Ordering a Stack with limited operations available.

This project focuses on data sorting on a stack using a limited set of instructions, aiming to achieve the sort with the minimum possible number of operations.

1. Introduction
2. Objectives
3. Common Instructions
4. Mandatory Part
5. Bonus Part

Push swap project focuses on creating a simple, straightforward sorting algorithm that uses two stacks and a set of instructions to manipulate both stacks. The primary goal is to write a program in C called push_swap, which calculates and displays the smallest program, composed of push_swap language instructions, to sort the integers received as arguments.

Writing a sorting algorithm is a critical step for a developer as it introduces the concept of complexity. The key learning objectives of this project are rigor, the use of C, and the application of basic algorithms with a focus on complexity.

• Project must be written in C.
• It should comply with the Norm.
• It should not quit unexpectedly (segmentation fault, bus error, double free, etc).
• All heap allocated memory space must be properly freed when necessary.
• Submission must include a Makefile to compile source files to the required output.
• If the project allows libft, its sources and associated Makefile should be copied in a libft folder.

The project involves two stacks named a and b, and a set of instructions to manipulate both stacks.

The goal is to sort the integers into stack a in ascending order using operations like swap, push, rotate and reverse rotate.

The program named push_swap should take stack a as a list of integers and display the smallest list of instructions possible to sort the stack.

Create a checker program to verify whether the list of instructions generated by the push_swap program actually sorts the stack properly.

The checker program takes stack a as a list of integers, waits and reads instructions on the standard input, and executes them on the stack.

If the stack a is sorted after executing the instructions and stack b is empty, the checker should display "OK". In any other case, it should display "KO".