Puzzle Rule

There is a board filled with pegs except center of the board first. You can jump up a single peg to the direction either horizontal or vertical, not diagonal. Then remove the peg which is jumped up. The goal of the puzzle is to eliminate as much pegs as possible.

You'll soon realize that it seems to be impossible to make it one left. Let's try to prove unpossibility of making this puzzle one peg left.

Play Peg Puzzle

Proof

First you can draw board with 3 colors(A,B,C) like below.

		C	A	B
	C	A	B	C	A
C	A	B	C	A	B	C
A	B	C	A	B	C	A
B	C	A	B	C	A	B
	A	B	C	A	B
		C	A	B

Then all 3 continous place always contains 3 colors. In the initial state, pegs in each colors will be 12. All possible move and the Increment/Decrement of pegs in each colors will be clarified as the table below.

Move	A	B	C
B->A	+1	-1	-1
C->B	-1	+1	-1
A->C	-1	-1	+1
C->A	+1	-1	-1
A->B	-1	+1	-1
B->C	-1	-1	+1

Next let's check out the change of the sum of pegs in 2 colors.

Move	B+C	C+A	A+B
B->A	-2	0	0
C->B	0	-2	0
A->C	0	0	-2
C->A	-2	0	0
A->B	0	-2	0
B->C	0	0	-2

The change of the sum of pegs in 2 colors are all 0 or -2 as you see. It means that any movement which this puzzle allows doesn't change the odd/even of the sum of pegs in 2 colors. This is the invariant characteristic of this puzzle.

In initial state, the sum of pegs in 2 colors are 24/24/24 (even/even/even). You can't change this characteristic with any move. In the state of one peg left, the sum of pegs in 2 colors must be 1/1/0 (odd/odd/even). The result is that

You can't make this puzzle with one peg left!!

Program & Proof by Koh Itoh (c)