The Hourglass puzzle is a Sudoku variant with additional constraints: (1) Normal Sudoku rules apply (1-9 in every row, column, and 3x3 box); (2) A blue line divided by 3x3 box borders has sections that must each sum to the same value (15); (3) On orange 'Dutch Whispers' lines, adjacent digits must have a difference of at least 4; (4) Cells marked with X must add to 10; (5) Cells marked with V must add to 5. The solving strategy involves first determining the blue line sum (15) by finding the minimum sum of five different digits (1+2+3+4+5=15) and the maximum sum of four different digits (9+8+7+6=30), then using the constraints to place digits systematically, with 5 being particularly important as it cannot be placed on orange lines and must be placed on blue lines to satisfy the sum requirements.
Deep Dive
Prerequisite Knowledge
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Deep Dive
A Beautiful Hourglass FigureAdded:
Hello. Welcome back to Cracking the Cryptic. Great to have you with us. Now, I'm a sucker for a bit of art that'll look good on the thumbnail. So, here is Hourglass by Sudoku Joker, which got recommended to us, and I'm keen to give it a try. Um it's a I think it's a region some lines and Dutch whispers puzzle bit of XV in it as well. Now we have apps that feature all those types of sedoku. So do check out our apps.
They are available at the well on all of um iPhone and Android and Steam. And you can check though the app store is what I mean by iPhone. And you can check those out on the link under the video and get the extra add-ons like classic Sedoku 2 and the worms including Blobs's Worm.
Now on Patreon, I think we've got something new either today or tomorrow.
Simon's putting up a video of him solving a cross word by Oscar from Finland, no less, who has created who's got really into cryptic crosswords and uh has created one and joined a highlevel Patreon here to persuade us to give it a go. and um I tried it. It's very good puzzle and Simon will be solving that in a Patreon video. So, do check that out. Um, if you're interested in the crossword content and if you are, we have Friday master classes and regular videos on Patreon. Um, as we do for gridoggram and connections and of course we have the monthly hunts, so you can still do Spider-Man Sodoku and defeat Spider-Man's eight evil enemies and save the city if you feel like it.
Um, Patreon is great fun. It really is the best puzzle club around. Do check it out. Um, we've got a bit of merch as well. That's probably the best merch around, but I haven't compared with others. Check out Marty Sears Rat Run.
Um, Rat Run mugs and mouse pads and hoodies and sweatshirts and stuff. Um, it's all on the links under the video.
As is this puzzle. This is by Sedoku Joker. I think Sedoku Joker created the remembrance puzzle. Was it the one based on poppies that I failed to get done on the right day?
Um anyway, this one another time connection as it is clearly an hourglass. The sands of time are running through this and we will look at the rules now and see how we get on with it.
So normal Sudoka rules apply one to nine going in every row, every column and every 3x3 box. A blue line, the blue line, there is only one is divided into sections by the 3x3 box borders. So each section of blue line has the same sum.
And for clarity, the region in box five counts as one section. So those five cells will add up to the same as those two, those two, those two, those two, those three, and those three. Now, on an orange Dutch whispers line, adjacent digits have to have a difference of at least four.
Cells on an X add up to 10. Cells on a V add up to five. That's it. That's the rules. Give it a try. I'm going to start now. Let's get cracking.
So, that clarification about the central X is useful.
That is useful because that has a minimum sum that we can definitely use.
The digits 1 2 3 4 5. The minimum five digits that are different from each other that you could put on it add up to 15.
Now 15 is not normally a maximum for two cells. But look in column seven for instance. There are two lots of these two cell pairs.
And what is the maximum for four cells in a sodoku puzzle that have to be different from each other. 9 8 7 and six add up to 30. Two lots of 15. So I think we've straight away found our n our blue line segment sum. It is 15. These digits are 1 2 3 4 five. We can't put fives on x's.
And we've got another of these pairs obviously. And this beautiful symmetrical hourglass in column three.
So one of these pairs is a 6-9 pair and the other one an 8 seven pair to add up to 15. The same is true over here.
doesn't necessarily imply that those two can't be the same pair as each other as long as the positions are swapped, I guess. Right? These digits are not five because they're on an X. They're 6 7 8 or nine. Now, can we use the whisper here?
