diff --git a/edit-distance/README.md b/edit-distance/README.md
index 5cd47d8..13bbf9f 100644
--- a/edit-distance/README.md
+++ b/edit-distance/README.md
@@ -1,4 +1,164 @@
+## [Levenshtein distance](http://en.wikipedia.org/wiki/Levenshtein_distance)
 
-## TODO 
-  * write down thinking
+This is a well known problem.
 
+
+Illustrating this problem using wikipedia's example
+
+Making a table like [Interleaving String](../interleaving-string)
+
+```
++---+---+---+---+---+---+---+---+
+|   | * | k | i | t | t | e | n |
++---+---+---+---+---+---+---+---+
+| * |   |   |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| s |   |   |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| i |   |   |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| t |   |   |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| t |   |   |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| i |   |   |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| n |   |   |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| g |   |   |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+
+```
+
+`*` here is representing an empty string
+
+each gird is the `edit distance` of `string[0..x]` and `string[0..y]`
+
+First row and column girds are easy to be filled:
+
+ * `edit distance` of an empty string to an empty string is `0`
+ * `edit distance` of an empty string to any string is the length of that string
+
+So the table should be filled like:
+
+```
++---+---+---+---+---+---+---+---+
+|   | * | k | i | t | t | e | n |
++---+---+---+---+---+---+---+---+
+| * | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
++---+---+---+---+---+---+---+---+
+| s | 1 |   |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| i | 2 |   |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| t | 3 |   |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| t | 4 |   |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| i | 5 |   |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| n | 6 |   |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| g | 7 |   |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+
+```
+
+### what's about `k` and `s` 
+
+in `k`'s view
+
+ * Insert: cant transform to `s`
+ * Delete: cant transform to `s`
+ * Replace: replace `k` to `s` use `1` `edit distance`
+
+in `s`'s view
+
+ * Insert: cant transform to `k`
+ * Delete: cant transform to `k`
+ * Replace: replace `s` to `k` use `1` `edit distance`
+
+so the number in the gird `k` and `s` should be `1`
+
+### what's about `ki` and `s` (in `ki`'s view)
+
+ * Insert: cant transform to `s`
+ * Delete: delete `i` then, transform to the problem `s` and `k` we solved before. 1 + `edit distance` of `s` and `k`.
+ * Replace and Delete: replace `k` to `s` use `1` `edit distance`, then delete `k` use `1` `edit distance`. `1 + 1 = 2`
+ 
+
+`Delete` and `Replace then Delete` are the same thing at last, `Replace` is just the `1` `edit distance` costed by `edit distance` of `s` and `k`.
+
+In this case, we can transform the problem by deleting one char: 1 + `edit distance` of `string deleted one` and `k`
+
+We can fill up the second column and row now:
+
+
+```
++---+---+---+---+---+---+---+---+
+|   | * | k | i | t | t | e | n |
++---+---+---+---+---+---+---+---+
+| * | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
++---+---+---+---+---+---+---+---+
+| s | 1 | 1 | 2 | 3 | 4 | 5 | 6 |
++---+---+---+---+---+---+---+---+
+| i | 2 | 2 |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| t | 3 | 3 |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| t | 4 | 4 |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| i | 5 | 5 |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| n | 6 | 6 |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| g | 7 | 7 |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+
+```
+
+### What if `si` and `ki`
+
+Obviously, `si` and `ki` equals to the `edit distance` of `s` and `k`, because `i` are both in two strings.
+
+```
++---+---+---+---+---+---+---+---+
+|   | * | k | i | t | t | e | n |
++---+---+---+---+---+---+---+---+
+| * | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
++---+---+---+---+---+---+---+---+
+| s | 1 | 1 | 2 | 3 | 4 | 5 | 6 |
++---+---+---+---+---+---+---+---+
+| i | 2 | 2 | 1 |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| t | 3 | 3 |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| t | 4 | 4 |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| i | 5 | 5 |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| n | 6 | 6 |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+| g | 7 | 7 |   |   |   |   |   |
++---+---+---+---+---+---+---+---+
+
+```
+
+then we can fill up the table using technology we used before.
+
+
+### More common case
+
+`P[word1][word2]` is the the table,
+
+so `P[i][j]` have 4 choices:
+ 
+ * `P[i - 1][j] + 1` : by deleting `word1[i]` from `word1` and convert to previous problem.
+ * `P[i][j - 1] + 1` : by deleting `word2[j]` from `word2` and convert to previous problem.
+ * `P[i - 1][j - 1]` : only if `word1[i] == word2[j]`
+ * `P[i - 1][j - 1] + 1` : only if `word1[i] != word2[j]`, then replace one char `word1[i]` to `word2[j]`, then this problem is converted to `P[i - 1][j - 1]`.
+
+Dont worry about `Insertion`, it is just the oppsite to `Deletion`. Just different view.
+
+
+So the work remaing is easy, find the MIN of those four is ok.