How Can This Sudoku Have No Digits At All?
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- เผยแพร่เมื่อ 30 พ.ย. 2024
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Today's puzzle is the first we've covered by the constructor Fluster - indeed it is apparently their first sudoku... Well two of the testers think it's a masterpiece and we echo those sentiments... what an absolutely extraordinary debut. You can play it at the link below:
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Rules:
Normal sudoku rules apply. Digits cannot repeat within a cage. Cages with the same size must have the same sum and cages with different sizes must have different sums.
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Thank you for featuring my puzzle! After the first step there seems to be at least a few ways to disambiguate the rest of the cage sums depending on how much math you use. And I see Simon also found a little shortcut to find out the value of one cell cages a bit sooner than I expected. I believe this is inevitable with this rule set as it needs to obey some rules of mathematics.
An alternative way to proceed after finding that 2-cell cages sum to ten is to notice that the 4-box region forces the 4-cell cages to sum to an even number (as Simon discovered a bit later) and therefore needs to be 18 or 22. This creates a 1,9 pair in the cage sticking out from box 1 and a 3,7 pair in the cage sticking out from box 6. Then you can use the 3-box region to determine that the one cell cages need to be an odd digit and as they see both sides of the both pairs mentioned above, the only remaining value is 5. This is propably not an easier way to go forward, but one that does not require much calculation.
With this rule set it is also seems to be mandatory to disambiguate some cage sums with the second part of the special rule. Otherwise you could swap all the digit pairs 1,9; 2,8; 3,7 and 4,6 with each other to find another symmetric solution and all the cages would still work. I began with having only the first part of the rule, but knew I would later need to add something to it to break the symmetry. In the end I was very happy of being able to do that with a rule that fits so nicely to the wholeness.
I didn't look to closely at what it would exactly be but it could be summed up in multiple equations and then solved for each number once you know one of the numbers
Kiitos Jouni and thank you SImon. This is the best puzzle I've seen on CtC so far! And the fact that Simon was able to solve it in less than an hour (or at all) is astonishing!
I struggled with this for hours and finally gave up. I constructed the set of linear equations to solve each cage total, knowing that with 5 variables and only 3 equations, I would be left with a Diophantine mess.
I did actually find the relation between 1-cages and 2-cages, showing that if the if the 2-cage were offset by anything (which had to be a multiple of 2), the 1-cell would have to be offset by a multiple of 11, which would be a problem because neither 16 or -6 is a sudoku digit. (I was using plus/minus offsets wrt 5 instead of the actual magnitudes). So both the 1-cages and the 2-cages had to be offset zero (5 and 10, correspondingly).
That was all I could do.
Absolutely brilliant!
Cool puzzle! Both the math part and the remaining Sudoku were enjoyable and posed just the right amount of challenge.
I clicked the link, read the rules, stared at the screen for 20 seconds, closed the tab, and am now back here to watch what I assume is black magic.
r3c3 is odd and that's as far as I got. Not without a lot of head-scratching though. Oh, and r6c9 is the same as the other 1 cell cages by sudoku.
I looked at the thumbnail and instantly got anixious
@@sportsjefe That's where I am after nearly an hour. I've also very slightly narrowed my 2 cell cages from 5-15 to 8-13!
I must admit that this makes me feel really glad it took me 10 minutes before I gave up... I was wrong in any case
not black magic, mathemagic.
Simon, we've had words about this. I don't want to hear you say "I'm so stupid" ever again.
Simon, you are so brilliant and I hope you hold your head up high!
No one should ever say this about themselves. I've made a mistake is perfectly fine and covers the mishap.
Yes, Mark is good at that, when ever he berates himself "I am such an idiot..." he almost immediately verbally says "no I am not, sorry" which is good for the audience and himself. Take after Mark's example! :)
Quite aside from anything else, if he's stupid it's a miracle that the rest of us can tie our own shoelaces.
@@Alex_Meadows Thats why I just slip my shoes off and on without untying my laces.
Simon is the Columbo of Sudoku. Just when you feel he's beaten, there's always a 'oh, I've had an idea'. Genius.
"Oh, sir..."
"Just one more thing..."
"Right. Mark is trolling me. There is no way this solves. Well, I can see I can place a one...."
When he hits you with the "Just one more thing..." you know he's already got you figured out
If Simon is Columbo, Mark is Sir Peter Wimsey.
Remember when Sudoku had numbers? Haha. Good times.
Wait until next year, when puzzles won't even have a grid.
@@captainufo4587 I'm going to keep the grid, but drop the digits 2-9 completely, as these are now just too widely understood.
Here's a sneak preview.
"Normal sudoku rules do not apply. Place the digit "1" once only in every row, column, and box of the grid: ⬜ "
I barely remembered that cages had their sum in the upper left corner
@@captainufo4587 I've wondered how little info is needed to get a unique solution and...I think this is it. Nothing 'specific' like thermos or 'dots between cages' to lead any specific numbers, except the single-cell cages and the 'psuedo single-cell cages' that are part of an overlap outside boxes. As a first-time setter myself (well, still working on that), I've come up with a couple that have unsolvable non-unique 'wings' that could go in any order, with less info, but nothing as meticulously perfect as this!
