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Re: To Findout Prime number

by poolpi (Hermit)
on Feb 08, 2008 at 07:38 UTC ( #666924=note: print w/replies, xml ) Need Help??

in reply to To Findout Prime number

See Abigail one-liners here

#Abigail perl -wle 'print "Prime" if (1 x shift) !~ /^1?$|^(11+?)\1+$/' %% #Abigail perl -wle 'print "Prime" if (0 x shift) !~ m 0^\0?$|^(\0\0+?)\1+$0' %% #Abigail perl -wle 'print "Prime" if ("m" x shift) !~ m m^\m?$|^(\m\m+?)\1+$mm'
Thanks Abigail !

'Ebry haffa hoe hab im tik a bush'. Jamaican proverb

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Re^2: To Findout Prime number
by ZlR (Chaplain) on Jul 17, 2009 at 12:31 UTC

    perl -wle 'print "Prime" if (1 x shift) !~ /^1?$|^(11+?)\1+$/'

    OK so I thought i'd elaborate a little on how this works since the monks from the chatterbox took time to explain it to me :)

    So you have a number, say 6, and "write" it as a sequence of 1 ie (111111) .
    If you write it (11)(11)(11) "you just found that 6 = 3 * 2", as moritz explained it.

    ELISHEVA explains the math behind this :
    if there is a repeating group of the same number of ones, then a factorization is possible and hence the number can't be prime.

    It means that if you can write the number as M groups of K ones, then it factorizes as :

    M * [ Sum(p=1->k) 1 ] which really is just M * K

    Which means that our number can be divided by M (or K), and therefore it's not prime. Shmem gives an example: i.e. for m = 1763, the group found would be 11111111111111111111111111111111111111111 repeated 43 times - not prime

    So how does the regexp implement that ? We can decompose it like this :

    m/ ^ # start of line 1? # the number 1, zero or one time $ # end of line | # alternation ^ # start of line ( # remember the match in \1 11+? # the number 1, then the number 1 once or more t +imes, but the less time possible ) # \1+ # the matched sequence, once or more. $ /x ;
    So the real trick is in (11+?). In a standalone context, it will only match 11. That's because +? means "once or more but the minimum number of times". On the other hand, ^(11+?)$ will match a whole sequence of ones from begining to end.

    Here (11+?) is followed by \1+$ which means : itself, once or more, until the end of the line .
    So what happens is that the "minimum number of times" that's contained in +? is seen from the \1+$ that is after it in the expression
    So when no match occurs, the engine will go back to the (11+?) and try again.

    I understand that's what backtracking is. Eventually the regexp engine will try every grouping of ones and fine none.
    Since it needs at least two groupings to match (enforced by the + in \1+ ), a prime number will not match.

    The only problem is that for some numbers you get a Complex regular subexpression recursion limit (32766) exceeded error. My guess was that it happens when the engine has to try more than 32766 number of times, ie the first fail appears for a number that needs K>32766. That's not the case though, since prime number 32779 does not yield the error. 65558 does, though. I went on and brutforced it, to find that the error appears first with 65536, which is 2 * 32768, which computes since it needs two groups to match...

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