Swaroop: the answer key for FIRE was 9#DO#XXXXXX---The third cell down the diagonal gets burnt at the last second. So: an actual error that several people seem to have made. (Obviously a small one, since you found 9 and DO and the rest should follow immediately from it if you are careful).Denis

sduran: Based on the PDF file that Deb posted, your submission's first row has a "20" clue, which can only be OOXOOOXOOO or OOOXOOXOOO or OOOXOOOXOO, but in which you gave a forced 'X' as the first cell. I didn't check the text-mode entry recorded and whether it was correctly converted to a PDF grid, but the PDF grid has no valid solution. Sorry!Denis

Edited by auroux 2014-04-14 3:10 AM

oh yeah just saw. Thanks . Have to be careful next time.

Answer Keys added in the score page.
Ratings also added. Astronomer is the best rated puzzle and Loop-between-Polyominoes best rated optimizer.

Thank you Riad, for a great set of puzzles.
The first nine were time-consuming but not too tricky. Not an issue for a week-long contest though.
The three optimizers were clever and enjoyable.

And congratulations to Stefano and Sebastien for amazing results, the top 3 placings could have been in any order.


I want to give an insight into how I approached the optimizers, I hope it's not boring:

10) Loop Between The Polyominoes was the best optimizer I have seen in a while.
Not many pieces to place, but many different options, and it's never obvious how to achieve a better solution.

I started with some maths. If you don't like maths then skip this bit!

If the grid contained an even number of cells then I would have to forget one of the three odd-sized shapes, but an odd-sized grid would allow all 11 pieces to be used.
Also useful is shading the grid black and white like a chessboard. 7 of the pieces would cover equal black and white cells, but the T-tetromino covers 3 blacks and 1 white, and the 1 and 3 and bent-3 cover 1 more black than white.

This is important to know when crossing large empty spaces. If you remove a black cell at one end of a 4x26 rectangle (there are 3 options in the first three columns), and a white cell at the other end, then you have a wiggling pair of lines that gives a unique solution.
Remove two black cells at the end of a 8x26 rectangle, 2 white cells at the other end and it is unique again (e.g. put the T-tetromino at one end and the 1 and bent-3 at the other end, as Ivan described in this forum already).

Theorem: In general, for any width strip, imagine slicing vertically between two columns in the middle of the grid. If there are B loop segments crossing this cut left-to-right and going to black cells, and W loop segments cross the cut left-to-right into white cells, then (B minus W) must be equal to 2*(b minus w), which are the number of missing black and white cells in the left section of the grid. The proof is not too hard (the loop must always alternate between black and white squares).

This explains why removing 3 black cells and 1 white cell from the left end of the 8x26 rectangle means that every column must have 4 lines going from white to black (left-to-right). This continues until a column is reached with more whites than blacks removed.
(Note: This method can also used for making solutions with long forced diagonals, instead of long horizontal wiggly lines.)

But we cannot automatically extend this approach to 12x26, because that would need 3 excess blacks at one end and 3 excess whites at the other end, and our pieces only give us 5 excess cells (T gives 2, the odd shapes give 1 each).
Although we can still use this effect locally. Imagine we have removed 2 blacks at the far-left and 2 whites at the far right, but want to do more. By placing a 1x2 domino horizontally (covering a black on the left, white on the right), then there must be 6 lines crossing from white to black in that same 'column' as the domino. To the left and right the white will cancel with the black, and we return to 5 blacks and 1 white for each crossing. The other tetrominoes (I, Lx2, Sx2) can give the same effect, and we have enough of them to carry a 3rd loop from one end to the other.

This is why I chose a 13xN rectangle (rotate my published solution anti-clockwise). The top 8 rows would travel automatically across, and then I placed pieces along the 9th row to keep the bottom loop separate and unique. This worked for 13x22 before I ran out of pieces. I see Sebastien (Semax) had the same idea.
It was only when I placed the long shapes along the bottom edge that they were more efficient. The bottom loop still was not able to join the loop above it, and they could now reach to the 26th column. I even discarded a 3-mino instead of the 1-mino to have more white-space, just in case!

I think it may be possible to go slightly further (e.g. 13x27 allows the 11th piece to be used), but I think that is about the limit for this puzzle. So having 26x26 maximums was good.
I'm very surprised that making M,N<27 was a last minute change for data entry reasons. 'Infinite' solutions are always unfair shortcuts. I think it was 2 years ago when Riad asked for the largest Tapa using limited clues. The best answer was a 1x10000 rectangle with a "1" at each end, which avoided many hours of work.
(Phew. I talk too much!)


11) Arrows with Voids

Here my submission was symmetric on all 4 sides. So once I chose 8 arrows along the top, I then reflected those arrows around the grid to the sides and bottom. So whenever I drew an arrow, I could imagine 4 lines going into the grid. In particular all the diagonals are double-lines.
It also meant the numbers-grid would be very symmetric. If I could find 4 different numbers in the top 4 rows (every second cell, checkerboard-style), then I knew all 8 rows and all 8 columns would be ok too.
Stefano (forcolin) had the same approach, but without using the full checkerboard.
The only remaining trap is to avoid grids with multiple solutions. This can happen if you have a loop of arrows, each pointing to the next one. Reversing all those arrows would give a second solution for the same loop.


12) Roman OX

My embarrassment. I had entered my solution (190) within 2 hours of the contest starting. I could see that no clues could be removed, and a couple of other options I tried didn't work. Even if there was a better solution it could only be slightly better, so then I just concentrated on the other 2 optimizers. Well done to the (many!) 160s and 180s who thought about this harder than I did.

Please note that at the most bottom-right cell there is one more X placed. There should be unique solution I guess.Could you re-check.







Attachments
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puzzle-12.zip (10KB - 2 downloads)
puzzle-12.png (10KB - 2 downloads)

There are several places with FOUR consecutive O's in your solution. This violates the rule of the puzzle.

ok. the problem is I didnot read the question maybe lack of time. I had concentrated on the restrict no more than 4 consecutive letters. but it was 3 letters. Now I knew this with your reply:)
Thanks

Bottom left 8x8 in Across 4 Grids. Am I correct in thinking that Battleships/Graffiti are on top and ML/DY are on bottom? If so, I get a contradiction in bottom left 8x8 as I can't get ML or DY to work in that region. Could there possibly be black squares in My Line? That's the only way I can think of to get it to work down there. Thanks.

Maybe this helps?:





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4 Grids sml.gif (41KB - 2 downloads)

Thank you for clarification. The rules say black cells cannot be "adjacent." I thought it was pretty standard that adjacent means including diagonals, but I guess I was wrong. Perhaps next time the rule should be "black cells cannot share a side."