Contest over. IB/PB still available for solving at leisure

Errata:
(added 22/8/12): #1 & #2 Easy as Penny Lane: Missing text: "The letters outside the grid indicate the first letter encountered in that row/column when viewed from that edge"
(22/8/12) Double Block: Every row and column contains 1-5 once each, plus 2 black cells.
(23/8/12) Easy as Penny Lane: ... to be read in any direction ... should say ... to be read in any orthogonal direction ...
To keep in line with UKSC scoring: In addition to the bonus for fully complete entries, bonus points will also be available for nearly complete entries:
"A bonus of 1 point per minute saved will be awarded if 22 or 23 puzzles are solved correctly."

---------------------------------------------------------------------------------------- UKPC Walk-through index: (Feel free to add your own solves to this thread - no harm in multiple versions!)

Are there extra constraints not listed or is LURUDLU an alternative solution to example 5? This path is shorter than the example but has an equal number of tilts so I think it is equally correct?

Or does the constraint-based nature mean there will be no unique solution to this puzzle?

GarethMoore wrote:Great range of puzzles!
Are there extra constraints not listed or is LURUDLU an alternative solution to example 5? This path is shorter than the example but has an equal number of tilts so I think it is equally correct?
Or does the constraint-based nature mean there will be no unique solution to this puzzle?

Thanks.
It's the number of tilts that is counted, so that's a perfectly valid alternative solution.
The competition puzzle was produced professionally (unlike the example), and has sufficient constraints that it only has one solution.
You can practice Andrea's other multi-goal tilt mazes at http://www.clickmazes.com/newtilt/ixtilt2d.htm

Just clarifying some rules:
Tren: "Each number in the grid must be part of a block, indicating how much length-wise movement that block can make."
- Should each block contain exactly one number? (i.e. you can't add blocks that don't contain any number, and you can't have two or more (identical) numbers on a block)
Double Block: "Fill the grid with numbers 1-5"
- Should the digits 1-5 occur exactly once in each row and column?

kiwijam wrote:Tren: "Each number in the grid must be part of a block, indicating how much length-wise movement that block can make."
- Should each block contain exactly one number? (i.e. you can't add blocks that don't contain any number, and you can't have two or more (identical) numbers on a block)
Double Block: "Fill the grid with numbers 1-5"
- Should the digits 1-5 occur exactly once in each row and column?

Tren: In your solution, every block should contain exactly one number.

Double Block: Correct. Every row and column contains 1-5 once each, plus 2 black cells.

Nice variety of puzzles, inclduing quite a few that I will be hopeless at. However, looking forward to it.

In Easy as Penny Lane, it needs to be clarified that the words PENNY or LANE can be read exactly once in either a VERTICAL or HORIZONTAL direction. In C-Note, can we assume that a zero cannot be added as the second digit of a 2-digit number?

david mcneill wrote:In Easy as Penny Lane, it needs to be clarified that the words PENNY or LANE can be read exactly once in either a VERTICAL or HORIZONTAL direction.
In C-Note, can we assume that a zero cannot be added as the second digit of a 2-digit number?

Re Penny Lane: Yes, row/col intended, Not diagonal. Will clarify in the 1st post errata section.

Re C-Note: No. '0' is a valid digit, so could be added to a given digit.

Any errata will appear in the first post of this thread. Please check here before you start the contest, for any last-minute information. The PB has now been uploaded, and contains 14 pages.

Separately, I've decided to award 1 pt/minute saved for nearly complete entries, as per the recent UKSC.

Not really an errata, but note that I will accept Flip Mirror Sums answers that contain comma separators. This is in line with the general policy that if admin are 100% sure the puzzle was solved correctly, then full marks will be given.
Due to potential formatting issues, you may not be awarded points immediately, but I will be checking periodically throughout the weekend for any fixups required. eg, use of spaces - there are none in the official answer keys, so please avoid using them. Use the comments section on your answer page at the end of the contest to notify me of anything you want to say to me.

Please do not discuss the puzzles anywhere until Tuesday, when the contest is finished. Thank-you and good luck!

Time's up! Thanks very much to Alan, the setters and the testers.

I bombed this like Peter Kay in the beer advert and am claiming a provisional eighty, which I suspect is going to prove so weak that I'm going to write it as a word rather than a number to try to be inconspicuous. Enjoyed the first hour and a quarter thoroughly, most notably the part where I made very, very heavy weather of the Double Block, but ended up learning an awful lot about the puzzle format that I didn't already know, even after doing dozens of the blighters on Croco. The second half saw me get Tapa #2 out (very neat!) and spend a long, long, fruitless time on the Either-Or Skyscraper and the Nobel Prize for Literature puzzle.