Pepperidge Farm remembers
Does it drive anyone else crazy that Simon never takes notes (like writing down on paper or a notepad app) things like the potential cage totals for each size, but rather forces himself masochistically to just remember EVERYTHING? It’s brutal to watch 😂
Yes, it drives me nuts too. Simon's always in a hurry but he wastes a lot of time by not using a jotter pad and also makes it much harder for the viewers to follow what he's doing.
I managed to solve this without resorting to pen+paper or notepad, but I did end up writing some of the box totals in as pencil marks: tens as a corner mark, units in the middle so I could even write 22 when it was still an option. I didn't think of using the three 4-cells plus two 5-cells to get rid of 22, instead I did something horrendous using the other two regions and working from the options for three 3-cells plus one 5-cell to see the 5-cell 22 forced the duplicate 4-cell total. Fun though and I'm pleased to say I didn't need to watch the video until after.
Sort of - though I've come to the conclusion that there are two types of thinkers, people who use paper and people who don't, it doesn't make one any better than the other. I have always had terrible short term memory, so I have to write stuff like this down. This puzzle was actually not too bad, I was able to solve without watching the video, by a completely different method.
@@anneharrison1849 I'd like to know what method had you used.
But the message was written so long time ago that you probably don't remember it?
I like to use the sudoku app itself to write notes on... I pick an area with no useful information in it, write things like cage totals as 2 digit numbers by using both corner digits and centre digits, and then give the cell a weird multicolored flash so I know it's just a note when I'm scanning the puzzle
Day one of appreciating Simon.
Quote of the day: "one-one and one-one waowaowaowao nori-nori"
That made my morning.
NoriNori to the rescue.
for anyone wondering: 20:40
I can't believe I watched a man do 5 variable linear algebra in his freaking head. 🤯
He was very clever, by merging all the types of cages into the X apart from one, he reduced the problem to 2 variables. If he had done the 5 variables math in his head he would have gotten all the totals right away without needing the even/odd trick. It's impossible tough.
With a piece of paper and 5 conditions on the boxes, you get a linear system of 5 unknowns and 5 equations, which is easily solvable and you get the totals after 5 minutes, but you NEED that piece of paper, you can't do it in your head
The 18 and 22 could hv been easily cleared cause the total was supposed to be divided by 3 cause there are three in 180
My thoughts exactly
@@Snowridertn overdetermined vs underdetermined systems of equations,, love it
@@Snowridertn How did you find 5 conditions? I see 3 conditions, corresponding to the three regions Simon uses.
„... one of the best sudokus they have ever seen.“
Ah, so the usual.
And, as usual.... he's not wrong.
@@TrackpadProductions Never doubted it :D
Has he ever said "Well, that was a crap sudoku. Boo. It sucked"? Can you even imagine??
@@ayannababii I'm sure he would say this in a parallel world where he would solve one of my imaginary crafted puzzle
'I really hope you solved it'
Oh Simon. Of course I didn't. It was about a light year above my pay grade. Absolutely extraordinary puzzle and remarkable solve
I find it rather touching that he has such high, and utterly unwarranted, expectations of me.
I can hardly solve hard level regular sudoku hahaha
What you did was basically solve diophantine equations. The way you found the 5 and 10 was different, but later, you did in fact solve diophantine equations. I just did the whole thing with diophantine equations.
Suppose A is the 1 cell cage total. B is the 2 cell total, etc. You get the 3 diophantine equations if you take boxes 1,2,3 as well as boxes 4,7 as well as boxes 5,6,8,9. Just count the number of cages of the same size in each of those groups of boxes. You had mentioned that the cages are fully contained in each of these groups of boxes.
You get the following equations:
1) 2E+2D+2C+1B+1A = 135
2) 1E+0D+3C+1B+2A = 90
3) 2E+3D+0C+6B+2A = 180
So 3 equations with 5 unknowns.
First thing is to get rid of E. Double the second equation to get 2E and subtract each pair of equations.
1) 2E+2D+2C+1B+1A = 135
2) 2E+0D+6C+2B+4A = 180
3) 2E+3D+0C+6B+2A = 180
Subtract equation 2 from 3.
4) 0E+3D-6C+4B-2A = 0
Subtract equation 1 from 3.
5) 0E+1D-2C+5B+1A = 45
Now get rid of D's. Multiply equation 5 by three.
5) 0E+3D-6C+15B+3A = 135
Subtract equation 4 from equation 5.
6) 11B+5A=135
While everything else was a diophantine equation. This one has the classical format.
We can see that B has to be a multiple of 5 exactly as you've said in the video, but using a different method here.
There are very few possible candidates here. B can go from 7 to 13 and A can go from 1 to 9.
The only thing that works is B = 10 and A = 5.
Put these numbers back in the original equations.
