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finished setting up page for 2521-fundamentals workshop (#248)
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‎data/articles/break-continue-ternary.mdx

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---
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title: Break, Continue, and Ternary
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date: 2023-09-27
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desc: None of this code makes any sense to me! Help!
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tags:
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- 'COMP2521'
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- 'Exam Revision'
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author: CSESoc Education Team
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coverPhoto: '/images/generic/alexandru-acea-XEB8y0nRRP4-unsplash.jpg'
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---
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# The break statement
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Break statements work by telling your loop to immediately stop. It will not execute the remainder of the current iteration, and will just immediately jump to the end of the loop.
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```c
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while (1) {
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if (1) {
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break;
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}
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printf("this will not be printed!");
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}
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```
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Oookay, but that sounds pretty useless... When would I ever want this!
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Sometimes you want to check conditions before running an iteration, but it would become clunky top have it as a condition (think flag variables). In this case a `break` statement is great!
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Consider the following clunky yucky code:
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```c
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int index = 0;
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int found = 0;
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while (index < SIZE && found = 0) {
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if (array[index] == something) {
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found = 1;
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}
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index++;
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}
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index--;
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```
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This can be simplified to:
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```c
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int index = 0;
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while (index < SIZE) {
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if (array[index] == something) {
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break;
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}
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index++;
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}
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```
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Nice, this makes things a lot tidier!
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In general, I like to use break as a "catcher", whether it be to catch errors before they happen, or otherwise catch some condition to end my loop!
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# The continue statement
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Continue statements tell a loop to stop executing the rest of the code for that iteration, and jump straight back to the top!
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```c
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for (int i = 0; i < 20; ++i) {
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if (i % 2 == 0) {
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continue; // skip to the next iteration if the number is even
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}
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printf("%d\n", i); // will not print even numbers, since it will not get to this point
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}
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```
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You might be thinking "this seems a lot more niche than `break`"...
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And you're right! Usually `continue` just lets us avoid one layer of if-nesting, but if you spot a place where you want to go straight to the next iteration without ending the loop entirely, maybe this would be a option to consider!
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In general, I don't use continue basically ever unless I get the lightbulb moment where the situation exactly matches one where I want to go straight to the next iteration.
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# The ternary operator
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This one is luckily quite simple, though daunting at first! Let's take a look at the general format of a ternary operator!
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```c
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variable = condition ? value_if_true : value_if_false;
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```
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This is exactly equivalent to:
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```c
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if (condition) {
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variable = value_if_true;
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} else {
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variable = value_if_false;
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}
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```
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Though note that a ternary doesn't necessary have to bind its value to a variable - you can just return it or otherwise use it if you want!
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This is thankfully a pretty simple one - any time you want to set a value based on whether something is true or not, you can just use a ternary! **You should, however, think about whether it actually makes your code more readable.**

‎data/workshops/2521-funda-23T3.mdx

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---
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title: CSESoc COMP2521 Fundamentals workshop
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desc: Resources to give you a solid foundation for algorithms later in the course 😄
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course: COMP2521 Fundamentals
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offering: 23T3
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---
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# CSESoc COMP2521 Fundamentals workshop
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This is a set of practical programming problems designed to help you develope a solid foundation for algorithms later in the course.
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To start, log in to cse servers via ssh and create a new folder:
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```bash:~
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$ mkdir 2521-fundamentals
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$ cd 2521-fundamentals
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```
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<br />
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## Miscellaneous resources
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<br />
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### [Abstract Data Types - The Why and How]
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[slides](https://docs.google.com/presentation/d/1wZ6S2RVDqLpSOQOlAyEl0lNuVgo39vG4VSE3BWUG0ww/edit?usp=sharing)
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<figure class="video_container">
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<iframe
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src="https://docs.google.com/presentation/d/e/2PACX-1vQyClrhYOlgJ2duoeYd2eBVn21P4-lZ53pR10yWVk-Ou8XbtIeMpL2lKLAdZTIeikowLHdfrLSRoYs1/embed?start=false&loop=false&delayms=3000"
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frameborder="0"
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width="960"
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height="569"
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allowfullscreen="true"
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mozallowfullscreen="true"
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webkitallowfullscreen="true"></iframe>
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</figure>
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<br />
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### [Time complexity analysis and Recursion]
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<span> Recording </span>
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### [How to use Break, Continue, and Ternary]
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Access the article [here](/articles/break-continue-ternary)

