I think this is the first chocobanana puzzle that I've managed to solve - very elegant! It took me 40 minutes, which I'm very happy with. The biggest surprise is that Simon actually used brown and yellow ... the right way round! 😜
I love watching these. Simon is the only one that says something nice to me every day, even though I know he says this to everyone but still. Keeps me going 😊
That was quite fun - thank you Simon! I think we had a chocolate and bananas puzzle sometime fairly recently, or something along these lines - and so I was somewhat prepared for the ruleset. I decided, though, that I probably can't do it myself, but I enjoyed watching you do it.
Thank you Marty for another wonderful puzzle, I was surprised just how powerful the banana/circle constraint became. Thank you Simon and Mark for bringing us these brilliant puzzles every day and thank you Sven for the amazing software so we can play them ourselves with just one click.
Man this is such an interesting puzzle. One realization I made before you was that the uneven chocolate bars are interesting because they can't have a width bigger than 1. Namely 3, 5 and 7. And because they can only contain 1 line, it can't contain more than 1 diamond (because they would be on the same row or column and thus can't both be 3, 5 or 7. Then I realized that the diamond in the top right was the only one that could create a 1x7 chocolate bar without touching other diamonds. But given the one square to the right of the diamond, I could only paint in 5 adjacent squares brown. But it did give me the 7.
→ 22:14 "The difference between 81 and 90 is suspicious." - You can put the 9 either in yellow or in brown, not both (as a yellow 9 means all 9s are yellow). Which means one of your two 45s is missing 9 and becomes a 36. → Simon looking for non-diamond chocolate bars is funny - it's difficult enough to give all 12 diamonds different-size chocolate bars, not even thinking of putting additional ones in. → 29:52 "So the question is: Is there a 9th chocolate bar"? - You only get to 8 if you have two size-9 bars. (The four diamonds in c1r6, c3r5, c4r7 and c5r6 can only form two bars if one of them is a 3×3, otherwise you get even more bars.) → 43:30 "There are no 9es in yellow" + puts the 9 pencil-marks in r4b6 - This makes clear that the size 5 bar doesn't go in row 3. → 51:39 "There is only one 1 in yellow." - And you can actually pinpoint now where it needs to be: both in column 2 and row 2 the 1 can't be in brown. (Found a bit later.) → 1:04:06 "Now these [r4c589] are a 569" - They actually are 459. I think Simon then looked at r4c7 and saw that it's not 5 or 6, so concluded it must be 9 (instead of 4).
Absolutely loved this puzzle. Very clever! Took me 25 mins, but I think I was lucky that an image of the chocolate bars jumped immediately into my mind, and somehow my brain did the logic for me
Yes for me too the logic is quite intuitive, but that’s just because my brain is pretty good at that specific kind of geometric visualisation, (while being awful at other types of logic like abstract mathsy stuff, or anything that involves thinking too many steps ahead. Things like modular lines always confuse me!) So I can completely get that other peoples brains might find this sort of thing more tricky to visualise… but hopefully still fun 😊
Big thanks to Simon for helping me find one of the four pangrams on the NYT Spelling Bee game today. There is a word he uses often that I only know because of him and it came in handy today. A8 - Cheerful readiness
I suspect that some of the people claiming this is a less difficult puzzle took a less analytical approach to the opening and tried a few combinations of chocolate bar placements instead of fully proving each step as they went.
I see a Marty puzzle, I try it (and inevitably need help, even if it's just to understand the rules🤣) Loved this as always. I was wondering (if remotely possible) Has/could Marty feature in a 'How to Set Sudoku' video? I would love to see how he gets this to work.🥰
My break-in was different: I immediately ruled out 3 from r9c4 as that would leave a 1 cell banana in r9c1 so r9c4 is 1 or 2. I also ruled out 3 from r8c1 as that would leave a 3 cell banana region in r9c123 so r8c1 is 1 2 or 4. Then I asked what if r6c4 and r7c5 are banana? Then the r7c4 chocolate bar would be max 2 cells, so is 1 or 2. Now we have a 1 2 4 triple on chocolate bar sizes. This spells trouble for the r5c3 chocolate bar as that is now forced to be size 3 but that creates two more 1 cells chocolate bars in r6c1 and r6c5. So r6c4 and r7c5 are chocolate and are either part of a 4 or a 9 cell chocolate bar. If it is 4, then again we have a 1 2 4 triple on chocolate bars and the r5c3 bar still has a problem. So the r5c3 must join to make a 9 cell chocolate bar.
Finished in 74:51. I had solved the banana and chocolate bar parts decently, but for some reason, I mixed up the 6 and 8 bars randomly. I solved for that correctly, but when it didn't finish, I re-checked everything and realized that I had messed up the 6 and 8 bars, and had to redo everything which took about 15 minutes to finish. Fun puzzle!
the way to approach this was to realise that there are 12 brown diamonds but only 9 numbers to put into them. so 3 of them must be extras in a bar that already has a diamond. and there are only 3 places where multiple diamonds can in fact share a bar. 1. there is the possible 9 cell bar in r8c1/r3x3 1. there is a 2x2 (that can be enlarged) down at r8x6/r9c7 3. there are multiple possibilitues to join diamonds at r5c3/r6c5/r7c4. this can either be a 2x2 that can't grow un r6c5 ans r7c4 or it is a 3x3 tgat encompasses all three diamonds. now, the second option is a 9 cell ans would rule out the first 9 bar i mentioned. either option of the 9 bar would "eat up" 3 extra diamonds. there is no way to do that with 4 extra diamonds. hence, there are exactly 9 bars containung diamonds. meaning, bananas can't have 9s and no bar is off the diamonds. it's beautiful logic and in my opinion not actually very difficult to see. it's why this has a 3 out of 5 rating. sadly, simon didn't think of the fact that he has to use up 12 diamonds yet (I didn't finish the video yet). that's what made him struggle quite a bit more with this puzzle than necessary. it happens sometimes, sometimes we are just sidetracked by one logical train of thought and ignore the much more productive one
Does the rule "Every chocolate bar has a different size." mean to also tell us "Every chocolate bar has a diamond with its size in it"? Because nowhere in the rules does it say that every chocolate bar needs to have a diamond in it. (I can see at a certain point that there's no room for a chocolate bar with size 10 or more anywhere on the grid, so I guess that's the deduction we're supposed to make.)
YES! this exactly the way I thought about it. Simon never quite said it in this exact way, but he got there in a bit of a different way which was also really interesting
"If i put nine in a banana..." Then you have 9 nines on the grid in bananas and so you can't put a nine on a chocolate bar. The flipped argument for this is... If you put a nine in a chocolate bar (doesn't even have to be a diamond) then you CAN'T put nines in any bananas! That's the EASY way to figure this one out, you don't have to think about 81, 45 or that there would be 9 different size bananas etc... Also, as per the rules, there can only be one chocolate bar each of N size cells. If there is no nine cell chocolate bar, then nines go in bananas. But you have 12 diamonds to fill. If no nines go in chocolate bars, then you have to find 4 diamonds that can be part of the SAME chocolate bar. That is, the diamonds cannot line up or be in the same box. I only see THREE such possibilities. * Therefore i claim that there are nine different size chocolate bars. Hence, nines do not go in bananas. EDIT: * - Ok, there is a fourth possibility of all of R5C3, R6C5 and R7C4 being the same digit. Which would make the 12 diamonds be selected from 8 different digits. BUT, in doing so you would most definitely make a 9 cell chocolate bar, so those cells would all be nines! Thus my claim still stands, there are NO nines in bananas. One chocolate bar is a nine cell chocolate bar. And if all the mentioned cells R5C3, R6C5 and R7C4 were nines, you'd have to find a configuration of chocolate bars than can have the other 6 nines in them. While also only making the chocolate bar at R8C6 and R9C7 the only chocolate bar that can have the same digit twice in diamonds. Now you run into a the scenario where you would have to find an chocolate bars of sizes 1, 2, 3, (4), 5, (6), 7 and (8). The ones in parentheses can be part of the chocolate bar in R8C6->R9C7. But do remember that any odd digit size chocolate bar is a straight line. 3, 5 and 7 all need to be straight line chocolate bars. The 7 size bar would be in row 1 as that's the only place it can go. ... In fact, now that i think of it... There's going to be a 7 size chocolate bar. It's going to be a straight line. There's only one such possibility isn't there? R1C8 IS a 7... EDIT #2: Or think of it another way, the three close cells R5C3, R6C5 and R7C4.... What can they be? All 9, thus you have a square covering all of them. Or, they are all selected from the digits 1234. Neither of them can grow to more than 3 cells without having to join another chocolate bar. And if R6C1 is not a 1, then R5C3 cannot grow to the left, or it would join that bar and the diamonds would have the same number in the same box. And if R5C3 grows down, it forces one of the other R6C5 or R7C4 to be a 1. And you run into the problem of your 12 diamonds being populated by too few numbers. I don't know all the possibilities here but i will claim that R5C3 is from the digits 1 and 9. Because if it grows to any other size, it will either join another diamond or force an impossible configuration where you will have two chocolate bars of size 2. It cannot grow to three or more cells without becoming a 9 cell chocolate bar, or you would have chocolate bars orthogonally adjacent to each other.
