European Peg Solitaire Solvability
Source: Wikipedia article on Peg Solitaire
Problem:
The European version of the popular game "Brainvita" (or "Peg Solitaire") looks as follows:
Background:
The game fills the entire board with pegs except for the central hole. The objective is, making valid moves, to empty the entire board except for a solitary peg. A valid move is to jump a peg orthogonally over an adjacent peg into a hole two positions away and then to remove the jumped peg.
Problem:
The European version of the popular game "Brainvita" (or "Peg Solitaire") looks as follows:
Y Y O O O Y Y
Y O O O O O Y
O O O O O O O
O O O T O O O
O O O O O O O
O O O O O O O
Y O O O O O Y
Y Y O O O Y Y
Prove that the initial configuration as shown in the representation is not solvable.
The game fills the entire board with pegs except for the central hole. The objective is, making valid moves, to empty the entire board except for a solitary peg. A valid move is to jump a peg orthogonally over an adjacent peg into a hole two positions away and then to remove the jumped peg.
Similar Problems:
Rubik's Cube
Sam Loyd Puzzle Solvability
Update:
Solution posted by Pritish Kamath (CSE IITB 2012 Alumnus, Assistant Researcher MSR Bangalore) in comments!
Update:
Solution posted by Pritish Kamath (CSE IITB 2012 Alumnus, Assistant Researcher MSR Bangalore) in comments!
If you cover your board with three colours in the following manner :
ReplyDelete..abc..
.abcab.
abcabca
bcabcab
cabcabc
.bcabc.
..abc..
Let A, B and C be the number of pebbles on colours 'a', 'b' and 'c' respectively. The relative parity of A, B and C remains conserved after every move. And initially, A = B = C = 12. Hence it is unsolvable.
PS : By relative parity, I mean [A (xor) B] and [B (xor) C] are conserved.
Basically, if initially, A:odd B:odd C:even, then after any number of moves, it'll either be A:odd B:odd C:even or A:even B:even C:odd
What is logic behind conservation of relative parity?
DeleteJust colour it like a chess board. Then we will have 16 pegs in black and 20 pegs in white (or vice versa). A valid move always gets a peg into the same colour and a peg is removed from the same colour. So, definitely one peg from each of the colour will remain no matter what.
ReplyDelete@Pritish, Thanks for the correct solution.
ReplyDelete@Crazy Dino, Your solution is not correct.