Today’s puzzle variant comes from the 2008 Sudoku World Competition Instruction booklet. I’ve renamed it Even Sum Sudoku for clarity. About a year ago, I published a cartoon called Odd Sudoku, where either Odd or Even contiguous Cells of at least size 2 were offered. This is not like that.
I’ve eliminated some starting numbers from the original puzzle and identified the cells in yellow as pairs with values summing to an even result.
One question that occurs to me is: what is the probability of having Even Sum Pairs for all the arrangements of this puzzle? Obviously, there are at least 8 Even Sum Pairs that have already been earmarked. From previous calculations (Domino Sudoku Cartoon), excluding the center cell, there are 40 pairs of contiguous cells in an arrangement (and there are 2**4 = 16 pair arrangements since:
- In Row 1, columns 1 and 2 [A = Across] or the other starting in Column 1, Rows 1. 2 [D = Down]
- In Row 2, columns 2, 3 [A] or Column 2, Rows 2, 3 [D]
- In Row 3, columns 3, 4 [A] or Column 3, Rows 3, 4 [D]
- In Row 4, columns 4, 5 [A] or Column 4, Rows 4, 5 [D]
Since each of these can be selected independently, there are 2*2*2*2 = 16 arrangements.
For any arrangement, how many are Even Pairs are there? It turns out, once you’ve solved the puzzle, you can count:
|Across [A]||Down [D]|
|Odd: 8 Even: 8||Odd: 10 Even: 6|
|Odd: 9 Even: 3||Odd: 7 Even: 5|
|Odd: 3 Even: 5||Odd: 5 Even: 3|
|Odd: 3 Even: 1||Odd: 3 Even: 1|
|Totals:||Odd: 23 Even: 17||Odd: 25 Even: 15|
P(Even = 19) = .125
P(Even = 17) = .375
P(Even = 15) = .375
P(Even = 13) = .125
A nice discrete, symmetric binomial distribution! Enjoy getting even.