PWC 215

PWC 215

I have attempted the Perl Weekly Challenge after a hiatus of several weeks, as I was too busy at work. Still busy, so I have just attempted the current challenge in Perl 5. Like I like to do, I have restricted myself to perl 4 syntax as specified in the first edition of Programming perl.

Update and disclaimer: These programs will not work on Perl 4 as I discovered later when I found and installed a Perl 4.019 executable. For one thing, the ability to assign a subroutine to a locally scoped typeglob was not available in Perl 4.

Challenge 1 (Odd one out)

In this challenge, we are given a list of alphabetic strings, and asked to return the list without the strings in which the characters are not sorted (I assume that this means sorted in lexical ascending order). We are also asked to print the number of strings that were removed (i.e., that were not sorted). 

I use a subroutine &odd_one_out(@words) with a nested helper sub-sub-routine to check if a string is sorted. Here is the helper sub-sub routine (not as compact as it could be, but this is easier to read):

 local *is_sorted = sub { #not idiomatic but I prefer readable
    local ($string)=@_;
    local $chk=join('',sort {$a cmp $b} split(//,$string));
    return ($string eq $chk);
  };

Here is the core of &odd_one_out:

  for $word (@words) {
    if (&is_sorted($word)) {
        push @retval, $word;
    } 
    else {
        $ctr++;
    }
  }
  print $ctr;
  return @retval;

Challenge 2 (Number placement)

In this challenge, we are given a list of single-digit numbers restricted to 0 or 1, as well as a counter say $count. We are asked to find if there are at least $count 0's in the list that can be replaced with a 1, where the rule is that a 0 can only be replaced by a 1 if it does not have a 1 on either side of it prior to any replacement. 

As with Challenge 1, I use a subroutine with a nested helper sub-subroutine that checks if a digit has an adjacent 1. Here is the helper sub-subroutine (it takes the position of the digit in the original list as input):

local *has_1_neighbor = sub {
        local ($indx)=@_;
        
        ( ($indx > 0) && ($numbers[$indx-1] =~ /^1$/) )
        || 
        ( ($indx < @numbers-1) && ($numbers[$indx+1] =~ /^1$/ ));
    };   

Here is the core of the &number_placement subroutine:

   local ($indx,$ctr);
    $ctr=0;
    
    for $indx (0 .. @numbers-1) {
        ( ($numbers[$indx] =~ /^0$/) && (!(&has_1_neighbor($indx))) )
        && ($ctr++);
        
        last if ($ctr >= $count); 
    }
    
    ($ctr >= $count) ? 1 : 0; #-- the last line or return value




Comments

Post a Comment

Popular posts from this blog

PWC 258

PWC 258 Last week (#257) I finished challenge 2 and my blog post for challenge 257 late last week, so had only submitted challenge 1. Here is my blog post for PWC 257 , including the full code for my challenge 2 solution. This week, I have done both challenges in perl 4.019 on DOSBOX. Both scripts also work fine on the latest Perl 5.38. I also did the challenges in Raku, where they are one-liners.  Challenge 1 (Count Even Digits Number) We are given a list of integers, and asked to count the number of elements whose base-10 representation has an even number of digits. In Raku , this is a one-liner. Here it is: raku -e 'say +@*ARGS.grep({$_.chars %% 2})' $@ It could be a one-liner in Perl 4 too, but since I run it on DOSBOX, my script would not be a bash shell script. Here is the short subroutine that does the job.  sub count_even_digits {  scalar( grep( ((length($_) % 2) == 0), @_) ); } The full code is here . Challenge 2 (Sum of values) We are given an integer k  and a list in
Read more

PWC 249

PWC 249 Challenge 1 (Equal pairs) We are given a list of integers with an even number of elements. We are asked to check if the list can be completely divided into sub-lists of size 2 with each sub-list containing the same integer repeated, like say (2,2). If so, we are asked to return the list of such sub-lists. This is straightforward in Perl 5. I use modern conveniences this time, though it could be done using only primitive Perl 4 syntax. Here is my subroutine. It returns a list of array references. local *equal_pairs = sub {     #-- return () if input is of odd length     ((scalar(@_) % 2) > 0) && (return ());          #-- count number of occurrences of each element using a hash     local %_ = ();     map {$_{$_}++} @_;          #-- if any element occurs an odd number of times, return ()     (grep {($_{$_} % 2) > 0} keys %_) && (return ());          #-- loop thru %_ keys, returning pairs of such keys     map {         my $k=$_;         my $num_pairs=( ($_{$k}
Read more

PWC 253

PWC 253 I do both challenges this week using old software on DOSBOX: perl 4.019, and for comparison, also python 1.4 beta. My solutions also work on the current versions of Perl 5 (5.38) and Python 2 (2.7.18) respectively. Challenge 1 (Split Strings) Links to code: perl4.019 python1.4b We are given a list of strings, and a single character say sep . We are asked to return the list, with the strings split into component substrings using sep as a separator. We should not include any empty strings in the output. In perl 4, I first stick a backslash in front of the separator character, so that it is escaped if entered in a regex. This is because the perl split function takes a regex as its first argument indicating the separator. If we have a character like '$' or ',', we want to escape it so that it is read literally when parsing the regex. Here is the key subroutine:   sub split_strings {  local ($separator, @words)=@_;  local (@retval)=(); #array to be returned  #  #--
Read more