1) 2E+2D+2C+10+5 = 135
2) 1E+0D+3C+10+10 = 90
3) 2E+3D+0C+60+10 = 180
Simplify...
1) 2E+2D+2C = 120
2) 1E+0D+3C = 70
3) 2E+3D+0C = 110
You mentioned a few of these totals in your video, but not quite like this.
If you divide equation 1, we get the 60 total you used to get the one cell cage total.
Subtract equation 1 from equation 3.
7) 0E+1D-2C = -10
7) D+10 = 2C
Again, a classic diophantine equation.
D can go from 18 to 22. And you can't use 20. And since odd numbers can't be divided by 2, we only have two candidates 18 and 22.
And 22 doesn't work because if you plug in 22 into equation 3 for D, you get 22 for E which isn't allowed. This part, you did exactly what is shown here.
So D is 18, E is 28 and C is 14.
This is what I did right away. Solved the whole thing in about 30 minutes.
So it was actually quite simple.
@@paulmdt1 yeah right
That looks to be what I did, only my "right away" was about 15 minutes, and I took 90 minutes to solve. 😂
Nicely explained. 👍
You still have to place 1-9 numbers in cells, right? If you're not a sudoku solver you're screwed
Nerd
I love it when early on in a video Simon bemoans his chances of finishing and releasing on time because it looks too hard, because we have the gift of hindsight and know he does. The excitement when he makes headway is so great!
Too bad he's said in the past that we'll never see all the blooper videos where he fails to find a solution.
What a debut on the channel. Welcome to the community, Fluster.
Simon reinventing linear algebra on the spot is just something else...
Would still like to see a solve of an anonymous puzzle with its creator revealed to Simon at the end.
Do not bifurcate while guessing author!
The Masked Setter
@@FOmar151 The Stig.
@@PuzzleQodec Some say he's the Devil, other claim he's the Flying Dutchman. All we know is, he's the Masked Setter.....
I had to take notes and calculate lots of stuff to crack this puzzle and yet it still took me 3~4 hours. It really amazes me that Simon doesn't even need any notes and realizes the critical points of the puzzle rapidly. It's just next-level skills. Keep up the great works!!!
When you noticed the grid was divided into 3 distinct regions, I saw that you could solve this like an algebra problem. Once you figured out the 1 and 2 cell sums, those 3 regions give 3 equations which is enough information to solve for the 3 remaining unknowns. [3G + B = 70], [2G + 2R + 2B = 120], and [3R + 2B = 110]. I couldn't possibly imagine trying to do all of this in your head without writing anything down, so hats off to you!
That's actually not enough information. Those equations are degenerate. (Adding the first to the third gives three times the second.)
That's exactly where I got to, but I got that the 3-cage had to sum to 5. :-( Did u manage to make it work?
I feel like Simon needs to keep a notepad (physical or virtual) handy while doing these types of puzzles so he's not constantly trying to recalculate the same possibilities.
He occasionally pulls up Notepad (the program) on screen for the absolutely brutal movie-length puzzles.
It would be good to display a small notepad on-screen for just the essence of the key info...1=5, etc., for this one. Needn't take up much screen space for a lot of these puzzles that call for recording info off-grid. It might take him stopping the recording once he's figured out what's needed, organizing the screen, and continuing.
I think that both Simon and Mark try to do calculations in their head as much as possible, as a matter of pride.
@@joadbreslin5819 It really works the brain, probably making memory and calculating better.
Merch suggestion: the Cracking the Cryptic notepad and pencil - invaluable aid to advanced problem-solving.
I would have expected Simon to color similar sized cages at the start. He never misses an opportunity to color cages.
That's what I did, it made it much easier to keep track of the different numbers.
Ditto.
Dear Simon,
That was the single most impressive sudoku arithmetic I've ever seen. I am half jealous and half still in awe.
This puzzle would make for a very interesting constructor video. I'm having a hard time imagining how one can come up with something so unique and well made.
Especially with the dictum about solving as you construct...
I'm guessing either some incredible brain formation, or some incredible drugs.
Either way, it's not happening for me.
It sounds like he wanted to challenge himself and go "I wonder if there is a way to combine simultaneous equations and sudoku"
Not sure which is more astonishing - the elegance of this puzzle or the fact that Simon solved it (and explained his solve) in only 45 minutes.
This is a nice puzzle. As a mathematician, I really enjoyed solving the system of linear equations in the puzzle :)
i am also a mathematician and it's really nice to see how Simon creates the subject of solving a system of linear equations where the unknowns are integers. all the while saying, "I'm so stupid, I'm so stupid".
he is the Carl Friedrich Gauss of Sudoku.
Same dude... Loved the amount of mathematics involved and the crack was satisfying in the end
Haha I noticed that too, beginning my math phd this year!
Glad to see this one on the channel! I remember solving this and it was one of my favorites! It had such a beautiful application of modular arithmetic.
"Imagine If I break this now, I will actually cry on camera I think". I am severely torn on whether I would want to see that or not.
