How do I solve this sudoku without guessing?
$begingroup$
I have filled all the possible singles. Also identified and shortlisted candidates, pointing pairs. How do I move ahead without guessing?
PS: I applied "Nishio", choosing one of the two possible candidate values on row 1, column 1. 7 will come in top left. Using which I can solve the rest of the puzzle. But it is "guessing". Any other way?
sudoku
asked Dec 10 '18 at 16:57
jerrymousejerrymouse
1112
$endgroup$
add a comment |
$begingroup$
I have filled all the possible singles. Also identified and shortlisted candidates, pointing pairs. How do I move ahead without guessing?
PS: I applied "Nishio", choosing one of the two possible candidate values on row 1, column 1. 7 will come in top left. Using which I can solve the rest of the puzzle. But it is "guessing". Any other way?
sudoku
asked Dec 10 '18 at 16:57
jerrymousejerrymouse
1112
$endgroup$
$begingroup$
Not sure if any of the advanced techniques to reduce the candidates here can be applied somewhere? I will also take a look if I can apply any of them.
$endgroup$
– Kevin Cruijssen
Dec 10 '18 at 17:07
add a comment |
$begingroup$
I have filled all the possible singles. Also identified and shortlisted candidates, pointing pairs. How do I move ahead without guessing?
PS: I applied "Nishio", choosing one of the two possible candidate values on row 1, column 1. 7 will come in top left. Using which I can solve the rest of the puzzle. But it is "guessing". Any other way?
sudoku
asked Dec 10 '18 at 16:57
jerrymousejerrymouse
1112
$endgroup$
I have filled all the possible singles. Also identified and shortlisted candidates, pointing pairs. How do I move ahead without guessing?
PS: I applied "Nishio", choosing one of the two possible candidate values on row 1, column 1. 7 will come in top left. Using which I can solve the rest of the puzzle. But it is "guessing". Any other way?
sudoku
sudoku
asked Dec 10 '18 at 16:57
jerrymousejerrymouse
1112
asked Dec 10 '18 at 16:57
jerrymousejerrymouse
1112
asked Dec 10 '18 at 16:57
jerrymousejerrymouse
1112
asked Dec 10 '18 at 16:57
jerrymousejerrymouse
1112
asked Dec 10 '18 at 16:57
jerrymousejerrymouse
1112
1112
$begingroup$
Not sure if any of the advanced techniques to reduce the candidates here can be applied somewhere? I will also take a look if I can apply any of them.
$endgroup$
– Kevin Cruijssen
Dec 10 '18 at 17:07
add a comment |
$begingroup$
Not sure if any of the advanced techniques to reduce the candidates here can be applied somewhere? I will also take a look if I can apply any of them.
$endgroup$
– Kevin Cruijssen
Dec 10 '18 at 17:07
$begingroup$
Not sure if any of the advanced techniques to reduce the candidates here can be applied somewhere? I will also take a look if I can apply any of them.
$endgroup$
– Kevin Cruijssen
Dec 10 '18 at 17:07
$begingroup$
Not sure if any of the advanced techniques to reduce the candidates here can be applied somewhere? I will also take a look if I can apply any of them.
$endgroup$
– Kevin Cruijssen
Dec 10 '18 at 17:07
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You can spot that
the 1, 7 cell cannot be a 4 (and therefore is an 8)
because
setting it to 4 forces an 8 into cell 5,7; which forces a 4 into cell 5,3; which eliminates 4 as a possibility from all of the remaining open cells in the top left block (cells 1,2 and 3,3).
answered Dec 11 '18 at 15:23
HellionHellion
1,901819
$endgroup$
add a comment |
$begingroup$
You could spot that:
In the top left block, the bottom right 4/6 cell must be a 6.
Why?
Because if it is a 4 then in the bottom left block the right side becomes 5/6,5/6. But now the solution isn't unique as setting either of these cells as 5 has no vertical effect on the board (from the bottom right block).
answered Dec 10 '18 at 17:49
JonMark PerryJonMark Perry
18k63786
$endgroup$
$begingroup$
I am still unclear. What is the vertical effect in bottom right block please?
$endgroup$
– jerrymouse
Dec 10 '18 at 19:00
$begingroup$
I'm confused. Using (column,row) notation: setting (3,3) to 4 would require (1,7) to be 4, then (2,9) to be 8, (9,9) to be 5, (3,9) would be 6, and (3,8) would be 5. No ambiguity.
