In Arrow Sudoku, where digits along arrows sum to the digit in the arrow's circle and double arrows connect two circles with equal sums, algebraic reasoning can solve complex puzzles. By assigning variables to unknown digits and using the fundamental Sudoku rule that every row, column, and 3x3 box must sum to 45 (the sum of digits 1-9), solvers can deduce that certain pairs of circles must add to 15. This algebraic approach transforms visual puzzles into mathematical problems, allowing solvers to systematically eliminate possibilities and find the solution through logical deduction rather than trial and error.
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Added:[music] >> Hello, welcome back to cracking the crypt tech and thank you so much for joining us and uh and for respecting the effort we go to to bring you two full-length Sudoku videos every day, minute cryptic, wordle in a minute, cryptic crossword masterclasses on Fridays, streaming most weeks, um we are just trying to bring you loads of extra content. One of the ways we do that is by having Sudoku hunts on Patreon and um crossword videos, some very hard cryptics there like Sabres Listener, some um gridogram and connections videos, loads of content.
We've also, of course, got our books. You can find them in links under the video. We've got um our apps.
They do feature arrow Sudoku and line Sudoku, which are relevant to today.
They're great fun. Uh there's also um the worms amongst the apps and they're clever.
So, check it all out. I mean, there's a load going on. We've got a bit of merch as well in case you wanted that, but um we really appreciate the support we get from the people who follow the channel and we cannot believe 700,000 of you have subscribed.
Um that's very good of you and not unsubscribed those 700,000. Incredible.
Anyway, um we got an email in from Andreas who has been looking at this puzzle on the screen and worrying that he can't get a break and he hasn't got a single digit after after quite a lot of work. Now, this is arrows and double arrows. What could possibly go wrong? Well, I I get in the same trouble as Andreas, quite clearly.
Let's do the rules. You can try it on the first link under the video. I'm going to go through the rules. Normal Sudoku rules apply. So, 1 to 9 will go in every row, every column, and every 3 by 3 box. Digits along an arrow sum to the digit in that arrow's circle.
Where a double arrow line connects two circles, the sum of the digits on the line is equal to the sum of the two circles.
Um so, those three will equal the sum of those two. This one will equal the sum of those two. These two will equal the sum of those two, etc., etc. Give it a try. I don't know what we'll find as we immerse ourselves in this.
Let's get cracking.
Um >> [clears throat] >> It feels like row one and nine ought to be helpful. In each of these, the pattern is the same.
An arrow, a circle.
Right. How How do we sort of Let's call this arrow I think I want to reduce it to algebra, cuz otherwise I'm not going to understand what's going on in the row. Let's call this A, this arrow total. And that circle must be A as well.
Let's call this B, and this circle must be B as well.
So, I think we've got two A there and two B there.
What's more, this group of cells must be A plus B.
So, the whole row is now 3 A plus 3 B.
I think that's right. It makes sense, because if you divide that through by three, then the whole row is three times a plus b. Now, we do know the sum of every row, column, and box in a Sudoku. It's always the secret number, 45, because that's the sum of the digits 1 to 9.
And that means if three lots of a plus b is 45, then a plus b is 15.
And these two circles must add up to 15.
And the pattern is, as I said, repeated in the bottom row.
So, this pair adds up to 15. This pair adds up to 15. This trio adds up to 15.
And these cells between them add up to 15.
And obviously that's 15 pairs represented in the circles either by 6 plus 9 or by 7 plus 8 in each row.
Now, having said that, where am I then?
Ah, this gets interesting.
Because this digit is the sum of that plus that. So, this can't be nine anymore, and that means this can't be six.
And this must be one, two, or three to allow this total to be seven, eight, or nine.
Oh, and that is the sum of these two on its arrow. Oh, this is very weird how the circles and arrow intersect. So, this can't be a one anymore, and these are either a one-one pair adding up to two, which is legal, or a one-two pair adding up to three.
Now, if that has been restricted to two or three, this can't be seven anymore, and this can't be eight anymore, I don't think. Eight plus at least two gives 10.
So, that has gone down to six or seven, and this is eight or nine.
Now, I'm looking at that and thinking it's not going to be wildly helpful.
One of these digits, at least, is a one, and they both see this arrow.
So, this arrow has to be at least 2 + 3, and this circle has to be at least five.
Yeah, I mean, it's not It's not a wonderful deduction, but nonetheless, [clears throat] I like truths.
Now, what about this column? Have we got anything interesting going? We've got a pair of sixes or sevens up there adding to 12 or 14.
That means the rest of the column adds up to 33 or 31.
