The Pips puzzle from the New York Times requires placing dominoes to satisfy mathematical equivalences across regions, where the key strategy involves analyzing digit distribution constraints and placing dominoes that connect regions without breaking numerical equivalences, with the most challenging puzzles requiring careful tracking of available digits and their placement in equivalence regions.
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Pips - NYT Domino Puzzle Game - 10 May 2026Added:
Hello. Let's solve pips. We've got two by Ian, one by Rulfo. Today, we'll start with the easy puzzle as usual. Let's place some dominoes. Okay, we need 10, which will be uh Okay, we don't have two five. So, this is 64. So, we have double six. So, this one needs to avoid the four. So, that goes there. Um, now this can't be 4 three because we have no one to finish the four. So, this is 4 0, which means four goes here. Uh, two goes here to be less than 10. And that's that. Let's move on to medium. Okay, we need a four connecting to something that doesn't break four. We only have one four, so that goes there. Now, we need a two connecting to something that doesn't break this larger four region, [clears throat] but we have less than one, which is obviously a zero, but that goes into a void. Oh, and so does this.
So, two of our three zeros uh are are spoken for.
And this one connects to something we have two of. So, it's I know it could be 01, could be 02, could not be 06. So, this is not 06. That's the only thing we know. so far. Actually, you know what? I think this probably goes here because if it doesn't go here, where else? Oops.
Sorry. What I meant was this.
If that doesn't go there, where else could it go?
The six needs to be put somewhere. Oh, I guess never mind. I guess actually it could go it could go here, for instance.
H. [clears throat] Okay, that's possible.
But actually, no. If we put that there, where does the five go then? Because the five can't connect to either of these less thans.
Yeah, there's nowhere to put the five.
We don't have we can't double it. So, I think this does I think this does go here. And and similarly again, where do we put the five? I mean, I think it needs to go here, but it could go probably in one of two directions.
No, it can't go here because we don't have a 03. So, it does go here. So, now the rest of this four needs to be a zero and a one. We have both possibilities for this two.
Then we need another zero or one pointing into a four.
Can we make the four out of two twos? I don't think we can because then we don't have a double for this. So, I think the four is going to be 3 1. Uh, we can't do that. So, this goes here, which means this is the two, this is a one, this goes here. Great. Okay, let's try the hard.
And this yet again, as I've continually been doing in recent days, I've left the timer on overnight. So, I'm going to have an annoyingly long time claimed as my soft time here. Okay, so this is interesting. 1 2 3 4 5 6 Six different inequivalences. Do we have all seven digits? I'm sure we do. 0 1 2 3 4 5 6.
Yeah. So all but one goes in there.
Um and then we have these very large equivalence regions. We have five, five and four. So what do we have five of?
Not zeros.
We have five ones.
Um, six threes, six fours, five fives, and four sixes. is wow. Okay, what on how do we get started with this twos? We have a two here and a two here. How many twos do we have?
Not very many, but still more than two.
We still have three of them.
So, I mean, potentially we put Oops, I need to sneeze. One moment. Okay, the sneeze passed, which means it may return. So, it may be the case that we have twos in those two enumerated cells and then one more in the in equivalence.
How do we How do we figure out what that's going to be?
This double five is interesting.
Obviously, no double anything can go over here. And this is going to break quite a few of our regions. It can't touch a two or a three.
I mean, it probably goes in one of these big ones. Is actually, is there anywhere else it could go other than one of these big regions? That's that's actually the question. Same will apply to double four3 maybe not because that could technically overlap. We'll look at that in a moment.
But it can never overlap any of these because we don't have enough of any digit to fill two of these large regions. The only other place it could go in theory would be here. Do we have enough fives to do that? We would still need four more of them. I forget how many fives we had.
No. So it can't go there. So this double five does go. So five is one of our large regions, but it could be the smaller one. It could just be four of them.
What about I mean I think it's the same with the four. How many fours do we have? I think we have lots of these. 1 2 3 four five.
Yeah, we have six. So that one could that one could even go here. I mean, no, actually it couldn't. It couldn't go here because we would then need a double four to go here. Sorry. And we don't have another one here or here or here.
here. It needs to hinge around around this. So, um so it couldn't go there.
That's fine. Um what about threes? I mean, in theory, threes could do something like that.
Do we have a two three?
