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I'm not sure what techniques you're familiar with, but here's some things that will help.
There's a Naked Triple in Row 7. >!R7C368 contain only the digits 679, so these digits can't appear anywhere else in this row.!< I don't think it breaks the puzzle open though.
I see two options to break the puzzle open.
There's an XY-Wing: >!R7C2, R8C3 and R8C8 form an XY-Wing. R8C3 is called the pivot. The two options for R8C3 forces either R7C2 or R8C8 to be a 2, so at least one of them has to be a 2. Therefore any cell that see both R7C2 and R8C8 can't be a 2, eliminating 2 from R7C7.!<
There's also a W-wing: >!R9C1 and R8C8 are both 24. Together they see all 2s in Column 5. Therefore they can't both be 2, as then there's nowhere to place a 2 in Column 5. This means that any cell that sees both R9C1 and R8C8 can't be 4 as then they would both have to be 2, eliminating 4 from R8C3 and R9C9.!<
I hope this is all clear.
Look more closely at your 9s.
In box 1 they can only go in r1, so the 9 in r1 of box 2 can be eliminated.
In box 5 they can only go in c4 so can be eliminated from c4 of b2 as well.
That only leave one place for a 9 in b2.
This is called locked (pointing) candidates.
Hodoku is where you can find everything you need for solving sudoku's. From the most basic tactics to the most advanced ones, all explained with examples.
They have a sudoku generator and solver as well. You can select the tactics you are proficient with or would like to train and it will generate a sudoku solvable with those tactics. And when you're stuck you can get vague hints, concrete hints or step by step solutions on how to remove possible candidates and solve the sudoku.
It's an amazing tool much better than any others out there, the only downside might be that it's an offline tool. So you'll have to download it.
There's a string of three W-wings can get you going all kinda involving the same cells.
1) R6C3 and R7C7 together see all 2s in Row 5, so they can't both be 2. So any cell that sees both cells can't be 4, eliminating 4 from R7C3.
2) R6C3 and R7C7 together see all 4s in Box 7, so they can't both be 4. So any cell that sees both cells can't be 2, eliminating 2 from R7C3. This makes R7C3 a 5.
3) Now R8C3 and R7C7 together see all 2s in Row 9, so they can't both be 2. So any cell that sees both cells can't be 4, eliminating 4 from R7C2 and R8C7. This leaves a single 4 in Box 7 and Box 9.
There could be easier things, but this was the first thing I spotted and W-wings are one of my favourite techniques.
The hodoku program help has a clearer explanation (plus lots of other techniques, but:
* only one of r1c4 & r1c7 can be the 2 (it's also possible that neither are).
* if r1c4 is the 2, then r1c7 cannot be and r7c7 has to be.
* if r1c7 is the 2, then r1c4 cannot be and r9c4 has to be.
* if neither r1c4 or r1c7 is the 2, then both r7c7 and r9c4 are 2.
* since either r7c7 or r9c4 (or both) must be the 2, then r7c6 cannot be 2.
that should help you finish this one.
*Edit: moved to be a reply.
You'll need to use some notation to see where numbers can go...
...once you do you'll find an XY-Wing (a.k.a. Y-Wing, Bent Triple) which will set r5c4 ("5D" here) and the puzzle will be singles form there.
Alternately, a W-Wing will set r6c5 and again, the puzzle solves from there.
It can't be BUG+1 because I think all the cells must be bivalue cells except one cell has 3 possible candidates. At least at this point in the puzzle. there may be one later.
Take a look at what kirlappeee is saying. It looks like a UR Type 1:
Candy_Lawn is pointing out a Unique Rectangle Type 2. Here's a very good wiki page on uniqueness techniques:
http://hodoku.sourceforge.net/en/tech_ur.php
Also, if you are interested, I did a tutorial on UR Type 2 which explains the logic in pretty good detail:
Totally. The short explanation is that you have exactly 3 cells in a column that contain exactly 3 numbers between them, and therefore those three numbers can't appear anywhere else in the column. Try putting an 8 in R6C9 and see what happens. Here's a longer and potentially more confusing explanation of the Naked Sets technique: http://hodoku.sourceforge.net/en/tech\_naked.php
It looks like you're missing some basic techniques.
http://hodoku.sourceforge.net/en/techniques.php
I'd say read up through x wing. From there you should be able to solve most generated puzzles. (Most phone apps never get more complex than this)
After that you would advance your repetuiore with tricks
You won't find techniques other than bifurcating, backtracking etc for the really hardest Sudokus. Even XY-chains and loops can be very difficult and complex.
Here is a pretty comprehensive list as implemented in Hodoku: http://hodoku.sourceforge.net/en/techniques.php
Sudokuwiki's list: https://www.sudokuwiki.org/Strategy_Families
Im also pretty new to the Skyscraper method so I would recommend taking a look at this site, it does a pretty good job of explaining how it works and gives a couple examples: http://hodoku.sourceforge.net/en/tech_sdp.php#sk
Why? Because BUG+1. Every remaining cell contains only two potential candidates, except r7c2, which contains three candidates. This implies it must be a 6.
