# RabbitFarm

### 2022-12-18

The Weekly Challenge 195 (Prolog Solutions)

*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 > 0. Write a script to print the count of all
special integers between 1 and $n.*

### Solution

```
code2digit(C, D):-
number_codes(D, [C]).
special(N):-
number_codes(N, NCodes),
maplist(code2digit, NCodes, Digits),
sort(Digits, DigitsSorted),
length(Digits, NumberDigits),
length(DigitsSorted, NumberDigits).
number_special(N, NumberSpecial):-
findall(I, (between(1, N, I), special(I)), Specials),
length(Specials, NumberSpecial).
```

### Sample Run

```
$ gprolog --consult-file prolog/ch-1.p
| ?- number_special(15, NumberSpecial).
NumberSpecial = 14
(1 ms) yes
| ?- number_special(35, NumberSpecial).
NumberSpecial = 32
(1 ms) yes
| ?-
```

### Notes

The definition of a *special integer* for this problem is an integer whose digits are
unique. To determine this specialness we split the number into its digits using
`number_codes/2`

and a `maplist/3`

which uses a small helper predicate to convert the
codes back to the corresponding digit.

After getting set with identifying special integers all the is left is to count up all the ones found in the given range.

## Part 2

*You are given a list of numbers, @list. Write a script to find most frequent even numbers
in the list. In case you get more than one even numbers then return the smallest even
integer. For all other case, return -1.*

### Solution

```
even(N, Even):-
(0 is mod(N, 2), Even = N);
(Even = nil).
frequency(ListNumbers, N, Frequency):-
delete(ListNumbers, N, ListDeleted),
length(ListNumbers, L),
length(ListDeleted, LD),
Frequency is L - LD.
most_frequent_even(ListNumbers, MostFrequentEven):-
maplist(even, ListNumbers, EN),
delete(EN, nil, EvenNumbers),
length(EvenNumbers, LengthEvens),
LengthEvens > 0,
maplist(frequency(ListNumbers), EvenNumbers, Frequencies),
msort(Frequencies, FS),
reverse(FS, FrequenciesSorted),
((
nth(1, FrequenciesSorted, F1),
nth(2, FrequenciesSorted, F2),
F1 \== F2,
nth(N, Frequencies, F1),
nth(N, EvenNumbers, MostFrequentEven)
);
(
nth(1, FrequenciesSorted, F1),
nth(2, FrequenciesSorted, F2),
F1 == F2,
findall(MFE, (member(FX, FrequenciesSorted), FX == F1, nth(N, Frequencies, FX), nth(N, EvenNumbers, MFE)), MostFrequentEvens),
sort(MostFrequentEvens, MostFrequentEvensSorted),
nth(1, MostFrequentEvensSorted, MostFrequentEven)
)
), !.
most_frequent_even(_, MostFrequentEven):-
MostFrequentEven = -1, !.
```

### Sample Run

```
$ gprolog --consult-file prolog/ch-2.p
| ?- most_frequent_even([1, 1, 2, 6, 2], MostFrequentEven).
MostFrequentEven = 2
yes
| ?- most_frequent_even([1, 3, 5, 7], MostFrequentEven).
MostFrequentEven = -1
yes
| ?- most_frequent_even([6, 4, 4, 6, 1], MostFrequentEven).
MostFrequentEven = 4
yes
| ?-
```

### Notes

The code here may look a bit more convoluted than it really is. Well my use of the
disjunction in `most_frequent/2`

may only be against my own personal sense of aesthetics!
Also, in balance the use of `maplist/3`

cleans things up a bit.

The main ideas here are:

Remove all odd numbers and check to see if any numbers remain.

Compute the frequency of each remaining even number.

Sort and see if there is a tie for most frequent.

If there is no tie in (3) then we're done in the first part of the disjunction. Otherwise, in the second part of the disjunction, find the smallest of the numbers tied for most frequent.

## References

posted at: 16:05 by: Adam Russell | path: /prolog | permanent link to this entry

#### Especially Frequent Even

*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 > 0. Write a script to print the count of all
special integers between 1 and $n.*

### Solution

```
use v5.36;
use boolean;
sub is_special{
my($x) = @_;
my %h;
my @digits = split(//, $x);
map{ $h{$_} = undef } @digits;
return keys %h == @digits;
}
MAIN:{
say q// . grep{ is_special($_) } 1 .. $ARGV[0];
}
```

### Sample Run

```
$ perl perl/ch-1.pl 15
14
$ perl perl/ch-1.pl 35
32
```

### Notes

The definition of a *special integer* for this problem is an integer whose digits are
unique. To determine this specialness we define `is_special()`

which splits any given
number into an array of digits. Each of the digits are added to a hash as the keys. If any
digits are not unique then they will not be duplicated as a hash key and the test will
return false.

Once `is_special()`

is set all we need to do is to map over the given range and count up
the results!

## Part 2

*You are given a list of numbers, @list. Write a script to find most frequent even numbers
in the list. In case you get more than one even numbers then return the smallest even
integer. For all other case, return -1.*

### Solution

```
use v5.36;
sub most_frequent_even{
my @list = @_;
@list = grep { $_ % 2 == 0 } @list;
return -1 if @list == 0;
my %frequencies;
map { $frequencies{$_}++ } @list;
my @sorted = sort { $frequencies{$b} <=> $frequencies{$a} } @list;
return $sorted[0] if $frequencies{$sorted[0]} != $frequencies{$sorted[1]};
my @tied = grep { $frequencies{$_} == $frequencies{$sorted[0]} } @list;
return (sort { $a <=> $b } @tied)[0];
}
MAIN:{
my @list;
@list = (1, 1, 2, 6, 2);
say most_frequent_even(@list);
@list = (1, 3, 5, 7);
say most_frequent_even(@list);
@list = (6, 4, 4, 6, 1);
say most_frequent_even(@list);
}
```

### Sample Run

```
$ perl perl/ch-2.pl
2
-1
4
```

### Notes

map and grep really do a lot to make this solution pretty succinct. First grep is used to extract just the even numbers. Then map is used to count up the frequencies. In the case of ties grep is used to identify the numbers with a tied frequency. The tied numbers are then sorted with the lowest one being returned, as specified.

## References

posted at: 00:53 by: Adam Russell | path: /perl | permanent link to this entry