Question: I'm unsure how to set up a dynamic programming solution for the following problem meeting the specified time complexity: Assume houses are located (in this

I'm unsure how to set up a dynamic programming solution for the following problem meeting the specified time complexity:

Assume I'm unsure how to set up a dynamic programming solution for the houses following problem meeting the specified time complexity: Assume houses are located (in are located (in this order) along a straight road. We treat the road as the real line, such that every point has a real number as coordinate. Let this order) along a straight road. We treat the road as the denote the point where house real line, such that every point has a real number as coordinate. is located; note that Let denote the point where house is located; note that . A . A vehicle stands at some point vehicle stands at some point at time 0. The vehicle must visit at time 0. The vehicle must visit all houses. However, every house all houses. However, every house must be visited before some individual deadline must be visited before some individual deadline . Nothing special is assumed about these deadlines. The vehicle can drive. Nothing special is assumed about these deadlines. The vehicle can drive at maximum speed 1, that is, at most one length unit per time unit. (We assume distances are large, such that we neglect acceleration, deceleration and time stopped at the house. That is velocity can change instantly, and visits cost no time.)

A problem instance is given by the numbers at maximum speed 1, that is, at most one length unit per, time unit. (We assume distances are large, such that we neglect acceleration, and deceleration and time stopped at the house. That is velocity can change. A solution computes a route for the vehicle such that every house instantly, and visits cost no time.) A problem instance is given by is visited before the numbers , and . A solution computes a route for the.

Beware: It is in general not enough to first visit all houses to the left of vehicle such that every house is visited before . Beware: It is, and then all houses to the right, or vice versa. It may be the case that customers with close deadlines await on both sides of in general not enough to first visit all houses to the left. A complicated zigzag route may be needed. You can hardly avoid dynamic programming for this problem.

Aim for an algorithm whose time complexity is merely polynomial in of , and then all houses to the right, or vice versa., regardless of the numbers It may be the case that customers with close deadlines await on and both sides of . A complicated zigzag route may be needed. You.

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