My WXYZ-Wing Tutorial is often recommended to viewers who watch this video, so I had to watch it again. Swami is a great teacher and I hope he posts again on RUclips with his advanced series.
I’ve been trying to get sudoku tutorials from google and RUclips and every single tutorial is either complicated and doesn’t make sense or too longwith a lot of talking and not much teaching. Then I found this channel and my prayers have been answered. Your the best sudoku tutor ever
I love the practice examples you do after you teach each strategy. You’re a real teacher! You realize that most people need to practice in order to make the strategy sink in. Thanks
I simply LOVE ur 'ha ha' comments. Till now always bring a smile to my face. Especially enjoyed the comment 'now isn't that brilliant or what ' in virtual pairs video. God bless you.
Thank you for this valuable series of videos. You do an excellent job of presenting and then cementing concepts in our minds. Sometimes I forget the many logical deductions one can make from UR patterns, so I am reviewing your videos to improve my game. Excellent information!!!!!!
For Beautiful Custom T-Shirts & Coffee Mugs featuring the Swami Logo, and also for Selected Classical Piano Pieces played by me, now available via Digital Download, please visit the Sudoku Swami Gift Shop! sudoku-swami.shopify.com
So, I hope this means that after watching my Tutorials on UR's, everything is clear to you now? if not, let me know if you have any questions; here in the Comments area, or at sudokuswami@gmail.com :-)) Good luck!
I love your style and explanations. I can't quite grasp UR Type 4 yet 😬 Eureka! Watched and LISTENED to explanation for Type 4 a few more times and the penny has dropped! Amazing instructional content, thanks.
I think I have reached my saturation point. I have watched this video 4 or 5 times now and I still can't remember what to do when I encounter these different types of URs... except maybe the most basic, with 1 extra candidate in 1 of the 4 UR cells.
I came across with the similar situation, 23:10 in the video. Only thanks to watch this I understood, that my UR at this moment doesn't eliminate any values ;-)
been trying to find a logical way to deal with puzzles when the solver says "finished with brute force" now i can do it without backtracking each possible number one by one. thank u
First I didn't see why there MUST be 2 solutions in any case of a UR, but then I realized: If, for example, in 10:11 R1C7 would be the digit 1, so IF it would lead to a valid solution, so we can say for certain that the digit 4 in that cell would lead to a valid solution TOO. And here's why: Because we can switch them without changing the content of the rows, columns, and blocks. I think It's a really important detail that wasn't said in the video.
Thanks for the video. I've previously been forgetting to check the block for type 4 rectangles. Hopefully I'll be spotting more type 4s now. I also didn't know about type 5 until now (I don't know how many of those I'll actually see in a puzzle though), still nice to know.
Type 4 and Type 7 (Hidden Rectangle) are probably the two most common. Then Type 1 and Type 2. Type 3, 5 & 6 are probably the most rare, but it's good to know how to solve ALL of them. Good luck.
Question about the relationship between an x-wing and a unique rectangle: If I have a Type 4 UR and look at just one of the two candidates, it can look like an x-wing, i.e. 2 candidates in the base set where the base set intersects with the cover set. Using x-wing principles, I can eliminate candidates in the cover set that are not in the intersections. But, with a Type 4 UR, I need one or more of these to allow the elimination of a candidate in the UR. Consider Tutorial #20 at 20:14, where the 6s form an x-wing and should enable you to eliminate all the other 6s in row 1. But, if you do that, you than can’t eliminate the 6s from the two candidate cells and end up with a catastrophic unique rectangle. (Compare this to Tutorial #10 at 9:05.) Am I missing something (not surprising) or is it just a matter of looking for URs before x-wings? Thanks.
First of all, you are talking about diagrams I have drawn BY HAND, specifically to demonstrate a particular point. They are NOT real puzzles. In those instances, please try to focus on what I am explaining, and do not look for other patterns. Second, if you have an X-Wing, use it. If you have a UR Pattern, use it. They will both work, and it doesn't matter which order you employ them. Nothing catastrophic is going to happen.
Excellent explanations! I have always had trouble with the logic of URs but I think I'm starting to grasp it thanks to your videos. In the Type 4 example at 25:00, is it not true that since there are no other 3s in block two, the two UR 3s are in fact a locked pair and the other instances of 3 in row 3 can be eliminated? In that case we can just reason the same way as before with the UR 9s. In fact, there had better be extra instances of digit 9 in *both* the row and the block, or the UR 9s would also become a locked pair and in fact, the UR 3, 9 in row 3 would then just be a hidden pair we would have the forbidden pattern. Am I thinking about this correctly?