‎data/workshops/2521-funda-23T3/palindrome.mdx

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---
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title: Palindrome
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desc: Determine if a line is a palindrome or not **without using loops**.
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class: COMP2521-funda
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difficulty: 2
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---
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# Palindrome?
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Create a new file called 'palindrome.c' and copy the starter code [here](https://github.com/Allynixtor/comp2521-fundamentals-23T3/blob/main/problems/palindrome/palindrome.c) into it.
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An palindrome is a string which contains the same characters going forward and backwards.
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Implement the `is_palindromic` function in `is_palindromic.c` to return true if a given input line is a palindrome, and false if it is not.
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**But we have a problem, our computer is a tsundere and doesn't want you to use any loops at all!**
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So, we must implement `is_palindromic` without any loops. Though you are free to make any extra functions you want (provided they also do not contain loops).
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The output from your program should look **exactly** like this:
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```bash:~/2521-fundamentals/palindrome
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$ gcc is_palindromic.c -o is_palindromic
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$ ./is_palindromic
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Enter a line: racecar
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The line is a palindrome!
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```
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```bash:~/2521-fundamentals/palindrome
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$ ./is_palindromic
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Enter a line:
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The line is a palindrome!
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```
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```bash:~/2521-fundamentals/palindrome
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$ ./is_palindromic
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Enter a line: racecars
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The line is not a palindrome!
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```
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```bash:~/2521-fundamentals/palindrome
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$ ./is_palindromic
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Enter a line: maam
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The line is a palindrome!
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```
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```bash:~/2521-fundamentals/palindrome
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$ ./is_palindromic
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Enter a line: eeteee
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The line is not a palindrome!
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```
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## Assumptions/Restrictions/Clarifications
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- The main function handles replacing the newline with a null-terminator
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- The input line string will be of length $N$ ($0 ≤ N ≤ 1024$)
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- You are allowed to make other functions, but **must not use loops**.
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## Hints
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<details>
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<summary> Hint 1 </summary>
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<br />
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if a string is a single character then it is always a palindrome, and you can
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easily figure out if a two-character string is a palindrome with a single
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comparison
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</details>
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<details>
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<summary> Hint 2 </summary>
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<br />
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if a string is a palindrome, what can we say about the substring that excludes
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the left-most and right-most letter?
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</details>
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<details>
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<summary> Hint 3 </summary>
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<br />
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for a string to be a palindrome, the left-most and right-most letter must be
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the same, and the substring excluding those two letters must be palindromic
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</details>
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## Solution
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You can view the solution code to this problem [here](https://github.com/Allynixtor/comp2521-fundamentals-23T3/blob/main/problems/palindrome/solution.c).

‎data/workshops/2521-funda-23T3/poggers.mdx

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---
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title: Poggers
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desc: Determine if a list is poggers or not **without using loops**.
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class: COMP2521-funda
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difficulty: 2
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---
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# Poggers?
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Create a new file called 'poggers.c' and copy the starter code [here](https://github.com/Allynixtor/comp2521-fundamentals-23T3/blob/main/problems/poggers/poggers.c) into it.
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An list is poggers if every node matches the following criteria:
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- If the node is even, then the next node must be of strictly greater value.
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- If the node is odd, then the next node must be of strictly lesser value.
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Implement the `listIsPoggers` function in `listIsPoggers.c` to return true if the input list is poggers, and false otherwise.
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**But we have a problem, our computer is a tsundere and doesn't want you to use any loops at all!**
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So, we must implement `listIsPoggers` without any loops. Though you are free to make any extra functions you want (provided they also do not contain loops). For this exercise though, extra functions should not be necessary!
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The output from your program should look **exactly** like this:
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```bash:~/2521-fundamentals/poggers
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$ gcc listIsPoggers.c -o listIsPoggers
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$ ./listIsPoggers
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Enter list size: 5
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Enter list values: 3 2 3 2 3
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List: [3, 2, 3, 2, 3]
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POGGERS POGU POG
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```
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```bash:~/2521-fundamentals/poggers
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$ ./listIsPoggers
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Enter list size: 0
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List: []
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POGGERS POGU POG
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```
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```bash:~/2521-fundamentals/poggers
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$ ./listIsPoggers
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Enter list size: 5
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Enter list values: 3 2 3 2 1
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List: [3, 2, 3, 2, 1]
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NOT POG
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```
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```bash:~/2521-fundamentals/poggers
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$ ./listIsPoggers
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Enter list size: 9
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Enter list values: 2 4 8 9 3 -5 -6 -4 0
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List: [2, 4, 8, 9, 3, -5, -6, -4, 0]
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POGGERS POGU POG
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```
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## Assumptions/Restrictions/Clarifications
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- You are allowed to make other functions, but **must not use any loops**.
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## Hints
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<details>
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<summary> Hint 1 </summary>
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<br />
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We may trivially say that an empty list or a list with a singular node is
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poggers.
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</details>
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<details>
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<summary> Hint 2 </summary>
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<br />
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Assuming the sublist from list->next onwards is poggers, what can we say about
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the requirement(s) for our list onwards to be poggers?
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</details>
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<details>
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<summary> Hint 3 </summary>
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<br />
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If the sublist from list->next onwards is poggers, then we know that we only
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need to check conditions between list->value and list->next->value.
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</details>
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## Solution
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You can view the solution code to this problem [here](https://github.com/Allynixtor/comp2521-fundamentals-23T3/blob/main/problems/poggers/solution.c).

‎pages/articles.tsx

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{/* Uncomment once we have content for workshops */}
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{/* <Flex
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<Flex
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as="main"
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css={{
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flexDirection: 'column',
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Explore the many workshops our Education Team has curated to become big brain.
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</Text>
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<WorkshopsContainerHomePage allWorkshopsOffering={workshopOfferingContent} />
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</Flex> */}
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</Flex>
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<Text
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size="headline"

‎pages/index.tsx

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</Flex>
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{/* Uncomment once we have content for workshops */}
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{/* <Flex
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<Flex
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as="main"
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css={{
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flexDirection: 'column',
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Explore the many workshops our Education Team has curated to become big brain.
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</Text>
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<WorkshopsContainerHomePage allWorkshopsOffering={workshopOfferingContent} />
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</Flex> */}
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</Flex>
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<Flex
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as="main"
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css={{

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