And that means this pair, which represents a little less than half the column, can't be more than 15, and that this digit is odd.
Going to mark that just in case it helps later.
This pair adds up to 15 at a maximum.
And that was 6 7 8 or 9. Uh, does that doesn't really limit this, I don't think. Well, this is limited a bit by this combination. This can't be eight or nine because the total of these two circles is here.
So, now this is at least seven on the circle.
And these two add up to 15 at the most.
What is the least they might add up to?
11?
That would give us 22. Let's say that was a six doubled.
34, no, that doesn't limit this cell at all.
Five, six, or seven though now we've limited this to and it doesn't have a one in.
Oh, it's so nearly has to have a two.
It doesn't quite, but if it did have to have a two, then we could limit both of these cells to ones and this digit would become a two.
Okay, the the alternative that doesn't say that there has to be a two here is this being a three-four pair that add to seven.
Then that digit becomes six.
So then they add up to 12. This lot adds up to 33.
The six Oh, I've just realized these could have added up to nine to I got the maths wrong cuz I didn't allow for 12, 33, one there, 32 here. These could add up to 16. I don't think it has changed anything about how we've limited them, but they they add up to somewhere between 11 and 16.
That that's that's by the by, but let's let's clear up something I might have misspoke about.
Now, >> [snorts] >> if this is a three-four pair and this is a seven, this must be a nine and this must be two. Aha!
If this is a three-four pair, this is a seven, which makes this a nine and makes this a two. And these become two ones by that means. Whereas if there's a two on this, which is the alternative, then these are two ones. So, I think they have to be two ones either way around, whether that's a three four because of these or whether it has a two.
So, that digit is a two in the circle.
That doesn't change any of these, but at least we've got a digit. I feel good about that. So, the two in the top row is either here in a pair adding up to six or seven or here in a pair adding up to eight or nine.
That's not very interesting.
This digit is now not one, although it is odd.
I suppose one way to look at this column is that these three digits multiplied by two plus orange are 45.
So, if orange is a three, these three digits add up to 21.
I mean, this is all true, but it doesn't doesn't help at all.
Oh, let's have a look here. These are at least two three making this digit at least five.
Right, I'm not going above seven in my candidates there because that plus this equal this.
And that can't be a one.
So, this is now two three or four making this one seven eight or nine.
Now, if this was Let's try it again with twos and threes. If this was a two three pair with a five here, that would have to be a four. Ah, this this blue cell is effectively a total of these green ones because you add those two to get that, then you add these two to get that. So these three add up to that and they don't include a one. So they must be two, three, and four. This is lovely and that blue is a nine. Isn't that fascinating?
And this is still five, six, or seven.
Literally it hasn't limited that at all, but we found a nine.
A very unexpected nine.
These are from five, six, seven, eight.
I mean, I know I know it's excessive pencil marking. I can't help it.
Uh these are two, three, four triple.
No, not necessarily from the point of view that this could just be a two, five pair.
If this isn't a two, five pair, these are a two, three, four triple in the row.
These circles aren't really considered.
We're obviously adding up somewhere between eight and 13.
Right, let's think again very briefly. I don't think it's going to do anything, but I'm going to think again about this being a three, four pair cuz that makes this seven and this nine and this eight and this six.
And this two.
Nah, it doesn't doesn't seem to lead to anything very obvious after that.
We've got a one by Sudoku in one of these three cells in row six.
I do fancy it could be on this line. I don't know. No, no, there's no evidence at all for that.
It's very interesting challenge this puzzle.
This digit is at least two, so this one is at most a five.
If that was a six, these cells would be 1 5 and 2 4 in some order.
Then these would all be 3 7 8 or 9, and that would have to be a three on this arrow.
If that was a six, this would have to be a three.
But then this could be 2 3 or 4 leading to any of these possible totals, although not six again here. So, three and two or four and five or seven.
If that was a seven, all of these digits have to be under seven.
Right.
Where is this digit in box one? I think it's got to be here.
It can't be on this arrow because it's eight or nine, and that would bust the arrow. It can't be on this arrow because that would bust this arrow. So, it must be there.
These two are the same.
They're probably here, but it could in theory red could be an eight with an eight-one pair here adding up to nine.
Then that would be 6 2 4 3 5 7.
I'm trying to remember that these two circles add up to 15.
Oh, maybe I maybe I should think about that.
I don't know. A bit more, maybe.
If that was seven, what did we say this We didn't say this would be a three.
This would be six or five in that case.
And this would be 1 3 or 5. And it would be from the pair that can add to seven that wouldn't be on that arrow or that arrow.