I love the way your mind works Simon. Great logic you have never seen before but you are still able to spot it and go with it.
Edit: Simon may get flustered but Fluster didn't get Simon. Brilliant puzzle, especially for a first one.
If you work up a sweat, the designer of the puzzle is very happy. That's the mission ...
The audience is having fun, and when you find a solution, everyone is happy and also very impressed! Keep it up!
Very pleased to have solved this one without watching first! I did use a spreadsheet to keep tally of the min and max values of each cage, and gradually narrow them down. Then the realisation that the 2 cell cage totals must be divisible by 5 helped a lot! At least the invisible cage total problem didn’t matter today!
for boxes 1,2,3 a+b+2c+2d+2e=135, for boxes 4,7 2a+b+3c+e=90, for rest of the boxes 2a+6b+3d+2e=180, also a,b,c,d,e are all different positive integers. You can solve without looking at the grid.
I went down the same algebraic path. Algebra rules! About six times I was thinking "get out a piece of paper and pencil!". Honestly, I don't know how Simon keeps all this stuff straight in his head.
from those equations you get the following:
a=5
b=10
c can be one of 17, 14, 13, 12
d can be one of 24, 18, 16, 14
e can be one of 19, 28, 31, 34
some sudoku restrinctions i applied, that without looking at the grid, to narrow the correct totals you got to look at the grid
Yeah, but how do you solve that?? There's 5 variables...
Sudoku is like life: you keep solving the same puzzle, yet you keep coming up with new ways to solve it.
Sudoku is like life: it keeps throwing puzzles at me that other people manage to solve but I don't.
DEEP 🌊🌊🌊🌊🌊
A bit late to this one, but I recently stumbled across this channel and this was the first puzzle I ever tried...This was a blunder I must say. I was a bit stubborn and took one or two breaks to get a drink. In the end, it only took me just under 9 hours! haha
But I struggled a bit because my exhaustion caused me to get stuck for a few hours on a silly addition mistake I had made half-way through. I can't express my disappointment in myself when I eventually found it. Though I am quite surprised at your method of solving it by dividing the sum of the board by the number of each cell. That would have saved me my first hour easily. Instead, I brute-forced the problem by noting all possible sequences that could occur between a certain range of numbers for each of the cage types (1-5). From there I used cells 1 and 6 to figure that the dominos are 10 and the only possible sequence for a 4 cell cage must contain a 3, 5, and a sum of 10. making it 18. And by deduction making the singles which I had marked out, 5. I hope this made sense.
This is arguably the best puzzle presented on this channel so far. It is truly brilliant. And Simon's solve was nothing short of brilliant also.
Congrats to fluster for making such a gorgeous puzzle! I’m in awe
That was a hard break in... so happy we got to see this one...
One of the greatest constructions of all time. One of the most pleasing solves of all time.
30:55 That is INCREDIBLY beautiful setting. That you need that restriction in that moment in the solve. That is amazing.
The ruleset is small and relatively easy to keep in mind, not a single given digit, only killer cages, this must be one of the greatest ever.
Beyond my capability to solve, although I may give it a try in a few months time when my brain has forgotten this solve.
Seeing you reasoning this out was a joy. But there was an (in my opinion) easier way to get the cage totals. You can construct 5 equations with 5 unknowns to calculate them. From there it's "just" sudoku 😁
So many equations relating to the cages. And they told me I would never find a use for linear algebra after school 😉
Isn't it the other way around? The teachers say math is everywhere and you don't believe them?
I only found 4. I couldn't find the 5th equation to get the 5 unknowns :(
@@TheMeanderingduck6 That's true. But if you eliminate 3 variables you get an equation in the remaining 2 which has very few integer solutions.
@@TheMeanderingduck6 I'm operating on very little sleep for 10 days in a row, so my brain is a bit foggy. But I was thinking the solution might (maybe) be helped along by the fact that the five unknowns must all be different values.
@@TheMeanderingduck6 - cone to think of it, the entire puzzle could be the fifth equation, couldn't it?
Simon I decided to play along with this sudoku and I had a wonderful time especially and only because when I got stuck which was often I could come back to this video and piggyback off of your working out, so thank you for giving me a wonderful and engaging 2ish hours solving this puzzle with you.
I found a different path to fix the value of the one cell cage that I'm quite proud of. Once you know the 2-cage is equal to 10, you can use boxes 1, 2, and 3 to determine the one cage has to be odd. Then you can use boxes 5, 6, 8, and 9 to determine the 4 cage must be even. Box 1 means the 4 cage must be 18 or 22, with the 2 cage being a 1/9 domino. The one cage sees that 1/9 pair and therefore has to be 3, 5, or 7. One last time look at boxes 5 and 6 to determine that r5c4 and r6c4 must sum to either 6 or 14. Whichever it is, that domino cannot contain a 3 or a 7 and therefore the one cage has to equal 5.
This is THE best Sudoko, ever. Amazing. Elegance, very hard but solvable. Wow!
I agree with Simons observation that after such a challenging start, it’s nice to have a downhill finish.
