How To Transform An Extreme Sudoku Into An Easy Puzzle!

[@nmorawski2210]

Next Episode: How to Transform an Easy Sudoku into an Extreme Puzzle.

[@patricgmuer2988]

nmorawski22 - Random Stuff let me guess, to change from easy to extreme you delete some numbers

[@sircheesethethird6179]

oh, i do that all the time. it’s called, *overthink the living daylights of yourself, the sudoku, and everyone within a ten mile radius.*

[@monolyth421]

It involves barbed wire and high voltages

[@Freeman4815]

Close your eyes

[@thereoncewasalimerickwriter]

I can already do that.

[@thanderhop1489]

At 17:13 when you say "huge number," of course, that number is just 6.

[@letMeSayThatInIrish]

Well, in general it's n! which is huge!

[@jlhidalgo]

I think the "huge number" he was referring to was the number of potential placements of nine sevens, nine eighths and nine nines on the 27 cells available for them, which applying simple combinatorics gives you exactly... a huge number, I guess, since I don't remember my combinatorics and I'm too lazy to look that up 😅!!

[@Gamesaucer]

Well yes, but the _valid_ potential placements number to 6. And he was talking about valid solutions only.

[@syndicatedpuzzles]

Embarrassed to see I'd left a display bug on my solver and it's now immortalised. The row highlight on a found strategy. All fixed now. It inspires me to add a permutation option on the solver. A very interesting technique well worth knowing about.

[@MyReligionIs2DoGood]

Love your solver, Mr. Stuart! Not sure if you already did, but could you make it possible to track back more steps than one, please? It doesn't let me go back more than one - or am I just too blind to find the option?

[@richardfarrer5616]

I like this technique, which seems obvious in retrospect. As I commented on the earlier video, I used colours for three cells (the bottom row in my case) and then used normal Sudoku rules on them. It's not possible to add notes to cells, admittedly, but that's not much of an issue with only three colours. The bonus is that once one of the existing 7,8,9 is coloured, you can go back and fill in the rest without having to relabel anything.

[@Lucasaraujo-um1pr]

i did that too!!!

[@Hichamnostik_YT]

nice idea!!!

[@teambellavsteamalice]

So cool, I took your idea and expanded upon it to make this trick helpful in many more situations: So you keep the 789 where they are, just work your way onwards with a yellow, green and blue label until any of them resolve into a 7, 8 or 9. If it doesn't help wipe them and start again with another set of three squares... OR ... add another set of three colors! Add orange, red and purple and continue with these temporary labels until any resolves into EITHER 7, 8 or 9 OR into yellow, green or blue!

[@DeborahJLeslie]

@@teambellavsteamalice I wish I could see the work in progress. Do you color the given or derived 7, 8, 9 in the central column? How do you begin with the colors? Is your technique a colorful way of bifurcating?

[@teambellavsteamalice]

@@DeborahJLeslie I think bifurcating is a proper way to describe it. It's like have unknown a, b and c variables that can be 7, 8 or 9. The given 7, 8 and 9 will only get a color once they match with one. I think it's equally fast as you only advance from the color side. In another sudoku, there may be more to work with on one of the numbers and you could branch from both sides until a number matches a color.

[@Qazqi]

Since you seem so curious, if you'd like to go down the rabbit hole in a non-sudoku setting, this is exactly an application of reduction (the term you actually used in the video) in computer science. One of the more famous applications is in NP-completeness, which requires a bunch of background, but is pretty interesting to get into. Suppose you have a map of countries bordering other countries. A common way to represent this in computer science is what's called a graph, where each country is a dot (vertex) and each border between countries is a line (edge) connecting those two dots. Suppose you want to colour the map with only three colours such that no countries bordering each other share the same colour. This is a well-known problem for graphs called colouring. For a graph, this means colouring each vertex such that no edge connects two vertices of the same colour. Rather than putting effort into colouring maps, researchers have put effort into colouring the more abstract graphs. Any solution that applies to the graph can be easily applied to the map. You could say we've reduced the problem to something more well-studied. As a more practical example, suppose you have 10 delivery cars and a predetermined set of deliveries where you know how long it will take to deliver to that address and return to HQ. You can have many different deliveries overlapping in various ways. How do you allocate cars most efficiently? How many cars would you need to do it perfectly before adding more cars becomes pointless? Both of these can be answered by making a graph where each delivery is a vertex and overlapping deliveries are connected with an edge. From there, you can colour the graph optimally, which is a very well-known problem with lots of research, and then apply the results back to your delivery service. The number of colours you need is the number of cars you need to make it perfect. Being efficient with 10 cars means colouring the graph as optimally as possible with 10 colours. We've reduced the delivery system to well-known graph problems. There are a *ton* of things that can be turned into graphs, making them an extremely useful abstraction for real-world problems. You can also turn sudokus into graphs where each cell is a vertex and each edge connects two cells in the same box, row, or column. Each number would be a colour. This comes remarkably close to the technique discussed here. The point is, you can reduce sudokus in general to problems in graph theory. You can go further the other way as well. There's a famous problem called the Boolean satisfiability problem (SAT). If you have a Boolean logic expression such as "(x AND NOT y) OR (z AND x)", the problem asks whether there is a set of true/false values for x, y, and z (or whatever other literals you might have) that makes the overall expression true. You don't have to know what that solution is, only whether it exists. It turns out that we know how to reduce this to a more specialized version called 3-SAT where you have a general structure "_ AND _ AND _ AND ..." and each _ contains 3 literals maximum and only OR and NOT. We know how to convert any expression to this form and 3-SAT tends to be easier to work with, so we can do work on 3-SAT and apply it to all expressions. Where NP-completeness comes in is that it turns out we can reduce that graph colouring problem from earlier (whether you can colour a graph with 3 colours, note: not giving an actual solution) to 3-SAT. We can also reduce 3-SAT to that graph colouring problem. That means any progress we make on either of these problems applies to the other one. What's more is that there is a whole slew of hard problems that are also in this set of problems that reduce to each other. If we figure one out really well, we can do the same for all of them. You might have heard of the P=NP problem, which has a $1 million prize. In short, to solve it, someone needs to either prove that any of these problems can be solved "fast" (for a precise definition of fast), in which case we can apply that to the others, or that any of them cannot be, in which case none of them can. That got long, but if it piques your interest, know that there's a whole bunch more and you know how to get started on looking for it.

