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_Rocket_

Count to 10 in the Most Complicated Way Possible

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Posted  Edited by _Rocket_

It is time to flex your program'r skilz and count to 10 like you learned in school.

 

I challenge everyone here to create the most complicated "count to 10" program possible. The goal is simple. All you need to do is make your program print numbers 1 through 10 into console OR by using file I/O. You can separate each number using a space or by inputting a new line. HOWEVER, here's the catch. You need to make it as complicated as possible. Each number should take a ridiculous amount of code to print.

 

Now please, don't just spam if else blocks. Try to be creative. More creative "solutions" will be loved more than quick and sloppy ones. Being able to make your code run in a loop would be really cool if you can pull it off. But you have the freedom of choice here.

 

When you make something, please provide the language you used. If you use assembly, don't just copy off a compiler like godbolt. Everything should be original code wirtten by you (besides the printing method or file I/O of course).

 

I'll pitch in a few when I get a chance 😄

 

You can paste them in a code block here. You also have the choice of GitHub and other platforms if the code gets REALLY complicated.

 

The obvious rule is the code MUST compile/run. Errors are not accepted.

 

And of course if you really want to, you can use a web designing language to print these numbers out onto a localhost page. This is also acceptable.

 

Have fun! And feel free to use a variety of languages! It would be super interesting to see other languages get utilized in this weird challenge haha

Edited by _Rocket_

I write programs and stuff.

 

If you need to contact me, here is my discord tag: Dustin#6688

 

I am a busy person. So responses may be delayed.

1840045955_Thicco(1).thumb.png.87c04f05633286f3b45b381b4acc4602.png

 

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Posted  Edited by Cyber_Spectre - Edit Reason: Put in 10,000,000 and crashed my computer, have fun!
##Crazy.py
#######
#Purpose: Count to the number 10


import sys
import time


def nummy(numbers):
    
   #@ return void
   #Take the binary amount and convert that into decimal -> charcters -> decimalOfCharacter -> hex of ASCII
   #Convert Hex back into decimal to print

   #Create an array to store information about our numbers
   nums = []
   # Create this array to show ASCII Chars 
   chars = []

   #convert decimal into ASCII chars
   decimal = int(numbers,2)
   #Loop from 1 -> n + 1  ie 1-10
   for c in range(1, decimal+1):
      # Have to append to an array to actually show this
      chars.append(str(chr(c)))

   #Print the cool ASCII 
   sys.stdout.write("ASCII Chars:" + str(chars) + "\n")
   
   
   # from 1 to (n + 1) convert this from decimal to char to decimal to hex
   for i in range(1, decimal+1):
      nums.append(hex(ord(chr(i))))

   #Print the final outcome of HEX
   print("HEX: " + str(nums) + "\n")

   #Use a 'with' statement, because I sometimes forget to close file
   #
   with open("readthis.txt", "w") as f:
      for i in nums:
         f.write(i + '\n')

   #Clear the array to allow for 'new' data
   #Not really...it's the same crap lol
   nums.clear()

   #Make it seem the program is doing something important
   sys.stdout.write("Doing some important ST0xFF.." + "\n")
   time.sleep(2)
   
   #Open our file and read it as a var of 'f'.
   #Read every single line and split them up, not have i.e. 0x\0x1\0x2 etc.
   with open("readthis.txt", "r") as f:
      x = f.read().splitlines()
      
      #Append this to our 'cleared' array nums
      for i in x:
         nums.append(i)
   
   #Loop through the nums array and convert all of the hex to decimal and show to the user
   for i in nums:
      sys.stdout.write(str(int(i,0)) + " " )

   
      
       
#main method
def main():
   #Give the option to put in however many numbers you want
    userNum = int(input("Enter number: "))
    #call the function nummy and plug in numbers converted to binary
    sys.stdout.write("Inserting: BINARY -> " + str(bin(userNum)) + "\n")
    nummy(bin(userNum))
    
    

main()

something.PNG.3ede91f04ef786bbfd77071f8b1f6368.PNG

 

Allowed the user enter an arbitrary amount of numbers to the program: arb.thumb.PNG.1d2722979b1e602c79e7ee005ddf78ae.PNG

 

Thought this was a great idea @_Rocket_! Thanks. Can't wait to see what others put here.

 

This is no way the most complicated way. I may change it

 

 

*If someone just does straight up assembly I would be amazed. ;)*

#Also I saw no reason to check for non-positive numbers, try it out ^#

Edited by Cyber_Spectre
Put in 10,000,000 and crashed my computer, have fun!

- Spectre

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Posted  Edited by Snowy

I've definitely outdone myself on this challenge. Using: https://github.com/kelseyhightower/nocode

 

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I've managed to create this monstrosity.

 


Build the app using 

Quote

 

 

 

Output:

Quote

 

 

/s

Edited by Snowy

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Posted  Edited by Joshy

What did you mean by making it run in a loop?  Like have it count 1 through 10 over and over again?

 

z5sCq1m.png

 

#include<stdio.h>

int firstNumber = 1, lastNumber = 10;

int main()
{

	void counter(int timerStart, int initialCount, int finalCount, int *bucket);
	void myPrinter(int firstCall, int lastCall, int *locationPoint);
	int myArray = int("Yes");
			
	counter(-5687, firstNumber, lastNumber, &myArray);	  	  
	myPrinter(firstNumber, lastNumber, &myArray);
	
	return int("Joshy");

}

int counter(int timerStart, int initialCount, int finalCount, int *bucket)
{

	/* Count top four bits of a 16-bit integer and stores into cell
	 * Cell increments each time a number is stored otherwise keep writing same number into old cell */
	*bucket++ =  ('J' - 'O' + 'S' + 'H' + 'Y' - 'r' - 'u' - 'l' + 'e' - 's' + '!') & (timerStart++ >> '$'/3 ) == initialCount ? initialCount++ : *--bucket;
	
	// Stopping criteria
	initialCount <= finalCount ? counter(timerStart, initialCount, finalCount, bucket): finalCount;

	return false;
	
}

int myPrinter(int firstCall, int lastCall, int *locationPrint)
{

	// Prints the array

	printf("%d\t", *(locationPrint++));
	firstCall++ < lastCall ? myPrinter(firstCall, lastCall, locationPrint) : int("Does it matter?") ;
	
	return true;

}

 

Edited by Joshy

PoorWDm.png?width=360&height=152

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