#### A Math Forum Project

Print This Problem

### 203: 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.

#### Submit your answer to  "Traffic Jam"

If you are under 13, you must have permission from your parent or teacher to participate in this web project. You will be asked to provide the email address of your parent or teacher when you register. At any time, parents or teachers may request that we remove personal information by writing to removal@mathforum.org or by contacting us via postal mail or telephone (800-756-7823).

Math Forum Home || Math Library || Quick Reference || Math Forum Search

© 1994-2009 Drexel University. All rights reserved.
http://mathforum.org/
Contact the Problem of the Week administrators
The Math Forum is a research and educational enterprise of the Drexel School of Education.