I was quite stumped at first how a brown diamond could indicate the size of the brown region if every region had a different size and there were more than nine diamonds, until I realized that a region could contain more than one diamond.
Not me thinking for the first half hour that the rule was a digit in yellow indicates how many banana regions contain that digit and being stuck because there is only 4 banana regions 😅
Correct me if im wrong, but the logic at 24:30 or so doesnt hold up when you consider multiple diamonds can be in the same square? i.e, you can fulfill the diamond requirement with only 8 chocolate bars, thus you could have a bar of size 9,7,6,5,4,3,2,1, and thats only 37 squares? You can put 9 9s in bananas with a minimum of 27 banana squares, i dont know why simon is talking about 45 squares in a banana. Much easier logic to me is, there are 9 9s in a sudoku, if there is a 9 in a banana there can not be a 9 in a chocolate bar, and therefore vice versa (since every square is in a region). I dont really understand the math he was doing at all
You can't put 9 9's in bananas if you're using one in a bar. Also, 81-37=44. That means you're not using 1's and no 9's because they're in the bars. That's 35 and doesn't add up. A better way to look at it is that the maximum cells for either bananas or bars is 45 with the digits 1-9. The minimum of the other set is 36 because 81-45=36. So the 9 can go in either set. Remove the 9's completely. We're now working with 36 cells in each set and they're both built with the digits 1-8 for both bars and bananas. There is no more degrees of freedom. If you remove cells from one set to add them to the other set, you'll have to add a digit and you're already using them all.
@@alienrenders but why does it need to be 45 if theres a 9 in a banana? it takes minimum 27 squares because there must be at least 9 bananas of at least size 3, and 27 fits into that 44 limit just fine. So the math logic doesnt make any sense
@@darkthunder312 Because if bananas have 9's, then the maximum you can make the bars is 1 to 8 that adds up to 36 leaving a minimum of 36 cells (81 - 9 - 36 = 36) that must be added to the 9 cells in bananas. 36 + 9 = 45.
The size of the banana is irrelevant - the only implication of that is that you can't trap a 1/2-cell banana between chocolate bars, but beyond that there is no restriction or information about the size of the banana(s), and there's nothing to stop it being one giant mutant 36-cell banana...
There's a mistake in the rules. The rules say that each region consists of orthogonally connected cells. Cells being plural indicates that there can't be a region of size 1, but there is.
I got one of my bananas intermixed with chocolate, so I decided have a peep at Simon before I dared to go further. Simon did come up with a quite penetrating discussion … before getting the banana split, him as well. But ... it turns out that you don’t need to exclude any occurrence of non-diamond chocolate to solve the puzzle.
Hi Simon, this was lovely, I didn’t expect another feature so soon after my birthday feature last week, but am very grateful that you chose to show this one, as it is definitely one of my recent favourites that I’ve made. You were right that counting circles are something that I came up with, in my puzzle Circular Reasoning which you solved last September… (although obviously I don’t know for sure if someone else did the concept before me… to be honest I would be surprised if not, because it seems too simple an idea to have not been thought of before…) But yes I was thrilled to see how they caught on and became popular. I’ve seen so many fantastic and varied puzzles from other constructors using counting circles, and it always brings me lot of joy. And I still like setting with them myself, especially when they are not just standard circles, but instead applied in other ways, as I’ve done in this puzzle. The inspiration came from Scojo’s birthday feature on the 9th of May… you solved his fantastic puzzle “Chocolate Banana Sandwiches” and it left me feeling that I really wanted to try setting a chocolate banana puzzle myself. As usual I’ll delve a bit into my setting process, for anyone who enjoys reading about such things... One thing that my brain seems to be drawn towards is including a “full set” of something… Eg: in “My Zipper is Broken”, the big broken zipper line has all possible sums from 2 - 18 along it, and so for setting a Chocolate Banana puzzle, my immediate thought was “what if all chocolate bar sizes from 1 - 9 appeared once each?”… I played around from there and found that the sizes from 1-9 fit into a 9x9 grid much more snugly than I had any right to expect. A good example of luck coming into the setting process. It could easily have been impossible to fit them all in, or equally it could have been too easy and not very constrained at all, requiring loads of other clues to force their placement. But the universe decided to be nice to me on this occasion and made it just the perfect amount of constrained - ie; possible, but only just. Initially my idea was to have the counting rule apply to chocolate bars instead of bananas, but this turned out to be way too problematic. It would mean that the size 9 chocolate bar would contain 987654321, the size 8 bar would contain 98765432, all the way down to the size 1 bar which would just contain a 9. But geometrically in the grid, sudoku always then ended up placing the wrong digits in places… it was all way too restricted, and actually even if it had worked it wouldn’t have been nearly as fun or interesting as what I came up with instead… putting the counting rule on bananas, and then having diamonds that indicate the chocolate size. Another stroke of luck was that with all 9 chocolate bar sizes, the total size of yellow would be 36 cells, the exact triangular number for 8, so that yellow would have to include all 8 digits apart from 9, the only digit that it couldn’t take anyway given that 9 appears in chocolate. It all seemed to fit together so perfectly. When I started experimenting with the diamonds, I enjoyed that their placement could force the positions of the chocolate bars so powerfully. The breakthrough was realising I could put more than one diamond within a chocolate bar, and noticing how this would force things even more and lead to some rather interesting global considerations. The way that I thought about this opening was to ask a particular question, which you never quite did, but you got to the same conclusion in a slightly different way which was also very interesting. My way was: “There are 12 diamonds, so what’s the minimum number of repeated digits you need to have in diamonds?” Clearly you need at least 3 repeats. And the only bars able to contain a repeat are 9, 6 and 4 (not 8 because eight 8s are needed in yellow... one of my favourite realisations in the puzzle.) You can then locate all the possible diamonds which are able to join with another, and it’s not many. It turns out you need all 9 chocolate bar sizes to be able to fill all 12 diamonds, and so you can’t waste an opportunity by having a chocolate bar with no diamond in it. Basically, all diamonds which CAN join to another diamond MUST do that, with only one degree of freedom. But for sure you need all 9 chocolate bar sizes, and as you spotted, the size 7 chocolate bar is the key to it all, only being able to go in one place, and thus removing the degree of freedom for the others, which all become quickly forced after that. I found quite a few different arrangements of the 9 chocolate bars that worked, some of which just resulted in one massive banana. But I preferred having at least 3 bananas, and this final layout proved by far the most interesting in terms of everything forcing itself, and more importantly it led to a solution without the counting rule breaking! Hooray! The 7 diamond was originally right in the corner in r1c9, but then the whole bar was immediately forced along the top. I decided I wanted the diamond moved to the left a tad, so that the 7 chocolate bar could potentially go to the left or to the right, and the only way to deduce which way is by using the fact that you can’t leave a tiny square banana in r1c1 (such things please me more than they should.) I realised that although the whole puzzle almost forced itself towards a unique solution, I needed one more clue type that would probably only need to be used just a few times in order to take it over the edge. I toyed with different clue types such as thermos and arrows, but I thought that aesthetically anything liney might distract during the earlier chocolate bar building logic, so I preferred a small arrow based clue. I spent the last couple of days trying to find the most potent place to put these clues so that I would need no more than 3 of them, and ideally each one would serve a few purposes, each giving you a lot of ‘bang for your buck’. Eg: the 6 arrow not only gives some useful information about the 2 and the 4 below it, but it also means it can’t be a 7 itself (the main reason for me putting a clue there), forcing 7 down into chocolate in the column, giving the 2nd brown 7, which means 7s can then be pencil marked in all kind of bananary places. Right, that’s quite enough of that. Thankyou once again for all the entertainment and lovely words of praise in the video, it means a huge amount to me to see how much you enjoy all the little details :)
Oh lovely the setter is here :) I wanted to just write a comment but I can say it to you then. I think the rules need a bit of rephrase.. I think saying "each region must consist of orthogonally connected cells" does not allow a one cell region. That threw me off a bit, I thought I broke it at one point :), but I just went with the one cell and it solved.. Beautifully :) It's a brilliant puzzle. Thanks for the detailed process too. I've always wondered how one would set something like this.
