u/chx_

I posted the strategies and notation helper here.

Different orders are possible but the logic remains the same. There are many solutions but they are all the same. I kept the counting to an absolute minimum.

1. >!3c0 is three 0s, the 1c0 is the fourth, 0s are booked.!<
2. >!With 0s booked, the 6c≠ contains one of each non-0.!<
3. >!We only have one more counting step ahead of us, instead of counting, we will rely on doubles: 2-2, 5-5 and 6-6 exists, while 0-0, 1-1, 3-3 and 4-4 does not.!<
4. >!If the middle tile of the 3c0 goes left or right it'd be the 0-0 which doesn't exist so it goes down into the 4c=.!<
5. >!The right tile of the 3c0 can't go down as it would be the same as the previous one so it goes to the right. It can't be the 0-2 because you'd need another 0 in the 2c2 to finish it so it's the 0-1.!<
6. >!The tile in the 4c= below the 2c2 now only has equal neighbours, it's one half of a double.!<
7. >!Out of the 0-[2/3/4] only the 2s have a double. Place the 0-2 to the middle of the 3c0, the 4c= are 2s.!<
8. >!The left tile of the 3c0 can't go left as it would be the 0-1 again, it goes down. There are only three 3s and so it can't be the 0-3 because there would be no 3 left for the 6c≠. It is the 0-4 marking the top 3c= for the 4s.!<
9. >!The left 1c1 also goes down, the 2c12 is 6+6, so it's the 1-6.!<
10. >!The last 0-?, the 0-3 can't go left because the other 3c= also can't be 3s so the 0-3 is on the 1c0-1c>2 border.!<
11. >!If the bottom right of the top 3c= goes left it'd be the 4-4 which doesn't exist so it goes down into the other 3c=.!<
12. >!The corner of the top 3c= can't go down as it'd be the same domino, it goes to the left, it's the 4-6.!<
13. >!Now the the left two tiles of the bottom 3c= is a double, the other tile is the non-4 half of 4-[1/3/5] and only the 4-5 has a double so place it and the 5-5.!<
14. >!The top tile of the 2c2 is a 1, the top 2c12 is also 6+6, so the top tile of the 2c2 can't go up because the 1-6 is used, it goes to the right into the 2c7.!<
15. >!If the 2c2-2c7 is the 1-5 you'd need the 2-2 on the 2c7-4c= border but we know the 2-2 is in the 4c=, that's how we found the 4c= is 2s in the first place, so it's the 1-4.!<
16. >!Place the 3-2 on the 2c7-4c= border.!<
17. >!Finish the 4c= with the 2-2.!<
18. >!Place the 6-6 in the top 2c12.!<
19. >!Place the 1-5, 2-6, 3-4 into the 6c≠ any way.!<

Alternatively, with a bit more counting, I tried to not repeat any arguments except the very beginning:

1. >!Same as before, we find the 0s are booked into the 3c0 and the 1c0 and the 6c≠ contains one of each non-0.!<
2. >!Then the 0-3 is not in the 6c≠ but where is it? There isn't enough 3s for a 3c= much less a 4c=, the only neighbour of the 0s that can take a 3 is the 1c>2.!<
3. >!Then where is the 4-3? If not the 6c≠ then the only place for the 3 half is the 2c7. If the 4 is in the 2c7 as well then the 6-6 is above it and the 2c2 can't be made because the 0-2 is booked and the 1-1 doesn't exist. If the 4 half is in the 4c= then all the 4s are booked into the 4c= and the 6c≠ and there isn't one more left to finish the 2c7. So the 4-3 is in the 6c≠.!<
4. >!There's a 2 in the 6c≠, one of 2-[0/2/3/6] but the 2-0 is booked, the 2-2 is a double, the 2-3 would collide with the 3 from the 4-3, it's the 2-6.!<
5. >!Then the last domino in the 6c≠ is the 1-5.!<
6. >!Still, the only place which can take a 3 is the 2c7 so it's 3+4 and then there are only three 4s remaining.!<
7. >!With that, the 4c= are 2s and all 2s are booked, nothing else has four of the same left.!<
8. >!There are three 1s, one is in the 1c1, two 1s remain.!<
9. >!The two 2c12 books all the 6s.!<
10. >!The two 3c= are 4 and 5: the 0s are booked, two 1s left, the 2s are booked, the 3s have been placed, the 6s are booked.!<
11. >!If the top 3c= are the 5s then the 5-5 is on the corner and then the middle tile of the bottom 3c= under it would be one half of a 4-4 which doesn't exist so the 5s are on the bottom and the 4s are on the top.!<
12. >!If the 5-5 is on the right then the left tile of the 3c= would go up but there's no 5-6 so it's on the left.!<
13. >!Finish the bottom 3c= with the 5-4.!<
14. >!The 4-0 has nowhere else to go but the top of the top 3c=: the 0 can't be in any 2c12 and placing the 4 in the 2c7 would put the 0 in the 2c2 and then the 2c2 would need another 2 but all are booked into the 4c=.!<
15. >!Place the 1-6 next to it.!<
16. >!Place the 6-4 under it.!<
17. >!Finish the 2c7 with the last 4-?, the 4-1.!<
18. >!Place the 6-6 above it.!<
19. >!Place the last 1-?, the 1-0 on the 1c1-4c0 border.!<
20. >!Place the 2-2 under it.!<
21. >!Finish with the 2-0.!<

I see multiple units available for rent in Carmana on Alberni. Any horror stories :) ?

I posted the strategies and notation helper here.

Nice puzzle, lots of traps and misdirections but we chart a path without those. A little trial-and-error is needed but luckily nothing deep.

