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What about this..!!3 weeks + 7 days + 11 hrs + 13 hrs + 1 day = 30 days..

Hmm… let’s try this in C, shall we?#include <stdio.h>  static const char homework_digits[] = "12526"; static const int homework_indices[][2] = {  { 0, 1 }, { 2, 3 }, { 1, 3 }, { 0, 3 }, { 5, 1 }, { -1, -1 } };  int main() {  int i = -1;  while (homework_indices[++i][0] >= 0) {  int j = homework_indices[i][0];  const int k = homework_indices[i][1];  const int d = 1 - 2 * (j > k);  while (j != k) {  putchar(homework_digits[j]);  j += d;  }  putchar('\n');  } } Yeah, that works: [Wandbox]三へ( へ՞ਊ ՞)へ ﾊｯﾊｯ

This 1 line, recursive decimal-to-binary converter I wrote in JavaScript just shows the “beauty” of the language(sarcasm)(_$=($,_=[]+[])=>$?_$($>>+!![],($&+!![])+_):_)(255); //"11111111" It doesn’t even have numbers or wordsSo how does it work?Here is the ordinary decimal to binary converter. A given integer is repeatedly divided by 2, and its remainder is saved to a string. This process continues until the number cannot be halved any further.function d2b(d){ //d is your base-10 number  var b = ""; //initialize a string that will eventually contain the binary form of the number  for(var i = d; i > 0; i = Math.floor(i/2) ){ //halve the number each iteration and round down    b = i%2 + b; //add the remainder to the beginning of the string  }  return b; }   d2b(255) //"11111111" Step by step, I’m going to obfuscate the code. Maybe you’ll learn some tricks along the way.1.) The >> operator performs a bitwise shift to the right by “n” bits.For example, 8 >> 1 would be 4. 8 in binary is 1000, and shifting over all the bits to the right by 1 gives us 0100, four.Thus, x >> n is the same as dividing an x by 2^n (rounded down). In our case, we want to divide by 2, so we perform a bitwise shift the right by one (i>>1) instead of Math.floor(i/2)function d2b(d){ //d is your base-10 number  var b = "";   for(var i = d; i > 0; i = i >> 1 ){ //halve the number each iteration and round down    b = i%2 + b; //add the remainder to the beginning of the string  }  return b; }   d2b(255) //"11111111" 2.) Let’s look further inside the parenthesis of the for loop. In JavaScript, 1 and 0 are synonymous with true and false, respectively. Additionally, every number besides 0 has a “truthy” value. We want to end our loop once ‘i’ cannot be halved anymore, in other words when i=0. So we can replace i>0 with just ‘i’ because ‘i’ will only be false if ‘i’ happens to be 0.Additionally, just as x+= 5 is the same as x = x + 5, this applies to every operator in JavaScript. For instance, x%= 5 is the same as x = x % 5. So lets replace i = i >> 1 with i>>=1function d2b(d){ //d is your base-10 number  var b = "";   for(var i = d; i ; i>>=1 ){    b = i%2 + b; //add the remainder to the beginning of the string  }  return b; }   d2b(255) //"11111111" 3.) Further, we happen to be using the % (modulus) operator with a power of 2. We can change this to a bitwise modulus by using the & (bitwise AND) operator with a number which is 1 less than the power of 2.For instance:x%16 is the same as x & 15x % 8 is the same as x & 7,x % 4 is the same as x & 3,x % 2 is the same as x & 1./*  Showing x % 4 is the same as x & 3   0000 & 0011 = 0000  0001 & 0011 = 0001  0010 & 0011 = 0010  0011 & 0011 = 0011  0100 & 0011 = 0000  0101 & 0011 = 0001  ... ... ...  */    function d2b(d){   var b = ""; //initialize as string  for(var i = d; i ; i>>=1){    b = (i&1) + b; // i&1 evaluates to 0 if i is even, and 1 if i is odd  }  return b; }   d2b(255) //"11111111" 4.) Now lets begin to make our function recursive, as the for loop takes up a lot of code. We will change our variable “b” to a second parameter which evaluates to “” if no value is given (When we first call the function).To make it recursive, we’ll just call the function again, but the new value of our decimal number, ‘d’, will just be half of our current d. So we’ll pass d>>1 as our parameter. Also, we need to pass in the string we have created so that the 1’s and 0’s can be added to the beginning. Since we have specified a value for b, it will not initialize as “”.Lastly, we want to return our string if d reaches 0, meaning we’re done. Instead of writing if(d==0), we can write if(!d), since 0 is a false value, as we discussed before in step 2.function d2b(d, b=""){ //if no value of b is given, b will initialize as ""   if(!d) return b; //if d is 0, return our string because we cant halve it any further, meaning we are done!    b = (d&1) + b; // d&1 evaluates to 0 if i is even, and 1 if i is odd  return d2b(d >> 1, b); }   d2b(255) //"11111111" Now there’s no point in havingb = (d&1) + b; and then passing in b immediately afterd2b(d >> 1, b) so we might as well just pass in b like this:return d2b(d >> 1, (d&1)+b); So now this is what our function looks like:function d2b(d, b=""){ //if no value of b is given, b will initialize as ""   if(!d) return b; //if d is 0, return our string because we cant halve it any further, meaning we are done!    return d2b(d >> 1, (d&1) + b); //substitute b for (d&1) + b; }   d2b(255) //"11111111" 5.) But wait! All we have left is 2 return statements and an if statement… we can make this into 1 line using Conditional (ternary) Operator// Syntax  condition ? expr1 : expr2 If d is 0, we want to return b, otherwise return d2b(d >> 1, (d&1) + b). So we can write this on one line:function d2b(d, b=""){    return d ? d2b(d >> 1, (d&1) + b) : b;  }   d2b(255) //"11111111" 6.) So now all we have left is a function with a return statement. This is perfect for an Arrow function.// Syntax (param1, param2, …, paramN) => returnExpression Now our function looks like this:d2b= (d, b="") => d ? d2b(d >> 1, (d&1) + b) : b;  d2b(255) //"11111111" 7.) In JavaScript, you can do some wacky things with type coercion. As you know, [] is an empty array. Because the array exists, it has a “truthy” value. now lets throw in some ! to convert it to a boolean. ![] evaluates to false and !![] evaluates to true. To convert a boolean to a number, we can use the + operator, so +![] evaluates to 0, and +!![] evaluates to 1.![] // false !![] //true  +![] //0 +!![] //1 Thought that was the end of it? We can concatenate the items of 2 arrays doing the following:["pen pineapple"] + ["apple pen"] // "pen pineappleapple pen" So to get an empty string, we can do [] + [];[] + []; // "" So now lets implement this in our code.d2b = (d, b=[]+[]) => d ? d2b(d >> +!![], (d&+!![]) + b) : b;  d2b(255) //"11111111" 8.) Lets make the function self-executing by wrapping the entire thing with parenthesis then calling it with (value);(d2b= (d, b=[]+[]) => d ? d2b(d >> +!![], (d&+!![]) + b) : b)(255); //"11111111" 9.) We can use the characters $ and _ as variable names, so lets replace d with $, b with _, and d2b with _$. Oh, and let’s also remove all the uneccesarry spaces because we’re terrible people (:(_$=($,_=[]+[])=>$?_$($>>+!![],($&+!![])+_):_)(255); //"11111111" Enjoy!Edit: I don’t advise ever using this line of code; It’s completely unreadable and unmanageable. Especially on a development team, you will probably get fired. The question asked for a piece of code to hang on the wall, not “What kind of code do you think is good”, so I wrote this answer to demonstrate some fun/obscure parts of JavaScript.Thank you for all the comments and upvotes:)