Puzzle creator here 🤓 wonderful solve Simon! And thank you for the kind words, it was a honor to be featured on the channel! The key to that puzzle was ruling out that R7C7 could not be low polarity, which you did beautifully and as intended. That was, truly and honestly, my first Sodoku puzzle I’ve created and I very much enjoyed making it and watching you solve it. I’ve loved the concept of the Phistomefel ring and wanted to put my own math twist on it. Hope you enjoyed! Here’s to hoping I’ll be back for more sometime soon…
I haven’t seen the video yet but having somehow made it through, I’m blown away by some of the logic in this. Particularly the interplay between boxes 3, 7 and 9 (which if intended was quite simply a masterstroke of a sudoku step). It was a puzzle where it was difficult to know what to focus on but looking from the right angle revealed some absolute magic. Thanks for coming up with it.
The moment Simon put those sixes pencilmark in column 8, saying that it is threes. I wonder when it will blow. But in the end it ended correct pencilmark. And yeah 10886400 / 10! = 3 :) Nice.
I knew someone else would notice the "so 3s has to be here" -> corner marks a 6. I also wondered when it would blow up. I don't know how things like that work out in the final solution for Simon and Mark. Whenever I do something like that I generally have to undo forever or restart.
I heard every word as you spoke it. Each digit you placed, I did poke it. Every move I did copy. I didn't get sloppy, Yet somehow or other, I broke it.
This is probably very elementary, but one of the most impressive things to me was your using prime factors to work out how to decide what digits were going to have to be on the ring. I realize, having watched you do it, that this is the most basic and easiest way to figure it out, but it was pure magic, in my admittedly non-mathematical eyes. Thanks so much, Simon, and thanks for the stream tonight, too!
@@leeway3739I don’t think it did. It was a 67 pair and there was a 7 in r9c9. Simon tends to ignore corner pencilmarks in his scanning so he was more likely to notice the actual digit
Though being Dutch, I often talk to myself nowadays in British English whenever solving a sudoku! By the way, I am wondering how someone from France would pronounce "bobbins"? ;-)
Great solve! I throughly enjoy back seat driving with Simon, luxuriating in him being in control and making the hard deductions, while retaining the right to shout at him along the way about small misses. Keep fighting the good fight ❤
My first comment on here and loving the content! This is a great solve Simon although I think that you were extremely lucky at 1:06:27 when you put 6s pencil marks into Box 6 (column 8) when you ment to put 3s. It transpires that 6 in column 8 was in box 6 but that was just luck anyway, loving the content as always
The sheer luck of having these wrong 6 pencilmarks instead of 3, but it works out because of the 7 having to be on the ring, meaning that R3C9 is indeed a 6. Grrrrr. That would never work out for me, and I would end up with a broken sudoku and restarting from zero.
To be fair, Simon is often very good at backtracking and figuring out where he went wrong and how to save it. Though in this case it doesn't appear he noticed.
A very well stated layman's definition of the Fundamental Theorem of Arithmetic: namely that every positive integer can be uniquely defined by its prime factors
18:01 "One in the Quintuple has to go here. And all of a sudden, we're actually getting a *Load of the Ring* filled in" This feels like a Lord of the Ring reference
Around 1:01:20, the way I progressed was using the logic that you almost figured out earlier. If the 1 goes on the whisper, it forces a 12 pair in box 3. Going a bit further, that pair forces a 34 pair at the bottom of the column, forcing r7c7 to be a 2, but it can't be because of the X-wing or arrow pencil mark, whichever you prefer.
37:00 Making that purple corner high doesn't change the number of other high digits required because it's forcing exactly one more high into the bottom right green square but also one onto the ring (if it's low there's still a high one at R9C8)
There was an incredible step in this puzzle which Simon didn’t notice (though his solve was brilliant). Once 6 could only go in one of 4 places in box 3 (2 in column 7, 2 in the red pair), this created a floating x-wing on 6s between boxes 3, 7 and 9 (the 6 in box 9 could only go in row 7 or column 7 and by transposing the pencil marks from the red pair, the same was true in boxes 7 and 3 respectively), ruling 6 out of the ring. Thus row 7 column 7 could not be low or the product would be at least one 2 or 3 short.
At the point you know to achieve atleast 10! With 5,7,8,9 and there are only 4 places where you can place digits. It must be 5789. Atleast i think. 5 and 7 cant be more divided and you cant make out of 2x2x2(8) and 3x3(9) two digits, well outside of 8 and 9. But all digits now cant be multiplicated with 2 or more, that would break Soduku.
I agree. Simon almost figured it out when he was counting high digits. If something goes down, something else goes up. The only wiggle room is the second digit on the R6C6 Arrow (which ends up being the multiplier of the factorial), and the only other high digit that might work is 6 in place of 9, but the 3 in column 7 nixes that.
Mindblowing. I got of with a really good start, having a good understanding in factorization. But then I got stuck, just as Simon did, but the irritating yet inspiring thing is that you constantly have the feeling of being very close to the next step.
Question for the people who watch both vids every day / do lots of variant sudokus on their own: Has there ever been a SET puzzle that required multiple applications of SET with different groups of cells? Seems like a cool idea, but I'm unaware of a puzzle that tries this.
I seem to remember them doing one that used both the standard Phistomefel ring and also the exploded Phistomefel ring in the outside columns/rows. But I don't know what it was called or who it was by, so there's little chance of me finding it.
48:21 I think was a repeat o a deduction from earlier. Just show how hard it is to keep all those deductions in your head while talking and making new deductions. So much easier to keep track when you are just watching.
