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ThisSpaceAvailable comments on Open Thread, Jun. 1 - Jun. 7, 2015 - Less Wrong Discussion

3 Post author: Gondolinian 01 June 2015 12:45AM

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Comment author: ThisSpaceAvailable 01 June 2015 03:40:57AM 2 points [-]

Suppose we have a set S of n elements, and we ask people to memorize sequences of these elements, and we find that people can generally easily memorize sequences of length k (for some definition of "generally" and "easily"). If we then define a function f(S) := k log n, how will f depend on S? Have there been studies on this issue?

Comment author: SolveIt 01 June 2015 08:39:08AM 1 point [-]

Why k log n? I imagine n would be largely independent of k, so f(S) would become arbitrarily large just by using bigger and bigger sets.

Comment author: gjm 01 June 2015 11:31:21AM 5 points [-]

k log n is the number of bits it takes to represent a random sequence of k elements from S. So f(S) is the number of bits one can remember when they're encoded as sequences of elements of S.

My guess is that f(S) will be smaller when n is very small (it's not twice as easy to remember 20 digits from {0,1} as to remember 10 from {0,1,2,3}, unless perhaps you explicitly convert them) and maybe when n is rather large (as it starts to require cognitive effort to distinguish elements of S). It will surely also depend in complicated ways on what sort of thing is in S (words? common first names? photos of faces? ...).

I'm sure there's been work on this in the context of computer security -- looking for things that are a bit like passwords but make a better memorability/bits-of-security tradeoff than passwords do. In that context it's also worth considering what happens if you allow the sequence to be remembered slightly wrongly.

Comment author: Clarity 01 June 2015 08:03:52AM *  0 points [-]

Sorry, this was an useless post so now it's gone

Comment author: philh 01 June 2015 01:03:59PM 1 point [-]

k and n are implicitly functions of S. So f(S) = (length of a memorizable sequence of elements of S) * (cardinality of S).