How to resolve the algorithm Count occurrences of a substring step by step in the Prolog programming language

Published on 12 May 2024 09:40 PM
#Prolog

How to resolve the algorithm Count occurrences of a substring step by step in the Prolog programming language

Table of Contents

Problem Statement

Create a function,   or show a built-in function,   to count the number of non-overlapping occurrences of a substring inside a string. The function should take two arguments:

It should return an integer count. The matching should yield the highest number of non-overlapping matches. In general, this essentially means matching from left-to-right or right-to-left   (see proof on talk page).

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Count occurrences of a substring step by step in the Prolog programming language

Source code in the prolog programming language

count_substring(String, Sub, Total) :-
    count_substring(String, Sub, 0, Total).

count_substring(String, Sub, Count, Total) :-
    ( substring_rest(String, Sub, Rest)
    ->
        succ(Count, NextCount),
        count_substring(Rest, Sub, NextCount, Total)
    ;
        Total = Count
    ).

substring_rest(String, Sub, Rest) :-
    sub_string(String, Before, Length, Remain, Sub),
    DropN is Before + Length,
    sub_string(String, DropN, Remain, 0, Rest).


?- count_substring("the three truths","th",X).
X = 3.

?- count_substring("ababababab","abab",X).
X = 2.


:- system:set_prolog_flag(double_quotes,chars) .

occurrences(TARGETz0,SUBSTRINGz0,COUNT)
:-
prolog:phrase(occurrences(SUBSTRINGz0,0,COUNT),TARGETz0)
.

occurrences("",_,_)
-->
! ,
{ false }
.

occurrences(SUBSTRINGz0,COUNT0,COUNT)
-->
SUBSTRINGz0 ,
! ,
{ COUNT1 is COUNT0 + 1 } ,
occurrences(SUBSTRINGz0,COUNT1,COUNT)
.

occurrences(SUBSTRINGz0,COUNT0,COUNT)
-->
[_] ,
! ,
occurrences(SUBSTRINGz0,COUNT0,COUNT)
.

occurrences(_SUBSTRINGz0_,COUNT,COUNT)
-->
!
.

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