Great solve, as usual, thank you, Simon! There is, however a glitch in logic at 1:10:06 when you put 6 into r3c9 following a typous pencil mark of 6 in row 8 of box 6 (that was a 3 pencil mark, not 6, see 1:06:09). The logic stays, however, as at this point it is known that red digits are a 67 pair and r9c9 is looking at r3c9, eliminating the 7 out of it.
1:03:03 finish. I got caught up for a few minutes going the wrong way when I tried to make the total exact, when all I needed was to make sure that it was at least the total. A fun puzzle!
It took me 222 minutes. I figured that at least two fives and at least one seven must exist. I also saw that r8c1 couldn't be big. I ended up having to bifurcate it at the end, but I feel like my logic leading to it was very solid. Fun Puzzle!
Brilliant puzzle. Although I initially misinterpreted the rule with the product and therefore struggled on the wrong path for a long time, I still managed it after ages.
Sudoku rules cannot apply then as one corner rules out 2 others. The Christmas puzzle last year, however, was a redundant puzzle (solve, erase some, solve again, erase some, etc) and ended up with 12 3s in corners...
67:56 Spent a bit of time doing set at the start (rows 4,6 & col 3,7) using the circles in box 5 and replacing them with their arrows. Gave the 1's in the ring early and helped later with there being some constraint on what numbers could be placed.
At 38 minutes is my favorite part of the solve: "I still need to put 7, 8, 9, and 5 in the ring" and also "I only have room for 3 high numbers and a 5." I'm excited to see how long Simon needs to put these 2 together!
It wasn't the digits 7, 8, 9 and 5 though. It was the prime factors of those numbers. For instance, the factors of 8 might have appeared as a 2 and a 4 in the ring.
There weren't enough digits left. There was the set (9,8,7,5,2,X), where X is the multiple of the factorial. 2 and X are needed for the arrow at R6C6 (X could be no bigger than 4), leaving 3 high digits for 3 spots. You *can* manipulate 9 and get 6, but 9 is needed in box 9 as soon as you place the 3. Once you need the 9, the rest is locked in.
@@crazypantaloons The clear inference in the way you stated your opening comment was that the "7,8,9,5" that still needed to be dealt with, exactly fit with "I only have room for three high numbers and a 5" statement, even though they relate to different things. You implied Simon was slow to match the two up. The X (multiple of 10!) could have been 1, so that wouldn't have had to go in the ring, and wouldn't have taken a spot on the arrow.
@@RichSmith77 But you need 3 high digits. You can't steal a factor of 2 or 3 from 7 or 8 and still have a high digit. You CAN remove a 3 and add a 2 to 9 to get 6, but you have to place a 9 long before that's ever important.
@@crazypantaloons At 38 minutes he's still some way from determining that he needs three high digits. The polarity along the German Whisper is still undecided.
I got really stuck with his deduction at 53:56 where he said the only place 3 would go in a 1-2-3 triple would be R2C7, which would break the consecutive clue. I could not for the life of me figure out his logic (and maybe he proved it elsewhere), but I want to explain what got me to accept this. For me, it was convincing myself that "2" could not be in R2C7, because it would break the ring. We know there's a 2 in each arrow on the right side of the ring, so we could not possibly put it there. Putting the 2 in the consecutive dot would be impossible, since it would need a 1 or 3, both of which are along the arrow. The only possibilty then would be to put 2 in the middle of the arrow, forcing the 1 at the top of it, and thus breaking the consecutive clue again. Therefore, a 1-2-3 combo is impossible, and 6 must be removed as a possible sum for that arrow. Anyhoo, great solve, and I love a puzzle like this were multiple deductions can be made!
Isn't his logic simply that he has r2c7 pencil marked with candidates that don't include 1 or 2? (1 was ruled out by 1 in the column, and 2 by the x-wing of 2s in columns 6 and 7 that he pointed out earlier.)
Is the Phistomefel rule shiftable? For instance, instead of the green c1/c2/c8/c9 plus yellow would it hold true for c2/c3/c7/c8 plus the same yellow with the result being a different set of matching digits? This wouldn’t produce a nice ring but would the logic hold true?
Yes. That is the whole idea of SET theory. Another example can be found in for instance ruclips.net/video/-srsmgIv2Bg/видео.html (and loads of other videos).
When Simon eliminated the 6 from the arrow's circle in box three, at that point there were now no 6's possible on the Phistomefel Ring. Not that it drives any value in the solve, but just found it curious to not use that.
Might it be possible to have a notepad overlay the rules-portion of the screen and then you could use that as the aide-memoir? It is very intriguing to see some of the thoughts and reasonings written down, but having it in the grid can be confusing. The rules are already displayed in the intro, as well as being written in the description of the video. Thank you for countless hours of enjoyment and entertainment.
Edit: just remembered what Simon pointed out early... Of course the product of the ring could be more than 10! as long as it includes all the factor. Forgot about that. Did we not know much earlier that the cell on the German whisper and in the ring is low? The two from the arrow goes with the 5 to make 10, and then the 3 or 4 makes one of the digits 8 or 9 together with the German whisper number and 7 goes with the second 8/9 in the ring
You totally forgot about title of puzzle at some point. It used to be 10! and you still had to deal with 7,8,9 to put in circle and there were only 3 spaces left. And you were thinking for over 20 minutes if bottom right corner can be low polarity or high while it HAD to be one of remaining 7,8 or 9. But somehow you solved it on your own way anyway.
It has to be divisible by 10! (10 factorial, so 10*9*8*7*6*5*4*3*2*1). It therefore has to be divisible by 10*5, or 2*5*5, so it has to be divisible by 5*5.
