'Obviously Sudokus Only Use Integers'... Hold My Beer!!

[@codingentity]

No one has done more for the advancement of unique puzzle rules than Simon baiting the setters to not try and do something.

[@brianmcadam443]

Never underestimate the motivational power of "Don't you dare!"

[@MichaelDusoe]

And we are all the better for it!

[@windybeach2184]

He’s the person who declares that the hospital ward is quiet before they get a sudden rush of patients.

[@listey]

No disrespect to Simon but the setters contribute far more than anyone who only solves.

[@chameleon_yura]

Thank you for trying the puzzle! :) And sorry if it messed with someone's head. I had a feeling that somebody could send the puzzle to troll someone else, but that wasn't my intention, thus the disclaimer at the start of Logic Masters page. Anyway, it was a big joy to listen to the explanations of "one plus something something", and remembering the formula for quadratic equations from the school program without practicing it the whole life? Wow, that's astonishing!

[@AllenMS828]

I thought it was amusing that a puzzle by a setter named Chameleon benefited from using colors to place the unknown numbers. 😄 I wonder if there are any new color-based tricks you could come up with to help solve a puzzle. Or maybe do one of those multi-phase puzzles (like the Advent calendar one I mentioned in another comment), where some numbers keep their colors between phases, but others change. 🤔 Anyway, I think I've solved some of your other puzzles before, and they're always fun to figure out (or watch Simon and Mark figure out 😆). Keep up the great work!

[@davidrattner9]

This is just absolutely mesmerizing from you!!

[@andy02q]

When I read the rules, I thought that there were 18 different sets of 6 distinct numbers, but there is actually just 1 set with 18 different permutations. Even when I realized that there's going to be many solutions I did not know what I had understood wrong. I assume, that this part is clear to people who do more solves on Sudoku variants.

[@delta3244]

Thank you so much for setting this puzzle, Chameleon. It was a ton of fun. ...ah, I know this is bad form, but I cannot resist bragging a little bit. I did not take a single note - including placing large or small digits in the grid - until after I knew every digit in the puzzle. My time was ~one hour ten minutes (I didn't start my timer until ~10 +2/-5 minutes into the solve, so I don't know the exact time). So thank you twice, once for the fun puzzle, and once more for this fiendish test of my mental mathematics. I appreciate it for both reasons.

[@HelloEurope2]

The rules were extremly poorly explained in the rules section of the video. I do not wanna search the internet for better explanations when trying to follow a CtC solve.😕

[@joostvanrens]

Imagine being Sven and writing this beautiful Sudoku tool that had more options than a normal Sudoku player would need and then Chameleon comes along with an idea...

[@onijester56]

Imagine when Chameleon starts using doublers and halvers with this non-integer format.

[@bluerizlagirl]

@@onijester56 When are we going to see the first j-doku, where each cell contains a complex number m + jn, where m and n are integers between 1 and 9?

[@JonnyPowell]

@@bluerizlagirlI’d hope given the relatively pure nature of sudoku rather than j from the applied sciences we’d use the mathematical notation i for the imaginary unit

[@chameleon_yura]

@@bluerizlagirl , hopefully it will have some product arrows...

[@QwDragon]

@@bluerizlagirl integers? floats!

[@mycroft64089]

Simon mentioned the square root of -1. I sense a Sudoku that needs to be solved with complex numbers incoming.

[@annek3296]

It could actually work, since multiplying a number by i (the square root of -1) can be interpreted as rotating it 90 degrees counterclockwise. (So if you multiply it twice by i, you've rotated the point on the number line 180 degrees, which is the same as multiplying by -1.) And we've seen many sudokus involving rotational symmetry.

[@bryanskiteflying]

Yup. When Simon declared "I wouldn't have even thought about putting "i" in a puzzle", my first thought was "next time on CtC, we're going to see complex numbers."

[@Vedvart1]

One way to do this could be to use the 8th roots of unity plus 0, rather than the digits 1 to 9. Any clues involving multiplication could equivalently be solved if you think of them as modular addition clues for the digits 0-8 (since the two are identical Abelian Groups), but clues involving addition on the roots (say, for a killer cage sum) would *not* be translatable. Now that I think about it, killer cages could be quite cool for roots of unity, since there would be a number of paths you could take to get there, and a solver could think either geometrically or componentwise. That being said, this would be a terrible idea. Unless a truly great setter comes along, I think this would just end up being no more complex in terms of solve logic than many other sudokus, but much less accessible to lots of people.

[@Mitrasmit]

"How would I already guess that they don't have to be integers? Do people just see it?" - Simon, staring directly at a 4 cell sum adding to a maximum of 9.

[@puceronie]

One day, Simon will say "sudoku are bound by the laws of physics" in a video, and some constructor will rise to the challenge. It will be beautiful.

[@metleon]

"Place the numbers 1 through 9 on the klein bottle..."

[@abcde_5949]

44:04 Simon starts deducing something about the last digits of the decimal part being even, but little did he know that all those numbers will end up being irrational

[@GuilhermeCarvalhoComposer]

right? I was like "what do you _mean_ the last decimal digit?? what if yellow is like sqrt(2)?"

[@roadrunner9843]

If Simon says the last digit of an irrational number is even I believe him!

[@Qazqi]

I think starting out with both green and yellow confuses it a little bit, but of course still works. What you can do to simplify further is to make x the decimal part of yellow, i.e., 0 < x < 1 and yellow = 1+x. Then you can reduce the other colours using that: yellow: 1+x green: 1+2x blue: 5+3x orange: 8+6x Note that orange < 9, so x < 1/6, and that's the key constraint to tossing one of the solutions. From there, it's a bit more straightforward to set up the equation: {1,2} + (1+2x)(5+3x) = 8+6x ... 6x²+7x-3 + {1,2} = 0 Note my equation is different because x is the decimal part of yellow, not all of yellow. Using the quadratic formula and discarding the negative values of x, you get ~2.85/12 and ~1.55/12. Clearly, the former is more than 1/6, so it's also inadmissible, meaning 2 (red) goes into the equation. That leaves x≈0.13, but you don't need to solve for it and I just approximated the answers rather than pop it into a calculator. You could also avoid doing the square roots by saying √97 will come out >9 and that lands you at x>2/12, which is >1/6, so you can pick red without finishing the math. In case you're wondering how I approximated, there's a neat calculus trick for square roots: Start at the nearest whole square and its square root, then add or subtract the difference over 2 times the starting square root. For example, 97 is near 100, so √97 ≈ √100 - 3/(2√100), i.e., √97 ≈ 10 - 3/20 = 9.85 (actual: 9.8489). It works better the closer the starting point is, but it's pretty quick, reasonable to do mentally, and produces a surprisingly close answer (even for √73, which is right in the middle of 64 and 81, 9 - 4/9 still only has an error of about 0.011, so the rounding happens to be 8.55 instead of 8.54, still surprisingly good for how quick and easy that was). The trick is actually doing one iteration of Newton's method starting from a nearby perfect square (multiplying by 2 here comes from d/dx x² = 2x). And then of course the admissible answer comes to ~1.55/12, which is 0.13 minus a bit from times tables, so that's easy to round.

