John Guttag's book - 2nd exercise. My attempt in Perl.

pritesh_ugrankar has asked for the wisdom of the Perl Monks concerning the following question:

Replies are listed 'Best First'.
Re: John Guttag's book - 2nd exercise. My attempt in Perl. by choroba (Cardinal) on May 20, 2017 at 20:32 UTC
> While this works fine for all the test cases I tried Does it? I'm getting `4 is the biggest odd number` [download] for the very sample you gave. Maybe the second `$z` on line 26 should be `$y` ? Update: Here's my solution. It first sorts the numbers, then finds the first odd one starting from the greatest one. `sub cmp3 { my ($x, $y, $z) = @_; ($y, $x) = ($x, $y) if $y < $x; ($z, $y) = ($y, $z) if $z < $y; ($y, $x) = ($x, $y) if $y < $x; if (1 == $z % 2) { return "$z is the biggest odd number\n" } elsif (1 == $y % 2) { return "$y is the biggest odd number\n" } elsif (1 == $x % 2) { return "$x is the biggest odd number\n" } else { return "All are even numbers\n" } }` [download] Tested against yours with Read more... (492 Bytes) ($q=q:Sq=~/;[c](.)(.)/;chr(-\|\|-\|5+lengthSq)`"S\|oS2"`map{chr \|+ord }map{substrSq`S_+\|`\|}3E\|-\|`7**2-3:)=~y+S\|`+$1,++print+eval$q,q,a, [download]	[reply] [d/l] [select]
Re: John Guttag's book - 2nd exercise. My attempt in Perl. by duelafn (Parson) on May 20, 2017 at 21:44 UTC
This will do it a bit more nicely (fewer branches) without modifying the variables (and using only if/elsif as requested): `use 5.010; my ($x, $y, $z) = (-11,-13,4); if ($x % 2 and ($x > $y or not $y % 2) and ($x > $z or not $z % 2)) { say "$x is the largest odd"; } elsif ($y % 2 and ($y > $z or not $z % 2)) { say "$y is the largest odd"; } elsif ($z % 2) { say "$z is the largest odd"; } else { say "All numbers are even!"; }` [download] Update: If I were doing this for real, I'd do: `my @nums = (-11, -13, 4); my @candidates = sort { $a <=> $b } grep $_ % 2, @nums; if (@candidates) { say $candidates[-1]; } else { say "They're all even!"; }` [download] Good Day, Dean	[reply] [d/l] [select]
Re: John Guttag's book - 2nd exercise. My attempt in Perl. by kcott (Archbishop) on May 21, 2017 at 06:29 UTC
G'day pritesh_ugrankar, Here's one way to do it: #!/usr/bin/env perl use strict; use warnings; use constant { INPUT => 0, EXPECT => 1, NO_ODDS => 'no odds found' }; use Test::More; my @tests = ( [[qw{-11 -13 4}], -11], [[qw{-12 -14 4}], NO_ODDS], [[qw{0 0 0}], NO_ODDS], [[qw{1 1 1}], 1], [[qw{1 0 1}], 1], [[qw{0 0 1}], 1], [[qw{2 3 2}], 3], [[qw{2 -3 2}], -3], [[qw{-2 -3 -4}], -3], [[qw{-1 -3 -5}], -1], [[qw{3 5 7}], 7], [[qw{7 5 3}], 7], [[qw{5 7 3}], 7], ); plan tests => scalar @tests; for my $test (@tests) { my ($x, $y, $z) = @{$test->[INPUT]}; my $u = ($x, $y)[$x % 2 ? ($y % 2 ? $x < $y : 0) : 1]; my $v = ($u, $z)[$u % 2 ? ($z % 2 ? $u < $z : 0) : 1]; ok((NO_ODDS, $v)[$v % 2] eq $test->[EXPECT]); } [download] See Test::More if you're not familiar with that module. It's very useful in situations like this where you want to test a solution against a variety of input data. All those tests passed: `1..13 ok 1 ... 2 to 12 all 'ok' also ... ok 13` [download] As you can hopefully see, it's very easy to add more test data by simply adding more elements to `@tests`. — Ken	[reply] [d/l] [select]
Re: John Guttag's book - 2nd exercise. My attempt in Perl. by shmem (Chancellor) on May 20, 2017 at 20:55 UTC
