The Phistomefel Ring Is INCOMPLETE
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- čas přidán 13. 06. 2024
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** TODAY'S PUZZLE **
If we ever make another book, this puzzle needs to be in it. Jeff Wajes' Circles And Sets is so spectacularly clever that it's almost sentient! Prepare to be dazzled if you have a go at this one.
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Rules:
Normal sudoku rules apply. In each cage, digits do not repeat and sum to the given total. A digit in a circle indicates how many times that digit appears in circles.
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▶ Contents Of This Video ◀
0:00 Theme music & Puzzle intro
1:19 Hexcells stream TONIGHT
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Amazing that sudoku solving has come to a point where understanding the phistomofel ring is common knowledge enough to be used as a logic step for circles. It doesn’t feel too long ago where Phistomofel was worried his discovery was too esoteric to be ‘found’ in a puzzle at all. Now it’s almost second hand. I simply love the sudoku evolution these last few years.
Sans serif fonts are generally easier to read.
I agree - it was mind-blowing the first few times I heard Simon explain it. But now I just automatically think about it without any steam coming out of my ears or anything!
How does this pertain to the comment you are replying to?@@markmichell6015
I totally agree with everything you said. But it also seems that a lot of puzzles that incorporate it lately, "highlight" it, such as today's. Of course, if you have no idea what it is, then having the circles mostly outline it doesn't really help at all. I honestly can't imagine I ever would have figured out the logic of it by myself like Simon originally did in the first puzzle that featured it.
That puzzle, with the missing circles in the Phistomefel ring and squares - what a mind-bender for me to get the fact that those two digits were NOT the same! Your discovery that it was all about parity, Simon, was again a lesson to me to not overlook simple and straightforward options and lines of thought. Thank you so much for this video.
I normally don’t watch these long videos but the appearance of the Phistomefel ring tempted me and I can resist anything except temptation! Glad I watched. What a clever puzzle! Brilliantly solved too.
I ‘got’ this one straight away. Realised the odd digit pairing and I was off. And I remembered the amount of circles in conjunction with that. I must be getting better at these as a year ago, I would have stared at this forever and then just watched Simon do it. 😊. Well done Jeff.
Yes! I loved the logic in this one. It's the first puzzle I've seen on this channel that I've been able to solve without help. Usually the logic needed to start (or get un-stuck at the end of) these is just too much for my limited memory. I expect this will be my favorite of the sudoku puzzles I've done for a long time.
I don't get to say this very often (basically never) but I actually intuited something almost immedialty that Simon took a good 40+ minutes to get to (with to be fair a lot of explanation from him in the meantime): The uncircled Phistomefel cells have to be two different odd numbers. And from there its a very short jump to it has to be 7 in the corner with 3 or 5 in the other spot.
Simon sort of backed into once he got the numbers but never really laid it out as he was solving.
Any number in a Phistomefel area has to appear an even number of times. Same number of times in green as in orange and any number added to itself is always even. So for any odd number to appear in Phistomephel area would need an odd number of instances in uncircled cells. In this puzzle, exactly one. An even number in Phisto couldn't be uncircled unless it was the same number in both, so there is either 0 or 2 odd numbers in the Phisto areas. Well, since the circle total needs to be 30 and 2+4+6+8 is only 20, the uncircled cells must be two different odd numbers with a total of at least 10. We know 9 is out because you can't have an uncircled 9, which means 1 is also out, or we cant get to 30 with only 2 odd numbers.
So we have to have a 7 and then either 2348 or 458 and for the reasons Simon gives 7 goes in the top and 35 in the other uncircled spot.
though you just mentioned the 2 odd numbers have to equal 10 so logically, if 7 must be in then the other has to be 3
@@flameritsu7631 2, 4, 6, and 8 aren't all guaranteed to be in at that point. You actually need to exclude 15 pts worth of an unknown number of digits. 9 and 1 are two of the digits so you're looking to exclude 5 more pts which must include another odd digit. Your choices at that point are either 2,3 or 5.
I loved this one. The realization about the odd numbers was just incredible.
This was what got me started. Then I added up the even numbers, which gives you 20 circles, so you need to achieve another 10 with odd numbers. As 9 was missing, the only possibility was 7 plus 3, which had to go into the two unmarked colored boxes, to complete the ring. Then I tried the same as Simon with on the inside ring, and had the 7 outside and the 3 inside in a quarter of an hour.
