arctan here! thank you very much to simon for solving this, to james for recommending this, and to everybody for solving it! a few notes below: --this is not my channel debut, technically, but it is my sudoku debut; i previously constructed the phistomefel ring crossword from september 2021. --the rules for pairing the circles technically allow for the squares to not be aligned with the grid, which is not the intent; apologies for that. the original wording did explicitly rule that out, but that technicality crept in when trying to rewrite the rules to be briefer and easier to understand. --apologies for no threes in the corner; as much as i wanted it to happen, the elegance of the sudoku had to come first.
Such a perfect puzzle, thanks!! 49 mins well spent from me. I enjoyed the challenge of the solve, but also marvelled at the construction - what a great piece of work!!
Lovely puzzle. Indeed that rule must be explicit about aliging the squares with the grid. I took the 45-45 in the name to imply that the squares' edges could be at a 45 degree angle and got stuck. Actually also sqares in which vertices are a knight's move away from each other are also possible.
Rows 4-6 can only have two circled sixes because they're all in two boxes. Rows one and nine can't have any sixes. Therefore, the circle in row 8 must have a six. From there, the puzzle unfolds a lot faster than how Simon solved it.
***IMPORTANT*** The "Squares" must be orthogonally laid out and NOT diagonally! 😊Hopefully i saved a few of you the thirty seven minutes i wasted unaware of this.
Interesting - I hadn't even considered the possibility of the square not being aligned horizontally/vertically 😮 - that would be brutal if they could be on the tilt, especially if they didn't have to stick to being at 45°
That's kind of huge. Someone needs to set a puzzle where the broader rule is used. I gather in this puzzle it was possible to actually rule out any possibilities there? In any case strange not to have it noted since there's even a whole tilted square pattern going on with the circles.
@@BenCragg1 I didn't even read your whole comment, just the first line was enough to give me a light-bulb moment and finish the puzzle within a few minutes. Incredible how some of these puzzles just need a particular way of looking at it and they become obvious!
Something I've noticed about these videos: Up until recently, Simon's video titles would capitalise the first letter of every word, including small connecting words (the, and, a, etc.), while Mark's video titles followed a more traditional method of capitalising just the important words, nouns and such. Interestingly, I've noticed that Simon's videos have begun capitalising nothing but the first letter of the title (with the exception of proper nouns and of course words capitalised for emphasis such as "PERFECT" in this particular video). I'm sure the conspiracy probably goes even deeper, and while I'm frightened of what I may unearth I believe it's for the good of all humanity that whatever cryptic message is being left by CTC is brought to light.
Wow did Simon place those 6's the hard way (at 40:20 ish). Here was my reasoning: There needs to be six 6's in circles, and there are only seven columns with circles. Columns 4, 5, and 6 will support only two 6's total, so you will need to have a 6 in circles in each of columns 2, 3, 7, and 8. This places the 6 in column 8, which places the 6 in column 2, which places the 6 in columns 7 and 5, and then in columns 1 and 9 by sudoku. I had that in my first 8 minutes.
I usually don't read comments before writing my own, but for some reason I did skim a lot of them this time. I see that there was a quicker way to do something or other - but it is entirely OK with me that you did it the way you did, Simon. This also goes for proving something "the hard way" instead of some easier way. I love watching you solve, and even when you are not as efficient as possible, I feel that I learn something that I can apply to my own solving. I love it that you do not (neither you nor Mark) solve puzzles on video that you have already worked out a perfect solve path for. I far prefer that you do it cold, live, whatever you want to call it, even if that means that the solve is not a "best" solve. Your amazing brain is a privilege to watch in action. Thanks for this video.
The break in for me was noticing that r7c7 couldn't be a 6 because then you could only place five 6s in the puzzle. Spotted it relatively quickly and from there it was an easy solve.
Classic Simon, he comes up with a wonderful but unnecessary deduction why 7 can't be circled, ignoring the fact that you have to put a 7 in box 5 and at least two 7s on the perimeter 🙂 Fascinating ruleset, I really enjoyed solving this one.
I think there is a loophole in the rules. At 25:35 Simon deduces the corners of square (greens), but there is multiple other options to form squares if you don't assume that the square is alligned with the sudoku grid. For example, R2C7 and R3C4 could be the circle corners and then R1C5 and R4C6 would be one of the "sum" corners. Forming a slightly tilted, but still a valid square.
A circles counting puzzle is always going to pique my curiosity, and this ruleset was an amazing twist on them. I agree with what Simon says at 20:45 - these are my absolute favourite types of puzzles to solve (and watch Simon solve) - the ones with meta deductions at the beginning - unwritten rules to discover, new logic to get your head round. This puzzle had all that good stuff. Absolute genius - creativity and innovation ooze out of this one
Delightful! This is the sort of puzzle that has me kicking myself at the end in a joyful way -- I spent easily twenty minutes narrowing down my digits in circles, placing most of my sixes, parsing out the base logic, and not able to find a single break point past that, before I gave up and resorted to the video, only to discover that I'd forgotten a rule! (When you think the pairs of digits only have to be rectangles, not squares, the ruleset gets MUCH HARDER.) And then as soon as I knew that, because of all the groundwork I'd already done, everything else snapped into place immediately -- what fun! If only I'd paid attention to the puzzle name... Thank you, arctan, for a lovely one.
I did exactly this, except I got stuck for 90 minutes before giving up and watching the video. It is surprising how far you can go without assuming squares!
Ah Cracking the criptic with Simon. You guys have no idea how you help me go through my anxiety of the last few weeks. Getting better every day now, but you helped sooth me when it was at its worst.
I finished in 98 minutes. This was such a unique and cool ruleset. It was cool to realize that there must be 5 unique digits. I saw that since there were 16 circles, 8 pairs must exist. If we use four cells with 6 and 5 guaranteed, then 6 must take up six of the eight pairs, leaving the 5s unable to take 6 as a partner, giving a total of 11 pairs. This breaks everything. The only one that works is five cells, which 64321 is the only one that works. The 6s must be paired with all 123s, leaving 4s to pair with themselves for a total of eight pairs. That was very cool to see. Great Puzzle!
18:31 finish. For my break-in, I looked at the columns. There are seven columns with circles, however only two of the same number will fit in the middle three columns (only circles in boxes 2/8). Therefore, you cannot use 7/8/9. You do need 6, to have enough digits to fill all circles. Since only two of them can go in columns 4/5/6, therefore columns 2/3/7/8 must have circled 6s. This places the 6 in column 8, which places the 6 in column 2, which leads to the 6s in boxes 3 and 8. The last two 6s (boxes 2/4) can be differentiated by considering the square rule. The 6s in r2c7 and r7c2 need opposite corners on the diagonal between the two, and therefore neither circle on that diagonal can be a 6. The other two circles in boxes 2/4 are the last two circled 6s. Since the maximum sum is 9, 6s have to pair with 1/2/3, and the 4s pair with themselves to make 8 sums. Since you have placed all of your circled 6s, the cells in r3c3 and r7c7 have to pair with each other, and must be 4s. Sudoku will place the other two 4s, and the square logic will identify all of the sum cells. From that point, it is basically classic sudoku, with a pair of short arrows thrown in. An excellent puzzle; fun fun fun!