[@samthomas3255]

This is also how I approached it, but your explanation is way more coherent than the mess in my head!

[@nakorbluerider]

This definitely feels like the way to go about it, yeah. Once you name the decimal part as a variable and think of the values as being "m+nx" in that fashion the rest of the math becomes a lot more tractable I think. Kind of wild that this has a unique solution among all the reals... and wouldn't if it were just the math equations, it seems like it certainly requires the sudoku component to enable that unique solution.

[@JohnRandomness105]

Even though I broke the puzzle, it's nice to see someone confirm the sqrt(97) that I got.

[@oliverinspace9252]

I also managed to disambiguate R3C1 by considering fractional parts. However, I also found a way to do so without needing to solve any quadratic equations, specifically by using the secret on rows 5 and 6. (For brevity, I will denote values using the first letter of each colour as used in Simon's solve) Using the clued equations, the row sums for rows 5 and 6 can be reduced to O + 2B (row 5) and 2O + R (row 6). By the secret, the sum of values in any two sudoku units (row/column/box) sum to a constant. Furthermore, since R = 2, we have the relationship O + 2B = 2O + 2 => O = 2B - 2 Orange is the right side of the equation in R3. Hence, we have k+G*B = 2B - 2, where k is either 1 or 2. Grouping similar terms, we obtain B*(2-G) = 2+k Now, we consider the fractional parts. Say Y = 1+x, then it follows that G = 1+2x (row 2) and O = 8+6x (row 6). Since O ≤ 9, we need x ≤ 1/6. Note that 2-G = 1-2x ≥ 2/3, and since B > 5, 2+k = B*(2-G) > 10/3 > 3. Consequently, k must be 2 for the inequality to hold.

[@kylefullerton9645]

That's pretty much how I did it, too. Thanks for the trick for approximating the square roots, though! I got to 7 < sqrt(73 or 97) < 9, and I know sqrt(73) is between sqrt(64) and sqrt(81) so that's the right answer, but I will have to remember that approximation :)

[@skasperl]

15:44 Well. There is a concept called 'triangle numbers'. And if you add up four different digits, you will get at least 10 (1+2+3+4). And you can't have 10 in a Sudoku cell. Also: if you ever do a 9x9 sudoku you will notice that the triangle number for 9 is 45. Maybe that'll come in handy sometime. Not sure, this is first video I'm watching.

[@bobblebardsley]

Shh, that's a secret 😉

[@scottrothbaum8296]

Thanks for sharing a secret you only share with your very favorite people with Simon

[@bbgun061]

"you can't have 10 in a sudoku cell" ...yet!

[@scottrothbaum8296]

​@@bbgun061 What a knowledge bomb

[@馬善萄]

I don't know why but I just burst out laughing when I hear Simon said "there are only 36 cells, how hard can it be ?!" Just a small point, to see that "not only integers are included" from the start of the puzzle, you can use the triangular number of 4 (1+2+3+4) to tell this must be true.

[@esti369]

Me too. Especially when I then looked at the video length.

[@Ardalambdion]

Have the setters joined a club with the sole purpose to defeat Simon?

[@abj136]

Simon doesn’t publish fails, so I can only imagine it’s happened a few times.

[@skasperl]

I would love some patreon bonus videos of failed attempts.

[@chameleon_yura]

Why would anyone do that? It's a common knowledge that Simon is undefeatable.

[@Anne_Mahoney]

@@skasperl I thought that too, but then Mark posted a failed puzzle (maybe to fill in a day around the holidays, was it two years ago?) and it was just miserable to watch. So I'm just as glad the failures get suppressed.

[@Ανδρέας-ΓεώργιοςΣκίννερ]

Simon opening Notepad and Calculator certifies this video as the craziest video on the channel

[@easternbrown]

I think the rules could be better stated, to make clear that the same set of six numbers must be in every row, column and box.

[@ronjohnson6916]

Simon leaning back and muttering his equation at the ceiling had me laughing out loud. Congratulations Chameleon on being effectively evil. Wonderful puzzle.

[@longwaytotipperary]

Simon, would you possibly mention there’s no way to obtain world peace so that someone will prove there is? You are mighty inspiration to sudoku setters, I think you could use that gift universally! 😄.

[@annek3296]

I agree! Simon clearly has the heart & brain of a mathematician. I'd love to see him do some actual math research. I think with a couple years of study he'd be good to go, in some area like combinatorics, which hasn't already been explored to death.

[@michaelmatter1222]

Considering how much he likes fog-of-WAR puzzles that might be an odd move...

[@ArloLipof]

The ‘scintilla’ is called an (infinitesimal) increment ε in maths.

[@Hetpust]

the difference in torque between idling and full throttle on my nissan king cab

[@tattoocyborg1480]

Once I worked out the cascade of the relationships, I reformulated all my numbers in relation to a fraction variable "F", ie Red=1+F, Green=1+2F, Blue=5+3F, Orange=8+6F. Once I plugged that into the r3 formula and solved for F, and discovered that F=(√73 - 7)/12, I opted to stick with my clean algebraic expression instead of muddle with all those nasty decimals.

[@Anne_Mahoney]

Yes, and thank you for that! Simon's approximations were driving me crazy. Yeah, I know -- I started out in pure math, and *he* was in economics and then banking: the rules of the game are different. Still, the square root of 73 is NOT equal to 8.5, 8.54, or even 8.544003745. C'mon, dude! 😺

[@vlad1209palovic]

Exactly! Starting from ' 8 + 6F

[@ianoz1]

Only attempted this puzzle today and hadn't read comments until after adding my own. I did muddle with the nasty decimals (d'oh) but got there using the same method as you.

[@michaellautermilch9185]

Yes, thank you!!! This is much easier

[@Vendavalez]

Would it be possible to ask Sven to add an "Arbitrary typing" feature to Sudoku pad? More and more puzzles seem to need, or at least benefit from, such a feature. I can't imagine that it should be very difficult to implement in and of itsefl. Coming up with the UI-UX for it that looks good and feels good however? That may be a challenge.

