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    Smithsonian | Elizabeth Landau | 3/9/20 | 9 min
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    • deephdave
      Top reader this weekReading streakScout
      2 weeks ago

      If P=NP, then it’s possible to solve every problem whose solutions are easy to verify, says Stephens-Davidowitz. So, if this inequality persists, the general knapsack problem will always be hard.

    • vunderkind
      2 weeks ago

      Knapsack problems are tricky. First time I read the traveling salesman problem, I didn't expect it to have a time complexity of polynomial time.

      Need to brush up my study into quantum computing.