I was honestly a little happy to see that, because I did the exact same thing. Some minutes later noticing the 3, and realizing that it being coloured meant it had been there the whole time was definitely a bit of a facepalm moment for me lol.
It really is fascinating how Simon can elegantly crack the toughest of puzzles with extremely complicated logic that would take most of us brute-force bifurcation, but when it comes to doing very simply sudoku, Simon's brain proceeds to malfunction.
I've spent hours on this and what I'm getting from the comments is that the rules do not specify the squares must be orthogonal but that they are. Damnit. After reading that comment it took me like 10 min to finish. That is a rule that really needed to be more clear.
Well, that's because the rules in Logic Masters do specify it. For whatever reason, when the channel simplified the rules to fit the format of the video, though, the specification was left out.
28:55 for me. Normally I struggle with puzzles where you need to sit with a global constraint for a while before you can start to think about pencil-marks, but this one felt perfectly pitched in difficulty, and I loved how it fell into place after I managed to get started. Really really elegant!
Same here - usually the puzzles that Simon loves, ie the ones where you need to do lots of theory before you can start attacking the grid - are beyond me, because I don't like going down blind avenues not knowing if I'm making any progress on the right path (this may be why I struggle to see SET break-ins) ... but this one was just right, the initial theory was accessible enough that I could work through it and figure out how to get started 👍🏻
Interesting thought. Seems like that’s the intent though the rules are ambiguous. If the intent is the cells make a square, then the answer to your question is yes. If you’re allowed to draw a line between cells and that line forms the square, then no.
@@dustpan5356Even assuming that cells make the square, there's still a question about whether squares rotated 45 degrees from the axes (so their corners are NESW) count as squares for the rules
That's the issue i was having. The definition of a square does not preclude it from being rotated any number of degrees. A square is a square is a square.
I had to watch the video because I didn't assume this. I don't think there's only one solution if you don't assume this (after an hour of me puzzling it out, this is my conclusion, but I may be wrong)
Arctan has confirmed that was the intention, and that the clarification was accidentally left out of the rules while trying to simplify them. (I hadn't even considered the possibility of the squares being at an angle so I'm very glad that that wasn't the case!)
At 23'55 it is mentioned (Simon discovers) that the squares' diagonal is a diagonal of the grid. The two short sides being horizontaly or verticaly. However, that is not necessary, you could have a square with the angles on top or down: The two cirkels in row 2 could form a triangle with R1C6.
tried it yesterday evening tired af and for some reason thought the rule was that it could be any shaped rectangle for some reason and that seemed way too complicated and i gave up. after reading in a comment that it was squares i facepalmed myself, went back at it (so it obviously doesn't have the time from yesterday added, so it's not _as_ great as it sounds) but i'm still proud to finish the whole thing in under 19 minutes then
when i got all the 6s in place, i had forgotten that the rules said SQUARES not rectangles. i spent 20 minutes trying to find a crumb of logic before i reread them, then finished in seven minutes after that LMAO
At 21:30 Simon makes an unproven assumption, I believe. The circle on the arrow between boxes 2 and 3 could be a two, were it's two digits both one. It turns out not to be a two later on, but it's still not obvious at this stage.
Hello ! I fell in love of a puzzle game, close to the “baba is you” style, it is very simple to understand and which you might like! It's called "Patrick's Parabox", I hope you will like it !
Tried this myself a few days ago but only just got the time to watch your solve, Simon. Really enjoyed the deductions, and I appreciated how you (unlike me) recognised how the 6s and the 1,2,3s were both necessary to make the pairs work - initially, I had considered 1,4,5 and ran into trouble, so such a deduction would have absolutely helped me, haha. Still, I very much enjoyed this style of puzzle and am hoping watching more of these vidoes helps me to better understand the intricasies of this great "circles" format!
Got it in 38:40! The beginning went fairly smoothly until I hit a major snag that had me staring at the grid for a good 5 minutes. Then I reread the rules and remembered that the pairs of circles had to make SQUARES, not just rectangles, and the rest flowed pretty smoothly!
How Arctan, the setter, he dangles Like carrots, the clues in triangles. I seek that reward But it's more like a sword For my brain is a wreck and in tangles.
I solved this one rather quickly, after figuring out that 7 could not be placed in circles. Then, you can easily place four of the six 6's in the grid, and the puzzle unravels quickly from there. The key to figuring out where to put the 6s is to realize that across row 4-5-6, you can only have two 6's, so every other row needs one. That immediatly puts a 6 in row 8, then by simple logic you can figure out the one in row 7, row 5 and row 2. It's only a few minutes from that point to solve the rest.
the box in the middle is also eliminating a digit in the circles. basically, if you can find n non overlapping sets of 1-9 with no circles, then the only digits that can be in circles are 9-n and bellow. it doesn't matter that there are 7 rows and columns because of the box, only 6 is possible from only looking at sets with no circles. this is simple enough to explain, non overlapping sets must have all the digits, and so 3 lots of all digits cannot be in circles because of the 3 non overlapping sets with no circles. :)
Another trick for circle rules you can use is checking the maximum number of circles in a single sudoku unit. For instance, R7 has 4 circles in it, so you need to make 16 up of at least 4 digits Similarly you can check the minimum number of sudoku units that contain circles, in this case 7 contain circles So you know that you need to make up 16 with a set of numbers whose count is between 4 and 7 inclusive, which eliminates the possibility of using something like 7 and 9 to make 16
If you split the sudoku grid into units comprising columns 1,2,3,7,8,9 and boxes 2,5,8, then you can get it down to just six units that have circles. (Also works with rows and boxes 4,5,6.)
I found the 6s very easy to place: You can only place 2 sixes between columns 4-6, which then means you only have 4 columns left for the other sixes, which makes it obvious from that point.
I agree. And the same is true for cleaning up pencil marks. Being tidy is arguable just as important, if not more, than focusing on the logic of the puzzle itself. The logic will only take you so far if you can't trust your pencil marks and your scanning. Besides, if you do to keep tidy, some of the logic actually becomes simpler anyway.
2:24:44 - I got the break in quite quickly but could get my head around the circles. Once I’d watched Simon for a bit I realised that the 4s had to be paired up.
I think the rules should say “different pairs cannot use the same cell for their sum” not “same square”? At least I think that’s what the example is saying, or else I’m being dense (likely!).
