10. Peg Solitaire
Robert Snapp
[email protected]
Department of Computer Science
University of Vermont
Robert R. Snapp © 2012
10. Peg Solitaire
CS 32, Fall 2012
1 / 19
The Puzzle
Five cylindrical holes in a row.
Two red pegs.
Two blue pegs.
The Initial State
The Goal State
Robert R. Snapp © 2012
10. Peg Solitaire
CS 32, Fall 2012
2 / 19
Four possible moves
1. A red peg can step one position to the right into the empty space.
Robert R. Snapp © 2012
10. Peg Solitaire
CS 32, Fall 2012
3 / 19
Four possible moves
2. A red peg can jump two positions to the right into the empty space.
Robert R. Snapp © 2012
10. Peg Solitaire
CS 32, Fall 2012
4 / 19
Four possible moves
3. A blue peg can step one position to the left into the empty space.
Robert R. Snapp © 2012
10. Peg Solitaire
CS 32, Fall 2012
5 / 19
Four possible moves
4. A blue peg can jump two positions to the left into the empty space.
Robert R. Snapp © 2012
10. Peg Solitaire
CS 32, Fall 2012
6 / 19
Questions
1
How many states (i.e. peg configurations) exist?
2
Which peg configurations can be obtained using a sequence of legal moves?
3
Is the goal configuration accessible? (Is it possible to solve this puzzle?)
4
If the puzzle can be solved, how many solutions exist?
5
If the puzzle can be solved, what is the shortest solution?
6
Can the puzzle be generalized?
7
Can you think of any other questions?
Robert R. Snapp © 2012
10. Peg Solitaire
CS 32, Fall 2012
7 / 19
Data Representation
We can use a list to represent each state of the puzzle.
A red R will represent a red peg.
A blue B will represent a blue peg.
An black E will represent the empty space.
A list of five symbols will represent the state.
The initial state is thus
(R R E B B).
The four successors are
(E R R B B), (R E R B B), (R R B E B), (R R B B E).
The goal state is thus
(B B E R R).
Robert R. Snapp © 2012
10. Peg Solitaire
CS 32, Fall 2012
8 / 19
Enumerating the states of the four-peg puzzle
For every unique state of the four-peg puzzle, there exists a unique list containing two
Rs, two Bs, and one E.
Thus there exists a one-to-one correspondence between the set of states of the
puzzle, and the set of lists (containing two Rs, two Bs, and one E); and thus, the
number of state equals the number five-letter anagrams that can be formed using two
Rs, two Bs, and one E:
5Š
D 30:
2Š 2Š 1Š
It is not clear at this point, however, if all 30 states can be reached from the initial
state by following a sequence of legal operations.
Robert R. Snapp © 2012
10. Peg Solitaire
CS 32, Fall 2012
9 / 19
The Four Operations
Each of the four legal operations can be expressed symbolically:
Assuming we begin in the initial state:
1
Red Step
(R R E B B) ! (R E R B B):
2
Red Jump
(R R E B B) ! (E R R B B):
3
Blue Step
(R R E B B) ! (R R B E B):
4
Blue Jump
(R R E B B) ! (R R B B E):
Robert R. Snapp © 2012
10. Peg Solitaire
CS 32, Fall 2012
10 / 19
The Four Operations: Generalization
Let the symbol ? denote a single stationary peg of either color, and
let the symbol * denote any number of stationary pegs.
1
2
3
4
Red Step
(* R E *) ! (* E R *):
(RS)
(* R ? E *) ! (* E ? R *):
(RJ)
(* E B *) ! (* B E *):
(BS)
(* E ? B *) ! (* B ? E *):
(BJ)
Red Jump
Blue Step
Blue Jump
Robert R. Snapp © 2012
10. Peg Solitaire
CS 32, Fall 2012
11 / 19
State Transistion Graph: A directed graph
(R R E B B)
(R R B E B)
(R E R B B)
(R E B R B)
(R B R E B)
(E R R B B) (R R B B E)
(R B E R B)
(R B R B E)
(R B E B R)
(E B R R B)
(R B B R E)
(E B R B R)
(E R B R B)
(R B B E R)
(B E R R B)
(B E R B R)
(B R E R B)
(B R B R E)
(B R B E R)
(B E B R R)
(B B R E R)
(B B E R R)
This is an example of a directed graph: edges can only be traversed in the direction
of the arrows, as each peg can only move in one direction. Solid edges denote steps;
dashed edges denote jumps.
Robert R. Snapp © 2012
10. Peg Solitaire
CS 32, Fall 2012
12 / 19
How many solutions are there?
Two solutions of 8 moves (4 jumps + 4 steps) exist for the 4-peg puzzle.
How many moves are required to solve an eight-peg puzzle?
How many moves are required to solve a 2n-peg puzzle?
These questions can be answered either by constructing the graph of the state space
(as we did above), or by delving a little deeper into the structure of the puzzle.
Robert R. Snapp © 2012
10. Peg Solitaire
CS 32, Fall 2012
13 / 19
Algebraic method for computing the solution length
Imagine that we have a special odometer that measures the total distance D that all the pegs
have traveled as we work through the four-peg puzzle. Thus for the initial state the odometer
reads D D 00. With every red or blue step, one unit is added to the reading. With every red or
blue jump, two units are added. What does is odometer reading once the goal state has been
reached? Proceeding sequentially along either solution you should be able to determine that
the final reading is D D S C 2J D 12, where S D 4 (the number of steps) and J D 4 (the
number of jumps.
However, we can also infer this value simply by adding up the net displacement require to
move each peg into its final position.
Note that each of the four pegs must move three spaces to reach their final destinations.
Hence D D 4 3 D 12.
