A masterclass in deductive elegance that transforms modular arithmetic into a high-stakes architectural feat. It proves that the most profound beauty often lies within the most rigid constraints.
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You Told Us This Was "Puzzle Of The Year"... You Might Be Right!
Added:Hello and welcome to Sunday's edition of Cracking the Cryptic on another roasting hot day here in the UK. It is absolutely baking. Um, but I am I've got a job to do today. You know, I'm going to be attempting this puzzle that so many of you have written to me about. It's called On and Off. It's by Cactus Love, the great Cactus Love, who is on an absolute tear at the moment. And lots of you uh have said that this must be the puzzle of the year. Um, which I believe you. Don't get me wrong. I I believe you. I've read the rules and they're slightly terrifying because yeah, because I can't my simple brain, you know, it reads the rules. It understands the English of the rules, but it doesn't actually understand the implications at all. So, that is going to be potentially a problem. But, I'm this is this is going to be my job in the heat to have a go at today. Um, so I'm very very much looking forward to this one. It is meant to be an absolute worldy. Um, before we kick off, what can I mention? We've got one day left if you'd like to enter the monthly competition over on Patreon. Um, and be in with a chance of uh let me see if I can show you. Is it that one? Yeah. Be in with a chance of um coming on to the channel and appearing in a video and solving a puzzle. So, if that might appeal, do get your entries in soon. Um, lots and lot thousands of you have been solving these puzzles and absolutely loving them. the feedback we've had has been so good. So, um yeah, basically do have a go. You will have fun. Um and that is after all the aim of the game.
Um what else? I've got one birthday to do today. Let me go to my um my birthday uh my birthday note. I want to wish Dom a very happy birthday for tomorrow actually. Um the same day as the moon landing. Oh, okay. So, the 20th of July was the date of the first moon landing, was it? I did not know that. But your Dom, your partner Kirsty wrote to me and told me that you were turning 34. And um I think you introduced Kirsty to the channel about a year into your dating lives together. And were you what? Were you shy about one of the great hobbies?
I don't understand that at all. Um, but I know you've been together for a few years now and apparently you both watch the videos now and sometimes um sometimes compete over the puzzles which I fully approve of too. Um anyway, uh Dom, I hope you have a great birthday today.
And um and uh I don't think there'll be a moon landing tomorrow, but hopefully there'll be a large slice of very heavily iced chocolate cake. Um and I think that's all the news. Please like, please subscribe. You know the you know the drill. Let's have a look at the rules to on and off by cactus lav. What do we have to do to to solve this puzzle correctly? We've got normal sudoku rules applying. So we're going to put the digits one to nine once each and every row, every column, every 3x3 box.
Then we have to draw a single path through the centers of cells. The path may not branch, cross itself or visit any cell more than once.
segments of the path alternate between diagonal segments joining two cells that touch at a corner. So, let's just stop there. I mean, even this part of the rules, I wasn't sure I understood when I read them before.
Segments of the path alternate between diagonal segments joining two cells that touch at a corner. So, if we find two cells that touch at a corner, let's pick those two.
Oh, no. Cuz they don't touch at a corner. They touch at an edge. I see.
So, I have to do those two. Okay. And the path goes through the centers of cells. Right. Okay. All right. I understand that. So, that's a segment. I think segments of the path alternate between diagonal segments joining two cells that touch at a corner and vertical segments joining two vertically adjacent cells. Oh, I see.
So, oops. I was going to do that and that. Is that Is that legitimate?
Any four consecutive cells along the path. Okay. Well, I've got four now. Any four.
In fact, let's let's just go a bit further. So, so then this would go diagonally again. And this would go diagonally again because I have to alternate every time.
And then I'd have to go straight again and straight again. So I can create weird like p patterns down the grid and then any four consecutive cells along the path contain digits with four different remainders when divided by four. So I mean I've seen quite a lot of puzzles in the past where we've been dealing with mod 3 division by three and making sure you've got three different remainders i.e. remainder 0 1 or two.
But here we've got presumably remainders 0 1 2 and 3.
So if this had a zero remainder, it could just be a four. If this had a it doesn't have to have a one remainder cuz it could like have a two. It could have a three remainder. Let's make it a three.
Have give it a three remainder. Then this could have a one remainder. That could be a five.
And then this would right. So if we were in this position then both this cell and this cell would have because we haven't had a remainder two digit have we? So these would have to be I see they'd have to be one of them would have to be a two and one of them would have to be a six because they could they couldn't both be two. There's there's two digits from the digits 1 to nine that have a remainder of two when divided by four I think and they're two and six it feels like so that that is a potential way we could organize the line and there's just one more paragraph. It says every square lies on the path. I've managed to get no squares on my path. So all of those would be on the path.
Every circle lies off the path. So these would all be off the path.
If a square contains the digit n, then digit n occurs exactly n times on the path. If a circle contains digit n, then digit n occurs exactly n times in cells not on the path. Right? So if that was a five that would be saying right that's off the path is it? Yeah that's a circle so it's off the path.
So so the digit five appears exactly it says exactly five times not on the path which implies I suppose that it would appear exactly four times on the path.
And what would happen if on the other hand this was a five? If this was a five, which it can't be, but let's if that was a five, then there's five fives on the path.