[@ahnowrong6457]

how do I save a comment

[@Qazqi]

@@ahnowrong6457 Haha, you could always copy it to a text file. You can even bookmark a text file on your computer if you want through `file://` URIs, or put it in Pastebin, or an email to yourself, or whatever else works best for you.

[@DemonLilyYT]

Now you just need to make sure that SETH our lord and savior is real, protecting us from insane speed at solving them parametrized NP-complete problems.

[@johnharriman85]

One of my first professors in university was Kenneth Appel, he of the '4 colors suffice' fame! very interesting read. You may ruin the internet with such intelligence if you're not careful.

[@dickybannister5192]

think you're under selling this, and I'm not sure "reduction" covers it either. this is actually pretty shocking. a random selection usually assumes a bifurcation: i.e. you have NO method to decide which might be right. so you might as well randomly try one. but, if that fails, you have to "go back" and try the others till you get the right one. one other possible use is to crank probability theory to reduce the likelihood of getting the wrong answer. this doesnt guarantee the correct answer, but with successive trials makes it less likely in a large way. these Random Algorithms are useful because checking any solution is right is easy, so the trade off is productive. if, here, you define a strategy as a linear collection of tactics (i.e. the list of things the computer might try), you can find an optimal solution in terms of the "best strategy" for a given sudoku instance in terms of the number and complicatedness of the tactics used. note: a strategy MUST work, be finite and run in finite time. again, random selection is never usually optimal because it is the worst possible! you have to try everything. note: "finding the solution" and then just saying "do" the solution... isnt a strategy... you havent explained WHY it is right: you have to DESCRIBE how to find it. this isnt a "proof" because we cant say, for sure, that there is a shorter tactic we havent discovered yet, but this is, frankly, bonkers. it basically says that the best strategy involves choosing at random. this is the first time that (albeit with backtracking, ie. actually undoing some part of the solution you know to be correct) that i've come across wherein random selection can actually be shown to be part of the best known strategy.

[@PiperWyattP4]

I got it right up until all the 7-8-9. I am kinda proud of myself (and I found it rather easy) but then omg I looked and looked and could not figure it out 😅

[@cmonkey63]

In second year integral calculus, we joked about "assault with a deadly substitution". I waited 38 years to use that line in real life. :)

[@biaberg3448]

What a wonderful technique! It makes me happy to see - and it`s actually very simple. Like changing the 7-8-9 with x-y-z and then putting them back afterwards. Beautiful !

[@Blubb5000]

3:44 When a Sudoku champion says "What the heck is that" then every average person should just give up on this puzzle.

[@Goryus]

The label at the beginning (Extreme, Diabolical, etc.) shows the type of techniques the solver needed to complete the puzzle. The score is a cumulative running total, where every technique on the right is worth a different number of "points." So, Extreme-183 means it required some Extreme techniques to solve, but not very many of them. Extreme-800 would mean it not only required Extreme techniques, but it also required very, very many of them. It's theoretically possible for a puzzle to be "tough" 200+ if it's requires many, many applications of "tough" level techniques. Also, when the solver finds a bunch of naked singles or something in a single step, it only counts it as "one" technique, whereas if a naked single reveals another, it adds points for that twice. This puzzle "lost" a whole bunch of points compared to your average puzzle because of that, and because starting with a lot of givens meant it was shorter than average as well.

[@jlhidalgo]

I think this is a version of a trick I use some times, using pencil and paper, but mine is (maybe) a bit more applicable and a bit less "intrusive": whenever you have two or more digits of which there are only one or zero (there can only be one digit not present in the puzzle, of course), and you see some cells where some of these numbers should go, and it seems that solving those cells would help advance the puzzle, start using "a" and "b" as new numbers, or "a", "b" and "c" as on this case would be required, and follow the normal sudoku techniques with those additional "numbers", as if the single cell with each of those numbers was empty. For some time you'll have cells in the puzzle with small numbers like "23a" or "pairs of 7 and a", or things like that. But whether you get to "solve" one the original cells that you removed with "a", "b" or "c" (which are their only options, by the way) you know the equivalence between that letter and the original number on the cell, so you replace that letter everywhere and keep on with the sudoku normally. It's like solving the X on an equation, that's my mental image for this trick. Sorry that my explanation is awfully, yet I hope it's somehow understandable!

