Скачать или смотреть Understanding Recursion in Prolog: A Guide to Successor Arithmetic

Explore how to effectively use recursion in Prolog with an in-depth solution to represent numbers using successor arithmetic.
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This video is based on the question https://stackoverflow.com/q/68362346/ asked by the user 'Ran suari' ( https://stackoverflow.com/u/10929963/ ) and on the answer https://stackoverflow.com/a/68362784/ provided by the user 'rajashekar' ( https://stackoverflow.com/u/4437190/ ) at 'Stack Overflow' website. Thanks to these great users and Stackexchange community for their contributions.

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A Journey Through Recursion in Prolog

In the realm of logic programming, Prolog stands out with its unique approach to problem-solving, especially when it comes to recursion. If you've ever tried to implement a recursive function to represent numbers in Prolog, you know how challenging it can be. In this guide, we'll delve into a common problem faced by learners and explore the correct way to implement recursion to achieve the desired representation of numbers using successor arithmetic.

The Challenge

Let's set the stage. Your goal is to create a Prolog decision that translates natural numbers into a successor format like so:

1 → s(0)

2 → s(s(0))

3 → s(s(s(0)))

Initially, you attempted the following code to achieve this:

[[See Video to Reveal this Text or Code Snippet]]

Upon running a query like retnum(A,2)., you encountered two major issues:

The result A=0, which is not the expected output.

A "stuck limit exceeded" error, indicating an infinite loop in your recursion.

Understanding the Mistake

Your original implementation had a critical flaw: the recursive call should reduce towards a boundary condition, meaning each step of recursion needs to actually make progress towards the base case. If this does not happen, you risk an infinite loop and excess resource consumption, leading to the errors you experienced.

A Corrected Approach

To fix your recursion, let’s break down the solution into manageable parts and understand the updated implementation.

Revised Code

Here's an improved version of your function:

[[See Video to Reveal this Text or Code Snippet]]

Explanation of the Fixed Code

Stop Condition: The line retnum(s(0), 1). serves as the base case. This indicates that when Prolog reaches s(0), it should return 1.

Recursive Call:

The second clause retnum(s(S), Z) handles numbers greater than 1. Here, S is designed to converge towards 0.

It calls itself with S, and combines the computed Z1 (where Z1 represents the integer value of S).

The expression Z is Z1 + 1 effectively translates the Prolog representation into a human-readable integer.

Simplifying Further

While the above revision works, there is an even better method to structure this recursion for clarity and performance:

[[See Video to Reveal this Text or Code Snippet]]

In this version:

Each call decrements the integer Z, guiding it towards 1 and simplifying the logic behind the recursion.

The condition Z > 1 acts as an additional check, ensuring you don't run into negative numbers or other undefined states.

Conclusion

Recursion in Prolog, when handled correctly, is a powerful tool that allows for elegant solutions to problems like number representation using successor arithmetic. By refining your approach and understanding how to implement proper base cases and recursive calls, you can not only achieve your desired outcomes but also hone your skills in Prolog programming.

If you're ever in doubt while coding in Prolog, take a step back to review your logic and ensure that you're making adequate progress towards your base case. Happy coding!