Re^4: Reverse Polish Notation Question

For a linear equation it could look like this (and yes, it took me more than 15 minutes...)

use strict;
use warnings;

my $equation = "(2*(4*x-8)+3*x)-(10*x+1-9)"; # " = 0" omitted
$equation =~ s/x/\$_[0]/g;
my $eval = sub { eval $equation };
my $val1 = $eval->(0);
my $val2 = $eval->(1);
die "No solution for this equation\n" unless $val1 - $val2;
my $result = -$val1 / ($val2 - $val1);

printf "Result is %.10f\n", $result;
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Re^5: Reverse Polish Notation Question by Laurent_R (Canon) on May 02, 2015 at 22:13 UTC
I tried to time your version and mine (removing the printouts). As I would expect yours is faster, but mine is converging fast enough to make the difference very small: Your solution: `$ time perl -e 'use strict; > use warnings; > my $val2 > = $eval->(1); my $equation = "(2(4x-8)+3x)-(10x+1-9)"; # " = 0" omitted > $equation =~ s/x/\$_[0]/g; > my $eval = sub { eval $equation }; > my $val1 = $eval->(0); > my $val2 = $eval->(1); > die "No solution for this equation\n" unless $val1 - $val2; > my $result = -$val1 / ($val2 - $val1); > > printf "Result is %.10f\n", $result;' Result is 8.0000000000 real 0m0.044s user 0m0.015s sys 0m0.031s` [download] and mine: `$ time perl solver.pl Result is: 8.0000000000 real 0m0.050s user 0m0.031s sys 0m0.015s` [download] 0.44 versus 0.50 sec. I might have been lucky on this one, but, frankly, my narrowing-down algorithm works fairly well. Je suis Charlie.	[reply] [d/l] [select]
Re^5: Reverse Polish Notation Question by Laurent_R (Canon) on May 02, 2015 at 17:44 UTC
Hi hdb, that's a pretty neat solution. It took me twenty seconds or so to figure out why you were replacing x by `$_[0]`. I must say that this is a nice clever idea. Je suis Charlie.	[reply] [d/l]