Re^2: Challenge: Simple algorithm for continuing series of integers


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Re^2: Challenge: Simple algorithm for continuing series of integers

by moritz (Cardinal)

on Oct 19, 2008 at 20:21 UTC ( [id://718084]=note: print w/replies, xml )

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in reply to Re: Challenge: Simple algorithm for continuing series of integers
in thread Challenge: Simple algorithm for continuing series of integers

Well, your reply is interesting, but not very helpful.

Actually some time ago I used to always say "17" when somebody asked me how to continue a a given series.

So I thought about a more strict formulation of the question. I came up with something along these lines:

Given a series @a = $a[0] ... $a[$n-1], find all pure functions f($n, @a) that only uses the operators +, -, *, /, % and **, the arguments and number literals (and parenthesis for grouping), and where the following identity holds: For all $k <= $n: f($k, @a[0..$k-1]) == $a[$k].

Of all these functions, use the simplest, ie the one that uses the smallest number of operators. Then calculate the next value with that function.

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Re^3: Challenge: Simple algorithm for continuing series of integers
by JavaFan (Canon) on Oct 19, 2008 at 21:42 UTC

a₁

a_k

f(x)

k-1

f(1) == a₁

f(2) == a₂

f(x)

x

Given that there's at least one function that uses only the given operations, and a finite number of such operations, there's also one that uses the smallest number of operators. But it's not necessarily unique.

I don't know whether this is helpful.

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