u/jxd73

Inventory ( "*" denotes the presence of a double)

0: 7*

1: 3

2: 6*

3: 6*

4: 3*

5: 1

6: 4

1. >!We'll start with the 1c4. We have 2-4 and 4-4, neither works with the adjacent tiles to the left and above, so it goes right into the 3c= and it must be the 2-4 as we don't have enough 4s. The orange 3c= is made from 2s!<
2. >!There are seven 0s, 1 is clearly marked, two are needed in the 3c1, one in the 2c1, leaving 3 extra. There can be none in the 2c8, none in the 3c14, at most one in the 3c9. Conclusion, the green 3c= must be filled by the remaining 0s. All the 0s are reserved!<
3. >!One possibility for the 0-5 is top left, bridging 2c8 and 2c1, this would force the tile to the left to bridge the orange 3c= and 2c1, and it must be the 1-2 which doesn't exist, therefore 0-5 cannot go there.!<
4. >!It'll have to bridge the 3c14 and 3c= in the bottom row, also placing the 0-0!<
5. >!Above 0-0 we have either the 0-6 or 1-6; over in the top left corner we also need one of the same two pieces, ruling them out for other locations!<
6. >!Thus we cannot have the 1-6 reaching from the teal 1c6 in the third row into the 3c9, which is one of two possible 4-4 placements. This means the only area that can accommodate the 4-4 is the 2c8!<
7. >!Since we do not have 1-2, the only way to connect the orange 3c= and the 2c1 is by using 0-2, then 1-6 to the right!<
8. >!Multiple pieces can now be forced, starting from the top row 1c3 going counter clockwise, 2-3, 0-1, 0-6, 0-3!<
9. >!Down below, the 2-2 is too low to make the 3c14, it has to be the 2-6 on the blue 1c2, 3-6 is forced to the teal 1c6 and 3c9!<
10. >!1-3 completes the 3c14, finish the puzzle with 2-2 and 3-3!<

Again there are multiple solutions, I'll do the easiest one which is different than NYT's.

1. >!The bottom 2c3 has to be the 0-3, and the >5 above that is our only 6, 3-6 which must extend into the discard to not trap the 1c3!<
2. >!At this point we can take stock of the pieces we have, we have three of 0s and 1s, five 2s and 4s, four 3s and 5s. 2-2 and 4-4 are our doubles.!<
3. >!Since we don't have another piece valued at 3, the outer tiles of the top 2c3s point down into the 3c= and 1c2 respectively.!<
4. >!There are two 1c2s, but we also need 2s for one of the 3c=s, therefore we have no 2s for the 2c3s, making both 0+3!<
5. >!This means, with the purple 1c0, all 0s are effectively marked, and both <2s are 1!<
6. >!What about the 5s, even if both discards are 5s, one must be in an =/= area because the 2c8 can accommodate just one; if the 2c8 is 4+4, then both discards are 5!<
7. >!in the 2c8 is 3+5 scenario, we would have all the 3s booked as well: one in the 2c8, one each in the 2c3s, and a 1c3. Furthermore, the 2-4 is forced to the right bottom corner, meaning 1c0 is the 0-5 since at least one discard must be 5!<
8. >!In the 2c8 is 4+4 scenario, the 4-5 is forced to the same corner, and the 1c0 is the 0-5!<
9. >!Ergo we know that no matter what the 0-5 covers the 1c0!<

Branching starts here.

1. >!Let's pretend the discards actually form a 2c10, it'll have to be split since we have no 5-5, both tiles "go up", placing the 0-5. The other tile has to be the 4-5 since it needs a digit with a double. 4-4 finishes up the right side 3c=, and 2-2 in the other one!<

2. >!2-4 cannot be in the red 1c2 extending into the purple =/=, because that would make the 2c3 a single domino, the only other place it can go is the green 1c2 and 2c8. This would in turn force the 3-4 above it, and 0-2 in the left corner!<

3. >!The 2-3 is forced into the red 1c2, and taking the 0-1 with it.!<

4. >!3-5 can go in either the purple =/= or the 1c3, doesn't matter. And the other two pieces can go where ever!<

Very unsatisfying IMO.