@21.58 talking about the type ur, i see the ur candidates are 6 and 7, and I see that one of the yellow cells has to be 7. And i see that both of the other ur candidates ccannot be 6, otherwise we have the forbidden pattern. But one can be 6. This would mean its not a forbidden pattern. So I dont get the logic of why you can eliminate both 6 candidates. Pls explain
In Row 3, one of the Cells must be 7, and the other must be 6. So in Row 1, one of the two yellow Cells must be 7, because it is a Conjugate Pair in that Row. This means that NEITHER of the yellow Cells in Row 1 can be a 6, because it would put a 7 in the other Cell, and would therefore cause the illegal pattern.
So, ultimately, we dont want the combination to be 7/6 or 6/7, because these numbers can be interchanged allong with row 3, giving us the 2 solutions. So thats why we eliminate both 6's (the cell that is 7 cannot be a 6 and the other cell cannot be a 6, because then we can interchange all 4 cells), got it now? So in sudoku, we only want to solve for the one unique solution where the numeric positions cannot be changed to give another solution, 👍Thanks
Right. Very important point to remember: The four Cells that make up the UR Pattern (ANY of them), must be UNSOLVED Cells at the beginning of the puzzle. If any of the four Cells is a "Given," then the UR Techniques do NOT work.
Thank you. I am actually a retired musician/song-writer. I have never been a teacher......not even of music. But I am glad you enjoyed the Video. Good luck! :-))
Wow I was able to apply UR"s in a puzzle right after watching the UR series 20, 21 and 21A. When I looked at UR's on the websites it was so esoteric I didn't bother to put them into my repertoire.
I have a sudoku app on my iphone and all puzzles were solved with this technique until i found one that actually had both answers and I don’t understand how!!! ,i spent 5 hours not to make a mistake and it blown in my face, They are 2 columns 2 rows and only 2 blocks!! 😭😭😭😭, do you think the app is wrongly configured?!
New to this and I have a quick question, on the second example for UR #2 & #5 at around 16:35 into the video, would you also be able use it on the 4s in r6c56 (if there were other 4s in that block and row 6 as well), or can you only use the technique once per set?
I think you mean COLUMN 6, Rows 5 & 6, yes? And if so, the answer to your question is YES. You could eliminate any other 4's that can see the two 4's in R5C6 & R6C6, but there aren't any. There are no restrictions as to how many times you can use the technique. If the Rules apply, then it will work.
The 9 in R5C2 is a Given. UR's cannot contain any Givens. The 4 Cells of a UR Pattern must all be UNSOLVED at the beginning of a Puzzle, or the UR Rules DO NOT apply.
At 16:40 for type 2. There is, as you pointed out, the 2,3,6, combination in R5 and R6, C2. However, the 2, 3, 4 combination occurs in C6 of R5 and R6. How does that effect the eliminations?
The Cells you point out, (containing 2, 3 & 4), combined with the two PURE UR Cells containing only 2 & 3 (R5C7 & R6C7) form a completely separate UR Pattern. Since one of the Cells R5C6 & R6C6 must contain a 4 (or you would have the forbidden pattern), you could eliminate any Candidate 4 that can see BOTH of those 4's. This would constitute another Type 2 configuration. But as you can see, no eliminations on 4 are available in Block 5 or Column 6. This is why I pointed out the other configuration instead.
Yes. Absolutely. Two separate UR Type 2 Patterns. But the one with the 6's is productive, allowing for seven eliminations, whereas the one with the 4's is dead, and produces nothing, as a UR Type 2. However, if you look at it as a Type 4, you could eliminate the two 3's from R5C6 & R6C6. But I was demonstrating a Type 2 at that moment.
I am so confused with Type 1 UR example at 10:13 of the video,. What I see is the 4s is def conjugate pairs and form an X wing. The 1s are not conjugate pairs, so maybe that is then a Type 1 UR. But why can't the value in R1C7 be a 4 for example if the value in R1C9 is a 1? If R1C9 =1 => R7C9=4 => R7C7=1 => R1C7 = 4 or 5. and since that is the ONLY 4 in C7, it has to be a 4, not a 5. If R1C9 = 4 => R7C9=1 => R7C7=4 => R1C7 = 1 or 5 which is allowed as conjugate pairs in R1 and block 3. It says nothing about that R1C7 can't be either 1 or 5.