You can see the arrangement of the sixes in a much more straightforward way: You have 7 rows to place the sixes. Notice that you can only put 2 sixes in rows 4, 5 and 6 alltogether because of Sudoku. Now you have to place sixes in every row out of the rows 2,3,7 and 8 and two rows out of the rows 4,5 and 6. This implies: R8C5 is a six => R2C7 is a six => R7C2 is a six and R5C8 is a six. Now since you still have to place a six in one of R4 or 6 and this is only possible in C3, you can't place a six in R3C3 hence R3C4 is a six. Finally, since sixes can't pair with eachother and R2C7 is already a six, you can place the final six in R4C3.
What I should have said is that there is a 6 in columns 1 and 9 and one in block 5 and the other six are in circles. So the only one in column 8 is in the circle.This then gives r7c2 as 6 and sudoku helps a bit from there
one of them use 5, and 5 and 6 together cannot work, because they sum to 11 and you would necessarily get 5+5, 5+6 or 6+6 all of which cannot satisfy the square corner rule. so we must use the first sum and the circle numbers are 1,2,3,4 and 6 :)
I started with the 6s. You can only put six 6s in the circles if you use the circles in box 3 and 7 This means you need to use the circles in row 8 and column 8 (they are the only left circles in box 8 and 6) Now you need to put one 6 in box 2 row 3 (r3c4 and r3c6) and one 6 in box 4 column 3. (r4c3 and r6c3). The 6s in box 3 and 7 need to pair up with a circle on the diagonal between them. This means that r3c6 an r6c3 can not be 6. And now you have your six 6s. The circles in box 1 and 9 must pair up with each other and since they do not contain a 6 and all 1s, 2s and 3s must pair with a six they must be 4. Now the only other 2 circles where a 4 is possible are r2c5 and r5c2 And you have your 4s with corresponding 8s From here you can easily pair up all the circles. And then I got stuck.
Very late to this one, which took me 2:20 (hours!) because I did not assume squares lined up with the grid and kept drawing 45 deg and other pinwheel-type tilts. I finally narrowed to 90 and 45 only and at least managed to find most of the 6s. FWIW, all but 2 of the circles are on the same bishop's color (required if you want a 45 degree square). The two rogue circles had to be 4s with a non-circled 8 sum, and I could work out the other pair of 4s and most of the 6s, and begin eliminating positions for 2s even with 45-deg squares in play. Finally narrowed to just 90-deg and solved fast, with more of an extended classic sudoku run-out than we're used to here. I wish I'd read the comments for the accurate ruleset, but it was interesting to see how far I got without the constraints.
Has anyone mentioned this? Normally with these circle puzzles, Simon adds them all up, then figures out their make up. There are 16 circles so there can be only 1 way: 1-2-3-4-6. Not sure why he's even considering 9-8-7. Yeah but now, I need help in the next steps! Ha!
At several points in the puzzle I missed Simon considering 45 degrees rotated squares! Where in the rules is it stated that, say r7c4 and r7c6 could not be two paired circles, being two opposite corners of the square r7c4 - r6c5 - r7c6 - r8c5? I was not at all able to solve the puzzle until I switched to assuming only orthogonal squares are allowed. I'd prefer this was mentioned in the rules (if it was intended).
Re: 24:15 - Why can't squares be tilted? Why can't r5c8 be paired with r2c7, with the other corners being r3c9 and r4c6? It doesn't seem to be disallowed by the rules, and I can't find any deduction within the puzzle that would suggest it doesn't work.
i think the title of the video disallows that. what you mentioned would be a trapezoid (assuming the center of each cell is used to "draw" the squares) and they wouldn't have 45-45-90 angles in between
I also think about it. And the only answer, thatI find is that we're talcking not about dots in the middle of the cells, but about the cells. So the sides of the square shoul be parts of row/column.
I was wondering the same. r5c8, r3c8, r2c7 and r4c6 would form a valid square. I'd guess there's huge amount of solutions if tilted squares are accepted.
Much easier way to prove that 7 can't be circled is to consider this. Row1, Row9 and Box5 is each a complete set of 1-9, and none of them have circles, ergo there would be at least 3 digits of a circled number that aren't actually in circles
51:20, I started really quickly, got the numbers in the circles right away, but then hit a wall. Eventually I looked at the video and saw that I made a really dumb mistake with the circles (I placed the 4's and 8's right away, then for some reason didn't rule out one of the 8's as a possible cell for one of the other sums and didn't notice that I was using it twice). Eventually finished after undoing that error. I was drawing boxes and marking cells as I was figuring it out, but I erased the markings as I figured them out, which is how I made that mistake. Also, I did some trial and error working on the numbers in the circles, but noticed a much faster way of ruling out some of the options when I was finished.
Hey Simon, I wasn't solving the new app puzzle #54 and when looking at the hints, it seems like there's a king's move constraint that isn't in the rules?
This took me like 5 hours. I don't think I did this the intended way. The puzzle fell into place once I realized that the the following pair-and-sum 3-tuples were the same sets: "Purple": R2C7 circle [6], R6C3 circle [candidates 1 or 3], R2C3 Sum [Candidates 7 or 9] "Yellow": R4C3 circle [6], R7C6 circle [candidates 1 or 3], R4C6 Sum [Candidates 7 or 9] Because There was a third "Blue" 3 tuple of: R3C4 circle (6), R6C7 circle [1 or 3], R6C4 sum [7 or 9]. With three colors and 2 candidates, two of the colors had to be the same. Blue and purple have a 1-3 pair on row 6 so they must be different Yellow and Blue have a 7-9 pair in box 5 so they must be different. The only combination that remains is Purple and Yellow. This tells us which color has the 1 circle (Blue), and which colors have the 3 circle (yellow and purple ), because you can't have two colored circles with a 1.
uh at about 21:45, the zell in row 2 column 7 could very well be a 2 becaude the two cells the arrow points to aren't in thr same box and could therefor be 1 1
41:00 To the beautiful logic by Simon to disprove the 6 in r3c3, we might add the following: suppose that r3c3 is indeed a 6, then r5c2 and r2c5 are also forced to be 6. Then in box 6 the 6 is forced into column 7 and in box 8 the 6 is forced into row 7, which rule out a possible 6 from r7c7. Now we have only 5 circled sixes, so we are one short. Starting from r7c7 the other way round, the logic is identical. So r3c3 and r7c7 cannot be 6 and both must be 4.
The other confusing rule example is having a 2x2 corners 68;82...where one of the 8s is sum for another pair... Doesnt the 2nd 8 allow for this configuration..., 62 pair is using the other 8 as its sum rather than the 44s 8. The rule only states the sum square cannot be used for 2 pairs, it doesnt say they cant be equal so long as the other sumcorner is the same.
Haven't started the video yet, but one thing I'm not quite clear about regarding the wording of the rules - maybe it will become clearer when I watch the rule part. Does "grouped into pairs" imply no digit can be in more than one pair? Or does it just mean each digit must be in at least one pair,? In other words: can one circle be a corner of several squares? Example: Can we have r2c5 pair up with r4c3 as well as with r3c4 and also with r2c5?
