Para Yokozuna
Joined: 08 Nov 2006 Posts: 384 Location: The Netherlands

Posted: Tue Apr 17, 2007 1:43 pm Post subject: Examples of Law of Leftovers. 


Hi all
There have been a lot of questions about Law of Leftovers. So i thought it might be handy to show some examples of the eliminations you can make with Law of Leftovers.
I used Ruud's puzzles 32 and 33 for the examples.
Example No. 1 (Puzzle 32)
Code:  Nonet Numbering
111233333
112222333
114442263
144555266
148555269
448555669
748866699
777888899
777778999 
This example would be called Law of Leftovers on Column 789. I have drawn a red line between Column 6 and 7 to make it clear. There are 3 nonets (i number them 3, 6 and 9) that are almost completely in Columns 7,8 and 9. They have 4 cells that are on the left of the line(not in column 7,8,9): R1C56 + R7C56. These cells contain the same digits as the cells that are in Column 7,8,9 but aren't in the 3 nonets, namely R3458C7.
First elimination(pink circles):
We know R7C6 = 8, so R3458C7 must contain an 8. The only place for an 8 is R5C7.
So we know R5C7 = 8.
Second elimination(blue circles):
After placing the 8 in R5C7, R3458C7 contain the digits 2,3,6,8 and 9. So R1C56 + R7C56 can only contain 2,3,6,8 or 9.
So we can eliminate 5 and 7 from R17C5.
Third elimination (purple circle):
R3458C7 are all in the same house(column 7). So they can contain each digit only once. So also R1C56 + R7C56 can contain each digit only once. Because R7C6 = 8, there can't be an 8 in any of the other 3 cells.
So we can eliminate the 8 from R1C5.
Example No 2 (Puzzle 33)
Code:  Nonet Numbering:
112222333
111225333
112225633
114455663
444456666
744556699
774588899
777588999
777888899 
This would be called Law of Leftovers on Row 789.
Again a red line is drawn, this time between row 6 and 7. 3 Nonets (7, 8 and 9) have almost all their cells in Row 7,8 and 9 except 3 cells that are above the line: R6C189. These cells are equal to the cells in Row 7,8 and 9 that don't belong to these 3 nonets: R7C34 + R8C4.
First elimination (pink circles + pink lines):
We know R7C4 = 1. So R6C189 must contain a 1. The digit 1 can only be in R6C89, so one of these has to contain a 1. Therefor we can eliminate 1 from all digits that see both these cells. So we can eliminate 1 from R6C7(same row as both cells), R8C8 and R9C9(same nonet as both cells).
Second elimination (blue circles + blue lines):
We know R6C1 = 2. So R7C34 + R8C4 must contain a 2. The digit 2 can only be in R7C3 or R8C4, so one of these has to contain a 2. Therefor we can eliminate 2 from any cell that sees both cells. So we can eliminate 2 from R4C4 (same nonet as R7C3 and same column as R8C4).
I hope this will help everyone in their search for Law of Leftovers moves.
greetings
Para 