[@michaellautermilch9185]

How's about floating text boxes, like Windows sticky notes feature, that can be dragged around or attached to a cell at a distance using a line?

[@wossaaaat]

Maybe the 'you probably guessed it' bit is because of rows 5 and 6, where you're adding 4 different numbers together and not reaching 10+?

[@terracottapie]

I think it's just because the rules emphasize that the entries have to be NUMBERS in capital letters (i.e., not digits)

[@Pathogenus]

Simon usually notices himself when he says unusual things, but "i don't like idea of green squared in there" certainly should be added to this list of quotes

[@robinblackwell8167]

I think by capitalizing the word "numbers" in the ruleset, the implication is that the numbers 1-9 are not typical sudoku digits, which makes people anxiously think about fractions. The point on real numbers reads like a joke, like "haha, it might include imaginary numbers but not complex numbers".

[@Jonathan_Corwin]

Student: Why do I need to learn quadratic equations? I'll never need to use them Teacher: Well one day you might make a living solving Sudokus on KZbin Student: ?!

[@Robert_H.]

1 < {Yellow, Green, Blue, Orange} ≤ 9 Row 5: Yellow + 2 + Green + 1 = Blue Row 6: Blue + Green + 1 + Yellow = Orange ⟹ Orange = 4 + 2 * (Yellow + Green) ≤ 9 ⟹ 2 < Yellow + Green ≤ 2.5 ⟹ 1 < {Yellow, Green} < 1.5 With Row 5: ⟹ 5 < Blue ≤ 5.5 With Row 6: ⟹ 8 < Orange ≤ 9 ---------------------------------------- Row 2: A + B = C * D With Sudoku: A * B = 2 * C CASES FOR EQUATION *A + B = {1, Yellow, Green, Orange} + Blue* ⟹ 6 < A + B = 2 * C ≤ 14.5 ⟹ 3 < C ≤ 7.25 There is no C ∈ {1, Yellow, Green, Orange}! There can't be Blue on the left side! *A + B = {1, Yellow, Green} + Orange* ⟹ 9 < A + B = 2 * C < 10.5 ⟹ 4.5 < C < 5.25 The number C has to be Blue, BUT: Orange = Blue + Green + 1 + Yellow + (2 - 2) = Blue + (Green + 1 + Yellow + 2) - 2 = 2 * Blue - 2 ⟹ A + B = 2 * C ⟹ {1, Yellow, Green} + Orange = 2 * Blue ⟹ {1, Yellow, Green} + 2 * Blue - 2 = 2 * Blue | - 2 * Blue ⟹ {1, Yellow, Green} - 2 = 0 | + 2 ⟹ {1, Yellow, Green} = 2 Contradiction, because 1 < {Yellow, Green} < 1.5. Therefore C can't be Blue and there can't be Orange on the left side! *A + B = Yellow + Green* ⟹ 2 < A + B = 2 * C ≤ 2.5 ⟹ 1 < C ≤ 1.25 There is no C ∈ {1, Blue, Orange}! It's not possible to have Yellow and Green on the left side! *A + B = 1 + {Yellow, Green}* ⟹ 2 < A + B = 2 * C ≤ 2.5 ⟹ 1 < C ≤ 1.25 The number C has to be {Yellow, Green} (the other one)! ---------------------------------------- 31:15 - 37:25 SUDOKU Row 2: 1 + Green = 2 * Yellow ⟹ Green = 2 * Yellow - 1 Row 5: Yellow + 2 + Green + 1 = Blue ⟹ Blue = 3 * Yellow + 2 Orange = 2 * Blue - 2 ⟹ Orange = 6 * Yellow + 2 Row 3: {1, 2} + Green * Blue = Orange ⟹ {1, 2} + (2Y - 1) * (3Y + 2) = 6Y + 2 ⟹ {1, 2} + 6Y² + 4Y - 3Y - 2 = 6Y + 2 ⟹ 6Y² - 5Y - {3,2} = 0 ⟹ Yellow = {1.2374, 1.1287} ⟹ Green = 2 * Yellow - 1 = {1.4748, 1.2574} ⟹ Blue = 3 * Yellow + 2 = {5.7122, 5.3861} ⟹ Orange = 6 * Yellow + 2 = {9.4244, 8.7722} Therefore Row 3 has to be 2 + Green * Blue = Orange, so that Orange < 9. ---------------------------------------- *Solution* Yellow = 1.1287 Green = 1.2574 Blue = 5.3861 Orange = 8.7722

[@annek3296]

Wow - what a labor of love on your part - thanks for spelling this out!

[@oreo3910]

Imagine there weren’t any nonintegers in the puzzle and it was entirely a trick

[@leojs5673]

never thought i would nope out of a 6x6 sudoku, but these mad constructors have done it 😂

[@briangarrett2427]

The first instruction is ambiguous. Does it mean each row etc must contain 6 of 9 numbers OR each row etc must contain the same 6 numbers?

[@tjtribble.]

The same 6 numbers are used in every row column and box. There’s infinite possibilities for each of the 6 numbers though, not just 9.

[@stevieinselby]

It's standard "mean mini" rules, ie there are exactly 6 different numbers used in the grid, each number goes once each in every row, every column and every 2×3 box. What is unusual is that they are not taken from the _digits_ 1 to 9, but from the real number line from 1 to 9.

[@AlonAltman]

Here are the formulas: orange = 9/2 + sqrt(73)/2, blue = 13/4 + sqrt(73)/4, orange = sqrt(73)/6 - 1/6, green = 5/12 + sqrt(73)/12

[@LandOfForeverSummer]