The rules aren't clear about how the squares must be oriented. I interpreted, because of the puzzle title, that squares could be oriented either orthogonally or diagonally (i.e. pair r2c7 and r4c7 sum at r3c8).
I think you can place there 6s right off the bat. If you have to place six 6s in the 7 rows with circles but you acknowledge that a 6 in box five will eliminate one of the middle three rows then all of the outer four rows have to have a 6 (R2 R3 R7 AND R8). In R8 there is only one circle so that placed the 6 in R8C5. Which makes the only place for a 6 in R2 to be R2C7. Those two 6s then rule out all but one 6 in R7 which is R7C2. Sometimes I make logical conclusions that are totally wrong so someone help me out with this because it seems right but I might be missing something
Just to make sure on the rules: where exactly does it say that the squares have to be orthogonally connected? Like, why can r3c6 not pair with, e.g. R7c4 with r4c3 as a 3rd corner of a square?
thanks so much for explanation!!! i see you have a guitar behind, it would be so cool if there were some kind of puzzle connected with music or pitches, idk would that actually work but i just got the idea it may be possible to make a puzzle related to music👀
Interesting puzzle. Similar to my "SqUare-DO-sKUw" puzzle I submitted on Logic Masters Germany a couple years ago. So far, solved by 10 or 11 people (that I know of), including the most recent Tetris world champion, fractal.
If you imagine the grid divided into the following sudoku 'units' (each a set if the digits 1-9), then it helps seeing 6 is the maximum that can go in a circle (and also helps with placing some of the 6s): Cols 1,2,3,7,8,9 + boxes 2,5,8 Or Rows 1,2,3,7,8,9 + boxes 4,5,6 (Edited, as had my rows and columns mixed up)
Fun puzzle, after I figured out it has to be 1,2,3,4,6 I started by looking at what circles absolutely must contain a 6 and what can not contain a 6 based on the fact the circles are only in 8 boxes and 4 of those boxes all have only 1 circle, so for example 6 can not go in 2:5 because if it did it would knock 6 out of box 3 for sure and then either boxes 7 and 9 which is already too many to box 8 but because 7 and 9 can only have 1 6 it knocks one out of at least one of them therefore 6 can not go in 2:5 which then means box 1 and 2 can not both have a 6 which then really restricts 6s into very few spots, and after figuring out which spots had to have 6s and which couldn't then I did the same with 4s since 4s can't connect to 6s after that there was a forced 7/9 pair in box 5 which allowed all the 1s to be removed from all but 2 possible circles, then after that the entire rest the puzzle was just normal sudoku.
I am unsure if Simon's conclusion of possible squares @ around 25:00 is correct. Could you not in theory have r2c6 pair up with r8c6 making either R6c2 or r6c8 it's right angle corner?
At 25:49 Simon deduced the green square partner had to be r6c3. It could have also been r6c7 (unless I missed why that was eliminated as an option). Great solve either way!
I think it was the setter's intention that the squares should not be tilted, but have sides that align with the grid. As others have pointed out, the rules don't explicitly state that though. Simon was solving based on this assumption, and the fact it solved so neatly this way makes me think he was correct to do so.
@@rcambo4818 I know it's not conclusive evidence, but both examples of squares given in the rules were aligned with the grid. I admit, I never even considered the possibility that the squares might be tilted during my solve. (I did get stuck thinking, for some reason, rectangles were okay though. I had to watch Simon's solve up to the point he first mentioned they needed to be squares during his solve. It suddenly became a lot more straightforward. My numpty moment 😅.)
Because the central boxhas no circles and the outer lines are blank, boxes 2,4,6 and 8 are sort of xwings of 6s. So column 8 must instantly be a 6 or all of them cant be placed. I am pretty sure anyway as I haven’t actually solved this one. 😊
arctan here! thank you very much to simon for solving this, to james for recommending this, and to everybody for solving it! a few notes below:
--this is not my channel debut, technically, but it is my sudoku debut; i previously constructed the phistomefel ring crossword from september 2021.
--the rules for pairing the circles technically allow for the squares to not be aligned with the grid, which is not the intent; apologies for that. the original wording did explicitly rule that out, but that technicality crept in when trying to rewrite the rules to be briefer and easier to understand.
--apologies for no threes in the corner; as much as i wanted it to happen, the elegance of the sudoku had to come first.
Fantastic puzzle!!
The Sudoku always comes first 😁
Is it possible to rewrite the rules again to exclude non-aligned squares? Because that omission made solving impossible for me.
Such a perfect puzzle, thanks!! 49 mins well spent from me. I enjoyed the challenge of the solve, but also marvelled at the construction - what a great piece of work!!
Lovely puzzle.
Indeed that rule must be explicit about aliging the squares with the grid. I took the 45-45 in the name to imply that the squares' edges could be at a 45 degree angle and got stuck. Actually also sqares in which vertices are a knight's move away from each other are also possible.
Rows 4-6 can only have two circled sixes because they're all in two boxes. Rows one and nine can't have any sixes. Therefore, the circle in row 8 must have a six. From there, the puzzle unfolds a lot faster than how Simon solved it.
This is how I saw it, and I assume is the intended logic
Also how I saw how to break into the location of the sixes.
If you spot that all the circles are in 6 sets of the digits 1-9 you can also rule out 789 as circle digits for the circles.
I saw it in the columns. Columns 4,5,6 can only have 2 Sixes, so Columns 2,3,6,7 must have a Six, therefore R5C7 in box 6 must be a Six.
I did the same with columns 4-6.
Love that the creators name was arctan, and had a puzzle named 45-45-90
That in itself is as meta as the puzzle!
At the start of the solve, Simon erroneously removes a 2 from R2C7. Luckily, this can be resolved immediately by realising you need a 6 (1+2+3+4+5=15
Glad I am not crazy. Could have been 2 with double 1 on the arrow since it's diagonal and in two different boxes.
***IMPORTANT***
The "Squares" must be orthogonally laid out and NOT diagonally!
😊Hopefully i saved a few of you the thirty seven minutes i wasted unaware of this.
Interesting - I hadn't even considered the possibility of the square not being aligned horizontally/vertically 😮 - that would be brutal if they could be on the tilt, especially if they didn't have to stick to being at 45°
thanks i was about to ask this
Or any other angle.
I lost hours to that.
That's kind of huge. Someone needs to set a puzzle where the broader rule is used. I gather in this puzzle it was possible to actually rule out any possibilities there? In any case strange not to have it noted since there's even a whole tilted square pattern going on with the circles.
I hope removing 2 from r2c7 doesn't lead to a problem
I'm wondering about that too...
This can be resolved early on by realising you need a 6 (1+2+3+4+5=15
@@BenCragg1 I didn't even read your whole comment, just the first line was enough to give me a light-bulb moment and finish the puzzle within a few minutes. Incredible how some of these puzzles just need a particular way of looking at it and they become obvious!
