The Simplest Extra Sudoku Rule

追加: 2026-06-06

[00:00:11][music] Hello and welcome to Friday's second edition of Cracking the Cryptic. At least second edition, I hope. Um I'm hoping Mark has done a crossword video earlier. I don't know this because uh I'll let you in on a secret. I'm actually recording this um in advance because I have uh a prior engagement on Friday for once which means I can't record on the day. Um so I'm just going by the fact that I've asked Mark to do the crossword video on Friday morning.

[00:00:40]So hopefully you will have seen that if especially if you're interested of course in our cryptic crossword content.

[00:00:46]Um but what am I going to be trying to do in this puzzle in this puzzle in this video? Well, I'm going to be trying to do a puzzle. Um it's called Counting Killers. It's by Yus and we've had this recommended to us. I I I mean I have to say I love the rules to this. Um I did ave done a couple of puzzles this week with really short rule sets, but this this is another of those. And [snorts] it's just an incredibly simple uh rule, which is that if the digits in a cage sum to n, there are exactly n cages whose digits sum to n.

[00:01:22]So imagine those two summed to eight.

[00:01:26]There would be eight cages summing to eight. Isn't that weird? It's like counting circles but for for killer clues. Um and I've done a couple of yonas puzzles on the channel in the past. I think both of those were little rinky dinks on um on Killer Sudoku. Um and I remember enjoying I remember I think I did one back in around January 2026. So earlier this year. I do remember that one and and liking it very much. I'm hoping I will enjoy this one, too. Uh I don't know how difficult it is. I forgot to look it up on lo on logic masters. Um so I I can't tell you.

[00:02:02]You'll have to judge somewhat by the length of the video, assuming this becomes a video at all. At least you'll be able to see how difficult I found it.

[00:02:11]Um but we'll we'll read the rules properly and fully together in a moment or two's time. Um, let me uh let me just think about a couple of other things.

[00:02:20]I've mentioned the crossword video earlier on. Um, over on Patreon, we have a brand new competition which, um, it's called Alien Invasion, I think. Well, I know loads of you um, have been enjoying the puzzles there. They're a little bit easier than last month's. Um, and the feedback we've had so far is that they are uh, even more enjoyable, if possible, than last month's. So, um, do get involved in that. We themed it um to coincide with the fact there's been all these um uh declassified papers in the US about um alien activity on Earth. Um so we thought it was a an opposite uh subject ma matter. And uh if you want to win the chance to come on to the channel and solve a puzzle and appear in a video, then that is the way to do it. Um other than that, what else can I mention? Um Zeta Math. Um we've got our new Zetamath app out.

[00:03:14]is well worth checking out. That's available on all platforms. Uh, Puzzles by the Great Man. Um, and oh, birthday.

[00:03:21]I have got a birthday to do. Let me just find it now. Okay, I'm going to go for Sarah, but it could be Sara from the spelling. This is one of the things if you're going to ask for a shout out, if there is any way of me getting it wrong, I probably will get it wrong. Um, I think Sarah is most likely, but it it is spelled without an H, so Oh, I it could be either way. It really could. Um, I'm going to stay with Sarah. And if it is Sarah, then apologies. But Sarah, um, your boyfriend Noah wrote to me and said, um, look, I think I think the two of you stumbled across the channel and have stayed for the logic. Why wouldn't you? Um, so many happy returns. Um, I I actually I don't even know exactly when your birthday is this. I just know it is this week. Noah's email didn't actually give me the day. So, I'm sorry if I am a bit late with this. Um, but I know you've just finished your second year of law school. And I can tell you that Noah is so extremely proud of you. So, very well done on um on getting through that.

[00:04:25]I think the two of you have been together for seven years uh or will have will have been together for seven years soon. um in the coming weeks. So I hope on your seventh year anniversary there is some chocolate cake and I hope today there is some very very heavily iced chocolate cake as well. Many many happy returns. That's all the news. Now we can do some solving. Shall we have a go at Counting Killers by Yonas? Let's read the rules properly. And um although they're not going to take long to read, we've got normal Sudoku rules applying.