For more in see http://hodoku.sourceforge.net/en/tech_ur.php#bug1
There is a W-Wing.
The two sets of >!45 on r7c5 and r8c9!< are connected by two >!4 in r9!< allowing the removal of >!5 from r8c6!< leaving a >!47 pair in c6!<.
There's a W-wing in R6C5+R9C4 linked by Row 3. R6C5 and R9C4 can't both be 8 as then there is no room for 8 in Row 3, so any cell that sees both cells can't be 9, eliminating 9 from R9C5.
There's a 2-string kite on 1 using Row 6 and Column 6. R6C5 and R4C6 can't both be 1, so at least one of R3C6 or R6C7 must be 1. This eliminates 1 from any cell that sees both these cells, in this case R3C7. From here on you get a whole lot of singles.
Check the explanation in the hodoku docs about hidden rectangles. If only 1 cell (R8C8) of a UR doesn't have extra candidates, then look at the cell opposite it in the UR (R5C9). If only one of that cell's UR candidates (7 & 8) has matching candidates (in this case the 7) in related houses (C9, R5, Box 6) outside of the other cells of the UR, then remove it (leaves 1 & 8 as in the example).
I haven't seen Andrew Stuart's solver, but I use the above technique regularly.
*Edit: forgot the why. If R5C9 were 7, then r8/c8 would be 7, and the other 2 cells of the rectangle would be 8. that would allow 2 solutions, making this an invalid sudoku. *Edit 2 - readability.
Once that all looks good, take a look at this. The idea of a swordfish is the exact same idea as an x-wing but one higher level of abstraction. An x-wing can kinda be summarized like this: there is a candidate in a set of rows or columns where considering the candidate as already existing in those rows or columns will exclude it as a possibility from cells outside those rows or columns no matter where you consider it. Essentially, you have a set of possibilities and every one of them results in a certain candidate being excluded from some cells. The definition is exactly the same for a swordfish, however the number of cells is 2 or 3 instead of 2. In a swordfish, rather than finding 2x2 rectangles, you need three rows or columns that have that candidate three or less times to form a grid (it won't always have to be a grid but it's easier to understand it this way). The 4s that are in the green circles form a swordfish. When you consider the possibilities of a swordfish's cells you end up with either a direct elimination (like the first scenario of the x-wing) or you'll end up with an x-wing (or one extra step to direct elimination). In the image, here's the way it plays out, I'll use row 2's possibilities for 4 and I'll show you how r6c3 can't be 4 (you can apply the same process to any of the cells with a 4 in a red circle). In row 2, the 4 can be in column 2, column 3, or column 5. If it's column 2, then the 4s that are still possible in row 4 and row 7 form an x-wing that excludes 4 from r6c3. If it's in column 3 then it directly excludes it from r6c3. If it's column 5 then r7c4 can't be 4 so the only possible 4 in row 7 is in column 3 so 4 can't be in r6c3. No matter where the 4 will be in row 2, it is impossible for there to be a 4 in r6c3 or any other cell that crosses the swordfish.
Here's a web page I created with a boat load of Sudoku links on all kinds of topics:
I really like these wiki pages from Hodoku:
I see a few things:
- Skyscraper on candidate 5 in R3C58 and R5C68 eliminates 5 from R12C6 and R4C5.
- Skyscraper on candidate 6 in R18C8 and R28C4 eliminates candidate 6 from R1C6 and R2C79.
- Y-Wing on candidate 4 in R3C4, R5C4, R6C6 eliminates 4 from R4C4.
X-Wings are part of an interesting class of puzzle-solving techniques known as "fish":
http://hodoku.sourceforge.net/en/tech_fishb.php
The one you really want to learn at some point is Skyscrapers. Skyscrapers are very similar to X-Wings with a slightly different configuration. But once you start seeing Skyscrapers you see them everywhere all the time.
You're kind of doing a w-wing, but somewhat in reverse. R3C4 and R9C6 together see all 4s in Box 5, so they can't both be 4. So any cell that sees both cells can't be a 7, eliminating 7 from R13C6 and R9C4, leaving R3C4 and R9C6 as the only 7s in Box 2 and Box 8.
There's an XYZ-Wing in block 5. Here is a tutorial on XYZ-Wings:
https://www.youtube.com/watch?v=IN71m6sXhqU
If you are still having troubles reply to this post.
