Ah I knew I recognised your name! I just did your six by six Islands puzzle a couple of days ago! Love this ruleset. The 9x9s were a bit too hard for me but I think I made a respectable start on the first one all the same😊 Really looking forward to this video now.
I’m not very good at these difficult puzzles but I love following along. It’s also great for my self esteem when I’m able to see things that Simon (who I hold in rather high regard) does not , even though alas it is most often sudoku logic. But when it isn’t, I feel even smarter 😊
Travis said he forgot to tell you where we were - but we're in Australia so this actually came out on my birthday! Thank you for reading the shout out :) On being part of the household - he doesn't watch you guys, but every now and then I forget to bring my headphones to his place and so I listen out loud. He can tell you guys apart by voice, and no longer turns around when I scream at the screen about where to look next in the puzzle - depending on what I'm shouting, sometimes he can tell you apart when I'm wearing headphones 😅
By the 21 cage, he may conclude the degree of freedom from the 170 cage is spent in either box 3 or box 6, so the cage contents in the remaining boxes must be maximized
Yess, Though I did do the whole maths thing of finding 29, working on the 21 cage was the key break in. Though I love long CTC video this could have been a lot shorter.
After sum the cages(376) and realizing that the sum of the islands is 29, you have two degrees of freedom. However, if you use 9 ones, 8 twos, and 1 four (or 8 ones, 9 twos, and 1 three), it becomes impossible to place two islands in each row, column, and box. Therefore, you need 9 ones, 7 twos, and 2 threes. With this in box 4, the maximum number of cells you can remove is 4 (the 1 and 3), not 5 as Simon was considering.
I got the islands sum to 29. Which means at most 11 in any three boxes. The 8 cage and the two 9 cages must have at least 11 in islands, if you minimise the digits and subtract 8+9+9. So we know this is the case. We can place 123 in the 9 cage in box 1 of which two are islands. And 1234567 in the 8 and 9 cages in box 4 of which two are islands, and place 12 in the 9 cage in box 7 both of which are islands. In every other box the islands are 1 and 2.
SamuPiano here - actually, an early version of the puzzle utilized this fact! That deduction I eventually deemed to be too obscure for the solve which would detract from the experience.
I was thrilled with the length of the video and I was nearly not trying this puzzle. But turns out it only took me around 39 minutes to solve. Edit: I think Simon was confused about the 29 for the island cells. Actually the easier way to continue at around 46:14 is to do Sudoku on box 2, and notice r1c45 can only be 345 by Sudoku, and they are both in 6-cell 36 cage. Then by "the secret", the 3 missing cells have to sum up to 9, but if those 3 cells are either 234 or 135, r1c45 will be double 5 or double 4, so the only thing left is to miss out 126. Then by Sudoku again, using the 67 pair in box 2, 6 in box 1 is on row 2, which forces 6 in box 3 on r1c9, remove the 6 in the quarduple in box 6, which remove the freedom in the 170 cage. From there, the puzzle just flowed like water!
I also summed up boxes 5,6,7,8,9 plus the 89 pair in box 3 (242 total) compared to the total cages in those boxes (227 total) to realize that the islands in boxes 5-9 all needed to be 1/2 pairs.
Ah yes, that's clever! The "6-cell 36 cage" means the 6 cells from the 36 cage which still count after the island is removed. And the "3 missing cells" are the three digits not used in the 36 sum.
I'd basically just posted something similar. (Delete that.) Simon got really stuck on the last missing islands. But after being led from one cage to the next by the setter, I didn't spend any time on that and instead looked at the options for the now-restricted 32- and 36-cages. This led reasonably quickly to forcing a 6 into the 32-cage in box 1, which put a 6 in R1C9. This immediately meant the degree of freedom was used up in box 6, and the rest went relatively smoothly, as you say. I suspect this was the path the setter meant to be followed.
Simon, that was my longest scream to date. LOL. At around 0:24:00 you note that r3c9 and r6c8 had to be both an 8 or 9 and the same number. You had already noted the 1 degree of freedom in the cage. Almost 62 minutes later, you found it again. Now I have no idea if the handful of 8s and 9s that produced would have made the solve any quicker, but it couldn't have hurt. Maybe tomorrow I'll try a little louder. 😃
Having both be 8 would have only brought down the degree of freedom by 1. The 8 being in r3c9 brings it down by 1, but then that allows the cells in the box beneath to be 9876, with 8 in r6c8. So no, that wouldn't have been enough to conclude that those cells had to be 9.
@@fatman3762 *tips hat* When I'm wrong I'm wrong. Man, I even rechecked against the 21 cage when the islands in box 6 were revealed but going back over them now, they work the other way also.
The Moment when Simon recounts all the cage totals to make sure he didn't made a mistake and murmurs the sums, as suddenly he hears Maverick flying by and like a reaction from his spine he has to mention him. Its hilarious ^^ Gosh I just love to watch you solve puzzles
I'm not convinced removing the 3 from the islands in box 7 was earned around 1:08:00 (even tho it was right). All Simon had proved at that point was that the islands in boxes 1, 4 and 7 had to total 11, but not how they did so, so the island in box 7 still could have included a 3 at that point in the solve. The way I worked it out was by noticing that the totals of boxes 3, 5, 6, 7, 8 & 9 was 270 by Sudoku but 252 by Ocean with the 6 island pairs in those boxes accounting for the difference of 18 (1+2 = 3 , * 6), meaning the islands in box 7 had to be 1 & 2. Still, a great puzzle and as usual an entertaining solve by Simon. Love the channel.
I always need to watch Simon explaining the rules and fiddling for a bit with killer sudokus before I can try them myself, I'm really terrible at finding where to start. After Simon did the min/max calculation for the 170 cage and found out there only was 1 degree of freedom, I stopped watching and started doing the puzzle alone. I managed to finish it in about 40 minutes then. I didn't do it all by myself but I'm still proud for having finished it.
O frabjous day! Callooh! Callay! I've been following the channel for years now and this might be the first time I've ever truly stolen a march on Simon. 32:10 solve time for me. Started right away by adding up all of the box totals and realized the extreme limitations on the islands, then worked through the logic on the 8/9 cages. Never even looked at the 170 cage, it just fell out at the end. 1 W 999 L vs. Simon times, but I'm sure I'm starting a streak! (Of course I think this is mostly about Simon being mesmerized by the 170 cage up front, I'm guessing many others will have a similar experience.)
For someone who loves Star Battle 🌠 so much, Simon spent a lot of time _not_ looking at which rows, columns and boxes already had their two island cells 🏝 ... would have been a lot quicker if he had added up all the cages 🧮 at the start and found that the islands totalled 29 across 18 cells, giving very little freedom (which is quite appropriate ... if you are in a cage on a desert island, you have very little freedom!)
yes, even if it was towards the end with the 9 cage in box 1 I basically lost where Simon was heading when the solution was right there in the colouring.
My immediate response to this was to work out there were 29 islands (so at least 8 each of the 1s and the 2s, with the last two either being a 2 and a 3 or a 1 and a 4). Still really early in the video, though. He's messing with the 170 box and I'm smiling :-)
Quite pleased with a 64 minute time on this. Helped that I started by summing all the cages and getting my break-in in box 4 with the 8,9 cages needeing to use the degrees of freedom. I also seem to remember deducing early that there had to be 2 islands in the 8 cage, and it used one or two of the degrees of freedom. (I never thought to look at the maximum for the 170 cage, as Simon did. Fortunately, I didn't seem to need to.)
This was me too! Getting the sum of 29 islands was incredibly powerful and immediately I ran into the trouble of the 8 cage in box 4 and trying to make the math work out. From that the 9 cage next door leaps out as extremely constrained and everything builds from there. I think that Simon stubbed his toe badly in looking at the 170 cage!
@@dralthoff1 Although, reading the comments from others, I see it was possible to use the 21 cage to pin down the one degree of freedom for the 170 cage to box 3 or 6, which would have helped Simon a lot. So I guess there were two ways to approach it. But yes, summing all the cages was my first thought (followed by, "do I have to" and "I guess I better" 🙂)
I did this when first published, and I'm surprised by the video length. I've only skimmed the video, but it seems Simon hasn't thought to apply basic arithmetic to the puzzle as a whole, and has also missed box 4 as the breakin. Which is a shame, because it was a really good breakin. Add all the cage totals, apply the secret, the 18 islands must add to 29. Look at just box 4 + the two 9 cages that protrude from it. They must add to at least 45 + 6 + 3 = 54. The most r4c23 can be is 17, so the 8/9/9 cages must add to at least 37 when the islands are included. The most the islands can add to is if all the box 147 islands are in those cells, and all of the other boxes have 12 pairs as their islands, leaving 11 for the islands in box 147. 11 for the islands + 26 for ocean in the 8/9/9 cages = 37, so now that is forced. r7c23 is a 12 pair, and island. r3c123 is a 123 triple, with 2 islands. r4c23 is an 89 pair & ocean, and r3c2 is 1 or 2. So now r56c3 are ocean, r3c3 is island, r4c1 is ocean, and the 8 cage has 2 islands. A quick look at the possibilities for the r3 islands: 23 means 8 in r4c1, that's impossible. 12 means ocean 3 in r3c1, 6 in r4c1, so now the 9 ocean in c3 is 45, the ocean in the 8 cage must be 17, the islands must be 23, but that means 3 islands in c2 and only 1 in c1, so that's broken. So the islands in r3 are 13, r4c1 is 7, the islands in box 4 are also 13, the 8 ocean must be 26, leaving 45 for r56c3. And then the rest of the puzzle is very easy, and flows really well. Easy because every cage now reaches its total without using 1 or 2, so islands get quickly forced. e.g. if the 32 cage only has 1 island, its minimum sum is 345689 = 35, so it has 2 islands, and only 2 spots for them. 1289 are already in the cage, 3 can't be in it, the remaining 3 cells add to 15, so must be 456, and so on. edit: Simon does half the breakin at 1:05:00, and then the other half at 1:16:30. If he started there, it would have been a 30 minute video. ;)
At 1:07:58, was Simon's removal of the 3 from box 7 premature? He says that the highlighted cells (which includes those highlighted in box 7) need to include both degrees of freedom, but then discounts the silent digits in box 7?