[@wizzolo]

yes it's basically the same trick, with pen and paper you can assign letters, or arbitrary symbols ( X, O + etc ) to a set of empty cells, and solve by those, then when you solve the cells with the initially given digits. you translate them back. It would be nice if the sudoku software allowed extra notation letters/symbols for this purpose, but i guess the use would be relegated to this particular trick, and very little else.

[@johnharriman85]

I wrote my long winded version of this above before I read your comment. Cheers.

[@Wouter10123]

After you highlight all the squares of one digit to colour them, you can also replace the number with one keypress as long as you keep them highlighted.

[@thorndeux]

Haha, yeah that drove me nuts as well :D

[@CodRampageHD]

9:15 is as far as I got on my own. Couldnt figure out how to solve 3/4 or 7/8/9

[@12tone]

Very cool! How did you determine which box was the most promising? That's where I got stuck: I kept trying different ones but couldn't figure out which box would lead to the best results.

[@nigelm5777]

I think you can get the same result mapping to any row, column or box with only three unknowns. There are several possibilities, try one.

[@CrimFerret]

It doesn't matter. You don't have to wait until the end. If I'm not mistaken, at any point you should be able to remove all instances of the original numbers you removed wherever they appear in the puzzle and place them again in a different column row or box without affecting the solution for all the other numbers in the puzzle. You'd still be solving one of the same subset of puzzles. As long as once you solve it, you do the final swap back, you should have the solution.

[@thereoncewasalimerickwriter]

It's the only box with three cell empty and no 7, 8 or 9.

[@tallness2001]

So how would one recognize when this is applicable? A puzzle where 2-4 digits are almost impossible to fill? Find a row, column or box where they exist and switch the values to empty boxes?

[@Gauteamus]

Great video as always, it is a joy to use your excellent website of interesting puzzles, and your series on new sudoku techniques is amazing! Thanks! With the well deserved praise out of the way, I got a smile out of this: "... a fixing out of the HUGE number of permutations ..." One handed counting: 1, 2, 3, 4, 5, huge, huge, huge, huge, ...

[@moshehim1000]

Hi, 1. This is essentially Mark's favorite method of bifurcation, in a more graphically palatable way, which is specifically tailored specifically for the situation of a plethora of most number with a dearth of a few others, with the benefit of telling you exactly which fork is the correct one with the benefit of fore-hindsight. 2. If you replace 7, 8 and 9 with stars, you'll get battle sudoku... 3. Which brings me the idea of, instead of confusing the issue of relabeling the numbers, to use symbols (say a star, an orange and a banana) or just letters (A, B and C) to potentially represent either of the 3 numbers (without deciding which), starting from another square, row or column by populating them randomly, and only once you populate the central column with them to accordingly ascribe a certain value to each symbol. This has the benefit of not only being far less confusing, preventing errors in the last stage of conversion and sparing the use of color-coding to anyone using pen and paper, but will also save you some work and time, because the moment you manage to populate 2 out of the 3 original positions with symbols or letters , you can decode their meaning, translate them into numbers in the cells you've filled with them, and continue solving with regular numbers as normally. 4. Which only begs the question of where to start. Now, you've mentioned the bottom-right square seemed like a suitable candidate. And were proven to be correct. But you've never explained what makes it - or any other - suitable!

[@JalebJay]

Can always just use A,B,C and you don't have to worry about people being picky. Just say A=8, B=9, C=7 after you solve it all.

[@shane8037]

Simon, I better never catch you knocking Mark's "guess and call it good" technique after this convoluted monstrosity...

[@stevelwall99]

Your solution is incorrect. The 9 and the 8 were givens, and yet you had them in different positions than the givens in your solution, as well as the 7 in the middle column being in a different position than it should have been. So in other words, you did NOT solve the puzzle as given but a different one.

[@arthurgigon2451]

This technique is fascinating ! Thanks for the demonstration

[@jaguh9069]

I'm new to Sudoku technique but seeing their names: 3D Medusa, Jellyfish, Death Blossom... What?

[@syndicatedpuzzles]

Where I know the origins - usually the person who discovered them coined the name, I try and record it. Otherwise it's 'evocative'. www.sudokuwiki.org/Strategy_Families

[@st.armanini9521]

This feels like a Kobayashi Maru workaround trick to me! :D

[@hugodiazroa]

To change the numbers at the end a way faster method is to delete a number first, then put the deleted number in in correct place and then keep doing this with the last swapped number. The last one will fill the blank spaces

[@waswerde1]

hey there :) i've been looking for a practical web-interface for solving sudokus. the problem is just that i would like to insert my own sudokus and solve them (no need to check if they're solved correctly tho). your website seems perfect for that but it seems like i can only solve those sudokus from your videos there. is there access to a blank version or a similar app? i would be glad to hear from you :)

[@Qazqi]

I'd love this, but there's a possibility it could open them up to DMCA takedown requests or other legal hassles if people can upload copyrighted puzzles. If this is a concern, a short-term solution of not being able to generate a link to the puzzle would be fantastic for me. Coupling this with a simple import/export system (e.g., "__2__1_4__51..." for this puzzle) would be great and might be a way around DMCA.

[@Aimlesswaves.]

@@Qazqi Are sudoku's really copyrightable? Seems like just an almost arbitrary numberset, of such small value.

[@Qazqi]

@@Aimlesswaves. The channel has demonstrated how sudokus can be a creative work. I can only imagine they would fall under regular copyright laws the same way as other creative work. In my country, that includes protection for work you don't explicitly copyright. Anyway, it's just a guess; absolutely don't trust a random KZbin comment on legal issues.