But later on, I got stuck for nearly two hours and needed this Schoedinger Cell hint from the video, to complete the rest.
what a creative breakthrough, love how it still feels intuitive and not just some kind of magic
simon's brain jumping through hoops to prove that 5 wasn't circled and 2&3 had to be when there was already a circled 2 in the puzzle is somehow a marvel seen time and time again on this channel but never fails to incite bewilderment
And those complicated chains are surely day by day increasing my frustration that they take away from the beauty of the puzzle ..
The constructors have to be generous and thank him for featuring their puzzle but deep down they would be HURTING!
@@ruckmanikrishnan4221 I hear you, but I think you're overstating a bit. ;) It's a rare puzzle (maybe close to nonexistant?) where anyone follows the setter's path perfectly.
That said, I _did_ want him to see the circled '2'.
Gotta say, that's one of the best-set puzzles I've seen in a good while. Excellent work, and a fun solve.
First-look thoughts - I think you can have at most two odd numbers in circles - since phistomefel digits are all doubled you can only have an odd number if it also appears in one of the blanks... :edit: Pleasing to see that is indeed the sort of place the solve goes
That is what I was thinking.
Definitely, and 7 is the one to start with (after 8 in box 5)
And the the question is if 1, 3 or 5 is the last odd in the circles. We are missing six in total, and two odd digits. 51 works, as do 123 so we know for a fact that the last odd digit is a 3 or a 5.
@@57thorns and then knowing that the only combinations that work are 45678 and 234678 and its easy to prove the set has to have greater than 5 digits, so its a 3 and the set is 234678.
@@calculatrguy Did I forget about the 9 going on vacation from 45?
Yes, but it was much easier to deduct 3 and 7 as all even numbers added are 20, so we miss 10 to get to 30 with two odd numbers without using 9. So it must be 3 and 7. qed.
I usually don't do puzzles on this channel that are over an hour long, but seeing the Phistomefel Ring had me intrigued with this rule set. I am happy that I decided to do it, because this was a great puzzle. I did it in 165 minutes, which is considerably longer than the video. Good Fun!
A brilliant puzzle. The most fun to solve in ages. Not super difficult, just elegant logic the whole way through. I loved it.
Sometimes im not sure why simon uses corner pencil marks 😂 the one in row 6 was totally ignored. What a lovely puzzle and lovely solve!
He NEVER sees the corner pencil marks! 🤬
You would think, as he's pencil marking 1245 as options at one end of the row, he might be scanning for those digits appearing anywhere in the row. But no, not in the corners of cells.
Actually I agree with you...his lack of scanning, and following up on his logic to the end make the solve path complicated ..taking away from the beauty of the puzzle itself
26:26 for me. I really liked the break-in on this one, what a great idea. Very nice puzzle!!
You, sir, are either a liar, very intelligent, or (perhaps) just very familiar with these techniques.
Such a great puzzle! This circles rule really makes for interesting logic with a simple rule set and a clean grid. The thumbnail on this one peaked my interest although I was afraid that the video being over an hour would mean I might not be able solve it. But I was able to solve it and am very proud for it. I feel I owe all my sudoku skills to CTC. This took me a while but I kept at it because I thought I had a good understanding of the restriction posed by the Phistomofel ring by doubling the number of appearances for each digit. Thank you CTC and thank you Jeff Wajes for a great puzzle!
Simon always impresses me by how much he can keep in his head without writing any notes! Once I figured out that the circled numbers need to add up to 30 without using the 9, I wrote out the (just four!) ways to do this: 45678, 234678, 135678, and 1234578, and then ruled out possibilities as I figured out more about the numbers. I couldn't solve most of these puzzles without my notes... well done, again, Simon!
Just under 50 min for this one. I found it to be very straight forward in what it wanted me to look at with how it restricted things via the circles and the cages. I really loved it.
Wow, 44:37, surprisingly good result for me.
I love how this puzzle not just using a Phistomofel trick, but also has really interesting logic based on it after you figured you should use it!
First Digit should come right away, thanks to The Secret!
We only can have 2 Odd Digits in Circles, because we only have two save spots in the Phistomefel Ring + Squares, that don't have a Circle!
The Odd Digit can't be 9, because we can only put four 9s in Orange and four 9s in Green!
Because 5 and 7 add to 12, we still need 18 worth of digits: So we need 4, 6 and 8 in the Circles!
One of the Odd Digits has to be 7, because all even digits only add up to 20, and no combination of odd digits without 7 is enough for 30 Circles!
