| ▲ | Reverse math shows why hard problems are hard(quantamagazine.org) | |||||||
| 58 points by gsf_emergency_6 4 hours ago | 7 comments | ||||||||
| ▲ | emil-lp an hour ago | parent | next [-] | |||||||
Whenever the pigeon-hole principle is name dropped, we should also drop Dijkstra's commentary The undeserved status of the pigeon-hole principle. https://www.cs.utexas.edu/~EWD/transcriptions/EWD10xx/EWD109... | ||||||||
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| ▲ | degamad 3 hours ago | parent | prev | next [-] | |||||||
Specifically, reverse math (a subset of metamathematics which looks at swapping axioms and theorems) allows us to show that some hard problems are equivalent to each other. EDIT: I think this line is the most telling: > But he cautioned that the reverse mathematics approach may be most useful for revealing new connections between theorems that researchers have already proved. "It doesn’t tell us much, as far as we can say, about the complexity of statements which we do not know how to prove." So, at this point, it helps us understand more about problems we already understand a little about, but nothing yet about new problems. | ||||||||
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| ▲ | letmetweakit 22 minutes ago | parent | prev | next [-] | |||||||
The Travelling Salesman Problem in 1 dimension, on a line, is trivial, I wonder what the connection is between the dimensions and the hardness of problems like this. | ||||||||
| ▲ | Bankq 3 hours ago | parent | prev [-] | |||||||
The approach reminds me of NP-Completeness (Computational harness vs mathematical-proving hardness). Am I over-simplifying? | ||||||||
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