@@tessabrisac7423 After a good night's sleep I woke up and realized I had been thinking the product needed to be divisible by 1-10 seperately, not all at once. Thank you for your response.
I got the "easy" part of the break in but could not make any headway in a reasonable amount of time and just watched the video. Really hard puzzle. Interesting idea though.
@The confusing thing is that 6 is a triangular number. So saying the triangular number 6 could easily mean 6. You probably meant the sixth triangular number, rather than the triangular number 6.
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@@RichSmith77 Kind of a puzzle itself. 6 times 6 factorial squared is less than 10 factorial. To be transparent I should have written 6th (non-zero) triangular number. I think it´s interesting that 6! * 7! = 10!
I feel so silly now, I kept messing up because I erroneously thought the whisper had to be low at both ends to get to 10! but really as long as the prime factors were accounted for, any further digits are therefore a multiple of 10! and I never considered r7c7 was high. I feel so foolish even though the rules never said = I thought it was being coy
I've been screaming about the restriction of having at most 3 digits repeating 4 times on the Phistomefel ring which limits the possible digits on the ring, and is especially limiting with a 2 or 3 being in the bottom left corner of the ring for a good 20 minutes of the video..
Is this some weird challenge.. ? ??Make Simon explain the maximum variant sudoko rules.. The Secret, and atleast 1 X in the corner to make him sing..(?)
haven't watched yet, but my thought is, that 2^8 and 3^ 4 and 5 5 7, that's 15 digits, and the Ring is at least 16. And you can put at most a single 2 and single 3, on every side of the square. So that's 11 primes at best, and you can use leftover 2222 to make 2 other, but there should also be at least three 1s.
Apart from it was divisible by 10!, not 10! exactly. So you could have up to four 5s and up to four 7s scattered around the ring. There may not have been any 1s, if just using the divisible by 10! rule.
Yeah, I solved it, but I skipped through the video to see r7c7 had to be a high digit and don't quite understand the logic. I'm going to re-watch that section of the video. Not recording my time for this one...
If R7C7 is low, then it forces a 234 triple in the bottom right corner of the Phistomefel ring. However, being the product must be divisible by 10!, if we divide out the digits we know are in the ring, we are left with 2x3x5x7, thus the only way to accomplish filling in the missing digits is 5, 6, and 7. This forces the 2 “high” cells in column 8 to be a 67 pair, but neither are able to be a 6, making it impossible. Hope this helps!
Simon, you spent a lot of time about which factors are on the ring, without thinking of two simple limits: 1. You can't put any digits more than 4 times on the ring, so there can't be that many 2s on the ring 2. You have only 16 cells on the ring That together with the digits you can easily place (like the 1-5 in row 3) give you very little wiggle room and if you think it thorugh once more you can easily see the German whispers line has 3 high digits, not 3 lows. I spare myself explaining the thought process in detail, but the 2 limits are helpful in it and the one thing you didn't do is count the digits.
Of course you're right that the whisper line has 3 high digits, but I don't see how what you've said in your comment proves this. If the line has 3 low digits, you can still get a product of exactly 10! by using 5,6, and 7 in remaining squares that must be high digits. Obviously this ends up not working in the end. If you fill out the two corner squares that must be high digits and follow some straightforward logic, you eventually run into a problem where the 4-length arrow has a 5 in its top-right square but cannot have a 2 in the top left square, so the arrow sums to at least 10 and you get a contradiction. But this is far from obvious, unless I'm missing some more straightforward deduction.
@@frankthompson9630 just 2 observatiopns to start with: You pretty soon find out you have four 1s on the ring. That only leaves place for 12 other digits. You have a 23 in a corner cell, which reduces the possible number of either 2s or 3s to only three of them at max. You need a 7 and two 5s, as Simon also very early on. Continue along these lines of thinking and you find r7c7 is from 5789 and of course no 5. I didn't just do the counting, I also put in candidates before I had the 12 digits and used a bit of sudoku scanning to get there. And yes, I already said this is just a sketch and so a tip on how to tackle this, no proof, but just htink about it and you get there much more straight foward and still thoroughly.
@@OlafDoschke I think you're leaving out some crucial steps. How do you know that r7c7 cannot be 3 or 4? You can't simply claim that it's 5, 7, 8, or 9. Consider this way of filling out the ring, where parentheses indicate pencil marks: (45) (23) 1 (2345) (2345) (567) xxxxxxxxxxxx 1 (23) xxxxxxxxxxxxx (567) 1 xxxxxxxxxxxxxxxx (23) (23) 1 (567) (234) (34) Here the (23), (234), (34) cells in the bottom right are all different, as are the three (567) cells. The product of all these cells is exactly 10!, and so it is divisible by 10! How do you know that this is not a valid way to fill out the ring? As I said in my previous comment, you can get there without too much hassle. Notice that you only have two high digits in the ring, and both r1c8 and r9c8 must be high digits. The only possibility is r9c8 = 7 and r1c8 = 6. From there you can eventually deduce that r3c7 = 5 and r7c6 = 2, which means that r3c6 >= 3, and so the r4c6 arrow is broken because it sums to at least 10. But this chain of logic is quite long and uses a lot of elements in the grid (even including the white kropki dot). I don't think it's at all straightforward to do it in your head. Unless I'm missing a separate way to disprove this configuration, I think your logic is not sufficient here.
@@frankthompson9630I agree. This is very close to the arrangement around the ring that Simon looks at around the 1:00:00 mark in the video (apart from he has his second 5 placed in r4c3). I don't think there was an easier way to rule this option out, than what Simon does here.
That's true, but there are other ways to get to a multiple of 10! as well, so you can't use that as part of the solve. The solution here didn't end up including a 6 in the ring!