[00:04:56]So we're going to put the digits 1 to nine once each in every row, every column. and every 3x3 box digits. Oh, digits cannot repeat within a cage. So, a normal killer sedoku. In fact, very few of these cages could have a repeat in them anyway because of normal sodoku rules, but I guess that one could have done. Maybe that one. But anyway, we can't repeat digits anyway. It says we can't. Um, and as I said before, if the digits in a cage sum to n, there are exactly n cages whose digits sum to n.

[00:05:29]And that is the end of the rules. Do have a go. The way to play is to click the link under the video as usual. But now I get to play. Let's get cracking.

[00:05:41]Now, normally in a counting circles puzzle, what I would do is I would count the circles. So today, I suppose I've got to count the cages. Do I? Oh, look at that. You can double click.

[00:05:56]Yeah, I see. So if you double click on a cage, I think it highlights not all cages of the same shape, but all cages of the same size. So that so double clicking on this three three cell cage here also highlights other three cell cages. So let's just give these colors and then we can count them. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15. I think there are 15 purples.

[00:06:25]16 17 18 19 Oh, hang on. Hang on. 16 17 18 19 20 21 22. I might just double check that, you know. 6 7 8 9 10 11 12 13 14 15. I've got 15 for the purples twice. 1 2 3 4 5 6 7 for the greens.

[00:06:48]Okay, so we are at 22, I think.

[00:06:52]So, the cage totals.

[00:06:55]So, we've got sort of 22 to play with, which could be um thing is you could do that in loads of ways actually, couldn't you? Because the minimum size of a two cell cage would be three if it was a one two pair. So, you could have if you had a one two pair, you'd have three cages summing to three.

[00:07:19]But you could also have I mean a three cell cage. Oh well okay there's you can't add these to 24.

[00:07:28]In fact the three cell cages there are only seven of them.

[00:07:33]Can't add to more than the maximum for a two cell cage. The maximum for a two cell cage would be an uh an 8 9 pair which is 17. So imagine we tried to claim R. We could make this three cell cage add up to 18. Well, then there would need to be 18 cages adding up to 18. And there you couldn't make 18 in more than seven of these cages because the two cell cages could only add up to 17. So, [snorts] we know actually the highest total is 17 and the lowest total is three.

[00:08:12]And we have to somehow do this so that so that 22 cages are used. So I I'm the way my brain is going is something like we could have a 985 or something. So you could have it cuz 98 and five I know add up to 22 for some reason. But um that would mean there were nine cages adding up to nine. eight cages adding up to eight and five cages adding up to five.

[00:08:44]Or you could have 9 six7.

[00:08:47]That would also be true. Or you could have but you could perhaps have um now you've got to be a bit careful here.

[00:09:00]I was going to say 12 and 10. So you could have 12 cages adding up to 12 and 10 cages adding up to 10. But I think we are going to have to be careful because of this box and actually that box as well which have exactly the same sort of geometric property. They have a three cell cage and then three two cell cages.

[00:09:21]Um but I'm thinking the secret might be relevant there. The secret is something I share with only my very favorite people of course but if you're watching this video you're definitely one of my favorite people. So I will share the secret with you. The secret of sedoku is that any box of a sodoku because of the rules of sodoku contains the digits 1 to nine once each and if you sum the digits 1 to nine you get 45.

[00:09:45]So whatever combination of cage totals we we use in this puzzle they sort of have to be capable of summing in combination to 45 somehow. So 12 and 10 won't work because even if we had sort of three 12 cages, then we'd need uh we'd need nine more, wouldn't we? Three 12 cages would be 36.

[00:10:13]We wouldn't need 10.

[00:10:18]So in fact, this is interesting.

[00:10:24]No mocking me. It's all been done before. all all of the mocking, all of the nerd calling me a nerd and a geek and it's all been done. I am immune to it. Um the what I was going to say is that 9 985 or 967 won't work because because the biggest cage total would then be nine. And even if all four of those cages sum to nine, that would only get us to 36 where we know the box total is 45.

[00:10:59]So we definitely need Okay, well I can I can get rid of more from this. So 10 won't work as as the highest if 10 is the highest total available for a cage in this puzzle.

[00:11:16]Even if all four of these cages sum to 10, you'd only get to 40. That's not enough. We need to get to 45.

[00:11:23]11 won't work. That would get us to 44.

[00:11:26]So, the absolute minimum size of the highest cage total is 12. 12 could work if you had if you had three 12 cages and a nine.

[00:11:48]Yeah. See, that doesn't work, I don't think.