So, the other bit of that pair would be here. Indeed, the other eight or nine is in one of these cells in the box.
Where's the other eight or nine in this row? Well, we described the possibility of eight one being here, and that does as far as I can see remain possible.
Otherwise, the other eight or nine that isn't red is in this group.
And they would have to add up to 15 without a two. So, if it was nine, it would be 915.
If it was eight, >> [sighs] [gasps] >> not 816 because if eight was here, that would be nine and that would be six.
So, if it was eight, it would be 834.
Now, what happens if this is 918?
That's 642, and this is 357. So, there These are limited. They They're either 357, 915, or eight.
something.
843 357 348 159. I mean that's a weird set weird grouping of possibilities. I I don't think I'm getting anywhere with that. Um right, I'm going to have to think of something else to do. Oh, this digit Oh, no. I was going to say it can't be a six cuz it's got three arrows. It doesn't.
It has two arrows and a double arrow.
So, if that was a six Oh, that would be very interesting in the column cuz that would force this to be seven and this to be five.
They would add up to 18.
This part of the column would therefore add up to 36.
And this would be a nine.
Well, that's not impossible.
So, I can't rule out a six from here. If it was a seven, it would be a very similar situation. 7 6 and 5 would add up to 18. Doubled is 36. That would be a nine.
So, it changes once this can be an eight.
If it was a six, where I was actually starting with that is these two would these these cells would be 1 2 4 5.
Six and five would be there. This would have to add up to 11 without using 1 2 4 5 or 6 or 7 in these cells.
It couldn't do it.
Three and eight are the minimum. Right, let me just explain that again. If this was six, you'd get seven here and five here. So, six and five would make this add up to 11.
And the six there would force these digits to be 1 2 4 5.
So these could not now be 1 2 4 5 6 or 7.
You'd have to get to 11 with three cells of which these two could be three and eight at a minimum. That can't work. So this isn't six. That's taken quite a while, but there we go.
I think seven Well, once that's not six, this isn't nine cuz these two at the bottom add up to 15.
Now, can this Can this be a six?
Yeah, I mean, I I think it can. I don't really want to get into what goes on if it isn't.
This Every every cell that emerges from a circle looks at first sight like an arrow to me, and some of them aren't.
Okay, if that's a five for this digit not to be two there has to be no two on this line. It has to be a three four line with a seven here.
And then two Sorry, two would be here.
So one of these two is definitely a two.
Okay, if that's three four, this is seven nine eight six.
Seven six eight would lead to a three here. Seven six nine just doing the maths would lead to a one there. That's not possible.
So if this if this is a six seven pair this can't be nine because we'd get a one here. So, if this is a 6-7 pair, this digit has to be eight.
And this is a 6-7 pair unless we put five here.
If that's a 6-7 pair with eight here, we get a three here.
>> [snorts] >> That's quite interesting. Um Ah, this is Ah, it's hard to think about all the possibilities at once. It really is.
And maybe maybe that's not what I should be doing.
Okay, if this is a five, cuz I know what to do if it's not. So, if this is a five, we've got two, three, four.
Now, what else happens then?
Five on this arrow could be three and two or one four.
I'd love it if something goes wrong with five, two, three. Five in this column.
>> [snorts] >> If that was six, this would have to be four, too. We'd have 6-4-2-5.
One would be on this line.
Uh, I don't know. It's doesn't completely unwind. I mean, there are clearly powerful forces at play, but I don't see the resolution straight away. So, let's let's just keep on keep on keeping on with what we're thinking about.
These two add up to five, six, or seven.
Five there, which leads to two, three, and makes this a definite seven.
>> [sighs] >> That pushes seven out of these cells.
[snorts] Okay, let's try and rule out this being a six-seven pair then, because that would make this a five and get us to start filling in those digits. That might be even better. Right, if this is a six-seven pair, this has to be an eight cuz it can't be seven anymore. And if it was nine, the arrow and circle cells in the column would leave that being a one. So, this would be eight, and this would be three.
We'd have six or seven.
Eight and three.
Nine would be in one of these in the column.
It's it's really interesting.
Six or seven here.
That would be eight or nine in its own right.
Now, I cannot quite see how that either breaks down or builds up into something we can use. So, again, I am I have to declare I'm missing the point.
Now, I worked out if that was a six, this has to be a three.
And then this is two or four with a five or seven here.
>> [snorts] [clears throat] >> Does that three do anything else?
I don't think it does.
>> [snorts] >> Six there would force nine here.
Ah, eight in this cell would reduce this to six or seven and make this definitely eight or nine.
So, eight here would make this a seven.
