Merge pull request #3872 from TwoTau/TwoTau-lzw-readme

Add documentation for LZW compression

code/compression/src/lossless_compression/lempel-ziv-welch/README.md

@@ -1,4 +1,46 @@
-# cosmos
-Your personal library of every algorithm and data structure code that you will ever encounter
+# Lempel-Ziv-Welch compression
 
-Collaborative effort by [OpenGenus](https://github.com/opengenus)
+The **Lempel-Ziv-Welch** (LZW) algorithm is a dictionary-based lossless compressor. The algorithm uses a dictionary to map a sequence of symbols into codes which take up less space. LZW performs especially well on files with frequent repetitive sequences.
+
+## Encoding algorithm
+
+1. Start with a dictionary of size 256 or 4096 (where the key is an ASCII character and the values are consecutive)
+2. Find the longest entry W in the dictionary that matches the subsequent values.
+3. Output the index of W and remove W from the input.
+4. Add (W + (the next value)) to the dictionary as a key with value (dictionary size + 1)
+5. Repeat steps 2 to 4 until the input is empty
+
+**Example:** the input "ToBeOrNotToBeABanana" (length of 20) would be encoded in 17.
+```
+[T][o][B][e][O][r][N][o][t][To][Be][A][B][a][n][an][a]
+ 1  2  3  4  5  6  7  8  9  10  11 12 13 14 15  16  17
+```
+The brackets separate the values (e.g. `[To]` is encoded as one value). The LZW algorithm replaces the second occurences of `To`, `Be`, and `an` with one value, eliminating data redundancy.
+
+Note that there is **no need to store the dictionary** because the LZW uncompression process reconstructs the dictionary as it goes along.
+
+### Encoding example
+![Table showing the LZW compression process by character](https://slideplayer.com/slide/5049613/16/images/11/LZW+%E2%80%93+Encoding+Example+T%3Dababcbababaaaaaaa.jpg)
+
+> Image credits: https://slideplayer.com/slide/5049613/
+
+## Decoding algorithm
+
+1. Start with a dictionary of size 256 or 4096 (must be the same as what the encoding process used in step 1)
+2. Variable PREV = ""
+3. Is the next code V in the dictionary?
+    - Yes:
+        1. Variable T = dictionary key whose value = V
+        2. Output T
+        3. Add (PREV + (first character of T)) to the dictionary
+    - No:
+        1. Variable T = PREV + (first character of PREV)
+        2. Output T
+        3. Add T to the dictionary
+4. Repeat step 3 until the end of the input is reached
+
+---
+<p align="center">
+	A massive collaborative effort by <a href="https://github.com/OpenGenus/cosmos">OpenGenus Foundation</a> 
+</p>
+---

code/online_challenges/src/project_euler/problem_007/problem_007.js

@@ -0,0 +1,15 @@
+/* Part of Cosmos by OpenGenus Foundation */
+
+const primes = [];
+let n = 2;
+
+while (primes.length < 10001) {
+  // if this number is not divisible by any prime currently in the array
+  if (primes.reduce((isPrime, prime) => isPrime && (n % prime !== 0), true)) {
+    primes.push(n);
+  }
+
+  n++;
+}
+
+console.log(primes[10000]);