[@jonpaxman]

@@Qazqi it wouldn't need uploading, just a blank grid in which you can insert whatever sudoku you are working on.

[@Qazqi]

@@jonpaxman Sorry, I should've said hosting, as in being able to go to a URL with that puzzle like they do with the ones in the videos. In a way, you have to manually upload the puzzle by typing in the numbers, but I realize that's not a very conventional use of the word. If I can enter a puzzle and can't share a URL to it, I'm more than happy.

[@CoenBijpost]

I suppose maths minded people would find this technique more convincing if you just removed the numbers 7, 8 and 9 and put x, y and z in the bottom right box. Then at the end replace the x, y and z in the middle column with the corresponding original value and you’re done. Sort of solving for 3 variables using a sudoku. Pretty cool. It says a lot about the underlying maths of sudokus. You could possibly create an algebra of sudokus and have homeomorphic groups of solutions. I’m getting more and more into sudokus by the minute, cause of your channel :)

[@Czeckie]

mathematically minded people have no trouble understanding the symbol 9 as a symbol for the number 8 :)

[@richardfarrer5616]

You can do the same by replacing then with {purple, green, yellow} too. But yes, each puzzle is effectively a representative of 9! different puzzles for all the permutations of numbers.

[@CoenBijpost]

Czeckie yeah, agree. It does get a bit funnier if the 9 is a symbol for 8 AND 8 is a symbol for 9. Because in this case you’ve got an infinite loop in your representation. It’s just a bit of formalisation if you start working with letters to represent the replaced numbers, bit easier on the eyes as well, and no need for color representation during replacement.

[@dolf370]

@@Czeckie Sure, it just is easier to mess upp when you switch digits instead of having some other symbols which just makes it visually easier to keep track of what the heck you're doing.

[@ib9rt]

OK, I tried solving it before watching your video. I found the colouring in your software very useful. I used "superposition" and solved all the 789 squares simultaneously without resolving which was which. After filling in all the blank squares with different colours and no collisions it then became obvious which numbers had to go where (including the 3's and 4's. Cute puzzle. Here is the "unique colouring" solution: imgur.com/a/O6syt5K

[@theatog]

This is essentially the same technique he was introducing. I think of it this way: Instead of having 7,8,9, you just replace them with XYZ. Then if you consider X, Y, Z can be any of 789, the XYZ is your "superposition" equivalent.

[@thephysicistcuber175]

15:15 now that's a good use of the "i PeEl oFF tHe StiCkErS HahaAAaa" thing

[@leojf3828]

Me too. I LoL'D!

[@Coyotek4]

I like to just use colors to represent 7,8,9 … and eventually see which color goes with which number.

[@hugo154]

Thats how I solved it

[@coombscharlie]

That is really clever! (I might even be able to remember it.)

[@Hichamnostik_YT]

i use this technique all the time and i thought it was a bit of cheating

[@Reb0rnLP]

Solved in 17min by my new trick: "luck".

[@ericveneto1593]

Has he gotten LESS efficient as the channel has gone on?

[@JohnSmith-rf1tx]

Simon, a great addition to this video would have been to see if removing the 7-8-9 from the middle column resulted in the computer solver showing that there were 6 possible solutions to that emptied puzzle (i.e. the number of possible permutations of the 3 numbers). Fixing the order in the middle column would therefore be shown to be choosing 1 particular solution.

[@zilalapsa]

rofl. That "Extreme sudoku" is the 1st sudoku I managed to solve in under an hour :) 31:52, but I didn't solve using those complex techniques. Just got to pairs and checked both

[@chrisengland5523]

What an incredible idea - and so simple. But if I were solving this on paper, rather than on an app, instead of putting 7, 8 and 9 into the last box, I'd use A, B and C. Doing the final swap back to numbers would then be less confusing.

[@thezdbailey]

My God. I feel like my face is melting off after opening the arc...

[@JMPDev]

It’s easier to wrap your head around by using other standins such as the letters A, B, C, etc. then it becomes really obvious that you are dealing with temporary standing that should not be confused for a final solution.

[@JamesOrr81]

This is absolutely mind blowing. I mean, my brain can see that the ones you're relabeling in the middle row are... well, labels. you could label them A B and C for all it mattered. What I'm NOT getting is how you settle on the 789 in the middle as being re-labeled.

[@chessw1]

Isn't it because the 7 8 and 9 occur ONLY THERE, only in that one column? And isn't that what makes the technique rare, not often available for use? You have to get into a situation where certain numbers occur in a column, row, or box ONLY ONCE -- so you won't find many times when you can use the technique, but when it IS usable, it's very good.

[@admthrawnuru]

Basically it's algebra-like. You're solving for x, y, and z, but representing them by 7,8,9. I think the real trick would be recognizing when this is useful and which ones you can fill in to get an easier solution.

[@Barteks2x]

I'm not all that good at it, but I think a different more useful way to look at it, would be to give them non-numeric labels, like A, B and C, and solve with some arbitrary A, B and C placed. Then when you get to the point where you can prove that A, B or C would have to equal a specific number, you would actually replace them all. This might require some special support from the software to type it in, but that would be a less confusing way to do it. And that way could actually very easily generalize to situations where this way seemingly doesn't apply as is. And would be way easier to do on paper.