This would be eight or nine.
Ah.
Right, I think this is impossible now.
So, I'm going to fill in these digits to show you why this is impossible.
Let me just I'm going to start with this cell and say what if that was eight and this this I think will fail.
This digit couldn't be eight. It would now be six or seven.
These two have to add to 15 in the row.
So that one is now eight or nine, but this has to be a seven to add to 15 in the top row, the two circles again.
Now these three cells add up to an absolute Oh, bother. I was going to say an absolute minimum of 22.
And I've got that wrong. I've got that wrong. So this is possible.
Indeed, it could well happen.
But if those had an absolute minimum of 22, we'd have an absolute minimum of 44 on those cells and this would have to be a one. So in fact, these add up to less than 22. That is a known fact.
21 or less.
The absolute minimum I thought was It's 18. Oh, it's much smaller than I thought.
Ah, it gets wildly Yeah, it gets really interesting. If these do add up to 18, this is a nine.
Then that can't be a nine and this becomes six or five.
Oh, if they add up to 18, that has to be a five in fact and that's a six and that's a seven.
So if they add up to 18, you've got 5 6 4 2 7 9 and these are 1 3 8 something to 12.
>> Oh my goodness. I mean, I can see how it it might keep going, but I can't I can't solve that.
Anyway, I I I was just speculating on what these circles add up to. And now we know it's between 18 and 21. And unfortunately, that's not quite as helpful as I wanted. So, let's go back to thinking maybe I there was something I spotted and I got it wrong. If that's an eight, that's a seven.
Then this these can't be seven if that occurs.
And to keep down to 21 or less, Wow, it feels so powerful, but it doesn't quite solve the puzzle.
If that's eight and that's seven, there are other issues. Like, where does nine go in the top row? It's got to be in this group with one and five. So, you've got eight, seven, one, five, nine, two, six over this side and three, four here. That would be definite.
So, if that's an eight, we've got seven, three, four here.
Then this pair is either two, five or one, six.
If this this is worth pursuing.
Definitely worth pursuing.
If that's eight, this is seven.
And we've got three, four there.
Now, if that's seven, the minimum total here is 20.
And that is either five or three as a result of that.
So, with one six there, if that's a five, this is No, that was three four.
If that's a five, this is one six.
If it's a three, it doesn't narrow this down at all.
That has surprised me.
Oh, dear.
Okay, let's go back to thinking of Oh, if that were Oh, no, hang on. Hang on.
Ah.
Right. Here is why this digit is not eight.
If it was eight, where would red be in row one?
Red is there and there.
It would have to be in one of these three cells. It couldn't be nine, so it would have to be eight, and it would have to be here.
But, if eight is red and there, that makes this a six on this arrow.
And then eight plus six does not reach the requisite total. So, this is not eight. This is a nine. There we go.
That's how to get to that number.
Doesn't resolve whether it's red.
But, it does mean this digit is a six, and that is worth having.
This is now not a six. Oh, now these are one two four five in some order, and this digit is definitely a three, and this one is now two or four. This is the sort of go forward we wanted. These are from 7 8 9 and form a triple with that.
And therefore there's a two. Uh no, not necessarily. That's not two.
I was going to say two in the row had to be here, but of course it could be here.
This pair can't be 6 3 now.
If that's 1 5, the two in the top row has to be here and it has to be 2 7.
>> [clears throat] >> Otherwise, this is 2 4.
And then the only even digit available to contribute here would be eight. So, this is either 2 7 or 1 8 up here.
Three must be in this group of cells adding to 15 along with a pair that is either 8 4 or 5 7.
And then there's also a three in these cells.
Oh, this has become eight on this arrow.
Double arrow. That's an eight. That's the red digit. That's an eight. Now one of these two is an eight. They're an 8 1 pair. Now this is not 1 5. That's a 2 4 pair. This is a 1 5 pair looking down here. That digit becomes a seven as a result of that.
This one is not a seven now.
Now, if we add these three, 21.
If we made that a nine, these three circles would be 22. This whole group would be 44 and that would be a one. So, we're going to make this an eight and this a three.
And that eight forces this to be a seven cuz these two circles in the bottom row add up to 15. Seven is now ruled out of couple of cells in box six and is now ending up there or there.
This is either 2 5 or 3 4.
Which means that two is used up in this group.
Yeah, it does. It doesn't That's interesting. Right, this is either 2 5 or 3 4 as well.
This group is 2 4 9.
This one can't be nine. Can't be one either. So, that's two or four.
>> [sighs] >> Um hmm.
There is a three in one of these cells.
