Merge pull request #4658 from pjcs00/new-branch

New branch

4 files changed, 145 insertions, 0 deletions

diff --git a/challenge-024/peter-campbell-smith/README b/challenge-024/peter-campbell-smith/README
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+Solution by Peter Campbell Smith
+
diff --git a/challenge-024/peter-campbell-smith/ch-2.pl b/challenge-024/peter-campbell-smith/ch-2.pl
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+#!/usr/bin/perl
+
+# Peter Campbell Smith : 2021-08-02
+
+use strict;
+use utf8;
+use warnings;
+
+use Algorithm::Combinatorics qw(combinations);
+
+# Required: You are given a set of $n integers (n1, n2, n3, ….).
+
+# Write a script to divide the set in two subsets of n/2 sizes each so that the difference 
+# of the sum of two subsets is the least. If $n is even then each subset must be of size $n/2
+# each. In case $n is odd then one subset must be ($n-1)/2 and other must be ($n+1)/2.
+
+# Method:
+# The aim is to find a subset of $n/2 (or ($n-1)/2) of the input numbers which sums as close as
+# possible to half the sum of the input set (the 'target'). The remaining subset will then sum to 
+# a value the same distance from the other side of the target.
+
+my (@in, @out1, @out2, $target, $count, $subcount, $best, $set, $c, $total, $gap, $i, $string);
+
+# input set - uncomment one of these
+# @in = (10, 20, 30, 40, 50, 60, 70, 80, 90, 100);   # example 1 given in challenge
+# @in = (10, -15, 20, 30, -25, 0, 5, 40, -5);
+# @in = (18, 25, 7, 39, 11, 17, 22, 1 ,41, 19, -5);
+@in = (23, 67, 43, 96, 23, 14, 85, -300, 43, 20, 87, 99, 0, 35, 21, 77, 88, 77, 54, 70, 34, 56, 500);
+
+# some useful numbers
+$target = total(@in) / 2;   	# target value for each subset to add up to (half the total of the input numbers)
+$count = scalar @in;        	# number of numbers
+$subcount = int($count / 2);  	# number in each (or the smaller) subset
+
+# look for subset that is closest to the target
+$best = 1e10;   	# the closest gap found so far between the chosen combination and the target.
+
+# iterate over all possible combinations of the first $subcount numbers
+$set = combinations(\@in, $subcount);   
+while ($c = $set->next) {
+    $total = total(@$c);           # total of this combination
+	$gap = int($total - $target);  # gap between that and the target
+	if (abs($gap) < $best) {       # is it the best so far?
+		@out1 = @$c;
+		$best = abs($gap);
+		print qq[best so far $best\n];
+		last if $best == 0;
+	}
+}
+
+# get the other subset
+$string = '!' . join('!', @in) . '!';   # make the imput set into a string like !11!22!33!
+for $i (@out1) {
+	$string =~ s/!$i!/!/;   # remove each one which is in the fist subset
+}
+@out2 = split /!/, substr($string, 1);   # split the string back into an array
+
+# print the results
+print qq[
+Input:        Set = (] . join(', ', @in) . qq[) sum ] . total(@in) . qq[
+Output:  Subset 1 = (] . join(', ', @out1) . qq[) sum ] . total(@out1) . qq[
+         Subset 2 = (] . join(', ', @out2) . qq[) sum ] . total(@out2) . qq[
+];
+
+sub total {    # total(@x) returns sum of all the elements
+	
+	my ($i, $sum);
+	for $i (@_) { $sum += $i; }
+	return $sum;
+
+}
diff --git a/challenge-124/peter-campbell-smith/README b/challenge-124/peter-campbell-smith/README
new file mode 100644
index 0000000000..83f9d790e0
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+Solution by Peter Campbell Smith
diff --git a/challenge-124/peter-campbell-smith/perl/ch-2.pl b/challenge-124/peter-campbell-smith/perl/ch-2.pl
new file mode 100755
index 0000000000..247e04a2f6
--- /dev/null
+++ b/challenge-124/peter-campbell-smith/perl/ch-2.pl
@@ -0,0 +1,71 @@
+#!/usr/bin/perl
+
+# Peter Campbell Smith : 2021-08-02
+
+use strict;
+use utf8;
+use warnings;
+
+use Algorithm::Combinatorics qw(combinations);
+
+# Required: You are given a set of $n integers (n1, n2, n3, ….).
+
+# Write a script to divide the set in two subsets of n/2 sizes each so that the difference 
+# of the sum of two subsets is the least. If $n is even then each subset must be of size $n/2
+# each. In case $n is odd then one subset must be ($n-1)/2 and other must be ($n+1)/2.
+
+# Method:
+# The aim is to find a subset of $n/2 (or ($n-1)/2) of the input numbers which sums as close as
+# possible to half the sum of the input set (the 'target'). The remaining subset will then sum to 
+# a value the same distance from the other side of the target.
+
+my (@in, @out1, @out2, $target, $count, $subcount, $best, $set, $c, $total, $gap, $i, $string);
+
+# input set - uncomment one of these
+# @in = (10, 20, 30, 40, 50, 60, 70, 80, 90, 100);   # example 1 given in challenge
+# @in = (10, -15, 20, 30, -25, 0, 5, 40, -5);
+# @in = (18, 25, 7, 39, 11, 17, 22, 1 ,41, 19, -5);
+@in = (23, 67, 43, 96, 23, 14, 85, -300, 43, 20, 87, 99, 0, 35, 21, 77, 88, 77, 54, 70, 34, 56, 500);
+
+# some useful numbers
+$target = total(@in) / 2;   	# target value for each subset to add up to (half the total of the input numbers)
+$count = scalar @in;        	# number of numbers
+$subcount = int($count / 2);  	# number in each (or the smaller) subset
+
+# look for subset that is closest to the target
+$best = 1e10;   	# the closest gap found so far between the chosen combination and the target.
+
+# iterate over all possible combinations of the first $subcount numbers
+$set = combinations(\@in, $subcount);   
+while ($c = $set->next) {
+    $total = total(@$c);           # total of this combination
+	$gap = int($total - $target);  # gap between that and the target
+	if (abs($gap) < $best) {       # is it the best so far?
+		@out1 = @$c;
+		$best = abs($gap);
+		print qq[best so far $best\n];
+		last if $best == 0;
+	}
+}
+
+# get the other subset
+$string = '!' . join('!', @in) . '!';   # make the imput set into a string like !11!22!33!
+for $i (@out1) {
+	$string =~ s/!$i!/!/;   # remove each one which is in the fist subset
+}
+@out2 = split /!/, substr($string, 1);   # split the string back into an array
+
+# print the results
+print qq[
+Input:        Set = (] . join(', ', @in) . qq[) sum ] . total(@in) . qq[
+Output:  Subset 1 = (] . join(', ', @out1) . qq[) sum ] . total(@out1) . qq[
+         Subset 2 = (] . join(', ', @out2) . qq[) sum ] . total(@out2) . qq[
+];
+
+sub total {    # total(@x) returns sum of all the elements
+	
+	my ($i, $sum);
+	for $i (@_) { $sum += $i; }
+	return $sum;
+
+}