The Best Arrow Sudoku Ever?
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- čas přidán 2. 10. 2021
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Today's video features a puzzle that will live long in the memory. It's called Bowser's Castle (please explain this to us!) and it's by Emmettcito. It's been recommended by just about everybody and with good reason. This is a masterpiece. Give it a try at the following link:
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Rules:
Normal sudoku rules apply. Digits along an arrow sum to the digit in that arrow's circle.
Sam Cappleman-Lynes has made an extremely clever 6x6x6 set of puzzles. We have released a solution video for this puzzle on Patreon and you can try to solve it yourself at these links (one to our software, one to penpa (which allows you to increase the size of the grids):
app.crackingthecryptic.com/su...
git.io/JuB3x
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![Pseudo Coup: Sudoku With a Difference! [Audio Fixed]](http://i.ytimg.com/vi/5speYaGTNcc/mqdefault.jpg)
I think it’s been long enough now since I made that puzzle that I mostly forgot how to solve it. This video is bound to be a treat :)
Why is the puzzle called Bowser's castle?
@@Zvoosh25 I thought the layout of the arrows looked like a castle, especially the ones on the top and bottoms three rows. And since it is a hard level I thought it would be appropriate to specifically call it Bowsers Castle, as that is always the final level of any Mario game and as a result the pinnacle of difficulty in it.
@@emmettnelson7260 You know what? This is a beautiful puzzle, but lately there have been so many videos by Simon and Mark about brutal applications of SET (Set Equivalence Theory) that the break-in of this puzzle did not seem very hard to me. I guess many of the 4K CTC subscribers would describe it as one of the most enjoyable, but not one of the hardest puzzles they have ever solved or seen.
This time, Simon was slower than usual. I bet Mark would have solved this puzzle in 30 minutes. For instance, see:
🔹czcams.com/video/DbVOzckHLUY/video.html
🔹czcams.com/video/McNmBk4wrh0/video.html
@@emmettnelson7260
And i thought it was the way to win the lvl... first go down than middle and so on.
@@Paolo_De_Leva well a hard puzzle doesn’t mean the hardest puzzle. I just considered hard enough to warrant that name
First step: Get an X-Wing
Second step: Ignore the X-Wing
Jokes aside, it was a very good solve, specially the second half :)
@@Wecoc1 No, it was not a very good solve, just because of what you wrote first. Get an X-wing, talk about it for minutes, and then forget all about it.
That's why I said "the second half", where he redeemed that mistake with the logic on that r9 arrow and so on. I don't think he got off the rails too much after that.
@@Wecoc1 I quit the video after watching him not even trying to look at the 9's for several minutes. I get way to involved to be able to handle my frustration in situations like this.
I agree that part was frustrating, not gonna lie, specially since one of the 9s was over an arrow's _tail_ 😂
29:41 Did Simon really said "the sum of the digits 1-9 is 45" WITHOUT SAYING THAT IT IS A SECRET? Now the whole world will know...
Exactly what I was thinking
The secret is no longer a secret. 😢
@@gustafbengtsson6666 sad times...
Noooooooo
Mark wouldn't even tell us the secret in the newest GAS video. I feel sad.
Classic Simon logic, immensely clever logic to prove that there must be two repeated digits in the corners and that a 5 went with double 8. Then immediately forgets and finds a much more complicated way to make progress
I was really hoping someone else noticed that!!! 🤣
@@92maitani 🤣🤣🤣 Exactly what I came to the comments section to mention.
Same! Imagine how good he would be if his memory was longer than 3 minutes lol. I say that as someone with shit memory myself lol
He also found the X-wing on nines and then immediately penciled nines into the spots they can't go in.
I was screaming at my monitor over this.
16.04 Simon: "What we cannot do is put any more 9s in row 1 or row 9"
Promptly puts two more 9s in each.
Hmm.
Moreover, r9c7 is on an arrow and he still pencilmarks 9 there 😅
@@sniper1a259 Mistakes must be catching after Mark's whoopsie last night...
@@sniper1a259 🙃
@petrus So true...
and that was literally 2 minutes earlier, not like he was nor thinking about it for an hour ...
Simon: “you can’t put a 9 on an arrow”
Simon: 4 minutes later, pencil marks a 9 on a 2 digit arrow. 🤦🏼♀️
Simon: "We have an x-wing on 9s in rows 1 and 9. Let me explain the implication for the rest of rows 1 and 9"
Simon: A few minutes later, pencil marks 9s in additional cells in rows 1 and 9. 😂
Rules: 03:29
Let's get cracking: 03:54
And how about this video's Simarkisms?!