Then after the XYZ-Wing there's an AIC. Here's doc on AICs:
The reason the UR doesn't work is that you have four 3 x 3 blocks. For a UR to be value it has to only exist in 2 rows, 2 columns, and 2 3 x 3 blocks. I call this the 2 x 2 x 2 rule which is reviewed in this video:
https://www.youtube.com/watch?v=RQmgJryCa-c&pp=sAQA
Also, here is text from Hodoku's wiki page: "A "Unique Rectangle" (UR) consists of four cells that occupy exactly two rows, two columns, and two boxes. "
Actually, I have another one. There's something called the x-chain (see here - http://hodoku.sourceforge.net/en/tech_chains.php )
Look at this image I've posted. What you do is you start with a number and you find connections to another number but you have to do it like this:
-starts with a strong link (only two in the row, column, or house)
-then you can do weak or strong link
-then a strong link
-another weak or strong link
-a final strong link
Then if the number that you landed on (8 for example) sees another 8, and that 8 also sees the original 8 you started from, you can eliminate it.
So when I'm doing this I start with a number and just go "strong, weak, strong, weak, strong--" and then see if it sees the same number as where I started. And by "sees" i mean if it is the same row, column, or house. You can see that in the image I posted the 8 that we started with is in the same column and the 8 we ended with is in the same row.
Thanks!
I was referring to the "Chain" section. The app doesn't (yet) know the special terms for some of the chains. The skyscrapers will be the chains of length 3 where the green lines are parallel :). So at that length, chains 2 and 6.
You can read about what an AIC is, how type 1 and 2 work, and also continuous nice loops at the Hodoku website here. There are probably other sources, too.
I keep slowly adding more explanations to the app. Maybe one day people won't need to ask anymore :). Until then, thank you for asking!
I agree with you assessments. :)
This is a grouped continuous nice loop. All of your weak links do convert to strong ones, but that doesn't end up giving you any eliminations in this case.
I can confirm what xemnosyst is saying. UR's are dictated by the constellations of starting givens as well as the values we pick for cells. So as long was one of the four you have mentioned is a given, then the UR situation will most likely not exist.
Here is really good wiki page on the 10 most common uniqueness techniques:
http://hodoku.sourceforge.net/en/tech_ur.php
Notice how the grids do not show any of the four cells having a given. I think having the given prevents the ambiguity from ever happening.
You kind of used a Skyscraper. It's at least the simplest way to represent the elimination. It's actually easier to look at it from Column 7 as working from Column 8 just adds an extra (unnecessary) step to it.
Row 1 and Row 6 both have two options for 6. As R1C7 and R6C7 can't both be 6 at least one of R1C6 and R6C5 must be a 6. Thus any cell that sees both R1C6 and R6C5 can't be a 6, eliminating 6 from R4C6.
I made an app to find strategies you could use when you're stuck. You can see this game in it here, and check out the help menu.
There are a couple unique rectangles to find on this board! Not the simple ones (type 1), but harder ones to spot. The other kind I look for are hidden rectangles, which I learned about from here.
There are also, of course, chains. There are (almost) always chains. So if you get good at those, I bet you'll get past those diabolical puzzles.
When I started learning a lot of these techniques, I used Hodoku. Shortly after I downloaded Andoku from the Google Play Store and started solving puzzles in that. It has a lot of really helpful tutorials for techniques. Those techniques (hints) available while solving scale to the difficulty, e.g. you won't learn about Jellyfish while solving Easy puzzles. Andoku also has the ability to compute candidates for you at the harder levels and solve singles as well. It's glorious.
I'm sure you can find docs. I use the hodoku app -> help is here. Below is a chain that works on this sudoku. It's a bit longer than average, so it may not be the right choice for learning about XY chains
if r4c5 = 9, then r1c5 = 7
if r1c5 = 7, then r1c6 = 8
r3c6 = 5
r3c4 = 2
r7c4 = 7
r7c6 = 4
r7c3 = 2
r1c3 = 4
r1c1 = 2
else
r4c5 = 2
r4c1 can see both ends of the chain so it can't be 2
The more I look at it maybe there's some easier technique to learn.
Naked Subsets and Hidden Subsets would be the first place to start.
In my opinion in general this is a technique you're better off familiarising yourself with in puzzles below expert level as they're usually not enough to get through many expert level puzzles. This puzzle does have a bunch of them at this stage of the puzzle, but I haven't looked further through this puzzle to see how far you get with them.
>I can’t figure out why the c9 eight eliminates the r4 eight and not vice versa. I hope that makes sense.
They do eliminate each other. The pattern you've identified can be seen as two separate x-wings depending on which one (green or red) is assigned as the fin. If red is the fin, green can be eliminated. Same is true when the colors are reversed.
Interestingly the 8s in rows 2&6 form two sashimi x-wings, with either r2c7 or green as the fin. This is also commonly known as a skyscraper.
There’s a w-wing of >!4/5 in r3c1 & r5c8!< connected via the strong link of >!4s in row 6!< that removes >!5 from r3c8!<.
Basically there can only be >!two 4s in row 6 (r6c1 or r6c8), whichever it is you will always end up with a 5 in either r3c1 or r5c8. So you can remove 5 from any squares they both see (r3c8). !<
You can read more about the technique here.