I think if a square if a square contains a five, then the digit five occurs exactly five times on the path. Yeah. So that is how the rules work. Um, incredible, incredible, incredibly original rule set actually. Let me just see if I can delete. Yes, I can. I can delete everything there. Um, do have a go. This is meant to be an absolute worldy. The way to play is to click the link under the video as usual. But now I get to play. Let's get cracking.
So, we don't actually know, do we? Where the path starts and stops. I sort of want it to start there because it looks like it wants to.
Is that the most illogical thing I've ever said on the channel? Um, now I'm sure I've done a cactus love puzzle before. I think it was cactus love, but it wasn't it wasn't it wasn't it was diagonal and it might have been diagonal and horizontal movements were possible. I can't quite remember, but there there is a puzzle I've done and I want to say about a year ago with some rule set that it I don't think it had I think it might have had mod 3 in it rather than mod 4.
And that puzzle, if it's the one I'm thinking of, resulted in some sort of diagonal diagonal implication.
I really hope I'm not going to have to work this out because that is going to be absolutely terrifying.
I probably am going to have to think about things like that, aren't I? I mean this is clearly not going to be if people are saying puzzle of the year it's they're saying that because this is going to require some some sort of you know unaloyed genius in order to solve it.
If you if if you put a five in a square, you can't put five in a circle, can you?
Because because there is an odd number of digits in a sodoku in in i.e. the digit one appears nine times, the digit two appears nine times, all the digits appear nine times. So, if you were to put whatever digit you put into a square or a circle, you couldn't match it. Um, because it's not saying at least four times. If you put a four into a circle, it's not saying the digit four appears at least four times off the path. It's saying exactly, which means it's feeling it is got to be exactly five times on the path. So there is no correspondence between circles and squares and there are at least four different digits in circles because of column one.
Now, in terms of squares, squares are horrible in in terms of trying to work out how many different numbers are in squares because I mean, I think in theory, they could just be two different numbers. Those two definitely have to be different. But the rest of them could be, imagine this was a nine.
I mean you could literally all of those could then be nine. No, no, that one.
Oh, hang on. There there is a repeat in row seven as well. Okay. But but yeah, I mean you could do a lot of sevens and eights, I think. 9 98.
So there's no restriction really in the squares. There's a small restriction in the circles.
And all right, I'm just looking at the rules again. And the rules say the path may not branch across itself or visit any cell more than once.
Segment. And then we've got this weird definition of segments.
any four consecutive cells along the path contain four different remainders.
So okay, so the digits that have a one remainder are 1, five and nine. The digits have a two remainder two and six.
The digits have a three remainder of three and seven. And the digits have a four remainder or zero remainder I should more precisely say are four and eight.
But but it's not like you know how do we even know where I mean let's say this was the start of the path.
How would I how am I meant to know what that is in terms of what remainder it has?
And then say the next cell was there.
How am I meant to know what remainder that has?
I don't know. I don't know. Okay, let's let's let's think a bit more about how this path is built then because the path has to have vertical segments separated by diagonal segments.
So let's go back to my earlier example.
Um, so if that's a vertical and it's got to go diagonally, but it from there from that cell, sorry, I'm just pausing because I'm just trying to make it in my mind.
H has it really got four all four diagonal options is is the question I've got in my mind. So if if it drops from here to here and it's in this cell, can it go in all four diagonal directions?
And from there it could go up or down off the sides of all of these, couldn't it?
Okay. So if it arrived So if it arrived in this cell, it could arrive if if it arrived from a vertical segment, it arrives either from the above or below. It can then go in one of four directions. just sort of explodes outwards and from there it goes it's sort of it's like a it's like it's a car on a motorway and it's it goes into an adjacent lane.
So after after two moves it it it's moved lanes in the motorway and so so okay so every two moves it either goes into its own l it goes back into its own lane or it goes further away doesn't it? So let's say it was here. Hang on. So it goes it goes here, moves diagonally, stays in its lane, and now it's got to do a diagonal. So it either goes back into its original lane or back over there or back into its original lane, or so it then goes, it can move one more lane over again.
And then once it gets there, it goes straight straight straight straight down or up there. But I can't do two colors and down or up there.
So it behaves a bit like it's a car on a motorway.
So sorry, let me just think about this. I'm trying to understand how that that car movement relates to the modularities because whatever the modularity is in this cell Is it par? Is that Is that what's going on? If I'm not sure if because like three remainder digits are both odd, three and seven. Two remainder digits are both even 2 and six. One remainder digits are 1 159. They're all odd. and zero remainder digits of both even again. So whatever the parity of this cell is, let's say it was even.
Then by the time you get to one of those positions, which is going to be your your next your place position after two moves.
That's quite okay. Here's a different point. Sorry, I've got distracted with myself now. But just just something I've just noticed as we've done that is that after two moves, I'm either a knight's move away from myself or I'm sort of adjacent to myself. So I've either sort of accelerated into the other lane or I've just moved exactly sideways or I've decelerated into another lane and drop backwards. But I don't drop backwards linearly after two moves.