It seems like you're making the "no jokes about me at parties" joke in every video these days. At this point it's not really funny anymore, just depressing
You no longer need to waste video time explaining the ring anymore. Link to the numberphile video in the description. You have a video proof that only requires you to provide a link. Lucky for you this means you don't need to waste anymore of your life explaining it anymore.
Or you could skip it and let them keep making content that's accessible to new viewers of the channel (telling a first time viewer to watch another video as homework to be able to watch the video they clicked on is a surefire way to lose that viewer)
@@HunterJE Agreed. Remember there are lots of folks not familiar with Sudoku at all and taking a moment to explain the basic rules and rules commonly found in variant Sudoku allows them to understand what is going on in solving the puzzle.
Puzzle creator here 🤓 wonderful solve Simon! And thank you for the kind words, it was a honor to be featured on the channel!
The key to that puzzle was ruling out that R7C7 could not be low polarity, which you did beautifully and as intended. That was, truly and honestly, my first Sodoku puzzle I’ve created and I very much enjoyed making it and watching you solve it. I’ve loved the concept of the Phistomefel ring and wanted to put my own math twist on it. Hope you enjoyed!
Here’s to hoping I’ll be back for more sometime soon…
Very nice first puzzle.
You are currently a one hit wonder.
I haven’t seen the video yet but having somehow made it through, I’m blown away by some of the logic in this. Particularly the interplay between boxes 3, 7 and 9 (which if intended was quite simply a masterstroke of a sudoku step). It was a puzzle where it was difficult to know what to focus on but looking from the right angle revealed some absolute magic. Thanks for coming up with it.
Wonderful debut!! Lovely setting and look forward to more from you!!
@@davidrattner9 ditto!! 👏🏻👏🏻👏🏻
Thank you very much for the ‘End of Phistomefel Theorem Proof’ timestamp 👏🏻
Co-signed
He just needs the link to Numberphile
The moment Simon put those sixes pencilmark in column 8, saying that it is threes. I wonder when it will blow. But in the end it ended correct pencilmark. And yeah 10886400 / 10! = 3 :) Nice.
Same here... I spent four minutes shouting at the screen, and Simon completely ignored me.
@@greghill7759😁I know the feeling well!
@@greghill7759 He also ignores corner pencil marks!
I knew someone else would notice the "so 3s has to be here" -> corner marks a 6. I also wondered when it would blow up. I don't know how things like that work out in the final solution for Simon and Mark. Whenever I do something like that I generally have to undo forever or restart.
@@gordmain5370well you might think so because you don't record your solves, I bet it happens to all of us sometimes
The wrong 6 pencilmarked at 01:06:11 gave me extreme anxiety lmao
I heard every word as you spoke it.
Each digit you placed, I did poke it.
Every move I did copy.
I didn't get sloppy,
Yet somehow or other, I broke it.
beautiful poem xd
Fun fact.😊 10! Seconds is exactly 6 weeks. Thanks for a great Channel.
Just another fun fact 🙂: 6 factorial times 7 factorial is exactly 10 factorial, which seems quite a happy coincidence!
@@penningmeestercgkdelft9159this means that there are 5! 7! seconds in a week and 5! 6! seconds in a day
And that's how long I'd take to solve it! Crazy coincidence.
This is probably very elementary, but one of the most impressive things to me was your using prime factors to work out how to decide what digits were going to have to be on the ring. I realize, having watched you do it, that this is the most basic and easiest way to figure it out, but it was pure magic, in my admittedly non-mathematical eyes. Thanks so much, Simon, and thanks for the stream tonight, too!
Wonderfully written from you as always!! Great to see you tonight on stream. 😀
I was worried when Simon typoed a 6 into column eight when he meant to type a 3, but that actually ended up being true, so no harm done. Phew!
It got him the 6 in box three earlier than I think is deserved, but I could be wrong.
It sort of did but the logic still held because the 9/8 in r1c7/8 ruled out everything except 6 in r3c9 which only left 7 for r3c8
Yeah, noticed this as well. He could've (and probably may have) resolved the 67 pair using the 7 in r9c9.
@@leeway3739I don’t think it did. It was a 67 pair and there was a 7 in r9c9. Simon tends to ignore corner pencilmarks in his scanning so he was more likely to notice the actual digit
42:15 we literally JUST said that square can't be 4 simon 😆 😆 😆
"The reason I think it matters, is because it doesn't matter..." *pause for dramatic effect* "...what polarity this cell has."
I am a victim of Simarkisms….I keep saying bobbins now in my daily life even though I m french🤣
It s so disturbing but very funny
Though being Dutch, I often talk to myself nowadays in British English whenever solving a sudoku!
By the way, I am wondering how someone from France would pronounce "bobbins"? ;-)
Great solve! I throughly enjoy back seat driving with Simon, luxuriating in him being in control and making the hard deductions, while retaining the right to shout at him along the way about small misses.
Keep fighting the good fight ❤
1:11:05 "Ah, that's a" = Alexa. Your phone is hearing the vowel sounds, and guessing.
i would never ever have one of those things active. So intrussive and a violation of my privacy...
Rules: 04:03
Let's Get Cracking: 06:09
Simon's time: 1h8m54s
Puzzle Solved: 1:15:03
What about this video's Top Tier Simarkisms?!
Phistomefel: 9x (00:35, 00:41, 05:56, 06:18, 06:22, 18:38, 35:36, 37:10, 1:01:16)
Bobbins: 7x (22:08, 22:08, 22:08, 22:15, 29:16, 36:56, 1:12:31)
Three In the Corner: 7x (1:09:10, 1:09:27, 1:09:39, 1:09:41, 1:11:20, 1:11:31)
The Secret: 1x (34:35)
And how about this video's Simarkisms?!