Robert R. Snapp © 2012
10. Peg Solitaire
CS 32, Fall 2012
14 / 19
Algebraic method for computing the solution length
Now note that a change in the order between a red peg and a blue peg requires a
single jump. Likewise, to change the order of each red peg with two blue pegs
requires two jumps. Therefore, the total number of jumps requires is J D 2 2 D 4:
To complete the calculation, we note that each jump moves a peg by two spaces, and
each step by one space. Since D D S C 2J , it follows that
S D D 2J D 12 2.4/ D 4.
Thus, the total number of moves required is M D J C S D 4 C 4 D 8.
As an exercise, repeat this calculation for the eight-peg puzzle, which has four red
pegs and four blue pegs, separated by a single empty space.
Robert R. Snapp © 2012
10. Peg Solitaire
CS 32, Fall 2012
15 / 19
A 2n-peg puzzle
More generally, consider the puzzle that consists of n red pegs, and n blue pegs, placed a horizontal row
of 2n C 1 spaces, which leaves one empty space between each group. The initial state of this 2n-peg
puzzle is represented by the list,
.R1 R2 Rn E B1 B2 Bn /:
Applying our algebraic analysis, each peg must traverse n C 1 spaces to reach its final position. Since
there are 2n pegs, the total distance is
D.n/ D 2n.n C 1/:
(Note that we have introduced mathematical function notation, D.n/, in place of D , to emphasize that
value of the distance depends on the number of pegs.) Likewise, a little thought should convince you that
number of jumps required is
J.n/ D n2 ;
as each of the n red peg must some how cross over (or under) the n blue pegs. From our distance
formula, D.n/ D 2J.n/ C S.n/ we obtain
S.n/ D D.n/
2J.n/ D 2n.n C 1/
2n2 D 2n:
Thus, the total number of moves required to complete the 2n-peg puzzle is
M.n/ D J.n/ C S.n/ D n2 C 2n:
Robert R. Snapp © 2012
10. Peg Solitaire
CS 32, Fall 2012
16 / 19
A 2n-peg puzzle (cont.)
Here we tabulate some values of S.n/ D 2n, J.n/ D n2 , and M.n/ D J.n/ C S.n/,
vs. n:
n
S.n/
J.n/
M.n/
n
S.n/
J.n/
M.n/
1
2
3
4
5
6
7
8
2
4
6
8
10
12
14
16
1
4
9
16
25
36
49
64
3
8
15
24
35
48
63
80
9
10
20
30
40
50
100
1000
18
20
40
60
80
100
200
2000
81
100
400
900
1600
2500
10000
1000000
99
120
440
960
1680
2600
10200
1002000
Notice that for sufficiently large n, M.n/ J.n/ D n2 . Looking at the behavior of a quantity as one or
more of its parameters steadily increases is called asymptotic analysis. In this case we would say that both
quantities M and J increase quadratically with respect to n, as both grow like k n2 for some suitable
choice of the multiplicative constant k . (In this case, k D 1.) The particular value of k is usually much
less important than the factor of n2 . Thus it is convenient to ignore it, and summarize the asymptotic
dependence by writing M.n/ O.n2 /. Likewise, one could state that S.n/ O.n/, as S increases
linearly with respect to n.
Fine Print: By convention, the notation M.n/ O.n2 / usually means that M.n/ has an asymptotic quadratic upper bound, i.e., for sufficiently large n, there exists a
constant k such that M.n/ k n2 .
Robert R. Snapp © 2012
10. Peg Solitaire
CS 32, Fall 2012
17 / 19
Two-dimensional peg solitaire
There are several popular versions of two-dimensional peg solitaire. A very simple
one is an extension of our previous puzzle. Consider the following top-view of a
puzzle with 8 red pegs, 8 blue pegs, placed in a board with 17 spaces, leaving
exactly one empty space in the middle:
As before, the goal of the puzzle is to interchange the colors, with the restriction that
red pegs can only move to the right or down, and the blue pegs can only move to the
left, or up. Likewise pegs can either step one space into the empty space, or can
jump over a single peg of either color into the empty space.
Can you find a solution?
Robert R. Snapp © 2012
10. Peg Solitaire
CS 32, Fall 2012
18 / 19
Study Problems
1
2
3
4
How many different ways can n red pegs and n blue pegs be placed on a board that contains
2n C 1 spaces?
Write a scheme function of one integer argument, called initial-state that will generate the
initial state for an n peg puzzle. Consequently, the expression (initial-state 1) should
evaluate to (R E B), (initial-state 2) to (R R E B B), etc.
Write a scheme function of a single argument, called legal? that returns #t if the argument
represents a legal state of a 2n peg puzzle, for some value of n; and #f, otherwise. Thus, (legal?
’(R E B R B)) and (legal? ’(B E R)) should both return #t, but (legal? ’(E R)),
(legal? ’(R R B B)), (legal? ’(R E R E B B)), and (legal? ’(R R E B R) return #f.
Note that a legal state is not necessarily accessible.
Consider the two-dimensional puzzle with 8 red pegs and 8 blue pegs.
(a) How many total states are there? (Include both accessible and inaccessible states.) Please
express your answer as an integer.
(b) This puzzle is known to have a 48 move solution. How many moves in this solution are jumps?
(c ) Can you find a 48 move solution? Can you find a shorter solution?
5
We define a two-dimensional jigsaw puzzle to be regular if and only if the matches are unique; that is
two pieces fit together if and only if the are neighboring pieces in the solved puzzle. Note that most
picture jigsaw puzzles have this property. Letting n denote the number of pieces in the puzzle, show
that the amount of time required to solve a regular jigsaw puzzle scales as O.n2 /.
Robert R. Snapp © 2012
10. Peg Solitaire
CS 32, Fall 2012
19 / 19