So, if I if we were looking at green German whisper lines, I'd be telling you things like five can't go on them. They must alternate between low and high meaning lower than five and higher than five etc. You can't do that with orange whispers. However, there are alternation issues.
Digits do alternate between high and low unless five intervenes with one on one side of it and nine on the other side.
So this being high, I feel like this has to be low.
And I'm not going to go all the way. I'm going to allow five to exist there with a nine here and then a one nine pair here. In fact, that couldn't occur because of the blue line. That's quite sweet. So, just see if you can work out why that is. If this digit was a five, this blue line will break in a very, very peculiar way because that would be five. The only digits five can touch on the orange line are one and nine. So this pair would be 1 and nine. But we know the blue line adds up to 15. So this digit would have to be five. But we started with a positive five there. So that can't be five. That is low. I am becoming more and more tempted to color the low digits and the high digits in this puzzle. And it's time. I'm going to do it now. Why not? Why not?
So, these ones are all orange. As long as orange doesn't obscure the orange lines, not too bad. Now, right, I'm just quickly going to do a count around this line. If if there were no fives on this line at all, would these two cells be the same, be the right colors, as it were? would go high, low, high, low, high, low, high, low, high, low, high. They would. Now, what that is telling me is that either there are no fives doing polarity switches on the lines because these work fine as it is, or there are two fives doing polarity switches along the lines.
one in row eight and one in row nine.
Uh the one in row eight couldn't be here actually because that would have to be nine 5 1.
I'm I'm Yes, sorry. I'm not actually assuming that there is one in row eight and one in row nine. However, it feels quite likely because otherwise the nine in row the five in row nine is going to be right out on the perimeter.
We've got actually we've got the same pattern up here, haven't we? Either two fives on the line or none. Um, and you may think, "Oh, he's forgetting all about the case where it goes 959 and you have a five and no polarity switch." But I'm not because the shape of this line actually uh militates against that occurring because if you had a five here, these are in the same box and they couldn't both be the same. If you had a five here, these are in the same row and you couldn't do it. And the same applies everywhere.
But all of that is lovely fascinating theory but doesn't advance the solve one iota. These two are blue. Oh, I have I'm just they're orange. I mean I have just realized that I've included five in my light blue shading.
Yeah, maybe I don't dare shade those then.
I'd like if I'm going to do this coloring, I'd like to keep it to light blue for 1 to four, orange for 6 to nine. Ah, this is Well, these are both blue cuz they're on a V. This is blue because it sees a 6 7 8 9 quad in the column and it can't be five on an X. So, it's definitely low.
And this is a pair of low digits, right?
This is also low. And so is this because they're on an X and in a column with four oranges.
So this I can definitely shade as blue.
And this one, this one, don't know cuz we've got the possibility of double five. Now that would now require the five in row two to be definitely here.
Oh, and I don't think it's going to work, right?
Yeah, I don't think it does work to have two fives on this line because if you were to put one here, then the digit one would have to sit here. If you were to put a polarity switching five here, this would be one.
Now, where could you put five here? You can't repeat it there because it's in the same box as five posited here.
Anywhere else you put it along these cells, it needs to have a one on one side or the other. But the one is over here. Can you possibly put it there? So the one is here. No, because then it's on an X. So you cannot have two fives on this line. One here and one on this because of the der of ones that creates.
The same must be true with the symmetry down here. If you were to put five there, you'd have a one here. And then you couldn't put five here. And anywhere else you put it, it would create a need for a second one. And these just are unavailable. So no, there are no fives on these lines. And therefore, five in the top row is in one of those cells.
Five in the bottom row is in one of those cells. And this is a classic X-wing.
So we know that the five in the top row is in one of those. We know the five in the bottom row is one of those. That's going to use up the two fives in columns one and nine and mean that nothing else in column one or in column 9 or in row one or row nine can possibly be a five.
And that is going to mean where is the five in row eight? Where is the five in row two? It's not allowed to be in the perimeter. It's not allowed to be on the orange line. We've just worked that out.
laboriously. It must be on the blue line inside it.