Looking forward to seeing the logic that led to the answers for most of these, but maybe I'm looking forward even more to seeing how many - and who! - tried out for the UKPC and USPC, to see if we can see whether/where publicity efforts are working well.

I really enjoyed the UKPC - overall I personally thought it was better than the USPC, and a really solid test. Thanks to Alan and all of the authors and testers.

I particularly enjoyed the Tilt Maze (definitely my favourite puzzle), and I thought the Crypted Fence and Heyajilin were particularly clever (although I've no idea if I did the latter by luck or skill! I saw how the bottom-left 3 must go and how the two smaller 4s must be split into two shaded diagonals, but beyond that I just solved on the basis of uniqueness and "surely it goes like this"...)

Least favourite puzzle was Tren, but only because I couldn't solve the flipping thing - I swore after losing half an hour to not solve the USPC one that I would get this one. But I didn't. Another half hour wasted. At least afterwards it took me an additional half hour to finally solve - the USPC one took under two minutes when I tried again later! On this one I think the possibility of a horizontal 4 kept eluding me.

I do have one question, though. Was the 25 points for Rock the 90s due to a last-minute puzzle change or some other twist of fate? It takes barely a minute to try all possibilities in that tiny 5x3 grid - it's equivalent to a similar-sized hidato in complexity I'd say. Above it the yin yang different neighbours wasn't hard if you'd taken the time to work out the implications of the rules in advance (although I think that adds a certain barrier), but just the mechanics of shading it and writing the numbers and extracting keys took me at least 5 minutes... for 10 points. But that's only a minor quibble and it was open to everyone to see the points and choose puzzles so it hardly matters.

There's a new announcement on the site's front page

I've been through and checked every answer key (without names attached, in case of potential bias), and am happy with the scoring. I decided that tilt-maze solutions that visited all the grey cells, but that weren't the optimal route, should get 5 points. This affected 7 people. In addition, there were only 4 cases (from 1368 entered answer keys) of a 5-point deduction for an obvious typo. Unexpected spaces and commas meant some marks weren't awarded automatically, but have been now. Aside from the above, every answer key received 0 or full marks.

@Gareth (and others elsewhere) Thanks for your kind words. I'm just glad you enjoyed the test!

GarethMoore wrote:I do have one question, though. Was the 25 points for Rock the 90s ...

Yes, that was a little overvalued - In retrospect, perhaps 15 was more accurate. Actually the example and puzzle were supplied the other way round, but the larger grid was quite trivial, and the smaller one not so - Spotting the cyclic loop was easy, but cutting that loop to incorporate the 3 missing hands stumped a few people. Potentially a real time-waster, hence the initial points allocation.

Once you know the method for YY-Neighbours, yes, it is trivial. I couldn't justify more than 10 points when no-one took more than 3 minutes to solve it before the contest.

Penny Lane probably worth 15 as a pair, 20 seemed too much. Perhaps slightly undervalued at 10 for the pair. Smart solvers would come back to do at the end, as last-minute stocking-fillers.

I was concerned that Sigma Snake was overvalued, since the answer key format meant you didn't have to actually place the whole snake. If you could figure out the path, the values were easier to deduce than the full placement. However, only Triangle Minesweeper was attempted by less people!

None of the answer keys were intended to trip anyone up - I just wanted enough information to be convinced that the puzzle had been solved correctly and (hopefully) fully. Unfortunately 2/3 of scoring competitors made at least one error in their submissions. I like the idea that LMI have implemented of being told whether an entered solution is correct immediately, and allowing multiple chances to re-enter a solution (for decreasing points each time). I'd have to ensure unexpected spaces and commas were automatically handled correctly every time before I could implement this. Next time will be better...

I tried to judge the difficulty so that there was a 50% chance of it being solved completely by at least one Brit. I don't think it was too far from that mark. I'm happy to hear any opinions, criticism (preferably constructive) or suggestions on how to improve the contest.

One more thing - I've added stats to the bottom of the results page. (They honour the country selection filter, by the way)
edit: The big empty spaces in the stats were meant to be bar-graphs - fixed.