If it's in one of those two, one of these has to be five.
Oh, two and seven there. We can finish that off with a nine.
Now there's a nine in one of these two cells by Sudoku.
That's it. That's what we want. We want more Sudoku.
Okay, I'm going to remove the red and the orange colors there. They're all kind of jobs done.
Nice jobs done. Look, there's a 2 4 pair. Oh, in fact, we can just add this up. That's a four. That's a two. We're left with 4 9 down here. That's annoying cuz neither of those digits can contribute to one of the 3 8 sums. So, one of them is there, one of them is there, and this pair must also add to eight. Actually, that means five must be on the left side of this box. Whichever pair it's going with with three must be over here. Oh, that four looks up to the top and does 2 4. Now, two has to be in one of these cells. And similar to two and to three and five, six must now be on the right side of this box.
Then one and seven will be the other pair. Here we have no 8. That's a 7 9 pair.
These are now 3 5 7.
8 2 9 has to be here. Look at that 7 9 pair.
And remaining digits in the box are 1 4 and 6.
That can't be 4 or 3 and it is 2 therefore in this row.
And this now can't be 7. So, it's 5 or 6 and this is the only place for 7 in the box.
That's a strange discovery.
2 7 1 9 3 4 I'm looking at row 4.
Everything else is 5 6 or 8.
Now, we've got a pair here. They could be 2 7 or 4 5. I've got no information. Right.
7 and 7 [snorts] is going to leave 7 definitely here by Sudoku. I've just spotted these two 7s. So, that becomes a 2.
And that does nothing else. It does do something else. It definitely puts two into one of these cells vertically. So, that's a 2 I don't know what it is cuz I don't know what we're meant to be adding up to.
I've got I've got this wrong. I've I've pencil marked these as though they are an arrow from the 7 and that's silly.
Sorry about that. I don't think that what I've done affects anything else.
That 8 got generated and created that 7.
What we've got here instead is a double arrow that adds to 10 or 11. This pair adds to 10 or 11.
And it's got a 2 in it now.
It can't be 2 9. So, it must be 2 8 adding to 10. That's a 3.
This is a 4. We add up the arrows.
That's a six.
This is now five cuz it's being looked at by the two eight pair. That's an eight. I hope I haven't done anything wrong here. I don't think I have. That four says this is a two five arrow.
Now the three in the row is there. This is a six nine pair. That's going to make this cell an eight and this one a six.
This pair is a one five pair.
And these are four seven eight in some order.
Now in this column we've got four and six to go. So now we can place three in the box.
And then fill in this as a five. And that sorts out the one five over here.
Stops this being a one and that is the place for one in box two.
Now three is ruled out. Five is ruled out.
Excellent.
Oh, this is great. Right. Eight four sort out one eight.
These digits are one five and nine.
These ones are three four and six. How do we make up the seven arrow? Not with three or four there if it can't go here.
So it's six one.
Now in the bottom row, this eight arrow can't be seven one or six two and must be three five.
These are two four nine which do indeed add to the correct total of 15.
And this arrow doesn't have a two on it.
And can it be three five or one seven?
Yes, I think it can.
This one is either two six or again 1 7.
>> [clears throat] >> Okay, something vertical must be helping. There's a 3 7 pair.
So, everything else in the column is 4 or 8, and this one can't be 4, so that's 8. This is 4, the central box is finished.
That 4 is dealing with the 6 4 pair at the top.
That can't be 7 anymore.
This column needs a 1 3 or 5. This column needs 5 6 or 7.
That 8 is looking at this cell, so that's 2 and 8. Now, we get a 2 here, and that means this is a 6.
And that gives us the 9 6 unwind and the 9 7 unwind.
And that places 7 down here, and the last digit, 1.
5 1 look right there now.
Something looking across, 7 6 looking across to this cell gives us 5 there. We can probably finish this whole row straight away. 4 and 3, then we can finish the row above, 5 4 9.
This is a 7 on the arrow. 3 and 6 still to go. That is a beautiful puzzle by Jo Bow.
5 3 and a 7 to finish.
Fantastic stuff. Very enjoyable, very clever.
And uh thank you, Andreas, for the recommendation.
I I hope you don't think you weren't being clever to struggle with the breaking. I don't even know if you would call that double 1 pair the breaking, but it was certainly very important to me. It got me my first digits. And it that was a complicated little triangle involving those, wasn't it? Lovely puzzle. I've really enjoyed that. Thank you so much for watching us on the channel. We love to have a bit of your time and we'd be thrilled if you'd join us again tomorrow. Thank you so much. Bye for now.
>> [music]
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