And that is very interesting.
That is going to go with a pair of digits that sum to 10 to make the 15.
That is the sum of every blue line. Oh, also five being there means we can just do our magic alternation between other digits on the orange lines in the boxes where five has taken place already.
Now that's going to be five + one blue plus one orange. So is this.
And I can alternate this line and this one. So lots of blue and orange cells.
I'm beginning to be glad I did the coloring. Now, these are all six, seven, eight, nine.
So are these ones at the top.
And these blues are all 1 2 3 4 everywhere they appear because I've been careful not to pencil mark the cells with potential fives in the center.
Now, where are we now? And what can we do at this point?
Right. Oh, no. I was going to say five in this row must be in one of those two.
That's not true because it can be here.
This really is all about the fives this puzzle.
It's very hard to guess how I'm disambiguating the other digits further when we get on into the puzzle.
Um, but this is excellent. Right now, that Oh. Oh, I thought we had a quad of 1 2 3 4 and could whack five in the middle.
We don't.
Um, I haven't mentioned that four and six are a little difficult on orange lines. They only like five, they only have two possible neighbors. Four can sit next to eight and nine. Six can sit next to one, two. I don't know if that's important in this puzzle and I haven't figured it out. Ah, where's five in column three and probably column seven?
Two possible cells in each.
So the remaining five in column 8 that isn't in a corner and isn't there must be in one of these cells. Same column two and one of those. It's another weird five x-wing. And I'm now beginning to despair of fives and think they're not actually going to tell me anything about the puzzle at all.
very mean of them. Very mean.
Okay. If this was a if one of these was a five, this digit has to be one.
I don't know what that does.
I did not re I mean I know the puzzle looked good. I didn't realize how much the symmetry actually attacks it all over the place. It really is fascinating.
Um, the remaining digits in the bottom row and the top row that aren't five are low.
That's just doing a count.
One of these two is low and one is high.
Or maybe five.
Yeah, I don't know. We're going to have to disambiguate digits somehow. I'm straining to see how that occurs.
I guess one of these four digits must be a six.
Oh, one of them is not even on an orange line though. It's this one.
So maybe six is presence in that column is not a clue.
Hm. Where do we go next?
It's in It's interesting, but I can't work out quite what to do. We've got three low digits there. If that was low as well.
This can't be high.
It sees 6 7 8 9. So, this is from 1 2 3 4 5. And that's a quenchable of the low digits and five. So one of these two is five. The rest of the row is all high.
Now does that apply here? Yes, it does as well. The symmetry dictates that. Oh, no. But this is not a quinchable. That's only a quadruple. It's because of that V not being the same as that X in symmetry terms. Oh, botheration or bobbins as some might say. Okay, this digit is not five because five is in one of those cells. It sees all four oranges in its column. That's low.
What's going on here? These see a five in the box, but they only see two orange and three blue in the column.
So, one of these two has to be low, the other has to be five.
this somewhat mechanical approach I've adopted of coloring by polarity and trying to figure everything else out in that way.
It's not being wildly successful if I'm absolutely brutal. I may have taken a poor a poor tack. Okay, one of these is orange because that ah maybe they can't both be orange cuz either of them being orange completes a set of four in the column.
Um actually maybe let's count the total oranges we have in columns four, five, and six. Three in each column so far is nine. There are 12 in total.
No, that's not helpful.
I do feel that one of these digits is likely to be five, but I have not got any proof of that.
No.
Oh, right. I'm going to have to think about other aspects of the puzzle and see if something else. So, we've got 69 and 78 pairs in these blue segments.
And I guess the X in the same box as one of those pairs has to be from the other group in the sense that if that is 69, this digit is seven or eight. And this digit pairs with a seven or eight on an X.
These two are from the same pairing.
Yeah. Okay. I mean, I'm That's a weird pl I I don't know. I don't dare to start there. I think it would fall apart very quickly. I was going to sort of color them based on being in one of the the V pairs, either one, four or two, three.
But I don't think it tells me, it doesn't relate to that pair what that pair are.
Oh, does it?