[00:11:52]But the reason that doesn't work is truly strange.

[00:11:59]And and it might not be true either.

[00:12:02]What the basic reason I was thinking 12 wouldn't work is if we even if we had three 12 cages and the the fourth cage summed to nine. That would seem to work and then we'd have 12 12 cages in the puzzle. Nine nine cages in the puzzle.

[00:12:20]But we're left with one cage cuz there are 22 cages altogether. 12 + 9 is 21.

[00:12:25]So there would be one cage that should add up to one, but you can't make a two-digit sum add up to one. So there couldn't be such a cage. So it wouldn't work. But I'm just pausing there because is it possible that in this box, for example, rather than having three 12 cages, could you only have two 12 cages?

[00:12:52]If you had two 12 cages, that's going to be awkward. [laughter] But that's not that's not the same thing as it being impossible.

[00:13:12]I don't think it would work because, [clears throat] and this isn't going to Oh, this is a bit handwavy. Uh let me just think about this. There might be a way of actually proving it demonstrabably.

[00:13:32]If you had two 12 cages, those would sum to 24, which means the other two cages, which could be these two, we don't know, but they would sum to those cages would have to sum to 21.

[00:13:52]in order for the um box to be full of digits that sum to 45.

[00:14:00]Now 21 is odd.

[00:14:05]So you couldn't do that 21 with just one cage. You'd have to do it with two different cages, an odd sum and an even sum that wasn't 12.

[00:14:22]But you'd have to do that in such a way that those two sums summed to no more than 10 because the 12 cage is going to the fact that there are two 12 cages means there are 12 12 cages. So I've only got 10 cages left to c to to cope with.

[00:14:44]And there's there's no way that can work if you had like a seven cage and a three cage for example that are meant to those those two cages are meant to sum to 20 21. It's just total nonsense. You can't get anywhere near in fact. Okay. So is that a general a generalizable point? Are we saying that whatever the the biggest cage is in this puzzle, there have to be three copies of it in these boxes?

[00:15:22]Because if there were only two copies of it, I'm not sure. I'm not sure I am going to say that.

[00:15:37]Oh, this is this is actually quite awkward, isn't it?

[00:15:43]If you only had two copies of it.

[00:15:48]So, whatever the biggest cage is, let's say it was 16.

[00:15:54]So if you had two 16 cages, you 16 is difficult as well because 16 you couldn't put 16 into two two cell cages in the same box because the only way of making 16 in a domino is with a 79 pair. So this would have to be a 16 and one of these would have to be a 16 and they would sum to 32 which means the other two cages would have to sum to 13.

[00:16:22]Yeah. And you couldn't do that because you'd only have six cages spare.

[00:16:32]Yeah, there's there's no way you can do it. It's It's actually This is really It's very constrained actually. I'm not sure 15 will work either. So if we had 15 and we had two 15 cages in here, again, they're always the whatever the if we're saying that there isn't a triple.

[00:16:53]I I if we're saying that the the the biggest cage total used in this box is not used three times.

[00:17:09]Let me just think this through. It's clearly not poss. So, I'm not I'm not articulating this very well. I know I am not. I'm I'm trying to get my head around it. A B C D. So, let's let's start let's do this diligently. Can all four cages be different? No, because A + B + C plus D would equal 45 and that would mean there would be 45 cages in the puzzle. There are not. There are only 22. So, that's total. and utter rubbish. Um, now what I'm trying to prove at the moment is that if say we had two cages the same and these are the biggest value cages, it won't work.

[00:18:00]And we can see that double A will always give us an even total for those digits used because it's going to be two lots of A. So even if a itself is odd, two lots of a will always be even. Which means that these b + c will always sum to an odd number.

[00:18:19]And therefore one of these is odd and one of these is even.

[00:18:24]But if if a was say 15, then we've only got seven cages free cuz we've got 22 cages altogether. 15 are used for the 15 cages. So these can't add up to more than seven and yet have to add up to 15.

[00:18:44]Yeah. Okay. And yes. So it it would get closer if you could do double seven, but you can never do double seven. You could never do that, for example, because this would 2 B + 2 A would it would would not be odd.

[00:19:02]It would always be an even total. So, this would get you very close if you did um 15 and seven. Say, that would get you very close because you'd have two lots of 15 for 30, two lots of seven for 14, and that 44 is close to 45, but it's not exactly right.