I don’t know if anyone has done the maths yet, but here is my working out: We start with the four equations that Simon finds for us: 1 + green = 2 x yellow ( Equation 1) [1 or 2] + (green x blue) = orange (2) 1 + 2 + green + yellow = blue (3) 1+ blue +green + yellow = orange (4) First, lets do some minor reworking to get our colors in terms of green: yellow = .5 + green/2 (1.1) Substituting in yellow into equation 3 blue = 3 + green + .5 + .5 x green = 3.5 + 1.5 x green (3.1) Substituting 1.1 and 3.1 into 4 orange = 1 + (3.5 + 1.5 x green) + green + (.5 + .5 x green) = 5 + 3 x green. (4.1) Lastly, because at this point we don’t know if equation has 1 or 2, lets break equation 2 into two equations 1 + (green x blue) = orange (2.1) 2 + (green x blue) = orange (2.2) Let’s deal with these one at a time, starting with equation 2.1 Let’s substitute all colors for their green equivalent using 3.1 and 4.1 1 + (green x blue) = orange (2.1) 1 + (green x (3.5 + 1.5 x green)) = 5 + 3 x green 1 + 3.5 x green + 1.5 x green^2 = 5 + 3x green Combine like terms by subtracting 1 and 3 x green from both sides to get 1.5 x green^2 + .5 x green = 4 We want green^2’s coefficient to be 1, so multiply all by 2/3 green^2 + 1/3 x green = 8/3 From here, we want to ‘complete the square’ (American name, don’t know how it differs across the world). To do this, note that (Y+#)^2 = Y^2 + 2x#xY + #^2, so if we take this pattern and apply it to the left half of the equation, we get Y = green, 2x#xY = 1/3 x green = 2 x (1/6) x green, and to finish the transformation, we need to add (1/6)^2 = 1/36 to both sides (adding the same thing to both sides doesn’t change the value of the unknowns). So we get: green^2 + 2x(1/6) x green + 1/36 = 8/3 + 1/36 We can condense that to (green + 1/6)^2 = 8/3 + 1/36 = 96/36 + 1/36 = 97/36 Square root everything green + 1/6 is roughly 1.641 green could be 1.475 if 2.1 is correct. We can do the same thing with equation 2.2 2 + (green x blue) = orange (2.1) 2 + (green x (3.5 + 1.5 x green)) = 5 + 3 x green 2 + 3.5 x green + 1.5 x green^2 = 5 + 3x green 1.5 x green^2 + .5 x green = 3 green^2 + 1/3 x green = 2 green^2 + 2x(1/6) x green + 1/36 = 2 + 1/36 (green + 1/6)^2 = 2 + 1/36 = 72/36 + 1/36 = 73/36 green + 1/6 is roughly 1.424 green could be 1.257 if 2.2 is correct. To see which of the two possibilities is correct, let’s plug in the values into equation 4.1 orange = 5 + 3x green orange = 5 + 3x1.475 = 9.475, which is too high, so 4.1 is wrong So we plug in the value from 4.2 into the three color expressions of green green = 1.257 orange = 5 + 3x green = 5 + 3x1.257 = 8.771 blue = 3.5 + 1.5 x green = 3.5 + 1.5 x 1.257 = 5.386 yellow = .5 + .5 x green = .5 + .5 x 1.257 = 1.129 It’s a bit easier than what Simon ended up doing, and hopefully, if you also worked it out, you came to the same answer. Thanks Simon for the excellent puzzle!

[@alberttatlock1541]

Sorry, I did not like that one.

[@simovihinen875]

The way in which you can guess the numbers don't have to be integers is the all caps word in the rules, NUMBERS, replacing the commonly used word "digits". That's probably what they meant.

[@victorfinberg8595]

simon, you had better get some chocolate cake for yourself, after getting through that

[@neil2796]

The maths term for a scintilla is infinitesimal, I think.

[@wrightn9]

Or episilon

[@carlscheuermann3684]

Usually denoted by "epsilon", I believe.

[@annek3296]

Yes - the conventional symbol for an "infinitesimal" is the greek letter "epsilon".

[@dcbuccieri]

This was by far my least favorite sudoku and CtC video I've ever seen.

[@TheFreeBro]

4 cells adding to a single cell would be the “obvious” indicator of non integers I think

[@LuProch]

This is absolutely the craziest sudoku I have seen in my life :-D I was expecting some nice numbers like 1.2 or so, not a proper real number with infinite decimals :-O Thumbs up to Chameleon and thumbs up to Simon, brilliant as always :-)

[@nathanmays7926]

Your setters were so preoccupied with whether or not they could, they didn’t stop to think if they should.

[@kellymills7346]

Great solve but something I would never attempt

[@HitchHitchHitch]

Ridiculous puzzle made even more ridiculous as Simon, who notoriously inputs sudoku with a keyboard, then inputs his calculator with his mouse 😅

[@nope.0.]

Speedcrunch is the calculator any keyboard lover should be using.

[@stella1913]

Great solve!

[@xChikyx]

yeah, didnt notice

[@simoonsingame3833]

For the final disambiguation (37:00) , I did something else which seems a lot easier than solving immediately the quadratic equation: Let say yellow = 1 + x, with 0

[@Madjerozkaz]

Brilliant!

[@Scum42]

I'm excited to see you play Hexcells, although honestly, I'm afraid it might actually be too easy for you. Some of the later puzzles can get tricky, but you'll probably fly through most of them!

[@laurasmith2173]

I love seeing Simon overjoyed by math working out correctly and being willing to solve these difficult puzzles.

[@janerobson2297]

This was so painful. But compelling viewing. I felt so bad for Simon. But also was so impressed that he kept going. Simon you are a superstar.

[@F1r1at]

It's pretty frustrating that puzzle doesn't include solutions. You don't need to include any quadratic equations, though. For 1 + g = 2*y you can say that decimal part of g equals 2 decimal parts of y. And decimal part of blue therefore is 3 decimal parts of yellow, decimal part of orange is 6 decimal parts of yellow. Also 6 decimal parts of yellow is less then 1, so yellow is less or equal then (1 + 1/6), green less/equal (1+1/3), blue less/equal (1.5) By that we have limited options for 9 >= green*yellow > 7 And if you check them a little bit, you'll find that minimal option for green will be about 1.29, and for blue it will be about 5.435, it will be slightly more then 7 (almost zero decimal part), but then yellow decimal part will be .87 When you check max options, it'd be 1.33 for green and 5.445 for blue, multiplied it'll be about 7.25, but then orange decimal part is almost 1. By that you can tell that blue*green is less then 7, therefore it needs 2 more, to be greater then 8.

[@spatulamahn]

1 + 2 + 3 + 4 = 10 :) So those equations wouldn't work with integers only! Straightforward as all get out :)

[@RicardoRibeiro1978]

Am I the only one that didn't really enjoy this puzzle? Maybe my problem is trying to think too hard of this as a sudoku while it is really much more just a maths problem. Anyway, Chameleon has lots of other great puzzles but this one just didn't really do it for me from the time I saw the rules.

[@Mujaki]

And by the title "MEANest" mini, I thought it would involve adding the digits together, and dividing by six. But it's definitely an above average setting, either way!

[@rubbish8522]

The link that supposedly includes a solution checker doesn't include a solution checker.

[@NIMPAK1]

The fact that the puzzle could've been even more evil by having the math only work in Base-6 and/or require complex numbers means we've only reached the tip of the iceberg in terms of puzzle design.

