Re: do u know ramanujan numbers algorithm

On Feb 29, 2:32 pm, Andreas Leitgeb <a...@xxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
Roedy Green <see_webs...@xxxxxxxxxxxxxxxxxxxx> wrote:
Just follow 4 for loops And test a^3+b^3 =c^3+d^3=a Number And print
those number.
you could improve slightly on that brute force approach this way:
1. you don't need to find all the permutations, just the ones where
a < b, a < c, and c < d (perhaps <=, I am not familiar with the exact
rules).

You could also iterate only over two variables and save the values
of a^3+b^3 as keys in a map. Before actually adding it to the map,
you'd check for it's previous existence in the map:  if it was
already there, you've found a solution.

Of course, when adding one, you need to store the value a as value
for key a^3+b^3, such that when you re-encounter the key, you also
know the previous a and (with simple math) b.

There's further potential for improvement by casually cleaning up
keys in the map that are small enough that re-encountering them
can be proven impossible. e.g. because it is smaller than current
a's cube.

Perhaps, it's even better to iterate the sum a+b in the other loop
and the difference (only till sign-change) in the inner loop, so you
do not have to decide on any maximum beforehand.

Has anyone paid the guy for a O(n^2) solution, yet?  ;-)

Pay and get the Solution The guy has not responded it means he just
ask for curiosity.

Any one else interested pay me $200 for the solution.

Bye
Sanny
.

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