So we have [3,7] U [2,4,6,8] or [5,7] U [4,6,8] in the Circles!
The 7 has to be four times in Orange and four times in Green, but only seven times in Cirles, so one 7 has to be uncircled! If we put the uncircled 7 in the green save spot, it will remove the possibility of a fourth 7 in Orange (Box 9). Therefore 7 has to be in the uncircled Orange!
The 8 has to be four times in Orange and four times in Green and can't be in the corners of the Green Phistomefel Ring. With the Cage Sum in Box 5 all possible positions of 8 can be marked. We can mark in the same way the 7 in all Boxes.
Because of the 24-Cage in Box 4, we have the only one possible combination left: 4569.
Can only have 2 odd digits IN circles (missing corner from phistomefel).
So, must have 3 odd digits OUT of circles.
9 isn't circled because of box 5.
That leaves (45 - 30 circles - 9 ) = 6 = 5 + 1 (the only way to break 6 into 2 odd digits)
5 & 1 must also be out
@@Mephistahpheles you don't have to break 6 into 2 odd digits, you have to break it into 2 odd digits plus any number of even digits. thus you can omit either 5+1 or 3+2+1. so you still need more logic to distinguish those possibilities.
He put a two in a circle at 45:00. So that resolves which digits aren't circled. But classic Simon, does a whole bunch of other logic to rule out 5.
It’s crazy that I instantly saw a very similar path to the break in as you. I saw that 9 couldn’t be a number because of box 5, and 1 couldn’t repeat in the orange area, because the empty square would need to be saved for another odd number. I was left figuring out whether it would be 9, 1, 2, & 3 or 9, 1, &5.
@@SeanTBarrettAs OP ends with 4, 5 and 6 in box 4 not being in a circle, we come to the conclusion that box 4 has 7 and 8 and either 2 or 3 on its circles, so we can eliminate 5 from all circles and conclude that the circled digits are 2, 4, 6, 8, 3 and 7.
Beautiful puzzle, such a great difference from other sudoku puzzles, and makes you understand the Phistomofel ring so much better.
That was lovely. I discovered your channel a few days ago, and now I'm really regretting not doing my senior paper (oh, 12 or so years ago) on set theory in sudoku.
Anytime a video runs for more than a hour and I am still able to solve it is pleasing.
I couldn't focus for 3/4 of the vidéo just because Simon didn't remove the 8 as options on multiple Cage since the beginning dam!!! This is a Nice solve thanks !
Today my CtC book finally arrived! I had been told earlier that shipping towards the Netherlands could be a bit delayed, but eventually Maverick dropped the mail package from his plane into my backyard this afternoon :-)
Extraordinary puzzle setting indeed. This had simple enough phistomofel ring rules but it was highly confusing but Im glad I fought through this. And Simon you have such clarity of thought in spite of getting interrupted mid solve.
I solved this one a few hours ago and it was fantastic! Definitely deserves a feature here
Just some polite feedback about the rules card - the text is very bold and seems to lack proper kerning, making it rather difficult to read quickly and clearly.
When it comes to set theory explanations and scrabble tiles, the first time that it was explained to me and I got it, was from the angle of the other person having the solution to the puzzle in their hands.
That way you can describe it as giving both bags to someone who knows the solution to the puzzle and asking them to remove from both bags the number that will go on that tile when the puzzle is solved.
I don't know why that made all the difference to me, but putting it from the angle of letting someone else who knows the answer remove the tiles clicked immediately.
1:39:09 - Wow! I loved that one; gorgeous break in and beautiful logic throughout.
My starting logic was, that apart from the two uncircled Phistomefel digits all the others had to be even. Given the total of 30 in the circles, less the 20 by virtue of all the evens less the 7 we put in the corner we are left with the 3 in the encircled green cell. I was really proud to spot that.
Edit - Of course that thought was unsound as a 5 would also have worked if the 2 wasn’t included.
encircled = uncircled. All the other digits appearing in the Phistomophel ring or corners. I'm not quite smart enough (at least right now) to follow your logic, but I am smart enough to recognize an intelligent thought.
33:34 with a really nice break-in and the solve naturally flows from there (although I pencil marked too much as per usual). Great feature!
This was the most fun sudoku puzzle I've solved!
What a nice puzzle, should have tried it! 😅 thanks for the solve Simon! 😊
I love that simon places red lines in replacement of 6 corner marks and it inhibits him from realizing that he has all of the digits in the ring.