>!We start with two 1s and five 2s, but the 1-2 contains both so it really counts as 0.5 for each digit!<

>!The 3c= requires a double, and since 2-2 is the only one, and the 1c2 next to this area, the 2-2 has to be at the very bottom!<

>!there are three pieces that can fit in the <8, 1-2, 2-3, 1-4!<

>!Putting 1-4 in the <8 means we not only have to use the 1-2 in on the <2, but also another 2 in the <8, depriving us of enough 2s to complete the puzzle!<

>!Similarly 2-3 will also deprive us enough 1s or 2s!<

>!Therefore it's the 1-2 that must be in the <8!<

>!The next piece with the most constrained placement is the 4-6, it cannot go entirely in the 3c11 because we need another 1, but if it goes across the 3c11 and =/=, then it forces the 1-4 above and we are back to needing another 1. The only other possibility is putting its 4 in the <8!<

>!The only way to make the 3c11 at this juncture, giving the remaining pieces and without 1s and 2s is: 3+3+5, this forces the 1-4 to extend down into the =/= from the <2!<

>!which forces the 3-5 to its left and the 2-3 below!<

Pretty straightforward.

1. >!the bottom of the E requires a 0 or a 1 connecting to a 5c= which requires a double, the 1-3 is the only possibility, thus forcing the 0-6 to its right. Also placing the 3-3.!<

2. >!the 1-1 to the = in the I!<

3. >!in the S, the 2c2 requires two 1s, obviously one is the 1-2, the other could be 1-5, but then none of the 2s left, with either 3 or 4 can fit into the 2c11, therefore it must be the 1-4!<

4. >!At this point we need at least three 3s to finish the puzzle, two in the E, and one in the 2c7. We have four left, therefore 3s cannot fit in the = area in the L!<

5. >!Neither can 2s, there isn't a 2 with a <3, therefore 2-3 is ruled out for the L. The 2-3 also cannot fit anywhere in the S because it's too small for the 10 and the 11, It has to go to the discard in the E!<

6. >!In the S, the only potential placement of the 3-4 is spanning the 2c10 and 2c8 (with the 3 in the 8), this would force the 6-5 to its left; however, this leaves us needing a 5-6 or 6-6 above to complete the 2c11, which is impossible. The only other place for the 3-4 is the discard and 2c= of the L. 2-4 is forced into the L also!<

7. >!1-5 to the 2c6!<

8. >!With all the 2s and 3s gone, 2c8 is 4+4, placing the 4-6, 4-5, 5-6, and 3-5 in the S!<

9. >!3-6 finishes the E!<

I found this one tricky and had to do a pip count.

1. >!first, the 2c6 has to be split since no pieces have a face value of 6, so left tile of the 2c6 goes down, a piece needs to span the 2c7 and the blue 2c=. A piece also needs to bridge the purple 2c= and the 2c2!<

2. >!At a glance, four digits are capable of filling the = areas, 0, 1, 5, 6!<

3. >!But since we need one 6 for the 2c11, and the rest of the 6s come from the 6-6, and that we know all the = areas are split, 6 cannot go into any of the = areas!<

4. >!If we do a pip count, we arrive at a difference of 19 (this diff includes the >3 and the discard)!<

5. >!the only two pieces that can fit the >3 and the 2c11 are 6-6, and 4-5!<

6. >!we can use the equation 2a+2b+c+d=19, c stands for the >3, d for the discard. Since c is even(4 or 6), and 19 is odd, we can deduce that d is an odd number. Also, one of the = area has to be 5s, as 19 is too large for 0s and 1s to tally up to, this uses up all the 5s!<

7. >!if the discard is 1, the adjacent = must be 0s, the only other 0 is the 0-5, too large for the 2c2, this leaves the 3-5 as the only choice for the discard and 2c=!<