$endgroup$
– user3294068
Dec 10 '18 at 19:31
$begingroup$
@user3294068; well it can't be a 4 according to OP's p.s.
$endgroup$
– JonMark Perry
Dec 10 '18 at 19:49
$begingroup$
@jerrymouse; if you follow my logic, we could have both 568:645 or 645:568 equally, which bears no impact on the rest of the grid.
$endgroup$
– JonMark Perry
Dec 10 '18 at 19:51
1
$begingroup$
I might prefer guessing (brute force all options, doesn't seem too difficult here) to this approach, though: if you base any deductions on the solution's assumed uniqueness, you can only reach "an answer" at best. If you want to reach "the answer", you can't assume uniqueness, you have to prove it instead.
$endgroup$
– Bass
Dec 10 '18 at 20:54
add a comment |
$begingroup$
You can notice
r2c1, r7c1, r3c5, r7c5 form a finned X-Wing for 4.
Thus
Consider r2c1. It must be either 4 or 8. If r2c1 is 4 that forces r3c3 to be 6. If r2c1 is 8 that forces r7c1 to be 4 which forces r7c5 to be 8 which forces r3c5 to be 4 which forces r3c3 to be 6. So if r2c1 is either 4 or 8 r3c3 is forced to be 6.
Hopefully you can get it from there.
answered 14 hours ago
Rocus HalbaschRocus Halbasch
112
New contributor
Rocus Halbasch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Hi Rocus, welcome to PSE. Please hide your answers using ">!"
$endgroup$
– Greg
13 hours ago
$begingroup$
Yeah. Just figured out how that works. This was my first answer.
$endgroup$
– Rocus Halbasch
13 hours ago
$begingroup$
No problem. Happy puzzling!
$endgroup$
– Greg
13 hours ago
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can spot that
the 1, 7 cell cannot be a 4 (and therefore is an 8)
because
setting it to 4 forces an 8 into cell 5,7; which forces a 4 into cell 5,3; which eliminates 4 as a possibility from all of the remaining open cells in the top left block (cells 1,2 and 3,3).
answered Dec 11 '18 at 15:23
HellionHellion
1,901819
$endgroup$
add a comment |
$begingroup$
You can spot that
the 1, 7 cell cannot be a 4 (and therefore is an 8)
because
setting it to 4 forces an 8 into cell 5,7; which forces a 4 into cell 5,3; which eliminates 4 as a possibility from all of the remaining open cells in the top left block (cells 1,2 and 3,3).
answered Dec 11 '18 at 15:23
HellionHellion
1,901819
$endgroup$
add a comment |
$begingroup$
You can spot that
the 1, 7 cell cannot be a 4 (and therefore is an 8)
because
setting it to 4 forces an 8 into cell 5,7; which forces a 4 into cell 5,3; which eliminates 4 as a possibility from all of the remaining open cells in the top left block (cells 1,2 and 3,3).
answered Dec 11 '18 at 15:23
HellionHellion
1,901819
$endgroup$
You can spot that
the 1, 7 cell cannot be a 4 (and therefore is an 8)
because
setting it to 4 forces an 8 into cell 5,7; which forces a 4 into cell 5,3; which eliminates 4 as a possibility from all of the remaining open cells in the top left block (cells 1,2 and 3,3).
answered Dec 11 '18 at 15:23
HellionHellion
1,901819
answered Dec 11 '18 at 15:23
HellionHellion
1,901819
answered Dec 11 '18 at 15:23
HellionHellion
1,901819
answered Dec 11 '18 at 15:23
HellionHellion
1,901819
1,901819
add a comment |
add a comment |
$begingroup$
You could spot that:
In the top left block, the bottom right 4/6 cell must be a 6.
Why?
Because if it is a 4 then in the bottom left block the right side becomes 5/6,5/6. But now the solution isn't unique as setting either of these cells as 5 has no vertical effect on the board (from the bottom right block).
answered Dec 10 '18 at 17:49
JonMark PerryJonMark Perry
18k63786
$endgroup$
$begingroup$
I am still unclear. What is the vertical effect in bottom right block please?
$endgroup$
– jerrymouse
Dec 10 '18 at 19:00
$begingroup$
I'm confused. Using (column,row) notation: setting (3,3) to 4 would require (1,7) to be 4, then (2,9) to be 8, (9,9) to be 5, (3,9) would be 6, and (3,8) would be 5. No ambiguity.
$endgroup$
– user3294068
Dec 10 '18 at 19:31
$begingroup$
@user3294068; well it can't be a 4 according to OP's p.s.
