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Sep 21, 2019 at 0:35 vote accept Ray Shadow
Apr 30, 2019 at 18:18 answer added Ray Shadow timeline score: 17
Apr 29, 2019 at 10:02 comment added SonerAlbayrak @Shadowray Of course, my bad! Thank you for pointing it out :)
Apr 29, 2019 at 9:54 comment added Ray Shadow @Soner There is a difference in your timings, because RandomChoice takes time to create the expression. In the first expression you include RandomChoice inside Timing, but in the second you don't. This is not related to FreeQ performance.
Apr 29, 2019 at 9:24 comment added SonerAlbayrak @DanielLichtblau's point can also be seen if we compare RepeatedTiming[FreeQ[RandomChoice[{a, c}, 1000000], b]] with With[{listac = RandomChoice[{a, c}, 1000000]},RepeatedTiming[FreeQ[listac, b]]] . Clearly, the standard $O(n)$ complexity reduces to $O(1)$ due to the way the variable listac is internally stored.
Apr 26, 2019 at 21:47 comment added Daniel Lichtblau See tutorial/SomeNotesOnInternalImplementation. "Each expression contains a special form of hash code that is used both in pattern matching and evaluation." So in some cases what would be an O(n) test is instead O(1).
Apr 26, 2019 at 17:10 history edited Ray Shadow CC BY-SA 4.0
fixed typo
Apr 26, 2019 at 3:36 comment added march This is bizarre. I tried this same thing with all the different (lower-case) letters of the alphabet in the second spot. All of the RepeatedTimings were really fast, except for t, u, and v!
Apr 25, 2019 at 22:37 history edited Ray Shadow CC BY-SA 4.0
improved formulation and examples
Apr 25, 2019 at 21:00 history tweeted twitter.com/StackMma/status/1121519388794994688
Apr 25, 2019 at 19:14 comment added ktm My guess is that it has something to do with some sort of hash table data structure, but I'm not sure in what capacity.
Apr 25, 2019 at 19:08 history edited Ray Shadow CC BY-SA 4.0
added tag
Apr 25, 2019 at 18:33 history asked Ray Shadow CC BY-SA 4.0