Sorry: 8 (19:51, 23:22, 23:24, 23:24, 35:14, 49:23, 54:06, 56:52)
Bother: 6 (34:02, 34:29, 34:32, 35:59, 36:01, 46:36)
Hang On: 6 (08:56, 09:00, 22:33, 32:48, 37:54, 39:56)
Clever: 5 (13:30, 30:38, 33:41, 41:27, 43:02)
Extraordinary: 4 (00:20, 02:29, 58:32, 59:43)
Incredible: 3 (01:13, 41:53, 42:00)
By sudoku: 3 (43:59, 44:48, 47:08)
The Answer is: 2 (15:43, 38:27)
Nonsense: 2 (19:51, 22:20)
Lovely: 2 (36:22, 47:01)
Gorgeous: 2 (53:57, 1:00:55)
Bobbins: 1 (51:23)
Good Grief: 1 (51:38)
Naked Single: 1 (51:55)
Beautiful: 1 (42:54)
Fascinating: 1 (01:27)
Ridiculous: 1 (42:59)
Come on Simon: 1 (49:12)
Approachable: 1 (03:00)
Of All Things: 1 (56:34)
FAQ:
Q1: What is a Simarkism?
A1: A Simarkism is something that Simon and Mark typically or frequently say.
Q2: How do you do this so fast?
A2: I'm not made of flesh and blood, but of sand ...
Q3: Why don't you include 'XX' and 'YY'?
A3: Probably it's already on the list ('Scooby-Doo' for example), but not mentioned in this video. But if you think it's not, tell me what you'd like me to include and there's a good chance I'll add it!
Q4: You missed 'XX' at 'YY:ZZ'!
A4: That could very well be the case! Human speech is hard to understand for computers like me, especially British sometimes! Point out the ones that I missed and maybe I'll learn!
Q5: Could you turn these statistics into videos?
A5: I've been playing around with the idea and I'm open to input as to what people would like to see. Let me know if you are interested in this and/or have suggestions.
I just realized that the time the video uploaded is the same as your comment…
@@fulgentiuswilly5043 they are not made of flesh and blood...
Thank you silicon man. Since you came along, I don't have to waste time sniping for the Actual start of the puzzle.
Only one "beautiful" ?! What is wrong with Simon? Is it a full moon or something?
Nice solve. We'll get past the ignored x-wing. It's always easy to criticize when you're watching. What amuses me the most and kind of annoys me a bit but in a funny way is when Simon gets excited he goes "that's a 5!" Keyboard goes "clic clac clac clic clac" puts a 6, deletes the 6 puts a 5.
I think it’s called Bowser’s castle because the arrows are similar to the spinning bars of fire in Bowser’s castle.
That's what I was thinking, too
When Simon announced that it took him 41 minutes to get the first digit, I said what? I’ve been listening that video for that long already? 😲
This is gonna be one of those "just watch don't attempt" videos, time to get the popcorn
i thought the same, when i saw the duration of the video. but i gave it a try and i solved it in 48 minutes. looking the video now, to see what i missed during my solve.
it is doable, though you have to sweat the available clues pretty hard.
my solve was different from Simons, so there are multiple ways to approach.
All CtC videos are those kinds of videos for me nowadays.
Wow, it is the first time I found a more elegant approach for the puzzle, than the one presented in the video :) color columns 2 and 8 (sums to 90 according to a "secret"). Then recolor 3 cell arrows instead of the colored circles (doesn't change the sum) as well as corresponding arrows of circles r4c2, r6c2, r4c8, r6c8, which will lead to isolated colored cells r5c2 and r5c8. And then just solve from there, the sum of colored cells is 90, but by putting minimum numbers in the colored cells of 9x9 boxes 1, 3, 7 and 9 you can reach 84 (the sum of digits 1 to 6 multiplied by 4), and in row 1 you cannot have digits 789 occupy 4 uncolored cells (therefore 1 degree of freedom goes away) as well as in row 9 you cannot have 789 in 4 cells, additionally there is a 2 cell arrow, the head of which cannot be 6789, without giving up another degree of freedom of uncolored cells, which in the end gives away 12 pair on isolated cells as well as an arrow 6=1+5 on row 9. Thanks for reading the long comment ;)
This was the way I solved it. Usually I take around about as long as Simon (much slower in reality as I he takes the time to explain things) but had this one done in 11:21 with that break-in. Lovely puzzle.
Can you help me understand what you mean by "and in row 1 you cannot have digits 789 occupy 4 uncolored cells"? I understand everything up until there, but that completely lost me. Did you mean column 1/ column 8? Because there are 7 uncolored cells in both row 1 and row 9 following this logic, not 4.
@@zach31194 He means 4 uncolored cells amongst the six cells in r1,box1/3. Imagine these are colored in blue, and the original twelve cells are colored yellow.