8. >!0-5 is forced to span 2c2 and 2c=, 1-2 is too low for 2c11, so it goes across into the other 2c=, and we need the 1-6 to reach into 2c7 as the 0-1 is too low!<

9. >!next 0-1 and 6-6 finish the upper left corner, finally the 4-5 fills the tip!<

>!no matter which 1c2 the 1-2 takes, it has to be vertical because the 1 is too low for the 2c8!<

>!we can systematically rule out possible placements for the 0-0: can't span the 5 because we do not have a 5, can't go in the 8 because it's too low, can't span the 7 from the 6 because we need the 2-4 for the 3c6 making the puzzle impossible to finish. The only valid placement for the 0-0 is thus entirely in 3c6!<

>!From here it's very straightforward, complete the 3c6 with 3-6, the right side 1c2 needs to "go up", it can't be the 1-2 because there isn't a second 3 to make the 7 by 1+3+3, so it's the 2-4. 0-4 is forced to the top, then 1-4 and 4-4 for the 2c8!<

1. >!0 is the only digit numerous enough for the 8c=, all eight pieces must interact with the 8c= in someway.!<
2. >!without pieces with a 0 and doubles, we have four pieces left for the =/=: 1-5, 2-6, 3-4, 1-3!<
3. >!Of these, 2-6 and 3-4 must be in the =/= since 2/4/6 are unique!<
4. >!1-3 conflicts with 3-4 meaning 1-5 is the last piece in the =/=!<
5. >!the 2c9 in the upper area could be 4+5, but the sole remaining 4 and 5 both come with a 0, they cannot both fit into this area, thus it must be 3+6!<
6. >!Our only 6 is the 0-6 so it must bridge the 8c= and 2c9!<
7. >!the 0-0 is forced by geometry to be in the middle of the 8c= area!<
8. >!there are three tiles that must be a 3 that cannot reach into the 8c= (where the 0-3 has to be), so the 3-3 must be 1c3 and 2c9, lest we run out of 3s.!<
9. >!1-1 must be on in the 1c1, and forcing 0-4 to be just below!<
10. >!the 5 in 0-5 is forced to be in the >3!<
11. >!The 3 of the 0-3 is forced to be in the 3c6, making the other two tiles 1 and 2!<
12. >!The only valid placement for the 2-2 at this juncture is above the 0-3!<
13. >!1-3 in the last 1c3!<
14. >!0-1 for the lower spot of the 2c3 and 0-2 in the <3!<

1. >!In the M, we have a horizontal 4c=, but all the pieces have to be vertical. Which means we need four separate pieces !<
2. >!immediately we can rule out 4s for this area as there aren't enough; 0 is out because we do not have a 0-0; though we technically have just enough 1s, 3s, and 5s, two of each are in doubles so it's impossible to make that area with them.!<
3. >!We are left with 2s (five) and 6s (six), suppose it's 2s, due to the 1c2 we'd have to use up all the 2s.!<
4. >!But this presents a problem in the A, without 2s, the 2c2 must be the 1-1 as we do not have a 1-3, then it becomes impossible to make the 3c3 (0-1 requires a 2, and then the next smallest piece is the 2-2)!<
5. >!Ergo the 4c= is 6s. We can place 0-6 and 2-6.!<
6. >!Now it's impossible for the 6-6 to go in the purple 3c= in the M, for that we would have another 6-6; it's impossible for it to protrude into the 3c9, we do not have a 1-2 or 0-3. Therefore it must go to the >9.!<
7. >!What about 5-5 then? It can't go in the 3c= in the A as we have no 0-5, it's too large for elsewhere in the A, too large for the 9, so it must be in the 3c= in the M. 5-6 and 3-6 are also placed.!<
8. >!Next up is the 4-4, the only place it fits is the 3c= in the A, also forcing 0-4.!<
9. >!Now 3-3 has only the 3c9 as a valid location!<.
10. >!1-5 must be in the =/=!<
11. >!our last 3, the 2-3 goes to the top of the A!<
12. >!Finish up the A with the 0-1 and 1-1.!<

A lambda? A duck with clown shoes?

