it will be sudoku or puzzle contest?

A Puzzle contest, but as a test-solver I can confirm that there will be something for the Sudoku solvers to attempt as well.

Is the April Contest document available in Russian?

Hi. I have a query on the Fire puzzle. In the example, when time = 1 second, R3C3 is lighted. Will this flame lights up the match in R3C2 in N=1 or N=2. As per the example, it suggests N=2 but I am confirming whether a match will light up another match in the same second or next second.

Vladimir, I will answer you by email.

If this second lights up the match in R3C3, the next second will light up cells R2C3 and R3C4.

Hi Riad, I want to ask you, whether all line segments which form the turn (in My Line - puzzle 4) have identical lenghts. In the example they have (number 3 and three ?).
Thanks in advance

a question on puzzle 1 fire

how many matches are lit at the beginning? Just one or may be more than one? does "some match" mean " a certain match" or "some matches"?

What are optimizer puzzles?

Hi Matus,
In each cell with a digit or "?" line turns. Segments comprising this corner have an identical length. In cells without digit or "?" where the line turning angle lengths may be different.

Just one, Stefano.

The last three puzzles.

To clarify in puzzle 2 (Minefields), the numbers are treated as empty spaces, right? (And hence a numbered cell in the small grid doesn't have to correspond to a numbered cell in the large grid; it just needs to correspond to an empty cell in the large grid, numbered or not.) Otherwise there's an obvious contradiction. But clarifying...

And for puzzles 10 and 11, are the bounds M,N included? Or to be more explicit, take puzzle 10 (Loop Between the Polyominoes): is the maximum allowed grid size 26x26 or 27x27?

Edited by chaotic_iak 2014-04-04 7:32 PM

With my knowledge of English is hard to understand a question about Puzzle 2. I hope that Prasanna understand it and respond.

In Puzzle 10 the three maximum allowed grids are 26 x 26, 25 x 26, 26 x 25. M, N in puzzles 10 and 11 are not related.

I mean, can a number inside the small grid correspond to an empty space in the large grid or must it correspond to another number?

For My Line of Puzzle 4, does the reverse rule apply? In the sense, if the line makes a turn at a cell, should that cell definitely contain a number?

In the Borderless Pentomino example, what significance do the rightmost 3 at the top and bottommost 3 at the right have? There are no pentominoes in those directions...

A number inside the small grid can correspond to an empty space in the large grid.

I answered a similar question from Matus:
In cells without digit or "?" where the line turning angle lengths may be different. But may also be identical.

In example pentominoes can be only into 4 x 4 area.

Then why does the last column contain 3 at the top when it is not part of the 4x4 area? What is the significance of that 3?

?????, ??????????, ??? ?????????? ??????? ???????? ?? ??????? ?????.

Post Russian text of the competition please

I have a question about Astronomers: Can there be an angle where the Astronomer sees no planets?

The Astronomer, as is normal vision, can see in (tending to) infinite angles. So the only thing to keep in mind is that there should be X number of visible planets and X number of invisible planets blocked by the visible ones. The angles in the example are just to get a better understanding of the visibility block.

Just as a formatting note, it would be awesome for puzzles like 3 if the numbers were printed smaller and in a corner so you could write the translation in the box without trying to overwrite the digits. Going well so far, but that will end when I reach the optimizers. :)

In puzzle 4 (Across 4 Grids), might the quadrant with black cells actually be My Line? Like, it's just a regular My Line with some removed squares, and the other quadrant is actually the Diayajilin.

Yes this is possible. You need to prove which is which.

Just saw this. That 3 would come into play IF you prove that the 4x4 area is inclusive of it. It can also act as counter-proof that it cannot be a part of the 4x4 area. The important part to remember in both cases is you first need to prove where that area is, so the added clues help in restricting that.

I have been participating in your sudoku contests since couple of years and love them. Actually they are the highlights of every month. I have a request please. Could you have the sudoku Classic and Variants contest every month in addition to whatever puzzle contests are on ? The puzzles are quite beyond my ken and almost all of them have to be solved offline which is a bit of a challenge for me. Could you please bring back the monthly online sudoku contests ?

A doubt in 10th Puzzle Optimizer.
It says maximize the are of rectangle. So does that mean if someone has area 26*26 and other has 25*26 but later one has more empty squares.
Does still 26*26 best ?

