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authorboblied <boblied@gmail.com>2023-01-25 09:10:18 -0600
committerboblied <boblied@gmail.com>2023-01-25 09:10:18 -0600
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Week 201 Task 2
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+#!/usr/bin/env perl
+# vim:set ts=4 sw=4 sts=4 et ai wm=0 nu:
+#=============================================================================
+# ch-2.pl Perl Weekly Challenge Week 201 Task 2 Penny Piles
+#=============================================================================
+# Copyright (c) 2023, Bob Lied
+#=============================================================================
+# You are given an integer, $n > 0.
+# Write a script to determine the number of ways of putting $n pennies
+# in a row of piles of ascending heights from left to right.
+# Example Input: $n = 5 Output: 7
+# Since $n=5, there are 7 ways of stacking 5 pennies in ascending piles:
+# 1 1 1 1 1
+# 1 1 1 2
+# 1 2 2
+# 1 1 3
+# 2 3
+# 1 4
+# 5
+#=============================================================================
+# This amounts to finding the partitions of a number
+# https://en.wikipedia.org/wiki/Partition_(number_theory)
+# "No closed-form expression for the partition function is known, but it has
+# both asymptotic expansions that accurately approximate it and recurrence
+# relations by which it can be calculated exactly."
+#
+# We will do a recursive expansion and count the results. For each pair of
+# possible sums, recurse if the second term is large enough to have sums
+# where both terms are bigger than the smaller summand.
+# n=7 n=8
+# 1 6 1 7
+# | 1 5 | 1 6
+# | | 1 4 | | 1 5
+# | | | 1 3 | | | 1 4
+# | | | | 1 2 | | | | 1 3
+# | | | | 1 1 | | | | 1 2
+# | | | 2 2 | | | | 1 1
+# | | 2 3 | | | | 2 2
+# | 2 4 | | | 2 3
+# | | 2 2 | | 2 4
+# | 3 3 | | | 2 2
+# 2 5 | | 3 3
+# | 2 3 | 2 5
+# 3 4 | | 2 3
+# | 3 4
+# Total = 15 2 6 Total = 22
+# 2 4
+# 2 2
+# 3 3
+# 3 5
+# 4 4
+#=============================================================================
+
+use v5.36;
+
+use Getopt::Long;
+my $Verbose = 0;
+my $DoTest = 0;
+
+GetOptions("test" => \$DoTest, "verbose" => \$Verbose);
+exit(!runTest()) if $DoTest;
+
+say pennyPiles($_) for @ARGV;
+
+sub pennyPiles($n)
+{
+ # We're accumulating all the possible orderings. We don't
+ # really have to do that to get the answer, but it helps debugging.
+ my @result = ( [ $n ] );
+ say "PUSH [ $n ]" if $Verbose;
+ _pile(1, $n-1, [], \@result, "");
+
+ say showResult(\@result) if $Verbose;
+ return scalar(@result);
+}
+
+sub _pile($p, $q, $soFar, $result, $indent)
+{
+ say "${indent}[$soFar->@*] _pile($p, $q)," if $Verbose;
+
+ # Take pairs of summands, but only in one order
+ while ( $p <= $q )
+ {
+ say "${indent}PUSH [ $soFar->@* $p $q ]" if $Verbose;
+ push @$result, [ $soFar->@*, $p, $q ];
+
+ # If the second term can be split into additions where both
+ # terms are greater than p, then recurse to that.
+ # For example, (2 6) can split the 6 into 2+4 or 3+3, but not
+ # 1+5 because the 1 would violate the ordering rule.
+ # For (3 4) we can't split the 4 in a way where both terms
+ # are at least 3.
+ if ( $q >= 2*$p )
+ {
+ _pile($p, $q-$p, [ $soFar->@*, $p ], $result, " $indent");
+ }
+ $p++; $q--;
+ }
+}
+
+sub showResult($result)
+{
+ for my $array ($result->@*) # ( sort { $#{$a} <=> $#{$b} } $result->@* )
+ {
+ say "[ $array->@* ]";
+ }
+}
+
+sub runTest
+{
+ use Test2::V0;
+
+ #is( pennyPiles(1), 1, "Test 1");
+ #is( pennyPiles(2), 2, "Test 2");
+ #is( pennyPiles(3), 3, "Test 3");
+ #is( pennyPiles(4), 5, "Test 4");
+ #is( pennyPiles(5), 7, "Example 1");
+ is( pennyPiles(6), 11, "Test 6");
+ is( pennyPiles(7), 15, "Test 7");
+ is( pennyPiles(8), 22, "Test 8");
+
+ done_testing;
+}
+