u/chx_

I have a ThinkPad P14s G5 on CachyOS, a Plugable USB4-HUB3A and a Macbook M4 connected to a downstream port for Thunderbolt networking. This works until I undock and dock again. Then I need to unplug the Mac and plug it again. Then it works again. (I also have some plain USB hubs chained but hopefully this doesn't change anything.)

Here's what journalctl says: https://termbin.com/xhiwp Linux is 7.2.6.

When I unplug the dock and plug it back I see:

1. the PCIe hierarchy comes back,
2. the USB4 hub comes back,
3. the retimer comes back,
4. the USB Mac device comes back,
5. and only the MacBook Thunderbolt router is missing

The docking starts at 11:15:11 and finishes at 11:15:14, I unplug the Mac at 11:16:58 and replug at 11:17:04.

enp0s20f0u1u3 and enp0s20f0u1u3i2 are Apple Ethernet adapters using cdc_ncm. The real, missing interface is thunderbolt0.

This is a reproducible failure.

https://www.kickstarter.com/projects/lakicome/laki-come-carry-on-suitcase-with-ride-on-platform/description

Where do we begin? The campaign is shitty because it's so thin on facts. How heavy is this thing? How much space is available on inside? Measurements and photos are a minimum, a packing video shouldn't be that much to ask.

I am eyeing the whole thing with great suspicion. Those wheels will be dirty, are you sure you want to handle them all the time. How will this contraption behave when a wheel gets stuck and it's loaded so very heavily. A normal four wheel suitcase can be titled into a two wheel one temporarily to help getting unstuck. This thing? Good luck.

I have no idea whether this is OK child safety wise. My gut feeling is "no".

I just moved back to Canada and needed something to put together my new desktop PC and so I picked up this cheap "Hyper Tough 4V Cordless Rotating Screwdriver" and much to my surprise when I plugged it into a proper USB C charger it immediately worked. My not at all scientific tool reports it is drawing 6W.

We all know how many of these things only work with the A-C cable it ships with. Not the case here.

Also, this is not an endorsement of the tool as a power tool. I have used it to drive like a dozen screws and it did the job. But: it's $19 in the US. Perhaps treat it as "you get what you pay for"?

https://www.kickstarter.com/projects/orcanyx/the-tiny-thunderbolt-5-egpu-unleash-rtx-power-on-mac-and-pc/

Campaign successful on Aug 13. Zero communication since. Buh bye.

Reminder: every electronics Kickstarter should be treated as a scam unless you have a good reason to believe it is real. Triple so for mobile phone, tenfold so for Thunderbolt.

Amikor Kanadából költöztem Európába akkor egy ukrán menekültek segítő szervezet egyik önkéntese vitt tőlem mindenfélét, ruhát, plüssmacit, elektronikát... mivel most ellenirányba adom elő ugyanezt és még sokkal drágább visszavinni, megint lenne adomány pár dobozzal ha van, aki elviszi.

https://www.kickstarter.com/projects/tengpai/tengpai-h1

There are two kinds of scam this can be:

1. reselling an existing product. That's a scam. That's not what crowdfunding should be.
2. You do not even intend to fulfill this.

The presumed positive outcome where the crowdfunding is successful and the rewards are shipped is not possible with a modern smartphone. It requires scale far beyond these campaigns.

Also, this is so bad of a slop there are typos. Many. Seriously, one of the section headers is Andriod 16. (Nevermind that Android 17 is out.) There's Negapixel , Surver (instead of Survey) and even more.

This is very easy. Many different ways to do this, I tried to pick a simple one.