But sometimes it's also nice to watch him suffering to the end.
of all the sudokus I've seen you do, this one was definitely my favorite to watch you solve!
I did this one as a not-too-hard linear algebra problem followed be a lovely little killer sudoku - which is basically what Simon did, but a little more formal. (actually converted it to some equations with 5 unknowns, from which the 2-cell and 1-cell cages fall out quickly, the rest take a bit more work but do fall out)
25 minutes after the start and I have the cage numbers! I can finally begin!
As a visual learner, I would really appreciate having a written formula for puzzles like this regardless keep up the good work simon
if this is Fluster's first ever puzzle, then he/she is another Phistomefel in the making. amazing puzzle!
It's extraordinary. How do you go from having never set a sudoku to setting this one, without passing through any intervening steps? It's as if Michelangelo wandered into the Sistine Chapel, picked up his first ever paintbrush and decided on a whim that he was going to have a bash at painting the ceiling.
32:06 for me. Was a pleasant ending to the week; I felt like I really struggled with the last few days' puzzles and ended well after our master solvers. Can't wait for more of Fluster's work!
It took me 3 hours, a restart, and way more math but I slogged through it. Normally I'm happily impressed when you find an easier path Simon, but this time I'm allowing myself to be salty about it. I had to write out systems of equations on paper. My algebra 2 teacher is snickering at me from beyond the grave. How dare you not suffer as I have suffered!
Started solving, stopped timer at 93:28 and went to sleep. Continued with cup of coffee after waking up and after more than 3hrs of solving, I feel like gone through my math final exam again after 30 years. Old fashion pen and paper for notes helped me a lot, so Simon, if you could add a small notepad on the side to the app, it would help solvers like me a LOT!
Word of the day again (04:08):
Panoply - an extensive or impressive collection
I'll add the etymology!
It comes from the ancient greek "pan", which means "all" and "oplos", which means "tools", "shield" or "weapons". So it means "with all weapons" or "with all tools".
In french, the word also exists, "panoplie". But it doesn't mean "an impressive collection". We use that word to mean "a complete set of something". For example, "la panoplie du pompier" (a firefighter's panoply) would mean the ensemble of all the tools and equipment a firefighter uses.
Concise solution to finding cage totals: Let the letters A, B, C, D, and E correspond to cages of size 1,2,3,4 and 5 respectively, by subtracting the mean cage (5,10,15,20,25 respectively) from the cage values. Doing this makes arithmetic easy.
From the whole grid, 8B + 5(A+C+D+E) = 0. So B = -5, 0, or 5 (cage is 5, 10 or 15). Doing suduko around box 6 we see only 10 is possible and B=0. Since 10 cages don't have 5s and by doing more sudoku we get 1s cage is 5, A=0.
Now box 1: D must be -2, -1, 1, or 2. Boxes 5689: 3D+2E=0. So D is even, and -2 or 2. So E is -3 or 3. But cages must differ so -3 + 25 = 2+20 = 22 is not allowed. So E=3, D=-2.
Boxes 47: E + 3C = 0. So C = -1.
Thus the cage totals are 5, 10, 14, 18, 28. Now you can sudoku.
Ive just started watching your videos and already I'm hooked xD
More than 50 years ago I enjoyed equations at school and then have spent a lifetime never finding a use for them. But now in my hour of need I couldn't fathom the right ones to get me going.
It's amazing how the logic of the killer cages was being used on a meta level. Really impressive! And a fun solve to watch. I'm impressed with your math memory
Simon: You must think my maths are horrible.
Me: *frantically scribbling down all the number just to keep up*
Simon, your ability to do these puzzles all the while keeping the numbers and logic in your head is truly staggering to me.
If that was Fluster's first(!), just wait till he gets more experience! Boy, was that something else, or what? Simon, you have my total admiration and respect!
This was seriously an amazing watch. I didn't even attempt it, but I can't imagine setting something this amazing, especially as a first puzzle.
There isn't an adjective I'm aware of that adequately describes the impressiveness of this.
Algebra works for this one.
Boxes 4, 7:
2a + 1b + 3c + 0d + 1e = 90
Boxes 1, 2, 3:
1a + 1b + 2c + 2d + 2e = 135
Boxes, 5, 6, 8, 9:
3a + 6b + 0c + 3d + 2e = 180
Therefore:
-a - c + 2d + e = 45
2a + 5b - 2c + d = 45
Therefore:
3a + 5b + e = c + d
Substitute this into boxes 5, 6, 7 ,8:
b + c + 4d + e = 180
Double boxes 4, 7:
4a + 2b + 6c + 0d + 2e = 180
Therefore:
4a + b + 5c + e = 4d
= 4 (3a + 5b - c + e)
= 12a + 20b - 4c + 4e
Thus:
9c = 8a + 19b + 3e
Triple boxes 4, 7:
6a + 3b + 9c + 0d + 3e = 270
14a + 22b + 6e = 270
7a + 11b + 3e = 135
And now that you're here, suffice to say you're just as bored as I am.