Oh I think I recognize it now. It's a disconcinuous nice loop, but where you have to use a group link for those two 7s in the top right. :)
You threw me for a loop (no pun intended) because I always start with considering that something is NOT a particular number, and you started with considering that it IS a 7. But I can see it as that discontinuous loop starting with "r8c8 is not 9".
I think that a good step would be to scan the puzzle for singles, where can a 1 go in box 1 and 5 what about a 6 in ROW 5? Some of your candidates are invalid as well, I noticed the 1 candidate in r7c4 (row 7, column 4). Once the invalid candidate is fixed you can solve all the 1's in the puzzle! I also recommend taking a look at the naked pair and naked triple in row 5 and row 1, these reveal a lot about the puzzle immediately!
Assuming all of your candidates are correct, I see a unique rectangle (type 1), which breaks the puzzle.
Candidates: >!4 and 9 in r2c1 and r2c2 and r4c1 and r4c2!<
Eliminations: >!4 and 9 from r4c1!<
I would NOT say it's the "wrong" way. I'm fine with it. I was just trying to understand how you figured it out. I think most of the chaining techniques are a bit of trial and error working down the puzzle with bivalues as starting points. I think the technique you used is called a "Forcing Chain":
The four corners of your rectangle are each located in different 3x3 boxes. So imagine an almost solved grid where only these four cells are remaining. It would definitely not be ambiguous, since in each of the boxes only a single number would be missing.
In order for such a combination of hinted cells to form a Unique Rectangle (the kind of deadly pattern you're looking for), they must also be confined to exactly two 3x3 boxes. Then even solving all other cells does not help you to determine the ordering of the two pairs and the Sudoku has multiple solutions.
In your specific case there is another, larger deadly pattern to avoid that you can use to solve the Sudoku: The Binary Universal Grave.
If you want to solve the puzzle without relying on uniqueness, look for a Y-Wing >!in boxes 1 and 4!<.
Hodoku is a great tool for practising with. It has a learning mode that lets you select which techniques you want to practise and it gives you partially solved puzzles where those techniques are immediately applicable. I also found that I rarely need to look outside of their technique descriptions to understand how they work.
Yeah that app suggestion is called bowmans bingo aka guessing and is a bad idea!
This website is a great place to learn beginning to advanced concepts.
http://hodoku.sourceforge.net/en/techniques.php
There are a lot of YouTube videos out there that explain them as well.
There is another way of looking at it. (since it causes contradiction)
If R3C3 is 8 then R3C2 is 9. Then R4C2 is 8. Also R6C2 will be 3. So the 9 in box4 can go only in R4C3. This forms 89 deadly combo.
As per hidden rectangle definition in http://hodoku.sourceforge.net/en/tech_ur.php, 8 can definitely be removed from R3C3.
If you put a 7 in either of R78C3, then the other cell will be an 8, and then you'll also have 7 & 8 in R78C6, which will form a 7-8-7-8 non-unique rectangle.
http://hodoku.sourceforge.net/en/tech_ur.php#u4 might help you understand it.
In rows 2 and 3 of column 3 (which are both in box 1), we don't know which cell will hold the 3 and which will hold the 7 - but we know for a fact that those two numbers will be in those two cells, so any other cell in that box (and in column 3, if there were any) that have a 3 or a 7 can have those 2 numbers erased as candidates.
A Remote Pair on four of the 2,8 pairs (starting at r6c3 --> r8c3 --> r8c7 --> r5c7) allow you to remove the 2 and 8 candidates from r5c1, leaving only one 8 in the row and the rest of the puzzle as singles.
Regardless of which number you pick in r6c3, r5c7 will always be the other, so any cell that sees both of those cells can have neither candidate.
There's a W-wing using the >!digit 8!< in >!Row 6!< and >!R3C9 and R5C5!<.
Full explanation: >!R3C9 and R5C5 together see all 8s in Row 6, so they can't both be 8. Any cell that sees both R3C9 and R5C5 therefore can't be a 5, eliminating 5 from R3C5.!<
The basic idea is to find two columns or rows where there are only two instances of one digit in each line. One digit in each line must be able to see each other. You can use the ends of the skyscraper to eliminate candidates both ends can see.
You can read about it here, let me know if you have any questions.
These are quotations taken from http://hodoku.sourceforge.net/en/tech_misc.php#sdc which is much better cited as source than screenshots. I have highlighted the screenshot sections on this hypothes.is annotation page.
That "can't be more than 2 cells" was dicta, and might be incorrect, I'd need to study the behavior of SDQ more to say. Basic can be three cells, it's explicit. The Hodoku explanation was not wrong, because of how they defined the "intersection cells," but confusing.
In that case, the Hoduku example, the annotation frame covers up the example being reviewed. To be able to see both at once, I opened up an additional window for the original page, and then reduced the two windows in size.