Ah: 24x (15:59, 15:59, 16:02, 18:22, 25:50, 27:06, 27:55, 28:58, 29:13, 36:53, 36:53, 36:56, 38:07, 38:07, 42:13, 43:40, 44:14, 49:07, 49:07, 51:45, 54:17, 1:02:51, 1:06:52, 1:12:31)
Obviously: 8x (13:38, 15:07, 20:17, 21:02, 29:29, 37:19, 43:07, 51:37)
By Sudoku: 7x (28:00, 37:21, 1:06:11, 1:10:05, 1:11:33, 1:14:27)
Sorry: 6x (28:22, 39:58, 43:04, 43:24, 59:35, 1:05:00)
Brilliant: 6x (00:51, 01:56, 03:34, 1:15:01, 1:15:12, 1:16:37)
Hang On: 5x (11:07, 11:07, 11:08, 43:40, 1:15:42)
In Fact: 5x (44:41, 48:54, 1:05:59, 1:06:52, 1:09:49)
Wow: 5x (34:01, 53:33, 55:49, 57:17, 1:15:15)
Nature: 4x (10:13, 10:16, 34:41, 34:45)
Weird: 4x (29:58, 30:32, 44:54, 1:02:16)
Bother: 3x (41:46, 54:24, 54:29)
Ridiculous: 3x (27:19, 1:06:28, 1:11:05)
Good Grief: 2x (1:01:32, 1:15:12)
Goodness: 2x (1:15:01, 1:16:21)
Naked Single: 2x (1:12:52, 1:14:01)
Clever: 2x (56:47, 1:05:13)
In the Spotlight: 2x (1:09:30, 1:11:26)
What Does This Mean?: 2x (09:41, 1:08:08)
Pencil Mark/mark: 2x (30:36, 50:47)
Cake!: 2x (03:09, 03:34)
Useless: 1x (27:00)
What a Puzzle: 1x (1:14:54)
The Answer is: 1x (12:55)
Missing Something: 1x (46:11)
I Have no Clue: 1x (43:38)
Lovely: 1x (22:53)
Beautiful: 1x (38:07)
Fascinating: 1x (1:16:25)
Incredible: 1x (06:54)
Going Mad: 1x (46:18)
Hypothecate: 1x (54:08)
Take a Bow: 1x (1:15:15)
Shouting: 1x (01:11)
Famous Last Words: 1x (1:10:12)
Magnificent: 1x (1:05:06)
Surely: 1x (1:06:23)
Box Thingy: 1x (1:02:08)
That's Huge: 1x (1:00:08)
Thingy Thing: 1x (1:02:08)
Triangular Number: 1x (16:43)
Most popular number(>9), digit and colour this video:
Ten (38 mentions)
Two (160 mentions)
Red (23 mentions)
Antithesis Battles:
High (54) - Low (32)
Even (4) - Odd (2)
Higher (11) - Lower (1)
White (4) - Black (1)
Row (16) - Column (11)
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!
Is your code/model published somewhere? I'd be very curious to have a look!
My first comment on here and loving the content! This is a great solve Simon although I think that you were extremely lucky at 1:06:27 when you put 6s pencil marks into Box 6 (column 8) when you ment to put 3s. It transpires that 6 in column 8 was in box 6 but that was just luck anyway, loving the content as always
I have scrabble bags under my eyes.
Debuts on this channel and Simon working his magic is always a treat for us!!
The sheer luck of having these wrong 6 pencilmarks instead of 3, but it works out because of the 7 having to be on the ring, meaning that R3C9 is indeed a 6. Grrrrr. That would never work out for me, and I would end up with a broken sudoku and restarting from zero.
EXACTLY
I was waiting for the monebt it failed because of that lol
To be fair, Simon is often very good at backtracking and figuring out where he went wrong and how to save it. Though in this case it doesn't appear he noticed.
“The 3 has to be here” *puts in a 6 pencilmark
A very well stated layman's definition of the Fundamental Theorem of Arithmetic: namely that every positive integer can be uniquely defined by its prime factors
What a puzzle... I would never be able to start this. Thanks for the solve 😀
18:01 "One in the Quintuple has to go here. And all of a sudden, we're actually getting a *Load of the Ring* filled in"
This feels like a Lord of the Ring reference
Around 1:01:20, the way I progressed was using the logic that you almost figured out earlier. If the 1 goes on the whisper, it forces a 12 pair in box 3. Going a bit further, that pair forces a 34 pair at the bottom of the column, forcing r7c7 to be a 2, but it can't be because of the X-wing or arrow pencil mark, whichever you prefer.
37:00 Making that purple corner high doesn't change the number of other high digits required because it's forcing exactly one more high into the bottom right green square but also one onto the ring (if it's low there's still a high one at R9C8)
“This digit is either 4 or 5”
There was an incredible step in this puzzle which Simon didn’t notice (though his solve was brilliant). Once 6 could only go in one of 4 places in box 3 (2 in column 7, 2 in the red pair), this created a floating x-wing on 6s between boxes 3, 7 and 9 (the 6 in box 9 could only go in row 7 or column 7 and by transposing the pencil marks from the red pair, the same was true in boxes 7 and 3 respectively), ruling 6 out of the ring. Thus row 7 column 7 could not be low or the product would be at least one 2 or 3 short.
At the point you know to achieve atleast 10! With 5,7,8,9 and there are only 4 places where you can place digits. It must be 5789. Atleast i think. 5 and 7 cant be more divided and you cant make out of 2x2x2(8) and 3x3(9) two digits, well outside of 8 and 9. But all digits now cant be multiplicated with 2 or more, that would break Soduku.
