Inspired by the example of that great Cornellian,VladimirNabokov,some of your friends have become amateur lepidopterists (they s
tudy butter- flies). Often when they return from a trip with specimens of butterflies, it is very difficult for them to tell how many distinct species they’ve caught—thanks to the fact that many species look very similar to one another.
One day they return with n butterflies, and they believe that each belongs to one of two different species, which we’ll call A and B for purposes of this discussion. They’d like to divide the n specimens into two groups—those that belong to A and those that belong to B—but it’s very hard for them to directly label any one specimen. So they decide to adopt the following approach.
For each pair of specimens i and j, they study them carefully side by side. If they’re confident enough in their judgment, then they label the pair (i, j) either "same" (meaning they believe them both to come from the same species) or "different" (meaning they believe them to come from different species). They also have the option of rendering no judgment on a given pair, in which case we’ll call the pair ambiguous.
So now they have the collection of n specimens, as well as a collection of m judgments (either "same" or "different") for the pairs that were not declared to be ambiguous. They’d like to know if this data is consistent with the idea that each butterfly is from one of species A or B. So more concretely, we’ll declare the m judgments to be consistent if it is possible to label each specimen either A or B in such a way that for each pair (i, j) labeled "same," it is the case that i and j have the same label; and for each pair (i, j) labeled "different," it is the case that i and j have different labels. They’re in the middle of tediously working out whether their judgments are consistent, when one of them realizes that you probably have an algorithm that would answer this question right away.
Give an algorithm with running time O(m + n) that determines whether the m judgments are consistent.
One day they return with n butterflies, and they believe that each belongs to one of two different species, which we’ll call A and B for purposes of this discussion. They’d like to divide the n specimens into two groups—those that belong to A and those that belong to B—but it’s very hard for them to directly label any one specimen. So they decide to adopt the following approach.
For each pair of specimens i and j, they study them carefully side by side. If they’re confident enough in their judgment, then they label the pair (i, j) either "same" (meaning they believe them both to come from the same species) or "different" (meaning they believe them to come from different species). They also have the option of rendering no judgment on a given pair, in which case we’ll call the pair ambiguous.
So now they have the collection of n specimens, as well as a collection of m judgments (either "same" or "different") for the pairs that were not declared to be ambiguous. They’d like to know if this data is consistent with the idea that each butterfly is from one of species A or B. So more concretely, we’ll declare the m judgments to be consistent if it is possible to label each specimen either A or B in such a way that for each pair (i, j) labeled "same," it is the case that i and j have the same label; and for each pair (i, j) labeled "different," it is the case that i and j have different labels. They’re in the middle of tediously working out whether their judgments are consistent, when one of them realizes that you probably have an algorithm that would answer this question right away.
Give an algorithm with running time O(m + n) that determines whether the m judgments are consistent.
1 answer:
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Answer:
fracture will occur as the value is less than E/10 (= 22.5)
Explanation:
If the maximum strength at tip Is greater than theoretical fracture strength value then fracture will occur and if the maximum strength is lower than theoretical fracture strength then no fracture will occur.
= 15 GPa
fracture will occur as the value is less than E/10 = 22.5