$endgroup$
– JonMark Perry
Dec 10 '18 at 19:49
$begingroup$
@jerrymouse; if you follow my logic, we could have both 568:645 or 645:568 equally, which bears no impact on the rest of the grid.
$endgroup$
– JonMark Perry
Dec 10 '18 at 19:51
1
$begingroup$
I might prefer guessing (brute force all options, doesn't seem too difficult here) to this approach, though: if you base any deductions on the solution's assumed uniqueness, you can only reach "an answer" at best. If you want to reach "the answer", you can't assume uniqueness, you have to prove it instead.
$endgroup$
– Bass
Dec 10 '18 at 20:54
add a comment |
$begingroup$
You could spot that:
In the top left block, the bottom right 4/6 cell must be a 6.
Why?
Because if it is a 4 then in the bottom left block the right side becomes 5/6,5/6. But now the solution isn't unique as setting either of these cells as 5 has no vertical effect on the board (from the bottom right block).
answered Dec 10 '18 at 17:49
JonMark PerryJonMark Perry
18k63786
$endgroup$
$begingroup$
I am still unclear. What is the vertical effect in bottom right block please?
$endgroup$
– jerrymouse
Dec 10 '18 at 19:00
$begingroup$
I'm confused. Using (column,row) notation: setting (3,3) to 4 would require (1,7) to be 4, then (2,9) to be 8, (9,9) to be 5, (3,9) would be 6, and (3,8) would be 5. No ambiguity.
$endgroup$
– user3294068
Dec 10 '18 at 19:31
$begingroup$
@user3294068; well it can't be a 4 according to OP's p.s.
$endgroup$
– JonMark Perry
Dec 10 '18 at 19:49
$begingroup$
@jerrymouse; if you follow my logic, we could have both 568:645 or 645:568 equally, which bears no impact on the rest of the grid.
$endgroup$
– JonMark Perry
Dec 10 '18 at 19:51
1
$begingroup$
I might prefer guessing (brute force all options, doesn't seem too difficult here) to this approach, though: if you base any deductions on the solution's assumed uniqueness, you can only reach "an answer" at best. If you want to reach "the answer", you can't assume uniqueness, you have to prove it instead.
$endgroup$
– Bass
Dec 10 '18 at 20:54
add a comment |
$begingroup$
You could spot that:
In the top left block, the bottom right 4/6 cell must be a 6.
Why?
Because if it is a 4 then in the bottom left block the right side becomes 5/6,5/6. But now the solution isn't unique as setting either of these cells as 5 has no vertical effect on the board (from the bottom right block).
answered Dec 10 '18 at 17:49
JonMark PerryJonMark Perry
18k63786
$endgroup$
You could spot that:
In the top left block, the bottom right 4/6 cell must be a 6.
Why?
Because if it is a 4 then in the bottom left block the right side becomes 5/6,5/6. But now the solution isn't unique as setting either of these cells as 5 has no vertical effect on the board (from the bottom right block).
answered Dec 10 '18 at 17:49
JonMark PerryJonMark Perry
18k63786
answered Dec 10 '18 at 17:49
JonMark PerryJonMark Perry
18k63786
answered Dec 10 '18 at 17:49
JonMark PerryJonMark Perry
18k63786
answered Dec 10 '18 at 17:49
JonMark PerryJonMark Perry
18k63786
18k63786
$begingroup$
I am still unclear. What is the vertical effect in bottom right block please?
$endgroup$
– jerrymouse
Dec 10 '18 at 19:00
$begingroup$
I'm confused. Using (column,row) notation: setting (3,3) to 4 would require (1,7) to be 4, then (2,9) to be 8, (9,9) to be 5, (3,9) would be 6, and (3,8) would be 5. No ambiguity.
$endgroup$
– user3294068
Dec 10 '18 at 19:31
$begingroup$
@user3294068; well it can't be a 4 according to OP's p.s.
$endgroup$
– JonMark Perry
Dec 10 '18 at 19:49
$begingroup$
@jerrymouse; if you follow my logic, we could have both 568:645 or 645:568 equally, which bears no impact on the rest of the grid.
$endgroup$
– JonMark Perry
Dec 10 '18 at 19:51
1
$begingroup$
I might prefer guessing (brute force all options, doesn't seem too difficult here) to this approach, though: if you base any deductions on the solution's assumed uniqueness, you can only reach "an answer" at best. If you want to reach "the answer", you can't assume uniqueness, you have to prove it instead.