If three of the four UR Cells are Bi-Value Cells, (i.e., containing ONLY the UR Candidates), then the fourth UR Cell CANNOT contain EITHER of the UR Candidates. Further, if there is only one other Candidate in the fourth Cell, it will be the soution to that Cell.
Thank you Swami. The key is the recognising the 3 bi-value cells in the specific formation, which then leads us to a solution based on prior knowledge about the formation. Earlier in video you said 2 cells with 76 cells in row 3 and again in 6 cannot exist. Why? Sorry for asking a stupid question, still learning all the tricks with Sodoku. I am assuming the answer is because they are just 2 Bi-Value cells, with more of the same candidates in their respective houses, and thus not 2 matching pairs (conjugate pairs) for example, Granted in the block with the cell containing extra candidates, we can't say for sure yet that it is the remaining cell for the matching pair in that block, but we can say definitely the other block has a matching pair/conjugate pair on the 76, in your example, for the row, block and column. The moment I see 2 pairs in the same house in 2 cells, like in your 1st example with the 76s in a block, they are not bi-value anymore, they are matching/conjugate pairs, and all other candidates in the house and corresponding houses can be eliminated. The formation you say can't exist, in my mind can exist, it is a double X wing using the bi-value as the corners with same values on the diagonals but alternating. It may be a unlikely formation, but a valid possibility. You need the look at the placement of the bi-value candidates in the remaining houses to rule that out. The formation which you say cannot exist, here is the valid example, assuming placement of the 7 and 6 in remaining houses is correct and does not invalidate the formation. E.g. In row 2 of your example, say a 7 and 6 and then in the bottom row a 6 and then a 7 in that order. Other possibility is a 6 and 7 in top row followed by a 7 and 6 in bottom row. So I am just trying to understand the basis why we can definitely say the cell with the extra candidates def can't contain ANY of the Bi-Value candidates. Your statement of the remaining cell can't contain EITHER of the Bi-Value candidates is only true, if you assume, that the cell in question is NOT a Bi-Value itself, with the same Bi-Values and thus has to be one of the other candidates. So you have to prove then, that it is not a Bi-Value. And in my mind only way to do that is using some Type 2 AIC on rest of the puzzle or some other formation in rest of puzzle to prove that. The 3 Bi-Value cells by themselves does not prove it.
Petrus, you are making this WAY more complicated than it is. In a UR Type 1, the fourth Cell CANNOT contain either of the UR Candidates, because it would lead to multiple solutions, and would render your puzzle invalid. It's as simple as that. This principle is FULLY explained in the Video. Please watch it again. Good luck.
@@SudokuSwami Thank you for explaining it, earlier in video you did explained that we are trying to avoid multiple solutions to the puzzle. I did not realise you can actually have a puzzle with all numbers filled in and only have the 4 UR cells and then lead to a unsolvable puzzle. So your explanation there makes sense and it is the basis of the whole UR solving technique. Thank you for your patience.
In order to be considered "valid," a Sudoku Puzzle must have One Unique Solution. If your puzzle is valid, then multiple solutions are impossible. This is the basis for ALL of the UR Techniques.
Hello Swami, in type 4 explanation at 26:00, since we say "in block 2, candidate 3 can only appear in the UR cells R3C4 and R3C6, so we eliminate candidate 9 from those cells", yes, but that also means we can eliminate candidate 3 from the rest of row 3, since that 3 can only appear inside the UR cells, isn´t it?
Yes, you are correct. But this is because the 3's in Block 2 are Locked Candidates Type I (see Tutorial #6). In this Video, I am only focusing on and demonstrating the implications and deductions made from Unique Rectangles. Good luck! SS
@@SudokuSwami Thanks for clarifying! I imagine I was blind focusing in the explanation of the UR technique and didn't think about the locked candidates. In a regular solving process I figure you won't normally find this configuration since locked and pointing candidates are solved early as very basic steps. (Sorry if I made big mistakes in my English, no time to check, it's time to go to sleep here ;-) )
I find the hardest UR type 4, it would be good if the master could demonstrate why in example 1 of UR type 4 , 6 could not be accommodate in either R1C1 or R1C6. Could someone help ?