Try as we might, we cannot fit 4 cells with numbers greater than 6, so one of the twelve yellow cells in box 1 and 3 have to be 7 or greater, bringing the sum up to at least 85.
Do the same with r9,box7/9(coloring the 4/12 cells red/green accordingly). But this time, we have the circle on r9c5 restricting this further.
Remember, the two cells on r5 (let's color these purple) sum to at least 3, so we're already at 85+3=88, so r9c7 cannot be lower than 5 .
If it is indeed lower than 5, putting the red cells as 4789, then the lowest possible sum of the yellow-purple-green regions would be (21+22)(yellow)+(1+2)(purple)+(21+(1+2+3+5+6+7))(green)=91, which contradicts the beginning of our logic that these cells sum to exactly 90.
Therefore, r9c7 is 5, the arrow is therefore 6-1-5, and the purple cells are a 1-2 pair, the yellow cells are 1-2-3-4-5-6 and 1-2-3-4-5-7 in some order, the green cells are 1-2-3-4-5-6 and 1-2-3-4-6-7 in some order, the blue cells are a 6-7-8-9 quadruple, and the remaining three red cells are a 7-8-9 triple.
I like when he says "That's so clever," as if he's not the clever one.
I got through this without doing any crazy set stuff. The secret for me was focusing on R9C7 after you figure out the arrows must be 789s. That means all the arrow cells need to be from 1-6. If you look at that cell, it sees all the arrow cells in the last three columns so that rules out 1-4 and that needs to be 5 or 6 (otherwise you''ll have problems summing the arrow in row 9)
Thank you, this is easier for me to grasp.
And how do you go from there? After Simon deduced the 56 in r9c7 I believe his next step was using his knowledge from SET to eliminate other 56's from the corner boxes, because those would cause the SET sum to be too low.
@@johabee Sorry, I did leave some logic out. So after you know that the 1 and 2 are the values that repeat and determine that the circles can be 7, 8, or 9. We know that one has to have (234=9, 125=8) because they don't share a 1. The question is can the other be (124=9, 135=7)? Well, if they both have the digits 1-5, then R9C3 and R9C7 also have to be >=6. Since R9C7 is on the arrow, R9C5 must be >=6. So that leaves cols 2,3,5,6 and 7 that would have to be >=6 which is not possible.
So both arrow pairs are 98. Now consider R7C3 and R7C7. Those also see all the arrow cells and with the stubby arrow now see 9 and 8. That means those must be 5,6 or 7 just like R9C3 and R9C7. Since each of the sets of the arrow cells must have a 5 or 6, row 7 or 9 must have a 7 in column 3 and 7. If you place a 6 in R9C7, that forces a 5 in R9C3 which means the 7s would be forced into R7C3 and R7C7 and you can't have 2 7s in R7. So R9C7 must be a 5.
Not all that more elegant, but no set theory involved. Hope that helps
Only took me 19 minutes! Break-in was everything here. I did it by shading columns 2 and 8, then moving them around so you have 6 cells shaded in boxes 1, 4, 7 and 9. If you take the remaining 3 cells in each box together, the max they can be horizontally is 2*6789 + 2*89 = 94, and so the min of the shaded cells in the boxes is 180-94 = 86. Since the remaining shaded cells r5c2 and r5c8 are min 3, and the total shaded area is 90, that leaves 1 degree of freedom, which you have to use on r9c7 which can never be 6.
did the same :) it also gives away an arrow on row 9, wrote a comment, but was not as fast as you :)
Also my break-in. Certainly wasn't as quick to find it as you.
I did something similar... Started with the same shading there, but then shaded those against boxes 7 and 9, so I was comparing 6 cells each in boxes 1 and 3 plus r5c2 and r5c8, versus three cells each in boxes 7 and 9 (r8c3, r9c2-3, and the mirror in box 9). Reasoned that I couldn't put 123456 in boxes 1 and 3, because then I have c1,2,8,9 all being 789, and I needed to skip a digit in r9c2,3,7,8 to make the arrow work down there...
Of course it took me 32 mins, not 19... I'm not as fast :)
Still, shocked to actually do one faster than Simon or Mark.
me too! i felt so proud of myself for solving it so much faster than he did
Very nice. Wish I'd seen this. I was able to limit a whole bunch of cells to 3 or 4 values using different sets. If I'd seen this, though, and the fact that if R9C7 was a 6 it would put a duplicate 7, 8, or 9 in the row which we are already assuming maximums of 7, 8, and 9 to make the sum work...wow. Good spot! I saw all the implications that got Simon where he was, but I got stuck until I watched the video to where he mentioned the mirror cells and the implications for boxes 4 and 6 in columns 2 and 8. That broke it for me, but I like yours better. It immediately throws that one cell under the bus so to speak, giving you the 5, and from there you know that there is no degree of freedom left, so you have to have 789 in the cells of row 9, an 89 pair in row 8, an 89 pair in row 2, and 6789 quad in row 1. The arrow in row 9 is now forced to be 6->15 because otherwise, duplicates. Pretty! :)
Was fun to hear my name called out, hadn't expected it, thanks! Only saw the puzzle challenges at the end of the day, ended up staying up til dawn to finish them. Had a few that slowed me down, but was still an excellent series of solves. Was surprised that I'm among the first handful, I'm only of middling skills, but have learnt much in the few years I've been subscribed.