1. >!Left edge, counting from bottom the 3rd and 4th tile are both one half of a double: they have equal neighbours.!<
2. >!Then there is a vertical domino for 5c=-2c10 and another vertical for 5c=-1c1.!<
3. >!1-4 has no double, so it's 1-0 marking 5c= for 0s.!<
4. >!Place 0-0 above it.!<
5. >!Further, there's a horizontal on 2c10-1c2 and that's 6-2 as 2-0 and 2-3 are too low.!<
6. >!2c10-5c= is 4-0.!<
7. >!There's a 0 at the top of the 5c=, there's two in the 2c= and there's a 1c0: the 0s are booked.!<
8. >!And also, there are two 2s and two 1c2 so the 2s are booked.!<
9. >!This means the 0-2 is the 5c=-1c2.!<
10. >!And the 2-3 is the top 1c2 and the 3 can't go left as it would need another 2 and can't go down as it'd be too low, it goes right into the 1c3.!<
11. >!2c5 is a whole domino, in theory 0-5/1-4/2-3 but the 0s are booked, the 2s are gone, it's 1-4.!<
12. >!The left of 2c0 can't go right it'd be the 0-0 again, it goes down into 3c16. 0-3 is too low, 0-5 would require a domino with a sum of 11 which doesn't exist, it's the 0-6.!<
13. >!Finish 3c16 with 5-5.!<
14. >!2c0-3c17 is 0-5, 0-3 is too low. The other two tiles are 6s.!<
15. >!1c5 is the last 5-?, the 5-3, goes up.!<
16. >!If the middle of 3c17 goes left it'd orpahn a tile so it goes down, it's a whole domino, the 6-6.!<
17. >!1c0-2c9 is 0-3, the last 0-?.!<
18. >!1c3-2c9 is 3-6.!<
19. >!2c=-1c3 is 3-3.!<

Ps.: I do not take days off. I write the guides several days ahead and posting them takes but seconds. There's no need to pretend someone else posts a guide so I can take a day off when we know all too well what it's about. Last time I simply stopped for a while. Let's not have this again, shall we.

I really like this puzzle, there are two key steps to the puzzle and both are a bit unusual. So the guide is very straightfoward but not trivial. Best sort of puzzle: it's the right level of challenge which doesn't fall easily but it doesn't need trial and error.

1. >!Four 1c0, four 0 halves, 0s are booked.!<
2. >!There's no 6-6 so 2c12 is made from two dominos, one on 2c12-2c4 one one 2c12-1c>2. 2c7 is a whole domino.!<
3. >!2c12-2c4 is one of 6-[0/2/3/4] 6-0 is booked, 6-4 needs a 0 so it's either 6-2 or 6-3.!<
4. >!If it's 6-2 then the 2c4-2c9 is 2-5, 2-0 is too low.!<
5. >!If it's 6-3 then the 2c4-2c9 is 1-5.!<
6. >!Either way, the top of 2c9 is a 5 the bottom is a 4, place 4-0.!<
7. >!1c0-2c10 is 0-6, both 0-2 and 0-3 are too low.!<
8. >!There are two more 1c0 for 0-3 and 0-2.!<
9. >!If the 0-3 is on 1c0-2c7 then 2c7-1c3 is 4-3 and if it's on 1c0-2c6 then 2c6-2c10 is 3-4.!<
10. >!Either way: the 3-4 is used.!<
11. >!2c7 in theory is 1-6/2-5/3-4 but there's no 6-1 and the 3-4 is used so it's 5-2.!<
12. >!Then 2c12-2c4 is 6-3 since the 6-2 would be followed by 2-5.!<
13. >!Place 1-5 on 2c4-2c9.!<
14. >!2c12-1c>2 is 6-5, 6-2 is too low.!<
15. >!1c>3-1c>3 is 4-4.!<
16. >!Then 1c0-2c6 is 0-3 as 0-2 would be followed by 4-4.!<
17. >!2c10-2c6 is 4-3.!<
18. >!1c0-2c7 is 0-2.!<
19. >!2c7-1c3 is 5-3.!<
20. >!2c8 is 2-6.!<

Writing a guide for this was unusually hard.

