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NOTE: use Perl; is on undef hiatus. You can read content, but you can't post it. More info will be forthcoming forthcomingly.

All the Perl that's Practical to Extract and Report

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  • You get one pair each for 7 and 11, but two pairs for 9, right? Same here.
    • I get more than that. They draw the prize next Wednesday, so I don't think it's fair to post code right now, but that's OK because mine is obviously buggy :)

    • Er, now that I stop to think about it, perhaps it's not fair for me to count a number and its reverse. D'oh!

    • Ok, the ambiguity is in the part of the solution they don't ask for. So their actual question has a unique answer.
      • I think you're seeing something I don't, then, because I have no idea what your first sentence meant :)

    • OK, now I get your results :(

  • #!/usr/bin/env perl

    my @candidate = grep { $_ % 10 } 101 .. 999;
    for my $N ( 7, 9, 11 ) {
        print join( ', ',
            grep { $_ % $N == 0 && reverse( $_ ) % $N == 0 } @candidate ),
          "\n";
    }
    That's my interpretation of the spec - but that's a hell of a lot more than six numbers as you say. Is that what you get?
    • Nope, not even close :) That tiny spec bears close rereading. For example, "using each of the digits 1 to 9" for 3-digit numbers means your first grep is off.

      my @candidates = grep { !/0/ } 111 ... 999;

      I also used &List::MoreUtils::uniq to pre-trim that list (since duplicate numbers are not allowed).

      Basically, you'll need a three stage process (I think). First, generate your candidate list. Second, find all three digit numbers which satisfy the reverse divisor requirement. Then construct your

    • First, you need a "grep { $_ !~ /0/ }" in there. Second, I think that your three numbers, concatenated, must use each of the digits only once all together. That is, a valid answer would be qw(123 456 789) if only those numbers divided properly. qw(123 331 882) is no good.
      --
      rjbs
      • That appears to be correct. You have to read the spec carefully. They ask you to use each of those numbers, not any. I got that wrong the first time.

  • i don't really think there is ambiguity. i got 3 different sets of numbers, but all of them have the same min/max numbers.

    you can check my code at: http://www.depesz.com/ovid.pl [depesz.com]

    of course i could have made mistake, but the numbers "look" right to me.
    • I get the same sets of numbers, but what does "six integers" mean in this context?

      • i think this is just an mistake on the side of new scientist.
      • You have three three-digit numbers, and their reverses. That makes six by my counting...
        • You have five three-digit numbers. One for 7, one for 11, but three for 9. If there was only one result for each of the three (which is what I was initially expecting), then six would make sense.

          • I think this mistake doesn't really matter.

            The request was:

            "What are the smallest and largest of your six integers?"

            even it we'll change it to "10 integers" the answer stays the same. i.e. the numbers i got for div/9 are neither min nor max.
  • Find a, b, c, d, e, f, g, h, and i where

    (a * 10**2 + b * 10**1 + c * 10**0) % 7 == 0
    (c * 10**2 + b * 10**1 + a * 10**0) % 7 == 0
    (d * 10**2 + e * 10**1 + f * 10**0) % 9 == 0
    (f * 10**2 + e * 10**1 + d * 10**0) % 9 == 0
    (g * 10**2 + h * 10**1 + i * 10**0) % 11 == 0
    (i * 10**2 + h * 10**1 + g * 10**0) % 11 == 0

    and where there exists an invertible function between the variables a..i to the numbers 1..9.
  • There are a total of 24 solutions to the puzzle. These 24 solutions are comprised of 10 unique integers. If you do not consider the reverse of an integer unique, there are 5 unique integers. No matter which way you slice this - there is no way to get to "six integers" unless the spec is incomplete.
    • 24 solutions? how did get them? can you show them?
        • They're not usually this unclear. I think it's just a fluke, but I'd have to work through some others to be sure.

  • I wrote a brute-force solution. My one little trick: next if $nine =~ qr/[$seven]/; A three digit number makes a fine character class inside of a regex when making sure that you're not reusing a digit.
    --

    --
    xoa

    • Hey, it's nice to see how different languages tackle the problem. Thanks!

          • Unsurprisingly, we used a lot of the same elements, including /(.).?\1/. I decided to recurse. Since these are all three-digit numbers, a simple (sort @nums)[0,-1] gives min and max, so as soon as we get a hit in 11 we have all the info we need. So yours could be shortened quite a bit further:

            for my $d_7 ( $f{7}->() ) {
                for my $d_9 ( $f{9}->($d_7) ) {
                    for my $d_11 ( $f{11}->($d_7, $d_9) ) {
                        printf "min: %d, max: %d