I agree. Simon almost figured it out when he was counting high digits. If something goes down, something else goes up. The only wiggle room is the second digit on the R6C6 Arrow (which ends up being the multiplier of the factorial), and the only other high digit that might work is 6 in place of 9, but the 3 in column 7 nixes that.
Mindblowing. I got of with a really good start, having a good understanding in factorization. But then I got stuck, just as Simon did, but the irritating yet inspiring thing is that you constantly have the feeling of being very close to the next step.
Question for the people who watch both vids every day / do lots of variant sudokus on their own: Has there ever been a SET puzzle that required multiple applications of SET with different groups of cells? Seems like a cool idea, but I'm unaware of a puzzle that tries this.
I seem to remember them doing one that used both the standard Phistomefel ring and also the exploded Phistomefel ring in the outside columns/rows. But I don't know what it was called or who it was by, so there's little chance of me finding it.
Same here: I remember there has been at least one, months ago, but have no idea how to find it.
48:21 I think was a repeat o a deduction from earlier. Just show how hard it is to keep all those deductions in your head while talking and making new deductions. So much easier to keep track when you are just watching.
Great solve, as usual, thank you, Simon! There is, however a glitch in logic at 1:10:06 when you put 6 into r3c9 following a typous pencil mark of 6 in row 8 of box 6 (that was a 3 pencil mark, not 6, see 1:06:09). The logic stays, however, as at this point it is known that red digits are a 67 pair and r9c9 is looking at r3c9, eliminating the 7 out of it.
Another amazing puzzle in a week of amazing puzzles!
The constant stream of amazing puzzles that we see is something to behold. 😁
@@davidrattner9 💕💕
Missing the scramble tiles in the bags ❤
Me, too.
@@ElizabethRoss-uj8rlditto!!
1:03:03 finish. I got caught up for a few minutes going the wrong way when I tried to make the total exact, when all I needed was to make sure that it was at least the total. A fun puzzle!
Not sure I can get through this video if Simon is going to keep saying "factorial" when he doesn't mean it
Okay, i feel way less bad on giving up on this seeing how much Simon struggled half way through not knowing what to do and where to continue either
It took me 222 minutes. I figured that at least two fives and at least one seven must exist. I also saw that r8c1 couldn't be big. I ended up having to bifurcate it at the end, but I feel like my logic leading to it was very solid. Fun Puzzle!
3 hours for me 😂 it's an easy start, but then I got stuck for about 2 hours! I'm super proud I got it done!!! 😁
Brilliant puzzle. Although I initially misinterpreted the rule with the product and therefore struggled on the wrong path for a long time, I still managed it after ages.
😱😱😱 I’ve never been this early!!! Thanks for the video Simon- your solves are always so entertaining while I work and cook
I want to see a constructor find a ruleset where there will be 3s in all four corners so we get to hear Simon sing 4 times in one video 😆
ruclips.net/video/LIiST_VvbCk/видео.htmlsi=k-l6im9C6p5Ih3TQ
Even Better
i don't remember which, but i definitely already saw a video where that was the case
Sudoku rules cannot apply then as one corner rules out 2 others. The Christmas puzzle last year, however, was a redundant puzzle (solve, erase some, solve again, erase some, etc) and ended up with 12 3s in corners...
@@icepyrox yeah that was it!
67:56 Spent a bit of time doing set at the start (rows 4,6 & col 3,7) using the circles in box 5 and replacing them with their arrows. Gave the 1's in the ring early and helped later with there being some constraint on what numbers could be placed.
At 38 minutes is my favorite part of the solve: "I still need to put 7, 8, 9, and 5 in the ring" and also "I only have room for 3 high numbers and a 5." I'm excited to see how long Simon needs to put these 2 together!
It wasn't the digits 7, 8, 9 and 5 though. It was the prime factors of those numbers. For instance, the factors of 8 might have appeared as a 2 and a 4 in the ring.
There weren't enough digits left. There was the set (9,8,7,5,2,X), where X is the multiple of the factorial. 2 and X are needed for the arrow at R6C6 (X could be no bigger than 4), leaving 3 high digits for 3 spots. You *can* manipulate 9 and get 6, but 9 is needed in box 9 as soon as you place the 3. Once you need the 9, the rest is locked in.
@@crazypantaloons The clear inference in the way you stated your opening comment was that the "7,8,9,5" that still needed to be dealt with, exactly fit with "I only have room for three high numbers and a 5" statement, even though they relate to different things. You implied Simon was slow to match the two up.
The X (multiple of 10!) could have been 1, so that wouldn't have had to go in the ring, and wouldn't have taken a spot on the arrow.
@@RichSmith77 But you need 3 high digits. You can't steal a factor of 2 or 3 from 7 or 8 and still have a high digit. You CAN remove a 3 and add a 2 to 9 to get 6, but you have to place a 9 long before that's ever important.
@@crazypantaloons At 38 minutes he's still some way from determining that he needs three high digits. The polarity along the German Whisper is still undecided.
I got really stuck with his deduction at 53:56 where he said the only place 3 would go in a 1-2-3 triple would be R2C7, which would break the consecutive clue. I could not for the life of me figure out his logic (and maybe he proved it elsewhere), but I want to explain what got me to accept this.
For me, it was convincing myself that "2" could not be in R2C7, because it would break the ring. We know there's a 2 in each arrow on the right side of the ring, so we could not possibly put it there. Putting the 2 in the consecutive dot would be impossible, since it would need a 1 or 3, both of which are along the arrow. The only possibilty then would be to put 2 in the middle of the arrow, forcing the 1 at the top of it, and thus breaking the consecutive clue again. Therefore, a 1-2-3 combo is impossible, and 6 must be removed as a possible sum for that arrow.
