Thanks so much for that feature! Rellik cages are such a wonderful invention by sujoyku. I played around with them back when sujoykus (second?) rellik puzzle took off but only set a few 6x6 puzzles with them back then. This was my attempt on a fairly linear introduction puzzle for that variant. Fantastic solve!
Have to say that I totally agree with you about the Rellik variant. It’s so satisfying the way the digits bounce backwards and forwards between cages that it actually makes me laugh. As for your puzzle, I enjoyed it on the whole but particularly the first half, which was a really lovely combination of Rellik and fog of war. Thanks for setting it. 🙂
This puzzle made me realize what the familiar feeling I was getting from Simon. It's like he's Bob Ross of sudoku puzzles, especially with the happy moods and the grid colors 😁
Hi Simon, I've been a lurker for a couple of months and this was the first puzzle i solved on my own without watching your video, ofc I still came back to watch but I'm very proud of myself. Solve time: 01:32:10
Way to go, @Winn3r__ ! Simon and Mark's lessons on here have helped me immensely with many of these puzzles, that I wouldn't have had a shot at solving a year ago. Hope you try more puzzles going forward
My argument for the 10 cage was that you can only 1 digit from each of the 4 pairs that add to 10 (1/9, 2/8, 3/7, 4/6). But the cage has 5 cells, so it needs 1 more digit and it can only be the 5. But then it actually needs 1 digit from each of those pairs, so the 5 knocking the 4 out means the 6 must be in, etc.
I thought about it in terms of the pairs we already know stuff about vs the ones we don't. We've got 2/8 split up already, and we already have one possible "escape square" for one of 4/6 in r4c3. That leaves 1/9 and 3/7, both of which need to have a digit outside the cage. The only two squares available for those digits are r6c23, so those two must be from 1379, and 6 is left with only one possible home in r6c1 - this then knocks the 4 out of the cage as well
@@mega9tales That's how I did it, 1 or 9 must be outside the 10 cage and the only place for them is r6c2 and r6c3 which gives you a triple at the bottom of the box with 6, 7 and then either 1 or 9 meaning that 3 must be in the 10 cage which knocks 7 out, once 7 is out the other outside cell (the top of the 7 cage) must now be 1 or 9 forcing the 6 into the cage and forcing the 4 out.
@@Qhartb Same here, getting me that 8 and 9 really early into the logic for box 9. Though it's clever to see that the same could be done elsewhere, and earlier!
@@MariaVlasiou Excellent! There were certainly a couple spots where I had to stop and focus on which digits were eligible, but it Was a fun puzzle! Congrats on making it through
That was a fantastic puzzle 😍 ... 24 minutes of joy! At 22:00, the way I articulated it was ... the red cage can't contain both 1 and 9, therefore one of them must join the 6 and 7 in r6 (outside the cage). That puts 3 _inside_ the cage in r5, and so 7 can't be in the cage with it, which places the 7, the 1/9 and the 6.
The way I got to it was the 10 cage can only contain one digit from 1-9 pair, one from 2-8, one from 3-7, one from 4-6 AND 5 to make 5 digits in the cage, so it definetly have the five from the 4-5 pair
The best way to describe what you're discussing at the 22:00 mark is by pencil marking those 2 cells. They can't be 245678, so the only options are 139. You can't include a 19 pair, so one of them must be a 3, and from there you can't put 7 in the box.
This. People seems to love overcomplicating it by thinking that the bottom row has to include a 1 or 9 so it must be a 1679 quad in the bottom three cells of box 4 therefore forcing a 3 into the box and eliminating 7, but its one of those situations where you dont need advanced logic of any kind. Just Pencil Mark the box like Mr Goodliffe, do the eliminations and bam, the rules take care of the rest.
How does Simon see (eventually) that 6 can’t go with 1 and 3 in the red cage, but doesn’t see that it can’t go with 4. Maybe it’s just me but I’d assume for most people it’s easier to see a 4 than the fact a combination of 1 and 3 also adds up to 4 😅
9:50 I don't know whether you get into this later in the video, but there is the pigeonhole principle. Obviously you can put 8 and 9 but not 7 in the 7 cage. Now we have the remaining six digits to contend with. I claim we can't put more than three of those into the cage. Consider the pairs (1, 6), (2, 5) and (3, 4). If you take four of the digits, you must necessarily complete at least one of the pairs, and that's forbidden, as they all sum to 7. In conclusion, it is impossible for a 7 cage to contain six digits.