@@martysears Best. Comment. Ever. Your puzzles are my favorites! Don't tell the other setters I said so please. Seriously I can't even wrap my mind around the creativity it must take to set puzzles like this. Thank you so much for doing what you do!
Thank you for featuring this Simon, it is an absolute stunner of a puzzle and I so enjoyed your emjoyment of it. However I do have a bone to pick with you! Did you HAVE to make purple flashes on the chocolate bars???! Now I neeeeeeeeeeed Cadburys!!? Marty.... you know I enjoy all your puzzles, but you have outdone yourself here!!! Just beautiful!
The ruleset is incorrect. Based on the rules there should not be a 1 cell chocolate region. "Each region must consist of orthogonally connected cells". The single cell chocolate region is neither cells (plural) nor orthogonally connected. I spent a lot of time trying to avoid a single cell region only to find it was impossible.
Thank you for showing your mistakes. It really comforts me when someone who is WAY smarter than me shows how they can still make mistakes ❤ Amazing solve Simon! Always a joy to watch!
The way Simon says "Oh, Marty!" at 45:01 tickles me pink. Like, the genuine appreciation of not just the puzzle but also Marty Sears the setter/person.
76:51 This felt like two puzzles for the price of one. Placing the colourings with a little help from the numbers was full of tricksy logic and evil ideas, then filling the numbers by reference to the colouring was an even bigger conundrum. Ingenious puzzle.
Child 1: Mom, why are you yelling? What is wrong? Child 2: Simon put the wrong number in. She's fine. Child 1: What? Me: It's just sudoku. Nothing's wrong. Husband: Oh, you know, she only yells like that when Simon missed something. Child 1: You all see how weird this is, right?
can you guys upvote this so they can see it. cracking the cryptic should have two channels, one for light mode viewing and dark mode viewing. i would subscribe to both, and i would watch videos on both depending on time of day. imagine falling asleep to a dark mode CTC video.
Much as I enjoyed this puzzle, but wouldn't want to do too many as it has filled my head with the idea that I must have chocolate. Wow, watching Simon's solve I realise I skipped a whole load of stuff at the start due to assumptions.
It seems I did something backwards. It was only after I'd fully coloured my chocolate bars and bananas that I counted how many banana cells I had. Early on, I found the three diamonds that ended up being in the 9 bar had to be part of a single size 9 bar, otherwise they forced too many size 1,2,3,4 sized bars. Then I saw I had to have eight other bars to cover the remaining diamonds. (Only the top one could be the 7 bar.) I then managed to place all the other chocolate bars before counting how many banana cells that gave me.
I did the same. Then I thought I'd broken the puzzle when I for some reason calculated that 81-45=46 (total cells minus chocolate cells equals banana cells) and one 1, two 2's and so on up to nine 9's is only 45, what's the last digit? Then I realised my mistake and since I already had chocolate 9's I knew there were no banana 9's.
Does anyone know why when Simon says words like “region” or “nature” he spends so long on the first syllable? Is this a feature of a certain regional English dialect? “Nature” even contains almost a glottal stop in it before the “ch” sound. Very odd to my American ears.
I feel like the difficulty disparity comes from the kind of logic this puzzle requires being more spatial than numerical. Or at least more than usual for most sudoku. I've been watching CTC for a while now and it's really illustrated to me how different my logical strengths are compared to Simon's. He picks up on numerical patterns and math things so fast and it's always things that never would have occurred to me. But then I watch him do a puzzle like this where he misses a lot of the spatial deductions that seem a lot more obvious to me and it really shows just how differently our brains work.
I have a puzzle collection book with chocolate banana in it and the rules in the book says something like "chocolate regions must be rectangles, including squares." Which is reassuring but also refreshing in that it acknowledges the definition of rectangle.
Stunning puzzle. Fisaco at the end was deserved, because after the break-in, which he always does well, he rushes round the grid in a blind panic making more and more of a mess and finally slipping over in it. More discipline and more calm thought would be very welcome.
I looked at the number of chocolate bars a little bit differently than Simon. I marked the 5 diamonds that could not join with any other diamond. (This includes the diamond in R6C1, which cannot join with the other diamond in box 5 no matter what Simon says. :P) Then, I looked at the remaining diamonds to figure out how I could limit them to only 4 chocolate bars. The key for me was recognizing the only diamond that could support the size-7 chocolate bar. Then, putting the last 3 bars into the remaining diamonds was quite limited. The 3-diamond size-9 bar, for example, was immediately obvious.
I tackled this so very differently ... via the sizes of chocolate bars: bars of sizes 12357 can only take one diamond, so I was working on allocating the diamonds to bars - (I couldn't take enough diamonds without a 9) and also the placement of 5 and 7 bars is restricted: 9 and 8 also. The logic overlaps in a way, of course.
@@MarkBennet10001 I had also tried to start like you did, but unfortunately abandonned this way. I should not have done so, as this logic (with 2 simple equations ) immediately leads to the result, that there has to be 1 bar with 3 diamonds, 1 with 2 and seven with a single diamond. (1+1+7 =9 (sizes), and 1*3+1*2+7*1 =12, (diamonds)). This relies on the fact, that there is only one place for a 3-diamond-bar, an no place for bars with more than 3 diamonds. This could have saved me a lot of time and I could have solved this puzzle much faster than I did.
32:36 Brilliant puzzle, but for some reason I cannot fully follow the logic here. I see that there must be a chocolate bar of size 9, but I don't see why they're 9 different chocolate bars. The bar of size 7 could have been omitted, e.g.. Can someone help me with what I'm overlooking?
Hi Mahoel... it's because there are 12 diamonds. That means at least 3 diamonds need to be on a chocolate bar with another. If you miss out any chocolate bar sizes, this number increases to 4 or more. However, most of the diamonds are unable to be on the same chocolate bar with another diamond. Once you've marked these, you see that you can only achieve 3 repeats in diamonds. It's impossible to achieve 4 repeats in diamonds. This means all 9 chocolate bar sizes are needed, including 7. And there's only one place the 7 can go, which means there's now only one place the 9 can go, and it all kind of gets forced from there...
Thanks! I'm getting it now. You can see quickly that you cannot put 4 or 4x2 together without having them touch each other. However, I'm missing or overhearing this in the solve at this point. It may be obvious to others...
36:29. After determining the chocolate bars and bananas counted the bananas. Realized it was either 9 or sum to 9 that had to be bananas. Saw that 9 was quite possible. Then started recolouring as I solved from 8. Down of course couldn't finish 8s fully but following the logic, as I fully solved a digit I changed all occurances to black. Once sudoku could solve it I was left with an oozy brownish mess (between brown yellow and black)
I finished in 119 minutes. I really liked this puzzle. The rectangular nature of the chocolate bars made filling the grid very satisfying. I can't even imagine how you would set this regarding the banana rule. Solving it felt almost impossible, so setting it seems inconceivable. Incredible work, as always, by Marty. Great Puzzle!
Everyone here should try normal chocobanana puzzles. This was a beautiful puzzle, but the logic was closer to circle sudoku than standard chocobanana, which should be enjoyed.