So yeah, but okay. But it's more four moves. That's it's four moves I should be thinking about, isn't it? Because four moves gets me back where it gets me back to the original. Yeah. Four moves gets me back to the original modularity of the blue digit. So let's say that this was a four or an eight. So it's 0 mod 4. Zero remainder when divided by four. Then yeah, if we look at the positions, we could I mean the thing is I I only did them for this this this string. If I went down here, look, and to this particular position, but but I but it's I I think the principles are common. So, if I was in blue and I moved to this position, after two moves, then after four moves, I'm going to be here, here, here, here, or here.
I'm going to be in one of those positions.
Or could I be back there? Oh, hang on.
Could I be back where I started? I didn't think of that.
I No. Well, I can't, but only because I'm not allowed to revisit cells. But otherwise, I could have been.
So, so, so the blue modularity which I'm saying is a 0 mod 4 modularity reoccurs.
It reoccurs either in my own column, my own lane of the motorway, or sort of two lanes away. So, if I'm in the fast lane of the motorway, it never reoccurs.
It It actually can never reoccur in the adjacent lane. It can't ever be in the middle lane. It's impossible.
It would h it has to reoccur either in the lane I'm in already or in the inside lane two lanes over and sort of missing out a lane is a better way of putting it cuz some people's motorways might not have three lanes they might have four or something but So sorry how has this taken us forward? We have learned that in this puzzle any path any path sorry I can't get this any path I don't know I think I think this I I've even stopped talking. I apologize. I apologize. I can't get this clear in my head. I I can see the the I can see the bold fact, which is that the the modularity I'm on at the moment repeats.
It must repeat either in my lane or skip a lane if you see what I mean or skip a column.
But what I can't see is what that means.
If indeed it means anything.
Um, this is this is very hard. This is okay.
That that's where we're up to. This is very hard. Let me look at the rules again. Draw a single path through through the squares basically, isn't it?
So, we are definitely we have to visit these cells.
It's not a loop or anything. It's just a path. It says, but it doesn't tell us where the path starts or stops.
Segments of the path. I understand about well, I understand in the loosest sense how segments work.
Any four consecutive cells have to have all four remainders. Okay. Well, I sort of understand that, but I don't really understand how that leads me to anything. And every square every square is on the path. All the circles are off the path. Those things I understand. And then we've got this sort of counting circles or counting squares type rule about uh this is I'm confused now. Sorry. I'm just going to get rid of this because I don't really understand what that's doing. Let's highlight all these.
All those are on the path. All these are off the path.
So, we do have to get to there. I mean, this might be the start or the end of the path. So, we can't what you can't do is go um that is a cell that's midway along the path cuz because if it was midway along the path, then you could never get to this cell or from this cell to this cell because horizontal movements are not possible.
So we would actually if we could prove this wasn't the end of the path. We would have that we could draw that straight in and then this couldn't close. So that would have to go like that and then this would have to not go down. So it would go like that. Then it has to do a diagonal. So you'd actually get loads drawn. If you knew that this was not the beginning or the end of the path, then you'd have to go down.
Then you could go up or down.
But that assumes, you know, that that isn't the end of the path, right? Or which it could be.
That's weird. So I was about to say and this disproves my earlier thing because it's quite clear that this could repeat its mo modularity in this in this position because I was sort of looking at this thinking that seemed a very long line but it actually can't cuz whatever this is modularity is it it it will repeat its modularity in four cells time. So it's going to depend where this goes and that's got to do a diagonal which is going to be there there there or there. So it does it does not end up in that column does it?
So the modularity of that one whether or not this is the start or the end of the path will either be in that one or in that one.
Oh, now I have just had a sensible thought.
Okay, I have just had a sensible thought. I think this is the first sensible thought I've had.
I might have just had a very sensible thought.
I'm not sure it's correct though if I know I'm not sure it is correct.
But it's the digit, not the modularity.
So if whatever the mod hang on, let's just give this one a color. I'm just going to give that one a color. And what I'm trying what I was going to say, and I'm I'm not sure it's correct, but I'll say it, and then we can think about it together, is how could that be a big number?
But because because what what I've just been looking at with my motorways is that whatever this number is whatever the mod is is it the modularity? It is the modularity.
Yeah. But the modularity sort of converts when we actually when you actually put something in. Yeah. Okay.
when you actually put something into a square, it matters.
So what I mean by that is we we have to sort of hybridize our concepts of modularity and counting.
Let me try to explain.
I'm not even sure this is right actually.
Okay, here's my thought. Let's imagine this is a six in the corner of the grid.
How How is that possible?
Because I just don't think it is. I just don't think Oh no, I might be wrong. I can't I can't get I can't get it clear in my head.
The reason I think that might be a problem is because what I said was is that this modularity, whatever it is, can only repeat either in my own motorway column or in a column that's sort of misses out the adjacent column, the adjacent lane. So whatever this is, it could repeat in this. It could repeat in this. It could repeat in this. And it could repeat in this.
So, so I think I think it is right, isn't it?
So, so okay, let me change the question because what sorry, what's confusing me is why couldn't I have a two in this column in this modularity?
But I don't I almost think that's a misnomer. The point is that if I put six here that that is saying because of the rules of the puzzle that is saying you must put six sixes on the path. You must Simon you must put six sixes on the path.
