RabbitFarm

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

Part 1: Circular

You are given a list of words. Write a script to find out whether the last character of each word is the first character of the following word.

We can do this in a single predicate which recursively examines the list of words, which we’ll take as a list of character code lists.

The rest of the code just wraps this predicate into a file.

Sample Run

$ gprolog --consult-file prolog/ch-1.p 
| ?- circular(["perl", "loves", "scala"]). 
 
true ? 
 
(1 ms) yes 
| ?- circular(["love", "the", "programming"]). 
 
no 
| ?- circular(["java", "awk", "kotlin", "node.js"]). 
 
true ? 
 
yes 
| ?-
    

Part 2: Subsequence

You are given two strings. Write a script to find out if one string is a subsequence of another.

This is going to be a quick one, seeing as GNU Prolog has a sublist/2 predicate which does exactly this! As in the previous part we’ll take the strings as lists of character codes.

Finally, let’s assemble our completed code into a single file.

Sample Run

$ gprolog --consult-file prolog/ch-2.p 
| ?- subsequence("uvw", "bcudvew"). 
 
true ? 
 
yes 
| ?- subsequence("aec", "abcde"). 
 
no 
| ?- subsequence("sip", "javascript"). 
 
true ? 
 
yes 
| ?-
    

References

The Weekly Challenge 316
Generated Code