Box 4 (45)+3 cells in box 1(6)= minimum 51 9+17+8+9+island cells(maximum 8)=51 Minimum=maximum tells us everything must fixed, so we already got 2 more degree of freedom of island cells in those highlighted cells
You did it the hard way by not seeing the 170 single degree of freedom was used up in box 3 or 6. If you put an 8 in r3c9, then you used it up, and if you put a nine in it, then you needed to use the six in the 21 cage, and used it up. Then you would have had most of the 12 islands and the maximum values in the rest of the blue 170 cages, which would have made things a lot easier.
I do wish Simon would apply his revelations of forgotten rules (in this case 2 islands per r,c,b) to the rest of the grid before fumbling about for 5 minutes trying to use it for the one occasion he's noticed.
My break-in. Cage totals = 376... subtract from 405 gives the number of "silent" cells = 29 within 9 boxes. If all silent cells were min (1+2), total silent would be 27. Gives only two degrees of freedom across ALL silent cells. Gives either 2 boxes with silent 1+3... OR... 1 box with silent 1+4 or 2+3. All other boxes have silent 1+2. Looked immediately at Box 4.
spoiler for 1:05:00 my break past that point was a bit different I focused on the 32 cage and the 36 cage, and after going through all the ways to get 32 in 5 digits and 36 in 6, I concluded that: 1. 6 was forced to be in the 32 cage 2: because 8 and 9 were also forced into the lower 3rd of the 32 cage by their inclusion in the 36 cage 6 had to be forced in the center domino of box 1 and the corner of box 3! this ate my only degree of freedom in the 170 cage because i could no longer use 6 in the box 6 portion of it. I kinda wish i thought to calculate the 18 island maximum, although I did have 16 of the islands placed at this point and (without starting over to prove it) I don't think that would've helped me
Really beautiful puzzle by SamuPiano! And a somewhat confusing, but still very clever solve by Simon! I'm a little confused myself about my own break in, after watching SImon's solve, and wonder if I just got lucky: I started immediately with the global sum of all cages, figured that I needed all 1's and most 2's on the islands, which pointed me to the 8-cage, as I couldn't figure any way to have less than 2 islands in it. This then kept pushing islands around the grid, before I even started looking at the 170 cage. What I failed to appreciate was Simon's theory of one island with a 4, or one 1 going in the ocean. I just couldn't make it work as I looked at the 8-cage, and claimed the islands had to be 9 1's, 7 2's and 2 3's, but in hindsight it makes me wonder...
It occurred to me to note the secret of a sudoku board, 405, and then sum the cages to 376, which leaves 29 on islands which have an average for rows, columns, or boxes of 3 remainder 2.
At 24:25, Simon remarks both those squares are the same, but they can't both be 8, as there is only 1 degree of freedom in the 170 cage. Simon only resolves this one hour later at 1:25:35! Edit: I was wrong about this. @BOTHLine explains it well below.
I thought the same for quite some time.. but actually if it was an 8 in the 3rd box, it would just move the 9 in the 6th box to the right column.. which wouldn't use up the other degree of freedom. Use either had 8 + 9 in box 6 anyway, no matter if the other digit was a 9 or an 8.
I usually don’t attempt puzzles where the video is over an hour, but I love killer cages so gave it ago. Finished in 51 min. Feel like I found the intended path by doing the math and realizing I only get two degree of freedom on island values, then focusing on the interplay of the 8 and 9 cage in box 4. Basically moved around the grid after that as each cage was filled in.
[Mild SPOILERS] If only Simon had spent a bit more time on the 21 and 170 cages, this would have gone a lot smoother. Though I admire his patience and resilience in the face of daunting puzzles because due to some quick faulty logic in box 1, I had to redo all the logic from scratch and the mistake I made was - I just assumed that 89 pair goes in r1c1 and r2c1 as it was seeing the 89 pair in r1c6 and r2c6. (1:21:55 see simon tackle it correctly) So solvers beware this ocean island sudoku is not that easy so tread carefully and don't pencil mark it to the brim. Keep on cracking.
I've never seen a more different solve path from my own and Simon's. I started on the fact that was discovered about 50 minutes in that there were 29 worth of non-counting cells in the grid. Then combined with the small cages in box 4 and box 1 adding up to 26 but having a minimum sum of 34, which means they need to use 8 worth of non-counting cells and given that it's in two of the boxes, that uses up the two degrees of freedom and proves all the other non-counting digits not in boxes 1 and 2 are 1/2. And then I basically completely ignored the 170 cage since I thought it wasn't useful at that point. So my grid started filling in in box 4 and went around the grid, but Simon started in box 6 and went the other way. Although, I think I would've solved it much faster by at least considering the 170 cage because it seemed pretty profitable to know that the digits in its cage were near-maximal. Kind of amusingly I thought that the 1/2 non-counting digits in the 30 cage box 3 would force the 6/7/8/9 in box 2, but ended up proving that the box 2 was 6/7/8/9, thereby forcing the 1/2 non-counting. And Simon proves it the way I thought it was supposed to be really early in the solve!
Wow! What an amazing puzzle. I worked out most of the 170 cage straight away but made the fatal error of assuming that R3C8 was a 9! Of course that didn’t then work and took me over an hour to work out why! Finally completed it in 2:36:32!
When he says I like comments especially when they are kind makes me sad cause it implies sometimes there's people putting unkind comments on these videos 😭 I hope they are very rare. I love these videos even though I don't often watch them, keep up the good work. :)
I took a look at this puzzle and knew that I wouldn’t be able to solve, and after watch it I know that it’s because I’ve never been any good at counting nuns.
I love watching the brilliancy of the solves, while also slowly dying inside because of the x clue haveing to be made of two counting cells. You got there, but seeing that you need two numbers to make 10 would have been a big help and saved about 20 min i believe ;)
At 38:00, Simon makes a small logic error. The 32 (or 31) totals in boxes 5 and 8 only force a 9 into them. They could be 97654 (where the loss of 1 degree of freedom occurs), with the 8 joining the 2 in the 10-cage. It doesn't end up mattering, of course, because the degree of freedom is lost elsewhere.
I wasn't going to watch this but there was a pile of washing up to do so I set my iPad up above the sink and watched as I washed. Glad I did. What a puzzle, and so brilliantly solve.
Just under 50 min for me - glad I tried it as it's a lovely puzzle with the interaction between the cages in boxes 3 and 6 taking away the last degree of freedom from the 170 cage.
Nice puzzle and very good video. Excuse my English, but I will explain part of my solving path. I started by analyzing boxes 5,6,7,8 and 9 (total sum: 5x45=225); together with cages 10,16,10,21 and 170 (oceanic sum (os): 10+16+10+21+170=227). So: the oceanic sum of boxes 5,6,7,8 and 9 is: OS(box5to9)=227- os(r3c8;r3c9), but also: OS(box5to9)=225-(islands U [r7c2;r7c3 ;r4c4;r4c7]) So: os(r3c8;r3c9)=2+(islands U [r7c2;r7c3;r4c4;r4c7]) The left part is max. 9+8=17. The right part is minimal when [r7c2;r7c3;r4c4;r4c7] is included in the set of islands and they have minimum values: 2+(1+2)x5=17. BINGO. Equality will hold only under those conditions. This not only immediately concludes that r7c2;r7c3;r4c5;r4c7 are islands, but that each of the islands in boxes 5,6,7,8, and 9 contains only the digit 1 or 2.
I think I did most of the logic Simon found near the end at the very beginning - after summing everything we notice that there's 2 degrees of freedom on islands, so neither 8 nor 9 can be 3 cells large! 125 means the lowest islands are 3/4, 134 2/5, 234 1/5, 126 3/4 and 135 2/4, all of which are bigger than the two degrees above 3 we're allowed. In that case - box 4 has two islands in 8 (the rest is ocean), box 7 has two islands in the 9 cage (the rest is ocean), X has to be ocean as one cell can't be 10, meaning that islands in box 1 are on position 7 and 9. That was my start to this puzzle, apart from noting that all islands had to be from 1/2/3/4 and were extremely constrained.
at 34:57 you put an 89 in the cage that sums up to 21. This removes the option of the 89 next to it (the one in the top of the 170 cell) to be an 8 (it would take at least 2 degrees of freedom, we only have 1). I believe the puzzle was supposed to be started from there.