1. >!Today every domino is horizontal instead of vertical.!<
2. >!Start from the bottom right 4c=, the right hand is an equal, above it is a horizontal into the 2c9 and above that is a domino on the 2c2-2c9 border.!<
3. >!Also, there'll be a domino on the 4c=-2c8 border and then this forces a whole domino into the 2c9 which is the 4-5 because the 3-6 doesn't exist.!<
4. >!The available doubles for the 4c=s are 1-1,3-3,4-4,6-6 but there aren't enough 4s now so the two 4c= can be 1,3,6.!<
5. >!If the vertical 2c9 is 3+6 then there are only three remaining of both which means one of the 4c= can't be made. So the vertical 2c9 is 4+5.!<
6. >!Then the 2c9-2c2 is one of 4-[4/6] which are too high or 5-[1/3/6] and only the 1 is low enough so it's the 5-1.!<
7. >!The top tile of the 2c9 is a 1, the bottom tile of the 2c9 is a 4.!<
8. >!The remaining 4-? dominos are 4-[4/6] and only the 6 is candidate for 4c=. Place the 6-6 to the right.!<
9. >!Place the 5-1 on the 2c9-2c2 border.!<
10. >!Place the 6-5 on the 4c=-2c8 border.!<
11. >!The bottom 2c8 is finished with a 3, three 3s remain.!<
12. >!The top 4c= is now 1s, all other candidates are either placed or booked.!<
13. >!Place the 1-1 to the left of the top 4c=.!<
14. >!There will be two more 1s on the top, the last one is the top tile of the 2c2, the 1s are booked.!<
15. >!The 2c8-3c8 border is one of 3-[1/3/5] but the 1 is booked, the 3-3 would need a domino with a sum of 6 but the non-1 dominos make 4 and 8 and not 6 so it's the 3-5.!<
16. >!Finish the 3c8 with the 2-2.!<
17. >!The 2c8 above it is the 4-4 now, the 3-5 is used.!<
18. >!The only remaining non-1 domino is the 3-3, place it to the bottom of the 3c9.!<
19. >!Place the 3-1 on the 3c9-2c2 border.!<
20. >!The 1-0 is too low for the 4c=-1c>0 border so it's the 1-2.!<
21. >!Finish with the 1-0 with the 0 in the discard.!<

I posted the strategies and notation helper here.

11-7=4. Nice. The puzzle is nice but not hard just a little tedious at the very end.

1. >!1c<2 is either 0 or 1 but neither of 1-[0/2] fits the 1c>2 much less the 1c>4. So both 1c<2 is 0s.!<
2. >!Then you have 0-[0/1/4/6] and only the 0-6 fits the 1c>4 and after that only the 0-4 fits the 1c>2.!<
3. >!2c>11 is 12 which is 6+6, place the 3-6 on the 1c3-2c>11 border.!<
4. >!2c11 is 5+6, place the 3-5 on the 1c3-2c11 border.!<
5. >!6-2 is too low for the 1c>2 so finish the 2c11 with the 6-4.!<
6. >!Finish 2c>11 with the last 6-?, the 6-2.!<
7. >!With all the 6s gone, the whole domino at the top of the 3c14 is the 5-5. The next largest sum would be the 4-4 which is too low.!<
8. >!The only 3c14 is finished with the only 4-?, the 4-4.!<
9. >!There is a 3 in the last 1c3, two 3s remain.!<
10. >!The top of the 3c= is a double, the available doubles are 3-3 and 0-0. There aren't enough 3s for a 3c= so it's the 0-0.!<
11. >!Place the last 0-?, the 0-1 to the bottom.!<
12. >!2c7 in theory is 1+6/2+5/3+4 but there are no 1s,6s,4s left so it's 2+5.!<
13. >!The only 5 is the 5-2.!<
14. >!If the 5-2 is on the 2c7-3c7 border then the 2c7 is finished with the 2-3 because the 2-1 is too low for the 1c>1 and the 3c7 needs a domino which has a total of 5 which is also the 2-3 so this is not possible.!<
15. >!Place the 5-2 on the 2c7-1c>1 border with the 2 in the 1c>1.!<
16. >!If the 2c7-3c7 is the 2-3 then you'd need a domino with a total of 4 to finish the 3c7 but the only ones are 3-3 making 6 and the 2-1 making 3. So the 2c7-3c7 is the 2-1.!<
17. >!Finish the 3c7 with the 3-3.!<
18. >!Place the 3-2 on the 1c3-discard border.!<

Note different orders are possible but they do not change the logic at all.

I posted the strategies and notation helper here.

Important realization before placing: >!Much as it was yesterday, today every domino is vertical too. Start with the 2c10, the forced placement cascades to every domino.!<

The rest is easy. There are a lot of solutions.

1. >!So there's a double in the right 3c= and another in the 2c=. The available doubles are 5-5 and 6-6.!<
2. >!If the right 3c= is the 5-5 then the 3c=-2c9 is the 5-6, the 5-[0/2] are too low for the 2c9 but then the 2c9-1c0 needs to be the 3-0 which doesn't exist. So the double in the right 3c= is the 6-6 and the 2c= is 5-5.!<
3. >!With the 5-5 gone, the 2c10 is the 4-6 either way.!<
4. >!With the 6-4 gone, the right 3c= is finished with the 6-5, the 6-[0/2] are too low for the 2c9.!<
5. >!The 1c0-2c9 is the 0-4.!<
6. >!The 1c4 goes up, the 4-1 doesn't have enough for the 3c= so it's the 4-2.!<
7. >!Place the 6-2 on the 1c6-3c= border.!<
8. >!Place the 2-5 on the 3c=-2c10 border, the 2-3 is too low.!<
9. >!Place the 5-0 on the 2c10-discard border.!<
10. >!We have two 6s left, they are too high for 2c5 so one is in the discard, the other is in the 6c≠, this means the 1 from the 1-6 is in the 6c≠.!<
11. >!The 1-4 now needs to be in the 2c5, it can go either way.!<
12. >!Place the 2-3 in the 6c≠. Place the 1-6 and the 0-6 as well with one of the 6s in the discard, all of them are vertical, no other restriction.!<

I posted the strategies and notation helper here.