❤
Nice solve!
A class act for sure. Very proud of myself to have gotten this one without looking at the video! As you rightfully say Simon, there is a beautiful to-and-fro between the phisto ring and the corners. This video is also probably the first I’ve looked at with the circle rule, and it helped me realise deductions I hadn’t considered before. Thank you as always for your insights, and bravo to the setter for a sublime puzzle. (Solve time was just under 110 minutes for me, but I also didn’t have to say birthdays or explain a phisto ring, haha.)
Thanks to Simon's previous explantions I could see the logic quite quickly , which was nice
Absolutely stunning puzzle!
My first thought on this one is that "digits that appear in an odd number of circles" and "digits that occur an odd number of times in r1c1 and r7c7" must be exactly the same set… and those odd digits must sum to at least 10 (30 minus all the evens) without using a 9; so 37 or 57.
that wasn't my first thought but it was my first useful thought, about halfway through writing down all the ways to add digits up to 30
They don't have to add up to 10... They can add up to 12 if we use only 3 even digits (4-6-8).
So there are two options and we work thru to see that 2 has to be in a circle in Box 4 to choose the option that you suggest.
But Simon doesn't follow up on his own logic and goes here and there to find complicated chains to prove these things.
@@ruckmanikrishnan4221 I said they have to add up to at least 10.
My approach to the 468+57 option was to realise that if the same 5 digits are in the circles in r37 and c37, 4 and 5 would need to be in 2 corners of the ring, and 6 would also need to be in a corner.
amazing puzzle and solve, thanks for sharing with us!
@ 15:46 “at the moment, we would agree, hopefully” made my day.
You could also argue at 50:00 that there is now a circled 2 so the missing digits cannot be 123 but have to be 15 :)
That’s what I was yelling at the video, too.
We could've come to this conclusion even earlier, at 34:40: by noticing that there must be either a 2 or a 3 along with the 7 and 8 on the three circled cells in box 4, it's clear that 1, 2 and 3 could never have been the combination of digits that are uncircled.
@@LuKaSGLLyes, and once he knew the 1 in box 4 was in row 6 he could have completed the 18 cage in box 6. He even had the pencil marks there. Instead, he used a long process to come to the same conclusion several minutes later. Just follow your marks! He drives me absolutely crazy sometimes. He can intuit stuff that takes me forever but misses what is right in front of him.
1:33;20. What an incredible puzzle. By far the hardest circle puzzle I have done, but loved it. Once I worked out 1 & 5 were absent from circles, I immediately thought about where the 3s go. c7r7 was, as Simon would say, " my favourite square in the puzzle". Only place for another 3 in the Phistomefel ring. Superb setting, thank you Jeff
Great solve Simon
Woooo! Solved in 53:22. Very satisfying to make that discovery about odd numbers on the ring.
I'll have to try and remember Phistomefel's theorem.
The majority of the work discovering which numbers are in circles can be done in one go:
* There are 30 circles, therefore the distinct circle numbers must sum to 30.
* 9 can't be a circle number, because all of the 9s would then have to be in circles. Which they can't be, since there are no circles in the centre box.
* 8 must be a circle number, as without it the highest sum we can get is 28 < 30.
* 7 must be a circle number, as without it the highest sum we can get is 29 < 30.
* An even sum of integers, in this case 30, must include an even number of odd numbers, therefore we have an even number of odd circle numbers.
* All numbers must be balanced between the green and orange areas, and therefore occur an even number of times in the two combined. To achieve this with an odd circle number, we will thus need to place the same number in one of the uncircled Phistomefel cells. We know one of these is 7, therefore this is in one such cell. Since we have only two uncircled Phistomofel cells, we can thus have only two odd circle numbers.
* So far we have 7 and 8 as circle numbers, so the remainder must sum to 15. To reach this total with only one odd number, 4 and 6 must be circle numbers.
* This leaves the remaining circle numbers as summing to 5. So either 5 is a circle number, or both 2 and 3 are. More work will be needed to discover which is the case.
* This furthermore means that one of the uncircled Phistomefel cells is 7, and the other is either 3 or 5.
The scrabble tile analogy can be saved if the name of each cell is on the tile instead of the digit in that cell. By name here I am referring to the row and column of the cell (r1c1 etc). Then it is clear how you find the tile in each bag to remove because you only need the name which is known and not the unknown digit in the final grid.
Or just have the person removing the titles from the bag be different than the person doing the reasoning.
this was a fantastic puzzle!