[00:19:21]So, you can't you can't use double. It just won't work, I don't think. Double 1428 is the worst problem. These have to add up to 17. Um, so I now believe it is true to say that in some combination there [clears throat] are and it might be that these three all add up to a. It might be those three. We don't know which three but the three there are three instances of some total in box 9 and box uh seven for that matter.

[00:20:00]Now, now okay. So, having concluded that, what is A?

[00:20:12]A is not 15 because then it just won't work. The final cage will be zero. So, A is 14 at most. And if A was 14, 3 14s are 42. This would be a B would be a three cage.

[00:20:30]Um, and well, that's interesting because that would mean there could be an a different cage again. So you could have a 14 cage, a three cage, that adds up to 17. So there would also be a five cage as well.

[00:20:44]So the three cages used would be 14, three, and five. Now, but is there another way of making five? 1 and four? No, you can't. We can't use one, can we? So, this five is indivisible. So, we couldn't make the five a one cage and a four cage because one cages don't exist in the puzzle. So, it would be 1435.

[00:21:10]Now, would 13 work? If we had 13, it would be 39 and this would be a six cage.

[00:21:22]And but we need overall 22. So then we'd we'd have another three cage. If so if 13 is the biggest value, we'd need a three cage. And if 14 is the biggest value, we'd need a three cage.

[00:21:39]That's weird. [snorts] Okay. And if 12 was the biggest value, then we'd have 36 here. This would be a nine. This is the one I don't don't think works. This is what got me thinking about this in the first place because then the final cage would be a one and we can't have a one cage and we know 11 is too small. So we're either in a world where we've got 14 as the big digit or 13 is the big digit but not the big digit but the big cage total. And [clears throat] there are definitely in this puzzle three cages that sum to three and therefore therefore they they are two cell cages cuz you can't make a three cell cage sum to three. and they are a one two pair.

[00:22:23]It's going to be something to do with this pointing at this, isn't it? There are going to have to be two ways of making this. I don't know how to do this. Um, so if it's 14, if it's 14, you are using a three total in in in the box in one of the two cell cages if it's so right. So let's look at this one because I've noticed something else about this.

[00:23:04]If we're looking at this one, we know that that this cage or this box, this 3x3 box and this 3x3 box are made up of three 14 cages and one three cage.

[00:23:17]Well, the three cage is always going to be a a dominoed cage. So, one of these will be a vertical one two pair, which means neither of these could be a horizontal one two pair. Because if this was say horizontal one two, which one of these is vertical one two?

[00:23:33]Neither could none of them could be. So this would be a one two pair and these would be those would be 14.

[00:23:44]So those would be 5 9 and 68 in some order.

[00:23:49]That's it. That's that doesn't work.

[00:23:52]That's this is brilliant. It's absolutely brilliant. Right. So this doesn't work for this reason. The it is it is what I said about these these strings um that string and that string because we have to hypothecate one of the two cell cages to be a one two pair.

[00:24:15]The other two two cell cages in these boxes are going to sum to 14 which forces their composition. There's only two ways to make Dominoes sum to 14 in Sudoku. 59 and 68.

[00:24:30]Now, what that means is that the three cell sums are going to be 347. And this is going to be 347. And that cannot work because this cell is not allowed to be 347 if those three are 347.

[00:24:44]So, we can get rid of this one. That's not correct. This is just this is brilliant, isn't it? Um, now that does imply though that this has a solution, which if it doesn't, I've made an error. Um, it's better immediately in that this is this three cell cage is not appearing in here cuz we're going to have three 13 cages and a six.

[00:25:17]Now, I don't know whether the six cage is going to be in a two. Could it be in the three cell cage? Let's think about that for a moment. Okay. No, I can immediately see it can. That's That's all. Okay. Well, that suggests it will work because the fact that you can Oh, no.

[00:25:37]I'm not sure actually. Maybe if they both have to be one, two, three cages, that won't work. That would be terrible.

[00:25:44]I mean, I can see one, two, three would work. with this being the six cage because these could be 4 9 58 and 67 in some order.

[00:25:56]So the only way So we need Okay, so what we need to do is to find a way where if the if is to have um have this not be the six cage. So one of these needs to be a six cage so that we can change the composition of that string. So let's think about that. If that was 2, four, could you do it?