This was a disaster for me. Just couldn't get going with things like the rock of the 90's - that others have pointed out was really so easy.
It's really annoying to myself when I know that I can do them, but just make these stupid mistakes. Really good to see so many Brits taking part - this can only go from strength to strength.
Congratulations to James for earning his place on the team alongside Neil and Steve. Looks like we're going to have a good team this year.

I too spent ages on the Rock-Paper-Scissors puzzle, which looked easy but completely stumped me. That and trying some of the harder puzzles meant I solved nothing in the last 40 minutes. Nevertheless, very enjoyable overall.

PuzzlerNickG wrote:I too spent ages on the Rock-Paper-Scissors puzzle, which looked easy but completely stumped me.

On that puzzle, if you start on the bottom-left square (which was also where the example solved from, so a natural place to start, and also by inspection the most constrained part of the puzzle) then the whole solution is pretty much forced just by following the rules.

Are we having a thread this year with walkthroughs to the various puzzles? If so, please accept the following as my contribution for one of the few puzzles that I could crack, and a particularly interesting example of the genre; thanks to the constructor, Serkan Yürekli. If my explanation is too quick, please shout up and I'll work through it a bit more slowly. Typoes excepted.

#19 DOUBLE BLOCK (20 pts)

First, work out how many digits must be used to make up each row or column.
Rows first: 9 is 2-3 digits, 6 is 2-3 digits, 11 is 3-4 digits, 2 is 1 digit, 2 is 1 digit, 11 is 3-4 digits, 3 is 1-2 digits.
Columns: 15 is 5 digits, 5 is 1-2 digits, 11 is 3-4 digits, 8 is 2-3 digits, 10 is 3-4 digits.

As 15 is 5 digits, mark R1C2 and R7C2 as blackened and mark R2C2 to R6C2 as numbers.
As R1C2 is blackened and R1 is a 9 which requires 2-3 digits, mark R1C1 as a number, R1C3 and R1C4 as numbers, leave R1C5 and R1C6 blank and mark R1C7 as a number.
As R7C2 is blackened and R7 is a 3 which requires 1-2 digits, mark R7C1 as a number, R7C3 as numbers, leave R7C4 and R7C5 blank and mark R7C6 and R7C7 as numbers.
(Tricky step!) As C7 is a 10 which requires 3-4 digits but R1C7 and R7C7 are numbers, blacken R2C7 and R6C7 and mark R3C7, R4C7 and R5C7 as numbers.
As R2C7 is blackened and R2 is a 6 which requires 2-3 digits, mark R2C5 and R2C6 as numbers, leave R2C3 and R2C4 blank and mark R2C1 as a number.
As R6C7 is blackened and R6 is a 11 which requires 3-4 digits, mark R6C4, R6C5 and R6C6 as numbers, leave R6C2 and R6C3 blank and mark R6C1 as a number.
Observe that R6C2 is a number and R6 has all five numbers marked, so blacken R6C3.
As R3 is a 11 which requires 3-4 digits and R3C7 is a number, mark R3C3 and R3C4 as numbers. As R3C2 is a number, blacken R3C1.
As C5 is a 11 which requires 3-4 digits, mark R4C5 as a number.
As C6 is a 8 which requires 2-3 digits, and because R6C6 and R7C6 are numbers, mark R3C6 as a number.
Observe that R3 has all five numbers marked, so blacken R3C5.
As C5 is a 11 which requires 3-4 digits and R3C5 is blackened, mark R5C5 as a number and blacken R7C5.
Observe that C5 has all two blackened squares marked, so mark R1C5 as a number.
Observe that R1 has all five numbers marked, so blacken R1C6.
Observe that R7 has all two blackened squares marked, so mark R7C4 as a number.
As C4 is a 5 which requires 1-2 digits, blacken R2C4.
Observe that R2 has all two blackened squares marked, so mark R2C3 as a number.
Observe that R4 and R5 are both 2s which require 1 digit each, so one must have C1 and C3 blackened and the other must have C4 and C6 blackened. As C6 is an 8 which requires 2-3 digits, there is not yet an obvious way to tell which is which, so no more numbers can be marked at this point.
Obverve that all rows or columns marked 11 are made up from 3 digits, so they must have 1 and 3 outside the blackened squares and 2, 4 or 5 inside the blackened squares.
Mark R3C6, R3C7, R6C1, R6C2, R1C5 and R2C5 as being "13".
Observe that R2 is a 6 made up of two digits, one of which is either a 1 or a 3. As 33 is impossible, mark R2C5 as a 1 and R2C6 as a 5. Also mark R1C5 as a 3.
(Probably the trickiest step!) Consider C6. As it is a 8 and contains a digit 5 followed by either a 1 or a 3, it is either made up of a 53 or a 512. However, if it is made up of a 512, R4 is impossible, as R4 is a 2, which requires the 2 to be between the two blackened squares. So mark R3C6 as a 3, mark R3C7 as a 1, blacken R4C6, blacken R4C4, mark R4C1 and R4C3 as numbers, blacken R5C1 and R5C3, then mark R5C4 and R5C6 as numbers.
As R4 and R5 are 2s, mark squares R4C5 and R5C2 as 2s as the only numbers between blackened squares in their rows. All squares are now marked as either numbered or blackened.