This took a while for me ... First I tried out several versions of how the chocolates could be, then figured out that there is only a single option. Then it was somewhat painful elimniation. (I think I didn't use the "exactly 5 × 5 in yellow", that just came out at the end. 124:35, solve counter 502.
Beautiful puzzle. I'm also satisfied with my time on it, 53:24. Oh, yeah, have to say I never used other palettes in Sven's software, but 2nd one has some really pretty colors by default and is perfect for this puzzle (as demonstrated in the video).
At 35:46 I'm confused why Simon is saying there must be a chocolate bar of size 7? None of the rules require it, and the sevens that must exist within chocolate bars (as there must be either 2 or 9 7s in the chocolate bars) would just not be in one of the diamonds? It's perfectly possible at this point in the puzzle, so I'm not understanding why he says there must be a 7 sized chocolate bar?
After solving the puzzle, I did go back and look at it again, and I found it to be a valid proof, but I still don't understand the logic that Simon used, I just tried to run through it with the upper right brown diamonds making a 9 sized chocolate bar and not using a 7 sized bar. I found that there was no possible way to create a layout of bananas and chocolate bars that fits the rules. That being said, if someone could maybe explain it in more detail for my smooth brain, I would appreciate it!
He explained it earlier. If chocolate bars are all different sizes, then the maximum number of cells is 45. The minimum banana cells is 36 and must be from 1 to 8. That means only the 9's can switch sides. And for the chocolate bars, you need every digit 1-8 to get to 36 cells as well. If it's 45 cells, you need all digits. So the digits 1-8 are always used in both sets. That includes the 7.
Maths. The split between chocolate bars and banana cells is 36-45, one way round or the other. But in both cases, they have to be made by adding up unique digits. So the only way to do that is 36=1+2+3+4+5+6+7+8 and 45=1+2+3+4+5+6+7+8+9 So there must be chocolate bars of every size from 1 to 8, plus maybe a 9 as well.
The way I thought about it was this… what’s the fewest number of repeats you need in diamonds? you need at least 3, as there are 12 of them. And the only ones that could contain repeats are 4, 6, 8 and 9. Looking at the layout you can see that a maximum of 4 diamonds are capable of being repeats… as Simon eventually pointed out, there must be 8 distinct chocolate bars at least. But actually it’s impossible to get the repeats on ALL of 4, 6, 8 and 9. Because to make r6c3 AND r1c8 both be repeats, there would need to be two 9 size bars. So actually only 3 of the diamonds can be repeats. This means all 9 size chocolate bars are needed, and you can’t have any chocolate bars without a diamond. It means 9 goes in brown, and the 7 chocolate bar must go along the top… and because 8 is in yellow it means that there can’t be a repeated 8 in brown diamonds either, forcing the 3 repeats in diamonds to be two repeated 9s and then a repeated 4 or 6 at the bottom, which is forced to be 6 shortly afterwards
@@martysears There is a difference between "A digit in a banana region tells you how many such digits are in banana regions" and "A digit in a banana region tells you how many banana regions contain that digit". The rules say one thing, and at least as far as I have gotten, Simon consistently says the other. The rules as written do not lend themselves directly to use of the secret, but Simon acts as if it does when he says there can be at most 45 banana cells.
@@MasterHigure Yes the former is correct although worded slightly differently from how you worded it. My wording was: "a digit in a banana cell indicates exactly how many banana cells contain that digit." So that does instantly make the secret relevant, and does mean there can be at most 45 banana cells. The latter ('a digit in a banana region tells you how many banana regions contain that digit') would indeed mean something very different... (Cells are individual yellow squares, whereas regions are the whole bananas.) Given that there's only 3 banana regions, you wouldn't be able to have more than a 3 in a banana haha. But this isn't what the rule says, and isn't how Simon interprets it either
@@martysears Oh, right, I misread the rules. The rules (as written in the upper quadrant of the video) say "how many banana cells", and I read that as "how many banana regions". Yeah, my bad. I was the one who misread and misunderstood. I even double and triple checked before making my original comment. And I still got it wrong.
man didn't Simon take a really long way round to break into this puzzle? I feel like that whole beginning wasn't needed because you can determine the bottom left diamond is a 1 (assuming you believe that's allowed)
I'm confused (haven't watched the video yet)... the rule says each region must consist of orthogonally connected cells... but I have a chocolate bar of 1... a chocolate bar of 1 is a 1 cell brown region... that's not a set of orthogonally connected cells.
a one element set is still a set. if you only have one cell in the set, it's always an orthogonally connected set (because there are no other cells that could break the connection)
@@surrendherify I guess but that’s not what it says. “Each region must consist of orthogonally connected cells” (with an s, meaning more than one connected cell). But maybe we always treat this to mean “including 1 cell not orthogonally connected to anything” There have been puzzles before where the word being singular or plural was the clue in solving the puzzle or not.
![Yailin La Mas Viral, @kreizyk.official - Dale 2 (Remix) [Video Oficial]](http://i.ytimg.com/vi/JaYHBzgqz94/mqdefault.jpg)
Simon is so kind. He tells me I'm one of his favorite people nearly every day!
It’s so 🥰
And I smile every time
Must be true then 😊😊
You are watching the video after all
He doesn't really know me but I'm pretty sure we're best friends.
I think this is the first chocobanana puzzle that I've managed to solve - very elegant! It took me 40 minutes, which I'm very happy with.
The biggest surprise is that Simon actually used brown and yellow ... the right way round! 😜
Simon, you know Maverick has your office bugged.
who is maverick 😭😭
@@chaelicopter Several different males and females
I’m convinced it’s Mark!
Maverick is the name Simon has given to the loud plane that flies over him in nearly every video
@@adaoarfishes9405 thank you!
Some real weird looking bananas in this market
I was gonna say, that's a strange bunch of bananas.
Simon's inability to do sudoku was so painful in this video. But as his absolute genius at doing the more complicated things was near miraculous!
I love watching these. Simon is the only one that says something nice to me every day, even though I know he says this to everyone but still. Keeps me going 😊
Simon, thank you very much for wishing my father a happy birthday! He really liked the surprise :)
- Daniel
That was quite fun - thank you Simon! I think we had a chocolate and bananas puzzle sometime fairly recently, or something along these lines - and so I was somewhat prepared for the ruleset. I decided, though, that I probably can't do it myself, but I enjoyed watching you do it.
I am sure you would be able to do it eventually yourself...with a little nudge from Simon. 🙂
So happy to see this puzzle featured. Just beautiful. Congrats to Marty on another stellar puzzle!
There's always money in the banana stand!! ...94:51
Thank you Marty for another wonderful puzzle, I was surprised just how powerful the banana/circle constraint became. Thank you Simon and Mark for bringing us these brilliant puzzles every day and thank you Sven for the amazing software so we can play them ourselves with just one click.
30 minutes in, and not a single digit written. The logic implications are so deep with this one...
Simon has to be one of the most likeable people there are
Man this is such an interesting puzzle. One realization I made before you was that the uneven chocolate bars are interesting because they can't have a width bigger than 1. Namely 3, 5 and 7.
And because they can only contain 1 line, it can't contain more than 1 diamond (because they would be on the same row or column and thus can't both be 3, 5 or 7.
Then I realized that the diamond in the top right was the only one that could create a 1x7 chocolate bar without touching other diamonds. But given the one square to the right of the diamond, I could only paint in 5 adjacent squares brown. But it did give me the 7.
→ 22:14 "The difference between 81 and 90 is suspicious." - You can put the 9 either in yellow or in brown, not both (as a yellow 9 means all 9s are yellow). Which means one of your two 45s is missing 9 and becomes a 36.
→ Simon looking for non-diamond chocolate bars is funny - it's difficult enough to give all 12 diamonds different-size chocolate bars, not even thinking of putting additional ones in.
→ 29:52 "So the question is: Is there a 9th chocolate bar"? - You only get to 8 if you have two size-9 bars. (The four diamonds in c1r6, c3r5, c4r7 and c5r6 can only form two bars if one of them is a 3×3, otherwise you get even more bars.)
→ 43:30 "There are no 9es in yellow" + puts the 9 pencil-marks in r4b6 - This makes clear that the size 5 bar doesn't go in row 3.