Astronomer: Must the astronomer be able to see the whole of 16 planets (as in the example), or just some part of them?

Some part is fine. The invisible ones must be fully invisible.

Hi. We will definitely continue with the Beginners Contest in Sudoku and Puzzles in the forthcoming months. In the meantime, you can participate in the Monthly contests so that you will be able to know the different types and difficulty levels in World Championships. Also, you can participate in April Contest by Riad, which is currently running. This has few Sudoku variants, which will definitely interest you.

In the quadrant with My Line, all cells (except the black ones if it happens to be the bottom-right) should be passed right? As in, there's no extra "black square" or anything to the quadrant with My Line?

Yes. No extra black square. All white cells are passed.

Hi

In the My Line quadrant, does a segment of turn could leave the quadrant ? For example, can the 2 on the last row have a segment of his turn crossing the quadrant border ?
Does crossing the border cuts the length of a segment ? For example, with the 1 near the median horizontal border, could it cross this border and still count as one segment ? (as there is only one segment IN the quadrant)

Nice puzzles anyway !

Yes, a segment of turns can leave the quadrant. In the example I specifically gave one such angle (in cell with "?").

It is very difficult to make out the different regions in Sudoku with Stars. Please make the borders more prominent in future.

Any 26x26 solution is better than any 25x26 solution.
Only if two solutions have the same area, then the solution with more white squares is better.
If two solutions have the same area and the same number of white squares then they will get the same score.

PS: I notice the first two answer fields for puzzle 10 are validating that M and N are between 1 and 19 only, but if you have M or N >19 then ignore the warning and submit anyway. The earlier announcement in this forum says that the warnings are not always 100% accurate.

In Diayajilin "cannot be adjacent" means can't touch even diagonally or cant't touch side to side, which one?

In Diayajilin the black cells can touch diagonally, but not orthogonally (side-to-side).
This is the same as normal Yajilin rules, the only difference is how the number clues behave.

Fixed this warning. It shouldn't show warning with numbers <= 26 now.

just a query
for answer key of Trid and sudoku on cube we have to enter 9,9,9,9.. or 99999 without commas ?

In Trid can the numbers be repeated in the diagonals. That is the bold diagonals of rectangle/sides of triangles.

Both formats are accepted (in this contest and in all other contests at LMI)

Small question about Battleships: When you say "Maximum length of horizontal / vertical line segments", do you mean "CONSECUTIVE horizontal / vertical line segments"? Otherwise, in the example, the top right corner '2' is actually 2 1/2 long.

Similar question for Battleships With Line: can a row or column have a longer segment which is partially outside the quadrant, if the portion within the quadrant is below the given maximal length?

There are 6 small Trid puzzles, each has bold borders that are 7 circles long.
Numbers can NOT be repeated along those 7-circle borders, or any other diagonal within the same small triangle.
(it's the same idea as the Sudoku On Cube puzzle, there are 6 little puzzles with individual solutions, but they have identical digits along their common borders)

No, cannot.

OK, thanks. Since the beginning, I had misinterpreted the statement of 'Battleship with Line'. When I finally properly translated, it became (almost) easy ...

Hi. Another question: For optimization problems, I guess you can without penalty propose a first solution, then return it to improve, right?

Yes. There is no penalty.
Your final submission for each puzzle will be considered for scoring.

according to diayajilin rules;

1-digit in a cell shows total number of black cells on its both diagonals
2-or shows total number of black cells on its any diagonals

for example;

digit 2 in a cell and there is 2 black cells on any of its diagonal and there is one black cell on its other diagonal. that is satisfied the rule?

In Diayajilin, if there is a digit "2" in a cell then this is the total of all black cells in both diagonals, added together.

i.e.
there is 1 black in NW-SE diagonal and 1 black in NE-SW diagonal
OR
there is 2 blacks in NW-SE diagonal and 0 blacks in NE-SW diagonal
OR
there is 0 blacks in NW-SE diagonal and 2 blacks in NE-SW diagonal

(but you can NOT have 2 blacks in NW-SE diagonal and 1 black in NE-SW diagonal)

In my line if there is a no 2 is it necessary to have the length of lines equal to 2 on both sides or on side is enough.

both sides its necessary

Hi.
For 'Roman OX', I thought it was possible to use 'X' AND 'O' inside the grid, but the submission page specifies to use only 'X' and '-' ?