Observe that C5 now has a 1, 2 and 3 placed. Mark R5C5 and R6C5 as being "45".
Observe that R7 has two numbers marked between the blackened squares. Mark R7C3 and R7C4 as being "12".
Observe that C6 has a 3 and a 5 marked, and R7 has two squares marked "12". Mark R7C6 as 4.
Observe that C6 has a 3, 4 and 5 marked, and R6 has two squares marked "13". Mark R6C6 as 2 and R5C6 as 1.
Observe that C4 has one number marked between the blackened squares and is the column is a 5. Mark R3C4 as 5. This leaves 2 and 4 unmarked in R3, but as R5C2 is 2, mark R3C2 as 4 and R3C3 as 2.
Observe that R6 has a 2 marked and two squares marked "13". As R3C4 is a 5, mark R6 as 4 and R6C5 as 5. As C5 now has a 1, 2, 3 and 5 marked, mark R5C5 as 4.
Observe that R5 has a 1, 2 and 4 marked. As C5 contains a 5, mark R5C4 as a 3 and R5C6 as a 5.
Observe that C7 is a 10, R3C7 is a 1 and R5C7 is a 5. Mark R4C7 as a 4. C7 is now missing a 2 and a 3. As R1 has a 3, mark R1C7 as 2 and R7C7 as 3.
Observe that R7 has a 3 and a 4 marked and two squares marked "12". Mark R7C1 as 5. As R3C3 is a 2, R7C3 is a 1 and R7C4 is a 2.
Observe that C4 has a 5, 4, 3 and 2 marked. Mark R1C4 as 1. As R1 is a 9, mark R1C3 as 5 and R1C1 as 4.
Observe that C3 has a 1, 2 and 5 marked. As R4 has a 4, mark R2C3 as 4 and R4C3 as 3.
Observe that R2 has a 1, 4 and 5 marked. As C2 has a 2, mark R2C1 as 2 and R2C2 as 3.
Observe that R6 still has two squares marked "13". As R2C2 is a 3, mark R6C1 as 3 and R6C2 as 1.
Only R4C1 and R4C2 are open; as C1 is missing a 1 and C2 is missing a 5, mark R4C1 as 1 and R4C2 as 5.
Finished; read off marked rows for answer.

OK, I'm game.
Here's the Heyajlin (which was probably my favorite).
Note that this is more about loop logic then classic Heyawake logic, and the "no long strings of white cells" rule isn't in effect.

As with Heyawake, 2xn boxes are important, including those on the edge. But the loop logic means that we can't have any long stretches of "checkerboard"
In particular, on the edge of the puzzle (like the 2x5 with a 3 in it at bottom left) you can't have two nonempty columns in a row, since they will create a little dead-end.
Similarly, in the 2x5 blocks with the 4's near the center, you can't have 3 cells in a row filled. That provides the break that gets us off and running.

As you might expect, it is very useful to mark known loop cells with a dot or something even if you don't know how the loop is going to go.

This walkthrough outlines the key logical steps, but somebody may want to add pictures for the parts where you just fill in all the obvious loop fragments based on block position, since they're a pain to describe verbally.