Many thanks to this channel, I greatly enjoy the work you do!
the trick for this one is to look at the sum of columns 1,3,7,9 then move as many coloured cells into boxes 4 and 6. you end up with 1 degree of freedom that gets resolved by the arrow in row 9. from there on its no that difficult to finish.
The trick was (very slowly) explained by Simon from time 25:42 to time 29:51. This should be immediately repeated on the right side of the grid (using boxes 3, 9 and column 8),...
eventually "you end up with 1 degree of freedom that gets resolved by the arrow in row 9."
Yeah, I instantly saw this break in because there was a puzzle not long ago that used set theory with arrows in the corners. I immediately colored in the columns with the stems of the arrows and the rows with the circles. Then moved some cells around with the circles around box 4 and 6 and the puzzle broken open afterwards. Looks like there were many ways to go about this one.
Or you look at columns 2 and 8 and shove numbers in boxes 1,3,7,9. Same dof, same resolve. Bam, first digit after 10 minutes.
@@schneebaer42 Oh, it didn't take even 5 minutes for me to get a digit. Those arrows in the corners were a dead giveaway. But yes, there were many ways to break in and they seemed well telegraphed to me.
Excellent puzzle indeed. I am in a state of total shock that I managed to solve it. 45 1/2 minutes to get there. Somehow I managed to find that break in and I confess to letting out a (quiet) shout of joy when that bottom arrow went in and all the sets and max/min started to pay off. So much fun.
Without learning ideas from past puzzles on this channel I would have had zero chance of solving it. None at all even if I'd kept trying until this time tomorrow. Nobody can say all the lockdowns didn't teach us anything!
I sneaked home in 57 minutes. Delighted to beat the hour. A wonderful puzzle, I really enjoyed it.
My time 30:39. I finished faster then the video length before but never this much faster. I was able to use set theory at the start. I was slower to see the restrictions on the 12 pair in row 5 and the circles in column 2 and 8. And my scanning is slow compared to Simon's. Anyway I feel awesome. I enjoyed Simon's solve. Really loved the puzzle, what a great channel
I love this but I think we all ought to know what the Voldemort emergency was
30:15
Wow, this was a crazy ride. I started colouring both sides (18 colours!) before spotting the break-in and made an incredible Pollockesque mess trying to reconcile the two sets of 9 colours, but got there in the end after a lot of fun.
Wow! Feeling really proud that I got a start within thirty seconds and was well ahead of Simon.
Fast forward to the end for the solution. Job's a good 'un!
you should make a playlist of favorites/essentials/core videos
Maybe even playlists by type, or by setter.
Took me 75 minutes after I saw the set logic in the video. Quite the difficult puzzle!
Took me just under 45 minutes. One of the few times I actually finish in less time than the video is.
Solved this in less than 28:00 minutes and impressed myself :))) had a different approach though
I did this one on my own, took me about an hour - thus close to Simon’s time - and I am so proud of myself! I usually have to look at the videos with those ones. It was lovely to solve, thank you, Emmett!
I love this one so much that even after weeks I still got impressed by it, and I recorded my solve of this puzzle in my ‘mini’ channel too :)
Simon, after doing your set theory break in on the right side, you were left with 2 trinominos containing double 9s, a 5, a 7-8 pair, and a 7 or 8 which had to add up 45 more than the 1 or 2 in row 5 col 8. The only way to do that is choosing the 8 and 1 giving 2 digits.
After a little bit of clean up, the same logic can be applied to the left side afterwards, where your left side where you force max the values in the trinominos because a 2 is forced into the r5c2. Follow through with your logic!
This was a ride! The rare occasion where I manage to muddle through the puzzle myself for an hour-long video; usually I just give up and watch the video first on these, but I'm glad I tried first -- even if Simon explained it more clearly.
I also watch playthroughs of various games in Nintendo's Mario series, and I must say, I love the use of Bowser in the thumbnail.
Loved watching you solve this Simon. Great video!
Oh, I've been expecting/waiting for this one for a while, this is going to be a treat!