→ 51:39 "There is only one 1 in yellow." - And you can actually pinpoint now where it needs to be: both in column 2 and row 2 the 1 can't be in brown. (Found a bit later.)
→ 1:04:06 "Now these [r4c589] are a 569" - They actually are 459. I think Simon then looked at r4c7 and saw that it's not 5 or 6, so concluded it must be 9 (instead of 4).
Absolutely loved this puzzle. Very clever! Took me 25 mins, but I think I was lucky that an image of the chocolate bars jumped immediately into my mind, and somehow my brain did the logic for me
Yes for me too the logic is quite intuitive, but that’s just because my brain is pretty good at that specific kind of geometric visualisation, (while being awful at other types of logic like abstract mathsy stuff, or anything that involves thinking too many steps ahead. Things like modular lines always confuse me!) So I can completely get that other peoples brains might find this sort of thing more tricky to visualise… but hopefully still fun 😊
Big thanks to Simon for helping me find one of the four pangrams on the NYT Spelling Bee game today. There is a word he uses often that I only know because of him and it came in handy today.
A8 - Cheerful readiness
ALACRITY
I suspect that some of the people claiming this is a less difficult puzzle took a less analytical approach to the opening and tried a few combinations of chocolate bar placements instead of fully proving each step as they went.
Deadset corker of a puzzle, had to use all corners of the brain. Very satisfying
The creativity of this puzzle is bananas!
🍌🍌🍌🍌🍌🍌
I see a Marty puzzle, I try it (and inevitably need help, even if it's just to understand the rules🤣) Loved this as always. I was wondering (if remotely possible) Has/could Marty feature in a 'How to Set Sudoku' video? I would love to see how he gets this to work.🥰
Marty commented on the video and explained the puzzle setting process. It's soo interesting :)
@@mmmmmmmatssss ❤
My break-in was different:
I immediately ruled out 3 from r9c4 as that would leave a 1 cell banana in r9c1 so r9c4 is 1 or 2. I also ruled out 3 from r8c1 as that would leave a 3 cell banana region in r9c123 so r8c1 is 1 2 or 4. Then I asked what if r6c4 and r7c5 are banana? Then the r7c4 chocolate bar would be max 2 cells, so is 1 or 2. Now we have a 1 2 4 triple on chocolate bar sizes. This spells trouble for the r5c3 chocolate bar as that is now forced to be size 3 but that creates two more 1 cells chocolate bars in r6c1 and r6c5. So r6c4 and r7c5 are chocolate and are either part of a 4 or a 9 cell chocolate bar. If it is 4, then again we have a 1 2 4 triple on chocolate bars and the r5c3 bar still has a problem. So the r5c3 must join to make a 9 cell chocolate bar.
ooh nice, that is a very cool alternative break-in
Finished in 74:51. I had solved the banana and chocolate bar parts decently, but for some reason, I mixed up the 6 and 8 bars randomly. I solved for that correctly, but when it didn't finish, I re-checked everything and realized that I had messed up the 6 and 8 bars, and had to redo everything which took about 15 minutes to finish.
Fun puzzle!
Gotta go out and buy more chocolate! 🍫
I have tons by me..what do you fancy? 🙂
@@davidrattner9 dark chocolate is my favorite! 😋
46:01 figured out the rectangles quite easily,
Brilliant 2 deductions.
the way to approach this was to realise that there are 12 brown diamonds but only 9 numbers to put into them. so 3 of them must be extras in a bar that already has a diamond. and there are only 3 places where multiple diamonds can in fact share a bar.
1. there is the possible 9 cell bar in r8c1/r3x3
1. there is a 2x2 (that can be enlarged) down at r8x6/r9c7
3. there are multiple possibilitues to join diamonds at r5c3/r6c5/r7c4. this can either be a 2x2 that can't grow un r6c5 ans r7c4 or it is a 3x3 tgat encompasses all three diamonds. now, the second option is a 9 cell ans would rule out the first 9 bar i mentioned.
either option of the 9 bar would "eat up" 3 extra diamonds. there is no way to do that with 4 extra diamonds. hence, there are exactly 9 bars containung diamonds.
meaning, bananas can't have 9s and no bar is off the diamonds.
it's beautiful logic and in my opinion not actually very difficult to see. it's why this has a 3 out of 5 rating. sadly, simon didn't think of the fact that he has to use up 12 diamonds yet (I didn't finish the video yet). that's what made him struggle quite a bit more with this puzzle than necessary. it happens sometimes, sometimes we are just sidetracked by one logical train of thought and ignore the much more productive one
Does the rule "Every chocolate bar has a different size." mean to also tell us "Every chocolate bar has a diamond with its size in it"? Because nowhere in the rules does it say that every chocolate bar needs to have a diamond in it. (I can see at a certain point that there's no room for a chocolate bar with size 10 or more anywhere on the grid, so I guess that's the deduction we're supposed to make.)
@@nonyobisniss7928 no. but I explained why there is literally no way to incluse all 12 diamonds in 8 bars. hence, 9 bars have diamonds.
I think I follow, but did you mean r1c8 in point 1?
YES! this exactly the way I thought about it. Simon never quite said it in this exact way, but he got there in a bit of a different way which was also really interesting
That was a huge banana if I have ever seen one. lol🎉 Just under 1h for me.
"If i put nine in a banana..."
Then you have 9 nines on the grid in bananas and so you can't put a nine on a chocolate bar.
The flipped argument for this is... If you put a nine in a chocolate bar (doesn't even have to be a diamond) then you CAN'T put nines in any bananas!
That's the EASY way to figure this one out, you don't have to think about 81, 45 or that there would be 9 different size bananas etc...
Also, as per the rules, there can only be one chocolate bar each of N size cells. If there is no nine cell chocolate bar, then nines go in bananas.
But you have 12 diamonds to fill. If no nines go in chocolate bars, then you have to find 4 diamonds that can be part of the SAME chocolate bar. That is, the diamonds cannot line up or be in the same box.
I only see THREE such possibilities. *
Therefore i claim that there are nine different size chocolate bars. Hence, nines do not go in bananas.
EDIT: * - Ok, there is a fourth possibility of all of R5C3, R6C5 and R7C4 being the same digit. Which would make the 12 diamonds be selected from 8 different digits.
BUT, in doing so you would most definitely make a 9 cell chocolate bar, so those cells would all be nines!
Thus my claim still stands, there are NO nines in bananas. One chocolate bar is a nine cell chocolate bar. And if all the mentioned cells R5C3, R6C5 and R7C4 were nines, you'd have to find a configuration of chocolate bars than can have the other 6 nines in them. While also only making the chocolate bar at R8C6 and R9C7 the only chocolate bar that can have the same digit twice in diamonds.
Now you run into a the scenario where you would have to find an chocolate bars of sizes 1, 2, 3, (4), 5, (6), 7 and (8). The ones in parentheses can be part of the chocolate bar in R8C6->R9C7.
But do remember that any odd digit size chocolate bar is a straight line. 3, 5 and 7 all need to be straight line chocolate bars. The 7 size bar would be in row 1 as that's the only place it can go.
... In fact, now that i think of it... There's going to be a 7 size chocolate bar. It's going to be a straight line. There's only one such possibility isn't there?
R1C8 IS a 7...
EDIT #2:
Or think of it another way, the three close cells R5C3, R6C5 and R7C4....
What can they be?
All 9, thus you have a square covering all of them.
Or, they are all selected from the digits 1234. Neither of them can grow to more than 3 cells without having to join another chocolate bar.
And if R6C1 is not a 1, then R5C3 cannot grow to the left, or it would join that bar and the diamonds would have the same number in the same box.
And if R5C3 grows down, it forces one of the other R6C5 or R7C4 to be a 1. And you run into the problem of your 12 diamonds being populated by too few numbers.
I don't know all the possibilities here but i will claim that R5C3 is from the digits 1 and 9. Because if it grows to any other size, it will either join another diamond or force an impossible configuration where you will have two chocolate bars of size 2.
It cannot grow to three or more cells without becoming a 9 cell chocolate bar, or you would have chocolate bars orthogonally adjacent to each other.
50:54 While thinking about the 8s could have gotten a few more by asking where the 8 goes in row 6...
60:22, a bit of trial and error to figure out the 2,3,4,5 regions.