Putting an 8 in r3c9 only removes one degree of freedom from the 170 cage, not 2, because it doesn't restrict what goes into the cage in box 6 ... it can still be maximum of 6-7-8-9 totalling 30, whether the caged cell in box 3 is an 8 or a 9.
@@stevieinselby No, because you must have 789 in the cage within box 6, so as soon as you put an 8 in r3c9 you would have to put 8 into r6c8, and you have now used 2 degrees of separation. Ergo they must both be 9s.
1:07:59 I do not understand why there has to be a 12 pair in box 7 and it can not be a 13 pair. Is the degree of freedom already used up somewhere else in the 170 cage at this point? Could someone explain?:)
From 1:03:35, Simon is using the minimum value for the cells highlighted as being 45 (box 4) + (1+2) (box 7) + (1+2+3) (box 1) = 54. The only way 54 works is with an 8+9 pair in r4c2-3 and using the two degrees of freedom available in non-counting islands (8+9+9 (cages) + 17 + 11 = 54). None of those can be increased. So 54 is also the maximum for the set of cells. Since maximum (54) equals the minimum (54), they're forced to be what he first hypothesised to reach his minimum total, i.e. a 123 triple in r3c1-3, and a 12 pair in r7c2-3. I had to pause the video to understand that point myself. It wasn't immediately obvious to me.
Box 4 (45)+3 cells in box 1(6)= minimum 51 9+17+8+9+island cells(maximum 8)=51 Minimum=maximum tells us everything must fixed, so we already got 2 more degree of freedom of island cells in those highlighted cells
@@khoozu7802 ? Look at timestamp 1:03:30. Simon has selected a region containing 14 cells. Those ARE the highlighted cells, the ones Simon is talking about, and it very much includes the orange cells in box 7. 🙂 They're part of the 9 cage, needed in the calculation of the maximum of 54, using two available degrees of freedom in uncounted cells, for the 14 cell region. Minimum for uncounted cells in three boxes is 3*(1+2)=9. So maximum uncounted cells, using two degrees of freedom, is 9+2=11. The maximum in r4c2-3 is 8+9=17. Cages sum to 8+9+9=26. So the maximum overall is 11+17+26=54. This is the same as the minimum, when the orange cells in box 7 are 1,2 and the three cells in box 1 are 1,2,3. I'm just explaining Simon's logic used in his solve. Maybe there's another way to do it, but Simon's way works.
@@khoozu7802 (Isn't one of your 9s in "9+17+8+9+island cells" from the 9 cage that includes the orange cells in box 7?) Edit: Nevermind. Orange cells are already known to be uncounted cells. That's why you can treat the 9 cage as if it were only in box 4. Your way works too.
I finished in 171 minutes. This was a tough one, but fun to work through. I was able to get the break-in, but I made a slight error in which ones were maxed and which ones had play, and forced the one in box 5 to be minimum. I forgot that the digits could shift to the other ones. That was a silly mistake, but I was happy with figuring out the break-in, otherwise. Once I started doing the puzzle properly, it was really fun to see the strange logic that forms in a grid like this. It was quite strange and worrying to keep getting 12's as the islands, only for me to accept them and be bamboozled at the end when they were not present together in boxes 1 and 4. I enjoyed this one, despite what my time may say. Great Puzzle!
Very enjoyable solve! I have taken to solving alongside Simon while he solves these days while trying to sneak ahead. Early on, you can use the other cages (aside from 170) to prove that there are no degrees of freedom for the 170 cage. In box 6, in order to satisfy the 21 cage it must take at least a counting 6, even with an 8 or 9 in r3c8, so the 170 cage must be 5789. From there, the logic cascades to the other cages as well. It quickens the solve just a bit, but mainly just at the end, since you can maximize all cells in the 170 cage. For example, you know 89 needs to go in r12c6, which forces 89 in the 36 cage to box 1, forcing r4c23 to be 89, accelerating the logic there.
I simply love Simon making super hard deductions and forget it's a sudoku puzzle. There was way easier breakthrough from box 2 if you simply did bit of sudoku and puzzle was flowing quite nicely. I actually did the part which Simon used for breakthrough last. 62mins for me, could have been even faster but I forgot some maths I did earlier in the puzzle and when I used the 1 degree of freedom in a 170 cage it took me like 20 minutes to realise I had already used it.
And very fair to say, that I would never have been able to do the deduction in box 1,4 and 7 as Simon did around 1h06m into orbit................. Chapeau!
1:17:09 if the 1 and 2 in box 1 are both islands, that means the 1 and 2 in box 4 both become islands. That is impossible. So, the 3 in box 1 is an island.
37:50 I don’t think I understand why we have to have 8 and 9 in c5 box 5. I think the orange cells in box5 have to be 13 or 12 pair (because of 1 degree of freedom) but in case they are 13 pair doesn’t it mean we can still have a 82 in the 10 cage?
80 minutes for me. I spent over half an hour for the break-in which had to be the 170, of course, I was too slow. Let me explain: sum(cages)=376 which is 405-29, so the island cells have 2 degrees of freedom (27+2=29), so the maximum value for an island cell is 4. The 170 cage has a maximum ocean cells of 9+30+42+32+32+26=171 so there is one degree of freedom which disappears by combining it with the 21 cage. After that, you can color a lot of ocean cells and the rest is easy. Thanks for the great puzzle despite there was no 3 in the corner!
The easiest way to think about the totals in boxes 5 and 8 are that the 10 cage plus two islands of 12 account for 13, therefore the maximum contribution to the 170 cage is 32 from each box. In box 7, the maximum is 45-16-3=26. It all adds up to an annoying 171. However, in box 6, if we use 6789 in the 170, we only have 15 to use on the rest, which, if it includes two islands of 12, there is just twelve left for the 21 cage, which would require a 9 in box 3 (using up the degree of freedom). This means the degree of freedom is entirely used up in boxes 3 and 6. This is the crucial deduction you missed, which would have saved you an awful lot of time. We know the degree of freedom is used up either by making R3C9=9, with the 170 cells in box 6 being 5789, or by R3C9 being 8, with the cells of the 170 cage in box 6 being 6789. This means all of the island cells in boxes containing part of the 170 cage must be 12s. We can place 12s in R4C4, R6C6, R4C7, R7C2/3, and we can put a blue 89 pair in R3C8/9. This forces the 25 cage in box 3 to be all counting 3-7, putting island 12s in the 30 cage, forcing it to be 6789. @ 29:58 "Maybe I can prove this is an island" - that's easy. You can't afford to have two more islands in the 21 cage or you'd never make the total. I found the 376 cage total was quite useful, because it stopped me running away with the idea that all islands were 12s. I was able to place most 12s, and whilst I knew most were islands, it was tempting to think they all were, which would have been a mistake. Knowing that not all 12s were islands alerted me to be careful. It was also useful, because it meant there couldn't be a non-counting 9 in the 8 or 9 cages. @ 1:13:22 - considering where the islands are in box 1 - if the 3 were not an island, you'd have 12s as islands in both boxes 1 and 4, which would break your degrees of freedom. The 3 therefore has to be island, with a 1, so there must also be a 13 island pair in box 4. Because I spotted the restriction with the 21 cage using up the degree of freedom, I found this considerably easier, however it was still no pushover. It was definitely a very nicely set puzzle, and using the 21 cage logic, it had quite a nice flow (e.g. the 89 pair in box 3 forcing the 25 cage, which in turn forced the 30 cage in box 2). What an auspicious debut puzzle.
The very first thing I did was sum the cages and compare it to the sum of the whole grid. This gives some constraints which I thought were important but which you did without for long time, about 45 minutes.
36:25 for me Found out those degrees of freedom breakins earlier Also added up everything in boxes 5-9 plus r3c8&9 to realize all of those were 1s and 2s
Slight error at 38:00 cause the 8 could go in the 10 cage in box 5, if the islands are 1-3, cause in that case 8-2 or 4-6 takes the same total from the 170 cage Edit: ok, he caught it later without using it for anything
4:56 PLEASE record a video of you solving the 100x100 crossword. Or do a live stream! It could be one of those eight-hour livestream marathons for charity! PLEASE.
I'm pretty sure the 8 corner marks in box 5 at 38 minutes is unfounded. The ten cage being 82 rather than 46 would use up the 1 degree of freedom by forcing the islands to 13, but wouldn't break yet.
All those 1s and 2s on islands make me think of the song "One" "One is the loneliest number that you'll ever do Two can be as bad as one, it's the loneliest number since the number one"
7:13 Happy Birthday to Heather, Holly and Hope Newbury! Hope y’all are doing well. It’s such a small world to hear about people you went to high school with from someone across the pond. Tell Hope that sudoku variants are not only great fun but also help people our age keep our brains stimulated. Although, fortunately some of us get to wait a year for the big 5-0.😉🎉 Andy
I didn't attempt this puzzle, but I did see a better way to start looking for islands at least. I just added up all the cages (376) and subtracted that from 45x9 (405) to get 29. So we know the 18 islands only add up to 29 which means they have to be mostly 1s and 2s.