But I can't do that because I could put a I can only put this modularity on the path on the path mind. Obviously there's going to be a six in this column and there's going to be a six in this column etc. But if I want to put that six on the path, then it has to be in the odd columns of which there are five.
So you can't put it's impossible that that that can be a six. I think that is I think it's true for all of these because in fact it's going to be better.
Look at this one.
This one is in an even column.
So its modularity is capable of repeating on the path only in alternating columns which are these positions. So whatever the modularity is of this I can put it in this column. I obviously can't put the whatever this digit is again in this column, but I can put it in this column, this column, and this column. If I want to put it on the path, I can't put it in this column. It just doesn't work because of the way the path moves.
So, this digit is actually even more constrained than this digit. This digit could be a five. This digit can't even be a five. It could be a four.
So I don't know what that me I mean that so I think I think it is legitimate to label the well to label these 1 2 3 4 5 not if they're in an even column which that one is It can't be a five.
You see, it's interesting, isn't it?
Loads of them are in odd columns.
Now, uh, sorry, I I know I've stopped talking. I I don't I don't mean to.
So does does that mean these are the digits 1 2 3 4 5 somehow?
I mean I mean the thing is you could there's nothing that prevents you having a same digit in a square. Every square lies on the path. Every circle lies off the path.
A digit in a square contains n.
It doesn't say. Yeah, I I think it's possible that these contain repeated digits. Some of them can't repeat.
If that one does repeat, it has to repeat in those cells.
because these cannot be on the path and have this modularity.
Um, I'm almost getting this. I'm almost getting this.
So, if they were one, if these were for some reason 1 2 3 4 5, are they 67 89?
Maybe if if okay so how long is the path then if the path ah no no you got to be very careful with that noa I've spotted spotted a lacuna in my logic right let's say loads of these were fives um and let let's say that these digits were not the digits 1 2 3 4 5 they were some subset of those and these were these are obviously all different but let's say that there is a ninth digit and there might be more than one digit but let's say that there exists in this puzzle a digit that is neither in a square ever or in a circle ever then what is to prevent that digit from appearing nine times on the line nothing don't think.
So I don't know quite how to do this. I'm afraid I'm going okay what I'm going to do instead I think is try to understand how this logic works in the rows does it does it work the same way or different ways so in other words I don't really I don't really want to look at that one because I think that might be constrained by being in the corner.
Let's pick a totally new cell. Let's pick this one. Let's say that this is on the path and it has a particular modularity.
I should have paid more. I didn't think about this when I did my motorways. But what I want to now imagine is that the motorway is running like this.
So, it's running east to west rather or west to east rather than up and down.
And I want to try and understand whether there is any pattern that's applying in the rows now.
So how are we going to do that? We've firstly got to decide how we ended up in this cell. So we could have ended let's say we ended let's say our next move is a is a diagonal. So if our next move is a diagonal, we are moving out of our lane, which I'm saying is running this way.
And then I've got to do a vertical. So I can never do a horizontal. So I'm then doing a vertical.
So So my pos my end positions are I do a vertical.
I do a vertical. I do a vertical.
So I I started here and after two moves I'm in one of those positions I think.
So that is none of those positions are in an adjacent lane.
Yeah. So, so I mean, and if I if I repeat that process, I'm clearly it's not like I'm suddenly going to end up in this lane or this lane, am I? So, from here, I have to do a diagonal move.
Let's just use this one as an example.
I But I think I I now can see what's going to happen.
So what I'm saying is I actually let let's say we just pursued this option.
So I started here after two moves I'm here. I then have to do a diagonal.
Whoops, not that.
And I then have to do a vertical.
Um let's try and make sure we keep the colors as pure as we can't do that. That would return us to a cell we've already visited. But I don't think it changes the principle. The principle now tells us Oh, no. That's Oh, no. No. Hang on.
Okay. I'm I cannot end there. I haven't done my vertical. So, I do need to extend that cuz that that immediately confused me. I thought I was like, "Oh, I can end there." But no, I can't.
I can't because the spots I've now ended up with in I went from here to here and the spots I've ended up in are those.
So it's the same in the rows.
In the rows it works the same way. I cannot going to give that back its color. I cannot Oh, sorry about this. Everything's going to mess up. I can restore it all. Let's restore these. 1 2 3 4 5.
So whatever whatever the modularity is of any number. Now, if I want it to be on the path, it can be.
But for that to be the case, Yeah. All right. Okay. I now want to think about these two.
Yeah. Okay. Uh, in fact, I'm not Oh, yes. Okay. No, I'm not going to think about those two. I'm not I'm going to think about those two. I'm going to change my mind. I'm going to think about these two if that's all right. And I want to make an argument about these two. I think neither of those two numbers moduluses is the same as either indigo or purple.
And I'm going to try and justify that for you.
Let's firstly start with this one where you might say, "Ah, well that's two lanes over, Simon.
If the motorway is running north to south, that's two lanes over. So that could be the purple modularity. No. No.
Because what we just showed is that from a row perspective, if you want to repeat this digit on the path, you can absolutely do it, but you have to repeat it in either its own row, which is a nonsense, or a gap of one. So this digit cannot be purple. It cannot be the purple modularity digit. It just cannot be.
Now, we know it's not the indigo modularity digit because the indigo modularity repeats in these columns, not in this column. So, that is a new modularity.