1. >!Because of all the size 2 cages of large values using up 5s and 6s, only 2s and 3s are eligible for the 5c=!<
2. >!there are three 1s, let's look at 1-1, it cannot be placed except somewhere in the 3c=, more precisely, the left side of that area otherwise an area of an odd number of tiles will be cut off!<
3. >!with all the 1s gone we need 2s for the 2c2 and at least one 2c4, we start with five 2s, which means the 5c= is 3s!<
4. >!furthermore, the lower right has to be a combination of 2-5, 0-3, or 0-5, 2-3. Since we do not have 0-3, it's the latter!<
5. >!1-3 is forced to overlap the two = areas!<
6. >!The top of the 5c= needs a 3-x reaching into the purple 2c10, lest we have another odd number of tiles area, and a 0-y where x+y=10. Since 0-5 is already used, they are 3-6 and 0-4.!<
7. >!2-4 is forced to the 1c4 and 2c4!<
8. >!the blue 2c4 is 2+2, both pieces vertical. The 0-2 is forced!<
9. >!The orange 2c10 is 5+5 since we are out of 4s, placing the 2-5 and 5-6!<
10. >!2-6 and 3-5 are forced into the lower left!<
11. >!both 2c11s are 5+6, to make it work the 5-5 connects the two lower tiles!<
12. the rest are forced.

Another fairly straightforward hard, have we weathered the storm, or are we simply in the eye?

1. >!the only natural 8 is the 2-6!<
2. >!five 6s, 4 are marked in the two 12s. Four 5s, one marked by the 1c5. There are two 2c10s, 5s alone is not enough for them, thus one is 5+5, the other 4+6. As a result all 6s are booked as well!<
3. >!this lefts 3s as the sole candidate for the 5c=.!<
4. >!and makes the 2c8 split, and locks in the 0-3 horizontally.!<
5. >!there isn't a 5-6, so the 1c5 extends down and needs a double, it's obviously the 1-5.!<
6. >!the 2c10s are bordered by a 2, the 5c= (3s), another = and of course each other, given we have three 5s but two are bound in the 5-5. If the 2c10 were split, a solo 5 would end up in the adjacent areas. Therefore, one of the 2c10 is not just 5+5, it is the 5-5.!<
7. >!if it's the left 2c10, the top left 1c2 would be stranded, hence the right side 2c10 is the 5-5.!<
8. >!This forces a double to the right of the 5-5, which has to be the 2-2, 6-6 is forced between 2-2 and 1-5, then 3-3 in the middle of all the other doubles.!<
9. >!4-6 and 3-6 fill out the remaining 2c12!<
10. >!3-4 and 2-6 round out the upper section.!<
11. >!two 2s left, but one is needed in the 1c2, the last remaining = is 4s.!<

P.S. I did this entirely from memory, apologies for any mistakes.

0. >!all 0s and 6s are booked from the start.!<

1. >!in the fat T, because we do not have 6-6 nor 0-6, the red 1c6 must go horizontal, this makes the bottom 2x2 a self contained grid.!<

2. >! 5-6 is ruled out for this spot because 5s are not paired with <3s !<

3. >! Note that though we have three 3s, two of which are bound with 0 and 6, which means they have to be in the fat T.!<

4. >!if the orange 2c6 at the bottom of the fat T is a single domino, it has to be the 2-4, the 3-6 is forced to its left and 2-6 forced above, then the 1-2 or 2-2 reaching into the <3 because we have no 0-2.!<

5. >!this forces the red 2c6 to be a single domino or we'd have an odd number of tiles at the upper right corner, we have already used our only natural 6, therefore we reached a contradiction and the 2-4 does not go in the orange 2c6 meaning the bottom 4 tiles are made from two horizontal pieces.!<

6. >!As such, given there is no 3-3, we must use the 2-6 and 3-4. (I originally thought this was the only valid 3-4 from the start but I discovered my mistake while double checking)!<