Point to note is, it should only use if/else/elsif loops and / or comparison operators. So no modulo operator, no division operator, no arithmetic operators, no bitwise foo... how would you determine oddness only with comparison? You would need an array which holds all odd values of `\|N`. My computer can't. perl -le'print map{pack c,($-++?1:13)+ord}split//,ESEL'	[reply] [d/l]
Re^2: John Guttag's book - 2nd exercise. My attempt in Perl. by Laurent_R (Canon) on May 20, 2017 at 21:29 UTC
I guess the OP meant to say that you should not use things like `sort`, `map`, `grep` or `List::Util`. The code in the OP is using the modulo operator, so we should assume this is allowed.	[reply] [d/l] [select]
Re^3: John Guttag's book - 2nd exercise. My attempt in Perl. by shmem (Chancellor) on May 20, 2017 at 21:37 UTC
I guess the OP meant to say Yeah, that leads to I know what I mean. Why don't you? perl -le'print map{pack c,($-++?1:13)+ord}split//,ESEL'	[reply]
Re^4: John Guttag's book - 2nd exercise. My attempt in Perl. by Laurent_R (Canon) on May 20, 2017 at 21:56 UTC
Re^2: John Guttag's book - 2nd exercise. My attempt in Perl. by pritesh_ugrankar (Monk) on May 21, 2017 at 10:58 UTC
Hi, Modulo Operator can be used as I've used in my code. Sorry I should've stated that earlier. Thinkpad T430 with Ubuntu 16.04.2 running perl 5.24.1 thanks to plenv!!	[reply] [d/l]
Re: John Guttag's book - 2nd exercise. My attempt in Perl. by KurtZ (Friar) on May 20, 2017 at 20:38 UTC
Well a generic solution is to loop over all variables and check against a $max which is initially undef and grows whenever a bigger odd variable is encountered. You may want to store them in a hash to keep the names associated (are hashes allowed? )	[reply]
Re^2: John Guttag's book - 2nd exercise. My attempt in Perl. by marinersk (Priest) on May 21, 2017 at 04:30 UTC
Coded as such: use strict; use warnings; # Initialize my ($x, $y, $z) = (-11,-13,4); my $maxodd; # Intentionally left as undef # Check $x (though $maxodd will be undef, write it consistently in cas +e code is revectored later) if ($x % 2) { if ( (!defined $maxodd) \|\| ( (defined $maxodd) && ($x > $maxodd) ) + ) { $maxodd = $x; } } # Check $y if ($y % 2) { if ( (!defined $maxodd) \|\| ( (defined $maxodd) && ($y > $maxodd) ) + ) { $maxodd = $y; } } # Check $z if ($z % 2) { if ( (!defined $maxodd) \|\| ( (defined $maxodd) && ($z > $maxodd) ) + ) { $maxodd = $z; } } # Report results if (!defined $maxodd) { print "All are even numbers\n"; } else { print "$maxodd is the greatest odd number\n"; } # Fini exit; [download] This screams for the use of a subroutine, though I suspect the author isn't at that point in the book yet.	[reply] [d/l]
Re: John Guttag's book - 2nd exercise. My attempt in Perl. by shmem (Chancellor) on May 21, 2017 at 08:29 UTC
This is to be tackled using only if/else loop and comparisons, because that is what the author has covered till that part. Well, the modulo operator should be permitted. We could shortcut with next and unless, but if it hasn't been covered, an inner `if()` block will do: `use strict; use warnings; my ($x, $y, $z) = (-11,-13,4); my $max; for($x,$y,$z) { # next unless $_ % 2; if ($_ % 2) { $max = $_ unless defined $max; $max = $_ if $max < $_; } } print "max odd: ", $max \|\| "not found","\n";` [download] update: darn, I wrote a `for()` loop, which ostensibly hasn't been covered up to that point. But why hasn't it? OTOH, you wrote if/else loop - anyways, to avoid the `for()` loop, let's unroll it: `use strict; use warnings; my ($x, $y, $z) = (-11,-13,4); my $max = 0; if($x % 2) { if ( ! $max ) { $max = $x; } elsif ($max < $x ) { $max = $x; } } if($y % 2) { if ( ! $max ) { $max = $y; } elsif ($max < $y ) { $max = $y; } } if($z % 2) { if ( ! $max ) { $max = $z; } elsif ($max < $z ) { $max = $z; } } print "max odd: ", $max \|\| "not found","\n";` [download] This would be a good example for introducing loop control, leading to the loop layed out above. update: made each 2^nd `if` block into `elsif`, as per choroba's comment below. perl -le'print map{pack c,($-++?1:13)+ord}split//,ESEL'	[reply] [d/l] [select]
Re^2: John Guttag's book - 2nd exercise. My attempt in Perl. by choroba (Cardinal) on May 21, 2017 at 11:36 UTC
As $x can never be less than $x, you can turn each second `if` in each block into `elsif` (provided it has been covered). ($q=q:Sq=~/;[c](.)(.)/;chr(-\|\|-\|5+lengthSq)`"S\|oS2"`map{chr \|+ord }map{substrSq`S_+\|`\|}3E\|-\|`7**2-3:)=~y+S\|`+$1,++print+eval$q,q,a, [download]	[reply] [d/l] [select]
Re^3: John Guttag's book - 2nd exercise. My attempt in Perl. by shmem (Chancellor) on May 21, 2017 at 12:13 UTC
~~Sure, but the point was to unroll the loop.~~ Ah, I see. Right. Fixed. perl -le'print map{pack c,($-++?1:13)+ord}split//,ESEL'	[reply]
Re: John Guttag's book - 2nd exercise. My attempt in Perl. by Laurent_R (Canon) on May 20, 2017 at 21:48 UTC
Maybe this subroutine: `use strict; use warnings; my @array = (-11,-13,4, 22, 17); print find_max_odd(@array), "\n"; sub find_max_odd { my $max; while ($max = shift) { last if $max % 2; } return "Not found" unless @array; while (my $current = shift) { next unless $current % 2; if ($current > $max) { $max = $current; } } return $max; }` [download] Update: As pointed out by AnoMonk below, these while loops have a bug when one of the array item is 0.I suggest a better for loop in another post below.	[reply] [d/l]
Re^2: John Guttag's book - 2nd exercise. My attempt in Perl. by KurtZ (Friar) on May 20, 2017 at 22:12 UTC
you can avoid the first loop if you change `if ($current > $max) {` to `if ( not defined $max or $current > $max) {` $max would simply stay undef if all values were even. no it's more complicated ... wait This should do: `use strict; use warnings; my @array = ( -11, -13, 4, 22, 17); print "Result: ",find_max_odd(@array) // "undefined", "\n"; sub find_max_odd { my $max; for (@_) { next unless $_ % 2; unless ( defined $max ) { $max = $_; next; } $max = $_ if $_ > $max; } return $max; }` [download] I was confused, but becaus eof short-circuiting of the or the first idea should do too: `use strict; use warnings; my @array = (-11,-13,4, 22, 17); print find_max_odd(@array), "\n"; sub find_max_odd { my $max; while (my $current = shift) { next unless $current % 2; if (not defined $max or $current > $max) { $max = $current; } } return $max // "undefined"; }` [download]	[reply] [d/l] [select]
Re^3: John Guttag's book - 2nd exercise. My attempt in Perl. by Laurent_R (Canon) on May 21, 2017 at 09:27 UTC
Yeah, KurtZ, I definitely agree with you that it can be made simpler (++), and I most likely wouldn't code this the way I have shown. I only suggested a solution that used only very basic Perl knowledge since it was what the OP was asking for. Also, having two loops did not bother me too much, since they're browsing the array items only once anyway.	[reply]
Re^2: John Guttag's book - 2nd exercise. My attempt in Perl. by Anonymous Monk on May 22, 2017 at 18:03 UTC