A new kind of puzzle, I liked it. Not too hard. Two solutions but not really.

Important realization before placing anything: >!every domino is vertical, start from say the bottom left, the 1c1 can only go up then that domino forces the 1c2 to go up and so on for the single cages up to the 1c5. Then the top half of the 2c= must go up which together with the domino on the 1c3-discard forces the domino above the 1c2-6c= to be vertical and so on.!<

The rest is easy.

1. >!Place the 5-5 to the middle on the 1c5-1c5 border.!<
2. >!The 6c= is 0s, nothing else has more than four much less six. One 0 remains.!<
3. >!Place the 4-0 on the bottom 1c4-6c= border.!<
4. >!Place the 2-0 on the bottom 1c2-6c= border.!<
5. >!Above the 2-0 place the 0-0 fully in the 6c=.!<
6. >!The bottom 1c1 is either 1-4 or 1-6 with the other half in the 2c=.!<
7. >!If it's 1-4 then the 2c=-1c<3 needs to be the 4-2 and then the top 1c4 is the 4-3 but the 3 can't be on a 1c<3 so this doesn't work. The bottom 1c1 is the 1-6.!<
8. >!The top 1c1 is the last 1-?, the 1-4.!<
9. >!The bottom of the 2c9 is a 5, the only 5-? which fits a 1c<2 is the 5-0. It was the last 0, the 0s are booked.!<
10. >!Place the 6-2 on the 2c=-1c<3 border, the 6-0 is booked.!<
11. >!The remaining 0-?, the 0-3 and 0-6 are the two last 0s in the 6c=, the are on the 6c=-6c≠ border, either can be left or right of the 0-0.!<
12. >!Above the 0-0 place the 2-5, the other dominos have a 3 half which is already present in the 6c≠.!<
13. >!Place the last 2-?, the 2-3 on the top 1c2-discard border.!<
14. >!The top 1c3-6c≠ can't be the 3-6 as a 6 is present in the 6c≠ already so it's the 3-4.!<
15. >!Finish with the 3-6 on the bottom 1c3-discard border.!<

Kéne utazáshoz néhány tucat egy eurós érme (36-40 db). Hol tudnék váltani? Esetleg valakinek van otthon egy kupac amitől szeretne megszabadulni?

I posted the strategies and notation helper here.

Not a particularly elegant guide today, I am afraid.

1. >!The 6c= is 1s.!<
2. >!All the neighbours of 1c4 are 1s there's no 1-4 so 1c4 goes up.!<
3. >!The left tile of the 2c10 then is one half of a 1-? only the 1-5 is high enough for a 2c10. We do not know which way it goes yet.!<
4. >!Place the 4-5 on the 1c4-2c10 border.!<
5. >!Just pip count the rest. We can skip counting the 1s, the 5 half of the 1-5 because we know the 5 is in the 2c10, the 6+6 has the same amount of pips as 3c12, 6+4 as the other 2c10, the 3+3 as the 2c6. So we have left 5+6 (11 so far) + 4+4 (19) + 5+3 (27) + 2 (29) + 3 = 32 pips on dominos vs 5 + 6 (11) + 9 = 20 in the known cages. This means the 3c= is 12, it's 4s and that's all the 4s.!<
6. >!Further, the 4-6 can't be on the right end because then the 6 would be the left of 2c6 but the right can't be a 0 since there is only one 0, the 0-1 and the 1 is in the 6c=. So it's on the left, with the 6 in the 3c12.!<
7. >!Finish the 3c12 with the 4-4 on the right.!<
8. >!The 2c9 in theory is 3+6/4+5 but there are no 4s left, it's 3+6.!<
9. >!The top tile can't go down because there's no 3-6 so it goes left, there's still no 6-3 so it's the 6-6.!<
10. >!The 3-? at the bottom can't be 3-5 because we'd need another 1 to finish the 2c6, can't be 3-1 because that's booked, it's the 3-3.!<
11. >!Finish the 2c6 with the 1-3.!<
12. >!Now the 1-1 is on the bottom of the 6c= horizontally as both bottom tiles only have 1 neighbours.!<
13. >!The 1-2 can't go into the 3c12 because we would need a 4 to finish it, the 2c10 has the 1-5 so it goes into the 2c5.!<
14. >!Now we know the 1-5 is vertical.!<
15. >!Place the 1-0 horizontal with the 0 in the 3c12.!<
16. >!The 3c12 is finished with the last 6-?, the 6-5.!<
17. >!Place the 5-3.!<

I posted the strategies and notation helper here.

Very nice puzzle. If you are completely stuck, read step 1 but then stop and enjoy solving it. The next major breakpoint is step 8.