If you listen carefully to what I say, it is fully explained why. One of those Cells must be a 7. They are a Conjugate Pair in Row 1. This means neither of those Cells can be a 6, or it would allow for multiple solutions, which would make it an invalid puzzle.
at 25:55, if only candidate(3) in cell R3C2 (1-3-4-6); and if only candidate(9) in cell R3C5 (1-2-6-9). the candidates (3 and 9) different does not exist. Can i be eliminate candidate(9) in cell R3C4 and R3C6
No. The point in this example, is that when the two UR Cells with extra Candidates appear in a Row (or Column) AND a Block at the same time, (i.e., they are LOCKED), you can apply the Type 4 Rules to the Block as well. If both the 3 and the 9 appear in the Row and/or the Block, OUTSIDE the two non-diagonal Cells, then there is nothing you can do.
I just solved a puzzle as a UR type 4 but after I was done I ran it through a solver and the solver solved it as a UR Type 3. Is it possible that you can "choose" between 3 and 4? I hope so otherwise I made a lucky guess.
There may be some discrepancy as to the Type #'s from one source to another. I tried to use the Type #'s that I thought were the most commonly-accepted. But no, a Type 3 cannot be mistaken for a Type 4. (My) Type 3 is where a virtual subset is created with the additional candidates in two non-diagonal UR Cells. (My) Type 4 is where one UR Candidate is eliminated from two non-diagonal UR Cells, because they appear in the House outside the Cells, and the other Candidate does NOT. The two Types are entirely different.
Why not? Because of this RULE: "In order to be considered VALID, a Sudoku Puzzle must have only ONE UNIQUE SOLUTION." This allows us to utilize all the UR Techniques. If a Puzzle is Valid, according to the Rule, it is IMPOSSIBLE to have multiple solutions. Being ILLEGAL has nothing to do with it.
Hi you forget Unique Rectangle Type 2 on 17:43 in the video. This cells R2C1, R2C2 have 23 and the cells R5C1, R5C2 have 237 one of this 2 cells have to be 7 so the 7 on this cells R5C8, R5C9 and R6C2, have to be eliminate. I find it myself :)
Yes, you are correct. I did not "forget" it. :-) I was making the point that the four UR cells cannot lie in four separate Blocks. In ALL of my examples, you will find many things going on. But please try to focus on the point I am trying to make. The goal in these tutorials is to teach the various techniques; not to solve the puzzles. I will sometimes solve puzzles, from start to finish, in my Random Tips Series. Thank you for your interest. Good luck.
Bud, I have degrees in both Math and Comp Sci, both of which require tons of logic. If you have no training in logic, the level of natural understanding you have is scary.
My WXYZ-Wing Tutorial is often recommended to viewers who watch this video, so I had to watch it again. Swami is a great teacher and I hope he posts again on RUclips with his advanced series.
I’ve been trying to get sudoku tutorials from google and RUclips and every single tutorial is either complicated and doesn’t make sense or too longwith a lot of talking and not much teaching. Then I found this channel and my prayers have been answered. Your the best sudoku tutor ever
I love the practice examples you do after you teach each strategy. You’re a real teacher! You realize that most people need to practice in order to make the strategy sink in. Thanks
Love your lessons, thank you!
Type 1 at beginning, Type 2 at 11:42, Type 5 at 13:31, Type 4 at 20:12
I simply LOVE ur 'ha ha' comments. Till now always bring a smile to my face. Especially enjoyed the comment 'now isn't that brilliant or what ' in virtual pairs video. God bless you.
If you enjoyed this Video, please don't forget to click the SUBSCRIBE button, and the Thumbs Up Icon. It will really help me out. Thank you!
Thank you for this valuable series of videos. You do an excellent job of presenting and then cementing concepts in our minds. Sometimes I forget the many logical deductions one can make from UR patterns, so I am reviewing your videos to improve my game. Excellent information!!!!!!
Thank you for your kind comment. I really appreciate it. Good luck, and stay safe!
I didn't realize how common these can be! I'm starting to see these in a lot of the puzzles now
For Beautiful Custom T-Shirts & Coffee Mugs featuring the Swami Logo, and also for Selected Classical Piano Pieces played by me, now available via Digital Download, please visit the Sudoku Swami Gift Shop! sudoku-swami.shopify.com
Please visit sudokuswami.com for news about upcoming Videos and an Outline of my Complete Course!