1. >!2c0 is made from 0-1 and 0-6 with the 1 in the discard as these are the only 0-? and 0-1 can't be in 3c15.!<
2. >!Without 0s, 1c<2 is a 1, 2c2 is 1+1, three 1s remain.!<
3. >!With only three 1s remaining the least five tiles can make are those three 1s and two 2s and that's 7 so that's what 5c7 is, the 1s are booked.!<
4. >!Now that we know 1s are only in 2c2 and 5c7 it's worth checking where the 1-5 can be: 2c2-3c15 and 5c7-1c>4 are the only places for it.!<
5. >!Three 1c>5 are 6s, two 6s remain.!<
6. >!Where are they? There can be one in 3c15, one in 1c>4, one in 2c8. But if both the 3c15 and 1c>4 has a 6 then the 1-5 has nowhere to go so there's a 6 in 2c8, the other tile is a 2!<
7. >!1c<3 without 0 or 1 is 2.!<
8. >!Two 2s in 5c7, one 2 in 1c2, one in 1c<3, one in 2c8, 2s are booked.!<
9. >!1c2-3c= is one of 2-[1/2/4/6], there are not enough 1,2,6 so it's 4s, place 2-4 on the 1c2-3c= border.!<
10. >!With the 2-4 gone, the 3c=-2c6 is not 4-2 so the 2 is on the right hand side of the 2c8, place 2-6.!<
11. >!Finish 2c8 with 6-4.!<
12. >!6-5 is on 1c>5-2c= border, we know where 6s can be and the 5 has nowhere else to go.!<
13. >!Finish 2c= with 5-4.!<
14. >!6-3 is 3c15-1c<4, we know where 6s can be and the 3 has nowhere else to go.!<
15. >!Place 1-1 next to it making the 2c2.!<
16. >!Since the teal 2c= is not 2-2 as those are booked, it's the 4-4.!<
17. >!1c4-3c15 is 4-3.!<
18. >!The two 2s in the 3c7 comes from 2-2 and 1-2 which means the three 1s come from 6-1, 5-1, 3-1, these are the top from right to left.!<
19. >!Place 2-2 on 1c<3-5c7.!<
20. >!Place 1-2 on 1c<2-5c7.!<

Well, that was easy.

1. >!There are no doubles so let's do a corner check A1-A2 is 5-1, D8-D7 is 2-1.!<
2. >!Also, D1 is 1c>0. So that can't be 0, that's sus. What can be 0? Hey, there are only two on the board. Then three of the discards are 0s.!<
3. >!Back to the 0s, the 0 on B8 has 1,2,3 neighbours, the 0 on A4 has 2,3,4 neighbours, there's nowhere for the 5-0 but C7-C6. There are five 5s so none of the unknowns are 5s.!<
4. >!D6-D5 is 1-3.!<
5. >!C5 is left: 4-3, down? double.!<
6. >!B8-C8 is 2-0.!<
7. >!There's a 3-? on A8-A7.!<
8. >!B7-B6 is 1-0.!<
9. >!A6-A5 is 0-4!<
10. >!A4 down, 0-3: : right? 0-2, used.!<
11. >!D3-C3 is 1-4, no more 1s.!<
12. >!D4-C4 is 5-4.!<
13. >!B4-B3 is 2-3.!<
14. >!C2-B2 is 4-2, no more 4s.!<
15. >!C1-B1 is 3-5.!<
16. >!D1-D2 is 2-5, nowhere else for the 2.!<
17. >!A8-A7 is 3-6.!<

Not too hard.

1. >!There's a domino on the top left discard-purple 2c10 border.!<
2. >!1c<1 is a 0 and also it goes down per the previous point.!<
3. >!Then there's a double in the middle which means the next tile can't go right, it goes up which means the last tile can't go up as it'd be the same domino it goes down into 1c<2.!<
4. >!1c<2 is 0 or 1 but if it were 0 then the leftmost and the rightmost domino would be the same so it's a 1.!<
5. >!Rightmost domino of the 5c= is then one of 1-[0/1/2/3], the 0 would require the 0-0 twice, the 1-1 would require the 1-1 twice, there are only two 3s. So the 5c= is 2s.!<
6. >!Leftmost is 0-2.!<
7. >!Next is the 2-2.!<
8. >!Last 2-? is the 2-5. Another 5 will be needed here.!<
9. >!Bottom of blue 2c= is a double, remaining doubles are 0-0, 1-1, 4-4.!<
10. >!If it's the 1-1 then above it is one of 1-[0/3] both are too low for 2c10.!<
11. >!If it's the 4-4 then above it is one of 4-[0/3] both are too low for 2c10.!<
12. >!So it's 0-0 followed by one of 0-[4/5/6].!<
13. >!If it's the 0-5 then the 2c10 is finished with the last 5-?, the 5-5 and there's no 5 left for teal 2c=. So 2c10 is 4+6, there's only one 6, it's booked.!<
14. >!With the 6s booked, the 1c>4 is a 5, the 5-5 would put a 5 in this 2c= using up all 5s and the teal 2c= needs one more 5. So 1c>4-2c= is the 5-0.!<
15. >!Out of 0-[1/4/6] only the 0-1 fits 1c<4.!<
16. >!Finish teal 2c= with the 5-5.!<
17. >!The only domino which fits 2c>7 is the 4-4.!<
18. >!With the 4-4 gone the top of bottom yellow 3c= which is a double is the 1-1.!<
19. >!Finish this 3c= with the 1-3.!<
20. >!2c4 is 0-4.!<
21. >!Finish 3c= with 0-6.!<
22. >!Place 4-3.!<