They don't have a category for it, but searching for "fog" works well. Seems a bit stubborn that they haven't created a category for one of the most popular variants.
At 21:50 the simple way to see the logic around the 10 cage is that there are 4 pairs that add to 10, and you can only have 1 of each pair in the cage. Therefore the 5th digit in the 10 cage is the 5 as it has no pair
A brilliant puzzle. It was a joy to watch SImon get so excited during the solve! It was a delight to partake in this enjoyment. Simon, please clean up your pencil marks more often please.
The tactic I used a lot on this one was to think of what pairs needed to be "broken up" by the Rellik cages. Once you think about that, those "pairs" with one being in and one being out - typically took up all the spare room either inside or outside the cage - and any digits left over that didn't have to be broken up, would fill up the empty spaces. That got me through a couple tricky cages.
The beginning of this puzzle is beautiful! Feels like playing a 1-1 level in Mario games: it introduces the mechanics of the game one by one in an intuitive way. Lovely puzzle!!
I love these extremely simple rules which look like nothing for a few seconds and feel miraculous once you get to work on them, well done, very enjoyable ❤❤❤❤
21:50 I think the intended way for box 4 is: there are 2 available cells which aren’t in the 10-cage, and still need to place 1,3,4,5,6,7,9. 1/9 must be separated, and 3/7 must be separated, so you need to put one from each pair in r6c2/3. Now, 6 must be in the 10-cage, which resolves the 4/5 pair
an easy way to see fast how many possible digits a cage can be is to look how many 2-combinations + itself has. like a 7 has 3 possible combinations (1-6, 2-5, 3-4) and it self. so one of each of those combinations + 7 itself = 4 digits that can not be in the cage = max cage size for 7 is 5. Your Welcome
31:50 And now you should make use of the fact you know a lot about the 6 cage in box 8. How many ways are there of making 6 in any number of cells? [15] [24] [123] You can only pick one of the digits in those pair combinations and possibly 2 from the triple. I.E, 5, 2, 3 would work. What are the other two cells in that cage? They are bigger than 6, from [789]. Well, you can't have 9 in the current 5 cells you have. If the cage were to grow, then that would be a 9. So you have a [78] pair in the cage. And you know where the 7 will go. R8C5. 2, 5 and 8 doesn't work because you don't have a 1 in column 4 yet. There is a 1 in the not-a-6-cage so there is no 5. There is a 3, so you can't have a 2. The combination of digits you can have in column 4 in the cage is 1, 3 and 4. And R1C4 is the 6 in the column. You are left with a [58] pair in the column looking at R3C4, along with the 4 in the cage this makes R3C4 a 2.
24:50 Maybe the trick is to think of how many ways there are to make a cage sum in 2 cells. Like a not-7-cage that has 5 cells. How many ways are there of making 7 in two or less cells? 16 25 34 And 7 itself. So there are 3 pairs of digits and seven itself that cannot go in a 5 cell not-a-7-cage. So there are only 5 possible digit combinations that can go there. 1 and not 6 or vice versa 2 and not 5 3 and not 4 8 and 9 There are 5 cells, you have to fill them with some digits. One from each pair and 8 and 9. 8 and 9 will ALWAYS be in a 5 cell not-a-7-cage. And you can't make a not-a-7-cage bigger than 5 cells or you'd definitely end up with a pair that adds to 7 or the digit 7 itself. Sure, you also have to avoid triples adding to 7. But that doesn't restrict the size or the fact that 8 and 9 must be in there. When you have a not-a-10-cage it is restricted much the same BUT you have 4 pairs that can make 10 and one digit that is not in a pair. A 5 cell not-a-10-cage is always going to have a 5 in it. Or you'd end up with a pair of digits that add to 10. For your not-a-5-cage, there are only two pairs that could make 5. And there are 4 digits bigger than 5 so you could make a 6 cell cage with those restrictions. If it is 6 cells large, then you'd know you will have all the digits above 5 in it [6789] and one from [14] and one from [23] and no 5's of course. In any case, yes the not-7-cage starting at R6C2 must have an 8 and a 9 in it. And exactly... One of [16] One of [25] One of [34] And it doesn't matter which ones you use. high or low. Except you can't use 1, 2 and 4 specifically. And please fix R4C2 and R4C3 already. You know what they are!
→ Simon not fully blue-coloring the second 7-cage is really triggering me. → Just like me, after finding the 6 in the 10 cage in box 4, Simon took very long to resolving the 45 pair in the same box (or using the 3 + 6 for ruling out the 1). (It took me 134:28, spread across two days.)