So funny. For once in a life time, I came 45 minutes earlier than Simon to the observation that the sum of all cages + non counting cells = 405, hence ..... (non counting cells not adding up to 27 but to a bit higher sum). This will never happen to me again :-)
1:07:58 why do these need to be a 12 pair? Can't they be a 13 pair and take one of the degrees of freedom from the sum of the island cells? The interaction of the 21 cage in boxes 356 and the portion of the 170 cage in boxes 36 actually takes away the degree of freedom in the 170 cage, meaning everywhere else the 170 cage must have 12 pairs in the islands (i.e. boxes 5789), but without doing this logic can Simon make the claim about box 7 when he does?
Box 4 (45)+3 cells in box 1(6)= minimum 51 9+17+8+9+island cells(maximum 8)=51 Minimum=maximum tells us everything must fixed, so we already got 2 more degree of freedom of island cells in those highlighted cells
Something about that puzzle made me screw it up multiple times in stupid ways and I had to redo stuff. I guess seeing the opener really quickly had me go too fast. The 170 cage was the right place to look, you just overlooked how the 21 cage removes the degree of freedom and lets you max out the rest of the cage.
I somehow got this in half the time. For me the key was looking at combinations of the 170 cage with other cages that would help fill in the rest of a box or row. I finally computed that, taken together, the 170, 25, 21, and two 10 cages have a total of 236, while comprising 7-9 cells in each of 5 boxes (thus having a max of 42 each) and a max of 3+4+5+6+8 in the lower-left box -- which, as it turns out, works out to 26+5*42 = 236. Meaning there's absolutely no degrees of freedom for any of that, giving most of the ones and twos. And, as it turns out, you can also do this without the 25 cage, because once the 1 and 2 are ruled out, the 25 cage has to be 3-4-5-6-7. Then I got carried away with some maths and I think my first digit was actually the top-right cell!
46:20 if the sum of all the islands was 21, then all boxes in the 170 cage would be maximized and then the 170 cage would be 171... good the sum was 29... :)
With the 1-2's in rows 1 and 2 the 36 cage has 6 cells that can only be chosen from 7 digits from 3 to 9. That means the 36 cage doesn't contain a 6 , and that places a 6 in row 1 column 9 getting rid of a 6 possibility in the 170 cage in box 6 . That removes your one degree of freedom for the 170 cage.
Even before the 12s were placed, the 21 cage used up the degree of freedom, either by needing 9 in R3C8, or by needing 6 in box 6. This meant that all the islands in boxes covered by the 170 cage were 12s.
After getting the total of the island cells, I ignored the 170 cage, focused on box 4, which completely restricted the islands in the other columns and spiralled out from there. Much simpler arithmetic that could've halved Simon's time.
36:17 This might be the most that I've ever beat simon by Haven't watched the video yet so I don't know what he does, but what helped me is to figure out early the maximum possible value for the 170 cage when taking into account islands and 10 cages and realizing that it was 171 Really let me fill in a lot of ocean tiles and let me mark down some numbers Beautiful puzzle that gives you exactly what you need when you need it
One degree of freedom - c9r3 & c7r6 are the same... Well then you can't make both 8, that would use 2 degrees of freedom... Yelling for an hour :D Love it :)
I thought the same for quite some time.. but actually if it was an 8 in the 3rd box, it would just move the 9 in the 6th box to the right column.. which wouldn't use up the other degree of freedom. You had 8 + 9 in box 6 anyway, no matter if the other digit was a 9 or an 8.
I just realized that the first deduction I made on this puzzle might have been very, very premature and only ended up being correct by accident :( . I marked both cells on the X as counting, but after re-reading the rules, I realize that it says the non-counting cells don't count toward cage totals, but doesn't mention them not counting toward the X...
Hope someone reads this and can answer; if Simon was able to put the 2 by the X clue in by simply having to put the 7 and the 2 as the maximum value for their cage, how come the X Clue is nececary at all? he didnt need the 10-sum at all to resolve it, how is it required in the puzzle then? i am sure i am missing something....
The sum of counted cells is 376, but the real total sum must be 405. That means that the 18 uncounted cells have to add up to 29. That works out to only a few combinations of numbers being uncounted: nine 1s, seven 2s, and two 3s; nine 1s, eight 2s, and one four; eight 1s, nine 2s, and one 3. I think that's all.
Very hard for me, but I cracked it! [...and I didn't even realize that you are supposed to calculate the sum of island cells; somehow never thought about the bleedin' obvious...]
You could've much more easily proven R4C1/2/3 by noting that R4C1 couldn't be an 8 or a 9. A 9 would break the cage total since we know there's another counting cell, and an 8 would require a 1 to it which we already know is a non-counting cell. Thus only 6/7 were left as options, making the other two the 8/9 pair we're looking for.
Great video, I think the trouble you had in box 4 came from the fact that you made them green, you would have seen it faster if you left them uncolored I bet
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
SamuPiano (pronunciation ambiguous) here - thank you so much for featuring my puzzle and I'm glad you enjoyed!
Great puzzle. Hope you will produce more soon
Ah I knew I recognised your name! I just did your six by six Islands puzzle a couple of days ago!
Love this ruleset. The 9x9s were a bit too hard for me but I think I made a respectable start on the first one all the same😊 Really looking forward to this video now.
Masterful!
Hi friend, does Island Cell count towards X totals? (I know they don’t count for killer totals)
@@wuorson5111yes! The rules only state that islands do not count for CAGE totals, but nothing else. Therefore the X clue may include an island.
Glad Simon is colouring the sea blue, especially after the grey bushes earlier this week.
Has happened before where the sea was orange and the islands blue.
@@provence8917 Good Lord!
@@provence8917that one was a funny one 😂😂😂
Not all of us have a green thumb! ;P
The sea is a wall therefore it should be Grey.
Rules: 09:30
Let's Get Cracking: 11:41
Simon's time: 1h17m52s
Puzzle Solved: 1:29:33
What about this video's Top Tier Simarkisms?!
The Secret: 6x (13:22, 13:39, 13:53, 45:33, 53:47, 1:03:34)
Bobbins: 2x (46:14, 46:14)
Maverick: 2x (37:17, 37:17)
Cooking with Gas: 1x (41:06)
And how about this video's Simarkisms?!
Ah: 27x (03:13, 15:03, 21:56, 24:31, 24:31, 24:31, 25:43, 25:50, 25:58, 31:08, 31:08, 34:56, 42:51, 44:07, 49:31, 49:55, 51:35, 51:38, 59:23, 1:09:43, 1:10:06, 1:10:08, 1:14:58, 1:16:13, 1:19:41, 1:21:11, 1:25:16)
Hang On: 23x (03:09, 14:41, 15:45, 29:01, 51:38, 51:38, 51:38, 51:48, 51:48, 51:48, 1:08:07, 1:08:07, 1:09:18, 1:11:20, 1:11:30, 1:13:28, 1:13:39, 1:14:03, 1:14:30, 1:18:49, 1:22:56)
Sorry: 13x (01:57, 06:33, 21:42, 33:08, 33:48, 43:03, 57:53, 58:27, 1:00:01, 1:07:29, 1:10:58, 1:12:00, 1:13:12)
By Sudoku: 6x (1:09:22, 1:12:43, 1:22:48, 1:26:47, 1:27:16)
In Fact: 5x (31:53, 53:12, 55:04, 1:02:37, 1:20:31)
Obviously: 5x (04:42, 18:35, 21:53, 45:23, 1:26:25)
What a Puzzle: 3x (1:21:02, 1:29:24, 1:29:32)
Lovely: 3x (06:00, 06:11, 1:26:41)
Beautiful: 2x (1:26:30, 1:26:30)
Incredible: 2x (01:50, 01:55)
Deadly Pattern: 2x (1:14:18, 1:14:21)
Puzzling: 2x (01:39, 05:37)
We Can Do Better Than That: 2x (12:36, 1:17:48)
That's Huge: 2x (20:51, 20:52)
Cake!: 2x (05:52, 07:02)
Naked Single: 1x (1:25:58)
The Answer is: 1x (1:11:26)
Clever: 1x (1:29:43)
Stuck: 1x (32:38)
Brilliant: 1x (1:30:49)
Ridiculous: 1x (27:45)
Astonishing: 1x (04:04)
Come on Simon: 1x (1:21:00)
Shenanigans: 1x (23:50)
Surely: 1x (27:30)
Marries Up: 1x (49:35)
Wow: 1x (1:05:43)
Sting in the Tail: 1x (1:21:09)
Fabulous: 1x (03:52)
Most popular number(>9), digit and colour this video:
Ten (24 mentions)
One (160 mentions)
Blue (10 mentions)
Antithesis Battles:
High (3) - Low (2)
Even (5) - Odd (0)
Column (12) - Row (10)
FAQ:
Q1: You missed something!
A1: That could very well be the case! Human speech can be hard to understand for computers like me! Point out the ones that I missed and maybe I'll learn!
Q2: Can you do this for another channel?
A2: I've been thinking about that and wrote some code to make that possible. Let me know which channel you think would be a good fit!
