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The Mathematics of
Professor Alan's Puzzle Square
part 5 – odd permutations
https://magisoft.co.uk/alan/misc/game/game.html
I said there were two
families of 3x3 square,
and I’ve labelled them
even and odd.
We’ll use Group Theory
to prove this is true!
1 2 3
4 5 6
7 8 9
Even and odd
+839
other squares
+839
other squares
2 1 3
4 5 6
7 8 9
even
odd
Permutations
Recall that the puzzle is a form
of permutation group – it swops
around (permutes) the tiles.
Let’s look at the permutation
caused by on the first column.
1 2 3
4 5 6
7 8 9
4 2 3
7 5 6
1 8 9
42 3 75 61 8 9
Cycles
Most of the tiles are left where they are;
so we can just write this as a cycle:
Where this is read as:
Tile 1 goes to the old position of 7
Tile 4 goes to the old position of 1
Tile 7 goes to the old position of 4.
42 3 75 61 8 9
1
4 71
1
=
=
Let’s see all the moves as permutations:
The down and right arrows are just the opposite
way round.
Cycles
2 31
1
=
8 97
3
=
5 64
2
=
4 71
1
=
6 93
3
=
5 82
2
=
Notice that all the puzzle square
moves make cycles of length 3,
because they are moving a row
or column of three tiles.
In the 4x4 square they are of length 4,
and in the 5x5 square all length 5, etc.
All threes
2 31
1 =
8 97
3 =
5 64
2 =
4 71
1 =
6 93
3 =
5 82
2 =
Let’s go even simpler a 2x2 puzzle.
It won’t take you long to solve this
one!
Like the 3x3 puzzle, look at the at
the permutation caused by
A simple swop.
Just two
1 2
3 4
3 2
1 4
31
All the basic 2x2 puzzle square moves are swops
All swops
1 = 31
1 =
2 = 42
2 =
1 = 21
1 =
2 = 43
2 =
= (1,3)
= (2,4)
= (1,2)
= (3,4)
1 2
3 4
You may have noticed I slipped in some new
notation – using (1,3) for the permutation that
swops position 1 and 3.
This is partly because it is easier to write (!) and
partly because it is important that it is the
positions, not the specific tiles, that are swopped.
New notation
1 = 31
1 = = (1,3)
Combining permutations
The tile vs. position distinction is
important when we combine
permutations.
Let’s look at followed by
(1,3) + (3,4) = (1,4,3)
Note that by the time the second
swop happens the old tile at
position 1 is now at position 3, so
swops with the 4 tile.
1 2
3 4
3 2
4 1
1 2 3 2
1 4
All the swops
There are six possible swops:
(1,2), (1,3),(1,4),(2,3),(2,4),(3,4)
The basic moves give four of these,
and we can combine these to give
the remaining diagonal swops.
1 2
3 4
4 2
3 1
3 2
1 4
1 2
+ +
1
= (1,3) + (3,4) + (1,3) = (1,4)
3 2
4 1
1 1
+ +
1
= (1,3) + (1,2) + (1,3) = (2,3)
Swops are everything!
If you can do every swop, you can build any
permutation. Let’s see how …
Start: ABCDE Target: AEBCD = (2,3,4,5)
position current =target? action result
1: ABCDE A OK ABCDE
2: ABCDE needs E swop (2,5) AECDB
3: AECDB needs B swop (3,5) AEBDC
4: AEBDC needs E swop (4,5) AEBCD ✔
You can’t always do everything!
Remember the symmetry group of the triangle.
A
B C
B
C A
rotational symmetry reflection symmetry
A
B C
A
C B
1 identity
2 rotations
3 reflections
Exactly 6 symmetries of a triangle
B
C A
A
B C
A
C B
C
A B
B
C A
C
A B
(A,C,B) (A,B,C)
(B,C) (A,B) (A,C)
permutations
underneath
This is called the rotational group of the triangle.
We only have the identity and cycles of 3:
(), (A,C,B) (A,B,C)
No swops!
What if we just have rotations?
B
C A
A
B C
C
A B
(A,C,B) (A,B,C)
In Group Theory, this is
called C3 – the cyclic
group of order 3
Can’t we be clever?
For the 4x4 puzzle square, we made a swop out
of other moves … can’t we be clever now?
Every 3 cycle is a combination of 2 swops:
(A,C,B) = (A,C)+(A,B) (A,B,C) = (A,B)+(A,C)
check it!
A permutation that is made of an even number
of swops is called an even permutation.
If you combine two even permutations you get
another even one, and in general:
even + even = even even + odd = odd
odd + even = odd odd + odd = even
Just like even an odd numbers!
Parity
Proving it
I won’t do the proof here (find a Group Theory
book!), but one way is roughly as follows:
(i) break the permutations into individual swops
(ii) sort the swops
(iii) sometimes two identical swops meet and
they annihilate one another!