Now, I'm going to look at this one. Now, this is in column 7 and it's in row four. This is in row one. So, so this one deals with odd rows. That's in an even row. So that's no use. This one deals with odd rows. So that's not the same. This one deals with even rows. So this one could or these two could both be orange until you look at the columns again. Whereas this one, if you want to repeat its modularity on the path, it has to also work from a column perspective. And it doesn't. So that one is a is a is a new modularity that we've never seen before. So now I've got four different modularities.
Now, there must be a way of mapping these ones cuz I can't have more than five modularities cuz I've only got five different digits in these things.
But do I know? I mean, these have to be different.
I thought I saw something before on these before I started thinking about those.
Yeah. Uh, no, no, no, that doesn't. Oh, dear, dear, dear. Now I'm Well, okay. One of those can definitely be indigo. That is true because indigo these are both in oddnumbered rows and columns. And this is in an oddnumbered row and column. So one of those but only one of them can be indigo.
Now if we look at the other three I don't think they work because this one.
Yeah, it's in an oddnumbered column but it's in an even row and this is in an odd row. So they can't be the same. Same is true for that one. This one again it's it's it's it's from a column perspective it's parity parity is the wrong phrase parity is implying this digits odd and even nature I don't mean that I actually mean purely reference to the column and the row numbers this should share the parity visav its column number or row number with anything with any equivalent ent digit I want to claim is this color. So if I want to claim that's purple, it doesn't work. It works from a motorway perspective that's going east to west. It doesn't from a motorway perspective that's going north to south. Let me remind you why. It's because this modularity just never appears on the path in this column. It just doesn't. And that's what we learned at the very start.
So this digit is not these two colors.
This digit is not these two colors. You can see that very clearly. It's in an adjacent lane. So, so I think one of these is a new modularity that we've never seen before.
Well, when I say a new modularity, it's it's a it Well, yeah, it sort of is. And it's sort No, I'm wrong. I'm wrong. what it is.
I've been dealing very much in modularities when we've just been speaking.
Now, obviously within a modularity, you can have different digits that have the same modularity.
And that's what I think might be going on here. What I mean by that is that it is demonstrably true to me that neither purple, orange or green are these digits because they do not share a modularity.
Now the purple's modularity can absolutely be in either of these and in fact therefore has to be in both. So these are the same modularity and that's fine because if they're one and five they are the same modularity. So we can see they're different digits but they are the same modularity and they they're the same modularity as this one.
So this is a five pair. I don't know what that I'm going to uncolor that one and we're going to come and think about that in just a I should be able to get it. Um right okay one thing we can do here I'm going to get a digit one thing we can do here is we can say that now because these share modularity with this digit and one of them is a one on the path that is the only time one appears on the path. So how could this be a one? It can't be. That would be two ones on the path because every square is on the path. So this is a five.
This these are not these are all these are all 2 three and four in some order.
And this one this one is a five I think.
Yeah. The way we have to Okay, I'm getting better at understanding this.
What we do is we like we say okay what's the cell and this is in row 7 column 9 odd odd. So we we search in the puzzle for odd odd modularity which is given by this one. So we're in this this type of modularity.
Um so it this must also be five. It could have been one. One would have worked but we can't have two ones on the path. So this is this is indigo modularity.
These are both indigo modu Oh. Oh.
Although these are both different digits, they're both indigo modularity from the perspective of now we're now what we're going to try and do is work out what this all means for drawing a path through this grid. That's gray.
Now this is very exciting as well because now I have demonstrated that I have all of the digits 1 to 5 in the squares. And if I have all of the digits 1 to 5 in the squares, they are not now available for my circle digits which are now 6 7 8 9 just by just because they have to be.
And now now we know exactly how long the path is that we're drawing because the path now consists of five fives, four fours. One of these is a four, three threes, two twos, a one and zero nines because nine is nine is off the path. So nine 9ines are off the path, 1 8 because 88s are off the path.
Two sevens because seven sevens are off the path and three sixes because six sixes are off the path. So we're adding three to we're adding six to 15 which is the triangle number for five. So, so the square digits are telling us they they account for 15 of the length of the path and the the gray digits are giving are adding in another six of digits in the form of high digits that that are necessarily on the path. So the length of the path is 21 cells.
Now, now this is this I mean this is a crazy puzzle. I mean it's crazy that you have to think about motorway lanes to solve a Sudoku puzzle.
But how do I actually know where the path starts?
And all right, here's a different question.
How do I put five fives on the path?
Because remember, whenever I I have to put five fives on the path, but whenever I do put a five on the path, it has to be in a weird latis. It's got to be there are loads of cells that it can't be in now because it has to be if we if we read out the row column, it's got to be two odd digits.
um because it's always jumping to to to rows and columns with the same parity as this corner digit.
So it see it can't be here that you know in theory that could have been but by Sudoku it can't be a second five.
So the digit the digits that are possible these can't be because they've already had a five.
That can't be that can be.
We can't well we know there's one in row five but there can't be a second one.
There can't be another one in row. So this is one by the way. That's what we just learned there.
Yeah. So look at that. The only places in the grid, the only cells in the grid that have odd odd sort of indexing if you like, i.e. an odd row number and an odd column number are those that I've just highlighted.