Anyhoo, great solve, and I love a puzzle like this were multiple deductions can be made!
Isn't his logic simply that he has r2c7 pencil marked with candidates that don't include 1 or 2?
(1 was ruled out by 1 in the column, and 2 by the x-wing of 2s in columns 6 and 7 that he pointed out earlier.)
1:06:09 sheer luck?
Is the Phistomefel rule shiftable? For instance, instead of the green c1/c2/c8/c9 plus yellow would it hold true for c2/c3/c7/c8 plus the same yellow with the result being a different set of matching digits? This wouldn’t produce a nice ring but would the logic hold true?
Yes. That is the whole idea of SET theory. Another example can be found in for instance ruclips.net/video/-srsmgIv2Bg/видео.html (and loads of other videos).
When Simon eliminated the 6 from the arrow's circle in box three, at that point there were now no 6's possible on the Phistomefel Ring. Not that it drives any value in the solve, but just found it curious to not use that.
Might it be possible to have a notepad overlay the rules-portion of the screen and then you could use that as the aide-memoir? It is very intriguing to see some of the thoughts and reasonings written down, but having it in the grid can be confusing. The rules are already displayed in the intro, as well as being written in the description of the video. Thank you for countless hours of enjoyment and entertainment.
71:35 for me. Nice puzzle!
1:11:00 Please don't say that you have to turn her off while she's listening. Seen too many horror films and shows where this goes wrong.
I always feel like he should use X in the box for 10. Seeing 1 in the corner and 0 in the box messes with my eyes, lol. #SuperMinorFeedback
Edit: just remembered what Simon pointed out early... Of course the product of the ring could be more than 10! as long as it includes all the factor. Forgot about that.
Did we not know much earlier that the cell on the German whisper and in the ring is low? The two from the arrow goes with the 5 to make 10, and then the 3 or 4 makes one of the digits 8 or 9 together with the German whisper number and 7 goes with the second 8/9 in the ring
You totally forgot about title of puzzle at some point. It used to be 10! and you still had to deal with 7,8,9 to put in circle and there were only 3 spaces left. And you were thinking for over 20 minutes if bottom right corner can be low polarity or high while it HAD to be one of remaining 7,8 or 9. But somehow you solved it on your own way anyway.
11:35 the fundamental theorem of arithmetic.
Why does the ring need two 5s for the product to be divisible by 10?
It has to be divisible by 10! (10 factorial, so 10*9*8*7*6*5*4*3*2*1). It therefore has to be divisible by 10*5, or 2*5*5, so it has to be divisible by 5*5.
It must be divisible by the product of 5 x 10 , so at least two factors 5.
@@tessabrisac7423 After a good night's sleep I woke up and realized I had been thinking the product needed to be divisible by 1-10 seperately, not all at once. Thank you for your response.
There seems to be a problem with the chapters
Rules: 4:02
rules @ 3:39
I got the "easy" part of the break in but could not make any headway in a reasonable amount of time and just watched the video. Really hard puzzle. Interesting idea though.
I struggled a lot with this puzzle because I thought the rule was that the box multiplied to 10! rather than "is divisible"
Note that triangular number 6 is 21=3*7.
Help! What am i missing?
4x3x2x5x7x2x8x3x9x2x3=
2177280 is not divisible by 10!
(3628800) what am i not understanding?
You missed a 5. The product of the ring totals 10,886,400.
10,886,400 / 10! = 3
You're missing a 5. From top left clockwise (ignoring 1s): 4x3x2x5x7x2x8x3x9x2x3x5
Also 6!=8*9*10 in value.
'panoply' and 'quorate' in the same sentence? Did you get a 'word of the day' calendar for Christmas? lol
Great puzzle! The product is (spoiler warning) Triangular number 6 times 6 factorial squared.
Uh - I don't think so - 6! does not contain 7. So 6(6!)(6!) cannot be 3(10!).
@@annek3296 Trianglar number 6 is 21 which contains 7.
Triangular it shall be above.
@The confusing thing is that 6 is a triangular number. So saying the triangular number 6 could easily mean 6. You probably meant the sixth triangular number, rather than the triangular number 6.
@@RichSmith77 Kind of a puzzle itself. 6 times 6 factorial squared is less than 10 factorial. To be transparent I should have written 6th (non-zero) triangular number. I think it´s interesting that 6! * 7! = 10!
3x10! indeed!
Help what am i missing!4x3x2x5x7x2x8x3x9x2x3=
2177280 is not divisible by 10! (3628800) what am i not understanding?
You missed a 5. The product of the ring totals 10,886,400.
10,886,400 / 10! = 3
the answer is 3,628,800
I feel so silly now, I kept messing up because I erroneously thought the whisper had to be low at both ends to get to 10! but really as long as the prime factors were accounted for, any further digits are therefore a multiple of 10! and I never considered r7c7 was high. I feel so foolish even though the rules never said = I thought it was being coy
What is 230 - 220 x 0.5? You may not believe it, but the answer is 5!
Note 6*6!*6!
I've been screaming about the restriction of having at most 3 digits repeating 4 times on the Phistomefel ring which limits the possible digits on the ring, and is especially limiting with a 2 or 3 being in the bottom left corner of the ring for a good 20 minutes of the video..
Is this some weird challenge.. ? ??Make Simon explain the maximum variant sudoko rules.. The Secret, and atleast 1 X in the corner to make him sing..(?)