7. >!since we already ruled out 5-6 for the red 1c6, it must be the 3-6, and also forcing the 0-3 above.!<

8. >!without 0s and 3s, the 2c4 must be 2+2, the lack of 0-2 means it's a single domino, ie the 2-2 and 0-0 must go just below that.!<

9. >!there are two doubles left and two places that require them.!<

10. >!in the bottom of the thin T, we need a chain of 1-x, x-y, y->3 without using a double or the 5-6. The only valid combo is 1-2, 2-4, 4-5.!<

11. >!we are out of 2s and 3s, the 2c6 has to be 1+5, that places 0-1 and 5-6.!<

12. >!only 4s are numerous enough for the 3c=, and 5-5 in the forced double in the thin T.!<

A simple puzzle, so I'll try to be brief.

0. >!all 0s, 5s and 6s are booked right from the start.!<
1. >!work out the orientation of the pieces, we don't have a natural 10 so the lower right corner needs the two x-5s, both vertical, 6-6 goes in the green 2c12.!<
2. >!the only >2 with a 0 is the 0-4.!<
3. >!switching to the lower left the 3c= needs a double and be linked to the other 12, and the red 2c0 requires the 0-0.!<
4. >!3-6 is the only x-6 with a double and comes with a 6, this also confirms 1 as the only digit possible for the large = area, the two discards are the left over 3s.!<
5. >!0-6 is to the left of 3-6.!<
6. >!there are five 1s, all are needed in the 5c=, thus the 3c6 is 2+2+2!<
7. >!0-2 in the top tile of the orange 2c0, into the 2c6!<

The rest are all forced.

1. >! Lower left corner is the place to start since we do not have 0-3 only 1-3 and 3-5.!<
2. >! we need at least five 5s and 6s to complete the 2c10, 2c11 and 2c>10, so we cannot use 3-5 for the 1c3 due to interacting with the blue 7, 1-3 it is!<
3. >! let's look at the 3-5, the only place that can accommodate it is the bridge between 2c7 and 2c11!<
4. >! what about the 0-2, if it's on the tiles marked as 2 and 0, it forces the only >10 piece we have directly above it. To its left is the 2c10, which has to be split and we'd need a 4-5,5-5, or yet another 5-6, none are available.!<
5. >! 0-2 could the 2c7 and 4c<2, but that fails also because there's no 1-5!<
6. >! the only remaining possibility is 2c2 and 4c<2, and forcing the 0-4 as well !<
7. >! what about 1-4? not in the blue 7 because no more 3s, so it has to go to the 2c10!<
8. >! 0-0 is forced into the purple 2c0!<
9. >! 5-6 is forced into the 1c5!<
10. >! 2-5 in the >10 !<
11. >!0-6 and 1-6 round out the top part !<
12. >!1-2 pairs up with 1-3!<
13. >!0-5 and 0-1 finishes the puzzle!<

I don't have time for a full guide, so here's something to get started on, hopefully someone else can refine or expand on it.

We have doubles of 3,4,5 and 6.

1. >!Thanks to no 0-0 most piece's orientation can be ascertained, until we get to the 5c=. The green 2c= is split, the lower tile links downward.!<

2. >!the purple 2c= can be a whole piece which forces a domino into the spaces to its right, then the right side of the 5c= must be a double; alternatively the purple 2c= can both horizontal, and that also forces the right side of the 5c= to be a double; however, if the upper tile of the 2c= goes into the 5, the 5c= now requires two doubles.!<

3. >!this means that the 1c5 needs to reach into the 5c=, therefore that area is not of 5s.!<

4. >!regardless, the tip of the 5c= links to the 2c7!<

5. >!also, this means the orange 3c= is completely split.!<

6. >!If the 0-3 goes up, we need another 3 to complete the 6, we now only have four 3s. In order to retain one digit that can go in the 5c=, we have to use the 3-5 here; OTOH, if the 0-2 goes up, we have no choice but to use 4-5. Ergo the purple 3c= is made from 5s. We can place 5-5 as the first domino!<