This pair has to be different in terms of 69 or 78 from this pair.
I don't know that that's just not quite enough information unless it works here as well. Right, I'm going to color here. Let's call that pair the violet pair. It's either 78 or 69. This is the other pair. We'll make them yellow. This digit is a violet digit and this one is a yellow. So that's going with a yellow blue digit which is if that's seven or eight this is two or three.
Now these are one of each in the column.
Can I relate those to these which I I thought shy of doing it earlier. I'm just not sure now.
Or or is there another approach to take?
There is the thought that I've not had before that one of the digits in the bottom row in orange is a six and that must be touching one and two either in the row or in the box that it's in.
If it's in one of these external cells, then the pair near it is a 78 pair, which gives us a 69 pair at the top.
Now, I don't know. Maybe it's about this trio that add up to 15 because they have to use one digit. Yeah, that's interesting. They have to use one digit from one of these sets.
of six, seven, eight. Well, okay. I They have to use a digit.
Oh, that's very interesting. Now, well, they have to use a digit from one of these, don't they?
That's my current thinking. No, maybe not. If this was 591, these could both be 78 pairs.
And as weird as that is, I don't think and as unlikely as that is, I don't think I've proved it impossible. I don't think I've tried to possibly, but certainly haven't proved it impossible.
Oh, one of these is low.
Oh, goodness. Hang on.
One of these. Oh, no it's not. I was thinking that was an orange line connecting. No, that's not right. One of these three digits is low, but that doesn't mean Yes.
Yes, that's true. That's an important point that I could have made about 10 minutes ago. One of these three digits is low is from 1 2 3 4. One of them is five. One of them is higher than five, but the other one is lower than five.
That can't go here or here on the orange lines of the sand spreading out in the hourglass. So, it must go here. That is the fourth low digit in the column.
One of these is high. One of them is five. But now we found one 2 3 four in the column. And we can stick a five in the center of the puzzle where it so often goes in brilliantly designed symmetrical puzzles. Oh my goodness.
Well, that's something. Now that gives us fives here and no fives there. We could color those blue.
Right now has that done anything else?
We can color this in the column blue and this one also.
And now come on let's see this can't be five. We can color it orange in this column for for what it must be. Now it's difficult to color these. One is five, but one is five and that must touch a one in this cell.
So none of the other digits in this column can be one.
Um um um um. What next?
This is two, three, or four.
That doesn't impact these being six, does it? No.
Oh goodness. This took a long time even to get to this. Right. Where's one?
That is a group of one, two, three, four. That's a strange group.
Uh, no. I was going to say these are from the same per violet or yellow set, but that that's not known to be the case. I can't claim that.
H one of these is five and one is high.
What of it?
The rest of the things in the row are a low and a high.
Oh, this is on an orange line. This five. Oh, goodness. Why can't I even see it? It's just behind all my all my coloring. That's a nine. And on the X, we get a one.
And now some other digits aren't ones, including on the V where we must have a two three pair. This digit becomes a four.
That's not on an X. We can take four off those ones and one off this one. So that's now a two three pair. This digit is one or four. This one is six or nine.
This x is seven or eight. But in contrast to that now, what's that? I've got more digits in the grid. What's it done? This can't be a four.
It hasn't done much more than that. Um, nine. These can't have a nine in. One of them is a six and the other is seven or eight. These can't have a nine in. So, one of these at the top is a nine. And now these ones on the external can't be. And that can't be a one. Therefore, um, does that do anymore?
I'm really running out of X's and V's a bit, which is a shame.
94. That one is going to Yeah, I don't know. That five was on an orange line.
Why couldn't I see that? That's so irritating. Okay, if this was a four, it would be surrounded by eight and nine.
Oh, we've got a pair here.
One of these two is a one. We've got a pair here. No, we've got a pair. A low digit and that add to 10. On this line, there's a five. So, there is not a one on this line because you there'd need to be a nine on it. And we've just established that there isn't. That nine is in one of those cells. So, this is not a one. The one in the box has to be there.
That's odd.
Not just literally, but it doesn't really Placing ones on Dutch whisper lines doesn't do much.
That's the trouble, right?