This was a great puzzle!!
"come on, brain. Come on, brain... I've got nothing..." I feel that. A lot. hahaha. Great job, as always!
OMG I solved it all by myself! Well, unless you count a lot of use of the reference killer cages guide in my CtC book - thanks Simon and Mark! 37:43 for me. I also didn't have to explain the phistomofel ring, which meant I didn't spend any time avoiding using it, and I identified the parity/only 2 odd digits trick very quickly. Then I focused on where 1s and 9s could go.
Missed logic by Simon at 34:46 as it was evident that the 5 was not circled, as 2 or 3 was circled, so they both had to be circled, so 5 was the digit not circled. He never did figure it out, using that you couldn't put enough 5's in the ring, but it was available for ages by different logic.
45:32 finish. I was able to immediately ascertain the significance of the missing circles, but was tripped up when I failed to consider the possibility of 1-2-3-9 being absent from the circles. When I realized it, I stopped to check (I was about halfway through the puzzle at the time). I was able to use 1s, 6s, and 9s to place 5s outside the circles, and I got back to work. A very enjoyable puzzle; excellent!
A great puzzle, and a great solve by Simon as usual!
One of the few where I found that my own path to solving a puzzle was quite different from Simon's, though I'll admit I may have missed some step of logic and had a lucky guess at some point.
You can figure out right away that 7 and 3 must be the phistomefel digits not in circles, and that 1, 5, and 9 do not appear in circles. There are 30 total circled digits, and if N is included in that set, it is included N times. So the set of numbers included must sum to 30.
We know we need odd numbers in the set to be able to reach that sum. We therefore need exactly 2 odd numbers in the set, since a different odd number must go in each of the two uncircled spots in the phistomefel ring. So all digits not in the set must be the 3 remaining odd digits. 9 cannot be in the set because it breaks sudoku. So the remaining digits not in the set must sum to 6 (secret 45 - 30 - 9), and the only way to do that with odd numbers is 1 and 5.
A quick way to resolve that cage in the corner around 1:04:09.
7 + (2 or 6 or 8) + (2 or 6 or 8) + (1 or 5) is EVEN. So r1c3 must be odd. So r1c3=1, r3c1=5, and the other two cells are missing the 6.
I was going to make a comment just like this one. You beat me to it. 👍
Brilliant puzzle.
I think a quicker way to do the breakin here is to realize you can only have 2 odd digits in the circles because there are only 2 holes in the ring providing asymmetry. 2468 add up to 20, so we need 2 odd digits in circles that sum to 10. Can't be 1/9 because 9 can't be circled, and you can't have 5 counted twice, so you need 3/7 in circles along with all of 2468.
Every even circled number will be in both the green and orange sets, and there can be only 2 unpaired odd numbers placed (one each in the uncircled orange and green phistomefel cells). Thus, there can only be two odd numbers in the circled set. It is already proven that 9 isn't circled. And there are 30 circled numbers.
Thus, the combinations that could exist are (4,5,6,7,8) and (2,3,4,6,7,8).
At 47:08 You weren't sure we could place a digit here, but we actually could at this point. R3C7 couldn't be odd because it would see the odd digit in the corner and the 7 directly below it. The other odds aren't in the ring either. It can't be 2 because there is already a 2 in the ring that has to match with one in orange. The 6 and 8 below rule them out, leaving only 4
A delightful book worthy puzzle.
I think once Simon got the circled 2, the 5 had to be uncircled?
Once he was certain that there was a 5 in the 24 cage it was already proven that 5 had to be uncircled because you can only put 2 fives in green
yes 2 odds must be uncircled and add up to 6, so either 1,2,3 or 1,5 are uncircled, so either 7 and 3 or 7 and 5 take up the uncircled phistomphel pair. He said this before the break, but unfortunately I think the break to chop the tree lost a lot of momentum
Simon made row 6 way to hard. He had 1’s marked in box 4. That means box 6 had to have the 24 to equal 6. And box 4 had the 13.
Instead Simon ignores his pencil marks, and deduces a floating triple to come up with the same result. So that is why he doesn’t like to pencil mark like Mark does because he ignores them anyway. Somehow that type of thinking makes it so he can solve puzzles that most can’t.
This is one of those things where it's hard to say whether it should be called a creation or a discovery. It's such a wonderfully natural idea once you've seen it, it's like 'of course it must be true', but to imagine being the person to conceive of the idea in the first place is mindblowing. Truly great puzzle.
