# RogerBW's Blog

 Perl Weekly Challenge 102: Rare Hash 03 March 2021 I’ve been doing the Perl Weekly Challenges. The latest involved an obscure class of number and a self-consistent string format. (Note that this is open until 7 March 2021.) TASK #1 › Rare Numbers You are given a positive integer \$N. Write a script to generate all Rare numbers of size \$N if exists. Please checkout the page for more information about it. I don't really recommend that you look at that page; the guy may be mathematically competent, but nobody should ever have told him how to change text colour or increase font size. But anyway, this is clearly a large search space, so I'll take him at his word and use the optimisations he gives. ``````sub rn { my \$d=shift; my @out; my \$mxm=10**(\$d-2)-1; `````` The first digit has to have one of these values. `````` foreach my \$a (2,4,6,8) { `````` And the final digit, one of these. `````` foreach my \$q (0,2,3,5,7,8) { `````` Then we can constrain on possible combinations. `````` if (\$a==2 && \$q!=2) { next; } if (\$a==4 && \$q!=0) { next; } if (\$a==6 && \$q!=0 && \$q!=5) { next; } if (\$a==8 && \$q!=2 && \$q!=3 && \$q!=7 && \$q!=8) { next; } `````` If we're dealing with just two digits, that's it for eliminating candidates, so on to the final check. `````` if (\$d==2) { my \$t="\$a\$q"; if (is_rare(\$t)) { push @out,\$t; } `````` I couldn't be bothered to do the three-digit case, and there aren't any valid answers of three digits anyway (by inspection at OEIS). So we move straight on to longer examples. `````` } else { `````` Iterate over all middle values. (I could have made this faster by taking only valid second and penultimate digits, then filling in any remaining middle digits for the 5+-digit case, but I was getting very bored by this point. So if you actually have a use for this code, there's an optimisation for you.) `````` foreach my \$middledigits (map {sprintf('%0'.(\$d-2).'d',\$_)} 0..\$mxm) { my \$b=substr(\$middledigits,0,1); my \$p=substr(\$middledigits,-1,1); `````` More constraints on these digit combinations. `````` if (\$a==2 && \$b!=\$p) { next; } if (\$a==4 && abs(\$b-\$p)%2 != 0) { next; } if (\$a==6 && abs(\$b-\$p)%2 != 1) { next; } if (\$a==8) { if (\$q==2 && \$b+\$p != 9) { next; } elsif (\$q==3 && \$b-\$p != 7 && \$p-\$b != 3) { next; } elsif (\$q==7 && \$b+\$p != 1 && \$b+\$p != 11) { next; } elsif (\$q==8 && \$b!=\$p) { next; } } `````` Finally, test for rare-ness. `````` my \$t="\$a\$middledigits\$q"; if (is_rare(\$t)) { push @out,\$t; } } } } } return \@out; } `````` And that rarity test is to reverse, check that the number is larger than its reverse, then check that the difference and the sum are both perfect squares. ``````sub is_rare { my \$t=shift; my \$d=join('',reverse(split '',\$t)); if (\$d >= \$t) { return 0; } foreach my \$c (\$t+\$d,\$t-\$d) { `````` No perfect square ends with these digits, and checking that is faster than actually taking the square root if we don't have to. `````` if (\$c =~ /[2378]\$/) { return 0; } my \$s=int(sqrt(\$c)); unless (\$s*\$s==\$c) { return 0; } } return 1; } `````` So that's that. It's basically the same in other languages, though Rust makes the transitions between string and integer relatively hard work and Ruby makes it very easy. This seemed like a long enough task to be worth a timing test, so I did. In each case, the test was run on the same unloaded system, running the program twice and timing the second run. Perl took 132 seconds; Python 151; Ruby 211; and naïve Rust (compiling at the start of the run and leaving all debug info in place) 177. When I set the Rust code to optimise for release, it still compiled in less than a second, and the execution time dropped to 17 seconds flat. I don't know how long Raku would have taken; I gave up after half an hour (1800 seconds). Even with all the optimisations turned on. Maybe modern Raku is faster than the version in Debian/stable (Rakudo 2018.12, compiled 9 January 2019). TASK #2 › Hash-counting String You are given a positive integer \$N. Write a script to produce Hash-counting string of that length. The definition of a hash-counting string is as follows: the string consists only of digits 0-9 and hashes, ‘#’ there are no two consecutive hashes: ‘##’ does not appear in your string the last character is a hash the number immediately preceding each hash (if it exists) is the position of that hash in the string, with the position being counted up from 1 It can be shown that for every positive integer N there is exactly one such length-N string. I'll take your word for it, but this does mean that a depth-first search is an obvious solution. I represent the string as the series of integers that will lie in it, with an initial "#" shown as a value of 1. In Raku: ``````sub hcs(\$n) { my @s; my @t; while (1) { @s=(); my \$l=0; `````` If we have a candidate stub, load it into `@s` and calculate its current length. `````` if (@t.elems) { @s=(pop @t).flat; \$l=sum(map {(\$_==1 ?? 0 !! chars(\$_))+1}, @s); } `````` If that's the target length, that valid answer must be the only one, so drop out with it. If it's longer than the target, no point in extending it further. `````` if (\$l==\$n) { last; } if (\$l > \$n) { next; } `````` If we're still here, it's shorter than the target, so work out possible extensions. Most of the time there will only be one, which keeps the search space conveniently small: if you have "#3#" then the only possible extension is "5#". But an empty string can be extended to make "#" or "2#"; a 7-character string can be extended by "9#" or "10#"; and so on at ~100 characters, ~1000, etc. `````` my \$c=\$l; while (1) { my \$tt=(\$c==1 ?? 0 !! chars(\$c))+\$l+1; if (\$c==\$tt) { my @k=(@s».List.flat); push @k,\$c; push @t,@k; } if (\$c > \$tt) { last; } \$c++; } } return join('',map {(\$_==1 ?? '' !! \$_) ~ '#'}, @s); } `````` and the other languages look basically similar. Full code on github. Posted by RogerBW at 10:45am on 11 March 2021 Given that people managed to get Raku to solve problem #1, clearly some of them have faster runtimes than I do. Good! Parallelising would have improved speed, and would have taught me more about the languages; maybe next time. Abigail clearly got bored and used the OEIS listing, which seems fair enough. For part 2, the trick which most people spotted but I missed was to work backwards recursively: for target length 5 the string must end with "5#", so hcs(5) = hcs(3) concatenated with "5#", and so on. Comments on this post are now closed. If you have particular grounds for adding a late comment, comment on a more recent post quoting the URL of this one. 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