We don't quite know about this pair, do we? Do we know they have to have a nine in? I don't think we do yet. Although I suspect they will do because that seems likely to be seven or eight now.
Oh, this digit can't go. It's either seven or eight and it can't go here. It must go in one of these cells. So, six is not in those.
Six is in one of these. Yes, I could have done that the other way around just by observing where six is in the column.
But now six on the top row is in one of the end positions surrounded by one and two there.
And forcing the vertical pair in the same box to be a 78 pair, which is going to force one of these vertical pairs down the bottom to be a 69 pair.
And I'm sure that's doing something. Oh, there's no This group again uses a five and a pair that add up to 10. So, it doesn't use a nine because there's no one available in that group. So, that's a similar situation to row two.
Now, does that prove that one of these digits is a nine? I don't think so. We haven't got enough done quite in the row for that.
H. So this is one of these two is a six.
If that's a six, this is five and four.
But if this is a six, well then this has to be a two.
Well, that's quite interesting. So we can't have 28 on this segment of line.
If that's the six in the box, these are five and four to make the line bit add up to 15. If that's a six by the whisper rule, that's two. And in that case, these aren't 28 or we never have eight on this line. So that's not an eight. That's all I found out there. But it's something. Now these are either 35 or 45 in these cells. One of these two is a two.
Very tempted to think it's not this one.
On the grounds that one of these corner cells is a six, but that could be touching. Oh, this is an orange line and an X. That can't be 64, you muppet. That's been available forever. It's available down here as well. X going through an orange line. Somehow in my head, the X overrode the orange line aspect of that.
Except when I was considering the row as a whole. Okay. Now, has that helped?
This is two or three.
It's definitely seven or eight now. Yes, that does help because now this can't be the 78 pair. That helps massively. This is the 69 pair. This is the 78 pair.
This digit is not seven or eight now.
This is 6 or 9.
Well, I think I announced proudly that it helps massively without having any idea. Seven or eight here. Seven and eight here. Now, unfortunately, that didn't immediately translate across and make this a 6-9 pair, which, you know, it kind of feels like it does, but I haven't worked out how. There's every possibility that's the case.
Five.
One of these is a six. We know that.
that two three pair is replicated there.
So two and three must be oh this digit can't be two or three. That's uh been available for a long time. Doesn't really do anything I don't think but just hadn't noted it before.
One of these is a four.
If it was at the top or bottom that would have to be surrounded by eight. Oh look here's a 7 8 9 triple in the top row. That's the kind of thing we want.
This digit is six.
That is the kind of thing we want. Six is surrounded by one and two. It's also forcing this not to be a 6-9 pair. So, I'm getting rid of the color now that we got that job done without ever using my scheme. 7 8 are looking down at these cells. That's a 69 pair. This can't be nine. That can't be one.
Now, is that going to translate across the grid somehow? I don't think it is.
Seven or eight there. Can that be touching four? Yes. How about this one, two pair in this box? We've had 61278.
These digits are from 349 and all of these from 3459.
If that was a two, this is three, seven, four, eight, and nine. Five, one.
Now, where's the problem with that?
There probably isn't one. Okay, how about 69 down here or 78 or this pair?
There is a six in this bottom row. It's not here because neither of these can be one apparently. In fact, look.
Okay. Well, it's not six. These This is a two three pair. Looking up at that cell at the top. That's giving me a few digits. Not many, but a few. That one can't be two. Now, now here.
Oh, where's one in this row? It's in one of those two cells.
Um, if that was nine, we'd have 6 7 8.
This group is either 654 or 753.
Oh, that two is looking back down the grid to where it began. 3 two we get three here with seven on the X. We take seven out of those cells. One of these two is a six. So this one can't be three or four. That's one or two.
Oh, what about this?
Don't know.
No, it can't be three. I know that much because of this.
If it was four, that would be an 8 n pair. This would be six. We'd have a one two situation over here. I expect that's what we are going to have just from the point of view of symmetry. But but I need to prove it not just guess it.
549123.
I can take nine out of a few cells there.
Um yes, I've got a 69 pair in this column that I hadn't seen.