Should `while (my $current = shift)` be `while ( defined( my $current = shift ) )` ? Otherwise it seems to fail if `@array` contains 0.	[reply] [d/l] [select]
Re^3: John Guttag's book - 2nd exercise. My attempt in Perl. by Laurent_R (Canon) on May 22, 2017 at 19:19 UTC
Yes, Anonymous Monk, you're correct, this `while` loop has a bug when one of the values in the array is 0. IMHO, this array traversal should really be made with a `for` loop, as it is usually better to use a `for` loop to iterate over an array, but the OP did not specify very clearly what was authorized and what not. This is what it might look like with a `for` loop. `use strict; use warnings; my @array = (-11,-13,4, 22, 17); print find_max_odd(@array), "\n"; sub find_max_odd { my $max; for my $num (@_) { next unless $num % 2; $max = $num unless defined $max; $max = $num if $num > $max; } return defined $max ? $max : "No odd number found\n"; # this co +uld be replaced with if conditional statements if needed }` [download]	[reply] [d/l] [select]
Re: John Guttag's book - 2nd exercise. My attempt in Perl. by johngg (Canon) on May 20, 2017 at 22:30 UTC
You could use grep and the core List::Util`::max()` to get what you want. `johngg@shiraz:~/perl/Monks > perl -Mstrict -Mwarnings -MList::Util=max + -E ' my @tests = ( [ 17, -3, 12 ], [ 2, 4, 6 ], [ -7, -4, -11 ], [ -5, 5, 16 ], ); foreach my $test ( @tests ) { my $maxOdd = max grep { $_ % 2 } @$test; say $maxOdd ? $maxOdd : q{No result}; }' 17 No result -7 5` [download] I hope this is helpful. Cheers, JohnGG	[reply] [d/l] [select]
Re: John Guttag's book - 2nd exercise. My attempt in Perl. by tybalt89 (Monsignor) on May 21, 2017 at 17:11 UTC
Here's a solution using only if/else, <, and %. No \|\|, &&, and, or, =, or anything else allowed. It's sort of silly for something that is a one-liner in normal perl. ( note: the comments following open braces show which variables are still valid at this point.) #!/usr/bin/perl # http://perlmonks.org/?node_id=1190754 use strict; use warnings; my ($x, $y, $z) = (-11,-13,4); if( $x % 2 ) { # xyz if( $y % 2 ) { # xyz if( $x < $y ) { # yz if( $z % 2 ) { # yz if( $y < $z ) { # z print "z of $z is the biggest odd number\n"; } else { # y print "y of $y is the biggest odd number\n"; } } else { # y print "y of $y is the biggest odd number\n"; } } else { # xz if( $z % 2 ) { # xz if( $x < $z ) { # z print "z of $z is the biggest odd number\n"; } else { # x print "x of $x is the biggest odd number\n"; } } else { # x print "x of $x is the biggest odd number\n"; } } } else { # xz if( $z % 2 ) { # xz if( $x < $z ) { # z print "z of $z is the biggest odd number\n"; } else { # x print "x of $x is the biggest odd number\n"; } } else { # x print "x of $x is the biggest odd number\n"; } } } else { # yz if( $y % 2 ) { # yz if( $z % 2 ) { # yz if( $y < $z ) { # z print "z of $z is the biggest odd number\n"; } else { # y print "y of $y is the biggest odd number\n"; } } else { # y print "y of $y is the biggest odd number\n"; } } else { # z if( $z % 2 ) { # z print "z of $z is the biggest odd number\n"; } else { # print "All are even numbers\n"; } } } [download]	[reply] [d/l]
Re^2: John Guttag's book - 2nd exercise. My attempt in Perl. by Anonymous Monk on May 22, 2017 at 22:26 UTC
Now write one that checks 26 vars: $a .. $z Also, nested ifs count as ORs and ANDs :D	[reply]
Re^3: John Guttag's book - 2nd exercise. My attempt in Perl. by tybalt89 (Monsignor) on May 22, 2017 at 22:38 UTC
No, they don't.	[reply]