In the end you can say that to every sudoku, every one is a discory of a unique solution
you can only have two odd numbers be circle numbers. 2+3+4+6+7+8 works, if we are to modify it into an option, we cannot reduce the number of even numbers because we already have all the even numbers, and so changing any one of them would means either duplicating one or adding another odd number, which doesn't work. for example if we substitute 3 with 5, then we must add 2 to one of the even numbers to fill 30 circles, and that simply cannot work without having two of the same even numbers in the sum, and the same goes for 1. and so 2+3+4+6+7+8 is from the start a unique solution given 30 circles and two empty squared in the sets creating the two odd number constraint. another solution is 2+4+7+8+9=30, but then 7 and 9 has to go in the uncircled boxes and that cant work either, because the uncircled box in the central ring sees the circles in 3 boxes, so out of the odd numbers only 1,3 and 5 can go in that box. given the second solution has only 7 and 9, that box becomes impossible to fill, leaving only 2+3+4+6+7+8, with nothing done other than the ring set trick. pretty neat.
When Phistomefel is mentioned, I have to fast forward.
Simon completely bypassed a set logic solution for row 4&5 and columns 4&5. These are available immediately after determining that the non-circle rings must contain odd numbers, one of them is a 7 and the other is NOT a 1.
Rows: 90 - 24 - 18 - 2x eights (in the ring) - 37 (the 2 cages in Box 5) = -5... Cells R6C4 and R6C5 add up to 5 more than the other "non eights" in the ring in those rows. For the rows, the "non eights" must include a 7, which means R6C4 and R6C5 are maxed at 5 and 9 (positions known). The other "non eight" is minimum at 2.
Columns: Similarly, for columns 4&5, cells R4C6 and R5C6 add up to 10 less than the other "non eights" in the ring in those columns. The columns are easier... R4C6 and R5C6 are mimimums and must be a 12 pair, and the ring must contain a 7 and 6.
Also at this point: Since 2 is on the ring, 1 and 5 are not.
Simon still got there, of course, but I found I had a lot more opening digits.
200 minutes, 3+ hours. I feel like a brialliant detective. I had to use everything. Absolutely insane!
Could you please include logic-masters link to the puzzle in every description? That would be very useful
Also, beautiful set of rules and very flawless solve, props to Jeff and Simon! 💖
I just had a real weird thought, how would Schrödinger's cells work with circles?
You still need the appropriate number of circles but then you can have some weird smuggling going on that would throw off the way to break into them, right?
I will never not be amused by Simon’s feats of logic in figuring out that 5 can’t be in a circle, when all he had to do was notice that 2 was in a circle and the only way to make up the missing 6 circles was 1 and 5.
"Is there any thought that isn't mental? Probably not. Sorry for the tautology."
I was chatting to a fellow involved with the Oxford Uni AI Programme recently who had a different opinion.
36:23 For me, what a creative use of parity
Using the Phistomefel Ring logic, I quickly cottoned onto the idea odds were limited in the circles. If you put an odd into a circle, you could put pairs of that odd into green and orange circles, but then you have one digit left over to place in a circle, but that number must still appear within the other group and the only way that can happen if it takes up one of the uncircled cells. And we only have two uncircled cells to use, limiting the number of odds we can use in circles to two.
Now to get to 30, I could use [246837], or [46857]. [2468] only adds to 20, so we need at least 10 to come from odd digits, and we can't use 9 as box 5 can't have a circled 9 in it.
I had initially made the assumption it was [246837], but 7 is still required and the next bit of logic followed on regardless. The 7[3|5] pair in circles must go in the 'gaps'. Can 7 go in r7r7? No, because it would remove the possibility to place seven 7s in circles, as you could no longer use box 6 or 8 to place a 7, so would only have six boxes to place seven 7s. This places the 7 in r1c1, and 3 or 5 in r7c7.
Very long solve for me with lots of checking the video. made a wrong assumption about 6 in green, but I kept going because it was such a fun puzzle and I learned lots!
Around the 49 minute mark, where Simon thinks about if 5 could be a circled digit. Another way of thinking about it is what are the digits adding up to 30? Now you can only have two odd digits, and one is already confirmed to be a 7. At this stage you also have 2678 in the circles, which add to 23. We can't use 5 as well, as that would leave us on 28 and require another pair of 2s in circles which isn't allowed. But 3 and 4 fit in nicely.