We actually don't. So, it couldn't it couldn't technically go here. Maybe actually the three can't spread out anywhere else because it we don't have enough. We don't have seven of them, do we? Two. Three.
four, five, six. Now we have six.
Um, oh no, but this would need seven. Yeah.
Yeah. So this So that's kind of funny actually. These doubles all must go into these regions. Actually, again, they would need to. They would need to if this did overlap here. Even if we had enough, even if we had seven threes, we'd still need a second double three.
And we're we're never going to have that. So yeah. So they're all sort of trapped.
I don't know what to do with that.
We We don't even have very many. We have threes, fours, and fives, and that's it.
But they do all need to go in those regions, and that is three of them.
What does that tell me?
Okay, zeros are interesting because either this three region is all ones or it uses our only zero. And if it uses our only zero, that means we need one of every other digit in the inequivalence region.
So, is it possible for me to demonstrate that the three can't use the zero or that it must?
How many ones do we have? I forget. One, two. We have five ones. I mean, we need at least one one for the three.
Okay, that is useful because we can't Oh, and this two region up here as well.
So, either this two region or this three region or neither of them has a zero. I mean, I know that's obvious, but what I mean is it's it's significant. Any one of those is a significant thing.
But yeah, we do need we do need at least one in the three region because even if we used our zero, we could have a zero and a two and a one or we have three ones.
If this is three ones, how could we do that? What are what are our ones connected to?
And we don't have a double one.
So we could do something like that.
But we could also do this.
Well, no, we couldn't because then we'd need a double, too. Actually, we couldn't do that. So, this would have to be this, I think. Then one of them would have to point out into this region and one of them would have to point up there. We'd also need a double here, which is a constraint. I mean this is difficult to achieve but it might it's you know not claiming it's it isn't possible it's just a little bit difficult right whichever way this points up it needs a thing that doubles which could be could be 14 and then a double here for instance I'm not saying it's that and then we need another double when it where it points over here which could be five or 13. Could be either of those. I think 1 three I think is difficult, isn't it? Do we have enough threes for that?
This could be 43. We don't have a 43. I think I already knew that. No, we do. We do have a 43.
No, we have enough threes. But do we?
Well, we don't in this specific situation cuz then this needs to be six and that doesn't work.
Yeah, the doubles is the doubles is actually very constraining.
I'm wondering if this is actually a problem or if it just happens to be a problem the way I'm doing it because this could also be four five.
We do have double fives. This could be one five. Yeah, I mean this is going to work. I think this then needs a double five somewhere.
It could be here.
This is then five connecting to something else which looks like it actually has to be three which does work because we need another double up here.
But no, but it doesn't work because we we need a double down here in this situation. I think that's the problem. I think that is the problem.
If No, it's not necessarily the problem.
Sorry. Sorry. Sorry. Sorry. I didn't think of No, that's not a problem. Be I was thinking this has to be one two. It only has [snorts] to be one two if it points down. It could be something else.
It could be um 14 pointing over here. And this could be 42.
That's fine.
This could even still be fours over here.
Actually, it couldn't because we don't have another one four. These have to be different. And that's interesting. So, we probably don't want to use something we have doubles of if that's how this works like this.
But this is all speculation. This just means I can't I cannot disprove the thing I was looking. Oh, no. This No, it does need to be something that has a double. Oh, maybe this doesn't work.
Sorry, I didn't see that.
No, no, that doesn't. That's fine. That doesn't m This doesn't matter. This could be four or so, sorry, double four.
This could be one four. That's fine. I don't know. I don't know what why I was thinking that was bad. That's fine in theory.
Then this connects to something that fours have, which could be threes or whatever. I don't know. Maybe there's a reason this one doesn't work. But in theory, these things work. Okay, that's that's too bad. That's too bad.
If Although, if we did that, we would then need to use our zero up here. Maybe that's the point.
Maybe that's the point. Do we not have enough ones to do all of that? Or do we have to to use ones to do all of that?
One, two, one, two, three, four, five. Okay, it would be all of our ones. Let's just again do this as an example. It would be all of our ones if we don't use a zero in the three region or the two region.
I don't know if this is correct. I mean, obvious I'm assuming it's not. I'm just putting it in just to to use things up.
So, then we would need to put both of our ones up in here. I mean, we could put this though. That's fine. Then this would have to be five also. Fine.