After yesterday's madness this is easier. A bit. But it's still quite hard.

Before placement:

1. >!The two 2c12 books the four 6 domino halves.!<
2. >!The bottom orange 4c= has a double: the third from the left has equal neighbours.!<
3. >!The leftmost tile then can't go right as it'd be the same double so it goes up into 2c12.!<
4. >!2c12 is 6+6 so 2c12-4c= is one of 6-[0/1/4/5] and only the 1 and 4 has a double so it's one of those.!<
5. >!There are two 3-? , the 3-1 and 3-4 and two 1c3.!<
6. >!If the 3-1 is the pink 1c3 then the 1 only fits the 2c=, the 2c=-2c12 is the 1-6, then 2c12-4c= is 6-4 so the bottom 4c= is 4s.!<
7. >!If the 3-1 is the blue 1c3 it only fits going up making the purple 4c= 1s and there aren't eight 1s so the bottom 4c= is 4s.!<
8. >!After the yellow 4c= takes four 4s there are two 4 and five 5s are left and there are three 1c>3 and a 1c>4 which uses these, overall three of 4 and 5 are left which means neither of the remaining two 4c= are either 4 or 5.!<
9. >!There are only two 0s, no free 3s, no free 6s, so the purple 4c= and the teal 4c= are 1 and 2.!<
10. >!There are five 2s, so after booking four 2s into a 4c= there's only one left which means none of the 2c= are 2s so the last 2 is in the 1c<3. It is, however, not the 2-1 because the potential neighbours are 4/5/6. !<
11. >!Let's get back to the 3-1 when it's in the pink 1c3.!<
12. >!We already noted in this case the 3-1 goes left followed by 1-6, the remaining four 1s are booked into either 4c=. So now the 1 half of the 2-1 is booked into a 4c= and so is the 2 half because 2s are in 1c<3 which is not 2-1 and the other 4c=. But the two 4c= are not neighbours so if the 3-1 is in the pink 1c3 then the 2-1 can't be placed.!<

Placement.

1. >!So the 1-3 is in the blue 1c3 going up marking the top purple 4c= for 1s which leaves the 2s for the teal 4c=.!<
2. >!Place 6-4 on bottom left corner 2c12-4c=.!<
3. >!Bottom of right purple 2c12 is the 6-5 as the 6-0 and the 6-1 doesn't fit anywhere: too low for 1c>3 and the 4c= is 2.!<
4. >!Then the right of 4c= goes up, it's 4-2.!<
5. >!The 1c<3 is a 2, there's no 2-6 so it goes down, with the 2-4 placed it's 2-5.!<
6. >!Second tile from the bottom in the teal 4c= can't go right because there's no 2-6, it goes up, it's the 2-2.!<
7. >!Finish teal 4c= with the only 2-? the 2-1, it can only go right.!<
8. >!Place 1-6 on 2c=-2c12.!<
9. >!Left 2c12-2c= is 6-0.!<
10. >!2c= is finished with 0-5 it can only go up.!<
11. >!Corner of 4c= goes right, it's the 1-1.!<
12. >!Pink 1c3 goes up, it's the 3-4.!<
13. >!Finish purple 4c= with 1-5.!<
14. >!Place 4-5.!<
15. >!Finish bottom 4c= with 4-4.!<

Auuuugh. That's one tough puzzle.