[00:26:21]Then you couldn't use nine in these because these would be adding to 13 and nine would require a four. So 9 would move over here. So this would be 139.

[00:26:35]Yeah. Okay, that's that's good cuz that does work cuz then you could have a 58 and a 67 into these and it works. I wonder if um 15 would work as well. If we had 15, then we can't use eight in here. So, this would be 8 2 3 cuz it couldn't be 814.

[00:26:57]Yeah. And that works as well. So, you'd have 4 9 and 67.

[00:27:04]That's okay. That's weird. I mean, this is actually quite it's quite amazing.

[00:27:10]But what we have just learned is I believe that [clears throat] we have just learned let's try and summarize it in a pathy manner that the 22 cages in this puzzle are 13 13 cages six six cages and three three cages. But in these two boxes, we're not using three cages.

[00:27:44]The three the three sum cages. We're using three in each case. We're using three 13s and one six.

[00:27:57]And that's going to allow these to be different so that we won't have a problem.

[00:28:10]Okay. And what that's going to give me is is is a pencil mark because in both the variants we just in all of the variants we just looked at for how these boxes could work, there was always a two cell 67 cage.

[00:28:28]And therefore that is a two cell 67 cage because if I put the two cell 67 cage in either of those cells it would clash with what must be a two cell 67 cage in box 7. This yus is this is really quite amazing.

[00:28:47]So that is a 67.

[00:28:55]Now what does that mean?

[00:29:00]Oh no.

[00:29:04]I don't know actually.

[00:29:12]I'm really not sure.

[00:29:18]Can I use that somehow on this box? Do they have to be um two three four five six 15? Yeah.

[00:29:27]Okay. Those this cage and this cage are also 13 cages. Those ones there.

[00:29:36]And the reason I know that is you might say, "Oh, no, no. They could be a 1 2 3 triple." Well, but the problem with that is if this was a 1 2 3 triple, what would this one be? It would have to be a minimum of 4 + 5 + 6 which is 15 which is too high. The highest cage is a 13 cage.

[00:29:55]So those are both 13 cages which is 26 by the secret. Therefore those add up to 19.

[00:30:07]So if that was a no that wouldn't work. So this is also um a 13 cage. Let's let's let's make Okay, maybe we change the coloring.

[00:30:21]Yeah. Okay, I'm going to change the coloring.

[00:30:24]Sorry about this. [clears throat] I'm going to say a yellow cage is a definite 13 cage. And the reason that this is definitely a 13 cage is that those three sum to 19 by mathematics. So imagine this was a six cage. The maximum value of those would be 2 + 3 which is five which means this would be a 14. I can't put 14 in a single cell. So nine. So those are 19 and those are 13.

[00:30:56]So the difference between these two is six with uh so if 13 if that was a one. So yeah with this being bigger. So this is a 7, an 8 or a 9. And this is a 1 2 or a three. If this is a 1, this is a seven. If this is a two, this is an eight. Let's explain that. So imagine this was a two. These would add to 11. So this would need to be an eight to make the maths work. So these are always six apart with this being bigger.

[00:31:33]Wow.

[00:31:38]Okay, that's not good.

[00:31:45]That is not good, is it? So, how on earth do we take that forward then?

[00:31:55]Uh, I've got a clue, I think. Okay, I'm going to have to go back and think about what these three cell strings can be.

[00:32:05]because there were I think there were three different options. 1 2 3 I remember that as being an option. Let's where should we write them? Up in the corner here. You could have a 1 2 3 if it was a six cage.

[00:32:20]But if these were 13 cages, it depended what the nature of the six cage was. If the if the six cage was 51, so when we did 51 there, then we have to put 823. So 238 was one of them. And if the six cage was 42, then nine nine got invested in there. So that would be 139 to give us 13.

[00:32:45]So So these cages, which I I don't know are 13 cages, but I know the digits selected for them are from quite a narrow selection of digits. 1 2 3 8 9. So four and five in row seven have to be in the middle. Can that be four or five? No. 4 + 5 + 3 is 12 only.

[00:33:15]So this digit is four or five cuz I cuz in let's just recapitulate that quickly in this row. What I'm saying is this this cage and this cage are only selectable from these digits.