Fortunately it doesn't, and I don't think the spurious elimination allowed Simon to make any further dodgy deductions, but it was definitely a ricket!
@50:26 An obscure rule would help Simon: Black digits also count as digits.
This is actually a SECRET
@@Quantris😁
21:30 The exact same thing
I feel like this should be a sign on Simon's wall.
Something I've noticed about these videos: Up until recently, Simon's video titles would capitalise the first letter of every word, including small connecting words (the, and, a, etc.), while Mark's video titles followed a more traditional method of capitalising just the important words, nouns and such. Interestingly, I've noticed that Simon's videos have begun capitalising nothing but the first letter of the title (with the exception of proper nouns and of course words capitalised for emphasis such as "PERFECT" in this particular video). I'm sure the conspiracy probably goes even deeper, and while I'm frightened of what I may unearth I believe it's for the good of all humanity that whatever cryptic message is being left by CTC is brought to light.
15:04 for me. Omg what a fantastic puzzle. The way everything suddenly falls into place is incredible. One of my favourite puzzles so far this year!!
Wow did Simon place those 6's the hard way (at 40:20 ish). Here was my reasoning:
There needs to be six 6's in circles, and there are only seven columns with circles.
Columns 4, 5, and 6 will support only two 6's total, so you will need to have a 6 in circles in each of columns 2, 3, 7, and 8. This places the 6 in column 8, which places the 6 in column 2, which places the 6 in columns 7 and 5, and then in columns 1 and 9 by sudoku. I had that in my first 8 minutes.
Agreed. Or by rows the logic is similar. Either way is vastly easier than what Simon did.
I usually don't read comments before writing my own, but for some reason I did skim a lot of them this time. I see that there was a quicker way to do something or other - but it is entirely OK with me that you did it the way you did, Simon. This also goes for proving something "the hard way" instead of some easier way. I love watching you solve, and even when you are not as efficient as possible, I feel that I learn something that I can apply to my own solving. I love it that you do not (neither you nor Mark) solve puzzles on video that you have already worked out a perfect solve path for. I far prefer that you do it cold, live, whatever you want to call it, even if that means that the solve is not a "best" solve. Your amazing brain is a privilege to watch in action. Thanks for this video.
Totally agree!!
The things is the way he did it used an assumption that 2 is not in r2c7 which made the puzzle much more approachable…
The break in for me was noticing that r7c7 couldn't be a 6 because then you could only place five 6s in the puzzle. Spotted it relatively quickly and from there it was an easy solve.
this this was brilliant!
That's how I did it, from there it fell into place quickly.
Classic Simon, he comes up with a wonderful but unnecessary deduction why 7 can't be circled, ignoring the fact that you have to put a 7 in box 5 and at least two 7s on the perimeter 🙂 Fascinating ruleset, I really enjoyed solving this one.
I think there is a loophole in the rules. At 25:35 Simon deduces the corners of square (greens), but there is multiple other options to form squares if you don't assume that the square is alligned with the sudoku grid. For example, R2C7 and R3C4 could be the circle corners and then R1C5 and R4C6 would be one of the "sum" corners. Forming a slightly tilted, but still a valid square.
Reminds me of a puzzle where there the rules referred to "square" and "diamond" clues but the "diamonds" had equal length sides meeting at 90°
The original rules do not contain this loophole. The rules were simplified down incorrectly for this video.
A circles counting puzzle is always going to pique my curiosity, and this ruleset was an amazing twist on them. I agree with what Simon says at 20:45 - these are my absolute favourite types of puzzles to solve (and watch Simon solve) - the ones with meta deductions at the beginning - unwritten rules to discover, new logic to get your head round. This puzzle had all that good stuff. Absolute genius - creativity and innovation ooze out of this one
Delightful! This is the sort of puzzle that has me kicking myself at the end in a joyful way -- I spent easily twenty minutes narrowing down my digits in circles, placing most of my sixes, parsing out the base logic, and not able to find a single break point past that, before I gave up and resorted to the video, only to discover that I'd forgotten a rule! (When you think the pairs of digits only have to be rectangles, not squares, the ruleset gets MUCH HARDER.) And then as soon as I knew that, because of all the groundwork I'd already done, everything else snapped into place immediately -- what fun! If only I'd paid attention to the puzzle name...
Thank you, arctan, for a lovely one.
I did exactly this, except I got stuck for 90 minutes before giving up and watching the video. It is surprising how far you can go without assuming squares!
this is why one of my core solving rules is "if you're stuck, reread the rules". the hours of my life i've lost lmao
I think that black 3 was my favourite digit in the puzzle. It was used in the logic of so many squares.
Ah Cracking the criptic with Simon. You guys have no idea how you help me go through my anxiety of the last few weeks. Getting better every day now, but you helped sooth me when it was at its worst.
So glad to hear you are getting better!!!
43:34 "Which two of these unknown yellows... are the 4s...?" I know you hate to hear this but Sudoku might help with that XD
I finished in 98 minutes. This was such a unique and cool ruleset. It was cool to realize that there must be 5 unique digits. I saw that since there were 16 circles, 8 pairs must exist. If we use four cells with 6 and 5 guaranteed, then 6 must take up six of the eight pairs, leaving the 5s unable to take 6 as a partner, giving a total of 11 pairs. This breaks everything. The only one that works is five cells, which 64321 is the only one that works. The 6s must be paired with all 123s, leaving 4s to pair with themselves for a total of eight pairs. That was very cool to see. Great Puzzle!
18:31 finish. For my break-in, I looked at the columns. There are seven columns with circles, however only two of the same number will fit in the middle three columns (only circles in boxes 2/8). Therefore, you cannot use 7/8/9. You do need 6, to have enough digits to fill all circles. Since only two of them can go in columns 4/5/6, therefore columns 2/3/7/8 must have circled 6s. This places the 6 in column 8, which places the 6 in column 2, which leads to the 6s in boxes 3 and 8. The last two 6s (boxes 2/4) can be differentiated by considering the square rule. The 6s in r2c7 and r7c2 need opposite corners on the diagonal between the two, and therefore neither circle on that diagonal can be a 6. The other two circles in boxes 2/4 are the last two circled 6s. Since the maximum sum is 9, 6s have to pair with 1/2/3, and the 4s pair with themselves to make 8 sums. Since you have placed all of your circled 6s, the cells in r3c3 and r7c7 have to pair with each other, and must be 4s. Sudoku will place the other two 4s, and the square logic will identify all of the sum cells. From that point, it is basically classic sudoku, with a pair of short arrows thrown in.
An excellent puzzle; fun fun fun!
It is stunning that this even resolves after all the circles are filled in. An amazing construct!
When is a chocolate teapot not a chocolate teapot? When there is a green three staring at it. 51:10
Yeah, the last five minutes of every Simon video is sudoku torture. given digits and regular sudoku are his Kryptonite.