It would be nice to have letters in the sudoku app
As an example, Id prefer to put an A or N or X instead of colours
Hey, Simon and Mark! So glad to see you all got to 400k subs! I've been here for a few years, and if I recall correctly, you had maybe 30 or 40k subs when I joined. It's really crazy to see how fast you all have grown! Anyway, in an effort to please the Algorithm, I do think there are a few things you can do to help your videos reach a wider audience.
A little bit of clickbait never hurts anyone. Honestly, something grabby like "It Took Me 40 Minutes To Get ONE DIGIT?" is a bit of a spoiler, but it doesn't ruin even an ounce of the logic you used to get there, or give away anything too important. The thing is, all the fans are probably going to watch the videos regardless, so you've really just gotta target the broader audience at this point.
Greetings from America! You both got me into sudoku puzzles again after years away, and I feel sharper than ever. So, cheers!
24:34 for me. That break-in was amazing, I was lucky to spot it quite fast. Awesome puzzle!!
Luckily I bought more popcorn - so prepared for tonight's entertaining puzzle solve! And more for Tuesday's Babba is You! Looking forward to video with answering questions tomorrow!! 😍
The logic was even more beautiful than you thought! You were so close! When you got the 5 in r9c7, you could have gotten loads of digits including the purple 6s and the blue/yellow 1/2 in row 5. All this could have been deduced from your Math between blue and orange.
I was not able to spot the set theory and watched the video. I was surprised that Simon gave up on the set logic once he started getting digits, the 5 means it is at most 5-8-9 (22) + 7-8-9 (24) = 46, which means r5c8 can't be 2 (2+45=47) and that unwinds the entire rest of the solution.
Wow. What a puzzle. Holy smokes. Insanely beautiful puzzle.
Brutally difficult yet incredibly enjoyable.
My set breaking. Blue Rows 1, 4,5,6,9
Red Columns 1,3 7,9
Remove a blue set from box 5. You now have equal sets. Cancel out everything, like the short arrows you can end up with 4 red cells that at most can be 9,8 pairs = 34. And 5 cells in row 1. 5 cells in row 9. And 2 cells in row 5.
Because of the really neat logic in row 9 With it's length 2 arrow. You get min (1+2+3+4+6=16). Row 1 you get a regular (1+2+3+4+5=15) and with a 1,2 pair in row 5 you get a total of 34.. this gives you the 6,1,5 in row 9. It gives the 1 in row 1. Leads to the 9 in row 5, leads to 34 pair leads to 2 in row six. The 9 decides the 9,8 pair in row 8. Then if you can follow the logic about repeated digest on the arrows in columns 1 and 3 can be in row 5 column 2. The puzzle quickly solves. Really lovely puzzle,.
33:37 for me. Fun little puzzle. Didn't hit any obstacles. Just slow entering digits.
When you got the 5 in r9c7, you could have used the logic for putting a 1,2 pair in the centers of Boxes 4 and 6 to figure out that r5c8 is a 1.
Simon: does very complicated logic and math to prove that certain cells need to sum up to a minimum value.
Also Simon: completely ignores the result of this logic and finds other complicated logic to fill in the grid 😀
I know!
At 44:00 point, he knows the reds in c7-9 sum to either 46 or 47. They can only reach 46 if r9c8 is 8, also giving yellow as 1. It was genius working out the break-in, but he could have completed it within 10 minutes from this point if only he'd carried on with the earlier logic. 🙂
Having great fun with padlock btw :D
Got through 2 of them and working on the 3rd now
16:03 for me, such a great puzzle!
28:33 - kind of proud of myself. Looking forward to see where Simon struggled, for me it was a very smooth solve without any complicated deductions (apart from the break in maybe)
Love the fact that Simon:
1. Spent 40 mins of video figuring out math of 45+1or2
2. Was being able to use that 5 minutes later (right)
3. Haven't used it ever :D
Solved this puzzle and then watched the video. Apart from the initial 1-2 pair in the center row, my solve was almost the exact opposite of Simon’s. I had the stubby cells down to 2 possibilities which then told me about the 789 cell blocks in the corners.
I got the reason for the arrows to at least have a 9 via another route.
We've said that the two arrows had to have one repeated digit and one only. However, no combination of numbers adding to 6, 7 or 8 would escape that fate. All the options would share a 1, but 6 is 1, 2, 3 - 7 is 1, 2, 4 - 8 is either 1, 2, 5 or 1, 3, 4. You just need a 9 to make sure you can differentiate in more than 1 digit.
I have to admit. I didn't really like the break-in. But the puzzle afterwards was an absolute masterpiece by any standard.
26:35 for me! Think that has to be my best arrow solve ever and a gorgeous puzzle from start to finish 😃
I consider mine a good solve because I watched the 40 minutes of video until the break in,
After which I started the puzzle and finished in 20 minutes.