$endgroup$
– Bass
Dec 10 '18 at 20:54
add a comment |
$begingroup$
I am still unclear. What is the vertical effect in bottom right block please?
$endgroup$
– jerrymouse
Dec 10 '18 at 19:00
$begingroup$
I'm confused. Using (column,row) notation: setting (3,3) to 4 would require (1,7) to be 4, then (2,9) to be 8, (9,9) to be 5, (3,9) would be 6, and (3,8) would be 5. No ambiguity.
$endgroup$
– user3294068
Dec 10 '18 at 19:31
$begingroup$
@user3294068; well it can't be a 4 according to OP's p.s.
$endgroup$
– JonMark Perry
Dec 10 '18 at 19:49
$begingroup$
@jerrymouse; if you follow my logic, we could have both 568:645 or 645:568 equally, which bears no impact on the rest of the grid.
$endgroup$
– JonMark Perry
Dec 10 '18 at 19:51
1
$begingroup$
I might prefer guessing (brute force all options, doesn't seem too difficult here) to this approach, though: if you base any deductions on the solution's assumed uniqueness, you can only reach "an answer" at best. If you want to reach "the answer", you can't assume uniqueness, you have to prove it instead.
$endgroup$
– Bass
Dec 10 '18 at 20:54
$begingroup$
I am still unclear. What is the vertical effect in bottom right block please?
$endgroup$
– jerrymouse
Dec 10 '18 at 19:00
$begingroup$
I am still unclear. What is the vertical effect in bottom right block please?
$endgroup$
– jerrymouse
Dec 10 '18 at 19:00
$begingroup$
I'm confused. Using (column,row) notation: setting (3,3) to 4 would require (1,7) to be 4, then (2,9) to be 8, (9,9) to be 5, (3,9) would be 6, and (3,8) would be 5. No ambiguity.
$endgroup$
– user3294068
Dec 10 '18 at 19:31
$begingroup$
I'm confused. Using (column,row) notation: setting (3,3) to 4 would require (1,7) to be 4, then (2,9) to be 8, (9,9) to be 5, (3,9) would be 6, and (3,8) would be 5. No ambiguity.
$endgroup$
– user3294068
Dec 10 '18 at 19:31
$begingroup$
@user3294068; well it can't be a 4 according to OP's p.s.
$endgroup$
– JonMark Perry
Dec 10 '18 at 19:49
$begingroup$
@user3294068; well it can't be a 4 according to OP's p.s.
$endgroup$
– JonMark Perry
Dec 10 '18 at 19:49
$begingroup$
@jerrymouse; if you follow my logic, we could have both 568:645 or 645:568 equally, which bears no impact on the rest of the grid.
$endgroup$
– JonMark Perry
Dec 10 '18 at 19:51
$begingroup$
@jerrymouse; if you follow my logic, we could have both 568:645 or 645:568 equally, which bears no impact on the rest of the grid.
$endgroup$
– JonMark Perry
Dec 10 '18 at 19:51
1
1
$begingroup$
I might prefer guessing (brute force all options, doesn't seem too difficult here) to this approach, though: if you base any deductions on the solution's assumed uniqueness, you can only reach "an answer" at best. If you want to reach "the answer", you can't assume uniqueness, you have to prove it instead.
$endgroup$
– Bass
Dec 10 '18 at 20:54
$begingroup$
I might prefer guessing (brute force all options, doesn't seem too difficult here) to this approach, though: if you base any deductions on the solution's assumed uniqueness, you can only reach "an answer" at best. If you want to reach "the answer", you can't assume uniqueness, you have to prove it instead.
$endgroup$
– Bass
Dec 10 '18 at 20:54
add a comment |
$begingroup$
You can notice
r2c1, r7c1, r3c5, r7c5 form a finned X-Wing for 4.
Thus
Consider r2c1. It must be either 4 or 8. If r2c1 is 4 that forces r3c3 to be 6. If r2c1 is 8 that forces r7c1 to be 4 which forces r7c5 to be 8 which forces r3c5 to be 4 which forces r3c3 to be 6. So if r2c1 is either 4 or 8 r3c3 is forced to be 6.
Hopefully you can get it from there.
answered 14 hours ago
Rocus HalbaschRocus Halbasch
112
New contributor
Rocus Halbasch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Hi Rocus, welcome to PSE. Please hide your answers using ">!"
$endgroup$
– Greg
13 hours ago
$begingroup$
Yeah. Just figured out how that works. This was my first answer.