😀
I was honestly a little happy to see that, because I did the exact same thing. Some minutes later noticing the 3, and realizing that it being coloured meant it had been there the whole time was definitely a bit of a facepalm moment for me lol.
It really is fascinating how Simon can elegantly crack the toughest of puzzles with extremely complicated logic that would take most of us brute-force bifurcation, but when it comes to doing very simply sudoku, Simon's brain proceeds to malfunction.
I've spent hours on this and what I'm getting from the comments is that the rules do not specify the squares must be orthogonal but that they are. Damnit.
After reading that comment it took me like 10 min to finish. That is a rule that really needed to be more clear.
Well, that's because the rules in Logic Masters do specify it. For whatever reason, when the channel simplified the rules to fit the format of the video, though, the specification was left out.
28:55 for me. Normally I struggle with puzzles where you need to sit with a global constraint for a while before you can start to think about pencil-marks, but this one felt perfectly pitched in difficulty, and I loved how it fell into place after I managed to get started. Really really elegant!
Same here - usually the puzzles that Simon loves, ie the ones where you need to do lots of theory before you can start attacking the grid - are beyond me, because I don't like going down blind avenues not knowing if I'm making any progress on the right path (this may be why I struggle to see SET break-ins) ... but this one was just right, the initial theory was accessible enough that I could work through it and figure out how to get started 👍🏻
can you assume the sides of the square align with that of the grid?
Interesting thought. Seems like that’s the intent though the rules are ambiguous. If the intent is the cells make a square, then the answer to your question is yes. If you’re allowed to draw a line between cells and that line forms the square, then no.
@@dustpan5356Even assuming that cells make the square, there's still a question about whether squares rotated 45 degrees from the axes (so their corners are NESW) count as squares for the rules
That's the issue i was having. The definition of a square does not preclude it from being rotated any number of degrees. A square is a square is a square.
I had to watch the video because I didn't assume this. I don't think there's only one solution if you don't assume this (after an hour of me puzzling it out, this is my conclusion, but I may be wrong)
Arctan has confirmed that was the intention, and that the clarification was accidentally left out of the rules while trying to simplify them. (I hadn't even considered the possibility of the squares being at an angle so I'm very glad that that wasn't the case!)
At 23'55 it is mentioned (Simon discovers) that the squares' diagonal is a diagonal of the grid. The two short sides being horizontaly or verticaly.
However, that is not necessary, you could have a square with the angles on top or down: The two cirkels in row 2 could form a triangle with R1C6.
Hmm.. nothing in the rules says that the squares have to be axis-aligned with the grid 🤔
tried it yesterday evening tired af and for some reason thought the rule was that it could be any shaped rectangle for some reason and that seemed way too complicated and i gave up. after reading in a comment that it was squares i facepalmed myself, went back at it (so it obviously doesn't have the time from yesterday added, so it's not _as_ great as it sounds) but i'm still proud to finish the whole thing in under 19 minutes then
when i got all the 6s in place, i had forgotten that the rules said SQUARES not rectangles. i spent 20 minutes trying to find a crumb of logic before i reread them, then finished in seven minutes after that LMAO
At 21:30 Simon makes an unproven assumption, I believe. The circle on the arrow between boxes 2 and 3 could be a two, were it's two digits both one. It turns out not to be a two later on, but it's still not obvious at this stage.
Hello ! I fell in love of a puzzle game, close to the “baba is you” style, it is very simple to understand and which you might like! It's called "Patrick's Parabox", I hope you will like it !
Tried this myself a few days ago but only just got the time to watch your solve, Simon. Really enjoyed the deductions, and I appreciated how you (unlike me) recognised how the 6s and the 1,2,3s were both necessary to make the pairs work - initially, I had considered 1,4,5 and ran into trouble, so such a deduction would have absolutely helped me, haha.
Still, I very much enjoyed this style of puzzle and am hoping watching more of these vidoes helps me to better understand the intricasies of this great "circles" format!
Got it in 38:40! The beginning went fairly smoothly until I hit a major snag that had me staring at the grid for a good 5 minutes. Then I reread the rules and remembered that the pairs of circles had to make SQUARES, not just rectangles, and the rest flowed pretty smoothly!
How Arctan, the setter, he dangles
Like carrots, the clues in triangles.
I seek that reward
But it's more like a sword
For my brain is a wreck and in tangles.
😁
Brilliant!
65:40 for me. I loved this puzzle. The logic was really intuitive to me. Just amazing arctan.
This was a complete joy to solve. The logic behind the contents of the 16 circles was simply beautiful.
I solved this one rather quickly, after figuring out that 7 could not be placed in circles. Then, you can easily place four of the six 6's in the grid, and the puzzle unravels quickly from there. The key to figuring out where to put the 6s is to realize that across row 4-5-6, you can only have two 6's, so every other row needs one. That immediatly puts a 6 in row 8, then by simple logic you can figure out the one in row 7, row 5 and row 2. It's only a few minutes from that point to solve the rest.
the box in the middle is also eliminating a digit in the circles. basically, if you can find n non overlapping sets of 1-9 with no circles, then the only digits that can be in circles are 9-n and bellow. it doesn't matter that there are 7 rows and columns because of the box, only 6 is possible from only looking at sets with no circles. this is simple enough to explain, non overlapping sets must have all the digits, and so 3 lots of all digits cannot be in circles because of the 3 non overlapping sets with no circles. :)
Another trick for circle rules you can use is checking the maximum number of circles in a single sudoku unit. For instance, R7 has 4 circles in it, so you need to make 16 up of at least 4 digits
Similarly you can check the minimum number of sudoku units that contain circles, in this case 7 contain circles
So you know that you need to make up 16 with a set of numbers whose count is between 4 and 7 inclusive, which eliminates the possibility of using something like 7 and 9 to make 16
If you split the sudoku grid into units comprising columns 1,2,3,7,8,9 and boxes 2,5,8, then you can get it down to just six units that have circles. (Also works with rows and boxes 4,5,6.)
I found the 6s very easy to place: You can only place 2 sixes between columns 4-6, which then means you only have 4 columns left for the other sixes, which makes it obvious from that point.
I adore this puzzle. Solving for the the boxes gave me so much serotonin when it clicked.
Can’t help but think that removing colouring once it doesn’t serve any purpose anymore would ease scanning the grid.
I agree. And the same is true for cleaning up pencil marks. Being tidy is arguable just as important, if not more, than focusing on the logic of the puzzle itself. The logic will only take you so far if you can't trust your pencil marks and your scanning. Besides, if you do to keep tidy, some of the logic actually becomes simpler anyway.
Was quite surprised to solve it in 19:25 given the video length. Very enjoyable puzzle with some novel, at least to me, ideas
This was one of the easier puzzles for me on this channel.
2:24:44 - I got the break in quite quickly but could get my head around the circles. Once I’d watched Simon for a bit I realised that the 4s had to be paired up.