Best puzzle ever! Haven't seen anything similar before. Congrats to the creator and the solvers for this
I never thought I would spend my Saturday evenings doing simultaneous equations for fun, but here we are!
That was an incredible construction - I got a reasonable way in on my own, but then just needed the hint to look at the top three rows and then I realised I could get some more equations, and got the rest of it from there (although not without a couple of backtracks when I found my maths had let me down). Definitely couldn't have done it without a pencil and paper, or without colouring the different size cages in different colours.
Yes! I did it! I really managed to solve it without any help from the video.
(I'm writing this having seen the video only until 2:18 and haven't read any comments.)
I coloured the different cage sizes with different colours.
We have 5 1-cell cage (Let's call it A), 8 2-cell cages (=B), 5 3-cell cages (=C), 5 4-cell cages (=D) and 5 5-cell cages (=E).
Whole grid: 5·A+8·B+5·C+5·D+5·E=405 ⇒ 5·(A+C+D+E)+8·B=405 [eq.1]
That means, B must be a multiple of 5, so it can only be 5, 10 or 15.
5 and 15 have only two ways in a 2-cell cage, which would cause a problem in box 6.
(Seeing this took some time staring at the grid.)
So B=10.
Then I looked for boxes or combinations that include complete cages and found these:
Boxes 1-3: A+B+2·C+2·D+2·E=135 ⇒ 2·(C+D+E)+A+10=135 ⇒ 2·(C+D+E)+A=125 [eq.2]
[eq.1]: 5·(A+C+D+E)+8·B=405 ⇒ 5·(A+C+D+E)+8·10=405 ⇒ 5·(A+C+D+E)=325 ⇒ A+C+D+E=65 ⇒ C+D+E=65-A
[eq.2]: 2·(C+D+E)+A=125 ⇒ 2·(65-A)+A=125 ⇒ 130-2·A+A=125 ⇒ 5-A=0
So A=5.
Boxes 4/7: 2·A+B+3·C+E=90 ⇒ 2·5+10+3·C+E=90 ⇒ 3·C+E=70 ⇒ E=70-3·C [eq.3]
[eq.1]: C+D+E=65-A ⇒ C+D+70-3·C=65-5 ⇒ D+70-2·C=60 ⇒ D+10=2·C
That means, D must be even.
In box 1 we have 2 4-cell cages and 1 cell left:
Box 1: 2·D+r3c3=45 [eq.4]
That means, r3c3 must be odd. But since it is in a 10-cage it can't be 5.
And D is even, so the quantity 45-r3c3 is a multiple of 4.
So r3c3 is either 1 or 9. (r3c3={1,9})
[eq.4]: 2·D+r3c3=45 ⇒ 2·D+{1,9}=45 ⇒ 2·D={44,36} ⇒ D={22,18}
[eq.1]: D+10=2·C ⇒ {22,18}+10=2·C ⇒ 2·C={32,28} ⇒ C={16,14}
[eq.3]: E=70-3·C ⇒ E=70-3·{16,14} ⇒ E=70-{48,42} ⇒ E={22,28}
At the first option D and E would both be 22, but they have to be different.
So now I know all the cage totals:
A=5, B=10, C=14, D=18, E=28.
From there it was normal sudoku and killer logic.
I was able to crack this one. Thanks a lot for these beautiful puzzles. Simon, you are an inspiration.
I started off by writing a set of simultaneous equations, but you only get 4 equations for 5 unknown cage values, and need to get tricky to get the effect of a 5th equation.
I wanted to do that, but it seemed like overkill. Would it have been "unsporting" to throw matrix algebra at it? Where you able to throw number theory at it then to get the solution?
1a + 1b + 2c + 2d + 2e = 135
2a + 1b + 3c + 1e = 90
2a + 6b + 3d + 2e = 180
a1=5, b1=10, c1=16, d1=22, e1=22
a2=5, b2=10, c2=17, d2=24, e2=19
1st has duplicate cage values, so must be 2nd. :)
via www.hackmath.net/en/calculator/integer-diophantine-equations-solver
I think the solution to that is to get values for all 5 in terms of some unknown x (which could just be the value of one of the cages). You can then go around placing restrictions on x to find it's value (effectively what Simon was doing), using sudoku requiremants
@@Codes__ You can add a 4th equation, 8b + 5(a + c + d + e) = 405, but you still get the same two solutions, which you can resolve by uniqueness
@@steveunderwood3683 That equation is not independent of the others, it is just the sum of them. So it definitely will still have two solutions, even if it is certainly convenient to work with.
3,6,9 the goose drank wine. Simon constantly puts the clapping song in my head 😂
I'm colorblind and Simon keeps using purple and blue together, and green and yellow together and I can't see the difference between both pairs.... He does it in every videos!! :)
One of the best sudoku puzzles I have ever seen. Great solve, Simon!