Step 0: May as well mark the "0" boxes as empty, start the loop in the lower right corner, and you can mark the cells adjacent to all four corners (r2c1 etc) as loop cells if you want, too.
Step 1: The 2x5 marked as a 3 in the lower left. As noted, you can't have two nonempty adjacent columns, so the blocks have to be in columns 1, 3, 5. We can mark r11c1 as a block immediately, and also (after sketching in a little of the nearby loop) immediately also see r10c3.
Step 2: The 2x5 marked as 4's near the center. Since neither can have 3 filled columns in a row, both have their outer four rows filled and center rows empty. In addition, the cells immediately above and below these regions in c4, c5, c7, and c8 must all be empty, since whichever way the little 2x2 regions go, the cells above and below will either be adjacent to a block or will create a dead-end cell. Filling in these loop cells guarantees that the last block in that lower-left region is in r11c5.
Step 3: The cells we have marked as loop cells in the central 3x5 region how completely determine the blocks in this region. The center can't be filled. The two cells immediately above and below (r5c6 and r7c6) must be blocks, and the loop going straight through horizontally in the middle pushes the other blocks to the outside in r6c4 and r6c8.
Step 4: Now, look at the 2x2 area in r8-9, c4-5. We know it contains 2 blocks. If the blocks were to slope downward, there would be a problem with the loop logic, since the loop segment coming up in c4 along the bottom would be forced both up and to the right simultaneously by that configuration. So the blocks slope upward, one in r4c9 and one in r5c8. The loop will sort of zigzag along this diagonal stretch of blocks on each side of it.
Step 5: Trace out the loop logic in the center of the puzzle and you will find that all three other 2x2 blocks symmetrically placed to the one in step 4 will ALL slope up to the right. Place all of these blocks and fill in lots of loop fragments.
Step 6: In the 3x3 region with the 2 towards the bottom right of the puzzle, you should have the left column of this region all marked as loop cells. In the remaining 3x2 region, the middle row must be empty. A little work with the loop logic at lower right should force the blocks to be in r9c11 and r7c10.
Step 7: Now let's look at the 2x5 region in the top right. With the a little bit from existing cells you have marked empty and the fact that no two adjacent columns can be nonempty, and the 1x3 region below it, we can narrow this down.
c10 of this region must be empty (the top cell is next to a corner cell, the bottom would block both remaining cells of the "1" region). We see that the two blocks must be in some combination of c7, c9, and c11, and furthermore the possible cells are the top cells in c7 and c11 and the bottom cell of c9 (at least one of the top corner cells is a block, and will force the loop through the top cell of c9). Furthermore, each of the two possible block cells in the "1" region below will conflict with one of the two right-most candidates. Thus there MUST be a block in r1c7.
Step 8: End counting. As usual with loop puzzles, there must be an even number of loop fragments entering any region of the puzzle. The region on the right is now getting pretty well nailed down, and we can "count ends" to figure out whether the little loop segment that pokes up from r3c7 to r2c7 will turn left or right. It turns left. That forces the rest of the loop logic on the right hand side of the puzzle as well as forcing a block in r1c4 if you follow that left-hand loop fragment.
Step 9: Now that you've filled in the loop on the right, we see that the loop can't close down near the lower-left corner of the puzzle, but must instead push upwards along the left hand side of the puzzle.
Step 10: Meanwhile, whichever of the two blocks in the 2x2 "1" at top right is filled, we can see that there must be a block in r3c2 to prevent the little loop at the top of the puzzle from closing off by itself. That will be the one block in the 3x3 "1" region. It forces a block in r1c1. Connect the two halves of your loop in the only possible way and you are done.

#5 Tilt Maze (not my favourite puzzle type, and nearly the last one that I've solved after the event)

First note that the target in R5C3 can only be collected with a horizontal roll in row 5. Once it's doing that, the ball can't escape from column 1 and/or row 5 - so this target must be collected towards the end of the route.

In similar vein, if the ball ever rolls horizontally in row 1 or vertically in column 5, it has no way to escape. Therefore, targets R1C2 and R1C3 must be picked up with upward rolls, and R2C5 and R3C5 must be picked up with rightward rolls.

In order to roll upwards and pick up R1C2, the ball must first roll leftwards in row 2 and come to a stop in R2C2.
In order to roll leftwards in row 2, it must first roll upwards in column 4 and come to a stop in R2C4.
In order to roll upwards in column 4, it must first roll rightwards in row 4 and come to a stop in R4C4.
In order to roll rightwards in row 4, it can either roll upwards to R4C2 - impossible, as it can't come to a halt in R5C2 - or roll downwards to R4C3, which is forced. Once the ball leaves column 3, it can never again stop there, so we're near the start of the route. Before rolling to R4C3 however, target R1C3 must be collected. So working backwards, our route so far must include:

Column 3: U (collect R1C3 and R2C3)
Column 3: D (collect R4C3)
Row 4: R
Column 4: U (collect R3C4)
Row 2: L
Column 2: U (collect R1C2)
Column 2: D (collect R3C2)

We still need to collect R2C5 and R3C5 before moving to the bottom left edges, each of which can be done by adding one extra roll:

Column 3: U (collect R1C3 and R2C3)
Column 3: D (collect R4C3)
Row 4: R
Column 4: U (collect R3C4) Row 2: R (collect R2C5)
Row 2: L
Column 2: U (collect R1C2)
Column 2: D (collect R3C2) Row 3: R (collect R3C5)

From here it's trivial to complete the route with:

Row 3: L (collect R3C1)
Column 1: D (collect R4C1)
Row 5: R (collect R5C3 and R5C4)

Giving a solution code of UDRURLUDRLDR. I suspect that the observation in italics - once the ball leaves column 3, it can never again stop there - may be the first step of a more elegant solving path.

(If this walkthrough is confusing I suggest looking at the solution image as you read it.)

The only word with R at one end is FOUR, so R is the end of FOUR, and T is the start of either TWO or THREE.
The R is beside a 6-clue, the only other number that can touch this clue is TWO.
This means the bottom-left 3-clue cannot be from TWO+ONE, it must be from THREE.
If T is the start of 3 then it cannot reach that bottom-left 3-clue, so T must be the start of TWO, which must travel two squares right to touch the corner of the 6. No other words can touch the 6 now, so R6C8, R6C9, R7C7, and R7C9 are empty.
There's only one remaining square by the 11 that the snake can use (above the O) so it must be the start of the NINE.
The 9-clue cannot be reached by two small numbers, so the NINE must visit it: write in "IN" travelling to the right.
Both the top 6-clue and the 15-clue need 6 more for their sums, but we know the TWO, THREE, and FOUR are down the bottom, and the 5-clue will use either the ONE or FIVE. So the 6 and 15 must have the SIX touching them (and nothing else).
If either of R2C6 or R3C6 are empty then the SIX cannot get between the 6 and 15 without letting other words touch them also. So both R2C6 and R3C6 are part of the snake, and must be "SI" or "IX".
But if it is "IX" then the 21-clue cannot be achieved (it can't fit four words around it, and the remaining unused numbers don't sum that high anyway). So the SIX can be written into R2C6, R3C6, R3C5.
Whichever way the NINE joins to the SIX, it will require either 4 or 6 letters. 6 cannot be done (TWO and SIX are used already), and the only 4-letter word remaining is FIVE. And there is only one way to write the FIVE without it touching the 15-clue.
[Note: Don't do what I did, and think there was an odd number of letters needed here rather than even. It makes the rest of the puzzle impossible to finish, and wastes a lot of time!]
What's left then? The 21 and leftmost-15 clues need both an EIGHT and a SEVEN. This is not possible if the snake travels down between the 15 and 21, so it will have to go down the left of the 15, then go across and down (between the 21 and 10).
EIGHT cannot touch the 10 though, so EIGHT will follow from the SIX, and the 10 will get SEVEN and then THREE. (So we know the final order will be 295687314).
If the EIGHT goes directly across then down, it forces the SEVEN to finish next to the 8-clue. But then neither the THREE nor FOUR can touch the 8 also, and a solution is not possible. Instead the EIGHT must wiggle up and finish in R2C1, then the SEVEN, THREE, ONE, and FOUR will take the shortest path past the final clues and finish at the R.

This one seems to have caused more difficulties than most, so I'll give this a bash.
Cell numbering notation: columns A-J; rows 1-10 top down (A1 = top-left)

I draw borders around cells when known. I presume everyone does this.

6 at A8 is horizontal. D8-H8 are empty.
5 at A4 is vertical, and size 2. A1/2 & A6/7 empty.
3 at D9 is horizontal, size 2. A9/B9 exmpty
3 at F9 is vertical, size 2 (inc.F10). F6/7 empty.
4 at F5 is vertical. Note 3 at F1 is horizontal, and nothing can occupy F2. Therefore this has to be a size 3: F3/4/5.
2s at D5 & D7 are horizontal
4 at J5, 3 at E3, 1 at B4 are vertical.
Mark known empties (cells that no block can reach): B1,E10,D10 (3@B10=Horiz, size 2), G2,I4,J2,J8,J10 (2@H10 horiz, size 2)

If 1 at B4 includes B5, the 2 at D5 cannot be satisfied. So, B4 extends up, and must be size 3. This now has its 1 free space, so the 2 at D5 must extend to B5. This makes A5/E5 empty, so 5 at A4 is size 2 and includes A3.