When I first came across this, it drove me crazy. I kept trying to find out where I had a number twice in a row, col. or block.
So, I hope this means that after watching my Tutorials on UR's, everything is clear to you now? if not, let me know if you have any questions; here in the Comments area, or at sudokuswami@gmail.com :-)) Good luck!
I love your style and explanations. I can't quite grasp UR Type 4 yet 😬
Eureka! Watched and LISTENED to explanation for Type 4 a few more times and the penny has dropped! Amazing instructional content, thanks.
Yes, it's a good idea to "listen."
I think I have reached my saturation point. I have watched this video 4 or 5 times now and I still can't remember what to do when I encounter these different types of URs... except maybe the most basic, with 1 extra candidate in 1 of the 4 UR cells.
I came across with the similar situation, 23:10 in the video. Only thanks to watch this I understood, that my UR at this moment doesn't eliminate any values ;-)
Ha ha. Good! Thanks for your comment. I always enjoy the examples you post on the Sudoku Forum. Nice stuff. :-))
been trying to find a logical way to deal with puzzles when the solver says "finished with brute force"
now i can do it without backtracking each possible number one by one.
thank u
Because of u I am able to solve hard sudoku. Thank you sir
Good exercise!.will watch again
This is a very powerful tool, thanks
You're welcome. :-))
First I didn't see why there MUST be 2 solutions in any case of a UR, but then I realized:
If, for example, in 10:11 R1C7 would be the digit 1, so IF it would lead to a valid solution, so we can say for certain that the digit 4 in that cell would lead to a valid solution TOO. And here's why:
Because we can switch them without changing the content of the rows, columns, and blocks.
I think It's a really important detail that wasn't said in the video.
I demonstrate and say precisely that between 3:35 & 4:15
@@SudokuSwami yea, but it's less convincing when all the digits are known except for these four, because it's a special case and not a general case
It's the Principle itself, that matters. It makes no difference whether it is a special case or a general case.
Brilliant presentation. I thought I understood this. Now I understand it a whole lot better.
Glad to hear it. I hope you will Subscribe to my Channel. Many more Videos to come!
Thanks for the video. I've previously been forgetting to check the block for type 4 rectangles. Hopefully I'll be spotting more type 4s now.
I also didn't know about type 5 until now (I don't know how many of those I'll actually see in a puzzle though), still nice to know.
Type 4 and Type 7 (Hidden Rectangle) are probably the two most common. Then Type 1 and Type 2. Type 3, 5 & 6 are probably the most rare, but it's good to know how to solve ALL of them. Good luck.
hell of a view back there Sr
Question about the relationship between an x-wing and a unique rectangle:
If I have a Type 4 UR and look at just one of the two candidates, it can look like an x-wing, i.e. 2 candidates in the base set where the base set intersects with the cover set. Using x-wing principles, I can eliminate candidates in the cover set that are not in the intersections. But, with a Type 4 UR, I need one or more of these to allow the elimination of a candidate in the UR. Consider Tutorial #20 at 20:14, where the 6s form an x-wing and should enable you to eliminate all the other 6s in row 1. But, if you do that, you than can’t eliminate the 6s from the two candidate cells and end up with a catastrophic unique rectangle. (Compare this to Tutorial #10 at 9:05.) Am I missing something (not surprising) or is it just a matter of looking for URs before x-wings? Thanks.
First of all, you are talking about diagrams I have drawn BY HAND, specifically to demonstrate a particular point. They are NOT real puzzles. In those instances, please try to focus on what I am explaining, and do not look for other patterns. Second, if you have an X-Wing, use it. If you have a UR Pattern, use it. They will both work, and it doesn't matter which order you employ them. Nothing catastrophic is going to happen.
Excellent explanations! I have always had trouble with the logic of URs but I think I'm starting to grasp it thanks to your videos. In the Type 4 example at 25:00, is it not true that since there are no other 3s in block two, the two UR 3s are in fact a locked pair and the other instances of 3 in row 3 can be eliminated? In that case we can just reason the same way as before with the UR 9s. In fact, there had better be extra instances of digit 9 in *both* the row and the block, or the UR 9s would also become a locked pair and in fact, the UR 3, 9 in row 3 would then just be a hidden pair we would have the forbidden pattern. Am I thinking about this correctly?
Yes, you are correct.