1. >!1c3-2c= is one of 3-[1/5/6].!<
2. >!If the bottom of the 3c5 sums to 0 which is 0-0 then you need a 5 on top, one of 5-[0/1/3] comparing to 3-[1/5/6] is 0-0, 5-1, 1-3.!<
3. >!There's no domino with a sum of 1.!<
4. >!If the bottom of the 3c5 sums to 2 which is 1-1 then you need a 3 which would be the same domino into the 2c= this can't be.!<
5. >!If the bottom of the 3c5 sums to 3 which is 1-2 you'd need another 2 on top which doesn't exist.!<
6. >!If the bottom of the 3c5 sums to 4 which is 1-3 then you need a 1 on top, one of 1-[1/4/5] comparing to 3-[5/6] is the 1-5 and 3-5.!<
7. >!If the bottom of the 3c5 sums to 5 which is 1-4 or 0-5 you'd need a 0 on top one of 0-[0/5] comparing to 3-[1/5/6] is 1-4, 0-5, 5-3.!<
8. >!Let's presume the trio is 1-4, 0-5, 5-3.!<
9. >!The only remaining 1c>4 is the 6-3 and the 5-1.!<
10. >!The 1c>4-3c9 is not the 5-1 because the most you can make from the remaining is 1-3 on 1c<2-3c9 and 4-4 on 3c9-2c4 which is not enough. 1c>4-3c9 is the 6-3.!<
11. >!You need to make 6 from two tiles, that's 0+6/1+5/2+4/3+3 but there are no 6 and 5 left and only one 3, it's 2+4.!<
12. >!The only 4 is the 4-4 and then there is no domino left for the 2c>5.!<
13. >!Let's presume the trio is 1-3,1-5,3-5.!<
14. >!Then the two 1c>4 is the 5-0 and 6-3.!<
15. >!If the bottom of 3c9 is a 0 then without 5 and 6 the most you can make from two tiles even in theory is 4+4=8 so the 1c>4-3c9 is the 6-3.!<
16. >!Then the bottom of 3c7 is the 5-0 so you need to make 7 from two tiles that's 1+6/2+5/3+4 but there's no 5 or 6 or 3. This really doesn't work.!<
17. >!The 3c5-2c=-1c3 trio is the 0-0, 5-1, 1-3.!<
18. >!The 1c>4 are 5-0/5-3/6-3 so the bottom tile in 3c7 and 3c9 is either 0 or 3.!<
19. >!If the 3c7 contains a 4 tile then since the bottom is 0 or 3 the remaining tile is also 0 or 3. The 0 can only come from 5-0 which only fits to the bottom but neither 3-5 nor the 3-6 fits the 2c4 or the 1c<3. So the 3c7 doesn't contain any 4 tiles.!<
20. >!The middle of 3c7 is not a 1 because the top would either be a 6 (from the 3-6) or a 3 (from the 5-3 or 6-3) and the fits 1c<3.!<
21. >!The 2c4 in theory is 0+4/1+3/2+2, there's only one 2 and if it's 0+4 then the 0 is from 0-5 which only fits 3c9 and so the left tile would be a 4 but neither the 4-1 nor the 4-4 fits the 3c7. So 2c4 is 1+3.!<
22. >!The 1 is not on the left: the 1-1 and 1-4 doesn't work, the 1-2 would need need two tiles making 5 while the bottom tile is 0 or 3, that's which is 0+5 or 3+2 but there's not a second 2 so it's 0+5. But the 0 can only be the bottom and then there's no 5-? left fitting 1c<3.!<
23. >!So the right of 2c4 is a 1.!<
24. >!If it's the 1-1 then the bottom is not the 0-5 as it's too low so it's a 3 here making the top the 0-5 making the bottom of 3c7 into a 3 and there's no 3 left for 2c4.!<
25. >!So it's either the 1-2 or the 1-4.!<
26. >!If it's the 1-2 then the bottom can't be 0 it's too low so it's a 3 and the top is a 4, only the 1-4 fits.!<
27. >!If it's the 1-4 and the bottom is a 0 then the top would need a 5 but the only 5-? fitting is the 0-5 which is used for the bottom 0 so the bottom is 3 and the top is 2, the 1-2.!<
28. >!So the 1-2 and 1-4 are interchangeable with the 2 and the 4 making the top tiles of the 3c9.!<
29. >!And one of the 3s is at the bottom of the 3c9, the other is in the 2c4 meaning the bottom of 3c7 is the 5-0.!<
30. >!The only 1c<3 left is the 1-1, place it.!<
31. >!Finish 3c7 with 6-3.!<
32. >!Place 3-5 on 1c>4-3c9.!<
33. >!Place 4-4.!<