I’m not very good at these difficult puzzles but I love following along. It’s also great for my self esteem when I’m able to see things that Simon (who I hold in rather high regard) does not , even though alas it is most often sudoku logic. But when it isn’t, I feel even smarter 😊
Travis said he forgot to tell you where we were - but we're in Australia so this actually came out on my birthday! Thank you for reading the shout out :)
On being part of the household - he doesn't watch you guys, but every now and then I forget to bring my headphones to his place and so I listen out loud. He can tell you guys apart by voice, and no longer turns around when I scream at the screen about where to look next in the puzzle - depending on what I'm shouting, sometimes he can tell you apart when I'm wearing headphones 😅
By the 21 cage, he may conclude the degree of freedom from the 170 cage is spent in either box 3 or box 6, so the cage contents in the remaining boxes must be maximized
That was the key for my solve as well.
I was wondering during most of the solve whether he would discover this. It would have saved him a lot of time.
The same here
@@blcmd me too
Yess, Though I did do the whole maths thing of finding 29, working on the 21 cage was the key break in. Though I love long CTC video this could have been a lot shorter.
After sum the cages(376) and realizing that the sum of the islands is 29, you have two degrees of freedom. However, if you use 9 ones, 8 twos, and 1 four (or 8 ones, 9 twos, and 1 three), it becomes impossible to place two islands in each row, column, and box. Therefore, you need 9 ones, 7 twos, and 2 threes. With this in box 4, the maximum number of cells you can remove is 4 (the 1 and 3), not 5 as Simon was considering.
Ah that's cute. Because if there were 8 x 2 Islands, since the Islands follow sudoku rules, the 9th non-1 Island would also go where the 9th 2 goes.
indeed
I got the islands sum to 29. Which means at most 11 in any three boxes. The 8 cage and the two 9 cages must have at least 11 in islands, if you minimise the digits and subtract 8+9+9. So we know this is the case.
We can place 123 in the 9 cage in box 1 of which two are islands. And 1234567 in the 8 and 9 cages in box 4 of which two are islands, and place 12 in the 9 cage in box 7 both of which are islands.
In every other box the islands are 1 and 2.
SamuPiano here - actually, an early version of the puzzle utilized this fact! That deduction I eventually deemed to be too obscure for the solve which would detract from the experience.
@@stephenmor66 Yeah, that was my way in as well.
I was thrilled with the length of the video and I was nearly not trying this puzzle. But turns out it only took me around 39 minutes to solve.
Edit: I think Simon was confused about the 29 for the island cells. Actually the easier way to continue at around 46:14 is to do Sudoku on box 2, and notice r1c45 can only be 345 by Sudoku, and they are both in 6-cell 36 cage.
Then by "the secret", the 3 missing cells have to sum up to 9, but if those 3 cells are either 234 or 135, r1c45 will be double 5 or double 4, so the only thing left is to miss out 126.
Then by Sudoku again, using the 67 pair in box 2, 6 in box 1 is on row 2, which forces 6 in box 3 on r1c9, remove the 6 in the quarduple in box 6, which remove the freedom in the 170 cage. From there, the puzzle just flowed like water!
But… wouldn’t that have required Simon to do Sudoku in a Sudoku puzzle? 😂
@@jensschmidt Yes, that's why it's hard for Simon 🤣
I also summed up boxes 5,6,7,8,9 plus the 89 pair in box 3 (242 total) compared to the total cages in those boxes (227 total) to realize that the islands in boxes 5-9 all needed to be 1/2 pairs.
Ah yes, that's clever! The "6-cell 36 cage" means the 6 cells from the 36 cage which still count after the island is removed. And the "3 missing cells" are the three digits not used in the 36 sum.
I'd basically just posted something similar. (Delete that.) Simon got really stuck on the last missing islands. But after being led from one cage to the next by the setter, I didn't spend any time on that and instead looked at the options for the now-restricted 32- and 36-cages. This led reasonably quickly to forcing a 6 into the 32-cage in box 1, which put a 6 in R1C9. This immediately meant the degree of freedom was used up in box 6, and the rest went relatively smoothly, as you say. I suspect this was the path the setter meant to be followed.
Simon, that was my longest scream to date. LOL. At around 0:24:00 you note that r3c9 and r6c8 had to be both an 8 or 9 and the same number. You had already noted the 1 degree of freedom in the cage. Almost 62 minutes later, you found it again. Now I have no idea if the handful of 8s and 9s that produced would have made the solve any quicker, but it couldn't have hurt.
Maybe tomorrow I'll try a little louder. 😃
Longest smile to date for me, because it made me say: "Well done, brain." :)
Having both be 8 would have only brought down the degree of freedom by 1. The 8 being in r3c9 brings it down by 1, but then that allows the cells in the box beneath to be 9876, with 8 in r6c8. So no, that wouldn't have been enough to conclude that those cells had to be 9.
@@fatman3762 *tips hat* When I'm wrong I'm wrong.
Man, I even rechecked against the 21 cage when the islands in box 6 were revealed but going back over them now, they work the other way also.
@@fatman3762 Hm... wasn't one degree of freedom all that was needed in the 170 cage?
The Moment when Simon recounts all the cage totals to make sure he didn't made a mistake and murmurs the sums, as suddenly he hears Maverick flying by and like a reaction from his spine he has to mention him. Its hilarious ^^
Gosh I just love to watch you solve puzzles
I'm not convinced removing the 3 from the islands in box 7 was earned around 1:08:00 (even tho it was right). All Simon had proved at that point was that the islands in boxes 1, 4 and 7 had to total 11, but not how they did so, so the island in box 7 still could have included a 3 at that point in the solve. The way I worked it out was by noticing that the totals of boxes 3, 5, 6, 7, 8 & 9 was 270 by Sudoku but 252 by Ocean with the 6 island pairs in those boxes accounting for the difference of 18 (1+2 = 3 , * 6), meaning the islands in box 7 had to be 1 & 2. Still, a great puzzle and as usual an entertaining solve by Simon. Love the channel.
I always need to watch Simon explaining the rules and fiddling for a bit with killer sudokus before I can try them myself, I'm really terrible at finding where to start.
After Simon did the min/max calculation for the 170 cage and found out there only was 1 degree of freedom, I stopped watching and started doing the puzzle alone.
I managed to finish it in about 40 minutes then. I didn't do it all by myself but I'm still proud for having finished it.
O frabjous day! Callooh! Callay! I've been following the channel for years now and this might be the first time I've ever truly stolen a march on Simon. 32:10 solve time for me. Started right away by adding up all of the box totals and realized the extreme limitations on the islands, then worked through the logic on the 8/9 cages. Never even looked at the 170 cage, it just fell out at the end. 1 W 999 L vs. Simon times, but I'm sure I'm starting a streak! (Of course I think this is mostly about Simon being mesmerized by the 170 cage up front, I'm guessing many others will have a similar experience.)
Surprised to see an ambiguous amphibian reference here haha
For someone who loves Star Battle 🌠 so much, Simon spent a lot of time _not_ looking at which rows, columns and boxes already had their two island cells 🏝 ... would have been a lot quicker if he had added up all the cages 🧮 at the start and found that the islands totalled 29 across 18 cells, giving very little freedom (which is quite appropriate ... if you are in a cage on a desert island, you have very little freedom!)
Yeah: Star Battle was clearly in the background here, though with no restrictions on putting stars/islands next to each other.
yes, even if it was towards the end with the 9 cage in box 1 I basically lost where Simon was heading when the solution was right there in the colouring.
My immediate response to this was to work out there were 29 islands (so at least 8 each of the 1s and the 2s, with the last two either being a 2 and a 3 or a 1 and a 4).
Still really early in the video, though. He's messing with the 170 box and I'm smiling :-)
Quite pleased with a 64 minute time on this.
Helped that I started by summing all the cages and getting my break-in in box 4 with the 8,9 cages needeing to use the degrees of freedom. I also seem to remember deducing early that there had to be 2 islands in the 8 cage, and it used one or two of the degrees of freedom.
(I never thought to look at the maximum for the 170 cage, as Simon did. Fortunately, I didn't seem to need to.)
This was me too! Getting the sum of 29 islands was incredibly powerful and immediately I ran into the trouble of the 8 cage in box 4 and trying to make the math work out. From that the 9 cage next door leaps out as extremely constrained and everything builds from there. I think that Simon stubbed his toe badly in looking at the 170 cage!
@@dralthoff1 Although, reading the comments from others, I see it was possible to use the 21 cage to pin down the one degree of freedom for the 170 cage to box 3 or 6, which would have helped Simon a lot. So I guess there were two ways to approach it. But yes, summing all the cages was my first thought (followed by, "do I have to" and "I guess I better" 🙂)
Box 4 definitely does the trick. Looking at the 8 cage and both of the 9 cages together solves it all. The rest is quite easy after that.
I did this when first published, and I'm surprised by the video length. I've only skimmed the video, but it seems Simon hasn't thought to apply basic arithmetic to the puzzle as a whole, and has also missed box 4 as the breakin. Which is a shame, because it was a really good breakin.