1. >!The 1c>10 and 2c10 can only contain 4/5/6 so the 1c1-2c>10 and 1c1-2c10 in theory are two of 1-[4/5/6] but the 1-4 doesn't exist so the 1-5 and the 1-6 are booked on these borders.!<
2. >!With the 6-1 booked, the standalone 1c6-1c<3 is the 6-2, there's no 6-0.!<
3. >!2c7 in theory is 1+6/2+5/3+4. The standalone 2c7 is the 3-4 because the 1-6 is booked and the 2-5 doesn't exist.!<
4. >!There are three doubles, the 2-2, the 4-4 and the 5-5. One is in the standalone 2c=, one is on the right of 3c=, one double remains.!<
5. >!2c10 can be 4+6/5+5. When it's a whole domino then it's the 5-5 because the 4-6 doesn't exist.!<
6. >!So if the horizontal 2c10 is a whole domino then it's the 5-5 and then the 2c= under it needs a double but there's none left so the horizontal 2c10 is two dominos, one goes left into the top vertical 2= the other goes down into the bottom vertical 2c=.!<
7. >!This makes the bottom vertical 2c10 a whole domino, that's where the 5-5 is.!<
8. >!If the 3c= is the 4-4 then it's finished with the 4-5 as there's no more 4s but if there's no more 4s then the 2c10 is 5+5 and on the top vertical 2c=-2c10 border you'd need another 5-5 which means the 3c= is 2s.!<
9. >!Place the 4-4 in the standalone 2c=.!<
10. >!Place the 2-2 to the right end of the 3c=.!<
11. >!Finish the 3c= with the last 2-? the 2-0.!<
12. >!Place the 0-5.!<
13. >!The 1c6-2c= can't be the 6-5 as you'd again need another 5-5 so it's the 6-3.!<
14. >!Place the 3-5.!<
15. >!2c7 again is 1+6/2+5/3+4 but the only 1 is from the 1-3 and the 3 can't be in 1c<3 or 2c10, there are no 2s so it's 3+4.!<
16. >!Place the 3-1 to the 2c7-1c<3 border, the 1 can't be in 2c>10.!<
17. >!Place the 4-5 on the 2c7-2c>10 border, this could've been the previous step as 5 can't be in 1c<3.!<
18. >!The 5-1 is too low, place the 6-1 on the 2c>10-1c1 border.!<
19. >!Place the 1-5 on the 1c1-2c10 border.!<
20. >!Finish with the 5-6.!<

I posted the strategies and notation helper here.

Usual chessboard notation: A1 is lower left corner, a 1, F1 is lower right corner, a 5, A5 is upper left corner, a 2, F5 is upper right corner, a 3.

I am unhappy with the guide today. These puzzles can be solved two ways: either arena driven where you look at a tile and its neighbours determine only one domino can fit there or domino driven where a domino has only one space.

I have two solutions, one is arena driven but it jumps all over the place, the other starts with placing a domino and then it's nice and tight arena driven where every placed domino touches the previous one. Both have its "deus ex machine" style moments, alas when I don't really have a good explanation why that's the next step for the first solution and the first step in the second.

1. >!We always look at corners first, if there's a tile with the same value next to it then that's a starting domino because there are no doubles.!<
2. >!This we find at F5, G5 is also a 3, so F5 goes down, it's the 3-0.!<
3. >!G5 goes down, it's 3-5: left? 3-0, used.!<
4. >!C2 is right, 5-2: up or left? 3-5, used. Down? 5-5, doesn't exist.!<
5. >!C1 is right, 5-0: Left? 5-2, used.!<
6. >!D5 is left, 0-4: Down? 0-5, used.!<
7. >!B5 is left, 4-2: Down? 4-0, used.!<
8. >!F2 is down, 4-5: Up? 4-0 used. Left? 4-2, used.!<
9. >!E2-E1 is 2-1.!<
10. >!G3-F3 is 2-0.!<
11. >!A1 is up, 1-4: right? 1-2, used.!<
12. >!B1-B2 is 2-3.!<
13. >!A3 is up, 1-3: right? 1-4, used.!<
14. >!Place the 0-1, 4-3, 5-1 in any order as long as the 5-1 is not the first.!<

Or

1. >!5-4 has only one place to go, F1-F2.!<
2. >!F3 is left, 0-2: Up? 0-0, doesn't exist.!<
3. >!E2 is down, 2-1: Left? 2-2, doesn't exist.!<
4. >!D2 is left, 2-5: Down? 2-0, used. Up? 2-1, used.!<
5. >!C1-D2 is 5-0.!<
6. >!A1 is up, 1-4: right? 1-2, used.!<
7. >!B1-B2 is 2-3.!<
8. >!A3 is up, 1-3: right? 1-4, used.!<
9. >!A5-B5 is 2-4.!<
10. >!C5 is right, 4-0: down? 4-1, used.!<
11. >!E5 is down, 3-5: right? 3-3, doesn't exist.!<
12. >!F5-F4 is 3-0.!<
13. >!Place the 0-1, 4-3, 5-1 in any order as long as the 5-1 is not the first.!<

You could also start with trial-and-error: >!The 4-4 near the A5 corner is promising so if A5 goes down then B5 does too and so does C5 and then A3 can't be made so A5 goes right and then E2 goes down and you have arrived to F1-F2.!<

Or: >!The same 4-4 but look at C5. It can't go left, if it's right then B5 is left (down would be 4-0 again). If it's down (4-1) then A3 must go up which forces A5 to go right. So no matter which way C5 goes, A5-B5 is 2-4. And then the same E2 goes down and we are good.!<

I posted the strategies and notation helper here.