Which is exactly the time it took Simon to finish it...
bowser's castle ia also full of platforms that move in arrows and then fall in the lava
I can't remember if I've commented before, but this certainly deserves a BRAVO!
That was a blast! I ended up taking a coloring based approach in contrast to simon's set based approach. Seemed maybe a little bit easier to follow, but I was almost entirely out of colors by the end :D
This is not a normal Bowser castle... this is a Bowser castle of a kaizo hack! Great solve Simon, you deserve it!
Another great puzzle!
Normally Simon is way faster than me, but today somehow I finished in 38:07.
Keep up the good work :D
Great puzzle. Great solve, despite the criticisms by some. It's always too easy to see the oversights, not so easy to see the logical path through the puzzle, which both Simon and Mark do with aplomb.
Every time Simon explains the xwing, God kills a kitten.
Been watching for a long time, and I couldn’t figure out why you always apologize so profusely when you’re having trouble seeing a step. I finally read the comments, and now I understand, some of these “fans” are a bit cruel. So I just wanted to say, my partner and I actually get quite a kick when we are able to spot something you’ve missed. Quite often, as amateurs ourselves, we are just barely following as you blitz through the logic, so the rare occasion we spot something that you don’t feels amazing! I finally understand why people seem to thoroughly enjoy yelling at a coach or player in a sports broadcast, something which had frankly always been alien to me.
So thank you for the fantastic entertainment. Also I feel like your explanations of your thought process have improved over time, or perhaps my understanding, because I find myself learning more as well. And finally thank you for helping me empathize with an experience that had I didn’t previously!
There's a way to prove this is not actually a 5-star puzzle with modern techniques: i solved it in 37 minutes, QED :) From the middle infernal ring with collapsed arrows i got 8 digits (4 in each of rows 1 and 9) adding up to 59 or 60 and the freedom of placing digits was taken away 30 minutes faster. Very fun puzzle and interesting approach to the solution, thanks!
Hi Simon,
I disambiguated the 12 pair in row 5 when you put the five in the set in boxes 3 and 9. That total that you got had to be 46 or 47. But one of them needed to be 46. And that one can't be the one with the "blue/yellow" square being two. And the only way to get the five in the set is to have the total be 46. Just one thing I spotted and you didn't. There are so many the other way around.....
Then: When explaining "set theory" I think people would find it easier to follow if at one point you'd say: "Now at this point in time we know that orange and blue are both two sets of the numbers 1 to 9. So we also know that those two sets are equal. Now if I remove this square, we don't know what digit it is, so we no longer know the exact composition, but what we /do/ know is that those sets are still equal".
31 minutes for me. One of the very few times I was able to beat Simon's time. Use set theory 4 cells R2C1+R2C9 + R8C3+R8C7 = 12 cells (eventually 13 cells) 5 in Rpw 1, 5 (or 6) in row 9 + R5C2+R5C8. After that's it's a breeze.
As someone who isn't great at suduko I love when Simon explains a concept to me like an x-wing then narrows something down like the 9s and then doesn't use the method he just spent a minute teaching me XD another great video!
Simon, you are sooo good! X
Incredibly rare for me that I finish a puzzle quicker than Simon. 55:07
Circle and arrow are reminiscent of the fire columns in Bowers’s castle
Leaving in the coloring would have sped up the solve... especially when likely insuring that pencil marked cells would not be ignored.
Recoloring 4x seemed to me at each step along the way as incredibly short-sighted.
Wow! a very fun brain twister!
Awesome. You are the logic master.
🤓
That was heaps of fun, thanks to Emmett for the puzzle and to Simon for the solve. I'm particularly pleased as I solved in 92 minutes, so only 50% longer than Simon, instead of the usual 100-200% longer. Also got (the same) first digits in 72 minutes instead of Simon's 42 minutes. I kind of solved the inverse problem to Simon, focussing on the 3-cell arrows and hence the cells that could only be 1-2-3-4-5-6 rather than the ones that could only be 5-6-7-8-9. But got there in the end :)
I needed a hint to look at the RHS... I found the first step pretty easy (thanks to frequent CTC), I Coloured in column 2 rearranged the colours and got 21 in the top and bottom box and 1/2/3 in the middle... I was able to basically colour almost all of the LHS... but was completely stuck
very clever puzzle. well done.
38 minutes for me. I got in using set on columns 1,3,7,9 and rows 1,9 (so that's a +90). Resolving the arrows the vertical set end up completely occupying box 4 and 6 except for the centre digits, allowing the +90 to be rewritten as minus the two centre digits. The final result thus becomes 4 cells remaining in the vertical set minus the two centre digits from box 4,6 being equal to 10 cells in the horizontal set. Applying maximum/minimum values this leads to 34 - 3 for the vertical set and 2 x 15 for the horizontal set, leaving 1 degree of freedom however in row 9 there is an arrow tip outside the set whereas the arrow circle is inside it and thus the arrow tip must be a 5 and the arrow circle a 6 and there is no degree of freedom left.