[00:33:30][snorts] So four five in this row are in the middle section of it. But this can't be four five pair because that yellow cage is meant to sum to 13 and won't. So this is a four or five and one of four and five is in there.

[00:33:48]Um only one of them and not six or seven. But we need a we need a big digit, don't we?

[00:34:06]It's nearly It's nearly very diff. Oh, hang on. We can't even use actually.

[00:34:10]Hang on. How am I going to do this? I can't actually find a way to do it in my brain. Bad brain.

[00:34:18]Uh, hang on. I'm so sorry.

[00:34:25]What's going on? Why doesn't my brain want to My brain doesn't want to compute this at all?

[00:34:33]These add up to 13.

[00:34:38]Yeah. Oh, I tell you the simple way to do it is you can't put five in there.

[00:34:44]The reason you can't put five on in there is that the other two digits would have to add up to eight.

[00:34:52]And therefore they would need to use either five again which they can't. You can't repeat five in the box or a six and a seven neither of which can go in there because because the way to get to eight in two digits is either 53 62 or 7. So I don't think five is in there.

[00:35:10]That's given me my first digit which is a five here and four is in here then.

[00:35:15]But I'm not even sure four can work. So the other two digits add up to nine with this being 1 2 or three. Oh 1 18 would work. Oh, that's that's such a relief.

[00:35:27]My brain for some reason couldn't compute a way of it working. But if this was one, that's the only way it works because four's in there. So the other two digits add up to nine. So they're not four, five. They can't be 63 or 27.

[00:35:44]They have to be 1 eight.

[00:35:47]So 1 or 8.

[00:35:51]This is now a seven because we said these two were six different.

[00:35:56]These two digits are not well this is a 13 cage. So these are adding up to eight. They're 2 six.

[00:36:04]They're not 17. They're not 35. So these digits are 139. One 39. So we can put the one in. And this is a 39 pair. We've actually got somewhere at last.

[00:36:18]Okay, let's do we get rid of Well, no. Let's be careful here. Let's be careful.

[00:36:29]Okay, it's nearly good, isn't it? This one now can't be 2 38 because because of the eight in here.

[00:36:46]So this is either 139. It's definitely got one in it. It's either 139 or 1 2 3.

[00:37:03]It's still it's still being recalcitrant, isn't it? Oh, but hang on.

[00:37:07]Hang on. No, I can get that digit though just by maths in the row. Oh, no. I can't. No, I can't cuz I don't know whether this adds up to six or 13. But I can I can probably limit this.

[00:37:19]Right, let's let's do maths. We know the whole row adds up to 45. These add up to 17. They add up to 13. That's 30. So, these add up to 15. So if this was 13, this would be a two.

[00:37:37]And if this was six, this would be a a nine.

[00:37:48]That's annoying. [laughter] That's annoying. So there's two ways that would work. Can this be 931? That this this pencil mark is not preventing that.

[00:38:01]This was 931 then.

[00:38:11]Oh, I've just seen something else I could do. Uh, but but if this was 931, we would know the order.

[00:38:19]One of these would have to be a six cage and it would have to be 24.

[00:38:24]But but here's the other thing I've just noticed.

[00:38:27]um seven in this box by Sudoku is in one of the bottom three cells. Now, we know there's a six seven in one of these. So, the six is above it. And that corrects this digit. Look, that's got to be two and that's got to be six.

[00:38:48]Now, if you're saying that doesn't do anything, Simon. Oh, well, no. I've got something else for you. Then the five here is down here somewhere. Now five could have gone with one to make six, but it can't because the one has to be up there apparently.

[00:39:08]I can't remember why we thought this Oh, it had to have one because of the options here. So the five here is going with an eight over here.

[00:39:17]So five and eight are a pair down there somewhere.

[00:39:24]And four is appearing down here either with nine or with two depending on what this is. Oh, that's annoying.

[00:39:36]Oh, that is annoying. Okay, so how do we do this then?

[00:39:49]I haven't got a Scooby-Doo. I [laughter] haven't got a Scooby-Doo.

[00:39:54]Um, okay. I'm going to go further a field.

[00:40:00]That cage I can do. In fact, I'm going to make it yellow because it can't, if we look at the options, we've got 13 cages, six cages, and three cages. This can't be a three cage cuz it can't be 1 two. It can't be a six cage cuz it can be neither 2 4 nor 51.