1. >!All the 1c>4 are 5s.!<
2. >!The 2c0 is 0+0.!<
3. >!The right tile of the 2c0 can't go right because there's no 0-5 and can't go left because there's no 0-0 so it goes up.!<
4. >!The 1c>4 next to it also goes up.!<
5. >!2c2 in theory is 0+2 or 1+1 but neither 0-0 nor 5-0 exists so these two are 0-1 and the 5-1.!<
6. >!The bottom left 1c1 can't go up or left because the 1-5 is used so it goes right, the 1-1 would need three more 1s but there's only one so it's the 1-2.!<
7. >!The bottom right corner goes up, it's the 2-2.!<
8. >!The bottom left 1c>4 goes up, there are only two 4s so it's the 5-3.!<
9. >!The 5 half of the 5-4 is either in the discard or the 1c>4: 4c= is 2 or 3, 0/2/3/4 cages are too low, the 4-5 obviously doesn't fit whole in the 2c6 and the 5 half can't be in the 2c6 either because the neighbours (0/2/3) are too low for the 4.!<
10. >!The only neighbour of both the discard and the 1c>4 which can take a 4 is the 1c4.!<
11. >!The tile above the 5-3 can't go right because the 1c4 is one half of the 5-4 so it goes up, it's the 3-3.!<
12. >!The last tile in the top 4c= can't be the 3-4 because it can't go right into the 2c3 and can't go down either because the 1c4 is one half of the 5-4. So it's finished with the 3-0.!<
13. >!The 2c3 is finished with the last 3-?, the 3-4 with the 4 in the 2c6.!<
14. >!The 2c6 is finished with the 2-0.!<
15. >!The 2c2 is the 1-1.!<
16. >!If the 4-5 is horizontal then the 2-5 is also horizontal under it, if the 4-5 is vertical then the 5-2 is also vertical next to it.!<

Note >!we could've figured out where the 5-4 can be before placing a single domino but I think it was easier to see this way.!<

I posted the strategies and notation helper here.

Today we have an easy puzzle.

1. >!All the 1c>4 are 5s.!<
2. >!The 2c0 is 0+0.!<
3. >!The right tile of the 2c0 can't go right because there's no 0-5 and can't go left because there's no 0-0 so it goes up.!<
4. >!The 1c>4 next to it also goes up.!<
5. >!2c2 in theory is 0+2 or 1+1 but neither 0-0 nor 5-0 exists so these two are 0-1 and the 5-1.!<
6. >!The bottom left 1c1 can't go up or left because the 1-5 is used so it goes right, the 1-1 would need three more 1s but there's only one so it's the 1-2.!<
7. >!The bottom right corner goes up, it's the 2-2.!<
8. >!The bottom left 1c>4 goes up, there are only two 4s so it's the 5-3.!<
9. >!The 5 half of the 5-4 is either in the discard or the 1c>4: 4c= is 2 or 3, 0/2/3/4 cages are too low, the 4-5 obviously doesn't fit whole in the 2c6 and the 5 half can't be in the 2c6 either because the neighbours (0/2/3) are too low for the 4.!<
10. >!The only neighbour of both the discard and the 1c>4 which can take a 4 is the 1c4.!<
11. >!The tile above the 5-3 can't go right because the 1c4 is one half of the 5-4 so it goes up, it's the 3-3.!<
12. >!The last tile in the top 4c= can't be the 3-4 because it can't go right into the 2c3 and can't go down either because the 1c4 is one half of the 5-4. So it's finished with the 3-0.!<
13. >!The 2c3 is finished with the last 3-?, the 3-4 with the 4 in the 2c6.!<
14. >!The 2c6 is finished with the 2-0.!<
15. >!The 2c2 is the 1-1.!<
16. >!If the 4-5 is horizontal then the 2-5 is also horizontal under it, if the 4-5 is vertical then the 5-2 is also vertical next to it.!<

Note >!we could've figured out where the 5-4 can be before placing a single domino but I think it was easier to see this way.!<

I posted the strategies and notation helper here.

There's a lot of solutions but that doesn't bother us. The only challenge here is you need to count a little in the middle but I kept it to the minimal possible.

1. >!The corner of the 4c= is one half of a double because both neighbours are equal, the base of the "T" in both 5c= are one half a double because all three neighbours are equal.!<
2. >!The available doubles are 3,4,5 so that's what the 4c= and the two 5c= are.!<
3. >!Then where are 6s? One can be in the top discard, one in the 6c≠, two in the 2c= and nowhere else, all the 3 and 2 restricted cages are too low. And that's exactly how many 6s we have.!<
4. >!The bottom of the 2c= can't go up because there's no 6-6 so it goes to the right, it's the 6-2.!<
5. >!Then the top discard can't go up instead it goes to the left. Out of the available 6-[0/1/3] only the 3 can be an 5c=.!<
6. >!Above it the 1c2-2c2 border is a 2-? and only the 2-1 fits.!<
7. >!Finish the 2c2 with the 1-3 either horizontally or vertically and place the 3-3 next to it in the same direction.!<
8. >!We have three 1s left and at most one can go into the 6c≠ and so at least two are in the 3c3 which means all three are, that's all the 1s so we know the 6c≠ doesn't contain a 1, it must contain one of everything else.!<
9. >!There are six 4s and five 5s, one of each is in the 6c≠ so the 5c= is 4s, the 4c= is 5s.!<
10. >!Place the 3-4 on a 5c=-5c= border either horizontally or vertically and the 4-4 next to it in the same direction.!<
11. >!The top of the 2c= can't go up because there's no 6-5 and so it goes to the right, it's the 6-1.!<
12. >!Place the 5-1 above it.!<
13. >!Place the 5-5 above it.!<
14. >!Finish the 3c3 with the 1-4.!<
15. >!Place the 5-4 above it.!<
16. >!The 6-0, 4-2, 5-3 go into the 6c≠.!<

I do not know what to do.