Very exciting puzzle.
The example digit at 10:34 turns out to be correct lol
Edit: Also the shape of the region (@10:50) is correct lol
Hahaha just noticed that myself with the 8 example. Uncanny how often he manages to do that!
I was quite stumped at first how a brown diamond could indicate the size of the brown region if every region had a different size and there were more than nine diamonds, until I realized that a region could contain more than one diamond.
Yes I didn't want to be more explicit about that in the rules because I thought it made for quite a nice realisation after an initial 'what?!?' moment
Not me thinking for the first half hour that the rule was a digit in yellow indicates how many banana regions contain that digit and being stuck because there is only 4 banana regions 😅
Correct me if im wrong, but the logic at 24:30 or so doesnt hold up when you consider multiple diamonds can be in the same square? i.e, you can fulfill the diamond requirement with only 8 chocolate bars, thus you could have a bar of size 9,7,6,5,4,3,2,1, and thats only 37 squares? You can put 9 9s in bananas with a minimum of 27 banana squares, i dont know why simon is talking about 45 squares in a banana.
Much easier logic to me is, there are 9 9s in a sudoku, if there is a 9 in a banana there can not be a 9 in a chocolate bar, and therefore vice versa (since every square is in a region). I dont really understand the math he was doing at all
You can't put 9 9's in bananas if you're using one in a bar. Also, 81-37=44. That means you're not using 1's and no 9's because they're in the bars. That's 35 and doesn't add up.
A better way to look at it is that the maximum cells for either bananas or bars is 45 with the digits 1-9. The minimum of the other set is 36 because 81-45=36. So the 9 can go in either set. Remove the 9's completely. We're now working with 36 cells in each set and they're both built with the digits 1-8 for both bars and bananas. There is no more degrees of freedom. If you remove cells from one set to add them to the other set, you'll have to add a digit and you're already using them all.
@@alienrenders but why does it need to be 45 if theres a 9 in a banana? it takes minimum 27 squares because there must be at least 9 bananas of at least size 3, and 27 fits into that 44 limit just fine. So the math logic doesnt make any sense
@@darkthunder312 Because if bananas have 9's, then the maximum you can make the bars is 1 to 8 that adds up to 36 leaving a minimum of 36 cells (81 - 9 - 36 = 36) that must be added to the 9 cells in bananas. 36 + 9 = 45.
Have you guys ever thought of doing a puzzle on live stream and get help (or distraction) from the comments?
I need a mars bar
28:46 for me.
_"orthogonal edge"?_ 🤪
_"rectangular or square"?_ 🤪
By the way, ethimologically
Rect = ortho = right (90°)
angular = gonal
Chapters are off a bit, actual "let's get cracking" 11:49
Thanks, I think I've fixed them :)
You missed there can not be a 1 yellow, it would be a square. so Bannanas can only be from 2 to 9.
Wouldn't that mean there also can't be a 2 since that's a rectangle? (Not done the video yet, sorry if that's answered later!!)
The size of the banana is irrelevant - the only implication of that is that you can't trap a 1/2-cell banana between chocolate bars, but beyond that there is no restriction or information about the size of the banana(s), and there's nothing to stop it being one giant mutant 36-cell banana...
yes, a digit in yellow does not indicate the size of its region like a brown diamond does
I'm sorry, but what a horrid puzzle! 40 minutes before any actual start.
Damn, I had the bananas and chocolates set. when filling in, I made a mistake where i switched the arrow from r7c9 to r8c9. stupid mistake.
NOt even close to 3 stars.
There's a mistake in the rules. The rules say that each region consists of orthogonally connected cells. Cells being plural indicates that there can't be a region of size 1, but there is.
I got one of my bananas intermixed with chocolate, so I decided have a peep at Simon before I dared to go further. Simon did come up with a quite penetrating discussion … before getting the banana split, him as well. But ... it turns out that you don’t need to exclude any occurrence of non-diamond chocolate to solve the puzzle.
Hi Simon, this was lovely, I didn’t expect another feature so soon after my birthday feature last week, but am very grateful that you chose to show this one, as it is definitely one of my recent favourites that I’ve made.
You were right that counting circles are something that I came up with, in my puzzle Circular Reasoning which you solved last September… (although obviously I don’t know for sure if someone else did the concept before me… to be honest I would be surprised if not, because it seems too simple an idea to have not been thought of before…)
But yes I was thrilled to see how they caught on and became popular. I’ve seen so many fantastic and varied puzzles from other constructors using counting circles, and it always brings me lot of joy. And I still like setting with them myself, especially when they are not just standard circles, but instead applied in other ways, as I’ve done in this puzzle.
The inspiration came from Scojo’s birthday feature on the 9th of May… you solved his fantastic puzzle “Chocolate Banana Sandwiches” and it left me feeling that I really wanted to try setting a chocolate banana puzzle myself.
As usual I’ll delve a bit into my setting process, for anyone who enjoys reading about such things...
One thing that my brain seems to be drawn towards is including a “full set” of something… Eg: in “My Zipper is Broken”, the big broken zipper line has all possible sums from 2 - 18 along it, and so for setting a Chocolate Banana puzzle, my immediate thought was “what if all chocolate bar sizes from 1 - 9 appeared once each?”…
I played around from there and found that the sizes from 1-9 fit into a 9x9 grid much more snugly than I had any right to expect. A good example of luck coming into the setting process. It could easily have been impossible to fit them all in, or equally it could have been too easy and not very constrained at all, requiring loads of other clues to force their placement. But the universe decided to be nice to me on this occasion and made it just the perfect amount of constrained - ie; possible, but only just.
Initially my idea was to have the counting rule apply to chocolate bars instead of bananas, but this turned out to be way too problematic. It would mean that the size 9 chocolate bar would contain 987654321, the size 8 bar would contain 98765432, all the way down to the size 1 bar which would just contain a 9. But geometrically in the grid, sudoku always then ended up placing the wrong digits in places… it was all way too restricted, and actually even if it had worked it wouldn’t have been nearly as fun or interesting as what I came up with instead… putting the counting rule on bananas, and then having diamonds that indicate the chocolate size.
Another stroke of luck was that with all 9 chocolate bar sizes, the total size of yellow would be 36 cells, the exact triangular number for 8, so that yellow would have to include all 8 digits apart from 9, the only digit that it couldn’t take anyway given that 9 appears in chocolate. It all seemed to fit together so perfectly.
When I started experimenting with the diamonds, I enjoyed that their placement could force the positions of the chocolate bars so powerfully. The breakthrough was realising I could put more than one diamond within a chocolate bar, and noticing how this would force things even more and lead to some rather interesting global considerations.
The way that I thought about this opening was to ask a particular question, which you never quite did, but you got to the same conclusion in a slightly different way which was also very interesting. My way was: “There are 12 diamonds, so what’s the minimum number of repeated digits you need to have in diamonds?” Clearly you need at least 3 repeats. And the only bars able to contain a repeat are 9, 6 and 4 (not 8 because eight 8s are needed in yellow... one of my favourite realisations in the puzzle.)
You can then locate all the possible diamonds which are able to join with another, and it’s not many. It turns out you need all 9 chocolate bar sizes to be able to fill all 12 diamonds, and so you can’t waste an opportunity by having a chocolate bar with no diamond in it. Basically, all diamonds which CAN join to another diamond MUST do that, with only one degree of freedom. But for sure you need all 9 chocolate bar sizes, and as you spotted, the size 7 chocolate bar is the key to it all, only being able to go in one place, and thus removing the degree of freedom for the others, which all become quickly forced after that.
I found quite a few different arrangements of the 9 chocolate bars that worked, some of which just resulted in one massive banana. But I preferred having at least 3 bananas, and this final layout proved by far the most interesting in terms of everything forcing itself, and more importantly it led to a solution without the counting rule breaking! Hooray!
The 7 diamond was originally right in the corner in r1c9, but then the whole bar was immediately forced along the top. I decided I wanted the diamond moved to the left a tad, so that the 7 chocolate bar could potentially go to the left or to the right, and the only way to deduce which way is by using the fact that you can’t leave a tiny square banana in r1c1 (such things please me more than they should.)
