Algorithm Question

Peripatetic Shipping Lines, Inc., is a shipping company that owns n ships and provides service to n ports. Each of its ships has a schedule that says, for each day of the month, which of the ports its currently visiting, or whether it's out at sea. (You can assume the "month" here has m days. for some m > n.) Each ship visits each port for exactly one day during the month. For safety reasons. PSL Inc. has the following strict requirement: (dagger) No two ships can be in the same port on the same day. The company wants to perform maintenance on all the ships this month, via the following scheme. They want to truncate each ship's schedule: for each ship Si, there will be some day when it arrives in its scheduled port and simply remains there for the rest of the month (for maintenance). This means that Si will not visit the remaining ports on its schedule (if any) that month, but this is okay. So the truncation of Si's schedule will simply consist of its original schedule up to a certain specified day on which it is in a port P, the remainder of the truncated schedule simply has it remain in port P. Now the company's question to you is the following: Given the schedule for each ship, find a truncation of each so that condition (dagger) continues to hold: no two ships are ever in the same port on the same day. Show that such a set of truncations can always be found, and give an algorithm to find them. Example. Suppose we have two ships and two ports, and the "month" has four days. Suppose the first ship's schedule is port P1: at sea: port P2: at sea and the second ship's schedule is at sea: port P1: at sea: port P2 Then the (only) way to choose truncations would be to have the first ship remain in port P2 starting on day 3, and have the second ship remain in port P1 starting on day 2