Add all the cage totals, apply the secret, the 18 islands must add to 29. Look at just box 4 + the two 9 cages that protrude from it. They must add to at least 45 + 6 + 3 = 54. The most r4c23 can be is 17, so the 8/9/9 cages must add to at least 37 when the islands are included. The most the islands can add to is if all the box 147 islands are in those cells, and all of the other boxes have 12 pairs as their islands, leaving 11 for the islands in box 147. 11 for the islands + 26 for ocean in the 8/9/9 cages = 37, so now that is forced. r7c23 is a 12 pair, and island. r3c123 is a 123 triple, with 2 islands. r4c23 is an 89 pair & ocean, and r3c2 is 1 or 2. So now r56c3 are ocean, r3c3 is island, r4c1 is ocean, and the 8 cage has 2 islands. A quick look at the possibilities for the r3 islands: 23 means 8 in r4c1, that's impossible. 12 means ocean 3 in r3c1, 6 in r4c1, so now the 9 ocean in c3 is 45, the ocean in the 8 cage must be 17, the islands must be 23, but that means 3 islands in c2 and only 1 in c1, so that's broken. So the islands in r3 are 13, r4c1 is 7, the islands in box 4 are also 13, the 8 ocean must be 26, leaving 45 for r56c3.
And then the rest of the puzzle is very easy, and flows really well. Easy because every cage now reaches its total without using 1 or 2, so islands get quickly forced. e.g. if the 32 cage only has 1 island, its minimum sum is 345689 = 35, so it has 2 islands, and only 2 spots for them. 1289 are already in the cage, 3 can't be in it, the remaining 3 cells add to 15, so must be 456, and so on.
edit: Simon does half the breakin at 1:05:00, and then the other half at 1:16:30. If he started there, it would have been a 30 minute video. ;)
This Sudoku Puzzle has more "Nun Counting" than a Catholic Census.
It took me 72 minutes and was one of the most enjoyable puzzles I've done in weeks. Absolutely loved it, thanks for sharing it.
At 1:07:58, was Simon's removal of the 3 from box 7 premature? He says that the highlighted cells (which includes those highlighted in box 7) need to include both degrees of freedom, but then discounts the silent digits in box 7?
The highlight cells do not include the orange cells in box 7 because the orange cells in box 7 do not contribute the sum(45) to box 4
Box 4 (45)+3 cells in box 1(6)= minimum 51
9+17+8+9+island cells(maximum 8)=51
Minimum=maximum tells us everything must fixed, so we already got 2 more degree of freedom of island cells in those highlighted cells
53:31 This was a fun puzzle, finding where to trim the fat off the 170 cage was quite a treat, and I definitely liked the surprise near the end
You did it the hard way by not seeing the 170 single degree of freedom was used up in box 3 or 6. If you put an 8 in r3c9, then you used it up, and if you put a nine in it, then you needed to use the six in the 21 cage, and used it up. Then you would have had most of the 12 islands and the maximum values in the rest of the blue 170 cages, which would have made things a lot easier.
This
Wow, that does make it much easier. Nice one!
I do wish Simon would apply his revelations of forgotten rules (in this case 2 islands per r,c,b) to the rest of the grid before fumbling about for 5 minutes trying to use it for the one occasion he's noticed.
My break-in. Cage totals = 376... subtract from 405 gives the number of "silent" cells = 29 within 9 boxes. If all silent cells were min (1+2), total silent would be 27. Gives only two degrees of freedom across ALL silent cells. Gives either 2 boxes with silent 1+3... OR... 1 box with silent 1+4 or 2+3. All other boxes have silent 1+2. Looked immediately at Box 4.
Same approach here.
I really enjoyed solving this one. The logic is fascinating and unique. Watching Simon was a joy as well. Thank you.
49:10 ... I stunned myself on this one, finishing in under an hour. Thank you, math!
Incredible puzzle!
spoiler for 1:05:00 my break past that point was a bit different
I focused on the 32 cage and the 36 cage, and after going through all the ways to get 32 in 5 digits and 36 in 6, I concluded that:
1. 6 was forced to be in the 32 cage
2: because 8 and 9 were also forced into the lower 3rd of the 32 cage by their inclusion in the 36 cage 6 had to be forced in the center domino of box 1 and the corner of box 3!
this ate my only degree of freedom in the 170 cage because i could no longer use 6 in the box 6 portion of it.
I kinda wish i thought to calculate the 18 island maximum, although I did have 16 of the islands placed at this point and (without starting over to prove it) I don't think that would've helped me
Really beautiful puzzle by SamuPiano! And a somewhat confusing, but still very clever solve by Simon!
I'm a little confused myself about my own break in, after watching SImon's solve, and wonder if I just got lucky:
I started immediately with the global sum of all cages, figured that I needed all 1's and most 2's on the islands, which pointed me to the 8-cage, as I couldn't figure any way to have less than 2 islands in it. This then kept pushing islands around the grid, before I even started looking at the 170 cage.
What I failed to appreciate was Simon's theory of one island with a 4, or one 1 going in the ocean. I just couldn't make it work as I looked at the 8-cage, and claimed the islands had to be 9 1's, 7 2's and 2 3's, but in hindsight it makes me wonder...
It occurred to me to note the secret of a sudoku board, 405, and then sum the cages to 376, which leaves 29 on islands which have an average for rows, columns, or boxes of 3 remainder 2.
At 24:25, Simon remarks both those squares are the same, but they can't both be 8, as there is only 1 degree of freedom in the 170 cage. Simon only resolves this one hour later at 1:25:35!
Edit: I was wrong about this. @BOTHLine explains it well below.
I thought the same for quite some time.. but actually if it was an 8 in the 3rd box, it would just move the 9 in the 6th box to the right column.. which wouldn't use up the other degree of freedom. Use either had 8 + 9 in box 6 anyway, no matter if the other digit was a 9 or an 8.
I usually don’t attempt puzzles where the video is over an hour, but I love killer cages so gave it ago. Finished in 51 min. Feel like I found the intended path by doing the math and realizing I only get two degree of freedom on island values, then focusing on the interplay of the 8 and 9 cage in box 4. Basically moved around the grid after that as each cage was filled in.
[Mild SPOILERS]
If only Simon had spent a bit more time on the 21 and 170 cages, this would have gone a lot smoother. Though I admire his patience and resilience in the face of daunting puzzles because due to some quick faulty logic in box 1, I had to redo all the logic from scratch and the mistake I made was - I just assumed that 89 pair goes in r1c1 and r2c1 as it was seeing the 89 pair in r1c6 and r2c6. (1:21:55 see simon tackle it correctly)
So solvers beware this ocean island sudoku is not that easy so tread carefully and don't pencil mark it to the brim. Keep on cracking.
I've never seen a more different solve path from my own and Simon's. I started on the fact that was discovered about 50 minutes in that there were 29 worth of non-counting cells in the grid. Then combined with the small cages in box 4 and box 1 adding up to 26 but having a minimum sum of 34, which means they need to use 8 worth of non-counting cells and given that it's in two of the boxes, that uses up the two degrees of freedom and proves all the other non-counting digits not in boxes 1 and 2 are 1/2. And then I basically completely ignored the 170 cage since I thought it wasn't useful at that point. So my grid started filling in in box 4 and went around the grid, but Simon started in box 6 and went the other way. Although, I think I would've solved it much faster by at least considering the 170 cage because it seemed pretty profitable to know that the digits in its cage were near-maximal. Kind of amusingly I thought that the 1/2 non-counting digits in the 30 cage box 3 would force the 6/7/8/9 in box 2, but ended up proving that the box 2 was 6/7/8/9, thereby forcing the 1/2 non-counting. And Simon proves it the way I thought it was supposed to be really early in the solve!
There is a subtlety of cage placement that I really enjoyed.
Wow! What an amazing puzzle. I worked out most of the 170 cage straight away but made the fatal error of assuming that R3C8 was a 9! Of course that didn’t then work and took me over an hour to work out why! Finally completed it in 2:36:32!
Funny, I did exactly what you did and made the potential error of assuming R3C9 was a 9. I got lucky without realizing it until after the solve.
When he says I like comments especially when they are kind makes me sad cause it implies sometimes there's people putting unkind comments on these videos 😭
I hope they are very rare. I love these videos even though I don't often watch them, keep up the good work. :)
Wow, what a puzzle indeed! (That 6 in the corner was absolutely mighty)
I took a look at this puzzle and knew that I wouldn’t be able to solve, and after watch it I know that it’s because I’ve never been any good at counting nuns.
I can't even count them as far as 1 ... I just get stuck on "nun" 🤣
I love watching the brilliancy of the solves, while also slowly dying inside because of the x clue haveing to be made of two counting cells. You got there, but seeing that you need two numbers to make 10 would have been a big help and saved about 20 min i believe ;)
At 38:00, Simon makes a small logic error. The 32 (or 31) totals in boxes 5 and 8 only force a 9 into them. They could be 97654 (where the loss of 1 degree of freedom occurs), with the 8 joining the 2 in the 10-cage. It doesn't end up mattering, of course, because the degree of freedom is lost elsewhere.
Ninety-minute sudoku variant videos with Simon light up my life!
Love that you explain your logic. So fun to watch thank you!