[@Tehom1]

"What's the mathematical term for a scintilla" - you usually use epsilon to represent a tiny nonzero quantity.

[@annek3296]

Actually, any positive non-zero quantity.

[@SvenCodes]

Hi Chameleon, how about a slider for each cell to input the value?! ;)

[@robertosecco983]

First time ever that I did not click "like" on a CTC sudoku. Sorry, but I did not enjoy it at all.

[@pwinn]

I suspect this video will not bring in new subscribers quite like the original miracle sudoku video did! Marvelous to watch Simon work it out, though.

[@StephenMarkTurner]

I believe that Simon said "why would I ever have to worry about the square root of minus one". Over to the setters...

[@highpath4776]

How did the first 2 2s get placed (I thought they could be in the other boxes and cannot work out why not

[@_-_-Sipita-_-_]

This sudoku is as understandable as rocket science, not even the rules explained by simon are clear.

[@lukasr5867]

Can you explain what parts of the rules and sudoku you don't understand? Because to me it seems like basic school maths. The only somewhat difficult part to me would be, that we cannot put the values into the grid properly, hence have to keep a lot of information in our heads. So I'd definitely not agree with your statement in this generic form (as in, apparently applying to everybody). Or is there a language barrier with these maths-specific terms?

[@_-_-Sipita-_-_]

@@lukasr5867 like what are the numbers related to their actual values, i thought 2 was going to become sqrt2, 3 could be 2.9 rounded, a scintilla or what was that exactly, there is absolutely no clear way i can see this like how is there something less than one what digit is in there???

[@lukasr5867]

@@_-_-Sipita-_-_ Ah, ok, I guess then the first part would be due to the notation difficulty that I also noticed. Though Simon did try to make it a bit clearer by having the 1 and 2 as big digits to show that they are proper integers, whereas the small centermark digits would be that plus some decimal places. As for the second part, I think that's just a misunderstanding. The "less than one" thing is never on it's own, it's just a way to refer to the decimal part of the non-integer numbers (like green and yellow were 1 + something between 0 and 1). And since we didn't know the decimal part, Simon often calculated e.g. minimum values by adding only the integer parts and saying that the actual value must be strictly larger than that (since we haven't added the decimal part yet and we know that it's not 0). Did that clear it up a bit?

[@SwordQuake2]

10:10 not true. It's not guaranteed that it will be there... You didn't understand the rules...

[@juliadefranco1130]

This is the most irrational sudoku puzzle I've ever seen. (Joking, of course.)

[@kf7137]

Great video. I wish you would have plugged the numbers into all the equations and checked the results.

[@inspiringsand123]

Let's Get Cracking: 07:46 What about this video's Top Tier Simarkisms?! The Secret: 1x (04:15) Numpty: 1x (58:03) And how about this video's Simarkisms?! Ah: 12x (13:22, 20:42, 23:25, 25:24, 29:01, 30:05, 32:57, 34:36, 57:50, 1:02:56, 1:02:56, 1:07:06) Sorry: 11x (11:01, 13:27, 20:14, 22:45, 27:26, 28:12, 35:16, 58:38, 59:32, 1:02:16, 1:09:22) Hang On: 9x (17:29, 33:01, 33:01, 33:01, 37:48, 56:21, 1:02:39, 1:03:26, 1:04:46) In Fact: 6x (19:48, 35:39, 37:48, 37:48, 1:11:08, 1:11:49) By Sudoku: 5x (18:39, 30:59, 33:18, 34:17, 34:59) Pencil Mark/mark: 5x (11:26, 12:13, 12:15, 14:25, 35:31) Cake!: 4x (04:59, 05:39, 05:42, 05:51) Wow: 3x (19:22, 25:38, 25:38) Have a Think: 3x (20:22, 21:49, 31:53) Weird: 3x (02:00, 29:17, 1:06:17) What on Earth: 2x (37:57, 50:02) Goodness: 2x (48:39, 1:01:25) Nonsense: 2x (55:53, 57:58) Clever: 2x (01:42, 33:53) Fascinating: 2x (1:07:11, 1:07:14) Unbelievable: 2x (1:08:47, 1:15:39) Obviously: 2x (10:58, 55:52) The Answer is: 1x (09:35) Naughty: 1x (18:52) I Have no Clue: 1x (1:17:48) Beautiful: 1x (1:07:11) Insane: 1x (01:29) Brilliant: 1x (1:17:29) Break the Puzzle: 1x (37:53) Incredible: 1x (01:50) Shouting: 1x (1:18:11) Magnificent: 1x (1:17:13) Surely: 1x (29:34) Nature: 1x (44:40) Unique: 1x (39:48) Most popular number(>9), digit and colour this video: Twelve, Twenty Five (9 mentions) One (190 mentions) Green (62 mentions) Antithesis Battles: High (12) - Low (1) Even (21) - Odd (1) Highest (2) - Lowest (0) Row (7) - Column (4) FAQ: Q1: You missed something! A1: That could very well be the case! Human speech can be hard to understand for computers like me! Point out the ones that I missed and maybe I'll learn! Q2: Can you do this for another channel? A2: I've been thinking about that and wrote some code to make that possible. Let me know which channel you think would be a good fit!

[@renedekker9806]

This is the first puzzle where I will have to watch the puzzle solve video to understand the rules. Even after watching the rules explanation, they are completely unclear to me. What does “numbers between 1-9” mean? Numbers consisting of digits 1-9? Numbers larger than or equal to 1.0 and smaller than or equal to 9.0? There is no rule about repeating digits, does that mean that is allowed?

[@renedekker9806]

After watching the beginning, here are better rules: Pick 6 decimal numbers with values between 1.0 and 9.0 (inclusive). These 6 numbers must occur once each in each row, column and box.

[@TheFreeBro]

@crackingthecryptic Scintilla -> infinitesimal?

[@Gonzalo_Garcia_]

49:27 for me... I guess?? I seem to have something that works. Anyways, this is a maths problem, not a sudoku puzzle. Edit: Yep, I got to the same solution as Simon. Can't believe I actually solved it.

[@zackplyler3226]

I think the reason I realized the noninteger thing right away was because i viewed the two bottom equations as essentially being arrow clues.

