The "English Sixteen" puzzle is peg jumping puzzle that is essentially a twodimensional analogue of the Leaping Frogs puzzle. The board has two 3×3 grids of holes which share one corner hole. Initially the shared hole is empty, and the rest is filled with eight pegs of one colour in one half of the board and eight pegs of a second colour in the other half. You are allowed to shift a peg to an adjacent hole, or you can jump a peg over another if the hole is immediately behind that. The aim is to swap the colours, moving every peg to the other half of the board in as few moves as possible.
This puzzle is described in Professor Hoffman's "Puzzles Old and New" from 1893.
In some versions of the puzzle you are restricted in your movements by one or both of the following rules:
These restrictions are essentially the same as draughts/checkers, so sometimes this version of the puzzle is played on a checkered board with pieces on the black squares.
A1  

B1  B2  
C1  C2  C3  
D1  D2  
E1  
F1  F2  
G1  G2  G3  
H1  H2  
I1 
Swap Squares is a version of the puzzle which starts with only 7 pieces in each half  the centre hole of each 3×3 grid is initially left empty. The same restrictions may be applied if you wish. Note that in this game it is possible to shift/jump one piece several times in succession. You may wish to count multiple consecutive shifts/jumps with one piece as a single move, or only count consecutive jumps as a single move.
If your browser supports JavaScript, then you can play the English Sixteen / Swap Squares puzzle by clicking the link below:
If there are no restrictions on the moves, then it is easy to calculate the number of positions. There are 17 holes, and 8 pegs of each colour, giving a total of 17!/8!^{2} = 218,790 positions. If only forwards moves are allowed, then many of these positions are unreachable. It is difficult to calculate the number directly, but a full enumeration by computer showed there are 133,864 reachable positions.
Any direction  Forward only Jump any colour  Forward only Jump opposite colour  







In the 14 peg game, 7 pegs of each colour and 3 empty holes, giving 17!/(7!^{2}3!) = 2,333,760 positions if there are no restrictions on the moves. If we restrict to forward moves only, then there are 2,071,392 reachable positions.
Every jump and shift counts as a move  

Any direction  Forward only Jump any colour  Forward only Jump opposite colour  






Consecutive jumps with the same piece count as one move  

Any direction  Forward only Jump any colour  Forward only Jump opposite colour  






Consecutive shifts and jumps with the same piece count as one move  

Any direction  Forward only Jump any colour  Forward only Jump opposite colour  






Without the restrictive rules, the optimal solution is 46 moves (20 shifts and 26 jumps). Unsurprisingly, there are no backwards moves, so the first restriction makes no difference. More surprising is that there are such optimal solutions where jumps are only performed over pieces of the opposite colour. The second restriction therefore also makes no difference to the length of the shortest solution (although there are more optimal solutions if you do not impose the second restrictive rule).
Here is one of many 46 move solutions:
D2E1 F1xD2 G1F1 E1xG1 D1E1 F2xD1 G3F2 E1xG3 C3xE1 B2C3 D1xB2 F2xD1 E1F2 C3xE1 D2C3 F1xD2 G2F1 F2G2 H1xF2 G1H1 I1xG1 G3xI1 E1xG3 C1xE1 B1C1 D2xB1 F1xD2 G1F1 E1xG1 C1xE1 A1xC1 B1A1 D2xB1 F1xD2 H2xF1 G3H2 E1xG3 C1xE1 D1C1 C2D1 D2C2 F1xD2 E1F1 F2E1 D1xF2 E1D1
Suppose we only apply the first restrictive rule, so moves are only forwards, and there is no
restriction on the colours of pieces we jump over. In this case there are solutions which are
optimal whichever way you count the moves. It has 38 shifts/jumps, 24 moves when only consecutive
jumps with the same piece are combined to one move, and 18 moves if consecutive shifts/jumps of
the same piece are also combined.
Here is one such solution:
B1C2 F1E1 H2G2 D2xF1xH2 G1F1xD2xB1 C3D2xF1 E1F2 I1xG1xE1xC3 C1xE1xG1xI1 A1xC1xE1xG1 G3xE1xC1xA1 D1E1 F2xD1C1 B2xD1xF2G3 H1xF2xD1xB2 C2D1xF2xH1 G2F2xD1 E1F2
Allowing backwards moves does not allow for shorter solutions, except if you combine consecutive shifts/jumps of one piece. In that case, a 17 move solution is possible, for example:
F2E1 D1xF2G2 H1xF2xD1xC2 B2xD1xF2xH1 C3B2xD1xF2 E1D1xB2 G1xE1F1 A1xC3xE1xG1 G3xE1xC3xA1 H2G3xE1xC3 D2E1xG3H2 I1xG3xE1D2 C1xE1xG3xI1 B1C1xE1xG3 F1E1xC1 G2F1 C2B1
On the other hand if both rules are in effect (forward moves only, jumps only over opposite colour), then the solutions are all slightly longer  41 shifts and jumps, 31 if consecutive jumps are combined, and 21 if consecutive jumps and shifts are combined. The solution below is optimal in the first two metrics, and the solution below that in the third metric. There is probably no single solution that is optimal in all three.
B1C2 D1E1 F2xD1 E1F2 D2E1 F1xD2xB2 F2G2 H2xF1xD2 G3F2 G1F1 E1xG1 C1xE1xG3 G3H2 C3xE1xG3 A1xC1xE1 B2C3 D1C1 F2xD1xB2 B2A1 H1xF2xD1xB2 I1H1 G1xI1 H1xF2xD1 E1F2 C3xE1xG1 D2C3 C2D2 D2E1 F1xD2 G2H1 E1D1
B1C2 F2E1 D1xF2G2 C1D1xF2 A1B1C1D1 E1xC1B1A1 F1E1xC1B1 G1F1E1xC1 D2E1F1G1 C3D2E1F2 B2C3D2 G3xE1xC3B2 H2G3xE1xC3 F2G3H2 D2E1F2G3 H1xF2E1D2 I1H1xF2E1 G2H1I1 D1xF2G2H1 C2D1xF2 E1D1