# RabbitFarm

### 2022-09-04

#### The Weekly Challenge 180 (Prolog Solutions)

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

## Part 1

You are given a string. Write a script to find out the first unique character in the given string and print its index (0-based).

### Solution

``````
index_first_unique(Words, IndexUnique):-
index_first_unique(Words, 0, IndexUnique).
index_first_unique(String, I, IndexUnique):-
succ(I, Index),
length(String, Length),
nth(Index, String, Character),
delete(String, Character, Deleted),
length(Deleted, LengthDeleted),
LengthDifference is Length - LengthDeleted,
LengthDifference == 1, !,
IndexUnique = I.
index_first_unique(String, I, IndexUnique):-
succ(I, Index),
length(String, Length),
nth(Index, String, Character),
delete(String, Character, Deleted),
length(Deleted, LengthDeleted),
LengthDifference is Length - LengthDeleted,
\+ LengthDifference == 1,
succ(I, X),
index_first_unique(String, X, IndexUnique).
``````

### Sample Run

``````
\$ gprolog --consult-file prolog/ch-1.p
| ?- index_first_unique("Long Live Perl", IndexUnique).

IndexUnique = 1

yes
| ?- index_first_unique("Perl Weekly Challenge", IndexUnique).

IndexUnique = 0

yes
| ?- index_first_unique("aabbcc", IndexUnique).

no
| ?- index_first_unique("Prolog Solution to Perl Weekly Challenge", IndexUnique).

IndexUnique = 7

yes
``````

### Notes

The main steps here are to check to see if after a character is deleted from the list if the new list length only varies from the original by 1. If so, then we are done. In the case the list is exhausted without finding a unique character the predicate simply fails.

Also note that GNU Prolog's `nth/3` assumes 1 indexing and so to get the correct 0 based answer we do an extra `succ/2`.

## Part 2

You are given list of numbers and an integer. Write a script to trim the given list when an element is less than or equal to the given integer.

### Solution

``````
trimmer(X, Y, Z):-
X < Y,
Z = Y.
trimmer(X, Y, Z):-
X >= Y,
Z = 'trimmed'.

trim_list(Numbers, I, TrimmedList):-
maplist(trimmer(I), Numbers, PartialTrimmedList),
delete(PartialTrimmedList, 'trimmed', TrimmedList).
``````

### Sample Run

``````
\$ gprolog --consult-file prolog/ch-2.p
| ?- trim_list([1, 4, 2, 3, 5], 3, TrimmedList).

TrimmedList = [4,5] ?

yes
| ?- trim_list([9, 0, 6, 2, 3, 8, 5], 4, TrimmedList).

TrimmedList = [9,6,8,5] ?

yes
``````

### Notes

`maplist/3` is always tempting to use when required to process a list. In order to get the effect that we want, however, requires that the predicate used in the maplist succeed for each value of the list to be processed. Here `trimmer/3` will either succeed with the numerical value that passes that comparison or succeed and provide the atom `trimmed` for values in the list that fail the comparison. The resulting list is then used with `delete/3` to get the final list containing only the numerical values required.

## References

Challenge 180

posted at: 16:18 by: Adam Russell | path: /prolog | permanent link to this entry