1. I do not want to use Windows 11. Even before they started vibe coding it was full of bugs and missing features and now it's a pile of garbage.
2. I do not want to use Wayland. The entire concept of it is completely wrong. Which is why there's still no screenreader and there won't be without hacks. Overall, it sucks.

I am fine with Windows 10. It works and WSL provides me with dev tools. And there's a version which is supported until 2032. But Slack will stop working on it in a few weeks. I am fine with Ubuntu 22.04, it's certified on my laptop and Canonical will support it just the same until 2032. The few games I want to play work more or less. But Slack will stop working on it about a year from now.

I do not want to use the browser version, notifications are a broken concept within a browser because I have zero desire to grant the entire browser notification permissions and also I would lose the taskbar icon and it's disastrous I already miss Viber messages because it doesn't keep the taskbar icon inverted when messages are waiting. I miss Skype.

Soooo what to do?

I posted the strategies and notation helper here.

This one is really challenging. I will mark the key points with a (*) so you can enjoy the solving yourself stopping after each. After the second you shouldn't need the guide, really.

1. >!3c0 is three 0s. Also the middle tile is one half of the 0-0.!<
2. >!2c9 in theory is 3+6 or 4+5. But it can't be 3+6 because the left tile has nowhere to go: there's no 3-6, 0-3 or 0-6. So it's 4+5. Two 5s remain.!<
3. >!The 1c>4 in the middle 1c>4-3c= domino can't be a 6: 6-5 is not possible as the 3c= now can't be 5s, 6-6 is not possible either because there's only one 6 remaining. So it's a 5.!<
4. >!There's a 1c5 so that's all the 5s.!<
5. >!The 2c10 in theory is 5+5 or 4+6 but the 5s are booked so it's 4+6.!<
6. >!If the 4 in the 2c10 is on the left then it's the 4-5, if it's on the right then it's the 4-4 because the 4-5 is booked and the 4-[0/1/2] are too low for 1c>2.!<
7. >!What is the 1c5-3c=? If it's the 5-4 then the 3c=-1c>3 is the 4-4 and now the 2c10 can't be made. If it's the 5-6 then there are not enough 6s to finish the 3c= as the other two dominos are vertical.!<
8. >!So the 1c5-3c= is the 5-1. (*)!<
9. >!The 3c=-1c>3 is the 1-4, the 1-[0/1/2/3] are all too low.!<
10. >!Now we can look at the 2-2, 3-3, 4-4, 6-6 together and try to place them. They fall almost all places together, here are the places where they can't be (*)!<
11. >!1c>2-2c2 and 2c2-1c<2 on the left edge, the 2-2 can't be in a 1c>2 or 1c<2 the rest are too large for the 2c2!<
12. >!horizontal 3c=-vertical 3c= border because the bottom of the vertical 3c= is a double and it'd be the same double!<
13. >!1c>2-1c<2 on the left top, 1c>3-1c<3 on the right bottom because the same number can't be larger and smaller than anything at the same time.!<
14. >!2c9 wholly, these are doubles so they are even, the 2c9 is odd!<
15. >!2c9-2c6 border that could only be the 4-4 as the 2c9 only contains a 4 and a 5 but there's no 0-5 to finish the 2c9.!<
16. >!2c9-1c<2 border, anything with the 3c0!<
17. >!So in the entire "M" there are two places: the 2c10-1c>2 and the bottom of the 3c=. In the "A" the only places are the 2c6 wholly and the 2c=. That's four places for the four doubles so that's where they are. (*)!<
18. >!So the 2c6 is made from the 3-3.!<
19. >!Then the 2c9 is made from the 5-4 whole, either way.!<
20. >!The 0-0 is on the bottom of the 3c0.!<
21. >!The 3c0-1c<2 is the 0-1.!<
22. >!With the 5-4 gone, the 1c5-2c10 is the 5-6.!<
23. >!The 2c10-1c>2 is the 4-4.!<
24. >!The vertical 3c= can't be 6s because there's no 1-6 for the horizontal 3c=-vertical 3c= so it's the 1-2 and the 2-2.!<
25. >!The 2c= is 6-6 then.!<
26. >!The 1-1 only fits the 1c<2-2c2 border: none of the 1c>2 nor the 1c>3 can take a 1.!<
27. >!The 2c2 is finished with the last 1-?, the 1-3.!<
28. >!The 4-2 doesn't fit the 1c>2-1c<2 so it's in the 1c>3-1c<3.!<
29. >!The 1c>2-1c<2 is the 4-0.!<

I posted the strategies and notation helper here.

Really, the dominos basically fly to the board.

1. >!If the upper right corner of the 4c= went down then the same double would need to be next to it so it goes to the right, it's a horizontal double.!<
2. >!Then the lower left can't go right because that'd be the same double it goes down into the 2c12.!<
3. >!The 2c12 is 6+6, the posssible dominos are 6-[1/2/3/6] and only the 6-1 has a corresponding double. Place it and the 1-1 to the top.!<
4. >!The last 1-? is the 1-3 is on the 4c=-2c5 border.!<
5. >!The 2c5-2c12 is the 2-6.!<
6. >!The 5c= are 4s, nothing else has enough, all the 4s are booked.!<
7. >!Without 4s, the 2c10 is 5+5.!<
8. >!The bottom tile of the 2c10 can't go up because there's no 5-5 it goes to the left, it's the 5-4.!<
9. >!The top is one of 5-[2/3] but the 5-2 would need a 4 to finish 2c6 so it's the 5-3.!<
10. >!Finish the 2c6 with the 3-4, can't go up because it'd orphan a tile.!<
11. >!The 5-2 is on the bottom 1c5-2c2.!<
12. >!The 4-4 has three places where it can go, vertical-horizontal-vertical, choose one. The remaining two tiles are made from the 2-4 on the bottom and the 0-4 with the 0 in the discard, it's easy to see all three 4-4 placement leads to one solution.!<
13. >!The 0-3 can only be on the 1c<3-2c6 border.!<
14. >!Place the 3-6 on the 2c6-1c6 border.!<
15. >!Place the 6-6 in the 2c>8.!<
16. >!Finish with the 0-0 in the top 2c=.!<

I posted the strategies and notation helper here.