I got stuck until Simon showed the SET of comparing two boxes to the two vertical lines. I, then, did it on both sides at once, and saw that the maximum for the orange was 94, but the minimum of the blues plus 90 was 93. The arrow pointing into r9 immediately reduces the orange total, and this can only be reduced by 1.
I did not even need set logic to complete this, though I did a lot of mirrored colouring for both the left and right sides reducing what places certain numbers could be with the two cell arrow along the bottom creating the final pinch to start locking things in. I did not try for speed, but I got this in about 2 hours just poking away.
Why did he not use his set logic sums to work out that r8c8 was an 8 (to get to 46 or 47)?
Nice one!
Trampolines and now Voldemort... Simon, you must be one cool Dad!
Blue+45=Orange
Yellow+45=Red
Blue+Yellow+90=Red+Orange
3+90=Red+Orange
Red+Orange=93
Red and orange can be divided by row sums:
R1: 6789
R2: 89
R8: 89
R9: *5* 789
30+17+17+29=93, 0 degrees of freedom.
That makes R1C1 a 6 and both R2C9 and R8C3 8s.
I would love to see either Mark or Simon attempt the Professor Layton games if they can figure out a way to do it. Or a general solving of the Myst or Drawn: The Painted Tower games (in order).
Also, I'm taking a wild guess that its called Bowsers castle because its fireball themed, and the arrows look like fireballs spreading throughout the puzzle. I really don't know though.
The Professor Layton Games are an awesome idea!!!
Especially since the arrows in the top and bottom rows of boxes resemble the top and bottom of the castle at the end of most levels in the original Super Mario Bothers game.
Awesome break-in but is it just me or did Simon work out a couple of minutes before that a 5 could only go with an 8 so he could have put an 8 in R9C8 the moment he put the 5 in R9C7
The SET deduction could have been obtained much more easily by sudoku, and the logic was even more beautiful for it. The vertical arrows had to contain 123456 minus either 5 or 6, with 1 or 2 being repeated in their stead. The orange cells you filled with 56789 see all cells of both arrows in their column of boxes, so they see 1234 so can only be the missing 5 or 6, or 789. In addition, the stubby arrows in boxes 4 & 6 could only be the missing 56 and 78 because their tips also see both vertical arrows. In the corner boxes, if 5 is the digit not in the vertical arrows, you have a 5789 quad, and if 6 is the missing digit, you have a 6789 quad. The two cells in the corner boxes between the circles had to be the pair of non-repeat digits that were on the other vertical arrow. Once the 56 was placed in R9C7, it had to be the missing digit for the right-hand side, so it had to be the upper of the two stubby arrows, and the lower was the 78. The same logic could be applied to the left-had set of arrows.
This was an extraordinary construction, with beautifully tight logic. The logic of the two sides on its own was incredible, and way the two sides linked up to make a cohesive set was cunning.
I snort-laughed at "short stubby arrow logic."
That's a crazy break in
One year later, if I had to guess it's because Bowsers Castle is always a maze of a level, and this seems like it was a maze of a solve.
Not sure the SET theory was entirely necessary. If you work out the permutations of overlapping digits on the four long arrows, the 2 can only come from the 8(125) vs 9(234) which leaves the stubby arrows in the column as 6&7, and so the empty columns in the long arrow boxes are populated by 6789. The 1 can be either 7(124)vs 9(135) (stubby 6&8) OR 8(134) vs 9(126) (stubby 5&7), but combining the two, the vacant column is still permed from only 56789. This then finds the logic with the horizontal arrow in row 9 having to be only 5 or 6 and permeates around the grid in 789 triples to complete that row 9 arrow.
Simon missed some easy logic by not asking "where do the non-repeating digits on the arrow go?"
You can see that for example the non-repeating digits on the arrow in box 1 have to go in column 1 in box 4, therefore column 2 in box 7.
Therefore you can work out where the 7,8,9s cant be placed in the corner boxes and note that 7,8,9 can't go in r9c7. And go from where Simon starts putting digits into the grid.
I used roping instead of sets to place most of the 7/8/9s.
From there, I found the 1/2 and 5/6 pairs because the 2 arrows are made up of 6 cells containing 1 repeated digit and 4 unique digits out of [1-6]. With any repeated digit greater than 2, the sum greater than 17. This also implies that the 2 arrows excludes one digit from [1-6]. If you exclude anything less than 5, the sum is also greater than 17.