[@Paolo_De_Leva]

Actually, there is a simpler method❗ The initial version can be easily solved by coloring with three different colors the cells restricted to 7-8-9. After pencilmarking all possible *7-8-9 triples,* I started coloring them from box 9. I did not need to delete the given digits 7, 8, and 9. I did not need relabeling. And it was fun❗ I guess coloring is not easy to implement in a digital Sudoku solver, whereas for a human solver it is one of the simplest techniques. See also the comment by *Richard Farrer* below.

[@laatt1997]

While this technique is interesting I don't understand when the technique should be used. Please more information as to when the transformation is appropriate.

[@PaulaZF]

Laatt This puzzle had only one 7, one 8 and one 9 filled in on the puzzle. So a puzzle that has only one answer for three different digits. I hope this makes sense.

[@stumbling]

Coming back to this via the CTC book. Wow, Simon's presenting ability has really flourished since this time.

[@rbettsx]

Calling any particle physicists out there... Ive got a feeling this video would make a very nice illustration in an introductory lecture on guage symmetry?

[@gazzaheadon]

Very illuminating. Difficult to use on a pen and paper version.

[@Kyrelel]

Use letters instead of numbers, it makes the substitution easier at the end

[@thanderhop1489]

I'm curious to see if someone will ever make a puzzle that uses this technique even more drastically. For instance, could there be a sudoku with an extreme solver rating that starts with exactly 25 clues: all the 1s and two of each other number? In this case, you could possibly solve the puzzle with this technique by erasing all of the givens that are not 1s and replacing those 16 numbers wherever you like in the grid. Then you are working with 8! = 40320 equivalent puzzles at once (and the "label cells" pick out which one is the true puzzle) instead of 3! = 6 as in this video. Probably this is impossible since the sudoku would very likely be underdetermined with these stipulations (there are sudoku's with a unique solution with only 17 givens, but these are far from having the hint multiplicities I'm describing: for instance, being given all nine 1s doesn't restrict the number of solutions as much as 9 generic hints tend to). In that case, what about a puzzle with all 1s given, all 2s given, and two of each other number given for a total of 9+9+2*7 = 32 givens? Then this technique would let you erase 14 givens and place them as you like, and you would be working with 7! = 5040 equivalent puzzles at once. Probably one more level is the point where I'm very confidant such puzzles with unique solutions exist, but I doubt they would be as difficult for the solver as I want them to be. In that case, it's 9+9+9+2*6 = 39 given, almost half the grid, and this technique makes you work with 720 equivalent puzzles at once. I would be happy to see a puzzle setter create any of these. Another thing I've wondered about is the possibility of making puzzles where doing this technique twice in parallel, or doing it recursively is advantageous. For instance, erase your given "label" digits and place them elsewhere in two different ways, working on both versions at the same time to glean information from both. Finding the value of some cell in both puzzles lets you translate what one number is represented by in the other. Alternatively, use this technique, and suppose you find all of the 7s for instance as a result. Then you can do the technique again on your original set of labels but without the 7 label included.

[@Nihimon]

@10:35, when Simon starts the new technique, wouldn't it be simpler to notice that whatever appears in R7C7 must also appear in R9C3. Using purple to mark those two cells, I was able to immediately identify all the other purple cells in the grid (in order - R6C2, R5C4, R4C8, R1C9, R3C6, R2C1) and see that they must all be 8.

[@brucepayan2845]

Why not just assume a 7-8 placement in box 2 (forcing box 5 & 8 placements) and work to a potential contradiction or completion. Removes the relabeling hassle.

[@stefanholbek2449]

🤣🤣🤣🤣 This is complete BONKERS ... and ABSOLUTELY GENIOUS !!!! I am speechless ... gosh, wow ... and all that! 👍👍👍 Cheating? No. Using computers is cheating. This is just yet another technique! Thanx for your videos! I only solve the classical ones every now and then, and I always return to this channel for the latest and greatest. This one was hilarious!

[@orsonzedd]

I paused halfway through when I knew what the secret was and I added ABC to the lower right-hand box and solved from there, and then when I figured out what a b and c were I just that way and it was probably easier and less confusing

[@Lappy486]

Found this from the greatest hits book and wow. This is so fascinating. This is actually a really nice application of the math field of group theory. I remember in my class on it we were able to take two seemingly completely different things and show that, when you look at the structure of it and ignore the details, they’re actually the exact same thing. Or “isomorphic” if you prefer math jargon. So like in this case, individual numbers don’t matter. The structure of the puzzle is what matters. You could replace 1-9 with A-I or even 9 pictures of cats and it the solutions would all be identical, in a way. So why not just swap the numbers around?

[@glennmelven3414]

And what do you do if the location you relocated the 789s to prove to be just as hard? Do you just try again in a different location? Isn't that equivalent to bifurcation?