[@RichSmith77]

I'm not sure which astonished me more. That I managed to reach a solution, or that that solution turned out to be correct (or, at least, the same as Simon's). 😂 I just kept thinking "I've misunderstood something here". There's no way the solution is supposed to include surds involving √73. I must have messed it up. 😅

[@joelstevens5670]

You are certainly not alone there!! Felt like a minor attempt to send people crazy, especially the fact there were initially two options until one was ruled out. I love Chameleon’s innovation but this was relatively insane! 🙃😉

[@tristans623]

First of all, good work as always. A year ago i would not have attempted this. Watching Simon and Mark struggle and share their thought process has given me tools and resolve to give it a go myself. Thank you for that. I too was confused at the start because the instructions seemed to say that there were 6 of 9 digits in each row column and box, but not that they were the same 6 digits. it implied that each number may not appear in each box. I had to come to the comments to sort that out. i was able to solve by taking the equations, doing substitution and simplification. no quadratic formula. spoilers I found a possibly more elegant solution stated as fractions, which consisted of 1, 7/6, 8/6, 2, 33/6, 9. the finished colored grid was identical. i think if i tried to use a calculator, it would have rounded off and made for scintilla discrepancies between values which would make all the equations close but not exact.

[@terryr2990]

7:26 - "I wouldn't have even thought about putting i into the puzzle. Why do I have to worry about that? That's completely ludicrous!" Oh, Simon, you should know better than to make comments like that. Can't wait for the next sudoku containing imaginary numbers!

[@philb2972]

i'm imagining a product cage puzzle where the numbers in cages multiply to negative numbers

[@Spicy_Coconut]

At this rate we'll be getting 5d sudoku with multiverse time travel

[@chameleon_yura]

Like "The inconvenient square root" by filuta from December 2021? ;) (warning, it's super-hard)

[@umchoyka]

Noticing 4 numbers adding together

[@averygaron994]

The word you were looking for instead of scintilla is infinitesimal, also represented by epsilon. I was doing the math in my head like 1+(5+e)+(1+e)+(1+e) = 8+e

[@patrickv.3979]

Thank you for the birthday wish! Ended up getting a chocolate cake and an ice cream cake, to be extra festive! I also got snow for my birthday, so I'm enjoying the day to the fullest!

[@WeaselOnaStick]

I think it was supposed to be obvious that the numbers aren't all integers because of either row 5 or 6. Because with normal sudoku you'd have 1+2+3+4 sum to 1 digit number

[@QRebound]

The term you were looking for at around 13:20 was Epsilon!

[@prototypesoup1685]

(5 + sqrt[73]) / 12 is the most complicated number I've ever needed to write into a sudoku cell

[@JediJess1]

12:45 I believe the word you're looking for is "Infinitesimal"; an indefinitely small quantity, or a value approaching zero. lim(x→∞) 1/x = _ε_ or an infinitesimal. I like to think of it *naively* as |1/∞|. I am aware of some of the reasons why this is incorrect, and encourage any and all mathematicians and enthusiasts to discuss them in reply!

[@JohnWayne-hq1ns]

Some notes: 1. "Infinitesimal" is indeed a quantity, but it does not "approach zero." It is a rigorous definition that represents some degree of "smallness" beyond what the usual number system represents. Similarly, we have the concept of "infinity" in set theory, which represents a "largeness" beyond the system of finite numbers. We then define a countable infinity and many other uncountable infinities, each of which is rigorously defined. By direct symmetry, this means that you can define as many infinitesimals as you may define any infinite cardinals. 2. The limit of 1/x as x approaches infinity is simply zero. By definition, infinitesimal is not a "real" quantity, and it becomes less sense to do any comparisons as if we deal with the real numbers (for example, if you were to see "dx" on its own in a calculus expression without any other differential/integral operators, what would it even mean?). The quantity 1/x does gets infinitesimally closer and closer to 0 as x increases, so by definition, the limit is equal to zero. 3. Similar to note 1 above, |1/∞| isn't a well-defined expression because ∞ itself is not a rigorous quantity; it is a concept of "infinity," which is a collection of all infinite quantities. To define an infinitesimal ε, it is important to decide of which infinity it is a "dual"; conventionally, it is the dual of ω, the countable infinity. Perhaps it would be more apt to say ε = 1/ω. As a side note, perhaps an interesting one, in the Riemann sphere model, ∞ is a unique "number" and 1/∞ is *defined* to be 0.

[@JohnWayne-hq1ns]

Perhaps some clarification on the "approach zero" idea. For example, the number 0.99999.... (with 9 repeating indefinitely) is equal to 1. It does not approach 1 because a number, by definition, is a fixed value. It cannot be changing in any way. When you subtract 1 - 0.99999..., you get 0.00000... which is indeed equal to zero because all of its digits are 0 (any finite decimal place is a zero). The sequence 0.9, 0.99, 0.999, 0.9999, 0.99999, ... (adding one 9 to the end every time) does indeed approach 1.

[@joemucchiello4542]

epsilon is the smallest delta between two numbers. Not sure if that's where he was going though.

[@JoshGavroy]

"You guessed it" referrs to the two bottom formulas being 4 unique digits that can't sum to less than 10 using whole numbers.

[@AllenMS828]

I love math, but I had zero chance at solving this puzzle, because I misunderstood the assignment. The only time I've seen a "mean mini" before was in the Advent calendar puzzle before Christmas, where 16 3x3 boxes were individual mean mini puzzles. So I thought that in this case, each 2x3 box had a unique set of numbers from 1 to 9. Is a mean mini just a puzzle where the pool of available numbers/digits is greater than the size of each row, column, and box? Or is there more to it than that?

[@veggiet2009]

I think you'd guess the non- integerness because of the sum of 4 digits equalling less than 10

[@chitraagarwal8259]

A big shout out to whoever taught Simon high school algebra for ingraining the quadratic formula in his head!

[@annek3296]

He seems to have also had some calculus, since he knows about "epsilon", which he couldn't recall the name for, so he called in "scintilla" in the video.

[@uigrad]

Deduction at 29:00, but explained in an easier way: First: Notice that orange = r5c4 + r5c5 + blue + yellow. This is because the lower three values in box 5 are the upper three values in box 6. Second: Assume that r2c4 and r2c5 are blue and 2 (red) (as he is assuming here). So we get the expression of orange + ?? = blue + blue. Substitute sum from first point in for orange, and subtract blue from both sides. We get r5c4 + r5c5 + yellow + ?? = blue. Now, we also have an equation for blue (from row 5). Substitute that in: r5c4 + r5c5 + yellow + ?? = yellow + 2 + r5c4 + r5c5. Subtract out terms that appear on both sides, and get: ?? = 2. But, we already have 2 in box 1, so ?? cannot be 2. Therefore, blue cannot appear in the equation in row 2!