[00:40:19]So it has to be a 13 cage and it's not 58 or 67. So it is 49.

[00:40:27]And the rest of this Oh, so one in the column can be placed. So this could be a three cage. That's so annoying.

[00:40:34]So annoying. But these digits are now known. They are three 78.

[00:40:44]That's very important for the following reason. What's this cage? That's not a 21 K. Ah, this is also a 13 cage because this one can't be 15 2 4 or 1 2. So that's a 13 and it's not 49 and it's not 67. So that one is 58 and that that gives us some more joy. 8 and four can go in.

[00:41:10]Three can go in here.

[00:41:13]So now I get these digits which are 2 6 7. Now these digits are five and something four four five pair.

[00:41:29]Now um okay this one can't be five. I'm going to claim if this was five, this would be either a six cage, which would require a one here. Why is my phone buzzing? Let me just check politics. I don't know. Um, it or or this would be an eight, which it can't be. So, that's got to be a four. That's got to be a five. This is either a two or a nine, depending upon whether this is a six cage or 13 cage. And this cage, this is either a one or an eight.

[00:42:13]And that's I don't think we can do that.

[00:42:15]So these digits are 2 9 and six.

[00:42:24]It's a most peculiar puzzle this, isn't it?

[00:42:30]Um, okay. I'm going to get rid of this now and hope I can remember. We're only dealing with 13 cages, of which there need to be 13. And I found five so far.

[00:42:48]Six cages.

[00:42:56]Um.

[00:42:59]Wow. Wow. How do we do this then?

[00:43:07]Three has to be over here. Look. Yeah.

[00:43:10]Okay. I hadn't I should have got that before when I was looking at the options for those. But these are either 139 or 1 2 3.

[00:43:22]So that that's the full paniply of digits [snorts] and four is definitely down here but I don't know whether it's with two or nine.

[00:43:44]Um, six. Six. Six is up there.

[00:43:56]Do I know if I knew that this one was I I don't know how we're going to do this. I really don't.

[00:44:08]I'm trying to figure it out. Oh, hang on. I've got something. Did I work out one of those was a 58? I did.

[00:44:17]So these ah that's it. I've done it right. One of those is a 58. No idea which one. But that means neither of those can be 58.

[00:44:28]Which means that um well it means lots of things, doesn't it? It means I have to split up the five and the eight in in this box somehow.

[00:44:44]So, and I can't put eight in here at all because 8 8 cannot sum to six or three.

[00:44:52]So, it would be forced to sum to 13 with a five and break whatever happens over there.

[00:44:58]So, in fact, 8 has to be in this string, which certainly can't go with nine. So this is two and a 38 pair which means that one is in one of these along with five adding up to oh this is good. This is good adding up to six and this is a 49 pair. There we go. Now that should that's doing things. So 9 3 8 is now not on the bottom bottom here. So eight is above and five is below. In fact, we can we can get rid of the corner marks now. This is 468.

[00:45:39]This is 139.

[00:45:43]And this is 57 2.

[00:45:52]Yeah. So, there's a there is a vertical 24 pair in column 7. Uh, it's just it's quite brilliant this it it's difficult, I think. Oh, that's a one now. Okay. So, that that needs a new color cuz that's a six cage. This is a 13 cage. This is a 13 cage. That is a 13 cage. That is a six cage. That's a 13.

[00:46:18]Um, that one we don't know. Now, what have we got down here? We have got a 24. So, we've got two 13 cages being the 85 and the 67. So, how many 13s have I now got?

[00:46:32]1 2 3 4 5 say 6 7 8 9 10 11. I've got 11. And I could have 12 if that's one.

[00:46:45]I've not found any one two cages.

[00:46:51]But that could be one.

[00:46:56]That can't be one actually. that one there.

[00:47:02]Okay. Okay. Where are my one two cages in this puzzle? That's I think a legitimate question.

[00:47:11]I can't make this a one two cage and this a one two cage. That clearly won't work. Those three digits would only be selected from the digits one and two. So I can have one of those.

[00:47:24]I could have that one. So I could have one of these, that one, that one. Which other one? That can't be. That's three cells. Can't be. That three cells can't be. And this can't be a 2-1 gauge. So it is it's some it's I don't know about that one. I don't know whether it's a two here or a one two pair there. But I do know this is a one two pair. And I know this is a that's a one two pair. So that gives me a one and a five here.

