Math 3124
Tuesday, September 6
September 6 Ungraded Homework
The first position is
8
10
7
9
6
12
5
11
4
14
3
13
2
1
15
. We first move the space to the bottom right hand
corner; there will be several ways of doing this, and it doesn’t matter which we choose. One
way will give the position
8
10
7
9
6
12
5
11
4
14
3
13
2
1
15
. The corresponding permutation is
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
8
10
7
9
6
12
5
11
4
14
3
13
2
1
15
Writing this as a product of disjoint cycles, we obtain
(
1 8 11 3 7 5 6 12 13 2 10 14
)(
4 9
)
The 12cycle is a product of 11 (
=
12

1) transpositions, so the whole permutation is a
product of 11
+
1 transpositions. 12 is even, so we conclude that the given position can be
obtained from the starting position.
The second position is
1
2
3
4
8
7
6
5
9
10
11
12
15
14
13
. We first move the space to the bottom right hand
corner; there will be several ways of doing this, and it doesn’t matter which we choose. One
way will give the position
1
2
3
4
8
7
6
5
9
10
11
12
15
14
13
. The corresponding permutation is
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 PARRY
 Math, Algebra, Group Theory, Position, Prime number, Symmetric group

Click to edit the document details