Two of these have to be fives. Well, they can't both be five. So, that is a five. Moreover, it's a five on the path.
Now, that is next to dark green. So, does that mean that we can assume that those connect.
Cuz if they did, I could always say dark green.
I don't know.
And one of these has to be a four. And it has to appear four times on the path.
I still don't understand how to do this.
You know, how are you ever going to know?
I don't understand how you know where the path starts.
Do you have to sort of just assume I don't let me let me just I'm just going to sketch out a path from here.
I'm just going to imagine that this is the start of the path.
And I am going to Are we going to go one, two?
You see, I can I can get very efficiently to that one. Just like that.
That would work, wouldn't it? Then I could go diagonal down. I could go want to get to that one. I've got to go diagonally next.
I could do that. That that is that a problem?
Uh don't think so.
I know that's on I know both of those are on the path. I've got to go I've got to go up. Oh, that that's actually really easy. From there to there.
I've got I've now got to do a diagonal.
Do that. Then I've got to do an up or down. I don't want to I want to do a diagonal. Um Okay. I I'll do that.
Then I'll do a diagonal down. Diagonal down.
Okay, so that is Oh, this doesn't work by the way. This is wrong. Um because I have not got enough fives on my path.
But how long is this? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19. So that is 21. That's that's the length.
Ah. All right. Okay. So, that that at least sheds some light on how how we're meant to know what's going on here because what that has told us is that there are it's very difficult to do this in 21 cells because I did that really I'd really tried to be efficient in doing that and I got I got it done, but I didn't cuz because I haven't picked up one of those cells. So maybe that maybe we can do something different there. So I need to do a diagonal. So I could Yeah, I could do that, couldn't I?
I don't think that's cost me too many. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 18 19.
See, that's now 21.
But how would you know whether this was the path?
That is an interesting question. I don't know the answer to it.
I mean that okay may maybe this is how you answer it. Maybe what you say is that whatever the path is meant to look like it we can I think agree that it's got to be exceptionally efficient because that path I've just drawn there is is uber efficient in the sense that I tried to go from box to box as as efficiently as I could and I just managed to do Two, three, four.
One, two, three, four, five, six, seven, eight.
So, uh, I don't know. I don't know how to I don't know how to justify this except that this I mean, this could be right, but that's not the same as it being forced to be correct.
So, I mean the other the interesting thing about this as well is how could the path start or end anywhere but these two cells?
I mean, if we say it starts there and it's got to go over there, this is introducing a whole heap of extra cells that's going to take the total way over 21.
Is it something to do with knight's moves? Because we did notice early on, didn't we? that in like I didn't do this for the rows when we looked at that but when we looked at the columns we either ended up just moving a sideways movement or we ended after two moves or we ended up doing a knight's move like this and I'm just seeing those are a knight's move but these are a knight's move but not in the right way cuz you can't get from there to there in two. It It doesn't work.
It's got to be sort of a a vertical knight's move because because of the need to have a vertical component.
So you can see in going diagonally I can go here or here. I can't go horizontally to complete a horizontal knight's move in this direction. So this to this is always going to take an extra step.
Yeah. Okay. Here is Okay. Here is a different point. Look at this cell and think about what would happen if the next move we had to make from here was vertical.
Well, then you're in a right pickle. So, let's look at this cell. And if if if instead of doing this diagonal move to row 8, column 3, let's say I had to do a vertical move, how many would it take me to get to here?
1 2 It's hor It's horrible. You can't do it cuz now I've got to do another vertical.
Three, four, and then I could close it.
So, five.
Five is the minimum. I think I'm just going to try that a different way. If I go up, it's easier. One, two. It's not the sort of counting is wrong. It's like those blueprints morai games.
So, you can do it in five or three if you go diagonally.
Well, we haven't got as many degrees of freedom. So, okay. All right. So that might be interesting in terms of in terms of okay how you get from there to there then because you could see that to get from the to get between these two there are variety of possibilities if if we'd have been going if we'd come into this one after a diagonal we could have gone vertical diagonal but we now know I think that's not what we do because if we had gone vertical diagonal the next step from here would be vertical and then we would take too many cells to get to this cell.
So I am now going to allege that this at least that portion of the path is correct and okay and that means this portion of the path is correct because what I've got to be doing is arriving here arriving into this cell from a vertical and and I'm going to suggest and I think it must also be true to Hey, that given I've got these 21 cells of of path to cater for, I can't be I can't be going over here. That that's just so inefficient. How am I going to get this one involved? I've got to be as efficient as I can here to do this in 21. So, I'm now going to claim that two vertical knights moves there is rather beautiful. that allows me to have the correct count arriving in this purple one where I don't have to do a vertical out of it. I can do a horizontal and get to this within three cells. So I think I do not think it is it is possible to change anything in that. Um well, okay, it would have been poss and we saw it was possible to change how we went from purple to orange, but it's not possible if you take into account that we need an extra five on the path. So, we have to pick up one of these cells. Now, if I was to go from here to try and pick up this cell, I mean, I'm it's going to be so inefficient. I'm trying to get to that one. I'm going the wrong way. That that's just not going to be right. I haven't got this much latitude to play with. So this this is a five. This is a five. This is a one.
