Thread - 7X7 — July Sudoku Test

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!


	@ 2014-07-21 4:29 PM (#16157 - in reply to #16146) (#16157) Top
Bram28 Posts: 35 Country : The Netherlands	Bram28 posted @ 2014-07-21 4:29 PM Thanks Richard! They were fun. I do like the 7x7 format!
	@ 2014-07-24 10:33 AM (#16178 - in reply to #15892) (#16178) Top
vjain9 Posts: 44 Country : India	vjain9 posted @ 2014-07-24 10:33 AM Attempted the test . Just loved it ..... but now want to solve the puzzles at leisure - PW to the PUZZLE BOOKLET PLEASE !
	@ 2014-07-24 10:34 AM (#16179 - in reply to #16178) (#16179) Top
Administrator Posts: 3581 Country : India	Administrator posted @ 2014-07-24 10:34 AM vjain9 - 2014-07-24 10:33 AM Attempted the test . Just loved it ..... but now want to solve the puzzles at leisure - PW to the PUZZLE BOOKLET PLEASE ! Removed the password.
	@ 2014-07-26 8:14 AM (#16187 - in reply to #15892) (#16187) Top
vjain9 Posts: 44 Country : India	vjain9 posted @ 2014-07-26 8:14 AM thanks... Will try to go thru the Solving tips now
	@ 2014-08-03 5:27 PM (#16262 - in reply to #15892) (#16262) Top
vjain9 Posts: 44 Country : India	vjain9 posted @ 2014-08-03 5:27 PM Am working slowly thru the 7*7 Sudoku tips. Vide Quad sums . A. could not understand R7C1=4; R1C7 = 4 ( 4 in C1 R 6 and 7 , similarly 4 in C 7 R 1 and 2 ) . B. How was 2 ruled out in R67C6 ( you mentioned 2 in C7 in R67; 3 in R6 in C67 ) - 3 is understood in R6C67 . Vide Consequtive . 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 ?
	@ 2014-08-25 9:10 PM (#16440 - in reply to #16262) (#16440) Top
Richard Posts: 191 Country : The Netherlands	Richard posted @ 2014-08-25 9:10 PM 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!