r/nytpips

P/p as in this puzzle is poop in my opinion. It somehow made the hard ones earlier in the week seem like a walk in the park. I just got absolutely stumped trying to come up with a strategy to attack this puzzle. Below is the mess of a walkthrough I came up with (bear in mind I tried to avoid pip counting at all costs, which may be the easier solution?). Would be very interested to know how others solved this (other than trial and error). Would also be curious to know how the former guide master would have attacked this (if he didn't hold a grudge against us all).

1. >!There are five doubles (0-0, 3-3, 4-4, 5-5 and 6-6). By forced placement, each of the three 3c= must contain one.!<
2. >!The only whole domino with a sum of 3 is the 1-2.!<
3. >!Very few tiles can fit in the bottom of the 3c9. The 6-6, 6-5, 6-4 and 5-5 are all too large, and the 0-0 and 0-1 are too small. The 6-3 and 5-4 would need to be followed by a 0-x (with x>4), which doesn't exist. The 4-4 would need to be followed by a 1-x (with x>4) which doesn't exist. The 2-5 would need to be followed by a 2-x (with x>4) which doesn't exist. This leaves three possibilities: 3-3 (finished with 3-6), 3-4 (finished with 2-5), or 1-2 (finished by either 6-6 or 6-5).!<
4. >!Note that in all of these combinations (3-3-3, 3-4-2 or 1-2-6), the 3c9 is composed of two numbers ≤3.!<
5. >!There are 11 total "small tiles;" three 0s, two 1s, two 2s and four 3s. The 2c<3 requires two. Each 2c3 require two. Two go into the 3c9. Three remain.!<
6. >!The only possible places that can take a number ≤3 are the 1c>0, discard, 3c= or [one more in the] 3c9.!<
7. >!Let's assume for a moment that the 3c= takes a low number. The only ones with enough are 0s or 3s.!<
8. >!If the 3c= is 0s, all 0s are booked, which means that both 2c3s are 1+2, and now all numbers ≤2 are booked and we can't complete the 2c<3.!<
9. >!If the 3c= is 3s, there's one 3 remaining. Which means that the 3c9 can only be 3-4-2 or 1-2-6. Either way, one 2 is booked. This means that one of the 2c3s has to be 0-3, which books the last 3, and forces the 3c9 to be 1-2-6 (with the 1-2 in the bottom). The other 2c3 would have to be 1+2 also. This books all 1s, all 2s and all 3s. So the 2c<3 has to be 0+0. And the 2c<3 can't be made of two tiles (because then the left half of either the 0-0/0-1 goes into the 2c>10, which doesn't work), so it would be the whole 0-0. BUT then this fails, because above this the 2c3 would need a whole domino, and we already used the 1-2.!<
10. >!So therefore a 3c= is not made from a number ≤3. So they are made of 4s, 5s and 6s.!<
11. >!This also means that the three remaining low numbers have to go into the 1c>0, discard and 3c9.!<
12. >!Which means that the 3c9 is made of three low tiles, and this has to be 3-3 with the 3-6 above it. Three 3s are booked, one remains.!<
13. >!The 2-5 is now more limited as to where it can go. It can go on the 1c>0-3c= border (on the left in the capital P), the discard-3c= border (on the top of the capital P), or the 2c<3-2c>10 border. No matter where, one 2 is therefore going into a place that is NOT a 2c3. So one 2c3 is a 0+3 (which books the last 3), and one is a 1+2 (which books the other 2). This leaves us with two 0s and one 1 to fill out the remaining unfilled spaces of the 1c>0/discard/2c<3.!<
14. >!If the 2-5 goes in either slot within the capital P, it would require the 2-1 to go into the 2c3. But then this would be followed with either a 0/1 going into a 3c=, and since there is none with a half of 4/5/6, this fails.!<
15. >!So the 2-5 goes on the 2c<3-2c>10 border in the lowercase p.!<
16. >!The 2c<3-2c3 border can be the 0-0 or 0-1. If it's the 0-1, the 2c3 is finished with the 2-1 going into the 3c=, which fails. So it's the 0-0, followed by the 3-4 and then the 4-4.!<
17. >!The 0-1 now only has one place to go, it's on the discard/2c3 border in the capital P with the 0 in the discard. This is followed by the 2-1 beneath it into the 1c>0.!<
18. >!There are two 4s remaining, and they must go into the two 1c>3 slots. The 4-6 is on the 1c>3-2c>10 border to finish the lowercase p, and the 4-6 is on the 1c>3-3c= in the capital P.!<
19. >!Finish this 3c= with the 6-6.!<
20. >!The 1c>4-3c= on the right of the capital P is therefore the 5-6, and finished with the 6-6.!<