[00:47:59]So what does that do? Maybe nothing.

[00:48:02]Maybe something though. Two is in those cells and two can't go there because then this this this would be a one. So two is in one of those cells.

[00:48:19]This has become six or seven.

[00:48:30]Um, okay. Here's an interesting point. I think I think I now know where the last two 13 cages are. This cage can't be a 1 2 3 cage. So, it can't add up to six. It obviously can't add up to three. So, that's a yellow cage.

[00:48:51]Now, what about that one? That can't be a six cage. Three here would require a one, two pair, which doesn't ex work now. So, that's definitely a yellow cage. And I think I've got all the 13 cages I'm allowed. Now, I'm going to double check again. 1 2 3 4 5 6 7 8 9 10 11 and two of these. That's 13. So, this is not a That's got to be a six cage.

[00:49:22]That's got to be a six cage.

[00:49:26]One of those is a six cage.

[00:49:30]And if this is a six gauge, it's got to be two. Four. Look.

[00:49:35]So that's two. That's one.

[00:49:41]Um, this isn't a one.

[00:49:46]This is a 13 cage. So Oh, that's that's beautiful. Yeah. How could this be eight or seven? This domino would add up to five or six. It can't with with the minimum these could add up to is seven if they're three and four. So that has to be three. These have to add up to 10 without using three. One or two. So they have to be four six. And I know the order. This is so clever. So that's eight. Now that's five to make 13.

[00:50:16]One of these is six. Oh, look. That's two now. So that's two. That's four.

[00:50:22]That's six. That's seven. And suddenly we've just filled in a whole load of the puzzle out of nowhere.

[00:50:29]This is not two. So the one two pair is not here. It must be this one.

[00:50:36]So that earns earns its stripes as being entirely green. This therefore has to be um has to be a one five pair adding up to six. It's the only thing it can be.

[00:50:51]This is a one now.

[00:50:58]And I think we're close to forcing the we're close to forcing this to be a sodoku puzzle, which is an outrageous thing we might have achieved. That's a nine.

[00:51:11]Um the way we're going to force it is by getting this. And if we know the composition of this, we would be well, okay, I do know the composition of that.

[00:51:22]It doesn't have a one or a two in it. It doesn't have these two can't be one or two. This can never be one or two either. So how do you make three cells add up to 13 if they don't involve one and two? There's only one way. 3 4 6 I think. And in fact the four is known. So that's four.

[00:51:41]This is three six. We know the order of that as well. So six seven go in. Um now these are a 7 eight pair.

[00:51:55]This is six. This is seven. This is four. This is nine. Beautiful.

[00:52:02]Right. So what do we need in column 9?

[00:52:04]258.

[00:52:07]Where's two? It's got to go there.

[00:52:11]There's a 58 pair left to place. But that gives me a 58 pair in row row one.

[00:52:17]Look. So 37 367 into those digits. Six.

[00:52:22]We can place 6 73. And that's the three in a corner.

[00:52:26]That's three in the corner. That's three in the spotlight losing its religion.

[00:52:29]Okay. So those go in. Nine goes here.

[00:52:32]Eight goes here. That's just sedoku. 855 58 87 uh something three maybe that look right? I think so. Um this column needs a something technical term for a five.

[00:52:50]Um now these digits are 7 8 9. So we can put the eight in and the seven. Whoops.

[00:52:56]And the nine. These digits are 35. 35.

[00:53:01]Um, we need 679. So, we can put the six six 97.

[00:53:08]That seems to have to be eight. And if we haven't made a total holix of this, we can put seven and nine in. And I reckon that's right cuz the logic was just gorgeous, wasn't it? How long has that taken? 53 minutes. It's very clever. You really have to or I really had to think about these boxes and how they could possibly work. And the fact they've got different ways of working does not make it easier.

[00:53:38]And there may be there may have been a mathematical way we could have got to what was it 1363 more quickly. In fact, let me just check it's right. Yes, it is right. That's a relief. Um but it's a beautiful puzzle. really interesting. Oh, oh, oh. Hang on, hang on. We have not finished, but now we probably have. Um, let me know in the comments how you got on with it. I enjoy the comments, especially when they're kind. And we'll be back later with another edition of Cracking the Cryptic.