# RabbitFarm

### 2021-07-11

#### Swapping Bits / Time Angle

The examples used here are from the weekly challenge problem statement and demonstrate the working solution.

## Part 1

You are given a positive integer \$N less than or equal to 255. Write a script to swap the odd positioned bits with the even positioned bits and print the decimal equivalent of the new binary representation.

### Solution

``````
use strict;
use warnings;
sub swap_bits{
my(\$n) = @_;
my \$bits = substr(unpack("B32", pack("N", shift)), 24, 8);
my @bits = split(//, \$bits);
for(my \$i = 0; \$i < @bits; \$i += 2){
@bits[\$i, \$i + 1] = @bits[\$i + 1, \$i];
}
my \$swapped_decimal = unpack("N", pack("B32", substr("0" x 32 . join("", @bits), -32)));
return \$swapped_decimal;
}

MAIN:{
my \$N;
\$N = 101;
print swap_bits(\$N) . "\n";
\$N = 18;
print swap_bits(\$N) . "\n";
}
``````

### Sample Run

``````
\$ perl perl/ch-1.pl
154
33
``````

### Notes

This code re-uses much of the code from last week's challenge solution. The only difference here is the for loop which swaps the even/odd bits.

## Part 2

You are given time \$T in the format hh:mm. Write a script to find the smaller angle formed by the hands of an analog clock at a given time.

### Solution

``````
use strict;
use warnings;
sub clock_angle{
my(\$h, \$m) = split(/:/, \$_[0]);
my \$angle = abs(0.5 * (60 * \$h - 11 * \$m));
\$angle = 360 - \$angle if \$angle > 180;
return \$angle;
}

MAIN:{
my \$T;
\$T = "03:10";
print clock_angle(\$T) . "\n";
\$T = "04:00";
print clock_angle(\$T) . "\n";
}
``````

### Sample Run

``````
\$ perl perl/ch-1.pl
35
120
``````

### Notes

Perhaps not a whole lot going on here: the time is broken into hour and minute parts and then the angle is computed directly from those values.

## References

Challenge 120

posted at: 17:41 by: Adam Russell | path: /perl | permanent link to this entry