45:04 finish. I actually wrote down the equations for each of the totals (box 1-3, box 4/7, box 5/6/8/9) using a-b-c-d-e as the unknown cage totals. This allowed me to notice the requirement for 2-cell cages to be a multiple of 5, and it went from there. Once I had determined each of the totals, the REAL work began.
As a huge math and soecifically number theory fan, the breakin to this one was so fun to watch!
This was completely awesome!
Absolutely brilliant construction. The break-in idea was completely new to me.
I was thinking "system of equations" and then I watched you do it all in your head. Magnificent puzzle and solution.
I'm out of practice so I made a spreadsheet for this. This is a beautiful math puzzle with two boxes on the 45-(1/2 of a total) and the 2, 3, and 4 box totals all coming in to play. You can variable this out with the early limits on what the cages can sum to. It's beautifully done. Recognizing the 1 and a 2 cell cages quickly makes this puzzle a dream.
Excellence.
An alternative way of looking at the top 3 cages that I did was 3x45 - the 2 cell cage is equal to (2 x the sum of the 1,3,4,5 cage) minus 1 x 1 cage which gives you the 1 cage of 5.
thanks for this - you finding that the 2 cell sum had to be a multiple of 5 really made me smile x
I usually watch solutions, but every time an opportunity to make use of linear algebra comes up, I have to try to beat Simon to the answer with my fancy mathematics package. At 10:05, I set up a constrained linear equations in the package and it said "one -> 5 + 11 x, two -> 10 - 5 x, three -> y, four -> 2 (-5 + 7 x + y), five -> 70 - 17 x - 3 y" with x and y as integers. Given the size of a one cell cage, I figured out that x is 0. By 21:20, Simon had stated enough other restrictions that I had y as 14 or 16 and with the rule restriction eliminated 16.
Welp.... ok. = )
I'd bookmarked this one to try on my day off. It ended up becoming my entire day off - after 10 hours I'd made all sorts of unbelievable deductions about the puzzle and the relationships between cells and boxes and cages and rows and columns...but still didn't have a single digit in the grid. I finally gave up and watched the video to see what I'd missed. I hadn't done the simultaneous equation for the puzzle as a whole; once I knew the 2-cage was divisible by 10, my earlier work allowed me to solve the rest of the puzzle in half an hour flat!
20:05 The logic that gets to the actual value is obviously even more useful, but did notice different logic that lets you deduce the parity of the one-cell region here -- boxes 2 and 3 add up to 90, and currently have two times the "three-cage" total (even) plus two times the "five-cage" total (even) plus the number contributed by the two-cage (odd) plus the "one" value. Since the whole of the two boxes has to be even, and you already have even + even + odd, the one-cage total must be odd to make it work.
I greatly admire Simon because while fully knowing that he is running short on time, he still takes time to explain what he is doing.
This puzzle was very challenging, but I thoroughly enjoyed it. I didn't really know where to start until you pointed out that there were 8 of the 2-cell cages and 5 of everything else. Once you did that, I found the 4 sets of equations, and then it was just a matter of using the same techniques for solving systems of simple linear equations to get what the potential sums were, then applying the rules to eliminate all but one of the possibilities (I used Excel for this). I've been playing a lot of the kind of Sudoku where it's just cages with the sums listed, no digits provided (Killer Sudoku app on Killer difficulty), which prepared me for the remainder of the puzzle. One technique I learned from watching this video is using sums within the same row/column.
Simon, you should connect dots u discover. You noticed grid can be split in 3 parts and whole grid is forth. When u found out 2 cell box is 10 you could easy write 4 equals from which u would know values of all cages. Best wishes from Poland :)
You had some of your greatest reactions on this amazing puzzle, Simon 😁
Simon's voice is so calming. It helps me a lot.
You should hear him on the golf course - it takes it all out of him.
Brilliant idea. It seems like it could be a style that has enough different possible puzzles that could be made to be viable as a really interesting new variant. Until the magic square ones that are great to solve once, but then all of them are only slight variations on each other.
1:17:51 A half an hour longer than the video, but I got it. I liked it.
For my solve, I kept doing math to figure out what could satisfy r3c3, to determine what numbers could fit in boxes 1 and 6, which are entirely composed of 4-cell and 2-cell cages. Then that helped me figure out what the 1-cell cages were, and also what the number that was in r6c7 could be. That helped me figure out the values of the 4-cell and 2-cell cages, and narrow down the position of the 9 in box 6. Then it gave me some 139 triples in boxes 5 and 6 that I used to figure out box 4, then I started bouncing around boxes 8, 9, and 2.
This deserves offerings to the algorithm-overlords, just because of the awesome thumbnail.
9:46 - grid is divided on 3 parts and sums are the same - hold my beer, I need to math the s**t out of this sudoku
Congratulations on an amazing solve of a fantastic puzzle!
Commenting at 19:41. In this position, I would take the deduction that the 2 felled cages are 10 and subtract all of that from the total (405-80=5x). We know from this that 1 set of 1,3,4,and 5 cell cages = 65. In the top 3 rows of the grid plus the 1 cell cage in row 4 has 2 sets of all of the “x” cages and a 2 felled cage (65*2)+10=140. The top 3 rows of the grid add to 135 #thesecret. Therefore the remaining cell and all 1 celled cages =5.
