Maybe someone who knows the rules, should post few easy example puzzles here to show what kind of solutions are acceptable and what kind of solutions are not.

As much as I understand the rules (english is not my mother language), I'm pretty sure that 2x2 white areas are NOT allowed in town puzzle. It's also a rule if you start solving it with given clues after you have created one. I mean that if you are solving the puzzle you can use it to make a progress.

Example: If you have 2x2 area where 3 cells are definitely unshaded, then the 4-th cell must be shaded.

Note: Please correct me if I'm wrong. I also agree if you delete my post and replace it with correct expanation. I also have solved the puzzle as well as I could and it would not be a problem if I actually got it wrong.

Note2: I really appreciate the aesthetics of the puzzles after I had solved them. They looked just so beautiful. I even photographed them :)

I am pretty sure that too. This is a difference between Clouds / Regenwolken and usual Town puzzle. The first one does not have the rule about the absence of 2x2 white area. The second one does. But of course sometimes there are some small differences in puzzles of the same name so it should have been clarified in puzzle booklet always.

And this is another argument why this approach should be considered.

Puzzle 10.
Condition:
- - - -
2 ? ? ? ?
- ? ? ? ?
In the description, only 2, without 0.
Is there a unique solution or not?


Edited by Mihalich 2020-04-17 10:33 PM

According to the actual rules and organisers' statement, there is not a unique solution.

I will just make a summary and this will be my last message here.

There are two different approaches for Problem #10.
One of them states that the absence of 2x2 white areas is considered as a rule of a Town puzzle.
The seconds one states that it is not and that the unique solution of a Town puzzle should meet this condition.

Arguments for the first statement:
- contents of puzzle booklet for the whole standard time of the contest (excluding addidional time due to decision of extending the contest) - there was no mention that the unique solution must meet this condition;
- the same approach has been made for other problems (e.g. "borders of cards cannot have common segments" is part of rules of a Diamonds puzzle in Problem #11);
- usual Town puzzle probably contains that rule (if someone is able to find the example to get ensured, it will be great);

Arguments for the second statement:
- Riad's reply to Forcolin's message posted on the discussion thread - after that there was no official statement and clarifying the contents of puzzle booklet until today.

Below it is my opinion.
The decision of clarifying the rules and extending the contest was too late and for not enough time to make a second approach. The contest lasts 9 days so everyone can find enough time to solve problems.
If someone didn't read discussion thread before and has no time today and tomorrow then he has no chance to do it properly.
In that case, both approaches or any of them should be valid.

The decision is up to organisers. I will accept any decision even if I do not agree with it.
After all I really liked all the puzzles so I do appreciate a professional work of Riad.

perhaps the organizers can activate a second answering page for puzzle 10 where a solution obtained according to a different principle/approach can be filed. I believe i have used the correct approach but if i am wrong I would be happy to submit a second alternative solution

on the same puzzle, how are roads counted? as kiwijam mentioned, is it the number of words in a crossword or the number of stretches of length 1x3 at least?
In other words, in the example below whereas O=white, X=black
XXOXX
XXOXX
OOOOO
XXOXX
XXOXX

how many roads are there? 2 of length 5 crossing in the middle or 4 of length 3 all departing from the central square?


Edited by forcolin 2020-04-18 12:51 AM

There are 2 streets of length 5.

sorry this was a stupid proposal just ignore it

For what it's worth, I feel the same way thus I also made this comment:



But of course, I was lucky enough to read the discussion pretty early. Nevertheless, IMO accepting both approaches is kind of unfair as most people were apparently moving to the second approach and redoing the puzzle including me.

I'm happy to share my solution for the first approach after the contest end tho as I'm also pretty glad how it turned out. Maybe we can have an unofficial ranking for that one -- out of the contest?

I would have not had any problem if organisers had clarified rules of this problem in puzzle booklet or had made an official statement that rules of this puzzle has changed. They knew there was a problem and they didn't clarify rules. Mistakes may happen and mistakes are happening for all people. I am okay with that and this is not a problem for me. I just want organisers to make a fair decision. For me accepting only second approach would be also unfair.

Hello!
Where can i find the answer form for the Riad contest on the site ?