@21.58 talking about the type ur, i see the ur candidates are 6 and 7, and I see that one of the yellow cells has to be 7. And i see that both of the other ur candidates ccannot be 6, otherwise we have the forbidden pattern. But one can be 6. This would mean its not a forbidden pattern. So I dont get the logic of why you can eliminate both 6 candidates. Pls explain
In Row 3, one of the Cells must be 7, and the other must be 6. So in Row 1, one of the two yellow Cells must be 7, because it is a Conjugate Pair in that Row. This means that NEITHER of the yellow Cells in Row 1 can be a 6, because it would put a 7 in the other Cell, and would therefore cause the illegal pattern.
So, ultimately, we dont want the combination to be 7/6 or 6/7, because these numbers can be interchanged allong with row 3, giving us the 2 solutions. So thats why we eliminate both 6's (the cell that is 7 cannot be a 6 and the other cell cannot be a 6, because then we can interchange all 4 cells), got it now? So in sudoku, we only want to solve for the one unique solution where the numeric positions cannot be changed to give another solution, 👍Thanks
Right. Very important point to remember: The four Cells that make up the UR Pattern (ANY of them), must be UNSOLVED Cells at the beginning of the puzzle. If any of the four Cells is a "Given," then the UR Techniques do NOT work.
Got it
Very well explained.
At 26:50 the next step would be to remove the three 8’s from Blocks 3 and 6 because of the Locked 8’s in Block 9.
THIS DUDE IS SUPER SHARP
Thank you for the next interesting video and next solving technic :-)
Hello Mirek. Glad you like the tutorials. Please spread the word. Thanks for your continued support. :-))
Your AWESOME! Must have experience teaching.
Thank you. I am actually a retired musician/song-writer. I have never been a teacher......not even of music. But I am glad you enjoyed the Video. Good luck! :-))
very clear explanation,thank you
Wow I was able to apply UR"s in a puzzle right after watching the UR series 20, 21 and 21A. When I looked at UR's on the websites it was so esoteric I didn't bother to put them into my repertoire.
Great. Yes, UR's are "all over the place." It's good to know how to employ them.......especially Type 6. :-)
The Hidden ones, too (my Type 7). Types 1, 2, 4 and 7 are by far the most common. You should try to master at least those four.
I have a sudoku app on my iphone and all puzzles were solved with this technique until i found one that actually had both answers and I don’t understand how!!! ,i spent 5 hours not to make a mistake and it blown in my face, They are 2 columns 2 rows and only 2 blocks!! 😭😭😭😭, do you think the app is wrongly configured?!
What's the name of the App? Some do allow for multiple solutions, even though it is technically invalid.
New to this and I have a quick question, on the second example for UR #2 & #5 at around 16:35 into the video, would you also be able use it on the 4s in r6c56 (if there were other 4s in that block and row 6 as well), or can you only use the technique once per set?
I think you mean COLUMN 6, Rows 5 & 6, yes? And if so, the answer to your question is YES. You could eliminate any other 4's that can see the two 4's in R5C6 & R6C6, but there aren't any. There are no restrictions as to how many times you can use the technique. If the Rules apply, then it will work.
Thank you so much!
9:42 in middle block and middle left block why the number 9&7 are in that position?
The 9 in R5C2 is a Given. UR's cannot contain any Givens. The 4 Cells of a UR Pattern must all be UNSOLVED at the beginning of a Puzzle, or the UR Rules DO NOT apply.
@@SudokuSwami Thank you.
At 16:40 for type 2. There is, as you pointed out, the 2,3,6, combination in R5 and R6, C2. However, the 2, 3, 4 combination occurs in C6 of R5 and R6. How does that effect the eliminations?
The Cells you point out, (containing 2, 3 & 4), combined with the two PURE UR Cells containing only 2 & 3 (R5C7 & R6C7) form a completely separate UR Pattern. Since one of the Cells R5C6 & R6C6 must contain a 4 (or you would have the forbidden pattern), you could eliminate any Candidate 4 that can see BOTH of those 4's. This would constitute another Type 2 configuration. But as you can see, no eliminations on 4 are available in Block 5 or Column 6. This is why I pointed out the other configuration instead.
Let repeat just to make sure in understand. You can treat them as two sperate UR patterns. Sounds good to me, Appreciate your response.