I am not 100% happy with this guide but it's still OK just not as elegant as I'd prefer. Still, there's no backtrack so it's fine.

1. >!2c10 is 4+6 or 5+5 and there's no 4-6 or 5-5, both 2c10 are two dominos.!<
2. >!There's no 1-5 so the bottom one is made from a 4 and a 6, there's no 0-4 so it's 0-6 and 1-4.!<
3. >!3c1 is 0+0+1, there's an 1c0, the rest of the 0s are booked.!<
4. >!Without 0s, 3c3 is all 1s, there's no 1-5 so the bottom left can't go up, it goes right, it's the 1-1.!<
5. >!And the top right can't go left it goes up, it's the 1-0.!<
6. >!The 0 half of the 0-2 is in 3c1 but where?!<
7. >!If the bottom, the 2 half has nowhere to go, too low for 2c9 and 2c12 and too high for 3c1.!<
8. >!If in the middle, the 2 can fit the 2c4 but then above it you'd need 2-1 which doesn't exist or 2-0 again.!<
9. >!It's on the top and the 2 only fits the 2c4.!<
10. >!The bottom of 2c4 goes down because going right would be 2-1 which still doesn't exist or 2-0 which is placed.!<
11. >!The 0 half of 0-5 is in 3c1 but where?!<
12. >!On the bottom the 5 has nowhere to go: it can't go left because that tile is covered with the domino from 2c4 going down it can't go right because 2c12 is 6+6 and it can't go up because 5 is too high for 3c1.!<
13. >!It's in the middle and goes right.!<
14. >!Top of 2c10-2c6 is one of 5-[2/4/6]. 5-6 would need a 0 but those are gone, 5-2 would require one of 4-[4/5] going into 3c= but there aren't enough 4s or 5s for that so it's 5-4.!<
15. >!2c6-3c= is one 2-[3/5] but there aren't enough 5s so it's 2-3.!<
16. >!Place 3-3.!<
17. >!Place 5-6.!<
18. >!Place 6-1.!<
19. >!Finish 2c4 with the last 2-?, the 2-5.!<
20. >!Finish 2c9 with the last 4-?, the 4-4.!<
21. >!Finish 2c7 with the last 3-?, the 3-1, 1 only fits 1c1.!<
22. >!Place 6-6.!<

This is a fun one.