[@hlocne]

I did this by hand and got so bogged down by the possibility of r3c2 or r3c3 being 2 that I failed to use sudoku to see that it's clearly impossible... Maybe I'm turning into Simon? In any case a fantastic puzzle that I look forward to finishing the video for soon. Did everyone else get (sqrt(73)-7)/12 for the smallest non integer part?

[@victorfinberg8595]

i "solved" this without a calculator, algebraically. and no, i did NOT do this "in my head" here's how: 1) start by labelling cells in the bottom 2 rows ACDE (B,F are quickly replaced by the integers). after some effort, we get that A,C are between 1&2, D is between 5&6, E is between 8&9. 2) after some blundering about, i realized that i needed MORE equations, for the "residues" a,c,d,e (all between 0&1) the relevant equations have the same form as before (well, similar): a + c = d a + c + d = e 3) now, for the step at 29:00 in simon's solve, we try A + a + E + e = 2 ( D + d ) --> a + e = 1 + 2d --> a + 2d = 1 + 2d --> a = 1 but we know a < .5 , so this doesn't work 4) now you can do a bunch of pretty easy steps, and most of the grid gets filled in, algebraically. 5) now we have to solve a quadratic equation in the residues. (actually, you still have to pick one of two possible equations. i suppose there are 2 solutions) not likely !! i estimated a solution, given that the square of a residue is going to be small. 6) result a = 1/8 (approximately) (and maybe double that as well) added later: i stopped at this point, but there is a restriction from earlier, which is d < .5, and that eliminates the second potential solution.

[@chris5619]

"It must be correct, because how could it not be correct?" One of the best lines I've heard on here, and definitely true in this situation (and quite frankly, many puzzles in the past). However...it's not correct. B was incorrect in Simon's final submission as the calculator gave 5.386 and Simon entered 5.586 into the notepad. O was correct because he kept what was in the calculator for that calculation.

[@ventsiR]

12:45 an infinitesimal? :D

[@simonmoore8776]

13 Minutes in and I still have no idea what you are supposed to do. The rules seem very poorly defined and not as well explained as usual. 😞

[@stevieinselby]

I fully understand the rules. I just don't have any interest in doing complicated algebra on numbers that we can't enter into the grid even if we can work out what they are 🤷🏻‍♂. I could have got on board with an non-integer puzzle that had some simple rational numbers, as long as it was in software that allowed them to be entered, but this is horrific.

[@SSGranor]

Okay. Now who's going to figure out how to get transcendental numbers into a sudoku?

[@crazypantaloons]

Fun stuff! Without R3C1 (or by setting it to 0), the quadratic solution is a simple fraction (as might appear in a high school textbook). R3C1 is what ratchets the quadratic solution down to the sudoku range of 1-9. Brilliant idea setting R3C1 to both be ambiguous and necessary to resolve sudoku!

[@HunterJE]

I think they "you probably guessed it" is from the fact that the first rule makes a point of emphasizing "NUMBERS" where in normal sudoku rules you'd say "digits"

[@MichaelLamparty]

I was going to say it is probably because you can't total 4 sudoku digits and come up with a number smaller than 10.

[@HunterJE]

@@MichaelLamparty That'll do it too, hah! I was thinking in the vacuum of reading the rules before looking too closely at the grid...

[@JediJess1]

​@MichaelLamparty I saw both. It seems somewhat reasonable to deduce based on our predefined Sudoku Vocabulary, namely Digit vs Number vs Value (going off the recent 2×2 sudoku featured on the channel). Adding the observation that 4 distinct Sudoku digits add to more than one digit, I bet there was a chance Simon could've thought of it if the hint wasn't there.

[@SUPER7X]

This was obviously it.

[@maljamin]

Yeah if there isn't a way without notepad algebra (or the equivalent hacking about but in one's implausible head), such as through inequalities and ranges (as I'd hoped), this just seems to break my hopes for what a puzzle would be.

[@ettoremanca9538]

12:48 in maths we used to write the greek letter ε to indicate a small quantity… or, differentiating, lim (h->0) where h is x2-x1. Always useful to refresh these memories from school😄

[@annek3296]

The Greek letter is an "epsilon", which stood for an arbitrary very small number > 0. Lots of proofs in calculus/analysis involve the phrase "Letting epsilon go to 0, ...". I recall one professor whose turn-of-phrase for something essentially worthless was "of epsilon value".

[@RescueMichigan]

I once had a math teacher who would have insisted Simon's answer was wrong because he expressed the solution as rounded decimals instead of the exact expressions with the square roots of 73.

[@colej.banning2419]

Most of the decimals aren't even rounded; they're just truncated.

[@MariaVlasiou]

11:35 i don't understand. Help? Why is he deducting that the sun is all least 5? Can't it comprise 1+2+0.5+0.4=3.9? All are distinct reals.

[@Qazqi]

The rules specify a range of 1-9.

[@MariaVlasiou]

It said numbers and not range, so I took it as digits, ignoring the leading zero. Thanks.

[@Maxuras]

Let o, y, b and g be the fractional part of orange, yellow, blue and green. We get the axioms... 1.y + 1 = 2* 1.g 8.o = 1.g + 5.b + 1.y + 1 5.b = 1.g + 1.y + 3 8.o = (5.b * 1.g) + 2 ...from the Sodoku part. Let's make one corrolary: 5.b = 1.g + ((2* 1.g) - 1) + 3 (By inserting axiom 1 into 3) 3.b = 3* 1.g 0.b = 3* 0.g 8.o = 1.g + 5+ 0.b + 1.y + 1 = (5+ 0.b * 1.g) + 2 (combining axiom 2 and 4) 1 + 0.g + 5 + 3* 0.g + 2 + 2* 0.g - 1 + 1 = (5+ 3* 0.g * 1.g) + 2 (replacing 0.b through the corrolary) 1 + 6* 0.g = 3* 0.g + 3*(0.g)² 1 + 3* 0.g = 3*(0.g)² (0.g)² = 0.g + 1/3 And ... this makes no sense. Since 0 < 0.g < 1, (0.g)² must be smaller then 0.g. Yeah, I used to like math, then I learned more about it.

[@ianoz1]

This is the first puzzle on here that EVER frustrated me to the point of fury. Digits 1 & 2 were given (duh). Two other numbers NEED to be between 1 & 2, one having its decimal remainder half of the other (from equation in row 2). I called them 1+A & 1+2A. The two final numbers are 5+3A and 8+6A from equations in rows 5 & 6. That means that the "remainder" increment A can only be a MAXIMUM of 1/6 (0.1666666 etc). Now, plug those numbers into the row 3 equation to derive the increment A. First subtract two from both sides, so... (2A+1) x (3A+5) = 6A+6. Giving "numbers" 1, 1.12866697877, 1.25733395755, 2, 5.38600093632 & 8.77200187265. This was utterly brutal.