And this is lovely. This is lovely because now I think I can do some modularity coloring because uh I'm not going to I'm not going to assume this green because I don't I haven't investigated this. I don't know how I don't know this side of the diagram whether it's forced or not but I'm going to claim that I'm absolutely going to claim that in getting from here to here in fact in fact I'm now going to claim this is the start of the path as well because you can't go from anywhere else it's just it would be a nonsense so I think you have to to get from here to here I I've done it as efficiently as it's humanly possible to do it.
And I'm still going to struggle to get this done in 21. Um, let alone taking into account I've got to get all these modularities right. But the exciting thing is that orange is after indigo. So I can now color that.
I every digit after indigo modularity is is orange. now. So, this is going to be helpful.
Oh.
Yeah. No. Okay, that's it. Oh, hang on.
No, I can't I can't do that cuz I don't know how I get from there to there.
I can Okay, orange. Orange lives on up here.
Well, right. And And purple is now there. That's fair. Purple is there.
That's fair. So that the green modularity by sort of a process of elimination is in these cells because we know that these modularities were all different. And after purple we always get an or oh but I don't I don't know.
Well, no I don't I've got to Okay, I'm going to delete this. I don't trust it because um in fact I'm going to delete all of this. That was just a sketch in to see what was possible. Um, but look from here. I've got to do a diagonal. I could go there or there.
And then I've got to do a straight.
I don't I don't know how to do that.
Okay. Let's also uh now the other Oh, hang on. Yeah, I was wondering about this.
Oh, no. No. This is lovely. This is This is so important. Right. Look at Look at those two now. This is sick, right?
These are not the these are the same modularity as this digit, but they are not this digit. So So what does that mean? Well, it means if this was a two, these would both be six.
Now six is appearing three times on the path. So that's possible. If this is a three, these are both seven, which seven is appearing twice on the path. So that's possible. But if this is four, these are both eight cuz they have to have the same modularity, but they can't be four. But eight only appears once on the path. So that is not four. It's not okay. It wasn't as good as I hoped, but it was still something.
So these digits are either six or seven.
That's extraordinary, isn't it? Um, okay. Let's try Sudoku on fives. No, that's not going to work.
Let's No.
Yes, I can get that one. That is a five.
Now, that five. Oh, that does give me that five as well. Okay. Okay. So, I can do a bit with fives.
No, I can do better. Actually, I can do better. I can get all the fives. I'm not sure what that means.
Okay.
Okay. That but we know these aren't on the path, don't we?
They can't be on the path. Um because we've got our fives on the path in the correct modularity positions.
So maybe that means this is the path then.
Yeah, cuz a horizontal move would be woefully inefficient, wouldn't it? So the path's going there. That's a vertical. So then we have to do a diagonal. Now we need to do a vertical, not into a five. So I'm going to go that way. Now we Oh. Oh, I see. I could go into that one.
In fact, that's rather beautiful. Okay.
So let's this is the end of the path and it ends on an indigo. So before indigos you get you get a green. If we work just work back on the path here then you get a purple then you get an orange and then you get another indigo which has to be this one doesn't it? Cuz we know that's on the path. We have to be efficient.
And then before the indigo we get a green. So that's going to be that one.
And then before the green we get a purple.
And we're doing from here we're doing a diagonal move. So it's there or there.
I don't know which of those is is likely. O but this digit is now that is the same modularity as this 2 3 or 4 but it's the high variant isn't it? So that's 6 7 or 8.
So this is 2 three or4 that that's lovely. So that's 6 7 or 8. And this one has two of that type on it. So these can't be eight because eight only appears once on the path. So this is not four. And now I've got a two three pair in purple and green, which means that orange is four. And if orange is four, these ones are all four and eight.
And now eight has appeared on the path in column 2. So that is a four because eight only appears once on the path. But four has to appear again on the path.
There's a four in one of those cells.
Maybe this is Well, it would have to be that one by modularity cuz it can't the four can't appear on the path in this position for all for for our motorway reasons.
Now, this is not a four.
This is a two or a three.
I'm very tempted to think that this is a four on the path before green. We need a purple.
So the purple is going to be here or here.
And then we need an orange. So it's going to be there, isn't it? It's going to be there and then a diagonal.
Okay. Okay. I mean I think we can prove this if if the path Yeah. I mean if this if if that if that is 21 long given we need another four. It would have to be here cuz we we we've got no other way of getting to it in the correct positions.
So 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19. I've got two cells left.
One of them has to be the special four, which can only be here. And the only way of getting to that is to do a diagonal to there. This is I mean, how on earth has this been set?
That's now orange. This is now purple.
So, these these are from We can just count now, can't we?
We must be able to count. We we we need one of these to be on the line three times.
This one that is a low digit that is on the line three times. So it's not two. So that's three, which means green is seven.
It means these it means well this is two. one of these. I don't know which way round, but there are two six pair and these are both six. I mean, how how clever is this? Let let me just ask you that question. How clever is this puzzle? And if your answer is off the charts clever, that is correct.
Four makes this eight, this four. Now, where's four in column? for is there.
I I don't really want to color anything else. I I forgive me for that, but I really don't because I I think I'll just get confused. Four.