[@jdyerjdyer]

Possibly, but in placing the 7-8-9 in a new location, you can look at the puzzle to identify better placements over others. For example, where he placed them in the bottom right box, you could place them in several different configurations (6 to be exact) and some of those immediately give you a digit somewhere else, but possibly not all of them. So you would look for ways to quickly eliminate possibilities. It would be unlikely (but not impossible) for a puzzle with this setup to have all the possible placements lead to extreme difficulty. This is because even though those permutations are interchangeable, the positions in other locations are not. In other words, if you had a box with 5 open spots and you are swapping out only 3, you couldn't use this box as a placement because you could be placing the numbers where they can't go. Because of this, when you identify a spot that you can put them in arbitrarily (a box with only those digits left), then when placing them, they face that box of 5 digits and could eliminate some of the possible placements allowing you to progress, where as the original placement didn't face the right (or enough) open cells to help. Sorry if this is too long of an answer, but it is the easiest way I could explain it without going into math and permutation theory. Trust me, you don't want to go there unless you like wrapping your brain into knots. (That's a math pun. :) )

[@jdyerjdyer]

@Fester Blats I think his question is that just like in bifurcation you might not make any progress, i.e. you pick an equally (or harder) permutation that doesn't lead anywhere or at best only helps a little. In both cases you might have to back up, or try another permutation, but as you say, it is different in that any permutation you pick, once solved can be swapped out for the correct permutation.

[@jdyerjdyer]

One interesting thing to note in using this technique, and this might be of particular use in speed solving, is that any digits gained besides the ones you are permuting can be kept. So say you place the 7-8-9 and only gain placing a 3 and a 5, but become stuck again. You can permute the 7-8-9 into another location, but the 3 and the 5 are still validly placed, so you don't have to back them out. This is because the different permutations of the 7-8-9 don't affect the placement of the 3 and 5. Once identified, they are valid regardless of the permutation. So unlike bifurcation where if you go down the wrong rabbit hole you have to back everything out, using this technique, even if you pick a harder permutation, you might gain value. :)

[@glennmelven3414]

@@jdyerjdyer Yes, this is exactly what I had in mind.

[@joek511]

My solver rates this puzzle at 3750. I use HoDoKu..... Stickers on a rubix cube,,,LMFAO!! I did solve it in about 1 hour by chaining first off 7 9. starting in grid 7... the 7 failed so the rest was a walk

[@SloopyJohnG]

I found this technique less unnerving once I twigged I could use 'A', 'B' and 'C' to substitute the mystery digits (just as others have used colours). After that it's an 'algebra', and I found it much less scary. BUT let's not kid ourselves, this is a disguised bifurcation (or trifurcation) method, which is resolved at the end by swapping the numbers (or colours, or letters) back to fit their starting positions. So it's a bit naughty...

[@johnharriman85]

Unlike Ken Ken for example, and many variations on Sudoku, in classic Sudoku, the digits 1 - 9 are merely labels themselves and can be replaced by the letters 'A' through 'I' for example, or the nine members of the 2016 Chicago Cubs starting lineup, or 9 distinct colors (colours if you will). So, maybe it would be more intuitive to color the 3 empty squares in box 9 and to consider the 7, 8 9 in the grid as 'not yet colored' and solve those colors (knowing they represent a 7, 8 or 9). Then you will find that 'Blue' for example represents 9, red 8 and green 7 and no substitution is required in the end. Still, I love it.

[@JohnRandomness105]

I tried the initial version of the puzzle, and solved it. I bifurcated twice. Also, I did a loop (my term -- Y-wing perhaps?) eliminating one digit from a cell with three digits. I haven't yet listened to this new technique. I take it, the issue is resolving the 789s. Removing the 8 and 9 means that it's harder to place 5s and 3s. They were occupying potential a possible 5 and 3.

[@JPRobinso]

I too found this from the Greatest Hits book - and I used colouring on the 789 in column2 to find that one of those must be 9 and then went back and then coloured the 789 in box 9 differently, and this worked itself out. not something you can easily do on paper in the actual book...

[@terrencejohnson8731]

The technique will only work under exactly two conditions: 1. Three numbers appear only once in the grid; and 2. The three numbers are placed in a row/column/box that has only those three missing numbers. I think the narrator should have made that clear...kind a cool, but not very useful.🤷‍♂️

[@airwalkalman]

Why would you start this video with the included 9 and 8?!? By doing this you get the 5 xwing in rows 7 and 8 resolved which you can't get in the "easier" version.

[@stitch5604]

So what would be the criteria for a sudoku to justify, or even enable, the usage of this strategy? As a beginner, I was simply blown away by the logic of the strategy, it's simply ingenious. Though, I fail to really understand when it should be used, or if you could even, theoretically speaking, always use this strategy to solve a sudoku? Please excuse my noobiness, I'm just trying to understand, thanks!

[@martindolak6972]

Hmm, Instead of "removing the numbers 7 8 9 completely from the grid, you can just keep them there and randomly choose a box to use colors for 7 8 9, and continue to resolve using colors until you see which color belongs to which number ...

[@WtF6007]

honestly didn't find this puzzle that hard took me 25 min to solve 1-6 then it was just resolving the 789 took another 14 min 39 total and I didn't use any techniques that I would consider extreme I don't know maybe just me.

[@canebro1]

If you remove the initial 8 and 9, then add 7, 8, 9 to box 9, the solver can't grade it since there are multiple solutions. I tried a few other things (like partially solving the puzzle, but it stayed on easy.

[@HotelPapa100]

I often find that handcrafted puzzles betray the creator's thoughts by assigning special roles to the low digits. This can easily be masked, and thus another level of difficulty added, by reshuffling the digits.

[@topilinkala1594]

There's no magical in it. Mathematics uses the same thing all the time and it's called transformation. For example differential equations are hard to solve so use transformation to change them into algebraic equations and solve that (Not trivial if order is higher than 2 but always solvable) and reverse transform the symbols back to the original function. First time you see your teacher to do it to a non-trivial differential equation you think it as magic but it's just applying rules of transformation and reverse transformation. The theory behind transformations is a beatiful piece of mathematics itself but you don't need to study it to use the tools. And I liked your comparision to solving Rubik's cube by taking the stickers out and putting them back in right order because essentially that it is what relabeling anything is. But concider this: If I make a statement 1+1=2 it is valid in any language, but if I make statement like: one plus one is two, it's only valid in english. Which one is the realabeled version of the same statement. And that is why spoken language is not same as written language.