1. >!1c>5 is a 6, four 6 halves: neither 4c= are 6s.!<
2. >!Three 1c5 on the board, three 5s halves: neither 4c= are 5s.!<
3. >!Only two 4s: neither 4c= are 4s.!<
4. >!Three 1c3 on the board, five 3 halves: neither 4c= are 3s.!<
5. >!Three 1c0 on the board, five 0 halves: neither 4c= are 0s.!<
6. >!The two 4c= are 1 and 2.!<
7. >!There's a domino on 1c>5-4c=, 4c=-4c= and then the right of yellow 4c= which is a double. There's no 2-2, place 1-1.!<
8. >!Place 1-2 on the 4c=-4c= border.!<
9. >!Place 2-6.!<
10. >!Place 2-5.!<
11. >!If bottom 1c5-1c0 is a domino then so is 2c=, 1c3-discard, 1c0-1c5 again, that doesn't work.!<
12. >!So 1c0 on the bottom goes right, there's a domino on the 1c0-pink 2c= border, pink 2c=-1c3 border, discard-1c0.!<
13. >!Bottom 1c5 can't go up because there's no 5-? which fits 1c>3. It goes left, it's the 5-3.!<
14. >!Top 1c5 is the 5-0 then, we already discussed the 1c0 goes right so the 5-0 only fits down.!<
15. >!There is no place for the 4-6 but the 1c>3-2c=, I won't type it up but it's trivial to check -- look for places which can fit a 6 and check whether they can have a 4 as a neighbour.!<
16. >!There's no 3-4 so the 2c= is made from 3-6 and 4-6 with the 4 on 1c>3.!<
17. >!Similarly, 1-6 is on top of 2c<3 with the 6 in the discard, reuse the knowledge about where 6s fit.!<
18. >!Pink 2c= is made from 0-[0/1/2] and 3-[1/2/3] but if it's 2s then the blue 4c= doesn't have a 2 left so it's 0-1 and 3-1.!<
19. >!Place the last 1-?, the 1-4 with the 4 in 1c<5.!<
20. >!If the pink 1c0-discard is the 0-0 then the 0-2 would go into 2c<3 which doesn't work so the 1c0-discard is the 0-2.!<
21. >!1c0-2c<3 is 0-0.!<
22. >!Blue 4c= is finished with 2-3.!<
23. >!1c3-discard is 3-3.!<

This very neatly unravels using our usual tactics.

Before placement >!except one!<.

1. >!To make this easier to describe for the first few steps I will use the chessboard notation: blue 1c4 in the lower left corner is A1. Bottom of green 5c= is D1.!<
2. >!B2 is one half of a double, both neighbours are equal. It doesn't matter which way it goes, A1 will go the same direction and in either direction its the same domino. They move together like flipper paddles.!<
3. >!So that's A1-A2-B1-B2 covered. C1 thus goes right into D1, the green 5c=.!<
4. >!D2 then goes right, it's also a double.!<
5. >!This makes 2c10 a whole domino, there's no 4-6, it's the 5-5.!<
6. >!Now we make a tally of 5s and 6s: 2c9 and 2c10 each needs one, there's two 6s and two 5s, so that's all the 5s and 6s.!<
7. >!2c10 in theory is 4+6 or 5+5 but there's no budget for two 5s so it's 4+6. That's all the 6s.!<
8. >!2c9 in theory is 3+6 or 4+5 but we are all out of 6s, it's 4+5. Each of the four 2c9/2c10 contains a 4, there's one in the 1c4, 4s are booked too.!<

Placement.

1. >!4-[1/2/3] only 4-3 has a double, the pink 5c= are 3s.!<
2. >!Place 4-3 and 3-3 either way.!<
3. >!5c=-5c= is one of 3-[0/5/6], the 5s and 6s are booked, it's the 3-0.!<
4. >!Place 0-0.!<
5. >!1c3 and the pink 3c= has all the 3s, the green 5c= has all the 0s, the 2c9/2c10 has all the 4/5/6, 2c= and 4c= are 1 and 2.!<
6. >!Right of 2c10 can't go left as there's no 4-6 it goes down, it's one of [4/6]-[4/5] out of these only the 4-4 exists.!<
7. >!Left of 2c10 goes into 4c= so it's one of 6-[1/2], there's only 6-2, this 4c= is 2s.!<
8. >!Bottom of 2c9 goes into 2c= so it's one of 5-[1/2], there's only 5-1, this 2c= is 1s.!<
9. >!Top of 2c= is 1-0.!<
10. >!Left of pink 2c9 goes down as there's no 4-5, it's either 4-2 or 5-2, only 4-2 exists.!<
11. >!Finish 5c= with 0-2.!<
12. >!Finish 4c= with 2-1, it only fits left, this 2c= is 1s as well.!<
13. >!Finish 2c9 with 3-5.!<
14. >!Place 1-4.!<
15. >!Place 6-3.!<

I am wondering whether anyone made a bed frame with a tall bookshelf integrated with the headboard as a room divider. This won't reach the walls as there are a nightstand on both sides and I can't drill the ceiling either but I recon the bed frame and especially the 86lbs mattress would work as an anchor against tipping.