That doesn't gives me a 78 pair without doing anything else.
If that was one, this would be two.
This would be four. Probably is four, but goodness me. Oh, this is high because two and three have to be in this group.
Uh, no. That that that conclusion doesn't follow. Two and three could both be there.
Yeah. Okay. Sorry. Trying to reach a conclusion that's not available. Always a risk. This can't be two or three by Sudoku. It can't be four by whispers.
So, it's one.
And now this isn't one.
71. This can't be three.
This pair sees 1 3 2 6 7 in the column, four five in the box. That's an 8 n pair. Leaving four and five in the other digits in the column. 8 9 pair removes eight from that cell.
One of these has to be an eight.
Um these are from 167.
Uh I don't know. Two and eight are down here somewhere.
I just want to figure out these blue line pairs with the fives.
Then I think we're actually motoring.
Okay, in the top row, this can't be nine cuz there's a 7 8 9 triple.
So, one of these is definitely nine.
So, is one of those.
Still nothing nothing will crumble.
These can't have a four in since we got that four.
Um, can I get a pair in this column? No, not yet.
Uh, right. So, we've got this group adding to 15 or if that's a seven, this isn't four.
That's not very good. Right. Something else. Two, three pair here. That digits one or four. I don't know what that means.
Six one. That two is going next to seven or eight.
that one. That's fine. If this was a four, this would be nine and eight.
That would put six here and four there.
That's quite interesting. We'd have fours in both of those positions.
I don't know. I don't know if that does anything. Three. If this was four, five, six, that would be nine.
Ah, this digit can't be four. This digit can't be four because it would be surrounded by 8 n.
So four is either there or in this row.
This would be four. That would make this a six. If that was four, this would be 8 9 and this would be 7 six.
That would make this eight and this three and this two.
In fact, one of these two is a three.
And if it's here, it's surrounded by five seven.
357 would make this an eight.
That's not putting any pressure on this cell. Again, we're sort of one cell away from information. Oh, there's only two places in this column where one can go.
One is indeed in one of those two cells in the bottom row. That's fit.
Weird again.
fours in one of those two in row four.
Okay. If this was a six, it would be surrounded by one and two.
This would be nine.
That would be eight.
Five and four on the outsides. Oh, maybe it is about the outsides. Maybe I should be somehow comparing these four digits based on what can go on the blue horizontals.
Or maybe I should just be finding a bit of Sudoku that I can't spot yet. 69371 2 and 8 are in this group along with either four or five.
Oh, in this bottom row we've had 689 37.
Everything else is 1 2 4 or five.
And that's two of those low digits.
If this was four, this would be an 8 n pair.
This would have to be three at that point.
Going with 57.
And then we'd have 2 six up here. 489 35726 7 9 I don't get it. I don't get it. Two three pair looking at this cell stopping it being a three. The three in the box has to be in one of those two cells.
If it's here, it's next to a five and a seven. I want to know what that means. I don't know what that means. Something on the whis something on the whispers somewhere is is doing the job, isn't it?
If one of those was a seven, this couldn't be four.
So, if one of these was a seven, and that's Yeah. If this is a seven, that's what happens. One of these is a seven and this is not a four.
So if one of these is a seven, we have a two three pair there in the top row.
7 and 8.
7 6 9 six there.
I don't know.
Ah, six here. Six here would force that to be seven and put three on this line and eight there.
And that would have to be a one at that point.
Whereas six here forces six there and four into one of these. And this can never be four. I mean, this is a pretty minor candidate reduction, but I'm going to do it. Six here puts six there, which puts four on this line, and that can't be four. But six here prevents it being four anyway. So, it's not four.
Now, what about three here? Maybe even that's difficult. Three here forces nine there.
That forces six here and six here.
This is now 456 93 2.
Did I say 456 932? And breaking that cell? I did.
Three there makes that a nine and puts four on that line and this cell is broken.
Wow. Right. That's a one.
Now, I'm hoping that does something, but I'm not convinced it will. It stops this being a one, right? Nine there always gives us a three four pair, a virtual pair.
makes this seven or eight. If that's a nine, this is seven or eight.
I want to know how that turns out. And I don't know.