[@QwDragon]

You should've guess because NUMBERS is highlighted by caps and because 4 integers from 1 to 9 can't sum to 1-digit total in rows 5 and 6.

[@QwDragon]

Mathematical term seems to be epsilon?

[@QwDragon]

About 2*5: Row 5 (order doesn't matter): (1+X) + 2 + 1 + (1+Y) = 5+X+Y Row 6 (order doesn't matter): (5+X+Y) + (1+Y) + 1 + (1+X) = 8+2*(X+Y) = 2*(4+X+Y) So in row 2: 2*(5+X+Y) is always 2 less then 2*(4+X+Y), but can't use 2 there. Then 1 + (1+Y) = 2 * (1+X) => Y = 2*X

[@QwDragon]

After coloring: Row 2: Green = 1+2*X, yellow = 1+X Row 5: Blue = 5+X+2*X = 5+3*X Row 6: Orange = 8+6*X Row 3: (either red 2 or purple 1) + (1+2*X) * (5+3*X) = (8+6*X) (either red 2 or purple 1) + 6*X*X + 13*X + 5 = 8 + 6*X 6*X*X + 7*X + (either red 2 or purple 1) - 8 = 0 red case: 6*X*X + 7*X - 6 = 0, D = 49 + 4*6*6 = 193 => positive X = (-7 + sqrt(193)) / 12 < 1 purple case: 6*x*x + 7*x - 7 = 0, d = 49 + 4*6*7 = 217 => positive X = (-7 + sqrt(217)) / 12 < 1 see no problem with both cases... maybe I've done smth wrong...

[@QwDragon]

Uh, Simon's version is wrong at least at first... Waiting him to fix. 5 should be 1+g+y (all cells in g and y) not 3+g+y (decimal parts only in g and y) if 1+g=2*y. UPDATE: No that's me who is wrong. Missed 1 and 2 there.

[@QwDragon]

Saw it! My equation was not wrong: I got X (-7 + 14) / 2 = 1/2 So the red case is correct, yellow = 1 + (-7 + sqrt(193)) / 12 = (5 + sqrt(193)) / 12 And green = 1 + (-7 + sqrt(193)) / 6 = (sqrt(193) - 1) / 6 Hope solved it right...

[@DemonikYT]

You can solve without calculator. I ended up with equations a + 2 + 1 + b = c; b + c + 1 + a = d; x + b * c = d; 1 + b = 2 * a. Then I got a bunch of inequalities but most important one was 2 < a + b < 2.5. Using these equations I ended up with quadratic equation 3b^2 + b - 10 + 2x = 0 where x is 1 or 2. Solving this equation and discarding obviously negative roots I got b = (-1 + t) / 6, a = (5 + t) / 12 where t = sqrt(97) when x = 1 and t = sqrt(73) when x = 2. Because 81 < 97 < 100 and 64 < 73 < 81 so 9 < t < 10 when x = 1 and 8 < t < 9 when x = 2. Then calculate a + b replacing t with 9 and 8 and get 2.5 and 2.25. Because a + b will be larger than these values and a + b must be smaller than 2.5 means that x = 1 with t = sqrt(97) is wrong solution. So final solution I got was a = (5 + sqrt(73)) / 12, b = (-1 + sqrt(73)) / 6, c = (13 + sqrt(73)) / 4, d = (9 + sqrt(73)) / 2, I checked and it was correct solution.

[@calculatrguy]

Fun little analytical solution: If the two decimals are A and B and B=2A, and 8 + 2A + 2B is less than 9, then 6A is less than one. For the two values of the remaining cell, call it C=1 or 2, The two (positive) quadratic solutions are A=(-7+sqrt(97))/12 when C=1 and A=(-7+sqrt(73))/12 when C=2. Because (-7 + sqrt(81))/12 = 1/6 or 6A=1, we know that 6A is greater than one for C=1 and less than one for C=2, so C=2 is the solution.

[@airhorn1216]

I got: y = (√73 + 5) / 12 ~= 1.129 g = (√73 - 1) / 6 ~= 1.257 b = (√73 + 13) / 4 ~= 5.386 (different than Simon, don't really know why) o = (√73 + 9) / 2 ~= 8.772 I have to admit, having to analyse quadratic function and check its values on {1; 7/6} domain wasn't something I would expect in sudoku, yet here we are now I'm waiting for hypercube sudoku

[@GuilhermeCarvalhoComposer]

What a phenomenal puzzle. I spent waaaay too long staring at it wondering about fractions and how many of those fractions would have to add to an integer, and how you could multiply them so that... and then I started watching the solve and right at the start Simon goes "well since they're all greater than one..." and THAT's what unlocked it for me. :) What also helped me a lot was to consider the "slightly greater than 1" numbers as 1+a and 1+b, with a and b

[@anaphysik]

A more consistent way of talking about the yellow/green/blue/orange numbers is that yellow is 1+x, green is 1+2x (this is what 1+green = 2*yellow means), blue is 5+3x, and orange is 8+6x (for clarity: x is the little-bit-above-1 that yellow owns). Also: It would be a real pain to do the math without some notes (and of course knowing or looking up the quadratic formula, but it IS actually quite possible to determine which of 1-or-2 is correct without a calculator! Since orange can't be higher than 9, x can't be higher than 1/6th (or equivalently, yellow can't be higher than 1+(1/6)). The break point for that is funnily enough obtaining a 9 from the square-root portion of the quadratic formula, i.e. sqrt(81). So anything that asks you to sqrt a higher number than 81 is impossible within the confines of the rules, even if you cannot work out close-approximations of the actual irrational numbers involved without a calculator. (1 in r3c1 leads to too-high sqrt97; 2 in r3c1 leads to low-enough sqrt73) It's almost sad that the puzzle doesn't actually work out with orange equal to 9, with 'merely' rational numbers for yellow/green/blue (which would be 1-and-a-sixth, 1-and-two-sixths, and 5.5 respectively). Though it's also pretty amusing to have irrational numbers in a sudoku!

[@KyleBaran90]

Simon's right. We need a puzzle that involves finding the derivative or uses imaginary numbers

[@chitraagarwal8259]

Maybe something with integrals with killer cages/regions that need to be worked out based on the the area that the integral suggests