This is another puzzle which looks real hard when randomly trying but careful application of logic solves this real quick. It's not easy but it's quick.

Before placement.

1. >!On the top the 3c3 contains a whole domino and only the 0-0, 0-1 and the 0-2 are low enough. (It contains a whole domino because whether the corner goes left or down it's a whole domino.)!<
2. >!There are four 1c4, one 4 remains: neither of the 3c= are 4s.!<
3. >!On the bottom, there's a vertical domino on the 1c<4-1c>4 border which puts a domino on the 1c<3-3c= border and also there's a double in the 3c=.!<
4. >!On the top, there's a double at the bottom of the 3c= and then the top tile can either go up or right.!<
5. >!Let's presume it goes to the right which is again a 1c<3.!<
6. >!There are five doubles, 0-0, 2-2, 3-3, 4-4, 6-6 but we already ruled out the 4-4.!<
7. >!We need to find other halves for them which are less than 3. I will write down all the other halves but when you are solving you only need to keep in your head the very few which are less than 3.!<
8. >!Let's enumerate the available other halves for them: 0-[1/2], 2-[0/4], 3-[4/5/6] so none for the 3s, 4- 6-[3/5] and none for the 6s.!<
9. >!So halves for the 1c<3 can only be found for the 0-0 and the 2-2. The 2-2 needs the 2-0 and then the 0-0 only has the 0-1 and now the 3c3 can't be made.!<
10. >!This means the top tile of the 3c= can't go right, it goes up.!<

Placement.

1. >!So the top 1c<3 goes right into the 1c4, the only 4-? that fits into a 1c<3 is the 4-2.!<
2. >!Then the top 1c4-3c3 is the 4-3, the 4-4 and 4-5 are too high.!<
3. >!And the whole domino is the 0-0.!<
4. >!On the bottom the 3c= is made from 2-2 and 2-0 because the 0-0 has been used.!<
5. >!The 1c3 here can't go to the left because the 3-4 is used so it goes up which makes the 1c4 also go up. The available dominos are 4-[4/5] and 3-[3/5/6] so it's the 4-5 and 3-5.!<
6. >!The 2c<3 is the 0-1.!<
7. >!With the 3-5 gone the top 1c3-1c>4 is the 3-6.!<
8. >!On the left, the 1c4 is the 4-4, no other 4-? is left.!<
9. >!The 1c3 is the 3-3, no other 3-? is left.!<
10. >!The top 3c= is 6-6 and 6-5.!<
11. >!Place the 5-1 on the 1c>4-1c<4 border. (We could've started placing with this one because it never had any other place but it wouldn't have helped at all and arguing for it against an empty board would've been tedious.)!<

I posted the strategies and notation helper here.

The guide is not the simplest today. Actually, it might be the longest guide I've ever posted. Owie.

Before placement.

1. >!2c4 in theory is 0+4/1+3/2+2.!<
2. >!The only 3-? is the 3-0 and it is not in 2c4 because the 0 is not in the 2c10 or the 2c7 and it can't be wholly in the 2c4.!<
3. >!So the 2c4 is either 0+4 or 2+2.!<
4. >!2c7 in theory is 1+6/2+5/3+4.!<
5. >!Let's presume the 2c7 is 3+4.!<
6. >!The only 3 is the 3-0 and 0 can't be in a 2c10 or a 2c7 so it is in the 2c4 so the 3-0 is on the right.!<
7. >!The bottom tile of the 2c4 is a 4.!<
8. >!The left tile of the 2c7 is also a 4 and it can't go up because that'd orphan a tile so it goes down.!<
9. >!So the 2c10 is made from two non 4 halves of 4-0/4-1/4-6 and those do not make 10.!<
10. >!So the 2c7 is not 3+4, it's 1+6 or 2+5.!<
11. >!We will count 5s and 6s together, the count starts from eight, there are five 5s and three 6.!<
12. >!The 2c>10 is either 11 or 12, that is 5+6 or 6+6. Six of 5s and 6s remain. Also note neither 5-6 or 6-6 exists, it's two dominos.!<
13. >!2c9 in theory is 3+6/4+5 so it also contains either a 6 or an 5. Five 5s and 6s remain. Also note neither 3-6 nor 4-5 exists so it's two dominos.!<
14. >!We now know the 2c7 is either 2+5 or 1+6 which means it contains either a 6 or an 5. Four of 5s and 6s remain.!<
15. >!2c10 in theory is 4+6/5+5 so each also contains at least one of 6 or 5. Two of 5s and 6s remain. (Can be less if either is 5+5 but it's enough to know no more than two remains of these together.)!<
16. >!The 5c= is either 5s or 1s but there are at most two 5s left so it's 1s. One 1 remains.!<
17. >!The 2c2 in theory is 0+2/1+1 and with only one 1 remaining it's 0+2.!<
18. >!The 2c= can't be 0 or 2 because the neighbours of the top tile would be 0 or 2 and there's no 0-0/0-2/2-2.!<
19. >!The 2c= can't be 1 or 3 because there's one left of those.!<
20. >!The 2c= can't be 6s because the top tile has 6/0/2 potential neighbours, so the top is 6-2 horizontally and then neither the 6-1 nor the 6-4 can be the bottom one.!<
21. >!The 2c= is either 4s or 5s.!<
22. >!Let's presume the 2c= are 5s.!<
23. >!That's the two extra out of all 5 and 6s which means both 2c10 are 6+4. We budgeted one of 5 and 6 into them, the 5+5 would need a second but all of them are booked. One 6 is left.!<
24. >!The 2c>10 contains a 6, so that's all the 6s.!<
25. >!Without 6s the 2c9 is 4+5. That's all the 4s.!<
26. >!Without 6s the 2c7 is 5+2.!<
27. >!The 2c4 is 0+4/2+2 but there are no 4s left so it's 2+2.!<
28. >!If the top tile of the 2c4 goes down it's 2-2 which doesn't exist.!<
29. >!If the top tile of the 2c4 goes left then the bottom also goes left which is the 2-6 and the 2-4 but the 2-4 doesn't exist.!<
30. >!So the top tile must go up and since the 2c7 is 5+2 and there's no 2-2 that's the 2-5. The left tile of the 2c7 is a 2.!<
31. >!There's a domino on the bottom 2c10-2c4 border, the 1-2 is too low, it's the 6-2. The top tile of the 2c10 is a 4.!<
32. >!The left tile of 2c7 can't go up because that'd orphan a tile so it goes down into the bottom 2c10 but there's no 2-4 so this doesn't work.!<
33. >!The 2c= are 4s.!<