If I8 is empty, the 6 at A8 cannot be satisfied. So I8 is connected to I9. (We don't know if I10 is empty or not)
Now the 6 at A8 is sized 2, and C3 is empty.

If the 3 in C6 is horizontal, the 2 in D7 cannot be satisfied. So C6 is vertical, size 2, with C9/10 empty. Blocks in D7, D9 & B10 can be trivially completed, with known empties marked.

3 at F1 needs 5 or 6 cells for block + movement. If it uses 6 cells (size 3), they would be A1-F1. This would mean the 2 at I2 extending to G1, which would make itself unsatisfiable. Therefore, F3 uses 5 cells, and they must be D1-H1. So C1 must be part of the block at C3, and C4 is empty.

E5 & E6 are empty, so the 1 at E3 cannot extend to E4. Therefore, 1 at D4 extends to E4. Now E3 extends to E2, with E1 empty.
F1 extends to G1, with H1 empty. 2 at I1 is now vertical.
0 at H3 cannot be vertical, as nothing can stop it moving upwards. So extends to G3. It can't extend to I3, as the I1 block would be trapped. Therefore, I3 is part of the 2 at I1 (size 3). I5 must be empty, and I6 blocked by the 4 at G6. Filling this size 3 block in, J6 is empty. 2 at I7 is horizontal, size 3, extending to J7 (nothing else can block J7).

I10 must be empty to satisfy I9, so H10 extends to G10 (size 2), and the 4 at J5 extends to J4 (size 2), and we're done.

Marking size 2 blocks with a bold line through the middle of 1x2s makes is easier to count: 8 horizontal, 6 vertical.

Notation: Columns A-D, Rows 1-4, top left cell is A1, clues are referred to as R1L (Row 1 Left, ie the 29 in grid 2) or C2T (Column 2 Top, ie the 22 in grid 1)

Numbers 1 to 10 total 55.

From grid 1, A1+B1 + C4+D4 = 31
From grid 2, A3+A4 = 13
A1+B1+C4+D4+A3+A4 = 44
The remaining 4 cells (D1+B2+C2+B3) sum to 55-44, equals 11; 11 in 4 numbers for those cells must be 1/2/3/5, leaving the six we first considered to be 4/6/7/8/9/10

Options for each pair of cells to give the correct totals: A1+B1 could be 7/8/9/10, A3+A4 could be 4/6/7/9, C3+C4 could be 4/6/8/10

From grid 1 R4R, C4+D4 = 14, therefore:
From grid 2 C2T, B1+B2+B3 = 13, therefore:
From grid 1 C2T, A4 = 9
From grid 2 C1B, A3 = 4

Options for A1+B1 are now only 7/10
Options for C4+D4 are now only 6/8

From grid 1 R3L, if C4 were 6, B3 would be 7. It cannot be, so C4 = 8, D4 = 6.
From grid 1 R3L, B3 = 5.
From grid 2 R1L, C2 = 3
From grid 2 C2T, if B1 were 10, B2 would be minus 2. It cannot be, so B1 = 7, A1 = 10.
From grid 2 C2T, B2 = 1
By elimination, D1 = 2

I'm surprised how few people attempted this.
It can be solved as a Yin-Yang, using Odd/Even markers, then all the values filled in at the end.
First note that at 4 point intersections, pairs of like 'colour' (type O/E) must be adjacent - they cannot be diagonally opposite, ie, diagonally opposite cells at these intersection must be the opposite type. So immediately we can mark in nearly half the grid with O/E from the given values.
Then note that pairs of 2x1 cells touching must be different types, so the empty 2x1 on LHS must be E. Repeat the O/E chaining as above, and you are now left with just 7 cells unknown.
Now knowing that the border of a Yin-Yang must contain 1 continuous length of each type, to avoid disconnected regions, the 4 unknowns at the top-left must be O, and the other 3 are E.
Finally, fill in values of all Evens and Odds starting from the givens. Done

Not a full walkthrough yet, though I think I can do one - but a big, big starting point.

Two triangles in edge-adjacent cells must between them touch every corner in that pair of cells. It then becomes impossible to place a further triangle in any cell edge-adjacent to those two. In short:

There cannot be a group of three edge-adjacent cells that all contain mines.