In column 3, there's a 14 pair (in rows 1 and 4...r14c3), so 1 and 4 can be removed from the rest of the column.
That leaves a 23 pair in r56c3. In box 5 (r456c456), the 2s are locked in to r56c5. Because of uniqueness, we cannot have a 3 in either of these places, which only leaves r6c6 for a 3 in that box.
It should fall apart from there.
​
Uniqueness: http://hodoku.sourceforge.net/en/tech_ur.php
BUG, as I understand it, is the end result when everything is solved except for a remaining “unique rectangle.” It means the puzzle has two valid solutions, which means the puzzle is poorly designed or isn’t a proper sudoku. http://hodoku.sourceforge.net/en/tech_ur.php
There's a Simple Colors link containing 3 single links. http://hodoku.sourceforge.net/en/tech_col.php
I typed it into Andoku to show it to you. Do not click if you don't want to be spoiled:
You might want to try Hodoku. Runs on a computer (I've run it on both Windows and MacOS). Has lots of bells and whistles, but most important it supports multiple difficulty levels, pencil marks, hints (vague, concrete, and walkthrough), reasonable docs for many techniques, and more.
As far as your current puzzle goes, since you have no pencil marks, I'm not sure how to explain a kite. I'll just say that r3c3 is an 8.
So, read about sashimi X-wings here, and here. A normal X-wing basically makes a 2x2 grid where all other candidates can be removed. In a finned or sashimi version, one of the corners of the 2x2 is missing.
Now, because the corner is missing, you can't use it normally. You can still make eliminations like it was a standard x-wing, but only in the block where the omission occurred. That's why it's called a sashimi; it's an X-wing with a piece hacked off. You still need one or more 5's in that row, but they can be in any arrangement.
So let's take the first sashimi. You have 5's in r5c5, r5c9, r8c4, r8c6, and r8c9. This is not an X-wing, because you're missing a corner in r8c5, right? Well, that's fine for a sashimi. Imagine a 5 was there, and you can make eliminations in block 8. The only removal in that case is the 5 is r7c5.
The same process works for the 8's in rows 3 and 8. Heck, the cells in row 8 are even the same, to make it easy. Put both of these together, and you've removed 5 and 8 from r7c5.
Also, consider that a sashimi X-wing is one of the easier advanced techniques. X-wings and their derivatives (kites, towers, mutants, etc.) are a cornerstone to advanced solving. The others are probably things like XY-wings, XYZ-wings, and XY-chains. To a certain extent, you'll need W-wings, BUG+1, X-cycles, and so on, but you get the idea.
I won't even mention ALS-XZs or AICs, because they're much more complicated and require a far deeper understanding to how puzzles operate. The techniques I mentioned should serve until you get to the hardest of puzzles, and then all bets are off.
I made a this picture that will hopefully explain a little better. We don't worry about the 8 in r5c8 because we can exclude the 4 and 7 from r5c7 leaving us with 8 as the only option in that cell. In case I still didn't do a good job explaining, the technique is called a triple and you can read more about it here
Techniques I recommend (easy enough to learn and generally the least annoying to look for):
Naked/locked sets
Swordfish
Skyscraper/siamese/sashimi fish
Remote pairs
Empty rectangles
2 string kite/turbot fish/small x-chains
Unique rectangles
BUG
W-wings
Xyz-wings
Xy-wings (sometimes annoying but important to know)
Good job! With harder Sudokus, you will need more advanced techniques to make progress. Refering to them by name is much more efficient than explaining their logic each time, since there are many of them and the descriptions can get lengthy. I really like the explanations on HoDoKu and SudokuWiki.
You cannot place BOTH 2 and 8 into the lower red box in the picture above as you would create a rectangle pattern of 2’s and 8’s and the puzzle would have more than one solution. This means that 7 must be placed inside the red box.
More about the technique here: http://hodoku.sourceforge.net/en/tech_ur.php
The sashimi swordfish is on 4s in columns 2, 3 and 6. It is almost a swordfish, except for two additional "fin" candidates (r7c2 and r9c2, orange highlighting) and some omissions (r2c6, r4c2, r8c2).
The logic of finned swordfish works regardless of omissions: Either one of the fins contains the 4, or the rest of the candidates in columns 2/3/6 form a swordfish in rows 2/4/8. In either case r8c1 cannot be a 4.
The second elimination in r7c4 follows directly from the first, because the 4 in row 8 is pushed into box 8. Alternatively you could also see it as part of the sashimi-fish logic: Either 4 is in the fins, which then form an X-Wing with column 8 and eliminate r7c4. Otherwise the swordfish has only one candidate left in row 8 because of the omissions, so r8c6 must be 4, which also eliminates r7c4.
You can make pencilmarks to see what numbers could possibly go where. And that is usually enough for beginners. I'm not going to give a big explanation in here so DM me if you need help. BUT, I recommend taking a look at this page:
Hodoku Sudoku Tactics
Pretty much every sudoku tactic there is, is explained on this page. I'd recommend to at least read the first 4 categories. Singles - naked subsets if you want to get a good grip on solving the average sudoku.