The 6 proving the location of the other 6 was very nice
What a great puzzle, easy enough for a beginner but still took a while even for an expert.
I appreciate that arctan took the time to take the key digits (6 and 4) and put them in opposite corners of the full sudoku grid
I think the rules should say “different pairs cannot use the same cell for their sum” not “same square”? At least I think that’s what the example is saying, or else I’m being dense (likely!).
Quite right! Your wording would have made things clearer.
Middle box is not chocolate teapoted. The 3 goes in the middle of first column, since boxes 4 and 6 rule 3 out of all of the other options.
You can see the arrangement of the sixes in a much more straightforward way:
You have 7 rows to place the sixes. Notice that you can only put 2 sixes in rows 4, 5 and 6 alltogether because of Sudoku.
Now you have to place sixes in every row out of the rows 2,3,7 and 8 and two rows out of the rows 4,5 and 6.
This implies:
R8C5 is a six => R2C7 is a six => R7C2 is a six and R5C8 is a six. Now since you still have to place a six in one of R4 or 6 and this is only possible in C3, you can't place a six in R3C3 hence R3C4 is a six. Finally, since sixes can't pair with eachother and R2C7 is already a six, you can place the final six in R4C3.
26:39 for me. This was so satisfying! It flowed so well after I figured out what went in the circles.
6x6 in cirlces, none in columns 1 & 9, one in block 5, so r5c8 must be 6 for col 8
What I should have said is that there is a 6 in columns 1 and 9 and one in block 5 and the other six are in circles. So the only one in column 8 is in the circle.This then gives r7c2 as 6 and sudoku helps a bit from there
Indeed. And a similar argument works for rows, making r8c5 a 6 also.
Finished in 25:58, clever puzzle!
one of them use 5, and 5 and 6 together cannot work, because they sum to 11 and you would necessarily get 5+5, 5+6 or 6+6 all of which cannot satisfy the square corner rule. so we must use the first sum and the circle numbers are 1,2,3,4 and 6 :)
I started with the 6s.
You can only put six 6s in the circles if you use the circles in box 3 and 7
This means you need to use the circles in row 8 and column 8 (they are the only left circles in box 8 and 6)
Now you need to put one 6 in box 2 row 3 (r3c4 and r3c6) and one 6 in box 4 column 3. (r4c3 and r6c3).
The 6s in box 3 and 7 need to pair up with a circle on the diagonal between them.
This means that r3c6 an r6c3 can not be 6.
And now you have your six 6s.
The circles in box 1 and 9 must pair up with each other and since they do not contain a 6 and all 1s, 2s and 3s must pair with a six they must be 4.
Now the only other 2 circles where a 4 is possible are r2c5 and r5c2
And you have your 4s with corresponding 8s
From here you can easily pair up all the circles.
And then I got stuck.
My time was 30:16. A very fun puzzle!
At 21:37 why can't R2C7 be a 2? Its arrow spans two boxes so could contain two 1's.
Very late to this one, which took me 2:20 (hours!) because I did not assume squares lined up with the grid and kept drawing 45 deg and other pinwheel-type tilts. I finally narrowed to 90 and 45 only and at least managed to find most of the 6s.
FWIW, all but 2 of the circles are on the same bishop's color (required if you want a 45 degree square). The two rogue circles had to be 4s with a non-circled 8 sum, and I could work out the other pair of 4s and most of the 6s, and begin eliminating positions for 2s even with 45-deg squares in play.
Finally narrowed to just 90-deg and solved fast, with more of an extended classic sudoku run-out than we're used to here. I wish I'd read the comments for the accurate ruleset, but it was interesting to see how far I got without the constraints.
Has anyone mentioned this?
Normally with these circle puzzles, Simon adds them all up, then figures out their make up. There are 16 circles so there can be only 1 way: 1-2-3-4-6.
Not sure why he's even considering 9-8-7.
Yeah but now, I need help in the next steps! Ha!
At several points in the puzzle I missed Simon considering 45 degrees rotated squares! Where in the rules is it stated that, say r7c4 and r7c6 could not be two paired circles, being two opposite corners of the square r7c4 - r6c5 - r7c6 - r8c5? I was not at all able to solve the puzzle until I switched to assuming only orthogonal squares are allowed. I'd prefer this was mentioned in the rules (if it was intended).
Re: 24:15 - Why can't squares be tilted? Why can't r5c8 be paired with r2c7, with the other corners being r3c9 and r4c6? It doesn't seem to be disallowed by the rules, and I can't find any deduction within the puzzle that would suggest it doesn't work.
i think the title of the video disallows that. what you mentioned would be a trapezoid (assuming the center of each cell is used to "draw" the squares) and they wouldn't have 45-45-90 angles in between
I also think about it. And the only answer, thatI find is that we're talcking not about dots in the middle of the cells, but about the cells. So the sides of the square shoul be parts of row/column.
I was wondering the same. r5c8, r3c8, r2c7 and r4c6 would form a valid square. I'd guess there's huge amount of solutions if tilted squares are accepted.
Much easier way to prove that 7 can't be circled is to consider this. Row1, Row9 and Box5 is each a complete set of 1-9, and none of them have circles, ergo there would be at least 3 digits of a circled number that aren't actually in circles
At 51:00 "chocolate teapotted", rechecked and verified. Nope. You know the drill. Sudoku fatigue is real.
51:20, I started really quickly, got the numbers in the circles right away, but then hit a wall. Eventually I looked at the video and saw that I made a really dumb mistake with the circles (I placed the 4's and 8's right away, then for some reason didn't rule out one of the 8's as a possible cell for one of the other sums and didn't notice that I was using it twice). Eventually finished after undoing that error.
I was drawing boxes and marking cells as I was figuring it out, but I erased the markings as I figured them out, which is how I made that mistake.
Also, I did some trial and error working on the numbers in the circles, but noticed a much faster way of ruling out some of the options when I was finished.
Hey Simon, I wasn't solving the new app puzzle #54 and when looking at the hints, it seems like there's a king's move constraint that isn't in the rules?
This took me like 5 hours. I don't think I did this the intended way. The puzzle fell into place once I realized that the the following pair-and-sum 3-tuples were the same sets:
"Purple": R2C7 circle [6], R6C3 circle [candidates 1 or 3], R2C3 Sum [Candidates 7 or 9]
"Yellow": R4C3 circle [6], R7C6 circle [candidates 1 or 3], R4C6 Sum [Candidates 7 or 9]
Because There was a third "Blue" 3 tuple of:
R3C4 circle (6), R6C7 circle [1 or 3], R6C4 sum [7 or 9].
With three colors and 2 candidates, two of the colors had to be the same.
Blue and purple have a 1-3 pair on row 6 so they must be different
Yellow and Blue have a 7-9 pair in box 5 so they must be different.
The only combination that remains is Purple and Yellow.
This tells us which color has the 1 circle (Blue), and which colors have the 3 circle (yellow and purple ), because you can't have two colored circles with a 1.