>!P.S. the delta is 68!<

there are 5 possible configuration of the "u"

>!a) 0-1,1-4,4-4,4-6 b) 0-2,2-3,3-3,3-6 c) 0-2,2-4,4-4,4-6 d) 0-3,2-3,2-2,2-6 e) 1-4,4-2,2-2,2-6!<

>!as we know the delta from pip counting is 48, we can plug each of the configuration into an equation of sorts!<

for example:

>!b) subtotal 16, not including the 6 because it's already accounted for. so 3x+3y+z=48-16=32, now we can plugin in 4/5/6 for z resulting in 3x+3y=32-4=28, 3x+3y=32-5=27, 3x+3y=32-6=26. Because the left side can be further simplified into 3(x+y), we know this number has to be divisible by 3, only 27 works, so x+y=9=4+5(since not enough 6s)? But no, because in order to get to 27 we had to assume the >3 is 5, so there isn't enough 5 at this point, thus this configuration is invalid.!<

>!applying this process we can narrowing down the choices to a) and c) and we can start by placing 4-4 and 4-6 in the "u", and know for sure the <2 is 0, thus locking down all 0s!<

>!furthermore, in both scenarios x+y=8, only 3+5 is possible!<

edit: an eureka moment

>!moreover, let's break down c). Since we know 5s are in an =, and 4-6 in "u", the 2c10 is split to 4+6. In order to make this scenario work, we had to assume >3 was a 4, this results in a contradiction because we start with five 4s, four are in "u" and we have none left for the 2c10. a) is the only possible configuration!<

In the "U"

1. >!Without 0s and 1s, the only way to make 3c6 is 2+2+2, and because that would takes all the 2s, the only way to make 3c9 3+3+3 !<
2. >! one of the pieces from the red 3c= protrude into a 3c6, only 2-3 is possible, 3s are in the red= and all 5s are in the green =!<
3. >! because the red = needs a double, the top of the 3c9 is definitely horizontal and only the 3-6 works here!<
4. >!3-5 is just below, 5-5 rounds out the green =!<
5. >!0-3 bridges 3c9 and 4c0!<
6. >!0-6 is only possible in the 2c10, and 0-0 the middle of 4c0!<
7. >!2-2 has to be below 0-0 to maintain even number of tiles on both sides, 0-2 forced!<

the rest are forced.

Let me just say I don't like this solve, it feels like I had tunnel vision but I can't find another method without counting and guessing.

We are now up to the letter S, there are three doubles, of 1s, 3s and 6s, as well as a size six =/= area.

0. >!we can easily see all size 2 areas are split based on the layout!<
1. >!without 1-1 there is a single 1, which is 1-3; without 3-3 there are two 3s, 1-3 and 3-5.!<
2. >!This means either 1-3 or 3-5 must be in the "s", otherwise there will not be sufficient unique digits to complete it!<
3. >!in the 3-5 case, this is quite constrained because the other two pieces need to be 0-2 and 4-6.!<
4. >!Going further, without the 4-6, it's impossible to put a 6 in the 2c10, because then you'd have to use the 0-4 to complete the 10, and need another 4 to make the 2c4. So we have to use 5s, but the only 5 piece that can bridge the 2c10 and 2c9 is the 3-5, which is used up!<
5. >!Thus 1-3 is in the "s".!<
6. >!Without 1-3, the 2c2 is not 1+1 since we know it's split by geometry. The two places 1-1 can conceivable go are: a) bridging 2c6 and 2c4, b) entire in 3c8!<
7. >!If it's in a), we have to use the 3-5 to make the 2c4, but we now run into the same issue as before where we need another 3-5 to complete the 2c10. 1-1 is therefore forced into the 3c8!<
8. >!We need a 6 to complete the 3c8, only 4-6 and 6-6 work because of the 9, but if we use the 4-6, we'd need a 4-5/5-5/5-6 but we don't have those, so it must be 6-6 here.!<
9. >!Now we need to bridge the 2c9 and 2c10 with the 3-5!<
10. >!3-3 has two possible placements, a) 2c6 and 2c4, b) 2c9 and discard!<
11. >!if a), we need a 1-5, so 3-3 must be 2c9 and discard!<
12. >!Now we have to use 0-6 to finish the 2c9, then 0-2 for the 2c2!<
13. the rest are forced.