Alexa at 1:04:14: "But let's, uh..." sounds like "alexa"
Incidentally, you can mute Alexa so she doesn't hear you. However, it's fun trying to figure out why she responds to you ;)
when you do set theory tricks, you have some overlap sure, but you start out with X sets of 1-9 and that means you can have a maximum of X of any number in your final set, you migth very well have X but for one or more of the numbers you must have a lower number than the original sets you picked out. but you can for sure never end up with more than 8 of any number with this phistomefel ring construction that used 8 sets :). and the same goes for any other constructions. the original sets set an absolute upper bound of how many of a given number is in the set you end up with.
Rules: 04:16
Let's Get Cracking: 05:25
Simon's time: 1h3m23s
Puzzle Solved: 1:08:48
What about this video's Top Tier Simarkisms?!
Phistomefel: 13x (01:00, 05:35, 07:06, 07:21, 07:23, 09:14, 11:16, 11:32, 12:15, 12:32, 14:53, 29:25, 29:54)
The Secret: 5x (05:40, 05:48, 05:48, 08:16, 08:53)
Three In the Corner: 2x (1:08:37)
And how about this video's Simarkisms?!
Hang On: 14x (28:28, 28:28, 28:35, 31:12, 31:12, 31:12, 40:57, 41:02, 41:14, 41:14, 41:14, 48:08, 1:02:28)
Sorry: 13x (12:29, 13:07, 23:01, 34:22, 35:06, 36:39, 36:39, 42:43, 43:08, 48:21, 54:19, 1:01:45, 1:04:39)
Brilliant: 9x (02:54, 48:28, 48:32, 48:32, 50:23, 57:51, 57:53, 58:39, 58:39)
Pencil Mark/mark: 8x (34:38, 41:07, 41:56, 44:26, 46:18, 55:42, 57:15, 1:04:07)
Beautiful: 6x (14:45, 14:49, 43:37, 43:46, 50:17, 1:09:30)
In Fact: 5x (04:35, 04:46, 11:54, 29:51, 57:57)
Obviously: 5x (08:32, 12:32, 17:57, 38:37, 55:03)
Fascinating: 4x (05:12, 17:09, 20:49, 29:46)
What Does This Mean?: 4x (34:33, 42:45, 51:36, 51:53)
Weird: 4x (07:08, 07:11, 20:27, 20:29)
Nonsense: 3x (53:01, 53:08, 53:12)
Incredible: 3x (37:02, 38:07, 1:00:17)
Gorgeous: 3x (24:15, 24:18, 50:35)
Scrabble Tiles: 3x (16:06, 16:14, 16:19)
Clever: 2x (25:50, 1:09:17)
Shouting: 2x (03:07, 03:45)
Proof: 2x (19:53, 20:07)
Ah: 2x (45:30, 1:06:15)
Cake!: 2x (02:56, 04:06)
Good Grief: 1x (24:18)
What on Earth: 1x (38:32)
Goodness: 1x (34:11)
Bother: 1x (27:18)
Naked Single: 1x (45:34)
The Answer is: 1x (26:00)
Out of Nowhere: 1x (1:00:56)
In the Spotlight: 1x (1:08:43)
Stuck: 1x (1:02:53)
Elegant: 1x (26:31)
Deadly Pattern: 1x (1:06:34)
Hypothecate: 1x (42:56)
By Sudoku: 1x (1:08:26)
Surely: 1x (1:00:17)
Triangular Number: 1x (08:53)
Most popular number(>9), digit and colour this video:
Fifteen, Thirty, Forty Five (8 mentions)
One (115 mentions)
Green (24 mentions)
Antithesis Battles:
Low (2) - High (0)
Odd (12) - Even (7)
Outside (2) - Inside (0)
Column (19) - Row (12)
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!
fan va snabbt. 57 seconds. 👌
22:50 I calcutated as follows: 2+4+6+8=20 circled even numbers possible. So, you another 10 to make it 30. So the two missing numbers in the green ring and the orange boxes have to be 3 and 7 making up for the uneven numbers in the rings. I hope, Simon will come to the same conclusion sooner or later. And I hope also, my deduction will prove to be correct.
Hey, I did the puzzle and my conclution was correct. Simon noticed that near 50:00. This was the first time, that I've beaten Simon at all in time!
After reading some comments, I realized, that my deduction only was accidentally right. You don't need to put all even numbers into circles, so using 7+5-2 would be also correct instead of 7+3.