Is this correct? This is correct.
Are you kidding? Okay. Well, [laughter] no, it's not. Okay. I'm almost relieved.
Uh, we doubled the six. We doubled the six.
Okay. That's interesting. Does that tell me something?
That is interesting, isn't it?
Where would we put one more six?
Could go in the void or the sixes could be used. The sixes can't be used in the four region because we always need a double there.
So sixes are kind of a problem, aren't they?
That is interesting.
There's definitely a way to think about this where you you consider the quantity of each of these digits that can fill these big regions and you allocate them appropriately.
And I'm not really doing that. I if you in theory, let's say we could just magically swap out this six into a void.
We then have an extra four, which makes me wonder if the four then maybe we know the four would have the same problem the six currently has. In that situation, the difference is that the four has doubles and the six doesn't.
Hm. [clears throat] Okay, I'm running out of time before I have to actually leave. So, how do I figure this out?
Do I think this means I don't know if what that means is that we do need a zero in the three or the two. I'm not really sure if it means that. If we did, I don't know. I'm not sure how to solve this. I'm sorry.
Might have to come back and do this one later.
Okay. What is the zero? The zero is 06.
Okay. So, where could it go?
It couldn't actually go up here. It couldn't be in there. There's nowhere for it to go. Okay. This is what I'm going to start thinking about now.
Yeah. Okay, that makes sense. So, what if it goes here?
That changes the balance of ones. But we know it can't point up here because we need a double six which we don't have.
Is there the same reason it can't point out of that two? This is the only other place this so if the sorry I'm saying if the zero is used in a two the two or the three region it must go here. In which case this must be a 62.
uh which we do have because we used it before. Yes.
Right.
Then another two is used in this three region. How many twos do we have?
Yeah, we have two more. So we use one of them there, one of them up here, and that's all of our twos. Then we need one of everything else in the inequivalence region, which leaves us with a six still. We still have an extra six. And that must go in the void.
And it must be 63 because if it were 61, we don't have a double one.
So that would go there. This would go here.
Now we need two more ones in the two.
And one more one. No, this isn't going to work still.
We're still going to have an extra one, aren't we?
This would have to be two. I feel I've still made a bad deduction because we need one more one up here.
One more one down here. Then we have an extra one. Uh, sorry. I thought I was on to something, but I'm I'm not.
What am I missing?
So, maybe the zeros don't go.
If the zero doesn't go in either of these, then it does definitely go over here.
And what does that mean? That means we have three sixes left. So either I mean that doesn't make any sense, does it? If we have three sixes left, it means one of the sixes goes in the void and the other two go over here. If the other two go over here, then this must be the 62.
We've used our zero, so these need to be all ones. So, we have no we have no double to put going up there because it would have to be one one. So, this would have to go across, which would make it one six.
I think this is the only other possibility. If this doesn't work, then I don't know how else I can't think of any other possible universes in which this puzzle can exist. But we still have one more six. So that must go here. So this must be threes. This must be 2 one.
I mean, everything I'm I'm saying must because it feels as though it must be true to me. But I'm worried I'm still missing a possible reality. Then this I mean I mean this could go here or here, but it's going to be ultimately the same. Um then we have one more three. The other three goes into the inequivalence region. We have two more ones here which are four and five. Now how many fours and fives do we have? We have fours 1 2 3 4 5 6. We have six fours.
One of them goes over here which means we have five fours left. The five fours must be the big five equivalence region that goes there. This goes down here.
Double five. Now the other region is fours. So this is five four.
This is double four.
This is one four. This doesn't work.
It's broken again. We have too many fours.
I'm sorry. This is embarrassing.
I don't understand where to put all these dominoes.
So it must be that the zero goes I can't think where else the zero would go. The zero can't go in this void because then it would put six in this region. We we can't manage that. We need a double in there.
So the zero goes either in the two region or the three region.
Now it can't go in the two region, can it? Because No, because it would need it would point into here and we'd need a double here somewhere.
Yes, we absolutely would cannot go in there. So, it must go here. Now, it can't point to this region or this region because those would require a double somewhere.
So, it must go here. I I'm really confused as to what I'm missing.
Obviously, I'm not there's a possibility I'm not thinking of. Clearly, I just don't know what it is. So, this needs a two and a one.
