Discover an efficient method to express a disjunction of inequalities in Prolog while eliminating redundant solutions and improving resource management.
---
This video is based on the question https://stackoverflow.com/q/66061202/ asked by the user 'false' ( https://stackoverflow.com/u/772868/ ) and on the answer https://stackoverflow.com/a/66070776/ provided by the user 'Duda' ( https://stackoverflow.com/u/8080648/ ) at 'Stack Overflow' website. Thanks to these great users and Stackexchange community for their contributions.
Visit these links for original content and any more details, such as alternate solutions, latest updates/developments on topic, comments, revision history etc. For example, the original title of the Question was: How to express a disjunction of inequalities compactly without redundant answers/solutions
Also, Content (except music) licensed under CC BY-SA https://meta.stackexchange.com/help/l...
The original Question post is licensed under the 'CC BY-SA 4.0' ( https://creativecommons.org/licenses/... ) license, and the original Answer post is licensed under the 'CC BY-SA 4.0' ( https://creativecommons.org/licenses/... ) license.
If anything seems off to you, please feel free to write me at vlogize [AT] gmail [DOT] com.
---
Mastering Prolog: Efficiently Expressing Disjunctions of Inequalities
In the world of Prolog programming, dealing with inequalities can sometimes lead to convoluted expressions that produce redundant results or inefficient choice points. This guide will tackle a common issue faced by Prolog developers: how to express a disjunction of inequalities compactly without encountering these repetitive or unnecessary outputs.
The Problem: Inefficiencies and Redundancies
When trying to define a simple disjunction of inequalities using the dif/2 predicate, developers often end up with redundant answers. Take a look at the original implementation for clarity:
[[See Video to Reveal this Text or Code Snippet]]
This approach can yield multiple unexpected solutions, as demonstrated in these test cases:
Test Case 1: When running dif_to_orto(A, B, C), A = 1, B = 2, C = 2, we see two identical results for A = 1, B = 2, C = 2, leading to redundancy.
Test Case 2: The query dif_to_orto(A, B, C), A = 1, B = 2, C = 3 also outputs duplicate answers, revealing inefficiency in the logic.
Test Case 3: Even when valid combinations are expected, leftover choice points can waste resources and complicate the program's execution.
The Solution: A More Compact Formulation
To overcome these inefficiencies and eliminate redundancies, we propose a more streamlined approach:
[[See Video to Reveal this Text or Code Snippet]]
This formulation improves the existing method by ensuring that the dif/2 predicate only produces necessary outputs, as shown in the revised test cases:
Example Test Cases
Case One: ?- dif_to_orto(A, B, C), A = 1, B = 2, C = 2.
Output:
[[See Video to Reveal this Text or Code Snippet]]
Case Two: ?- dif_to_orto(A, B, C), A = 1, B = 2, C = 3.
Output:
[[See Video to Reveal this Text or Code Snippet]]
Case Three: ?- dif_to_orto(A, B, C), A = 1, B = 2, C = 1.
Output:
[[See Video to Reveal this Text or Code Snippet]]
Case Four: ?- dif_to_orto(A, B, C), A = 1, B = 1, C = 2.
Output:
[[See Video to Reveal this Text or Code Snippet]]
General Query: ?- dif_to_orto(A, B, C).
Output:
[[See Video to Reveal this Text or Code Snippet]]
Case of Same Values: ?- dif_to_orto(1, B, B).
Output:
[[See Video to Reveal this Text or Code Snippet]]
Conclusion
In summary, by modifying the dif_to_orto predicate, programmers can significantly reduce redundancy and improve the efficiency of their Prolog code. This approach allows for cleaner outputs and better resource utilization, making it an excellent alternative to the more verbose and less efficient original definitions. Whether you are a seasoned Prolog developer or just starting, implementing this solution can save you time and frustration. Happy coding!
Информация по комментариям в разработке