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Location: The Netherlands | I only saw vjain's questions today, so here is a belated response:
vjain9 - 2014-08-03 5:27 PM
Quad sums . A. could not understand R7C1=4; R1C7 = 4.
There is a 4 in R3C2. Because of that, the 4 in R1 must be somewhere in R1C3-7. Those cells are all in the same irregular area, so the 4 is ‘locked’ there. Hence, the 4 is not in R23C7. There is also a 4 in R5C6; because of being in that shape, no 4 in R4567C7. The only possibility for the 4 in C7 is now R1C7. This works exactly the same for R7C1.
vjain9 - 2014-08-03 5:27 PM
Quad sums . B. How was 2 ruled out in R67C6.
The 2 in C7 must be in the irregular area bottom right, more specific, in R4567C7. The sum-dot in that irregular area must contain the series 1-2-3-6 since the 4 is already placed elsewhere in that area. Combining both previous conclusions, the 2 in that irregular area goes in R67C7.
vjain9 - 2014-08-03 5:27 PM
Consecutive . R2 contains 2 dots which must have pairs {12} and {45} . How were pairs 56 , 67 ruled out ? Then next how was [ R2C13 = {67}; R13C2 = {15} ] derived ?
There is a given 3 in row 2 without a dot next to it. This means that the 2 and the 4 in row 2 have to go in two of the four remaining cells in that row. There are two dots there, both connecting two different cells. The 2 and 4 cannot be placed next to the same dot, so this means that in those 4 cells you find the pairs {12} and {45}, without knowing which pair goes where at this stage.
The 6 and 7 in row 2 have to go in R2C13.
Since the 2 and 4 also cannot be placed above or below the given 3 (no dots there too), the 1 and 5 have to go there.
Hope this helps, and sorry for the delay. Must be my holiday that I didn't notice your questions before! |
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