Strategy to capture piece on a graph I found the following PDF full of interesting games: http://www.ms.uky.edu/~lee/visual05/notes/games2.pdf
I'm currently working on Game $14$, Penny Chase. I've summarized the game below:

Place a penny and a dime on the indicated spots of Figure 12. Two
players alternate turns, one moving the penny, the other the dime.
Moves are made along a solid black line to an adjacent spot. The penny
player always moves first. Their object is to capture the dime by moving
onto the spot occupied by the dime. To win they must do so before they
make their seventh move. If after six of their moves they have failed to
catch the dime, they lose.

Here's the figure in reference:

One of the players has a winning strategy, but I'm not able to figure out what it is.
I'm wondering if anyone can come up with a winning strategy for either player. A lot of these problems involve rearranging the graph into a better form to work with, but I haven't been able to do so. I thought about using the cycles to the dime's advantage, but I haven't managed to do so.
Any help is appreciated.
 A: Here is how the penny can win.
First of all, parity issues prevent the penny from winning just by directly heading for the dime. Fortunately, by going around the triangle, the penny can break the parity.
Let me number the nodes as shown below:

In almost all cases, if the penny's first three moves are to 14, then 15, then 1, then the last three moves are enough to win. The dime's third move can be to 3, 4, 6, 10, 11, or 12, and here is a winning tree for five of those cases:
first three moves: 14, then 15, then 1
    dime to 3 => capture on move 4
    dime to 4 => capture on move 4
    dime to 6 => penny to 3
        dime to 2 => capture on move 5
        dime to 7 => capture on move 5
    dime to 10 => there's not enough time to catch it!
    dime to 11 => penny to 3
        dime to 7 => capture on move 5
        dime to 8 => capture on move 5
        dime to 13 => penny to 8
            dime to 11 => capture on move 6
            dime to 12 => capture on move 6
    dime to 12 => penny to 4
        dime to 8 => capture on move 5
        dime to 9 => capture on move 5
        dime to 13 => penny to 8
            dime to 11 => capture on move 6
            dime to 12 => capture on move 6

Unfortunately, the dime does escape by going to 10. So instead, if the dime's first two moves are to 12 then 9 (heading for 10) then we modify our strategy; instead of the penny going up to 1 as the third move, the penny goes to 5. This gives the following winning tree:
third move: penny to 5
    dime to 4 => capture on move 4
    dime to 10 => capture on move 4
    dime to 12 => penny to 4
        dime to 8 => capture on move 5
        dime to 9 => capture on move 5
        dime to 13 => penny to 8
            dime to 11 => capture on move 6
            dime to 12 => capture on move 6

So now we have a six-move winning sequence no matter what the dime does.
Furthermore, if we had seven moves to win instead of six, we could begin with the 14 to 15 to 1 sequence in all cases. If the dime's third move is to 10, the penny can counter with 15. This forces the dime to retreat to 9 or lose immediately; at that point, the penny can make the fourth move to 5, and transpose into the alternative line above with a one-move delay.
