BrettNortje
Banned
- Joined
- Jul 14, 2016
- Messages
- 793
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- Location
- Cape Town
- Gender
- Male
- Political Leaning
- Centrist
"The P versus NP problem is a major unsolved problem in computer science. Informally speaking, it asks whether every problem whose solution can be quickly verified by a computer can also be quickly solved by a computer...
...Consider the subset sum problem, an example of a problem that is easy to verify, but whose answer may be difficult to compute. Given a set of integers, does some nonempty subset of them sum to 0? For instance, does a subset of the set {−2, −3, 15, 14, 7, −10} add up to 0? The answer "yes, because the subset {−2, −3, −10, 15} adds up to zero" can be quickly verified with three additions. There is no known algorithm to find such a subset in polynomial time (there is one, however, in exponential time, which consists of 2n-n-1 tries), but such an algorithm exists if P = NP; hence this problem is in NP (quickly checkable) but not necessarily in P (quickly solvable).
An answer to the P = NP question would determine whether problems that can be verified in polynomial time, like the subset-sum problem, can also be solved in polynomial time. If it turned out that P ≠ NP, it would mean that there are problems in NP (such as NP-complete problems) that are harder to compute than to verify: they could not be solved in polynomial time, but the answer could be verified in polynomial time." ~ wikipedia
So, we need to be able to solve and check these algorithms before we set the computer up to 'do' them. this means that we need to make an algorithm that solves them quickly, quickly enough for us to verify.
If p = np, then the obvious answers are n = 1, or, that the equation is 'a whole one.' this could not be though, as, n equals a prime number, and one is not a prime number, thus the prime working out to be equal to 'one' would be where p = 0.5, or, [p] = anything to the power of or divided by one.
Seeing as how [p] needs to equal [np] and anything times by [n] or [p] needs to equal [p], then the answer comes with a 'to the power of,' or 'divided by' attached to it.
This means that we need to find the equation that [p] hides in. this would mean that it might be a remainder too! well, i am considering all aspects of it at this point. you could also say [n] = [1], times [1] or even divided by [1]. saying that all together will mean making a new symbol though.
...Consider the subset sum problem, an example of a problem that is easy to verify, but whose answer may be difficult to compute. Given a set of integers, does some nonempty subset of them sum to 0? For instance, does a subset of the set {−2, −3, 15, 14, 7, −10} add up to 0? The answer "yes, because the subset {−2, −3, −10, 15} adds up to zero" can be quickly verified with three additions. There is no known algorithm to find such a subset in polynomial time (there is one, however, in exponential time, which consists of 2n-n-1 tries), but such an algorithm exists if P = NP; hence this problem is in NP (quickly checkable) but not necessarily in P (quickly solvable).
An answer to the P = NP question would determine whether problems that can be verified in polynomial time, like the subset-sum problem, can also be solved in polynomial time. If it turned out that P ≠ NP, it would mean that there are problems in NP (such as NP-complete problems) that are harder to compute than to verify: they could not be solved in polynomial time, but the answer could be verified in polynomial time." ~ wikipedia
So, we need to be able to solve and check these algorithms before we set the computer up to 'do' them. this means that we need to make an algorithm that solves them quickly, quickly enough for us to verify.
If p = np, then the obvious answers are n = 1, or, that the equation is 'a whole one.' this could not be though, as, n equals a prime number, and one is not a prime number, thus the prime working out to be equal to 'one' would be where p = 0.5, or, [p] = anything to the power of or divided by one.
Seeing as how [p] needs to equal [np] and anything times by [n] or [p] needs to equal [p], then the answer comes with a 'to the power of,' or 'divided by' attached to it.
This means that we need to find the equation that [p] hides in. this would mean that it might be a remainder too! well, i am considering all aspects of it at this point. you could also say [n] = [1], times [1] or even divided by [1]. saying that all together will mean making a new symbol though.