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@@ -7,15 +7,21 @@ The basic idea is that you can generate your own data without really inflating a
3. Number of each unique data group occurance in your data
For example, say you have the data bytes "abc"
+
You know the length of the file is 3 bytes
-You know the hash (in this example I'll use sha256) is ba7816bf8f01cfea414140de5dae2223b00361a396177a9cb410ff61f20015ad
+
+You know the hash (in this example I'll use sha256) is `ba7816bf8f01cfea414140de5dae2223b00361a396177a9cb410ff61f20015ad`
Finally, for the unique data groups, I'll use the example of group of {1} and find the total number of binary 1's in the data. I believe this would be the most difficult to regenerate.
To do this we take our data "abc"
-which in binary would be 01100001 01100010 01100011
+
+which in binary would be `01100001 01100010 01100011`
+
So there will be 10 uinique groups of {1}
+Now we generate possible combinations untilt the hash of it is correct and we have our data! :D
+
So the the total number of possible combinations would be 24 choose 10 which == 1961256 combinations. Out of the total combinations of 2^24 == 16777216 the 1961256 brings the number of hashes to try to only 11.69% of all possible.
![LaTex Image](http://mathurl.com/ycgnob6r.png "quick mafs")