I wasn't going to watch this but there was a pile of washing up to do so I set my iPad up above the sink and watched as I washed. Glad I did. What a puzzle, and so brilliantly solve.
Just under 50 min for me - glad I tried it as it's a lovely puzzle with the interaction between the cages in boxes 3 and 6 taking away the last degree of freedom from the 170 cage.
I'm just intelligent enough to follow Simon's logic, and just dumb enough to not be able to discover it on my own.
Nice puzzle and very good video. Excuse my English, but I will explain part of my solving path.
I started by analyzing boxes 5,6,7,8 and 9 (total sum: 5x45=225); together with cages 10,16,10,21 and 170 (oceanic sum (os): 10+16+10+21+170=227). So: the oceanic sum of boxes 5,6,7,8 and 9 is: OS(box5to9)=227- os(r3c8;r3c9), but also: OS(box5to9)=225-(islands U [r7c2;r7c3 ;r4c4;r4c7])
So: os(r3c8;r3c9)=2+(islands U [r7c2;r7c3;r4c4;r4c7])
The left part is max. 9+8=17. The right part is minimal when [r7c2;r7c3;r4c4;r4c7] is included in the set of islands and they have minimum values: 2+(1+2)x5=17. BINGO. Equality will hold only under those conditions. This not only immediately concludes that r7c2;r7c3;r4c5;r4c7 are islands, but that each of the islands in boxes 5,6,7,8, and 9 contains only the digit 1 or 2.
I think I did most of the logic Simon found near the end at the very beginning - after summing everything we notice that there's 2 degrees of freedom on islands, so neither 8 nor 9 can be 3 cells large! 125 means the lowest islands are 3/4, 134 2/5, 234 1/5, 126 3/4 and 135 2/4, all of which are bigger than the two degrees above 3 we're allowed.
In that case - box 4 has two islands in 8 (the rest is ocean), box 7 has two islands in the 9 cage (the rest is ocean), X has to be ocean as one cell can't be 10, meaning that islands in box 1 are on position 7 and 9.
That was my start to this puzzle, apart from noting that all islands had to be from 1/2/3/4 and were extremely constrained.
at 34:57 you put an 89 in the cage that sums up to 21.
This removes the option of the 89 next to it (the one in the top of the 170 cell) to be an 8 (it would take at least 2 degrees of freedom, we only have 1). I believe the puzzle was supposed to be started from there.
Putting an 8 in r3c9 only removes one degree of freedom from the 170 cage, not 2, because it doesn't restrict what goes into the cage in box 6 ... it can still be maximum of 6-7-8-9 totalling 30, whether the caged cell in box 3 is an 8 or a 9.
@@stevieinselby No, because you must have 789 in the cage within box 6, so as soon as you put an 8 in r3c9 you would have to put 8 into r6c8, and you have now used 2 degrees of separation. Ergo they must both be 9s.
1:07:59 I do not understand why there has to be a 12 pair in box 7 and it can not be a 13 pair. Is the degree of freedom already used up somewhere else in the 170 cage at this point? Could someone explain?:)
From 1:03:35, Simon is using the minimum value for the cells highlighted as being 45 (box 4) + (1+2) (box 7) + (1+2+3) (box 1) = 54. The only way 54 works is with an 8+9 pair in r4c2-3 and using the two degrees of freedom available in non-counting islands (8+9+9 (cages) + 17 + 11 = 54). None of those can be increased. So 54 is also the maximum for the set of cells. Since maximum (54) equals the minimum (54), they're forced to be what he first hypothesised to reach his minimum total, i.e. a 123 triple in r3c1-3, and a 12 pair in r7c2-3.
I had to pause the video to understand that point myself. It wasn't immediately obvious to me.
The highlight cells do not include the orange cells in box 7 because the orange cells in box 7 do not contribute the sum(45) to box 4
Box 4 (45)+3 cells in box 1(6)= minimum 51
9+17+8+9+island cells(maximum 8)=51
Minimum=maximum tells us everything must fixed, so we already got 2 more degree of freedom of island cells in those highlighted cells
@@khoozu7802 ?
Look at timestamp 1:03:30. Simon has selected a region containing 14 cells. Those ARE the highlighted cells, the ones Simon is talking about, and it very much includes the orange cells in box 7. 🙂
They're part of the 9 cage, needed in the calculation of the maximum of 54, using two available degrees of freedom in uncounted cells, for the 14 cell region. Minimum for uncounted cells in three boxes is 3*(1+2)=9. So maximum uncounted cells, using two degrees of freedom, is 9+2=11. The maximum in r4c2-3 is 8+9=17. Cages sum to 8+9+9=26. So the maximum overall is 11+17+26=54. This is the same as the minimum, when the orange cells in box 7 are 1,2 and the three cells in box 1 are 1,2,3.
I'm just explaining Simon's logic used in his solve. Maybe there's another way to do it, but Simon's way works.
@@khoozu7802 (Isn't one of your 9s in "9+17+8+9+island cells" from the 9 cage that includes the orange cells in box 7?)
Edit: Nevermind. Orange cells are already known to be uncounted cells. That's why you can treat the 9 cage as if it were only in box 4. Your way works too.
"Not all X's are given..." sounds like something a divorce lawyer would say.
I finished in 171 minutes. This was a tough one, but fun to work through. I was able to get the break-in, but I made a slight error in which ones were maxed and which ones had play, and forced the one in box 5 to be minimum. I forgot that the digits could shift to the other ones. That was a silly mistake, but I was happy with figuring out the break-in, otherwise. Once I started doing the puzzle properly, it was really fun to see the strange logic that forms in a grid like this. It was quite strange and worrying to keep getting 12's as the islands, only for me to accept them and be bamboozled at the end when they were not present together in boxes 1 and 4. I enjoyed this one, despite what my time may say. Great Puzzle!
Very enjoyable solve! I have taken to solving alongside Simon while he solves these days while trying to sneak ahead.
Early on, you can use the other cages (aside from 170) to prove that there are no degrees of freedom for the 170 cage. In box 6, in order to satisfy the 21 cage it must take at least a counting 6, even with an 8 or 9 in r3c8, so the 170 cage must be 5789. From there, the logic cascades to the other cages as well.
It quickens the solve just a bit, but mainly just at the end, since you can maximize all cells in the 170 cage. For example, you know 89 needs to go in r12c6, which forces 89 in the 36 cage to box 1, forcing r4c23 to be 89, accelerating the logic there.
I simply love Simon making super hard deductions and forget it's a sudoku puzzle. There was way easier breakthrough from box 2 if you simply did bit of sudoku and puzzle was flowing quite nicely. I actually did the part which Simon used for breakthrough last. 62mins for me, could have been even faster but I forgot some maths I did earlier in the puzzle and when I used the 1 degree of freedom in a 170 cage it took me like 20 minutes to realise I had already used it.
And very fair to say, that I would never have been able to do the deduction in box 1,4 and 7 as Simon did around 1h06m into orbit................. Chapeau!
1:17:09 if the 1 and 2 in box 1 are both islands, that means the 1 and 2 in box 4 both become islands. That is impossible. So, the 3 in box 1 is an island.
37:50 I don’t think I understand why we have to have 8 and 9 in c5 box 5. I think the orange cells in box5 have to be 13 or 12 pair (because of 1 degree of freedom) but in case they are 13 pair doesn’t it mean we can still have a 82 in the 10 cage?
Wow, that was worth staying up for! Great solve!
24:07 for me. What an awesome puzzle, loved it!!
80 minutes for me. I spent over half an hour for the break-in which had to be the 170, of course, I was too slow. Let me explain: sum(cages)=376 which is 405-29, so the island cells have 2 degrees of freedom (27+2=29), so the maximum value for an island cell is 4. The 170 cage has a maximum ocean cells of 9+30+42+32+32+26=171 so there is one degree of freedom which disappears by combining it with the 21 cage. After that, you can color a lot of ocean cells and the rest is easy. Thanks for the great puzzle despite there was no 3 in the corner!
The easiest way to think about the totals in boxes 5 and 8 are that the 10 cage plus two islands of 12 account for 13, therefore the maximum contribution to the 170 cage is 32 from each box. In box 7, the maximum is 45-16-3=26. It all adds up to an annoying 171. However, in box 6, if we use 6789 in the 170, we only have 15 to use on the rest, which, if it includes two islands of 12, there is just twelve left for the 21 cage, which would require a 9 in box 3 (using up the degree of freedom). This means the degree of freedom is entirely used up in boxes 3 and 6. This is the crucial deduction you missed, which would have saved you an awful lot of time. We know the degree of freedom is used up either by making R3C9=9, with the 170 cells in box 6 being 5789, or by R3C9 being 8, with the cells of the 170 cage in box 6 being 6789. This means all of the island cells in boxes containing part of the 170 cage must be 12s. We can place 12s in R4C4, R6C6, R4C7, R7C2/3, and we can put a blue 89 pair in R3C8/9. This forces the 25 cage in box 3 to be all counting 3-7, putting island 12s in the 30 cage, forcing it to be 6789.
@ 29:58 "Maybe I can prove this is an island" - that's easy. You can't afford to have two more islands in the 21 cage or you'd never make the total.