Step (iii) always takes two out, so the even
/oddness of the number of swops stays the same.
So, if we only have
rotations as moves,
we have two worlds of
triangles …
… an even world and an
odd world.
Parallel universes that
never meet!
Even and odd even
odd
B
C A
A
B C
A
C B
C
A B
B
C A
C
A B
In the 3 x 3 puzzle all the basic moves were of
length 3 – they were even permutations.
That means no matter how many moves we
make, we will never get an odd permutation.
However, swopping the first two
tiles is an odd permutation.
It is forever unreachable
using the arrows alone.
Back to the 3x3 puzzle?
2 1 3
4 5 6
7 8 9
Just like the rotations
of a triangle, there are
two worlds of 3x3
puzzle squares …
So not every square is
solvable … but the
‘scramble’ button in
the online puzzle
square only chooses
ones that are!
1 2 3
4 5 6
7 8 9
Two worlds
+839
other squares
+839
other squares
2 1 3
4 5 6
7 8 9
even
odd
Why is it OK for 4x4
For the 4x4 square we were able to
make a simple swop by combining
a simple row right move:
With the series of moves to rotate
three tiles:
The latter is an even permutation
just like for 3x3), but the simple right arrow
cycles 4 tiles, it is an odd permutation .
In general …
For even sided squares (2x2, 4x4, 6x6, …)
The simple arrows (the generators) are odd
permutations, and any square is reachable.
For even sided squares (3x3, 5x5, 7x7, …)
The simple arrows are even permutations, and
so there are two families of squares: odd and
even, forever cut off from each other.
Other puzzles
Group Theory can help with the sliding tile
puzzles you can get in a Christmas cracker.
This one has its last two tiles
the wrong way round …
an odd permutation.
It turns out all the moves for the
4x4 sliding puzzle are even permutations.
Don’t bother – this puzzle is unsolvable!
By Micha L. Rieser - Own work, Public Domain, https://commons.wikimedia.org/w/index.php?curid=5016278
Lots that is new
We have covered a lot of ground!
1. We’ve seen the way permutations can be written as
cycles and the (A, B, C) notation for them
2. The simple swop from which you can build anything
3. Counting swops to give odd and even permutations
4. The rotational group of the triangle
5. Which has the same elements as the symmetry
group, but not all of the moves In Group Theory
terms, a subgroup

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Odd Permutations - Part 5 of The Mathematics of Professor Alan's Puzzle Square

  • 1. The Mathematics of Professor Alan's Puzzle Square part 5 – odd permutations https://magisoft.co.uk/alan/misc/game/game.html
  • 2. I said there were two families of 3x3 square, and I’ve labelled them even and odd. We’ll use Group Theory to prove this is true! 1 2 3 4 5 6 7 8 9 Even and odd +839 other squares +839 other squares 2 1 3 4 5 6 7 8 9 even odd
  • 3. Permutations Recall that the puzzle is a form of permutation group – it swops around (permutes) the tiles. Let’s look at the permutation caused by on the first column. 1 2 3 4 5 6 7 8 9 4 2 3 7 5 6 1 8 9 42 3 75 61 8 9
  • 4. Cycles Most of the tiles are left where they are; so we can just write this as a cycle: Where this is read as: Tile 1 goes to the old position of 7 Tile 4 goes to the old position of 1 Tile 7 goes to the old position of 4. 42 3 75 61 8 9 1 4 71 1 = =
  • 5. Let’s see all the moves as permutations: The down and right arrows are just the opposite way round. Cycles 2 31 1 = 8 97 3 = 5 64 2 = 4 71 1 = 6 93 3 = 5 82 2 =
  • 6. Notice that all the puzzle square moves make cycles of length 3, because they are moving a row or column of three tiles. In the 4x4 square they are of length 4, and in the 5x5 square all length 5, etc. All threes 2 31 1 = 8 97 3 = 5 64 2 = 4 71 1 = 6 93 3 = 5 82 2 =
  • 7. Let’s go even simpler a 2x2 puzzle. It won’t take you long to solve this one! Like the 3x3 puzzle, look at the at the permutation caused by A simple swop. Just two 1 2 3 4 3 2 1 4 31
  • 8. All the basic 2x2 puzzle square moves are swops All swops 1 = 31 1 = 2 = 42 2 = 1 = 21 1 = 2 = 43 2 = = (1,3) = (2,4) = (1,2) = (3,4) 1 2 3 4
  • 9. You may have noticed I slipped in some new notation – using (1,3) for the permutation that swops position 1 and 3. This is partly because it is easier to write (!) and partly because it is important that it is the positions, not the specific tiles, that are swopped. New notation 1 = 31 1 = = (1,3)