37:12 - At this stage you struggled to find the next step. You already found that the two arrows in boxes 1 and 7 only have one common digit, the digit in the middle cell of box 4. Dito the middle cell of Box 6 is the only double digit on the arrows in Box 3 and 9.
And since the arrow bulb cells in boxes 1,3,7, and 9 have an x-wing on 9s you can't have 9 in the corners of the boxes, because an x-wing, as you explained in detail, too, rules out the x-wing digit from the rest of the 2 rows and the rest of the two columns involved.
Now you'd only need to realize the arrow digits must be 1 to 6, in one case 122345 with a double 2 and in the other 112346 with a double 1. Well, at least you could have limited them to be from the digits 1 to 6.
At 38:15 you realised how useful r9c7 is. If you first had noticed, that the arrow cells needed to be from 1 to 6 and the arrow bulbs in columns 2 and 8 form a 6789 quadruple and also push 6789 into the columns on the other side of the arrow cells within their boxes, you'd realized that you already have a 6789 quadruple in row 9 in c2,c3,c5, and c8. So actually, the only remaining digit for r9c7 is 5. And then you know5 doesn't go on the arrows in column 7 and 9, so 6 has to be in them.
As result of that you don't know where 6 goes, but you know you have 1+2+3+4+6=16 and so the repeated digit must be 1, 2 would be too high. Which pushes the sum to 16+1=17 and you know you have a 89 pair in the bulbs of column8. You get r5c8=1, which turns r5c2=2. and there you must therrefore have 1+2+3+4+5+2=17 and the bulbs in row 1 and 9 of column 2 also are an 89 pair.
That turns several 6789 cells to 67 cells, and it becomes easier to solve the rest.
Since you also know the arrow compositions need to be 234,125 where the 2 is repeated in boxes 1 and 6 and they need to be 126 and 134 in boxes 3 and 9, where the 1 is repeated.
@ 29:00 I would have kept it as the sets, rather than going to the sum.
You previously established the arrows needed to be 1, 2, 3, 4, an extra 1 or 2, matching the blue cell, and a 5 or a 6.
That means they collectively contain the digit in the blue cell, and the digits 1-4 and a 5 or 6.
That means eliminating them and leaving just the other 4 cells means those 4 cells contain the digits 7-9, and a 5 or 6, with no repeats.
I think that would be much easier than trying to use the sums and thinking of the different possible sums to show you can't have the 5 or 6 twice.
And this can then feed back once you work out that one of them is a 5, as it being a 5 means the 2 arrows on that side of the grid cannot use a 5, meaning they need to be 1,2,3,4,6,1.
This gives 1 as the repeated digit which must go in the middle of the right side, making the circles for the arrows be 8 and 9.
This means the other repeated digit has to be a 2.
Which means the digits along those arrows on the left need to be 1, 2, 3, 4, 5, 2.
This means the other low digit needs to be a 6, not a 5.
This also means the sums need to be 8 and 9, giving you another x wing, this time on the 8s.
This gives you the 7s from those sets, and leaves you with 8,9 pairs.
Congrats! I thought I would try this one without watching the video first and could only get as far as coloring some of the single-digit arrows 😅
No idea how even to start approaching this stuff, I can't just count and subtract in the same way I can as in picross
Wowzers for Bowser's!
43:06 Once we get r9c7 is 5,cells colored in box3&9 should be 589+789 to sum to minimum 46. r9c8=8. It is the easier way for me.
Simon, I love love love your solving, but I know a secret that I’ve shared before and would like to whisper to you again. Every valid 3-digit arrow without repeats MUST include 2 digits from 1, 2, 3. Knowing this would have helped in previous solves, and in today’s solve it helps you easily eliminate 6 from the four arrow bulbs, because 6 uses all three of 123 and the other arrow must have two of 123, so there would be w digits that would need to squeeze into r5c2 (and r5c8). This secret often helps in other ways, too. Enjoy
I'd just assumed that it was called Bowser's Castle because the set coloring during the break-in resembled a castle's walls or ramparts. And Bowser... has castles...? Ok, look I'm just here to watch Simon say Bobbins.
at min 49 ,you can put a 6 in blok 9 in a central position that solves blok 3
Very rare solve for me, 31:07. I'm gonna sit down and watch now, but I think I spotted some fun SET that Simon may have missed given the length of the video! It sets up R5C2,8 as being minimum and a bunch of cells in boxes 1, 3, 7 and 9 absolute maximum. There's some funkiness to resolve everything involving an additional arrow. Now who's the clever one!? I'm already speaking in an English accent!
Looking through the comments, I have a feeling Simon will go over the break-in tomorrow! Can't find em all, live solving must be scary!