Project Euler with Befunge


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2014-09-16

Problem 023: Non-abundant sums

Description:

A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.

A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.

As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.

Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.


Solution:
v
# ... #
. . . .
.  .  .
. . . .
# ... #

                                   >v       >#        v#   <
>"d"4*:10p355***00p00g>1-::20p0\2/:|>:20g\%#^_:30p+30g>1-:#^_$v
                      |:p+1/g01g02% g01g02`g02<               <
v                $p100<            >         1^ vp+1/g01+g05g04%g01+g05g04+2%2g+1/g01+g05g04%g01+<
$                >v                   >v                  >v                   >v       v   <    g
>00g40p>40g1-:40p|>40g10g%40g10g/1+g2%|>40g1+50p>50g1-:50p|>50g10g%50g10g/1+g2%|>40g50g+00g`|    0
       ^                              <         ^                              <       $<   >40g5^
v                <                    ^                   <
>080p00g20p>20g1-20p20g10g%20g10g/1+g2/#v_v
           |g02               <p08+g02g0 8<
           >80g.@             ^         <
This program is too big to display/execute here, click [download] to get the full program.

Explanation:

The algorithm here is pretty ineffecient. First I brute force all the abundent numbers and then calculate every possible product of these numbers.
I used a little trick to store 2 boolean values per grid cell: The lowest bit represents if the value is abundent and the second-lowest if it can be written as two abundent numbers


Interpreter steps: 10 667 174 483
Execution time (BefunExec): 32min (5.42 MHz)
Program size: 400 x 88
Solution: 4179871
Solved at: 2014-09-16



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