Placement.

1. >!The top tile of the 2c= is a 4. If it's vertical then it's the 4-4, if it's horizontal then either 4-0 or 4-2. Out of all these only the 4-0 exists, place it horizontally.!<
2. >!If the bottom tile is vertical it's 4-5 or 4-6, if it's horizontal it's the 4-2. Out of all these only the 4-6 exists, place it vertically.!<
3. >!The 2c2 is finished with the 2-1 vertical.!<
4. >!The 3 half of then 3-0 can't be in the 2c4 either because the 0 is not in the 2c10 or the 2c7 and it can't be wholly in the 2c4 so the 2c4 is either 0+4 or 2+2.!<
5. >!The 2c4 is 0+4 or 2+2 and if it's 0+4 then the only 4-? one is the 4-1 which is too low for 2c10 and can't be wholly in the 2c4 so it's in the 2c7 and then there is a domino on the 2c4-2c10 border which can only be the 0-5 as the 0-3 is too low and then, as before, there is a vertical on the 2c10-2c7 border which wants to be the 5-6 but that doesn't exist.!<
6. >!So the 2c4 is 2+2.!<
7. >!Only the 2-5 and 2-6 remains, if the 5 is in the 2c7 then you'd need a 2 to finish it so the 2-5 goes into the 2c10 and the 2-6 goes into the 2c7.!<
8. >!This puts a whole domino in the top 2c10, the 4-6 is gone so it's the 5-5.!<
9. >!The 2c7-2c10 border is the 1-5.!<
10. >!With the 1-5 gone, the bottom of the 2c>11 is the 6-1 either vertical or horizontal.!<
11. >!The 1-1 goes next to it in the same direction.!<
12. >!Out of the remaining, the 2c9 is 4+5 and that books both of those.!<
13. >!Place the last 1-?, the 1-4 on the 5c=-2c9 border.!<
14. >!Place the 0-5 on the discard-2c9 border.!<
15. >!Place the 3-0 on the 1c>0-discard border.!<

I posted the strategies and notation helper here.

This is very very easy.

1. >!The middle of the 4c= on the bottom is a double.!<
2. >!The bottom 1c5 is either the 5-3 or the 5-2. The 5-2 doesn't have a corresponding 2-2, so it is the 5-3 followed by the 3-3 and the 3-1. !<
3. >!The last 5 is on the 1c5 on the right bottom corner.!<
4. >!Without 5s the 2c9 is 6+3.!<
5. >!There's another 3 in the 1c3 on the top right corner.!<
6. >!The last one is in the 1c<4 (lower right): it's a single 3 so it can't be in any equal cages. The rest obviously can't take a 3.!<
7. >!The 3-2 and the 2-0 doesn't fit either 1c>3 or 1c5 so it's the the 3-4 with the 4 in 1c>3.!<
8. >!The 5-2 is vertical next to it.!<
9. >!5c0 is all 0s, the 0s are booked.!<
10. >!The 1c3 must be the 3-2 going down, left would be the 3-4 which has been used, the 3-0 is booked into the 5c0.!<
11. >!Then next to it the 4 can't go to the left because there is no 4-0, it goes down so it is the 4-2.!<
12. >!The last tile in this 3c= can't go up because there's no 2-0, so it's the 2-1 going left into the 1c1.!<
13. >!The 0-0 is above it.!<
14. >!The 6-0 and the 3-0 are horizontal, two solutions, either can be top or bottom.!<
15. >!The 2c2 without zeros is 1+1, the 3c= are 4s and the 2c= is going to be 6s.!<
16. >!Place the 6-6 to the bottom.!<
17. >!Place the 0-1 to the 5c0-2c= border.!<
18. >!Place the 4-4 to the top of the 3c=.!<
19. >!Finish with the 4-1.!<