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February 12

The Turing test

Isn't it easy to tell a computer from a human, if it can't tell that 0.1+0.2=0.3? 134.74.85.3 (talk) 00:10, 12 February 2019 (UTC)[reply]

Babies can't tell you that either, but they are still human. Count Iblis (talk) 03:43, 12 February 2019 (UTC)[reply]

I'm not sure of your point. Do you mean that floating-point arithmetic on a computer will usually use rounding which says they are different? The Turing test assumes the computer is using an advanced program. Such a program could be coded to work with base-10 decimals in the software and detect that 0.1+0.2=0.3. The program would be slower at calculations than floating point hardware so it's rarely done. PrimeHunter (talk) 10:42, 12 February 2019 (UTC)[reply]

IIRC Turing's original work had the computer being tested giving an incorrect answer to an arithmetic question - a wrong answer was a sign of intelligence. Smart man. I think he passed his own test. ;-) John Z (talk) 14:56, 12 February 2019 (UTC)[reply]

Google already knows the correct answer. AndrewWTaylor (talk) 16:24, 12 February 2019 (UTC)[reply]

Dedekind psi function

The OEIS describes the Dedekind psi function (OEIS: A001615) as the "Number of primitive sublattices of index n in generic 2-dimensional lattice." An equivalent condition is give later, but it's not clear where this comes from. I'm pretty sure that a 'lattice' means free abelian group here, but 'primitive' has any number of meanings depending on context so Google turns up a lot of unrelated results. I'm guessing 'primitive sublattice' means a subgroup with cyclic quotient group; can anyone confirm this? --RDBury (talk) 06:41, 12 February 2019 (UTC)[reply]

Ok, I massaged my search string and found [1], p31. Most of the rest of the paper is over my head but that definition seems clear enough. --RDBury (talk) 15:14, 12 February 2019 (UTC)[reply]

Partitions of the 24-cell vertices into Hexagons?

I noticed that the 24-cell can be partitioned into 4 sets, three of which are ~~Octagons~~Octahedra and one of which is a Hexagon. The discussions at 24-cell refer to to disjoint sets of four of these 6-cell great circle rings, which given that the 24-cell is self dual would seem to indicate that there is a way to partition the verticies of 24-cell into four hexagons, but I'm not seeing it. Can someone please help? (I'm fine with either of the standard methods of denoting the 24-cell vertices (either permutations of (+/-1,+/-1,0,0) or ((+/-1,+/-1,+/-1,+/-1) union permuations of (+/-2,0,0,0))Naraht (talk) 19:36, 12 February 2019 (UTC)[reply]

(Taking the vertices to be permutations of (±1, ±1, 0, 0).) A possibility for the first partition is the intersection of the vertices with the plane x+y+z=0 for the hexagon, and the planes x=y, x=z, y=z for the three octahedra. The planes x+y+z=0, x-y+w=0, x-z-w=0, y-z+w=0 intersect the vertices in four disjoint sets of six, each of which forms a hexagon. Not sure if that's exactly the partition you were looking for, but it is "a" partition into four hexagons. --RDBury (talk) 17:41, 13 February 2019 (UTC)[reply]