This project aims to sort data(Integer numbers) on a stack, with a limited set of instructions, using the lowest possible number of actions.
Given a set of integers, sort it using a predefined set of instructions using two stacks. The instruction set is defined below. This program outputs a program in push_swap instructions that sorts the input integers.
For the following instructions, if the instruction is not possible, the part of it that can't be executed won't.
| Code | Instruction | Action |
|---|---|---|
sa |
swap a | swaps the 2 top elements of stack a |
sb |
swap b | swaps the 2 top elements of stack b |
ss |
swap a + swap b | both sa and sb |
pa |
push a | moves the top element of stack b at the top of stack a |
pb |
push b | moves the top element of stack a at the top of stack b |
ra |
rotate a | shifts all elements of stack a from bottom to top |
rb |
rotate b | shifts all elements of stack b from bottom to top |
rr |
rotate a + rotate b | both ra and rb |
rra |
reverse rotate a | shifts all elements of stack a from top to bottom |
rrb |
reverse rotate b | shifts all elements of stack b from top to bottom |
rrr |
reverse rotate a + reverse rotate b | both rra and rrb |
If you're gonna use radix algorithm you just need to use 'sa' 'pa' 'pb' 'ra' and 'rra'
For the stacks with size 5 or less I wrote a simple sort which can be found in the src folder.
In this project I used Radix sort as the main algorithm. Radix sort is an efficient algorithm to sort non-negative integers
with time complexity O (n). For example, we can sort following list of integers with this algorithm
87 487 781 100 101 0 1
Imagine there are 10 boxes labeled 0, 1, 2, …, 9
Start from the least significant digit (which is the digit in 1’s place), we put each number into the box which its digit corresponds to.
In the example, 87 has 7 in 1’s place, hence we put it in box 7. 487 also has 7 in 1’s place, so it should be placed in box 7 too (right behind 87) … And we repeat this process until every number is in one of the boxes.
box 0 100 0
box 1 781 101 1
box 2
box 3
box 4
box 5
box 6
box 7 87 487
box 8
box 9
After that, we connect every number according to the order of boxes.
100 0 781 101 1 87 487
As we can see, the numbers are sorted according to the digit in 1’s place. For those with the same digit in 1’s place, they’re sorted according to their order in the original list.
We repeat this procedure n times, whiere n is the number of digits of the largest number in the array (In this case 783 => n = 3).
After doing it n times and connecting numbers after each cycle we will have array sorted.
As we mentioned before, this algorithm is for non-negative integers. However, we’ll have negative numbers in this project, so we should simplify the numbers before we start.
To do so I gave indexes to each number in a stack. The smalles number gets index 0, the next smallest gets 1 and so on... With this idea, we can simplify any list of integers to make them in the range [0,N) ( ≥ 0 and < N, N is the size of the list).
After simplification we need to do something with the boxes. We have only two stacks instead of 10 boxes. Hence, I sorted the number in base 2 (to use 2 stacks instead of 10).
As in radix sort, we need two boxes for 0 and 1 respectively. Here we treat A as box 1 and B as box 0. Then, we start from the rightmost bit to the leftmost bit.
At the i-th digit from the right, if the i-th digit of the top number of A is 0, we perform pb to put this number in stack B. Else, we perform ra to leave it in stack A. After we perform one operation on each number, each of them is in the box that corresponds to its digit, as how we put numbers in the boxes in radix sort.
After that, we perform pa until there are no numbers in stack B, as we connect the numbers in radix sort.
Repeated the same procedure for every bit and after that got the sorted numbers in the stack a.
My push_swap sorts
3 numbers with maximum 3 instructions,
4 numbers with maximum 7 instructions,
5 numbers with maximum 11 instructions,
100 numbers with maximum 1084 instructions => 3 points,
500 numbers with maximum 6785 instructions => 4 points.
The algorithm is good enought to pass the project. If the Bonus part is also done the project could get more than 105%.
You can find some links and books below that might be useful during the project. You can find all the books in resources folder. Note that you do not have to read the books completly but you will find a lot of useful information there.
Books
Links