There are wayy more in-depth tactics, but start out simple.
In hard Sudoku puzzles, progress is measured better in candidate eliminations than in final digits entered. Sometimes to have to apply multiple techniques each eliminating one or two seemingly random candidates until the next digit reveals itself. But in this case the triple in column 8 also reveals a pointing pair of 7s in box 9, which leaves only one position for 7 in box 8.
After you've eliminated 3, 7 and 8 from r9c8 because of the naked 378 triple in column 8, the 3s in row 9 can only be in box 7. This means that r8c1 cannot be 3 (and therefore must be 2), otherwise there is no place left for 3s in the last row (Locked Candidates Type 2.
The area that has the hidden triple will also have a matching naked pair/triple/quad/etc.
If you look at the first hidden triple example at http://hodoku.sourceforge.net/en/tech_hidden.php notice how there is also a naked quad, and in the second example, there is a naked set of five.
So I find it easier to try and spot the naked sets, rather than the hidden ones.
Imagine if a 4 was in r4c7. The 6s would be forced into the adjacent corners of the r14c78 rectangle and a 4 would be forced into the opposite corner. This is a non unique arrangement since the 4s and 6s could be swapped to get another valid solution. So to avoid that the 4 cannot be in r4c7. Same goes for the 4 in r4c8.
Your complaint ought to be about a foul word unless you are considering chickens, but I'd agree, it was uncalled for.
Your problem with this puzzle is that your generator isn't excluding unique rectangle cases. It's not explicitly in the rules for solving a sudoku, but if you are of the opinion that the puzzle should have a unique solution then the generator needs to exclude patterns such as this.
There are a whole bunch of variants you need to catch (e.g. A/B, B/C, A/C in 3 rows).
You could remove the ambiguity by ensuring one of the 4 locations is a "given".
See
http://hodoku.sourceforge.net/en/tech_ur.php
for starters.
Definitely. If you have any specific questions just ask here. I really like the descriptions of Sudoku techniques on SudokuWiki and HoDoKu. If you like videos, Sodoku Swami on YouTube has a very comprehensive Sudoku course that takes you through the well-known techniques from simple to complex and also explains the lingo.
The Y-Wing pattern involves three cells A, B and C in the grid that all have two candidates each, where A has candidates x/y, B has candidates x/z and C has candidates y/z. The “middle” cell A (the “pivot”) sees both B and C, but B and C don't see each other. In your case A, B and C are the cells I pointed out in the previous comment and x/y/z are 1/5/6. If you find such a configuration, then you know that no matter which of the candidates x or y is the correct one for the pivot cell A, it forces one of the pincer cells B or C to contain a z. So every cell that sees both pincer cells sees a z in both cases and therefore cannot ever be a z.
Good explanation. Note that you can see that last paragraph visualized in the same app, in this case as Net 3 in the help menu (link here).
I have not heard this called a "type 7" unique rectangle before. I know it as a "hidden rectangle", which is not-so-clearly indicated by the word "hidden" in that screenshot.
Hodoku has a similar grading system. And you can set the values for each puzzle-solving technique. In other words, Hodoku's rating system is fully configurable which is pretty cool.
http://hodoku.sourceforge.net/en/docs\_cre.php#rating
The 2nd puzzle is cool with three XY-Wings.
How is this one for gnarly:
4.2...9.7.......2.97.5..8......71.....78524.....34......8..7.46.6.......7.9...2.3
The candidate 2s and 6s in boxes 1/2 and 7/8 are the keys.
2-String Kites or Finned X-Wings on either/both will take you to the end.
Here's a really wiki page on uniqueness techniques and the different types:
http://hodoku.sourceforge.net/en/tech_ur.php
Here is the way I like to think about the logic:
Hodoku's explanation for fishes is very general and abstract which can make them every difficult to learn. They do have a page for simple fishes which is more geared towards beginners:
http://hodoku.sourceforge.net/en/tech_fishb.php
You can also check out sudokuwiki's page on swordfishes:
Yes you're right about the other 9 but it doesn't matter in this case because it only matters that the 9s in row row 7 and column 6 are limited to the UR cells. If you put a 4 in r7c6 then the adjacent cells in the UR would have to be 9s and the opposite cell would be a 4, leading to non uniqueness. Hidden rectangle.
> Is it ok to use more if they do not cross?
Definitions of fish say that you must have the same number of base and cover sets. However, don't let that stop you from swapping one of the cover sets for an extra one to make more eliminations if you want! Just consider them 2 different fish with the same base sets 🙂.
> Do I have to have two cells in a box in the same row or col in order to make a cover set with another base set? And two parallel rows or cols with more than two cover sets-?