Re^4: John Guttag's book - 2nd exercise. My attempt in Perl. by Anonymous Monk on May 22, 2017 at 22:53 UTC
Re^2: John Guttag's book - 2nd exercise. My attempt in Perl. by Anonymous Monk on May 22, 2017 at 22:24 UTC
I wish I could downvote this 100 times.	[reply]
Re^3: John Guttag's book - 2nd exercise. My attempt in Perl. by tybalt89 (Monsignor) on May 22, 2017 at 22:34 UTC
I'm just following the posted rules :) Let's see your solution following the posted rules...	[reply]
Re^4: John Guttag's book - 2nd exercise. My attempt in Perl. by Anonymous Monk on May 22, 2017 at 22:54 UTC
Re: John Guttag's book - 2nd exercise. My attempt in Perl. by BillKSmith (Monsignor) on May 22, 2017 at 20:49 UTC
I am probably to late to be of any help, but I can add one significant detail. I find the logic more intuitive if we initialize the result with a large negative number. You can get the largest possible number from POSIX, but with only a small loss in generality, we can choose any number that is larger than any we expect in our input. `se strict; use warnings; #use POSIX qw(DBL_MAX); my $x = $ARGV[0] // -11; my $y = $ARGV[1] // -13; my $z = $ARGV[2] // -17; #my $MAX_NEG = -DBL_MAX; my $MAX_NEG = -99999; # Better to use DBL_MAX from POSIX my $max = $MAX_NEG; if ( $x % 2) { $max = $x ; } if ( $y % 2 and $y > $max ) { $max = $y ; } if ( $z % 2 and $z > $max ) { $max = $z ; } die "There are no odd values.\n" if ($max == $MAX_NEG) ; print "the largest odd number is $max.\n";` [download] Note: I took the freedom to use the disallowed operator (//) to override the default values of $x, $y, and $z with values from the command line. I do not consider this part of the solution. Bill	[reply] [d/l]
Re: John Guttag's book - 2nd exercise. My attempt in Perl. by Anonymous Monk on May 22, 2017 at 22:23 UTC
Not 3 vars but any cli args: `my $largest_odd; for (@ARGV) { $largest_odd = $_ if $_ % 2 && $_ > $largest_odd; } print $largest_odd ? $largest_odd : 'nothing odd', $/;` [download]	[reply] [d/l]
Re: John Guttag's book - 2nd exercise. My attempt in Perl. by pritesh_ugrankar (Monk) on May 25, 2017 at 19:18 UTC
Respected Monks, Thank you for all your answers. After seeing all the replies here, I too had the urge to try it out my way, so I threw away all the restrictions that the example had put and tried it my way, so that it will work for any numbers, positive or negative, decimal/hex/octal/binary, and came up with this. `use strict; use warnings; my @numbers = (10, 110, 24, 30, -17,-9,12,14,-011,-0xF,0b111,0); sub biggest_odd { my @odd_only = grep {$_ % 2 != 0} @_; if (!@odd_only) { print "No odd numbers at all\n"; return; } my $biggest_odd_num = shift @odd_only; foreach my $odd_num(@odd_only) { if ($odd_num > $biggest_odd_num) { $biggest_odd_num = $odd_num; } } print "Biggest odd is $biggest_odd_num\n"; } &biggest_odd(@numbers);` [download] And here is the output: `C:perlscripts>perl jg_find_biggest_odd.pl Biggest odd is 7` [download] Like I said earlier, I really love Perl, but I find the John Guttag book so interesting that I've started learning Python just so that I can understand the concepts given in the book. I don't know how much successful I will be learning two languages simultaneously, especially given the fact that I love the Perl way of solving problems... Thinkpad T430 with Ubuntu 16.04.2 running perl 5.24.1 thanks to plenv!!	[reply] [d/l] [select]


Pathologically Eclectic Rubbish Lister
	PerlMonks