I found the 376 cage total was quite useful, because it stopped me running away with the idea that all islands were 12s. I was able to place most 12s, and whilst I knew most were islands, it was tempting to think they all were, which would have been a mistake. Knowing that not all 12s were islands alerted me to be careful. It was also useful, because it meant there couldn't be a non-counting 9 in the 8 or 9 cages.
@ 1:13:22 - considering where the islands are in box 1 - if the 3 were not an island, you'd have 12s as islands in both boxes 1 and 4, which would break your degrees of freedom. The 3 therefore has to be island, with a 1, so there must also be a 13 island pair in box 4.
Because I spotted the restriction with the 21 cage using up the degree of freedom, I found this considerably easier, however it was still no pushover. It was definitely a very nicely set puzzle, and using the 21 cage logic, it had quite a nice flow (e.g. the 89 pair in box 3 forcing the 25 cage, which in turn forced the 30 cage in box 2). What an auspicious debut puzzle.
At 48:07 you included maveric into your calculations which as a long time fan of the channel made me laugh
Nice debut Samu!
Thank you Blocksy! :)
The very first thing I did was sum the cages and compare it to the sum of the whole grid. This gives some constraints which I thought were important but which you did without for long time, about 45 minutes.
36:25 for me
Found out those degrees of freedom breakins earlier
Also added up everything in boxes 5-9 plus r3c8&9 to realize all of those were 1s and 2s
gotta love how Simons relies its global Suduku rules 44:35 into the puzzle in a sudoku named Global Ocean
Slight error at 38:00 cause the 8 could go in the 10 cage in box 5, if the islands are 1-3, cause in that case 8-2 or 4-6 takes the same total from the 170 cage
Edit: ok, he caught it later without using it for anything
Fantastic puzzle. Absolutely loved it 😊
I believe sudoku solving is typically very convergent, so it is interesting to me that Simon didn't get my first digit within his first ten.
4:56 PLEASE record a video of you solving the 100x100 crossword. Or do a live stream! It could be one of those eight-hour livestream marathons for charity! PLEASE.
this is a beautiful and sad rule set and i love it. thank you
I love how Simon only shares the secret of sudoku with his favorite people...every. Single. Video.
I'm pretty sure the 8 corner marks in box 5 at 38 minutes is unfounded. The ten cage being 82 rather than 46 would use up the 1 degree of freedom by forcing the islands to 13, but wouldn't break yet.
Amazing puzzle! I can’t believe the structure honestly
All those 1s and 2s on islands make me think of the song "One"
"One is the loneliest number that you'll ever do
Two can be as bad as one, it's the loneliest number since the number one"
7:13 Happy Birthday to Heather, Holly and Hope Newbury! Hope y’all are doing well. It’s such a small world to hear about people you went to high school with from someone across the pond. Tell Hope that sudoku variants are not only great fun but also help people our age keep our brains stimulated. Although, fortunately some of us get to wait a year for the big 5-0.😉🎉 Andy
Simon said it so much that I started to consider the logistics of counting nuns . . .
I didn't attempt this puzzle, but I did see a better way to start looking for islands at least. I just added up all the cages (376) and subtracted that from 45x9 (405) to get 29. So we know the 18 islands only add up to 29 which means they have to be mostly 1s and 2s.
So funny. For once in a life time, I came 45 minutes earlier than Simon to the observation that the sum of all cages + non counting cells = 405, hence ..... (non counting cells not adding up to 27 but to a bit higher sum). This will never happen to me again :-)
1:07:58 why do these need to be a 12 pair? Can't they be a 13 pair and take one of the degrees of freedom from the sum of the island cells?
The interaction of the 21 cage in boxes 356 and the portion of the 170 cage in boxes 36 actually takes away the degree of freedom in the 170 cage, meaning everywhere else the 170 cage must have 12 pairs in the islands (i.e. boxes 5789), but without doing this logic can Simon make the claim about box 7 when he does?
The highlight cells do not include the orange cells in box 7 because the orange cells in box 7 do not contribute the sum(45) to box 4
Box 4 (45)+3 cells in box 1(6)= minimum 51
9+17+8+9+island cells(maximum 8)=51
Minimum=maximum tells us everything must fixed, so we already got 2 more degree of freedom of island cells in those highlighted cells
Something about that puzzle made me screw it up multiple times in stupid ways and I had to redo stuff. I guess seeing the opener really quickly had me go too fast.
The 170 cage was the right place to look, you just overlooked how the 21 cage removes the degree of freedom and lets you max out the rest of the cage.
I somehow got this in half the time. For me the key was looking at combinations of the 170 cage with other cages that would help fill in the rest of a box or row. I finally computed that, taken together, the 170, 25, 21, and two 10 cages have a total of 236, while comprising 7-9 cells in each of 5 boxes (thus having a max of 42 each) and a max of 3+4+5+6+8 in the lower-left box -- which, as it turns out, works out to 26+5*42 = 236. Meaning there's absolutely no degrees of freedom for any of that, giving most of the ones and twos. And, as it turns out, you can also do this without the 25 cage, because once the 1 and 2 are ruled out, the 25 cage has to be 3-4-5-6-7. Then I got carried away with some maths and I think my first digit was actually the top-right cell!
46:20 if the sum of all the islands was 21, then all boxes in the 170 cage would be maximized and then the 170 cage would be 171... good the sum was 29... :)
100:09 for me. Marvelous puzzle!
59:02 X waves heavily ^v^
edit: Simon does the right thing .. as usual ;-)
Why?
(Island cells don't count towards cage totals. There's nothing in the rules that says they don't count towards X totals.)
you r right! my fault 🧐
I want to join the Discord Server but when I tried it told me to do some stuff I didn't quite understand. Could someone please tell me what to do?
Funny 45:50 that is the first thing I thought of. Not that I could complete it without help even knowing that.
Oh god, I solved this the other day and it never even occurred to me to do the global sum step. As soon as you said "Oh no..." I was like wait...CRAP
With the 1-2's in rows 1 and 2 the 36 cage has 6 cells that can only be chosen from 7 digits from 3 to 9. That means the 36 cage doesn't contain a 6 , and that places a 6 in row 1 column 9 getting rid of a 6 possibility in the 170 cage in box 6 . That removes your one degree of freedom for the 170 cage.
Even before the 12s were placed, the 21 cage used up the degree of freedom, either by needing 9 in R3C8, or by needing 6 in box 6. This meant that all the islands in boxes covered by the 170 cage were 12s.
@@Raven-Creations Ah, right you are. That probably would have halved Simon's solve time.
Took me 2.5 hours, one of the hardest I've done on this channel tbh
After getting the total of the island cells, I ignored the 170 cage, focused on box 4, which completely restricted the islands in the other columns and spiralled out from there. Much simpler arithmetic that could've halved Simon's time.
36:17
This might be the most that I've ever beat simon by
Haven't watched the video yet so I don't know what he does, but what helped me is to figure out early the maximum possible value for the 170 cage when taking into account islands and 10 cages and realizing that it was 171
Really let me fill in a lot of ocean tiles and let me mark down some numbers
Beautiful puzzle that gives you exactly what you need when you need it
One degree of freedom - c9r3 & c7r6 are the same... Well then you can't make both 8, that would use 2 degrees of freedom... Yelling for an hour :D Love it :)
I thought the same for quite some time.. but actually if it was an 8 in the 3rd box, it would just move the 9 in the 6th box to the right column.. which wouldn't use up the other degree of freedom. You had 8 + 9 in box 6 anyway, no matter if the other digit was a 9 or an 8.
That is not true. If both are 8 you can still put 6,7,8,9 in box 6 inside the 170 cage. Therefore making both an 8 only reduces the total by 1.
63:08 of pure joy!!
I just realized that the first deduction I made on this puzzle might have been very, very premature and only ended up being correct by accident :( . I marked both cells on the X as counting, but after re-reading the rules, I realize that it says the non-counting cells don't count toward cage totals, but doesn't mention them not counting toward the X...
Hope someone reads this and can answer; if Simon was able to put the 2 by the X clue in by simply having to put the 7 and the 2 as the maximum value for their cage, how come the X Clue is nececary at all? he didnt need the 10-sum at all to resolve it, how is it required in the puzzle then? i am sure i am missing something....
The sum of counted cells is 376, but the real total sum must be 405. That means that the 18 uncounted cells have to add up to 29. That works out to only a few combinations of numbers being uncounted: nine 1s, seven 2s, and two 3s; nine 1s, eight 2s, and one four; eight 1s, nine 2s, and one 3. I think that's all.
Very hard for me, but I cracked it!
[...and I didn't even realize that you are supposed to calculate the sum of island cells; somehow never thought about the bleedin' obvious...]
So Simon, when will we see you tackle the crossword?!! I expect it should only take you a couple videos!!
You could've much more easily proven R4C1/2/3 by noting that R4C1 couldn't be an 8 or a 9. A 9 would break the cage total since we know there's another counting cell, and an 8 would require a 1 to it which we already know is a non-counting cell. Thus only 6/7 were left as options, making the other two the 8/9 pair we're looking for.
Great video, I think the trouble you had in box 4 came from the fact that you made them green, you would have seen it faster if you left them uncolored I bet
@1:28:49 Enter the finalé!