  • 10. Combining permutations The tile vs. position distinction is important when we combine permutations. Let’s look at followed by (1,3) + (3,4) = (1,4,3) Note that by the time the second swop happens the old tile at position 1 is now at position 3, so swops with the 4 tile. 1 2 3 4 3 2 4 1 1 2 3 2 1 4
  • 11. All the swops There are six possible swops: (1,2), (1,3),(1,4),(2,3),(2,4),(3,4) The basic moves give four of these, and we can combine these to give the remaining diagonal swops. 1 2 3 4 4 2 3 1 3 2 1 4 1 2 + + 1 = (1,3) + (3,4) + (1,3) = (1,4) 3 2 4 1 1 1 + + 1 = (1,3) + (1,2) + (1,3) = (2,3)
  • 12. Swops are everything! If you can do every swop, you can build any permutation. Let’s see how … Start: ABCDE Target: AEBCD = (2,3,4,5) position current =target? action result 1: ABCDE A OK ABCDE 2: ABCDE needs E swop (2,5) AECDB 3: AECDB needs B swop (3,5) AEBDC 4: AEBDC needs E swop (4,5) AEBCD ✔
  • 13. You can’t always do everything! Remember the symmetry group of the triangle. A B C B C A rotational symmetry reflection symmetry A B C A C B
  • 14. 1 identity 2 rotations 3 reflections Exactly 6 symmetries of a triangle B C A A B C A C B C A B B C A C A B (A,C,B) (A,B,C) (B,C) (A,B) (A,C) permutations underneath
  • 15. This is called the rotational group of the triangle. We only have the identity and cycles of 3: (), (A,C,B) (A,B,C) No swops! What if we just have rotations? B C A A B C C A B (A,C,B) (A,B,C) In Group Theory, this is called C3 – the cyclic group of order 3
  • 16. Can’t we be clever? For the 4x4 puzzle square, we made a swop out of other moves … can’t we be clever now? Every 3 cycle is a combination of 2 swops: (A,C,B) = (A,C)+(A,B) (A,B,C) = (A,B)+(A,C) check it! A permutation that is made of an even number of swops is called an even permutation.
  • 17. If you combine two even permutations you get another even one, and in general: even + even = even even + odd = odd odd + even = odd odd + odd = even Just like even an odd numbers! Parity
  • 18. Proving it I won’t do the proof here (find a Group Theory book!), but one way is roughly as follows: (i) break the permutations into individual swops (ii) sort the swops (iii) sometimes two identical swops meet and they annihilate one another! Step (iii) always takes two out, so the even /oddness of the number of swops stays the same.
  • 19. So, if we only have rotations as moves, we have two worlds of triangles … … an even world and an odd world. Parallel universes that never meet! Even and odd even odd B C A A B C A C B C A B B C A C A B
  • 20. In the 3 x 3 puzzle all the basic moves were of length 3 – they were even permutations. That means no matter how many moves we make, we will never get an odd permutation. However, swopping the first two tiles is an odd permutation. It is forever unreachable using the arrows alone. Back to the 3x3 puzzle? 2 1 3 4 5 6 7 8 9
  • 21. Just like the rotations of a triangle, there are two worlds of 3x3 puzzle squares … So not every square is solvable … but the ‘scramble’ button in the online puzzle square only chooses ones that are! 1 2 3 4 5 6 7 8 9 Two worlds +839 other squares +839 other squares 2 1 3 4 5 6 7 8 9 even odd
  • 22. Why is it OK for 4x4 For the 4x4 square we were able to make a simple swop by combining a simple row right move: With the series of moves to rotate three tiles: The latter is an even permutation just like for 3x3), but the simple right arrow cycles 4 tiles, it is an odd permutation .
  • 23. In general … For even sided squares (2x2, 4x4, 6x6, …) The simple arrows (the generators) are odd permutations, and any square is reachable. For even sided squares (3x3, 5x5, 7x7, …) The simple arrows are even permutations, and so there are two families of squares: odd and even, forever cut off from each other.
  • 24. Other puzzles Group Theory can help with the sliding tile puzzles you can get in a Christmas cracker. This one has its last two tiles the wrong way round … an odd permutation. It turns out all the moves for the 4x4 sliding puzzle are even permutations. Don’t bother – this puzzle is unsolvable! By Micha L. Rieser - Own work, Public Domain, https://commons.wikimedia.org/w/index.php?curid=5016278
  • 25. Lots that is new We have covered a lot of ground! 1. We’ve seen the way permutations can be written as cycles and the (A, B, C) notation for them 2. The simple swop from which you can build anything 3. Counting swops to give odd and even permutations 4. The rotational group of the triangle 5. Which has the same elements as the symmetry group, but not all of the moves In Group Theory terms, a subgroup