49:30 As 2 is a circled digit already, 3 must also be circled and 5 cannot be circled, as otherwise the number of circled digits cannot add up to 30 anymore...
38:21, just flowed for me. I knew I needed SOME odd digits, but only two would work because of the two cells without circles. Some math and figuring out I couldn't fit nine 9's and away we go.
Do more sudoku mr Simon itnwill definitly help you on multiple of your solves . But man what a brain you have to think all those sudoku résolutions👏🤟🤜👍
at @40:48, and I can no longer stop myself: THE EIGHT SIMON, THE EIGHT IN BOX FIVE. It might not do much for the solve, I can't tell yet, but it eliminates 8 from row and column 6.
I love how the 8 was basically a given digit so Simon was blind to it for about 40 minutes! Never change Simon 😂
46:43 that's a missed opportunity to say "too many tutu's in the grid".... Darn I really thought you were heading in that direction 😂
34:48 as soon as the 1 is not in a circle, there is either a 2 or a 3 in the circles in box 4. So therefore the non circled cell in the ring is the 3.
Too bad you had to take a break from the solve, although you got back into it really well! It would have taken me forever to pick up the pace.
I think this solve was a bit slow in general from simon. I instantly jumped to the conclusion of even/uneven digits in the phistomefel ring after counting. It could have easliy been a 40 minute solve, even before the break. Normally he is quite fast tho
I put a lot of pressure on myself to solve puzzles much easier than this one. This puzzle is out of my league. I can't imagine the pressure Simon must feel when he is solving a puzzle while recording. Clearly he is brilliant at puzzle solving and it is so much fun to watch him work through a complex puzzle.
We could have had a song when the 3 is the corner of the phistomphel!
2+3+5+6+8+9=30 is also a solution but 7 has to go in the orange corner, and so it is not possible to fit both 5 and 9. it is also possible ofc to add 2 to 3 and 2 to 7 and change from the correct solution to this option, but it is still not possible because 9 cannot go in either uncircled box. i think that is the best way to solve it probably, to determine that 9 cannot go in either box, and that only 2 odd numbers can be circled, and then 3 and 7 are the only options, and that leads to 2,4,6,8 as even compliments.
It took me one look at the Phistomefel of circles I realized that all counts in circles had to be even except only two (the one missing in full Phistomefel set). So all even digits are circled (which gives 20 circles), and remaining 10 circles need to be defined as two odd digits. As 9 was excluded, these need to be 3 (r7c7) and 7 (r1c1). Simon is at 25 minutes and got number 7, and at 30 minutes it is still not finding r7c7. Pity that I am not good enough with Sudoku, to solve any further 😞
Fun puzzle! Not too difficult but did take some noodling. 43:30 solve time for me.
6 can be done only in 3 or 2 cells, once the 1 is not circle there are only 2 way, 123 or 15, since the 2 is circle the option 123 is not available anymore and that mean 1 and 5 are not circle.
The way I deduced the numbers in the cage was asking “how do we make 30 using even numbers and only two odd numbers excluding 9?” There are only two ways, and you can only use 1 odd number bigger than 3 because of that bottom-right corner in the phistomefel ring. Great puzzle!
i solved it! honestly cant believe it cnsidering how long it took simon i wouldnt normally have got very far. got bogged down in the bottom right corner cage for a long while and realised i'd made a logic mistake much earlier on and had to backtrack all the way. (i'd thought all the middle of the ring in cages 2,4,6, and 7 were 678 triples) and then right at the end i found myself in a deadly pattern and realised i'd added the top left cage to 22 instead of 23 and had to unwravel AGAIN! but i made it
At 50:00, since there is a circled 2, the uncircled digits that sum to 6 are 1 and 5 (because they cannot be 123 anymore). So R7C7 is 3.
It was so much easier for me to see that once the went in, he needed 1&5 to make 6. He still got there in the end. I would have never seen it the way he did.
Your logic with having not enough sudoku units to put 5 in is brilliant, but once you put a 2 in the green area you had already ruled out 1,2,3 as your missing digits and therefore it had to be 5
Awesome , when horrible dd leads to correct solution(gonna have to solve this again).
Took sum37's D8 to mean the sum of the cages endpoints...which gave a correct DD on B4 and correct solution.
Like how the design involed parity play
5:26 Let's Get Cracking!
For people who want to skip the intro
56:51 for me. Nice puzzle!
At 1:03:50, Mark would have solved the 23 cage using parity! Which is not how I solved it. I worked through options like Simon.