Wow. The puzzle and you, are both brilliant.
@Jouni Karhima I think that was the most fun solving a puzzle I've had in ages. A brilliant idea done impeccably. Thanks so much
That moment when I'm solving the puzzle as Simon is also working it out, and we both have the same epiphany at the same time... priceless.
That is a thing of beauty that could be turned in to a flip chart lasting 60 seconds, the first 20 of which would be the blank board and then numbers getting filled in.
Beautiful.
Quote of the day: "you musr be looking at me thinking 'your maths is so bad'"
Yup, Simon is so crap at maths! The way he sometimes takes a whole TWO SECONDS to do sums that it'd take a calculator a hundredth of the time?! Shameful!
You know... he got about 95% of my thought right. The only thing that was wrong was the pronoun, because I actually thought: "My maths is so bad" :D
Great puzzle and great solve!
I worked it out myself in 417 minutes. And with pen and paper and a short program, to do most of the hard work.
I didn't work out the number of 1, 3, 4, 5 - cages. I did work out some lower and upper limits for cages. Then I wrote a script to add up the solid box sum equations of boxes (1,2,3), (4, 7) and (5, 6, 8, 9). I colored the 5-cell-boxes with the color where the 5 lives, purple, and 2 where the 2 sits (yellow) and so on, let (b)lue iterate from 12 to 18, (g)reen from 18 to 22 and so on and checked the equations, I could make up for the grouped boxes and for the most inner loop and branch, I printed `if ((2*r + 3*b + y + t == 90)); then echo -e "
r:$r y:$y b:$b g:$g t:$t"`.
Yes, dirty tricks, surely not allowed on a competition, but still some fun.
I was still presented with 2 solutions, 5, 10, 15, 20, 25 and 5, 10, 14, 18, 28. I found the contradiction for the first one pretty quick in box 9, too, leading to two 5s.
Then I marked in the beginning the 3-cell-boxes, for instance, with "014", using the 0 as a marker, that this is a sum and not two possible digits. I still had to use paper and pencil, to figure the ways out, how to sum to 28 with 5 digits.
If I had the idea to subtract and divide all those numbers, I would still need to use pencil and paper, to memorize all the steps. :) But the program to solve the equation system is of course not fair play.
I did manage to solve it, though only after watching you work out the 2-cell cages. But I got to the 1-cell cage on my own, after working out that there were two two 5-cell cages, 4-cell cages, 2 3-cell cages and a 2-cell cage in the top three rows along with a single 1-cell cage. That let me calculate 135 minus a 2-cell cage and two lots of the other cages aside from the 1-cell cage as being equal to the 1-cell cage. From there I worked out the rest, though it took me like 45 minutes to do what you did in, like, 10 -- and that was after probably two hours of trying to figure out the break-in to the sudoku.
Managed to solve it in 103 minutes. I first went for solving for the 5 different cage values. Doing that I managed to pull out:
1 cell - 5
2 cells - 10
3 cells - 14
4 cells - 18
5 cells - 28
I began my work hunting for single digits, not just the single squares but also the bits that stick out. I ended up having to bifurcate and resort to trial and error up there in boxes 2 and 3. But other than that there was no problem.
Never have I solved sudoku in my life until yesterday when I accidentally clicked on an easy-level sudoku tutorial which made me download one of the apps to start doing it myself... and now, I never would´ve thought I´d watch this at 1AM and be like:
dammmmm...how is this possible =)
"I was told that there would be no math on this sudoku." - Gerald Ford (Chevy Chase)
This method worked for me.
Let's consider x = r3c3, a = 1-cage, b = 2-cage, c = 3-cage, d = 4-cage, e = 5-cage.
We have:
All boxes: 5a + 8b + 5c + 5d + 5e = 405;
Boxes 4, 5, 6, 7, 8, 9: 4a + 7b + 3c + 3d + 3e = 270;
Boxes 5, 6, 8, 9: 2a + 6b + 0c + 3d + 2e = 180;
Boxes 1, 2, 3: 1a + 1b + 2c + 2d + 2e = 135;
Boxes 1, 4, 7: 2a + 1b + 3c + 2d + 1e = 135 - x, x = {1, 3, 7, 9}.
Then we have the matrix:
5 8 5 5 5 | 405
4 7 3 3 3 | 270
2 6 0 3 2 | 180
1 1 2 2 2 | 135
2 1 3 2 1 | 135 - x
Solving for x, the only value the allows integer results is x = 9.
Resolving by Jordan-Gauss, we have the following:
a = -72 + 11w/4;
b = 45 - 5w/4;
c = 63 - 7w/4;
d = 18;
e = w;
We have then that w is divisible by 4.
Knowing a is less or equal 9, the upper bound for w is 30 (-72 + 11w/4
Simon you should do a short video of you reviewing your top 10 favorite puzzles up to this point.