Welcome Guest [Log In] [Register]
Viewing Single Post From: Sometimes Things Are Easier Than They Look
jdege Jun 9 2008, 06:27 PM
Member Avatar
Elite member
[ * * * * * ]
 I was trying to explain this to someone, and noticed something else, when trying to pick out the row digits.

We know that the row digits will never appear doubled, and that they'll never appear adjacent to each other.

There are 10*9 = 90 possibilities. In the cryptogram I was trying to solve, there were four digits that appeared doubled, leaving 6 * 5 = 30 possibilities.

If you create a grid:
Code:
 
* 9 8 7 6 5 4 3 2 1
0 _ _ _ _ _ _ _ _ _
1 _ _ _ _ _ _ _ _
2 _ _ _ _ _ _ _
3 _ _ _ _ _ _
4 _ _ _ _ _
5 _ _ _ _
6 _ _ _
7 _ _
8 _
, then for every digit that appears doubled, put an X in all the cells for the row and the column for that digit (1,3,4,5):
Code:
 
* 9 8 7 6 5 4 3 2 1
0 _ _ _ _ X X X _ X
1 X X X X X X X X
2 _ _ _ _ X X X
3 X X X X X X
4 X X X X X
5 X X X X
6 _ _ _
7 _ _
8 _
, then for every pair of digits, fill in the cell for that pair (ignoring order, either 4+5 or 5+4 in the cipertext will X the cell for 4+5:
Code:
 
* 9 8 7 6 5 4 3 2 1
0 X _ X X X X X _ X
1 X X X X X X X X
2 _ X X X X X X
3 X X X X X X
4 X X X X X
5 X X X X
6 X X _
7 _ X
8 X
. In my case, there were only five unchecked cells left. Odds are that of these cells, it'll be the pair with the highest frequency that is correct, as it was in my case. But when it's not, there won't be many others you need to look at.
When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl.
Offline Profile Quote Post
Sometimes Things Are Easier Than They Look · General