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| author | Mohammad S Anwar <Mohammad.Anwar@yahoo.com> | 2023-03-04 12:08:02 +0000 |
|---|---|---|
| committer | GitHub <noreply@github.com> | 2023-03-04 12:08:02 +0000 |
| commit | 0f4871ccdeaf6b48eec36e5c36c6d23a2e4d1ae9 (patch) | |
| tree | 23584d970d8ae690ebe270641375809c04517ffc | |
| parent | e1779b58b69af9f64e1c8d3c21ae01c8367143c4 (diff) | |
| parent | 8656b1e35c15e3ca565de5570bf423697bd24ca8 (diff) | |
| download | perlweeklychallenge-club-0f4871ccdeaf6b48eec36e5c36c6d23a2e4d1ae9.tar.gz perlweeklychallenge-club-0f4871ccdeaf6b48eec36e5c36c6d23a2e4d1ae9.tar.bz2 perlweeklychallenge-club-0f4871ccdeaf6b48eec36e5c36c6d23a2e4d1ae9.zip | |
Merge pull request #7659 from jo-37/contrib
Solutions to challenge 206
| -rwxr-xr-x | challenge-206/jo-37/perl/ch-1.pl | 63 | ||||
| -rwxr-xr-x | challenge-206/jo-37/perl/ch-2.pl | 91 |
2 files changed, 154 insertions, 0 deletions
diff --git a/challenge-206/jo-37/perl/ch-1.pl b/challenge-206/jo-37/perl/ch-1.pl new file mode 100755 index 0000000000..923a8f1088 --- /dev/null +++ b/challenge-206/jo-37/perl/ch-1.pl @@ -0,0 +1,63 @@ +#!/usr/bin/perl -s + +use v5.16; +use Test2::V0 '!float'; +use PDL; + +our ($tests, $examples); + +run_tests() if $tests || $examples; # does not return + +die <<EOS unless @ARGV > 1; +usage: $0 [-examples] [-tests] [H1:M1 H2:M2...] + +-examples + run the examples from the challenge + +-tests + run some tests + +H1:M1 H2:M2... + list of times in 24h-format HH:MM. + +EOS + + +### Input and Output + +say shortest_time(@ARGV); + + +### Implementation + +sub shortest_time { + # Parse times, calculate minutes, convert to "long" piddle and sort. + my $times = long(map {/^(\d{2}):(\d{2})$/ ? 60 * $1 + $2 : ()} @_)->qsort; + # Take the minimum over the difference of cyclic consecutive time + # pairs modulo one day. + min +($times->rotate(-1) - $times) % 1440; + + +} + + +### Examples and tests + +sub run_tests { + SKIP: { + skip "examples" unless $examples; + + is shortest_time("00:00", "23:55", "20:00"), 5, 'example 1'; + is shortest_time("01:01", "00:50", "00:57"), 4, 'example 2'; + is shortest_time("10:10", "09:30", "09:00", "09:55"), 15, 'example 3'; + } + + SKIP: { + skip "tests" unless $tests; + + is shortest_time(qw(23:00 18:00 01:00)), 120, 'carry minimum over'; + } + + done_testing; + exit; +} diff --git a/challenge-206/jo-37/perl/ch-2.pl b/challenge-206/jo-37/perl/ch-2.pl new file mode 100755 index 0000000000..79c301738a --- /dev/null +++ b/challenge-206/jo-37/perl/ch-2.pl @@ -0,0 +1,91 @@ +#!/usr/bin/perl -s + +use v5.16; +use Test2::V0 '!float'; +use PDL; +use PDL::NiceSlice; + +our ($tests, $examples); + +run_tests() if $tests || $examples; # does not return + +die <<EOS unless @ARGV && !(@ARGV % 2); +usage: $0 [-examples] [-tests] [N1 N2...] + +-examples + run the examples from the challenge + +-tests + run some tests + +N1 N2... + even sized list of numbers + +EOS + + +### Input and Output + +say max_pairsum(@ARGV); + + +### Implementation + +# We are given an even sized list of numbers (N₁,...,N₂ₙ). The list +# shall be partitioned into pairs. There is a target function defined +# as the sum over the minimum of all the pairs. This target function +# shall be maximized, i.e. the task is to find a partitioning into pairs +# where the target is maximal and to provide this maximum value. +# +# For the sake of convenience we assume all numbers being distinct. We +# partition the list into pairs +# ((P¹₁, P¹₂),...,(Pⁿ₁, Pⁿ₂)) +# Neither the order of pairs, nor the order of the paired numbers are +# relevant for the target function. Thus we can chose an arbitrary pair +# P = (a, b), Q = (c, d) and arrange it such that +# a < b, c < d and a < c +# by exchanging the pairs and/or the numbers within the pairs without +# altering the value of the target function. The contributions to the +# target function are min(P) = a and min(Q) = c. +# +# Now suppose b > c, i.e. the two pairs "overlap". +# We build a different pairing where b and c are +# exchanged and all other pairs remain unaltered, i.e. +# P' = (a, c) and Q' = (b, d) +# The new contributions to the target function are +# min(P') = a = min(P) (from the precondition a < c) and +# min(Q') = min(b, d) > min(c, d) = min(Q) (from our assumption b > c +# and the precondition c < d) +# All summands in the target function remain unchanged with the +# exception of Q', which increased from Q. +# +# There is only one partitioning into pairs where none of the pairs +# overlap: Taking the pairs as adjacent elements from the sorted list. +# From the above considerations, this partitioning maximizes the target +# function. +# +# A long introduction to a short solution: +sub max_pairsum { + # Sum over the first elements of successive pairs taken from the + # sorted list. + sum pdl(@_)->qsort->(0:-2:2); +} + + +### Examples and tests + +sub run_tests { + SKIP: { + skip "examples" unless $examples; + + is max_pairsum(1,2,3,4), 4, 'example 1'; + is max_pairsum(0,2,1,3), 2, 'example 2'; + } + + SKIP: { + skip "tests" unless $tests; + } + + done_testing; + exit; +} |
