Rohan Rao - 2015-11-12 10:51 PM

Not sure if this is the best start, but here's what I did. It took me a while though.

R5C1 & R6C1 = 5 & 6
R5C2 + R6C2 = 5
R5C3 + R6C3 = 5

So, R2C3 + R3C3 = 5 OR 6 OR 7
It cannot be 7 because there is a 7-sum-pair below it (R3C3 + R4C3).
If it is 6, then R2C2 & R3C2 = 4 & 6 and R5C2 & R6C2 = 1 & 4 which is not possible (Basically, you can't fill up the 16-sum and 10-sum of column2-column3 together).

So, R2C3 + R3C3 = 5
R2C2 + R3C2 = 11 = 5 & 6
R2C1 + R3C1 = 3 = 1 & 2
R1C3 + R4C3 = 11 = 5 & 6
R3C1 & R3C3 = 1 & 2
R4C1 & R4C2 = 3 & 4

So,
R5C1 + R5C2 = 8
R6C1 + R6C2 = 8

Hence,
R5C2 & R6C2 = 2 & 3

R4C2 = 4 !

From there, it should get solved.

kishy72 - 2015-11-13 2:29 AM
I am wondering the source of the example .

debmohanty - 2015-11-13 5:26 AM


I picked it from my IPC2011 folder - so it had to be bit puzzle-y in nature.
However, I'm not able to see this puzzle in any IPC2011 booklet, so this may be one of the rejects because "too-hard-to-start"

kishy72 - 2015-11-13 2:29 AM

Thanks Rohan ! Excellent chain of logical thinking .But seriously,were all those meant to deduced during a test and if anyone would wait for logic to put in a digit ? I am wondering the source of the example .

RameshLMI - 2015-11-13 2:26 PM

Thanks Rohan.

But I am unable to decipher this logical deduction,

R5C1 + R5C2 = 8
R6C1 + R6C2 = 8

Can someone make this clear?

ghirsch - 2015-11-15 4:47 AM

I've been working my way through Thomas Snyder's Art of Sudoku book, and I'm pretty stuck on this one. Can anyone help me out (preferably with just a step or two to get me on the right track)? I'm probably just missing something obvious but I can't seem to come up with any ideas for what to do.

kishy72 - 2016-04-05 1:38 PM

Anti-diagonal from the Russian GP .I solved this particular sudoku for a long time and had to guess to finish it .Could someone tell me how to continue logically here ?

xiao01wei - 2019-10-26 7:23 PM

CSOC 63 P13's rule:
if there is two same shape, for example, two vertical domino shape, there will be 4 numbers in these 2 shapes:
A        C
B        D
and
if A>B, and then the result is C>D at the same time
if A if C>D, and then the result is A>B at the same time
if C the same inequality between the four numbers in two shapes with corresponding position

Puzzle_Maestro - 2019-11-23 6:15 PM

If you add up the clues on the top and bottom rows, you get 369. Adding up the clues in the left and right gives 327. The difference between the two is 42.

Observe that the arrows on the left and right point to 6 cells which are not pointed to by the arrows on top and bottom (with minimum sum 3*(1+2)=9), and similarly the arrows on top and bottom point to 6 cells not pointed to by arrows on the left and right (maximum sum 3*(8+9)=51). The difference between them must be 42, and since the maximum difference between them is 42, we can establish a few 12 and 89 pairs.