I realised that although the whole puzzle almost forced itself towards a unique solution, I needed one more clue type that would probably only need to be used just a few times in order to take it over the edge. I toyed with different clue types such as thermos and arrows, but I thought that aesthetically anything liney might distract during the earlier chocolate bar building logic, so I preferred a small arrow based clue. I spent the last couple of days trying to find the most potent place to put these clues so that I would need no more than 3 of them, and ideally each one would serve a few purposes, each giving you a lot of ‘bang for your buck’. Eg: the 6 arrow not only gives some useful information about the 2 and the 4 below it, but it also means it can’t be a 7 itself (the main reason for me putting a clue there), forcing 7 down into chocolate in the column, giving the 2nd brown 7, which means 7s can then be pencil marked in all kind of bananary places.
Right, that’s quite enough of that. Thankyou once again for all the entertainment and lovely words of praise in the video, it means a huge amount to me to see how much you enjoy all the little details :)
Thank you for this fascinating insight Marty! oh and as I said elsewhere, thank you for another delightful puzzle!
Just continue loss for words Marty. Love how you set. Wonderful insight as usual above from you. Exceptional puzzle!!
If you had your own signature thumbnail art like phistomefel with his devil what kind of character/picture would you like to be?
Oh lovely the setter is here :)
I wanted to just write a comment but I can say it to you then.
I think the rules need a bit of rephrase..
I think saying "each region must consist of orthogonally connected cells" does not allow a one cell region.
That threw me off a bit, I thought I broke it at one point :), but I just went with the one cell and it solved..
Beautifully :)
It's a brilliant puzzle.
Thanks for the detailed process too. I've always wondered how one would set something like this.
@@FairNuff Thanks for the comment mate, I have fixed this wording on LMD now. Unfortunately I'm not able to edit the link above
Marty Sears taking color choices out of Simon's hands is the funniest part of my day (I'm no fun at parties either)
100% - so glad he said, or we'd have purple chocolate and black bananas!
lol I'm not gonna lie, this did cross my mind while I was writing the rules haha
@@martysears Best. Comment. Ever. Your puzzles are my favorites! Don't tell the other setters I said so please. Seriously I can't even wrap my mind around the creativity it must take to set puzzles like this. Thank you so much for doing what you do!
@@martysears 🤣🤣🤣
Thank you for featuring this Simon, it is an absolute stunner of a puzzle and I so enjoyed your emjoyment of it.
However I do have a bone to pick with you! Did you HAVE to make purple flashes on the chocolate bars???! Now I neeeeeeeeeeed Cadburys!!?
Marty.... you know I enjoy all your puzzles, but you have outdone yourself here!!! Just beautiful!
Another example of how Simmon the Cat’s brain drives through puzzles by solving new logic before resolving old logic.
It’s cuz you said 5/6/9 instead of 4/5/9
The ruleset is incorrect. Based on the rules there should not be a 1 cell chocolate region. "Each region must consist of orthogonally connected cells". The single cell chocolate region is neither cells (plural) nor orthogonally connected.
I spent a lot of time trying to avoid a single cell region only to find it was impossible.
I usually put your videos on my second monitor or in the background. Just listening to all that talk about chocolate and banans made me smile a lot
Thank you for showing your mistakes. It really comforts me when someone who is WAY smarter than me shows how they can still make mistakes ❤ Amazing solve Simon! Always a joy to watch!
Every time you said “chock-let”, I would think of Moira Rose saying it. 😄
More exceptionalism from you Marty!! Continue to amaze!
Your joy, love and praise during your solve Simon shines bright.
The way Simon says "Oh, Marty!" at 45:01 tickles me pink. Like, the genuine appreciation of not just the puzzle but also Marty Sears the setter/person.
Me too 😀
It's definitely harder than 3 stars but i am so happy that i solved it, amazing puzzle. 100% definitely.
76:51
This felt like two puzzles for the price of one. Placing the colourings with a little help from the numbers was full of tricksy logic and evil ideas, then filling the numbers by reference to the colouring was an even bigger conundrum. Ingenious puzzle.
Child 1: Mom, why are you yelling? What is wrong?
Child 2: Simon put the wrong number in. She's fine.
Child 1: What?
Me: It's just sudoku. Nothing's wrong.
Husband: Oh, you know, she only yells like that when Simon missed something.
Child 1: You all see how weird this is, right?
It made for an easier start by working out the maximum number that could go in each diamond...
can you guys upvote this so they can see it. cracking the cryptic should have two channels, one for light mode viewing and dark mode viewing. i would subscribe to both, and i would watch videos on both depending on time of day. imagine falling asleep to a dark mode CTC video.
This is a great puzzle! It requires the same amount of effort throughout the entire solve, and there are plenty of beautiful deductions hidden within!
Much as I enjoyed this puzzle, but wouldn't want to do too many as it has filled my head with the idea that I must have chocolate.
Wow, watching Simon's solve I realise I skipped a whole load of stuff at the start due to assumptions.
It seems I did something backwards. It was only after I'd fully coloured my chocolate bars and bananas that I counted how many banana cells I had.
Early on, I found the three diamonds that ended up being in the 9 bar had to be part of a single size 9 bar, otherwise they forced too many size 1,2,3,4 sized bars. Then I saw I had to have eight other bars to cover the remaining diamonds. (Only the top one could be the 7 bar.) I then managed to place all the other chocolate bars before counting how many banana cells that gave me.
Same here
I did the same. Then I thought I'd broken the puzzle when I for some reason calculated that 81-45=46 (total cells minus chocolate cells equals banana cells) and one 1, two 2's and so on up to nine 9's is only 45, what's the last digit? Then I realised my mistake and since I already had chocolate 9's I knew there were no banana 9's.
I love the ultimate televance of one of Simon's examples... he must have been intuiting while reading the rules!
Does anyone know why when Simon says words like “region” or “nature” he spends so long on the first syllable? Is this a feature of a certain regional English dialect?
“Nature” even contains almost a glottal stop in it before the “ch” sound. Very odd to my American ears.
We do love a glottal stop 😊 Simon is very well-spoken… if you heard my more Londony accent you’d find it even weirder probably 😅
Now I want banana and there aren't any :(
I feel like the difficulty disparity comes from the kind of logic this puzzle requires being more spatial than numerical. Or at least more than usual for most sudoku. I've been watching CTC for a while now and it's really illustrated to me how different my logical strengths are compared to Simon's. He picks up on numerical patterns and math things so fast and it's always things that never would have occurred to me. But then I watch him do a puzzle like this where he misses a lot of the spatial deductions that seem a lot more obvious to me and it really shows just how differently our brains work.
Can we put this to bed? Squares are rectangles.
I have a puzzle collection book with chocolate banana in it and the rules in the book says something like "chocolate regions must be rectangles, including squares." Which is reassuring but also refreshing in that it acknowledges the definition of rectangle.
42:05 for me. The implications of the rules is half the puzzle. Still... not bad.
Stunning puzzle. Fisaco at the end was deserved, because after the break-in, which he always does well, he rushes round the grid in a blind panic making more and more of a mess and finally slipping over in it. More discipline and more calm thought would be very welcome.
Manic Miners with a touch of Battletoads?
Where my NES gamers at
Extremely clever puzzle; very enjoyable, too. 74:02
26:29 for me. What a fantastic puzzle, loved it!!
15:00 the chocolate can never be a three. It would leave a 1x1 yellow.
1:04:46 you say these squares must be a 5,6, and 9, when they were a 4,5, and 9. That is how you messed up.
178 minutes here 😊
I looked at the number of chocolate bars a little bit differently than Simon. I marked the 5 diamonds that could not join with any other diamond. (This includes the diamond in R6C1, which cannot join with the other diamond in box 5 no matter what Simon says. :P) Then, I looked at the remaining diamonds to figure out how I could limit them to only 4 chocolate bars. The key for me was recognizing the only diamond that could support the size-7 chocolate bar. Then, putting the last 3 bars into the remaining diamonds was quite limited. The 3-diamond size-9 bar, for example, was immediately obvious.
Yes this was pretty much how I intended it 😊
Stopped the video to give it a go and was glad I did.
