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in reply to Re^5: list of four digit lock combinations without repeated digits
in thread list of four digit lock combinations without repeated digits

If you don't want the spaces, don't put them in to start with. Rather than "@$_" use join'',@$_ or pack 'A*', @$_, or set $"=undef.

Of course I picked the slowest way(s) not to do it. I've got a bad habit of overthinking problems and overengineering solutions and shoving a regex into map like that when I know all about join is typical! Your keisaku is appreciated, I can see:

map + regex: ~0m0.950s
time perl -MAlgorithm::Combinatorics=:all -wle' my $i=variations_with_repetition(["a".."z"],$ARGV[0]); my @x; push @x, map {s/ //} qq[@$_] while $_=$i->next; print scalar @x' 4
map: ~0m0.830s
time perl -MAlgorithm::Combinatorics=:all -wle' my $i=variations_with_repetition(["a".."z"],$ARGV[0]); my @x; push @x, map {$_} qq[@$_] while $_=$i->next; print scalar @x' 4
pack: ~0m0.820s
time perl -MAlgorithm::Combinatorics=:all -wle' my $i=variations_with_repetition(["a".."z"],$ARGV[0]); my @x; push @x, pack "A*", qq[@$_] while $_=$i->next; print scalar @x' 4
join: ~0m0.720s
time perl -MAlgorithm::Combinatorics=:all -wle' my $i=variations_with_repetition(["a".."z"],$ARGV[0]); my @x; push @x, join "", @$_ while $_=$i->next; print scalar @x' 4
$"=undef: ~0m0.720s
time perl -MAlgorithm::Combinatorics=:all -le' my $i=variations_with_repetition(["a".."z"],$ARGV[0]); my @x; qq[$"=undef]; push @x, qq[@$_] while $_=$i->next; print scalar @x' 4
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Re^7: list of four digit lock combinations without repeated digits
by BrowserUk (Patriarch) on Jun 22, 2018 at 06:12 UTC
     push @x, pack "A*", qq[@$_] while $_=$i->next;

    That would probably match if not beat join, if you weren't stringifying the array twice.

    Ie. Use  push @x, pack "(A*)*", @$_ while $_=$i->next;

    Of course, if you want to put all the sequences as strings into an array, this is easier and probably quicker;

    my @x = map join( '', @$_ ), variations_with_repetition( ["a".."z"], 4 + );; print scalar @x;; 456976

    And if all you want to do is print the count, then:

    [ 7:10:21.87] C:\test>perl -MAlgorithm::Combinatorics=:all -le"print s +calar( () = variations_with_repetition( ["a".."z"], $ARGV[0] ) )" 4 456976 [ 7:10:27.12] C:\test>

    BTW, that is one seriously quick machine you're running. What is it? (6x faster than my (admittedly ancient) box.


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      That would probably match if not beat join, if you weren't stringifying the array twice.

      Actually you did that previously and I carelessly cargo-culted it into my code.
      However upon testing it's actually very slightly faster! I don't know why:
      perl -MBenchmark=:all -MAlgorithm::Combinatorics=:all -wle' for ("1".."4") { my $t0 = Benchmark->new; my $i = variations_with_repetition(["a".."z"],$ARGV[0]); my @x; push @x, pack "(A*)*", @$_ while $_=$i->next; my $t1 = Benchmark->new; my $ts = timestr(timediff($t1,$t0)); print $ts }' 4
      1 wallclock secs ( 0.77 usr +  0.01 sys =  0.78 CPU)
      1 wallclock secs ( 0.77 usr +  0.01 sys =  0.78 CPU)
      1 wallclock secs ( 0.75 usr +  0.00 sys =  0.75 CPU)
      1 wallclock secs ( 0.75 usr +  0.01 sys =  0.76 CPU)
      
      
      
      perl -MBenchmark=:all -MAlgorithm::Combinatorics=:all -wle' for ("1".."4") { my $t0 = Benchmark->new; my $i = variations_with_repetition(["a".."z"],$ARGV[0]); my @x; push @x, pack "A*", qq[@$_] while $_=$i->next; my $t1 = Benchmark->new; my $ts = timestr(timediff($t1,$t0)); print $ts }' 4
      0 wallclock secs ( 0.74 usr +  0.01 sys =  0.75 CPU)
      1 wallclock secs ( 0.72 usr +  0.01 sys =  0.73 CPU)
      1 wallclock secs ( 0.73 usr +  0.00 sys =  0.73 CPU)
      1 wallclock secs ( 0.74 usr +  0.01 sys =  0.75 CPU)
      
      
      
      perl -MBenchmark=:all -MAlgorithm::Combinatorics=:all -wle' for ("1".."4") { my $t0 = Benchmark->new; my $i = variations_with_repetition(["a".."z"],$ARGV[0]); my @x; push @x, join "", @$_ while $_=$i->next; my $t1 = Benchmark->new; my $ts = timestr(timediff($t1,$t0)); print $ts }' 4
      1 wallclock secs ( 0.65 usr +  0.01 sys =  0.66 CPU)
      1 wallclock secs ( 0.64 usr +  0.00 sys =  0.64 CPU)
      0 wallclock secs ( 0.63 usr +  0.00 sys =  0.63 CPU)
      1 wallclock secs ( 0.64 usr +  0.01 sys =  0.65 CPU)
      
      
      
      BTW, that is one seriously quick machine you're running. What is it?
      
      MacBook Pro i7
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