[@2000stevestu]

I was pondering the question of how many unique sudoku (basic) puzzles existed and the question arose was it a unique puzzle if two numbers were simply swapped. This technique seems an extension of the observation that the numbers are significant as positional identifiers rather than values.

[@matthewkaseman7457]

This feels like it's entirely a solution by bifurcation so I don't know if it's a reasonable technique for me!

[@DJMayjority]

am i the only one that finds this to be cheating? just like he said at 15:15 .. the setter of the puzzle wants it to be a hard puzzle, and you change the parameters of that puzzle by simply picking numbers yourself, and changing them back afterwards. I like the technique, and the elegance of it, it just seems wrong to me..

[@44yvo]

The solution is basically to guess the placement of the three difficult numbers 7,8 and 9. And test the outcome. If the first guess is not the right one, try a different placing of 7,8 or 9. Easy peasy. No need for all that mumbo jumbo.

[@topilinkala1594]

At about same situation as 9:33 I had a little bit more marked and noticed that there's only two possibilities on 8's at the bottom row. I selected the first eg. 8 in r9c3 and the whole puzzle collapsed. That might be called bifurgiation but I don't actually know how that differs from what relabeling is. I assumed that the 8 is false and I need then to relabel the 8's etc. but it actually was the correct one and I got lucky.

[@СергейСмирнов-и2ъ]

Hi, Simon. Thanks for your great job. Funny. After I started this video, I as usually clicked on the puzzle page to try it myself. When I finished the easy part, I used the technique I invented myself (or re-invented, most probably). I call it 'traffic light'. I assigned colors: red, yellow and green to digits 7, 8 & 9 and started coloring the grid. At some point I was able to disambiguate 3's and 4's and completed coloring. The last step was very easy: to assign each color to a particular digit by noticing which colors match the open 7, 8 and 9. The whole task took me just several minutes and I was very proud of myself ...just before I started to watch the video and realized that your method is absolutely the same, just executed in a different way. I still believe that coloring of the grid is easier and more stylish than putting arbitrary digits and I hope you use 'my' method in a future video. Sorry for my English - I am not a native English speaker.

[@craftypam9992]

No need to apologize for your English - I don't think I could have expressed myself that clearly!

[@Feds_the_Freds]

I solved it, please give it a go aswell, it's not that hard, you can do normal sudoku techniques up to 3:10 then look at the 4th row, there can only be a 7, 8, 9 color in the first of these fields, with normal sudoku technique (with an x-wing), you can find out, that in the lowest row, the second to last field has to have the same color (number) and in the top row also the second to last then you can continue in the 4th row, because you foud out that the second to last in the lowest row has to be a different number, you can color in the last number in that row and so on, I hope you get what I mean. EDIT: Ok, you can also use the bottom right quadrant first like you, but what I'm saying is that you don't need to think about taking any numbers away

[@richardmead9785]

When you showed the solver, it would have been informative, if not instructive, to leave the drawings on the screen long enough for the viewer to appreciate their complexity. You kept moving the mouse and clicking its button so rapidly that it was not possible to really appreciate the complex operations the solver was finding and displaying. It is often difficult to understand the purpose of your video discussions.

[@deucedeuce1572]

Can only imagine how AI would solve extremely hard sudoku puzzles.

[@grenvillephillips6998]

I would imagine that Edward de Bono would have enjoyed that solution.

[@MeriaDuck]

If I start off with those two left out from the start, my computer solver ends up with a three in the top-left corner losing it's religion. So that variant (number eleven in the Greatest Hits book) has more solutions than just a permutation of the seven, eight and nine.

[@MG-XYZ]

Messy if you try this technique on a piece of paper ! I think you could yield the same net effect just by guessing a cell. Not sure what the advantage is.

[@Lucasaraujo-um1pr]

instead of creating a new sudoku without the 7,8,9; i just used the colours in the sudoku and solved it with than, i think it is easier to solve this way. link with the image below: prnt.sc/pdgxee

[@solfeinberg437]

What a quirky idea. I put colors in box 9 and then was able to deduce everything one step at a time. It seems to depend on what box you start with. I had used another box and gotten nowhere.

[@jaredlamb7495]

how would you do this using the CTC software? i dont think its possible to remove the black given digits

[@douglashoward346]

column5 has a 9 row2 and an 8 row8 that are locked on the grid that are not on your puzzle making it impossible to do

[@alessandro2075]

Isn't this technique equivalent to brute forcing the puzzle?

[@Hpmanenz]

Didn't watch the video... just solved the sudoku, it took me 2 hours and 8 minutes haha

[@tombrouwer95]

I don't like this way, for me it's just trying something. I think it's not fair logic in my opnion.

[@patricepederson9211]

This was amazing. The logic helped me to figure out a way around a roadblock in my career. Thanks Simon!

[@jonpaxman]

It's a brilliant technique, but could I suggest a different way of implementing it, at least for paper solves? Call it "symbol substitution", and at the point where you can't make normal progress, you use the symbols a b and c to represent 7 8 and 9 (but not specifying which is which). You continue the solve in the normal way until you are able to place a b and c on top of the 7 8 and 9, and at this point the symbols "resolve" to the correct values of 7 8 and 9.