This was enjoyable. The ability of the logic to constrain a cell covered by fog was brilliant. 62 minutes, but for about 15 minutes I forgot how to resolve white dots and stared trying to find some way to make progress.
I love watching you solve a puzzle you especially enjoy, and this was at the top of that category, Simon. You enjoyed it and it shows. I will give it a try soon. I was unsure of the Rellik rule (through not reading carefully - entirely my fault) so I watched the video to see how it works. But I get it now, so I will put this puzzle on my list. Thanks!
28:28 I got to this stage by myself, but then got stuck. The watched the video until 24:24, where you had said something about the 7-cage on top, that suddenly unlocked my brain, for the 7-cage bellow. Something about the different ways to not make 7 in 5 digits. Because I had the (1) , and also the 4-8 pair, where 4+1=2, the 2 had to be outside, and the 5 inside, and placable, then the 9 That revealed more fog, and I was back on track. Wonderful puzzle, loved it!❤
I’m not cross with you for the 10 cage with 631 because I was stuck on that exact point for literal hours. I’ve given up and walked away and come back a few times I was so blind to it so THANK YOU! For getting stuck were I did!
It took me 1.5 hours to complete and I was about to give up, but I did it! What a lovely puzzle, I'm stoked I managed to see it through until the end :D
I love when things like this happen. I ordered new brake discs for my car the other day. They arrived today, the same day this puzzle comes up. The model of the brake discs is DBA617 and I just finished placing the bottom row of box 4.
21:50 The other way of seeing that it can't be a 7 is through plain old sudoku. R5C1 and R5C2 consist of 1-3-9. If 3 isnt in either of those two cells, you end up with a 1-9 Pair and that breaks the puzzle. This means 3 must be in the cage on R5, and therefore 7 can not be in the cage because that also breaks the puzzle, which gives you the 7 in the grid.
22:59 A lot more in box 4 can be resolved here-the 10 cage cannot have 4 and 6, which unravels the 45 pair, and the ten cage also can't have 136 which means the 139 cells contain a 9, which places a 1 in r6c2
Hah the fact that he eventually saw the latter point but not the simpler former point (which he seemed ultimately to resolve using the 48 pair in column 2) is so Simon
Interesting. The way I thought about that is that you need a 5 in the cage to get to a size of 5 because you can only use one digit from 19, 28, 37, 46. This then forces the 4 out (by sudoku) and the 6 in, the 7 out (by sudoku) and the 3 in. Finally, spotting the 6+3+1 option finishes the cage. But box 4 is certainly one of the hardest parts so I'm not surprised it took Simon a bit to see that, even if it looks "easy" in hindsight.
22:11 There are four distinct pairs that add up to 10: 1-9, 2-8, 3-7 and 4-6. Thus, the red 10 cage in box 4 must contain a 5 and one digit from each oc those four pairs. That means that digit to the right of 8 is 5.
I was thinking of the 10 cage in B4 like this: one of the digits 1 or 9 have to be in R6 because if both were in the cage, it would sum to 10. So R6 C1,C2,C3 have a triple of 6 7 and either 1 or 9. So 3 is forced into the box, where 7 cannot live. Hence, R6C3 is a 7.
Sknaht, ylevol elzzup. The iterative element, as it were, was miraculous and I only got a bit stuck right at the end resolving the 7 cage. Drat I missed some obvious stuff in box 9.
21:49 - there is a different way. By goodliffing first two cells in row 5, we see that they have only 3 options - 1,3 and 9. Since 1 and 9 can't go together, we certainly have a 3 there. Now 3 eliminates 7 as a possibility in R6C1 and shifts it into R6C3 (can;t go in 7 cage). Now we have just 1 cell outside the red cage, it must be either 1 or 9 to avoid having both in red cage, that forces 6 into R6C1, forcing 4 out of red cage and also in combination with 3 resolves the 1 and 9. Box 4 is complete.
Here is how I explained the box 4 problem @ ~21m. There are only 4 cells outside the 10 cage in that box. We know one has a 2, another has a 4 or 5, and another has a 6 or 7. This leaves only 1 and we need to have a 1 or a 9 somewhere outside because they cant both be in the cage. That leaves no room for a 3 outside the cage so it must be inside, which forces the 7 outside and the 6 into r6c1.
Did this one a while back on LMD, ended up finishing in 30:03 (conflict checker off); a bit slow in some places where I missed obvious things, but still a wonderful puzzle! Many thanks to gdc for it!