You should see the "Everything is Wrogn" and "2 truths and a lie" series of puzzles. Those boards can be filled with variant rules stuck together.
haven't watched yet, but my thought is, that 2^8 and 3^ 4 and 5 5 7, that's 15 digits, and the Ring is at least 16. And you can put at most a single 2 and single 3, on every side of the square. So that's 11 primes at best, and you can use leftover 2222 to make 2 other, but there should also be at least three 1s.
Apart from it was divisible by 10!, not 10! exactly. So you could have up to four 5s and up to four 7s scattered around the ring. There may not have been any 1s, if just using the divisible by 10! rule.
There were multiple moments where I assumed the product has to be exactly 10!, which isn't true. Really annoying.
49:30 for me. I got completely stuck for way too long, but at least I managed to finish it eventually.
Yeah, I solved it, but I skipped through the video to see r7c7 had to be a high digit and don't quite understand the logic. I'm going to re-watch that section of the video. Not recording my time for this one...
If R7C7 is low, then it forces a 234 triple in the bottom right corner of the Phistomefel ring. However, being the product must be divisible by 10!, if we divide out the digits we know are in the ring, we are left with 2x3x5x7, thus the only way to accomplish filling in the missing digits is 5, 6, and 7. This forces the 2 “high” cells in column 8 to be a 67 pair, but neither are able to be a 6, making it impossible.
Hope this helps!
38:40 for me
bobbins
KIND
Simon, you spent a lot of time about which factors are on the ring, without thinking of two simple limits:
1. You can't put any digits more than 4 times on the ring, so there can't be that many 2s on the ring
2. You have only 16 cells on the ring
That together with the digits you can easily place (like the 1-5 in row 3) give you very little wiggle room and if you think it thorugh once more you can easily see the German whispers line has 3 high digits, not 3 lows.
I spare myself explaining the thought process in detail, but the 2 limits are helpful in it and the one thing you didn't do is count the digits.
Of course you're right that the whisper line has 3 high digits, but I don't see how what you've said in your comment proves this.
If the line has 3 low digits, you can still get a product of exactly 10! by using 5,6, and 7 in remaining squares that must be high digits.
Obviously this ends up not working in the end. If you fill out the two corner squares that must be high digits and follow some straightforward logic, you eventually run into a problem where the 4-length arrow has a 5 in its top-right square but cannot have a 2 in the top left square, so the arrow sums to at least 10 and you get a contradiction.
But this is far from obvious, unless I'm missing some more straightforward deduction.
@@frankthompson9630 just 2 observatiopns to start with: You pretty soon find out you have four 1s on the ring. That only leaves place for 12 other digits.
You have a 23 in a corner cell, which reduces the possible number of either 2s or 3s to only three of them at max. You need a 7 and two 5s, as Simon also very early on.
Continue along these lines of thinking and you find r7c7 is from 5789 and of course no 5. I didn't just do the counting, I also put in candidates before I had the 12 digits and used a bit of sudoku scanning to get there.
And yes, I already said this is just a sketch and so a tip on how to tackle this, no proof, but just htink about it and you get there much more straight foward and still thoroughly.
@@OlafDoschke I think you're leaving out some crucial steps. How do you know that r7c7 cannot be 3 or 4? You can't simply claim that it's 5, 7, 8, or 9.
Consider this way of filling out the ring, where parentheses indicate pencil marks:
(45) (23) 1 (2345) (2345)
(567) xxxxxxxxxxxx 1
(23) xxxxxxxxxxxxx (567)
1 xxxxxxxxxxxxxxxx (23)
(23) 1 (567) (234) (34)
Here the (23), (234), (34) cells in the bottom right are all different, as are the three (567) cells. The product of all these cells is exactly 10!, and so it is divisible by 10!
How do you know that this is not a valid way to fill out the ring? As I said in my previous comment, you can get there without too much hassle. Notice that you only have two high digits in the ring, and both r1c8 and r9c8 must be high digits. The only possibility is r9c8 = 7 and r1c8 = 6. From there you can eventually deduce that r3c7 = 5 and r7c6 = 2, which means that r3c6 >= 3, and so the r4c6 arrow is broken because it sums to at least 10.
But this chain of logic is quite long and uses a lot of elements in the grid (even including the white kropki dot). I don't think it's at all straightforward to do it in your head.
Unless I'm missing a separate way to disprove this configuration, I think your logic is not sufficient here.
@@frankthompson9630I agree. This is very close to the arrangement around the ring that Simon looks at around the 1:00:00 mark in the video (apart from he has his second 5 placed in r4c3). I don't think there was an easier way to rule this option out, than what Simon does here.
As long as the ring has at least one of each of 2-9, and an extra 2 and 5, then its product will be divisible by 10!
That's true, but there are other ways to get to a multiple of 10! as well, so you can't use that as part of the solve. The solution here didn't end up including a 6 in the ring!
Technically, *any* number is divisible by 10!. :) In this case, *evenly* divisible is assumed.
It seems like you're making the "no jokes about me at parties" joke in every video these days. At this point it's not really funny anymore, just depressing
You no longer need to waste video time explaining the ring anymore.
Link to the numberphile video in the description.
You have a video proof that only requires you to provide a link.
Lucky for you this means you don't need to waste anymore of your life explaining it anymore.
Or you could skip it and let them keep making content that's accessible to new viewers of the channel (telling a first time viewer to watch another video as homework to be able to watch the video they clicked on is a surefire way to lose that viewer)
@@HunterJE Agreed. Remember there are lots of folks not familiar with Sudoku at all and taking a moment to explain the basic rules and rules commonly found in variant Sudoku allows them to understand what is going on in solving the puzzle.
Garbage take. Especially when they timestamp the end of the Phistomefel explanation!