The basic rules are:
- You have the same number of base & cover sets
- All the base candidates are contained in cover sets.
- No base set "overlaps" with another (i.e. cells they have in common are not allowed to have the candidate)
- No cover set "overlaps" with another
As you look for variants, like finned fish, there are FEWER rules, but eliminations become harder. Fins drop rule (2). You can also drop rules (3) and (4) to look for fish with "endo fins" and "cannibalism". However, as you delve deeper into variants the less useful they become, in my opinion, because they are harder to find and understand. It sounds like you're getting comfortable with finned fish, so maybe try to keep the other 3 rules for now?
My source of info about fish (which I consulted when writing these replies 🙂) is hodoku's docs. If you're willing to read slowly and think hard to understand the language used, it's a great resource.
It's a hidden rectangle because the 1's are limited to those 4 cells in an x-cross formation. If either of the 13 cells was a 3 then it would force a 1 into the 2 adjacent cells in the rectangle and a 3 into the opposite cell. So both of the 13 cells have to be 1s.
There's a Finned Xwing of 4s in r59c39, the fin is r7c3. The logic says, either the fish (Xwing) is true, or the fin is true.
If the fin cell didn't contain a 4, there'd be a true Xwing, and candidate 4 in r9c2 could be eliminated. Alternatively, if the fin is a 4, r9c2 can still not be a 4, and can therefore be eliminated.
I think you have a Sue de Coq in box 5, candidates 2,3,4,5. Should eliminate 2,4 from r6c5.
http://hodoku.sourceforge.net/en/tech_misc.php#sdc
I am just learning about these, so I could be wrong. Here's a great video tutorial on them:
W-Wing on the 1,9 pairs in r4c8 and r6c4, connected by a Conjugate Pair (Strong Link, both cannot be false) of 1s in r9. Any candidate 9 that sees both of the 1,9 pairs (the 9s in r4c5 and r6c7 can be removed) cannot be true, and the puzzle solves with a naked pair (1,9 in r6) and then singles from there.
How does this W-Wing work? Consider the 1s in r9. Both of them have a 1,9 partner cell, and one of them (the 1s in r9) has to be true. Whichever 1 in r9 is true will set its 1,9 partner = 9, so one of the 1,9 pairs will = 9. Any candidate 9 that sees both of the 1,9 pairs will see a 9 in the finished solution, and thus cannot be 9.
OK, here's a bone for you, using your 4,9 pair in r2c3.
Combine that cell with the 4,9 pair in r7c2 and the Strong Link (both cannot be false) pair of 9s in c5.
Note that each of the 9s in c5 see one of the 4,9 pairs. Any cell that sees both of the 4,9 pairs cannot be a 4.
This is called a W-Wing and can come in handy.
I agree there is something very close to a uniqueness problem with the cells R289 and R8C89. However, I don't see your logic for why you would eliminate a 7 from cell R8C8. The pattern with the 8 is almost exactly the same with cell R2C9. Obviously, there has to be a 4 in cell R8C9 or a 3 in cell R8C8 to avoid the UR problem. The 3 or 4 must be set in one of those two cells but it's not clear to me how you choose which 7 or 8 to remove.
Here's a list of UR techniques:
http://hodoku.sourceforge.net/en/tech_ur.php
Which number is the one you think it is?
This is a TYPE 3 UR that, unfortunately, doesn’t lead to any eliminations. All we can conclude is that in row four,
either of the 3 or 5 in column three is true
or any of the 2, 3 or 5 in column nine is true.
Yeah they're really fun. Many basic techniques like skyscrapers, w-wings and x/xy/xyz wings are pretty much AICs, it was awesome to run chains through them and figure out how each one worked.
But wait till you try passing AICs through ALS, that's where things get really interesting (and powerful). I think my favorite discovery was when I figured out how to form an AIC within a Sue de Coq, which is really two ALSes in disguise.
This wiki page is similar to what you are saying:
http://hodoku.sourceforge.net/en/tech_chains.php
I also like this page on this topic:
I don't see any Skyscrapers. Just a bunch of Hidden Singles everywhere.
I guess what you mean by "Fun Skyscraper" is just the title. I just assume I would be looking for a Skyscraper pattern in the puzzle as described here:
>Curious if this is something worth looking out for in the future...
Unique Rectangle Type 1 is very important. The above is a pretty advanced version of it (shown here for informational purposes only).
On this occasion, I did not check for X-Wing. I thought this particular pattern was noteworthy.
This is known as an XY-Wing. Yatopia is pointing out a very nice solution for this puzzle.
Here is a tutorial on XY-Wings:
https://www.youtube.com/watch?v=j0KyhIO8ICU
Here's a nice Wiki page on it:
1-3 Naked Pair row 5. Here is a list of puzzle-solving techniques with good commentary:
http://hodoku.sourceforge.net/en/tech_intro.php
Look at "Singles" and "Intersections" is a good place to start.
