= 0 mod 2 is an identity transformation.
Lemma 10 There are at least 2nidentity transformations.
Proof:
There are 2ndistinct ways to select column HIs. For half of them we have a
multiple of two. We can select zero row QIs for every row or one row QI for
every row which gives us a factor of two and hence, there are at least 2nidentity
transformations.
Corollary 11 There are at most 4mn/2n= 22mn−nsolvable positions.
Theorem 15 We can cycle any two distinct tiles in the same row forward by
two.
11
Proof:
Click once on one tile of the pair and click three times on the other tile. For the
tile that was clicked on once, click on all other tiles in the same column three
times. For the tile that was clicked on three times, click on all other tiles in the
same column once.
We can solve the puzzle when m= 3 mod 4 and n= 2 mod 4 using the
solution method for the m= 3 mod 4, n= 0 mod 4 case except that we instead
use the transformation described in the proof of Theorem 15 to cycle a pair
of tiles forward by two. As before we can solve 22mn−npositions which by
Corollary 11 is the total number of solvable positions and hence, this constitutes
a solution method.
Corollary 12 There are exactly 22mn−nsolvable positions and exactly 2niden-
tity transformations.
Theorem 16 A pattern of clicks is an identity transformation if and only if it
is composed from either X column HIs where X = 0 mod 2, or a row QI for
every row plus X column HIs where X = 0 mod 2.
Proof: There are 2nidentity transformations of these forms and there are ex-
actly 2nidentity transformations in total, hence these are all of the identity
transformations.
7 Obtaining a Short Solution
Suppose we want to actually solve a puzzle on the on-line site. Solving the
puzzle by repeatedly applying the patterns of clicks needed to transform the in-
dividual tiles is very inefficient in terms of the number of clicks and is very
tedious to perform in practice. But each solution has at least one equiva-
lent short solution and it is straightforward to compute one. Since we know
which tiles have to be cycled forward and by how much, we can simply add
up the necessary patterns of clicks and then for each tile reduce the total mod
4 to get a corresponding short solution. We can write a computer program
to automate this process. Writing such a program is a straightforward ex-
ercise and in fact, the author has written a program for the square boards
that are used at the Alien Tiles site. This program is available from the page
http://mysite.verizon.net/vze16nctz/GamesPuzzles/gp.html
8 Acknowledgements
The author is indebted to the following people for their helpful discussion on
the rec.puzzles newsgroup during May of 2000 which was shortly after the Alien
Tiles puzzle was put online. Dennis Yelle for Fact 2, Theorem 4, and Corollary 2,
and Nis Jorgensen for the his version of the proof of Theorem 3.
12