Oh, I got a one here. I was so proud of that. It's done nothing, has it?
That's annoying. Um, right. This being a three sits next to seven on the X.
Then that's an eight. If that occurs, that's not doing anything.
Okay. If that's a two, that's an eight.
Then that's a seven. That's a six.
That's an eight. Then we have eights here and one of them here and there and there.
and one of those cells. And there we get eight kind of pencil marked everywhere, but not in a manner that concludes anything.
Okay, that didn't work either.
Six. One of these two is an eight. If it was here, then that puts eight here and in one of those cells.
I don't think that's going to do anything.
Right. This being a three would stop this being a six and that would have to be surrounded by a 78 pair if it was a three.
That would put two into one of these cells. So, three there would make that a three in a rather surprising gesture, right? What else is happening to threes?
We get a three there.
Okay, this is quite interesting because that does happen. Three here puts three here. that puts seven here but it also puts three here which puts seven here. So in that case we'd have sevens.
Ah you cannot have sevens in both of these highlighted positions or you can't put one in box two.
That three that I'm postulating would force a three here.
And because it would put a two out here somewhere, it would put a three there, which would create sevens in those cells, which would negate seven in box two entirely. So that's not a three.
That is an even more complicated step, but it might get something done this time, which would be a massive relief.
Two and eight there. This puzzle has been much harder than I was led to believe. Six and four.
I mean, maybe I've missed a big thing, but I think it I think it's really clever. This is a one. It can't be a seven because of that 78 pair. This is a nine. I think we might be on the track trail this time.
Those are just getting done now.
Um, now this is where six looks down to the bottom here. That's a nine. That's eight. That's six. On the whisper, this has to be two. And that's one.
The two is looking up at that cell and indeed at this one.
Um, is that doing anything up at the top? Not instantly. Not instantly. But at the bottom here, we can take one and two out of those cells.
We're left with a four, five pair.
on the whisper.
This could still be four just I think.
Um, okay. Let's do some coloring because I haven't done that for a while where we've got known digits. I can color um four five pair there leaves a 28 pair in this column.
That's doing nothing, is it? What about this row then?
Uh, yes, nine in this box has to be in one of those cells now. 628. Oh, six has come out of there. This adds up to 15.
That needs a three in the middle. This is actually not just a nine, but a 94 pair. And that's going to sort out quite a lot of lovely old Sudoku.
That's a four. That's a five.
That's not a five. That's a five. That's not a five. That's a five. That's those fives all done, I think. Not quite. No.
Still got columns four and six to go.
These aren't fours, though. That's a 39 pair. Now, down here, I've got five seven pair. That resolves 87 there.
587. That eight fixes 28. Let's get rid of the corner marks because they're not really doing anything now. Uh, this is four with a six beside it. This is a 23 pair.
Just getting my coloring up to date.
Okay. Um, in this column perhaps 1564.
No, not in that column.
Oh goodness. Um whispering up here.
Eight is looking at that cell. So this is a six seven pair. That's an 8 N pair.
This is seven on the X. That gives us a three which sorts out our two threes.
The seven resolves 87. This group of digits is 2 48. And the top one can't be eight. In fact, the next one isn't eight because the bottom one has to be eight.
And that fixes 769 in that column. And I can place a three.
76.
Okay.
Okay. Um, this is good. This is good.
This is now no four on that because it's 735 adding up to 15. This is four and two.
That's become a four surrounded by eight. Nine is fine. Now something's got to resolve these last. Oh, we've got a three there to do 35. That does 57.
That does 76.
Six. I could have done before apparently. Now I could do 94. And then just before I do the last bit, let's complete the coloring based on what I was based on my earlier color scheme.
Hope I haven't got anything wrong in the color scheme. I think it's all there.
Nine at eight. And there we go. What a lovely puzzle. An hourglass solved just under an hour appropriately just before the sands ran out. And it's been popular, but it's a very clever puzzle.
That'll be why Sudoki Joker's absolutely nailed it again there. And thank you for the recommendation, whoever that was. I will see you again on the channel soon.
Um, absolute pleasure to be with you as always. Bye for now.
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