Fixed. Your submission was already accepted nevertheless.

For the puzzle 10;

I cannot figure out by equal dimension of figure and grid cells on the given example.
Could you explain pls.

It means that the shapes must cover grid cells exactly. e.g. You cannot place the 1x1 shape to cover half of one grid cell and half of another grid cell.

Another question, about Battleships with Line :
Does the maximum line length have to exist or is it just a maximum ?
In other words, for a row with a 2, do we need to have at least one line of length 2 or only 1 length lines is ok ?

It is needed to have such a lenght.

Hello,
I really enjoyed this contest, with dense and varied problems. As a beginner, I always lack of time to look for problems in the usual competitions. This time, I only missed a little time to improve my optimizations ...
By the way, when can we expect to have the final results?

We still are working on the optimizers (to check correctness of submission). Riad is going to solve all of them soon.

I like how there are essentially only six distinct submissions among those in Roman OX with score less than 200.

The 26x13 was amazing. I tried something similar but couldn't get 6 strands to run zigzagging like that because I ran out of minoes. But I tried it on 26x12; I should have tried 26x13 indeed. My first submission was a 26x8 with a T-tetromino at D1-E1-F1-E2, a monomino at A26, and a L-trimino at G26-H25-H26; something like the 26x8 on the booklet but with less minoes used. And obviously it can be extended at will, so a good thing there's that 26x26 limitation, too. (My submission is different, improved from the above.)

Meanwhile, time to guess the author!

Mine is the second one, 13x23. I could not place the last piece to make the rectangle longer, but even if I could have, the 13x26 would still be so much better. I just didn't think of putting the pieces directly at the left border...

and I am the third with 25*11. did try the 6 loops but could not figure it out.

Sebastian was very good. The top solution is amazing, the idea of leaving out the trimino to respect the parity is fantastic.

At first glance it seems to me that the first two solutions of the arrows are wrong as they do not respect the rule that white cells cannot share an edge. Which may reshuffle substantially the leaderboard for the top positions......

As usual, Riad has pulled out from his hat a set of puzzles of fantastic quality. The astronomer is in my opinion the best puzzle. of the optimizers I like the Pentomino, it has so many possible strategies for getting the extra step. Thanks to Riad and the other organizers.








Edited by forcolin 2014-04-13 7:24 PM

We added the restriction of 26X26 almost at the end because we thought there will be confusion over answer keys entering. But it looks like there would have been much confusion without it.

Yes, those are invalid grids. I missed that rule while making the pdf.

my guess is Tomoaki

The limit at 26 has also another function, to make illegal the solution (infinity, 2) which came to my mind immediately. Pity.

For Roman OX, the second listed solution with 160 (middle of top row on first page) is incorrectly formatted, with an extra ";-" in the horizontal clues and a missing vertical clue. I assume this is because the contestant separated the two halves with a ";" instead of a "," and in fact submitted the same solution as the 3rd listed one (immediately to the right).

(As for myself, I feel stupid for finding the solution scored at 180 and managing to convince myself that none of the clues could be removed, when the two distinct solutions scoring 160 are in fact obtained precisely by doing so. Duh).

Thank you Stefano,
You're in the top three winners, congratulations!
Soon there will be preliminary results.

I'd have suggested an infinity x 8 to score better than an infinity x 2. :P

yes but i use fewer polyominoes....

btw my guess was wrong. Congrats to James

Edited by forcolin 2014-04-13 10:07 PM

This fantastic decision gave James and once again won the April Contest.
Congratulations on your victory, James, Stefano and Sebastian!

Thank you Deb for your support!
With your pdf-files check solutions has become fast. Previously, I had to draw dozens of puzzles, and then check.

Thank you for a really fun contest! I enjoy being able to take the time to work on challenging problems. I made a stupid error while submitting my entry for #5, but I still did better than I usually do on the speed tests. Plus I usually skip optimizers but thought I'd at least try this time!

what is the answer key for first puzzle FIRE.
Because i see in wrong submission many have put same answer keys so it means have we interpreted answer key in wrong way, because the solution was satisfying all outside numbers ?

For the 12th puzzle, my submission was not included in the results.
there is still being checked or my submission violates rules?

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."