Yes. Absolutely. Two separate UR Type 2 Patterns. But the one with the 6's is productive, allowing for seven eliminations, whereas the one with the 4's is dead, and produces nothing, as a UR Type 2. However, if you look at it as a Type 4, you could eliminate the two 3's from R5C6 & R6C6. But I was demonstrating a Type 2 at that moment.
Thank you. OK. I am still learning, but slowly. I is retired--I not sure if it means I can now stay home or if I have new tires on me. HAHA.
I am so confused with Type 1 UR example at 10:13 of the video,. What I see is the 4s is def conjugate pairs and form an X wing. The 1s are not conjugate pairs, so maybe that is then a Type 1 UR. But why can't the value in R1C7 be a 4 for example if the value in R1C9 is a 1? If R1C9 =1 => R7C9=4 => R7C7=1 => R1C7 = 4 or 5. and since that is the ONLY 4 in C7, it has to be a 4, not a 5.
If R1C9 = 4 => R7C9=1 => R7C7=4 => R1C7 = 1 or 5 which is allowed as conjugate pairs in R1 and block 3. It says nothing about that R1C7 can't be either 1 or 5.
If three of the four UR Cells are Bi-Value Cells, (i.e., containing ONLY the UR Candidates), then the fourth UR Cell CANNOT contain EITHER of the UR Candidates. Further, if there is only one other Candidate in the fourth Cell, it will be the soution to that Cell.
Thank you Swami. The key is the recognising the 3 bi-value cells in the specific formation, which then leads us to a solution based on prior knowledge about the formation.
Earlier in video you said 2 cells with 76 cells in row 3 and again in 6 cannot exist. Why?
Sorry for asking a stupid question, still learning all the tricks with Sodoku.
I am assuming the answer is because they are just 2 Bi-Value cells, with more of the same candidates in their respective houses, and thus not 2 matching pairs (conjugate pairs) for example,
Granted in the block with the cell containing extra candidates, we can't say for sure yet that it is the remaining cell for the matching pair in that block, but we can say definitely the other block has a matching pair/conjugate pair on the 76, in your example, for the row, block and column.
The moment I see 2 pairs in the same house in 2 cells, like in your 1st example with the 76s in a block, they are not bi-value anymore, they are matching/conjugate pairs, and all other candidates in the house and corresponding houses can be eliminated.
The formation you say can't exist, in my mind can exist, it is a double X wing using the bi-value as the corners with same values on the diagonals but alternating. It may be a unlikely formation, but a valid possibility. You need the look at the placement of the bi-value candidates in the remaining houses to rule that out.
The formation which you say cannot exist, here is the valid example, assuming placement of the 7 and 6 in remaining houses is correct and does not invalidate the formation.
E.g. In row 2 of your example, say a 7 and 6 and then in the bottom row a 6 and then a 7 in that order. Other possibility is a 6 and 7 in top row followed by a 7 and 6 in bottom row.
So I am just trying to understand the basis why we can definitely say the cell with the extra candidates def can't contain ANY of the Bi-Value candidates.
Your statement of the remaining cell can't contain EITHER of the Bi-Value candidates is only true, if you assume, that the cell in question is NOT a Bi-Value itself, with the same Bi-Values and thus has to be one of the other candidates. So you have to prove then, that it is not a Bi-Value. And in my mind only way to do that is using some Type 2 AIC on rest of the puzzle or some other formation in rest of puzzle to prove that.
The 3 Bi-Value cells by themselves does not prove it.
Petrus, you are making this WAY more complicated than it is. In a UR Type 1, the fourth Cell CANNOT contain either of the UR Candidates, because it would lead to multiple solutions, and would render your puzzle invalid. It's as simple as that. This principle is FULLY explained in the Video. Please watch it again. Good luck.
@@SudokuSwami Thank you for explaining it, earlier in video you did explained that we are trying to avoid multiple solutions to the puzzle. I did not realise you can actually have a puzzle with all numbers filled in and only have the 4 UR cells and then lead to a unsolvable puzzle. So your explanation there makes sense and it is the basis of the whole UR solving technique. Thank you for your patience.
In order to be considered "valid," a Sudoku Puzzle must have One Unique Solution. If your puzzle is valid, then multiple solutions are impossible. This is the basis for ALL of the UR Techniques.
Hello Swami, in type 4 explanation at 26:00, since we say "in block 2, candidate 3 can only appear in the UR cells R3C4 and R3C6, so we eliminate candidate 9 from those cells", yes, but that also means we can eliminate candidate 3 from the rest of row 3, since that 3 can only appear inside the UR cells, isn´t it?