[@jonpaxman]

You could even do multiple variable substitutions for the same numbers. For example, put a b and c in the bottom right box, and x y and z in row four. Then keep solving until you can resolve each of them to a 7 8 or 9. So many possibilities!

[@ghosttwo2]

What a hassle. Better off just scrapping the damn thing and doing something else.

[@crazzzymick1]

when I try this puzzle I there is a 9 in row 2 and column 5 also a 8 in row 8 and column 5

[@joelyfish4216]

As a former Physics student, I must agree that 6 is an incredibly large number.

[@ianturnbull985]

Probably is if you're 1

[@Magetank4445]

Hello CTC! Ever since using your methods (the notations) I've had an easier time tracking down very hard Sudoku's. However, due to the high amount of small numbers and informations, I tend to need more time at the end of a Sudoku to fully complete it then if I didn't use your methods. This is because of the amount of numbers (pencil marks) distracting me from looking at what I can actually do. Any tips?

[@ratchet1freak]

practice, and don't be afraid to just delete your pencil marks when they start obscuring things.

[@ryokousha]

Instead of deleting the numbers 7,8,9 one could just use say A, B and C as variables in the bottom right box (as pairs are hard to notate with colors) and solve the puzzle with 1-9, A, B, C instead. As soon as we hit one of the remaining 7, 8, 9 with a variable (or exclude all but one possibility from a variable), that variable is resolved. Also I cannot see why this solving by variables technique should not be usable if the corresponding numbers occur more than once or not in the same house. This is a general method.

[@Adam-rt2ir]

Initial numbers don't have to be in the same box, but you need to make sure that you place them somewhere where you know they lie, but possibly in a switched order. Also, I am almost certain that this doesn't work if we remove more than one of the same initial number. The set of solutions will contain the original one, but might expand, you are risking that the sudoku might not be solvable afterwards.

[@ryokousha]

@@Adam-rt2ir There is no need to remove any numbers with this method. It is, in principle, applicable to an soduku at any stage. Just fill any house up with variables and carry on with the solve as if these variables were numbers. Of course, since each variable occurs exactly once, things have to be such that the variables (or at least one of them) "propagate" a bit through the grid during the next steps, so introducing variables to a sparsely populated grid will typically not be useful. Come to think of it, even assigning just one variable to any field can be useful, if that variable propagates. For another explanation, José Luis Hidalgo described the exact same method in the comments below.

[@ryokousha]

There are of course some pitfalls to avoid when using variables. For example a "7B pair" occuring in the solve could actually not be a pair if B can be 7. Similarly, a X-wing of A's could actually not be an X-wing, if candidate numbers for A are present in the rows or colums and so on. Edit: The X-wing example is of course nonsense, since candidate numbers in the rows or columns would just eliminate this candidate from the variable. After having played around a but with this technique, there are some quirks to get used to for sure. Sometimes fields with candidate numbers have to be treated as empty, sometimes they can be used to narrow the possibilities for the variables. Overall it can be said that the fewer candidates there are for a given variable, the fewer occurances of said candidate numbers there are in the rest of the grid and the more of all the other numbers are already filled in, the easier and more useful the method gets. i.e. in the Sudoku of the video, the method works very well not because 7,8,9 do only occur in one column, but because there are only 3 possibilities for each A, B and C, the numbers 7,8,9 occur only once outside the bottom right box and all of the numbers 1 to 6 are mostly filled in.

[@kirklorange]

Very cool, but I was able to solve it normally in just under 19 minutes.

[@Socrates...]

let 7 be x ; let 8 be y ; let 9 be z .....easier to plug in and change at the end...I think the method you are showing is genius

[@gunnarparment5050]

Or A, B and C. Can also be done by using pen and paper or a generic software like a spreadsheet. (MS Excel, Google Calc, etc)

[@poorest]

what "rediculous" technique!

[@jks-tp5pt]

Wasn't hard at all, solved in under 5 minutes looking at 789 relations in cells 4, 7, 8, 9 mainly. Maybe I just got lucky!

[@jks-tp5pt]

oops i meant boxes 4,7,8,9

[@mz.peachez321]

So without this software how would you solve the puzzle?

[@ssesf]

Got to 9:30 when I gave up and watched the video.

[@bernardherman9887]

Just use colors to begin with rather than renumbering

[@freddykabuffke461]

I managed to solve this on my own. I noticed that there was a single 9 in the upper row possibilities, and that solved it. Thanks for all the tips and tutorials

[@FazalFariz]

Could you please explain it? I'm still struggling to find a 'realistic' method of solution.

[@freddykabuffke461]

@@FazalFariz I just filled in what I could and came to a point, where i had three spots in the upper row wit 7/8, 7/9 and 7/8 Since two of the fields had to be a 7 or 8, the third one (or better, the second one) could only be a 9 Same with the right side i - 1 to i - 9, if you want. At one point theres only one square, where the three could go

[@freddykabuffke461]

@@FazalFariz So I tried again, and after one and a half hour, I'm still stuck at the remaining threes and fours as well as the sevens, eights and nines. Xwing and skyscrapers don't help here. So, I probably took a 7, 8 or 9 back then and pulled a "what if". Only thing I remember is, once I found out where it had to go, the rest was easy.