I loved the logic around box 4, when you were at 21:51: Once you have 6 and 7 in row 6, row 5 columns 1 and 2 is limited to "139" which means 3 must be in it since you can't have both 1 and 9, then 7 gets placed outside the 10 cage and the opposite of the 19 pair needs to be in the 7 cage / outside the 10 cage as well giving the 6 in r6c1.
I find it so reliably endearing how Simon will always move on from some piece of logic he’s half finished to some new shiny thing just revealed. I was chanting “finish box 4” like a loon.
I finished in 68 minutes. Rellick rules are such a fun type to work through. Although, for some reason my brain defaulted to single and double digits for ruling them out. Anytime I got stuck, it was nearly a guarantee that I forgot a three digit possibility existed. I got stuck on one part that really threw me for a loop. I was so focused on trying to figure out the next step on why a 7 cage appeared in r7c7, that I completed missed a naked single on the complete opposite side of the board. That was crazy and for someone to purposefully construct that, must be incredible. Beyond that, it just took some time for me to spot things and I was able to finish. I really enjoyed this one, especially the break-in with how barren it looked. Great Puzzle!
15:06 The flow is delightful. 40:57 if another Constrained Cages puzzle comes out like this again, I want to try it before watching if it's just as difficult or easier. 41:35 I enjoyed it, BUT not as much as you, you really did. There are some great ones we've seen on this channel, that I'd love to revisit.
Lovely puzzle, lovely solve. It reminded me of a tennis match, with some very gentle baseline hitting to start, before things began to heat up a little bit. I found some beautiful logic to unravel box 9, but having watched the video I realise that adequately doing sudoku on box 8 would have been a more straightforward way to proceed. Many thanks to both Simon and setter.
This is such a beautiful puzzle! It’s so fun to see the logic play out and clear the fog. I’ve never encountered rellik cages, so it was extra interesting to me.
In that 10 cage, the two empty cells in the middle can only be selected from 1-3-9 as they see all other digits, and with the 1-9 being out, that forces a 3, removes the 7, forces a 6 and thus removes the 1.
My shortcut at 18:00->28:00. In box 4, if 9 is in row 6, then row 5 will be 3-1 and you cant put a 6,7 or 9 in the last red digit. So 9 has to be in row 5, 1 in row 6 (but not red), then 3 in row 5, so last red digit can only be 6, white digit is 7 and blue is 1.
Took a lot of trial and effort for me, not being used to rellik-cages. I'd love to see an extra little feature that would allow me to turn off the clearing of fog! I know that sounds ridiculous, but... when I was busy in the bottom-right box I didn't have space left in the grid for notes and dabbles and I was sure some combinations did not work, but would like to be able to double-check by filling in the numbers and see for myself where the puzzle gets in a bind. However, some combinations of faulty numbers do lead to certain other digits being limited to one specific box and turns out that they were the right digit for the box. Since I would like to dabble without spoilers, being able to turn off fog clearing for a bit would help.
71:04, super enjoyed the break in, then got stuck for like 20 minutes on the 7 cage in the lower left and ended up looking at the video and seeing that I didn't think about three cells adding to 10 in box 4.
Fun fact: this cage constraint obeys the excluded summand rule just as regular killer cages do: that if the "total" is s and the cage has a digit n and other digits, then the other digits obey the same constraint with respect to a "total" of s-n (with the sum of no digits being zero). The complement of killer cages (that all digits must not sum to the same as the given total) also obeys this rule, but the complement to the special rule here doesn't (that there is a subset of digits that sums to the given total. It's possible that the sum s exists for a subset of the cage excluding n while the sum s-n doesn't, for example a cage like 1:345 with a sum of 7)
solved in 42 mins! it’s been really exciting being able to solve more and more of the brilliant puzzles featured on the channel as i improve. this was one of the smoothest solve paths i’ve encountered and my first time with rellik cages. wasn’t as tricky as i feared and had a really good time
I loved this puzzle. It was my first time with relik and fog, and it was a great intro to both. I feel confident and excited to try other reliks especially. One trick i worked out is that a lot of the cages were so large that you needed to put almost every number you can in, especially where the total is below 9.
It wasn't the 631 in the ten cage that was annoying so much as the 4 you had in there with the 6. Never noticed it until you removed it by other means. Classic Simon.
My logic for the 10 cage at 22:00 was that 1 and 9 still need to go in the box, so one of them must be in row six with the 6 and 7, and the other must go in the cage in row five with a 3. 3 in the cage knocks 7 out of the csge.