Four is there by Sudoku. Four is there by Sudoku. Four is here by Sudoku. All your fours are belong to us. Now, I haven't thought through like sixes and eights. So, that's not six. That's not six. That's not seven or eight.
Okay, I can't do anything with that's a three by Sudoku for what that's worth, which is quite a lot. That's a three by Sudoku.
Please, I will be so thrilled if I can solve this. And I know that probably sounds weird. I just How long am I? I've been going an hour and a quarter.
Come on.
Okay, in this column I need 179 79.
Um, so it's amazing cuz we have got two sevens, we've got one eight, we've got three sixes. It's the way it's worked and it's balanced and this is 21 long. I mean, it's sort of magic. It's sort of absolute magic now. Okay, this one is not six.
I still got to do this 2 six. I don't know how we're going to get the ordering of that done either.
Seven by Sedoku is there. So, that's not seven.
Let's double click. So, the Well, we don't know very much about sevens at all. We just got a small win for the goodies there.
Phone is buzzing. Let's uh not worry about that. Got bigger things, bigger fish to fry. Ah, okay. In this box, I've got two, six, and nine to place, but this sees a 69 pair somewhere there. So, that's got to be a two.
Ah, so in this column, I haven't put a one. So, I'll put a one down there. So this becomes 7 9. That becomes eight then. So that must be six. That's nine.
That's seven. So we've fixed some of this column nonsense. That's a one at the top of the grid. This now is not six or seven. So it's an 8 n pair. This digits very achievable. That's a six.
So okay, those digits are 289.
um probably does something.
Um nine in this box is placed. So that's nine. That's eight.
It's so it's so spectacularly Oh, whoopsie. Whoopsie.
It's so brilliant, isn't it? Now, what about what about that column then? One, two, three to place.
One, two, three. One, two, two, three bobbins. That's what we say about that.
That was not as good as I was hoping.
Um, okay. Let's try this row then. Eight and nine to place.
So, these are these are the old bus I used to get from Beckenham Junction Station. They're a 126 triple. That's got to be a one by Sudoku.
126. It was never there on time. Well, go back to the Swany where palagra makes you scrawny and the honeysuckle clutters up the vine.
I really am a fixing to go home and start a mixing down below that Mason Dixon line.
Um, Pax, how I love you. How I love you, my dear old Pax.
Sorry, but you can't you can't go wrong with a bit of Tom Lara. Um ah can't do it. Please, please, please don't be anything too horrendous.
Maybe we've got to color twos and sixes or something.
Possibly true. What about eight? Eight is in one of those two cells.
What what sort of where do we think that the digit is that's going to help me here? Nine is in one of those in this row. I need 1 2 3 and eight.
Okay, so that is a 1, two, or a three.
That does see eight up here, but that could be eight.
Is there some rule I've forgotten about counting how many thingies appear on the thingies?
Might be six. Oh, near that was nearly thought.
Six is in two positions in Okay, so six I can't get it. But six six in column um thingy column seven is restricted now.
I don't think I can get it anyway. What about six? Let's see if that was a six.
That would be a six. That would be splendid.
Splendid. Milk. 1379.
So this digit is a one or a seven.
Believe it or not, sees three and nine in the column.
That digit 1 three. That's a seven or a nine.
So that's almost interest. Okay, here is a totally weird question. Where's three in column 9?
I've just if we actually scan down there's a three here so it's not there's a three there so it's not there so there's a three in the corner and that's three in the corner that's three in the spotlight losing its religion that's actually doing work you know that is doing work this six makes that the two so the two and the six are resolved now which means that is the six that we were looking for that is a six so this is a nine in the corner of the puzzle That's a six. I feel like we just did all the sixes in the last couple of seconds. This is a 28 pair.
Unresolved.
Never mind. Um, okay. What about the bottom row then? 7 8 9 into those cells. That's not nine.
Okay, that didn't quite work, did it?
What about Okay, let's try C. Oh, there's a nine here. That's good. So, 8 and 9 can go in. That makes this a seven. Makes this a one.
So, let's let's go back to this column.
278 into this cell. It's not two. So, that's seven or eight. The two in the column goes here. 28.
This is now a nine. That should be a something. Technical term for a three.
Oh, threes have been done, I think, forever. Oh, yeah. Threes. Threes have all been done. Could have done them ages ago. Didn't didn't notice it. Nines then.
Yeah. Yeah. Yeah. There we go.
Two in that row. So that's two and one done. That becomes a one. I need here.
We need two and eight. Two and eight.
Eight goes here. We need a seven up there. And we need a something.
Technical term for a two. Okay. This is a one. That's a seven.
Yeah, we've done it. We've done it. 8 7.
You can't. That cannot be by accident.
That's one of the great puzzles that I have ever solved in my life. On and off by Cactus Love. 1 hour and 22 minutes of sheer unbridled unaloyed joy at the hands of a complete Sudoku genius. Cactus Love, take a bow.
I That's mad. That is that is mad. That is a mad a mad puzzle that required motorwayesque solving in order to under to be able to understand it. But isn't it beautiful? It's absolutely amazing.
Isn't it amazing that that rule set yields that path and it all somehow works? I think it's amazing. That's an incredible one.
Absolutely incredible. 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.
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