Yes, you are correct. But this is because the 3's in Block 2 are Locked Candidates Type I (see Tutorial #6). In this Video, I am only focusing on and demonstrating the implications and deductions made from Unique Rectangles. Good luck! SS
@@SudokuSwami Thanks for clarifying! I imagine I was blind focusing in the explanation of the UR technique and didn't think about the locked candidates. In a regular solving process I figure you won't normally find this configuration since locked and pointing candidates are solved early as very basic steps.
(Sorry if I made big mistakes in my English, no time to check, it's time to go to sleep here ;-) )
Your English is fine. Just know that the 3's in Row 3, (in Block 1 and Block 3) are False because of the Locked Candidates, and NOT because of the UR.
I find the hardest UR type 4, it would be good if the master could demonstrate why in example 1 of UR type 4 , 6 could not be accommodate in either R1C1 or R1C6. Could someone help ?
If you listen carefully to what I say, it is fully explained why. One of those Cells must be a 7. They are a Conjugate Pair in Row 1. This means neither of those Cells can be a 6, or it would allow for multiple solutions, which would make it an invalid puzzle.
at 25:55, if only candidate(3) in cell R3C2 (1-3-4-6);
and if only candidate(9) in cell R3C5 (1-2-6-9).
the candidates (3 and 9) different does not exist.
Can i be eliminate candidate(9) in cell R3C4 and R3C6
No. The point in this example, is that when the two UR Cells with extra Candidates appear in a Row (or Column) AND a Block at the same time, (i.e., they are LOCKED), you can apply the Type 4 Rules to the Block as well. If both the 3 and the 9 appear in the Row and/or the Block, OUTSIDE the two non-diagonal Cells, then there is nothing you can do.
I just solved a puzzle as a UR type 4 but after I was done I ran it through a solver and the solver solved it as a UR Type 3. Is it possible that you can "choose" between 3 and 4? I hope so otherwise I made a lucky guess.
There may be some discrepancy as to the Type #'s from one source to another. I tried to use the Type #'s that I thought were the most commonly-accepted. But no, a Type 3 cannot be mistaken for a Type 4. (My) Type 3 is where a virtual subset is created with the additional candidates in two non-diagonal UR Cells. (My) Type 4 is where one UR Candidate is eliminated from two non-diagonal UR Cells, because they appear in the House outside the Cells, and the other Candidate does NOT. The two Types are entirely different.
I cant grasp y is it considered illegal? If there r multiple solutions and it works then why not?
Why not? Because of this RULE: "In order to be considered VALID, a Sudoku Puzzle must have only ONE UNIQUE SOLUTION." This allows us to utilize all the UR Techniques. If a Puzzle is Valid, according to the Rule, it is IMPOSSIBLE to have multiple solutions. Being ILLEGAL has nothing to do with it.
Thank you. UR awesome....!
Ha-ha! Thanks. That is very funny, and VERY witty. :-) I wish I had 100,000 subscribers who think like you......
What software / app are you using?
Please send me an email. sudokuswami@gmail.com Thanks.
Are you the grandson of Albert Einstein?
Yes, I am. How did you know?
@@SudokuSwami O.O thanks for the tips.
What app are you using in laptop
Please tell me it will be useful to learn sudoku
For info on software, please write to me at: sudokuswami@gmail.com
Hi you forget Unique Rectangle Type 2 on 17:43 in the video. This cells R2C1, R2C2 have 23 and the cells R5C1, R5C2 have 237 one of this 2 cells have to be 7 so the 7 on this cells R5C8, R5C9 and R6C2, have to be eliminate. I find it myself :)
Yes, you are correct. I did not "forget" it. :-) I was making the point that the four UR cells cannot lie in four separate Blocks. In ALL of my examples, you will find many things going on. But please try to focus on the point I am trying to make. The goal in these tutorials is to teach the various techniques; not to solve the puzzles. I will sometimes solve puzzles, from start to finish, in my Random Tips Series. Thank you for your interest. Good luck.
you make the good work :) I see it is beautiful and I understand it
Thanks, Moncoeur. Yes, you certainly do understand it. That's great!
Bud, I have degrees in both Math and Comp Sci, both of which require tons of logic. If you have no training in logic, the level of natural understanding you have is scary.
No training in logic. Algebra came easy to me..... :-))