$endgroup$
– Rocus Halbasch
13 hours ago
$begingroup$
No problem. Happy puzzling!
$endgroup$
– Greg
13 hours ago
add a comment |
$begingroup$
You can notice
r2c1, r7c1, r3c5, r7c5 form a finned X-Wing for 4.
Thus
Consider r2c1. It must be either 4 or 8. If r2c1 is 4 that forces r3c3 to be 6. If r2c1 is 8 that forces r7c1 to be 4 which forces r7c5 to be 8 which forces r3c5 to be 4 which forces r3c3 to be 6. So if r2c1 is either 4 or 8 r3c3 is forced to be 6.
Hopefully you can get it from there.
answered 14 hours ago
Rocus HalbaschRocus Halbasch
112
New contributor
Rocus Halbasch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Hi Rocus, welcome to PSE. Please hide your answers using ">!"
$endgroup$
– Greg
13 hours ago
$begingroup$
Yeah. Just figured out how that works. This was my first answer.
$endgroup$
– Rocus Halbasch
13 hours ago
$begingroup$
No problem. Happy puzzling!
$endgroup$
– Greg
13 hours ago
add a comment |
$begingroup$
You can notice
r2c1, r7c1, r3c5, r7c5 form a finned X-Wing for 4.
Thus
Consider r2c1. It must be either 4 or 8. If r2c1 is 4 that forces r3c3 to be 6. If r2c1 is 8 that forces r7c1 to be 4 which forces r7c5 to be 8 which forces r3c5 to be 4 which forces r3c3 to be 6. So if r2c1 is either 4 or 8 r3c3 is forced to be 6.
Hopefully you can get it from there.
answered 14 hours ago
Rocus HalbaschRocus Halbasch
112
New contributor
Rocus Halbasch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
You can notice
r2c1, r7c1, r3c5, r7c5 form a finned X-Wing for 4.
Thus
Consider r2c1. It must be either 4 or 8. If r2c1 is 4 that forces r3c3 to be 6. If r2c1 is 8 that forces r7c1 to be 4 which forces r7c5 to be 8 which forces r3c5 to be 4 which forces r3c3 to be 6. So if r2c1 is either 4 or 8 r3c3 is forced to be 6.
Hopefully you can get it from there.
answered 14 hours ago
Rocus HalbaschRocus Halbasch
112
New contributor
Rocus Halbasch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 13 hours ago
answered 14 hours ago
Rocus HalbaschRocus Halbasch
112
New contributor
Rocus Halbasch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 14 hours ago
Rocus HalbaschRocus Halbasch
112
answered 14 hours ago
Rocus HalbaschRocus Halbasch
112
112
New contributor
Rocus Halbasch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Rocus Halbasch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Rocus Halbasch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Hi Rocus, welcome to PSE. Please hide your answers using ">!"
$endgroup$
– Greg
13 hours ago
$begingroup$
Yeah. Just figured out how that works. This was my first answer.
$endgroup$
– Rocus Halbasch
13 hours ago
$begingroup$
No problem. Happy puzzling!
$endgroup$
– Greg
13 hours ago
add a comment |
$begingroup$
Hi Rocus, welcome to PSE. Please hide your answers using ">!"
$endgroup$
– Greg
13 hours ago
$begingroup$
Yeah. Just figured out how that works. This was my first answer.
$endgroup$
– Rocus Halbasch
13 hours ago
$begingroup$
No problem. Happy puzzling!
$endgroup$
– Greg
13 hours ago
$begingroup$
Hi Rocus, welcome to PSE. Please hide your answers using ">!"
$endgroup$
– Greg
13 hours ago
$begingroup$
Hi Rocus, welcome to PSE. Please hide your answers using ">!"
$endgroup$
– Greg
13 hours ago
$begingroup$
Yeah. Just figured out how that works. This was my first answer.
$endgroup$
– Rocus Halbasch
13 hours ago
$begingroup$
Yeah. Just figured out how that works. This was my first answer.
$endgroup$
– Rocus Halbasch
13 hours ago
$begingroup$
No problem. Happy puzzling!
$endgroup$
– Greg
13 hours ago
$begingroup$
No problem. Happy puzzling!
$endgroup$
– Greg
13 hours ago
add a comment |
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$begingroup$
Not sure if any of the advanced techniques to reduce the candidates here can be applied somewhere? I will also take a look if I can apply any of them.
$endgroup$
– Kevin Cruijssen
Dec 10 '18 at 17:07