There is a Skyscraper on candidate >!3 in R48. Candidate 3 can be eliminated from R8C9 and R9C5!<. https://i.imgur.com/voziUWr.png
I think so. The 4's in column 5 form a set of locked candidates so the 5s must be removed. I believe this is a UR Type 4 as listed in this wiki page:
http://hodoku.sourceforge.net/en/tech_ur.php#ar1
Here's a tutorial on UR Type 4:
There are plenty of videos out there on the Yoututbes, just search "sudoku" and the technique's name.
You also might find this wiki page helpful on puzzle-solving techniques:
http://hodoku.sourceforge.net/en/tech_intro.php
For some reason it's currently not available. It does this from time to time. Try it in a few hours. But on this wiki page is a very good explaining of chaining techniques and uniqueness techniques. Try looking at it in a few hours.
Looks like you can uses the >!5s!< to make a finned x-wing with that rectangle. The fin is >!r1c8!<, and it eliminates >!r2c6!<. You can see it visualized as Net 4 in the help menu here.
This definitely has all the signs of a TYPE 3 UR. However, there are no obvious eliminations!
If there had been a cell in column seven with only 79 as potential candidates, there may have been an elimination or two. As it presently stands, it’s a Type 3 UR to nowhere.
Hodoku has a pretty clear explanation of Type 3 UR.
Generally, here is the algorithm I use for solving puzzles:
- Look for Naked Singles, if found, go to step 1.
- Look for Hidden Singles, if found, go to step 1.
- Look for Naked Pairs, if found, go to step 1.
- Look for Locked Candidates, if found, go to step 1.
- Look for Hidden Pairs, if found, go to step 1.
- Look for Naked Triples, if found, go to step 1.
- Look for X-Wings, if found, go to step 1.
At first glance it may seem a little robotic and mindless. But it isn't. What happens over time is as you get good and finding these patterns you develop new intuitions into the layout of the puzzle in general. These new intuitions will allow you to proceed in solving more difficult puzzles.
If you are not sure of what any of the seven puzzle-solving techniques are about listed above here is a pretty good set of wiki pages on them:
http://hodoku.sourceforge.net/en/tech_intro.php
The two techniques mention below in the other posts are "Hidden Singles."
If the two squares circled in blue are 3, then there's no place to put a 3 in box 2. If row 4 column 2 is a six, both the blue circled cells become three, and the puzzle is broken. Therefore you can remove the 6 from r4c2 and place a 3!
It's a "meta" trick that assumes a sudoku should only have one solution. if R9C6 was a 2, then R9C5 = 4, R1C5 = 2, and R1C6 = 4. (And vice versa if R9C6 was 4.) But in that case, you'd be able to just switch the 2s & 4s and the whole thing would still be valid (every row, column, and 3x3 box would still have 1-9), so the sudoku would have 2 solutions. But since we assume there should only be 1 solution, R9C6 can't be either 2 or 4.
This link can explain it better (see Unique Rectangle Type 1): http://hodoku.sourceforge.net/en/tech_ur.php
Once I can’t find pairs, triples, etc I move on the conjugate pairs. X wings are just the start. So many techniques using conjugate pairs. 2 string kites, skyscrapers, empty rectangles (which are all types of turbot fish), and all kinds of chaining and coloring.
I use this site as a reference a lot
> I see that and already eliminated any other candidates that were seen
The 1st UR lets you remove the 6 in E7. The 2nd removes the 6s in H1 and H3. This leaves a single 6 as candidate in G3.
Why? Sadly it was easier to learn to recognize URs to solve them than to understand exactly why. If you want a description of why URs work, check here
http://hodoku.sourceforge.net/en/tech_ur.php
or ask in this subreddit for people's favorite sudoku docs site.
There are several other eliminations possible using the 1 candidates. You should also look up Two String Kite. Any one of them will break this open.
I can provide more detail if needed.
This link provides good information about these strategies.
It's a bit of a "meta" trick. Basically, a proper sudoku is only supposed to have one solution. If R8C8 was a 2, then R7C8 = R8C4 = 7 and R7C4 = 2, but then even if you switch them around and let R8C8 = R7C4 = 7 and R7C8 = R8C4 = 2, it would have still been a valid solution (all rows, columns, and boxes would still have 1-9) - and this means that there would be two solutions. The same thought process would apply if R8C8 = 7 (the 7-2-7-2 would be interchangeable).
So under the assumption that there should be only one solution, R8C8 = 5.
I'm not very good at explaining, but you can read it here under the heading of Unique Rectangle Type 1: http://hodoku.sourceforge.net/en/tech_ur.php
Here is an article. It’s an alternating inference chain with an even number of digits and alternating strong and weak links, though strong links can replace weak links. At each end of the chain, one or both of the digits must be true so any digit that sees both ends cannot be true. Follow the logic through on the examples I have. If one end of the chain is true, then the other end cannot be (since these are all strong links).