I tackled this so very differently ... via the sizes of chocolate bars: bars of sizes 12357 can only take one diamond, so I was working on allocating the diamonds to bars - (I couldn't take enough diamonds without a 9) and also the placement of 5 and 7 bars is restricted: 9 and 8 also. The logic overlaps in a way, of course.
@@MarkBennet10001 I had also tried to start like you did, but unfortunately abandonned this way.
I should not have done so, as this logic (with 2 simple equations ) immediately leads to the result, that there has to be 1 bar with 3 diamonds, 1 with 2 and seven with a single diamond. (1+1+7 =9 (sizes), and 1*3+1*2+7*1 =12, (diamonds)). This relies on the fact, that there is only one place for a 3-diamond-bar, an no place for bars with more than 3 diamonds.
This could have saved me a lot of time and I could have solved this puzzle much faster than I did.
32:36 Brilliant puzzle, but for some reason I cannot fully follow the logic here. I see that there must be a chocolate bar of size 9, but I don't see why they're 9 different chocolate bars. The bar of size 7 could have been omitted, e.g.. Can someone help me with what I'm overlooking?
Hi Mahoel... it's because there are 12 diamonds. That means at least 3 diamonds need to be on a chocolate bar with another. If you miss out any chocolate bar sizes, this number increases to 4 or more. However, most of the diamonds are unable to be on the same chocolate bar with another diamond. Once you've marked these, you see that you can only achieve 3 repeats in diamonds. It's impossible to achieve 4 repeats in diamonds. This means all 9 chocolate bar sizes are needed, including 7. And there's only one place the 7 can go, which means there's now only one place the 9 can go, and it all kind of gets forced from there...
Thanks! I'm getting it now. You can see quickly that you cannot put 4 or 4x2 together without having them touch each other. However, I'm missing or overhearing this in the solve at this point. It may be obvious to others...
36:29. After determining the chocolate bars and bananas counted the bananas. Realized it was either 9 or sum to 9 that had to be bananas. Saw that 9 was quite possible. Then started recolouring as I solved from 8. Down of course couldn't finish 8s fully but following the logic, as I fully solved a digit I changed all occurances to black. Once sudoku could solve it I was left with an oozy brownish mess (between brown yellow and black)
I finished in 119 minutes. I really liked this puzzle. The rectangular nature of the chocolate bars made filling the grid very satisfying. I can't even imagine how you would set this regarding the banana rule. Solving it felt almost impossible, so setting it seems inconceivable. Incredible work, as always, by Marty. Great Puzzle!
Everyone here should try normal chocobanana puzzles. This was a beautiful puzzle, but the logic was closer to circle sudoku than standard chocobanana, which should be enjoyed.
This took a while for me ... First I tried out several versions of how the chocolates could be, then figured out that there is only a single option. Then it was somewhat painful elimniation. (I think I didn't use the "exactly 5 × 5 in yellow", that just came out at the end.
124:35, solve counter 502.
Beautiful puzzle. I'm also satisfied with my time on it, 53:24. Oh, yeah, have to say I never used other palettes in Sven's software, but 2nd one has some really pretty colors by default and is perfect for this puzzle (as demonstrated in the video).
Cool puzzle and its logic
32 min 42 sec solve for me, got the logic of the bars and was able to place them quickly. Super fun puzzle in a puzzle.
36:28 for me I think I got extremely lucky finding the break in so quickly
I’ve said it before, 3 in the corner is much more fun when it’s not alluded to earlier on in the solve
Lovely puzzle again from Marty! 26:12 for me today.
57:26 - Loved it! Some gorgeous logic throughout
At 35:46 I'm confused why Simon is saying there must be a chocolate bar of size 7? None of the rules require it, and the sevens that must exist within chocolate bars (as there must be either 2 or 9 7s in the chocolate bars) would just not be in one of the diamonds? It's perfectly possible at this point in the puzzle, so I'm not understanding why he says there must be a 7 sized chocolate bar?
After solving the puzzle, I did go back and look at it again, and I found it to be a valid proof, but I still don't understand the logic that Simon used, I just tried to run through it with the upper right brown diamonds making a 9 sized chocolate bar and not using a 7 sized bar. I found that there was no possible way to create a layout of bananas and chocolate bars that fits the rules. That being said, if someone could maybe explain it in more detail for my smooth brain, I would appreciate it!
He explained it earlier. If chocolate bars are all different sizes, then the maximum number of cells is 45. The minimum banana cells is 36 and must be from 1 to 8. That means only the 9's can switch sides. And for the chocolate bars, you need every digit 1-8 to get to 36 cells as well. If it's 45 cells, you need all digits. So the digits 1-8 are always used in both sets. That includes the 7.
Maths.
The split between chocolate bars and banana cells is 36-45, one way round or the other. But in both cases, they have to be made by adding up unique digits. So the only way to do that is 36=1+2+3+4+5+6+7+8 and 45=1+2+3+4+5+6+7+8+9
So there must be chocolate bars of every size from 1 to 8, plus maybe a 9 as well.
The way I thought about it was this… what’s the fewest number of repeats you need in diamonds? you need at least 3, as there are 12 of them. And the only ones that could contain repeats are 4, 6, 8 and 9. Looking at the layout you can see that a maximum of 4 diamonds are capable of being repeats… as Simon eventually pointed out, there must be 8 distinct chocolate bars at least. But actually it’s impossible to get the repeats on ALL of 4, 6, 8 and 9. Because to make r6c3 AND r1c8 both be repeats, there would need to be two 9 size bars. So actually only 3 of the diamonds can be repeats.
This means all 9 size chocolate bars are needed, and you can’t have any chocolate bars without a diamond. It means 9 goes in brown, and the 7 chocolate bar must go along the top… and because 8 is in yellow it means that there can’t be a repeated 8 in brown diamonds either, forcing the 3 repeats in diamonds to be two repeated 9s and then a repeated 4 or 6 at the bottom, which is forced to be 6 shortly afterwards
I'm just at the start of the video, but I'm watching in dread as Simon misreads and misunderstands the banana digit rule.
he seemed to read and understand it perfectly?
@@martysears There is a difference between "A digit in a banana region tells you how many such digits are in banana regions" and "A digit in a banana region tells you how many banana regions contain that digit".
The rules say one thing, and at least as far as I have gotten, Simon consistently says the other. The rules as written do not lend themselves directly to use of the secret, but Simon acts as if it does when he says there can be at most 45 banana cells.
(I do realize I might sound... brave... arguing the details of the puzzle rules with the actual setter, but still.)
@@MasterHigure Yes the former is correct although worded slightly differently from how you worded it. My wording was: "a digit in a banana cell indicates exactly how many banana cells contain that digit." So that does instantly make the secret relevant, and does mean there can be at most 45 banana cells.
The latter ('a digit in a banana region tells you how many banana regions contain that digit') would indeed mean something very different... (Cells are individual yellow squares, whereas regions are the whole bananas.) Given that there's only 3 banana regions, you wouldn't be able to have more than a 3 in a banana haha. But this isn't what the rule says, and isn't how Simon interprets it either
@@martysears Oh, right, I misread the rules. The rules (as written in the upper quadrant of the video) say "how many banana cells", and I read that as "how many banana regions". Yeah, my bad. I was the one who misread and misunderstood.
I even double and triple checked before making my original comment. And I still got it wrong.
52:03 for me. Delicious puzzle!
Hello there
62:08. Skadoosh....
man didn't Simon take a really long way round to break into this puzzle? I feel like that whole beginning wasn't needed because you can determine the bottom left diamond is a 1 (assuming you believe that's allowed)
why couldn't it be a 2 or a 3?
I'm confused (haven't watched the video yet)... the rule says each region must consist of orthogonally connected cells... but I have a chocolate bar of 1... a chocolate bar of 1 is a 1 cell brown region... that's not a set of orthogonally connected cells.
a one element set is still a set. if you only have one cell in the set, it's always an orthogonally connected set (because there are no other cells that could break the connection)
@@surrendherify I guess but that’s not what it says. “Each region must consist of orthogonally connected cells” (with an s, meaning more than one connected cell).
But maybe we always treat this to mean “including 1 cell not orthogonally connected to anything”
There have been puzzles before where the word being singular or plural was the clue in solving the puzzle or not.