Q.E.D. I say good riddance to this puzzle.

Lately I've seen a lot of comments from people about the nyt puzzles not really doing it for them. And ultimately, instead of hijacking a thread, thought i'd get some feedback!

So - blankly: what are you specifically after from a hard puzzle?

When I build the puzzles for my dotio app, I typically focus on ensuring that everything is locally deducible, and that each puzzle has at LEAST 1 semi-obvious starting point. Then, any forks/branching points go somewhat deep (3-5 moves before you're blocked) But I've equally found this to be super subjective; NYT pips have some puzzles that APPEAR entirely random, and I find it really hard to feel satisfaction when I've solved it; or when i try other pip puzzle games; they've felt like its just a pure randomisation? But its clear that people also enjoy this style of puzzle. So; what works for you?

Full transparency: any suggestions or thoughts people have, I will attempt to improve dotio with!

Lately the hard puzzles (and sometimes even medium) make me feel less clever and more like an antiquated sorting algorithm that just tries various pip combinations until something works.
These puzzles should require critical thinking. There should be at least one space that can be solved without any ambiguity, which should reveal another space, and another, etc. Instead, the only way to solve recent puzzles has been guessing and checking.
Very let down lately by the recent puzzles in this regard and will likely stop doing them if this keeps up. Anyone else feel the same?

Nice one today IMO. Hope you all don't "take the L" and solve it yourselves!

Pre-placement:

1. >!The 3c<2 is theoretically either 0+0+0 or 0+0+1. We only have two 0s, so it's the latter. All 0s are booked, and one 1 remains.!<
2. >!The 6c≠ (the lowercase l) therefore contains one of each pip from 1 through 6. All 1s are booked, and one 2 remains.!<
3. >!The 1c<3 is therefore a 2. All 2s are now booked.!<
4. >!This means that the 2c7 has to be 3+4, and since there isn't a whole 3+4 domino, it's two tiles, with the bottom one going horizontally into the 3c<2.!<
5. >!We started with six 3s, four 4s, four 5s, six 6s, but have so far booked two 3s, two 4s, one 5 and one 6. This leaves us with four 3s, two 4s, three 5s and five 6s.!<
6. >!The only number left with enough to satisfy the 5c= is 6s. All 6s are booked.!<
7. >!With the 6s gone, the 2c10 is therefore 5+5, and since there isn't a 5-5 tile, it's made of two vertical tiles. One 5 remains.!<
8. >!We haven't yet accounted for the two 2c= or the three 1c>3 cells. With what we have remaining, clearly both 2c= are made of 3s, and the 1c>3s are 4, 4 and 5.!<

Placement:

1. >!Place the 5-3 on the 2c10-2c= border.!<
2. >!To its left, there is no 5-5, so it's the 5-4.!<
3. >!The 3-0 can only go on the 2c7-3c<2 border.!<
4. >!The 5-0 can only go on the 1c>3-3c<2 border.!<
5. >!Place the 1-6 to its left on the 3c<2-5c= border.!<
6. >!There's no 4-3, so the 2c7 is finished with the 4-6 horizontally into the 5c=.!<
7. >!The only place for the 3-3 to go is wholly within the 2c= above this.!<
8. >!The 6-6 goes to its right.!<
9. >!We have two dominos with 4 left to use, with the 4 slotted for the 1c>3. But the 1-4 doesn't work because the 1c<3 is a 2. So it's the 4-2 vertically. (If you tried a different arrangement here, you'll see that you can't finish the 6c≠).!<
10. >!Finish the capital L with the 6-3.!<
11. >!The 1-4, 2-3 and 5-6 all go in the 6c≠ (the lowercase l) in any order and any orientation (hence, as u/harlows_monkeys pointed out, there are 48 possible "different solutions" to this puzzle).!<

Happy 1 year anniversary to this game! We get an appropriately-themed puzzle today.

1. >!There are six 1s. These are all booked into the five 1c1s, and the 3c1.!<
2. >!Each 2c>11 is 6+6. The 2c11 is 6+5. All 6s are booked.!<
3. >!The 2c>11-1c>1 in the "r" must be the 6-2.!<
4. >!There is no 6-5 whole domino, so the 2c11 is made of two dominos. This forces tile placement: there is a domino on the 2c11-3c1 border, the 1-1 above it, and the 6-6 to the right wholly in the 2c>11, and then a domino on the discard-2c11 border.!<
5. >!The only place where the 1-6 can go, therefore, is on the 3c1-2c11 border.!<
6. >!Finish the 3c1 with the 0-0.!<
7. >!There are six 0s remaining. Where can they go? If they went in either of the 3c11s, we'd need to finish those cells with 6+5 (but all 6s are already booked). And we just used the 0-0 in the bottom left of the "A", so the 2c= on the bottom right of the "A" can't be 0s. So this means that both two 2c=s in the "E" are 0s, and both discards are 0s.!<
8. >!Finish the "r" with the 6-0.!<
9. >!The 2c11-discard is the 5-0.!<
10. >!The 1c1-2c= is the 1-0.!<
11. >!The other two 1c1s in the "Y" are the 1-4 and 1-5 going into the 3c11 (in either position).!<
12. >!This 3c11 is finished with the 2-2 into the 1c>1 (the 2-0 doesn't work).!<
13. >!The 5-5 is in the 2c= on the bottom right of the "A".!<
14. >!The 3-5 is the only tile that works in the middle of the "E", but which way does it go? If the 3 is in the 3c11 half and the 5 in the 1c>1 half, then we'd need two halves with a sum of 8 to finish the 3c11, which we don't have.!<
15. >!Place the 3-5 in the middle of the "E", with the 3 in the 1c>1, and the 5 in the 3c11.!<
16. >!The 3c11 is finished with the 4-0 and 2-0 (can be in either position).!<
17. >!The 2c=-1c>1 at the top of the "E" is therefore the 0-3.!<

Noticed there wasn't a guide yet for today and this is not a nice one by any means, but it's the best I got. Happy to hear other people's strategies in the comments.

Pre-placement:

1. >!2c>10s can either be 6+6 or 6+5. 2c>9 can be 6+6, 6+5, 6+4 or 5+5. There are three 6s, two 5s and three 4s available.!<
2. >!There are no dominos with both halves large enough to satisfy any of these as a single domino, so they're all made of two dominos. Which also means by forced placement that all the 2-celled regions are made of two dominos.!<
3. >!This also places a double in the bottom of the 3c=, which is either 1-1 or 2-2.!<
4. >!If either 2c>10 is 6+6, the other would need to be 6+5, and then we can't make the 2c>9. So both 2c>10s are 6+5, and the 2c>9 is 6+4. This books all the 6s, all the 5s and one 4.!<
5. >!One 4 is booked for the 1c4. One 4 remains.!<
6. >!There are five 0s. Two are booked for the 2c0. Three remain.!<
7. >!There are two 3s, five 2s and six 1s.!<
8. >!2c<3 can theoretically be 0+0, 0+1, 0+2 or 1+1.!<
9. >!The only places that the last remaining 4 or either 3 can go are in the 2c4s or in the 4c4.!<
10. >!Let's start with where the 4 can go. If the 4c4 was in fact 0+0+0+4, then all 0s and 4s are gone. So the 2c<3 has to both be 1s. This leaves four 1s. There aren't enough 3s for all four of them to go into the four 2c4s, so the 3c= is 1s, leaving one 1 to go with one of the 3s in a 2c4. But then we'd have a spare 3 with nowhere to place it, so this fails.!<
11. >!So at least one of the 2c4s is 4+0. All 4s are booked, and two 0s remain.!<
12. >!Now, where can the 3s go? If the 4c4 took one of the two 3s, it would be made of 3+1+0+0, which would book all the 0s, and leave us with five 1s and one 3. Once again, this would force both sides of the 2c<3 to be 1s, leaving three 1s, which would have to go into the 3c= and again leave behind a spare 3.!<
13. >!So therefore, both 3s go into a 2c4 somewhere. This books both 3s; four 1s remain.!<
14. >!To recap, there are four 2c4 spots; we already know three of them (0+4, 3+1 and 3+1). The last one must be 2+2. Three 2s remain.!<
15. >!We haven't yet determined what completes the 2c<3, 3c= and 4c4, and we have three 2s, four 1s and two 0s.!<
16. >!Where can these three 2s go? Let's assume for a moment they don't go into the 3c= (meaning that the 3c= is 1s, and would leave one 1 remaining).!<
17. >!The most the 4c4 can take is two 2s (along with the last two 0s). But with only a 2 and a 1 left over, the 2c<3 can't be completed.!<
18. >!If the 4c4 took only one two, 4c4 would be 2+1+1+0, but there are not enough 1s left to make this combination.!<
19. >!So, the 3c= is 2s, which books all the 2s.!<
20. >!This leaves us with four 1s and two 0s yet to be placed. Which means that the 4c4 has to be 1+1+1+1, and the 2c<3 are both 0s.!<

Placement:

1. >!Place the 2-2 on the bottom of the 3c=.!<
2. >!Above this can either be the 2-0 or 2-4 (the 2-6 is too large), which means that the this 2c4 is 0+4. The tile above into the 2c>9 has to be either a 4 or 6. There is no 4-4 or 4-6, so it has to be the 2-4 followed by either the 0-4 or 0-6 (both can work).!<
3. >!Whichever was not placed continues into the 2c<3 (either the 6-0 or 4-0).!<
4. >!With the 0-6 gone, the 2c<3-2c>10 is the 0-5.!<
5. >!The last 4 is on the 1c4-2c4 border, its the 4-1.!<
6. >!The 3-0 goes next to it.!<
7. >!The last 0, the 0-2, goes above it.!<
8. >!Finish the left side (the capital O) with the 2-6.!<
9. >!Let's move to the lowercase o. Place the 1-1 in the middle of the 4c4.!<
10. >!The 1-3 is the only domino small enough for the 4c4-2c4 border.!<
11. >!The 1-5 and 1-6 can be placed in either configuration to finish it up.!<

For majority of the other games (wordle, connections, strands, etc) you can play past ones and look back at ones you've solved but not for pips. I always do pips at the end of the day but sometimes when I'm busy I wont get to them until just a few minutes before 12am and I dont always get to finish even if I'm one peice away from solving it, and I would love to look back to see old ones that ive finished but I cant; why😭.

Hi everyone, with everything going on this sub should have a bit more mod coverage.

If you would like to mod please reply with;

- Why you think you would be a good fit

- Your timezone

- Your past modding experiences

- If you’re comfortable using a third party app for modding discussion

- Anything else you think is relevant

Thanks for your interest
-Straz

I obviously know what the word means, but several times I've thought "That hard puzzle had no rhyme or reason" or "that puzzle was exactly straightforward" and came to this sub to see people saying the opposite. So, for you, what does a logical solution entail?