Editing 2821: Path Minimization
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==Explanation== | ==Explanation== | ||
+ | {{incomplete|Created by WAITING AN HOUR BEFORE SWIMMING - Please change this comment when editing this page. Do NOT delete this tag too soon.}} | ||
In this comic, it appears that Cueball, standing on shore, is observing a swimmer who is possibly (but not obviously) in distress. The comic illustrates five potential paths that can be taken to reach the swimmer, each with a different reason to make them viable, in the manner of demonstrating different optimal strategies that can be chosen. | In this comic, it appears that Cueball, standing on shore, is observing a swimmer who is possibly (but not obviously) in distress. The comic illustrates five potential paths that can be taken to reach the swimmer, each with a different reason to make them viable, in the manner of demonstrating different optimal strategies that can be chosen. | ||
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You could also fulfill the criteria of reaching the target in finite, but arbitrarily long, time by following a {{w|random walk}}(+swim) or even follow a {{w|space-filling curve}} carefully chosen to be the maximally finite scenario. Or you could simply choose any path, and stop for an arbitrarily long time, or travel at a speed approaching zero. In the comic, however, a requirement for simplicity of path may dictate the use of something close to the opposing {{w|great-circle distance}}, or a variation that has a maximal swim-time even without ''undue'' time-wasting detours, and assume equal speeds of travel on all routes. | You could also fulfill the criteria of reaching the target in finite, but arbitrarily long, time by following a {{w|random walk}}(+swim) or even follow a {{w|space-filling curve}} carefully chosen to be the maximally finite scenario. Or you could simply choose any path, and stop for an arbitrarily long time, or travel at a speed approaching zero. In the comic, however, a requirement for simplicity of path may dictate the use of something close to the opposing {{w|great-circle distance}}, or a variation that has a maximal swim-time even without ''undue'' time-wasting detours, and assume equal speeds of travel on all routes. | ||
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The comic pokes fun at two famous physical/mathematical problems that are usually stated as happening on a beach. The first is the Lifeguard problem, which Richard Feynman, in his book ''QED'', uses to illustrate {{w|Fermat's principle}}, or principle of least time, which states that the path taken by a light ray between two given points is the path that can be traveled in the least time. This is closely related to {{w|Stationary-action principle}} for mechanical systems. In Feynman's words: | The comic pokes fun at two famous physical/mathematical problems that are usually stated as happening on a beach. The first is the Lifeguard problem, which Richard Feynman, in his book ''QED'', uses to illustrate {{w|Fermat's principle}}, or principle of least time, which states that the path taken by a light ray between two given points is the path that can be traveled in the least time. This is closely related to {{w|Stationary-action principle}} for mechanical systems. In Feynman's words: | ||
''"Finding the path of least time for light is like finding the path of least time for a lifeguard running and then swimming to rescue a drowning victim: the path of least distance has too much water in it; the path of least water has too much sand in it; the path of least time is a compromise between the two."'' - ''Richard Feynman, QED - The Strange Theory of Light and Matter (1988, Princeton University Press), Chapter 2.'' | ''"Finding the path of least time for light is like finding the path of least time for a lifeguard running and then swimming to rescue a drowning victim: the path of least distance has too much water in it; the path of least water has too much sand in it; the path of least time is a compromise between the two."'' - ''Richard Feynman, QED - The Strange Theory of Light and Matter (1988, Princeton University Press), Chapter 2.'' | ||
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The second problem referenced in this comic is the [https://gametheory101.com/courses/game-theory-101/hotellings-game-and-the-median-voter-theorem Beach Vendor Problem], which is stated as follows. Suppose that on a long beach there are two ice cream vendors. Customers are uniformly distributed on the beach and each person will go get the ice cream at the closest vendor. Each vendor wants to maximize the number of customers that buy at their place. To minimize the customer's walking time, the optimal configuration would be to have one vendor at 1/4 of the beach length and the other at 3/4, but {{w|Hotelling's law}} predicts that the two shops will converge to the middle of the beach, in an attempt to steal as many customers as possible from the competition. This is a case of {{w|Nash equilibrium}} that is also related to the {{w|Median voter theorem}}. If the number of vendors is larger than 2, the problem may become [https://gametheory101.com/tag/hotellings-game/ considerably more complicated]. | The second problem referenced in this comic is the [https://gametheory101.com/courses/game-theory-101/hotellings-game-and-the-median-voter-theorem Beach Vendor Problem], which is stated as follows. Suppose that on a long beach there are two ice cream vendors. Customers are uniformly distributed on the beach and each person will go get the ice cream at the closest vendor. Each vendor wants to maximize the number of customers that buy at their place. To minimize the customer's walking time, the optimal configuration would be to have one vendor at 1/4 of the beach length and the other at 3/4, but {{w|Hotelling's law}} predicts that the two shops will converge to the middle of the beach, in an attempt to steal as many customers as possible from the competition. This is a case of {{w|Nash equilibrium}} that is also related to the {{w|Median voter theorem}}. If the number of vendors is larger than 2, the problem may become [https://gametheory101.com/tag/hotellings-game/ considerably more complicated]. |