0. >!Only 0, 3 and 5 have sufficient numbers for the 4c=, interesting 3 and 5 have identical connections, with 0 and 4.!<
1. >!we have three 1s, one is set for the 2c1. !<
2. >!we have two 2c2s, one of which has to be 0+2, the other 1+1. This rules 0 out for the 4c=!<
3. >!Let's examine the 2-2, can it be in the 2 in the lower left corner?!< 3a) >!if it's vertical, the >8 becomes impossible!< 3b) >!if it's horizontal, we'd run out of 2s, rendering the puzzle impossible.!<
4. >!both the 2c2s are bound by = areas that the 2-2 cannot overlap, except a single spot - the void, that's where 2-2 has to be.!<
5. >!the lower left corner is the 2-6, which way can it go?!<
6. >!If the 2-6 is horizontal, !< 6a) >!the 2c= adjacent to the 2 must be a double!< 6b) >!the >8 is a single piece, the 2c1 is a single piece, and the upper 2c10 must be a single piece, and must be 5-5 !< 6c> >!Most crucially the 2c= in the lower right corner must be a double (3-3) and we would have spent too many 3s and 5s, rendering the 4c= impossible.!<
7. >!Therefore, the 2-6 is vertical.!<
8. >!the red 2c2 is 1+1, this means only one tile, the bottom one links to the light blue 2c=, while the upper tile links to the 2c10. There is no 1-5 so it's 1-4 or 1-6. !<
9. >!The upper tile of the light blue 2c= also links to the 2c10.!<
10. >!the double in the 4c= must be on the lower two tiles, the left tile links to the blue 2c2, this would be the 0-3 or 0-5. Since the last 3 and 5 both come with a 4, the upper red 2c= are 4s.!<
11. >!this area in turn links to a 2c10, we need 4-4, 4-5, 4-6. We only have 4-5, as a result the 4c= is made from 3s.!<
12. >!5-5 goes to the blue 2c=!<
13. >!0-5 is the only tile that can bridge the 2c10 and 2c1!<
14. >!0-1 is forced to the bottom of the red 2c2!<
15. all remaining pieces have obvious placements.

pEsO

1. >!in the O, the middle two tiles of the = must be a double.!<
2. >!The >10 is then forced by geometry to be a single piece, only the 6-6 works here.!<
3. >!the >15 can only be [5+5]+6, given the remaining pieces.!<
4. >!we need a digit with a double that links to a 6 and a <2, only 3 fits.!<
5. >!the 0 in 0-6 is a singleton, it could go in the upper <3 in the E, but we need a x-6 where x>3 and we do not have such a piece, so the only place for 0-6 is in the =/=.!<
6. >!the 1-6 can go in the >3 in the E, but there is no 1-y where y<3, so 1-6 goes to the only other place it can be, bottom of the p.!<
7. >!in lower part of the E, 3 needs to be with a double (2 or 4) that is paired with a <3, it's not possible with 2s so 4s are in the 4c= in the E, though we can't determine whether it's 1-4 or 2-4 filling the last spot yet.!<
8. >!in the p, the = needs another 1, we have 1-4 and 1-5, if this piece goes vertical, we have to complete it with a 4-5 or 5-5. We no longer have those thus the 1 goes across and it's the 1-5.!<
9. >!2-2 is then forced to the = in the p.!<
10. >!back in E, 2s cannot fill the 2c=, because it's not paired with a <3, neither can 5s because no >3, neither can 4s because we need one of them for the 4c=, so it's the 3s, 2-3 vertical and 3-5 cross.!<
11. >!we have two 2s and two 4s, we need to use up one of each, so 2-4 completes the 4c= and the left over pieces to the =/=!<