24:05 ... it's hip to make squares! (the creator's name is also rather apt)
Nice puzzle!
uh at about 21:45, the zell in row 2 column 7 could very well be a 2 becaude the two cells the arrow points to aren't in thr same box and could therefor be 1 1
Pleased to get through this one, found it quite tricky in places, but it's all there, you just have to find it. Enough platitudes already.
41:00 To the beautiful logic by Simon to disprove the 6 in r3c3, we might add the following: suppose that r3c3 is indeed a 6, then r5c2 and r2c5 are also forced to be 6. Then in box 6 the 6 is forced into column 7 and in box 8 the 6 is forced into row 7, which rule out a possible 6 from r7c7. Now we have only 5 circled sixes, so we are one short. Starting from r7c7 the other way round, the logic is identical. So r3c3 and r7c7 cannot be 6 and both must be 4.
The other confusing rule example is having a 2x2 corners 68;82...where one of the 8s is sum for another pair...
Doesnt the 2nd 8 allow for this configuration..., 62 pair is using the other 8 as its sum rather than the 44s 8. The rule only states the sum square cannot be used for 2 pairs, it doesnt say they cant be equal so long as the other sumcorner is the same.
Has a name been coined for this type of puzzle? I vote for Xircle.
Help Mark and Simon to know where we are watching from.
Guyana🇬🇾
Haven't started the video yet, but one thing I'm not quite clear about regarding the wording of the rules - maybe it will become clearer when I watch the rule part. Does "grouped into pairs" imply no digit can be in more than one pair? Or does it just mean each digit must be in at least one pair,? In other words: can one circle be a corner of several squares?
Example: Can we have r2c5 pair up with r4c3 as well as with r3c4 and also with r2c5?
43:35 Simon studiously not removing 4s from central pencil marks affected by placement of 4 in R7C7 whilst pondering six 4s in remaining circles.
My brain sees those with ease but struggles to chain together the multiple steps of logic that Simon does so well.
Very clever and elegant puzzle!
Amazing puzzle and idea! It's not difficult at all once you know the break-in! 00:12:09 for me
The rules aren't clear about how the squares must be oriented. I interpreted, because of the puzzle title, that squares could be oriented either orthogonally or diagonally (i.e. pair r2c7 and r4c7 sum at r3c8).
I think you can place there 6s right off the bat. If you have to place six 6s in the 7 rows with circles but you acknowledge that a 6 in box five will eliminate one of the middle three rows then all of the outer four rows have to have a 6 (R2 R3 R7 AND R8). In R8 there is only one circle so that placed the 6 in R8C5. Which makes the only place for a 6 in R2 to be R2C7. Those two 6s then rule out all but one 6 in R7 which is R7C2.
Sometimes I make logical conclusions that are totally wrong so someone help me out with this because it seems right but I might be missing something
Just to make sure on the rules: where exactly does it say that the squares have to be orthogonally connected? Like, why can r3c6 not pair with, e.g. R7c4 with r4c3 as a 3rd corner of a square?
42:13 for me
nice puzzle
love it
thanks so much for explanation!!! i see you have a guitar behind, it would be so cool if there were some kind of puzzle connected with music or pitches, idk would that actually work but i just got the idea it may be possible to make a puzzle related to music👀
22:30 all sums are different than digits allowed in circles. So that rule about sum being allowed in a circle is redundant. or just to put you off?
The setter just called himself 45 in a posh way
Interesting puzzle. Similar to my "SqUare-DO-sKUw" puzzle I submitted on Logic Masters Germany a couple years ago. So far, solved by 10 or 11 people (that I know of), including the most recent Tetris world champion, fractal.
Could you please include not only sudoku pad link to the puzzles, but also to their logic master's page?
If you imagine the grid divided into the following sudoku 'units' (each a set if the digits 1-9), then it helps seeing 6 is the maximum that can go in a circle (and also helps with placing some of the 6s):
Cols 1,2,3,7,8,9 + boxes 2,5,8
Or
Rows 1,2,3,7,8,9 + boxes 4,5,6
(Edited, as had my rows and columns mixed up)
Fun puzzle, after I figured out it has to be 1,2,3,4,6 I started by looking at what circles absolutely must contain a 6 and what can not contain a 6 based on the fact the circles are only in 8 boxes and 4 of those boxes all have only 1 circle, so for example 6 can not go in 2:5 because if it did it would knock 6 out of box 3 for sure and then either boxes 7 and 9 which is already too many to box 8 but because 7 and 9 can only have 1 6 it knocks one out of at least one of them therefore 6 can not go in 2:5 which then means box 1 and 2 can not both have a 6 which then really restricts 6s into very few spots, and after figuring out which spots had to have 6s and which couldn't then I did the same with 4s since 4s can't connect to 6s after that there was a forced 7/9 pair in box 5 which allowed all the 1s to be removed from all but 2 possible circles, then after that the entire rest the puzzle was just normal sudoku.
And I thought it was impossible to square a circle. ;)
26:11 very happy with myself. Good Sudoku puzzle
Simon made the grid look like a bowl with non-stop (candy).
21:30 Simon says the two circles with arrows cannot be 2, but the one in box 3 could be 2. How can we eliminate 2 properly?
I am unsure if Simon's conclusion of possible squares @ around 25:00 is correct.
Could you not in theory have r2c6 pair up with r8c6 making either R6c2 or r6c8 it's right angle corner?
At 25:49 Simon deduced the green square partner had to be r6c3. It could have also been r6c7 (unless I missed why that was eliminated as an option). Great solve either way!
I think it was the setter's intention that the squares should not be tilted, but have sides that align with the grid. As others have pointed out, the rules don't explicitly state that though. Simon was solving based on this assumption, and the fact it solved so neatly this way makes me think he was correct to do so.
The rules state that each pair form opposite corners of the square, so they can't be in the same column.
@@RichSmith77 Agree now, but using an assumption…. is not the best way to solve… which I have learned several times over.
@@zondebok980 That is not correct if the square looks like a diamond square.
@@rcambo4818 I know it's not conclusive evidence, but both examples of squares given in the rules were aligned with the grid. I admit, I never even considered the possibility that the squares might be tilted during my solve.
(I did get stuck thinking, for some reason, rectangles were okay though. I had to watch Simon's solve up to the point he first mentioned they needed to be squares during his solve. It suddenly became a lot more straightforward. My numpty moment 😅.)
35;31 for me. Fantastic puzzle!
I'm sure your explanation of the puzzle title is succinct. I just thought 45-45-90 was arc tangent.
good 35 min solve here
Because the central boxhas no circles and the outer lines are blank, boxes 2,4,6 and 8 are sort of xwings of 6s. So column 8 must instantly be a 6 or all of them cant be placed. I am pretty sure anyway as I haven’t actually solved this one. 😊
It was "That's a" rather than "high digit" that likely activated Alexa.