There's no zero over here, which means this needs 1 2 3 4 5 and six in it.
Okay, maybe that's what I need to think about.
It needs one of everything we have left.
So, how many digits of how many of each digit do we need to place? 1 2 We have five ones. We need to place four ones.
We put two of them up there. One of them in here. We have one more four to place.
Sorry, one more one to place. Which means I guess it would have to go in this dead region, this void.
Now it can't point up in that case because then we would need a double well no double one wouldn't work. We'd need two different ones pointing to the same region. That's impossible. So this two cannot point that way which means it must point this way because the two needs to be made up of two ones.
So that's what we do with ones.
Then the other one points into this double region. Sorry, this five cell region which could be um fours or threes or fives.
That's ones. What about twos? We have only a single two left, which means it must go in here.
And in the current situation, I don't see any other possibility for that.
threes. We have 1 2 3 4 5 six threes.
Sorry, I'm already This already feels wrong.
Let me look at what I've done again.
Ones. One. We have four ones.
One, two, three. Yeah, I it all We already need one. One one in the void and also one one.
Sorry. We We have one one in the void.
We also need a six in the void.
So, this is wrong. This is wrong because we can put one more one in the two. I'm not I'm just going to put it there to get out of the way. We need one more one in the three.
We need a one in the equivalence region.
And we still have one more one.
It could Yeah, this one could never be over here because it would need to be the same digit as what that is. This could So that could never be a one.
So one of our ones needs to go in here obviously. But then we have one more that can't go anywhere.
So, this just doesn't work. It just doesn't work.
Which makes me think the zero needs to go over here. But then what do we do with our sixes?
We have three sixes now. What do we do with them? One of them must go into the void.
And it can't point up like this because we have the same problem we had before. These need to both be ones that point into the same thing. So, it must point across. It's not going to be a two, but um it's going to be six. It's going to be 63 or It has to be 63.
If it's not 62, this can't be twos.
Can't be ones because we don't have a double one. So, it would have to be this. It would have to be this. I don't understand what I'm not seeing. I'm very sorry about this. It's very embarrassing. Um, this has to be 2, one.
This has to be 1 three.
So now one of our threes goes in here.
One of our threes in in the void. So far so good.
But I think we know from experience that does not it doesn't say at all. Now one of our twos goes down here. The other two goes in the in the inequivalence region. The 26 cannot go in the inequivalence region, which is a problem we've ended up with before. So that means this two must be the 26. I mean, which makes sense that we have landed on that before, which means this would have to be 61, which means the other two.
Wait a second.
We don't know that two goes in here. We don't know what goes in here yet because we've put a zero in there.
But we know that six can't. So that's fine. I think what I've said still makes sense.
We still Yeah, we still have our full complement of sixes in the equivalent region. So the sixes still do have to go somewhere else. So I think so far this is fine. Now this is one pointing to 14 or 15. Now how many of each of these do we have?
15. We have 1 2 3 4 five of them. So one goes presumably in the inequivalence region and the other has four that goes in there. The other possibility is fours. 1 2 3 4 5 six.
So that needs to go in the five cell region. So the fives go in the four cell region. Then I think what I'm saying is right, but I'm really worried it's just going to end up being wrong again. And then fours go over here.
Three four.
Um4 45.
No, it worked this time. I'm sorry. I missed something before.
I don't know what [clears throat] I did.
Yeah. And it's again it's it's mocking me with this artificially long time. I mean, you'll know. It's it's it's easier for you to see where I went wrong because you've been watching the video, you know, potentially already having solved this puzzle. So, you might have seen where I went wrong. and I'll be curious to know. But I'm I can't believe that took me as long as it did. It felt like I was on the right track much earlier.
[sighs] I should have been thinking about I should have determined that the six goes in that void much earlier.
That should have been available very early just logically speaking because we knew the six could never fill any of the large equivalence regions. So where else could the sixes possibly go? Those are the only places the sixes could have gone. That's what I should have determined from the beginning and it's actually solves fairly smoothly once you see that. So that was that really ultimately was my mistake. The rest of it, you know, comes down to sort of counting the instances of digits digits that can fit in each particular region, but ultimately those these are where the sixes have to go. Okay, well that was the stray. Those are the pips. Sorry.
Back tomorrow. Bye for now.
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