## Math Forum - Problem of the Week

### Traffic Jam

There are eleven stepping stones and ten people. On the five lefthand stones, facing the center, stand five of the people. The other five people stand on the five righthand stones, also facing the center. The center stone is not occupied.

Everyone must move so that the people originally standing on the righthand stepping stones are on the lefthand stones, and those originally standing on the lefthand stepping stones are on the righthand stones, with the center stone again unoccupied. They follow these rules:

• After each move, each person must be standing on a stepping stone.
• People move in one of two ways:
• by stepping to an adjacent unoccupied stone.
• by "jumping" a person who is facing the opposite direction if there is an empty stone on the other side. People may not "jump" more than one person at a time.
• Only one person can move at a time.

Questions:

1. What's the fewest number of moves for 10 people (5 people on each side) to end up on the opposite side from where they started?
2. What hints or strategies would you give someone to help them find the fewest number of moves?

Use this Java applet to try this with fewer people: Traffic Jam Applet

Use the submit link below to get hints and also chances to revise.

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