# RabbitFarm

### 2021-10-17

#### A Couple of Brute Force Computations

The examples used here are from The Weekly Challenge problem statement and demonstrate the working solution.

## Part 1

Write a script to generate first 5 Pandigital Numbers in base 10.

### Solution

``````
use strict;
use warnings;
##
# Write a script to generate first 5 Pandigital Numbers in base 10.
##
use boolean;

sub first_n_pandigitals {
my (\$n)         = @_;
my \$found       = false;
my \$pandigitals = [];
my \$x           = 1_000_000_000;
do {
my \$test = \$x;
push @{\$pandigitals}, \$x
if ( \$test =~ tr/0//d ) > 0
&& ( \$test =~ tr/1//d ) > 0
&& ( \$test =~ tr/2//d ) > 0
&& ( \$test =~ tr/3//d ) > 0
&& ( \$test =~ tr/4//d ) > 0
&& ( \$test =~ tr/5//d ) > 0
&& ( \$test =~ tr/6//d ) > 0
&& ( \$test =~ tr/7//d ) > 0
&& ( \$test =~ tr/8//d ) > 0
&& ( \$test =~ tr/9//d ) > 0;
\$found = ( @{\$pandigitals} == \$n );
\$x++;
} while ( !\$found );
return \$pandigitals;
}

sub first_5_pandigitals {
return first_n_pandigitals(5);
}
MAIN: {
my \$pandigitals = first_5_pandigitals;
for my \$x ( @{\$pandigitals} ) {
print "\$x\n";
}
}
``````

### Sample Run

``````
\$ perl perl/ch-1.pl
1023456789
1023456798
1023456879
1023456897
1023456978
``````

### Notes

From the definition we know that we will need at least 10 digits and, intuitively, the first five pandigital numbers will start with `1`. So then, we start with `1_000_000_000` and iterate upwards testing each candidate until we find the first five. The test used here is to determine if `tr` finds all the required digits.

## Part 2

You are given 2 positive numbers, \$m and \$n. Write a script to generate multiplication table and display count of distinct terms.

### Solution

``````
use strict;
use warnings;
##
# You are given 2 positive numbers, \$m and \$n.
# Write a script to generate multiplcation table and display count of distinct terms.
##
sub compute_print {
my ( \$m, \$n ) = @_;
my \$distinct = {};
print " x | " . join( " ", ( 1 .. \$n ) ) . "\n";
print "---+-" . "-" x ( \$n * 2 - 1 ) . "\n";
for my \$i ( 1 .. \$m ) {
print " \$i | " . join( " ", map { \$i * \$_ } ( 1 .. \$n ) ) . "\n";
for my \$j ( 1 .. \$n ) {
\$distinct->{ \$i * \$j } = undef;
}
}
return \$distinct;
}
MAIN: {
my \$distinct = compute_print( 3, 3 );
print "Distinct Terms: "
. join( ", ", sort { \$a <=> \$b } keys %{\$distinct} ) . "\n";
print "Count: " . keys( %{\$distinct} ) . "\n";
print "\n\n";
\$distinct = compute_print( 3, 5 );
print "Distinct Terms: "
. join( ", ", sort { \$a <=> \$b } keys %{\$distinct} ) . "\n";
print "Count: " . keys( %{\$distinct} ) . "\n";
}
``````

### Sample Run

``````
\$ perl perl/ch-2.pl
x | 1 2 3
---+------
1 | 1 2 3
2 | 2 4 6
3 | 3 6 9
Distinct Terms: 1, 2, 3, 4, 6, 9
Count: 6

x | 1 2 3 4 5
---+----------
1 | 1 2 3 4 5
2 | 2 4 6 8 10
3 | 3 6 9 12 15
Distinct Terms: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15
Count: 11
``````

### Notes

This is a perfectly Perl shaped problem. The computations can be handled in a straightforward way, especially with `map`. Getting rid of duplicates is done using the idiomatic method with hash keys. Finally, formatting the output cleanly is done without much undo stress. Compare what we do here to format the table with what was necessary to represent the same table in Prolog.

## References

Challenge 134

Pandigital Numbers

posted at: 13:03 by: Adam Russell | path: /perl | permanent link to this entry