Krasnosulinec 1797.

the clock has stopped submissions again

Wow, lots of entries, and lots of great optimizer submissions.
Congratulations to those that found the optimal answers, and a big thank you to Riad and LMI for a fun and successful contest. :)

I will share my observations from optimization puzzles. Maybe someone will find it interesting :)


Problem #10

Maximizing N, the number of streets, is the same for both approaches (including or excluding the absence of 2x2 white areas as a rule of Town puzzle).
Observation 1. N <= 18. The proof is by checking how streets may be distributed and checking some cases. I will just give a sketch without technical details. It is quite intuitive but formal details are very long. So there are at most 12 dark rectangles in the grid (at most 4 rectangles in one row and at most 3 rectangles in one column). Hence, all horizontal streets must touch (from above or below) the rectangle which "touches" an upper or lower border of the grid (I say that the rectangle touches a border when it is at most 1 cell apart from the border). From that you can deduce that there are at most 12 horizontal streets and how they may be distributed. Similarly, there are at most 12 vertical streets. Analysing the distribution, some of them cannot occur together or they must be "glued" together. Here comes checking some cases and after that you obtain 2 limit cases. One with 3 vertical streets coming through the whole grid (but then easily it comes that there are at most 17 streets) and one with 2 long horizontal streets (there are at most 6 horizontal and 12 vertical streets). And this is our case. All eight middle rows are determined here. After that, by distributing streets over the upper and lower row, we are getting 5 different solutions (accurate to symmetry).

Observation 2. (including the absence of 2x2 white areas as a rule of Town puzzle) When N = 18, the sum of given values >= 26. The proof is by drawing 5 possible patterns and modifying them by some small changes. If you find another solution which differs only at e.g. row #1, column #1 and column #2, then you know that you must use at least one of these values. After that you get some sets of rows or columns from which you must use at least one. And actually we are getting a few disjoint sets so the number of cases is really limited here. Finally, there are exactly 2 different solutions accurate to symmetry (4 different solutions generally) meeting (N, sum) = (18, 26).
The same method works for the second aproach without the rule of white areas. I did not check it but I think that there are a few times more cases to analyse to get the most optimal solution but still I think it can be done with a pencil and rubber during the contest.


Problem #11

I do not have any formal observation here. I came here with 36 diamonds and area = 143 but maybe it can be done better. How did I get it? There are so many cases that I must have decided for some assumptions. We want to get as most as we can common diamonds. Cards #10, #9, #8, #7 have the most number diamonds so I just checked many ways of joining them together. When these cards had totally <= 25 diamonds, I was trying to join next cards (#6, #5, #4, #3, #2) consecutively in the best possible way locally.


Problem #12

Firstly it is provable that you cannot obtain the value of 60. It is because there are no 2 disjoint distributions of full flotilla. Hence, the value of 59 is the greatest. When I found one, I did not analyse this problem any longer. But I obtained it by drawing the random set of full flotilla as fragments of the sea (all horizontal fragments - of length 4 and 2 in the first row, of length 3 and 3 in the second row, of length 2 and 2 and 1 in the fifth row, of length 1 and 1 and 1 in the seventh row). I checked it. It didn't work but I noticed that moving two 1-cell elements from the seventh row to the right side is enough to fix it. So the way how I did it makes me believe that there are at least hundreds of solutions with that value. Also there are 8 different submissions with that value :)


Edited by lukasz6500 2020-04-19 8:35 PM

Preliminary results will be published in the next few minutes. We decided not to include points for problem 10 in the final table. The reason is the complexity or ambiguity in the rules, and from a fairness perspective.

Because flash problems I did not submited my solutions, but I really enjoyed problems. Thanks.

Riad, thank you so much for putting this together! I loved the puzzles and had a lot of fun working on them. Build a Dominoes and Diatapa were my favorites. Thank you!

Thank you Walker, thanks to all participants!

Congratulations to Lukasz, Hugo and Tomoya!

Riad! Thank you for the most powerful and interesting contest! This year was a lot of fun! It’s fun because this contest showed that we are all people, not just machines that create and solve puzzles!

Riad Thanks a lot for this contest which is the "Paris-Roubaix" of puzzle, the most enduring contest of the year. a monument. a classical event. Glad that the number of participants is increasing showing thesuccess of the event. Already looking forward for next year

This is my first time to join this contest series and I really love it, so thank you very much Riad for the puzzles! :D

Btw, I really like (9) CHAIN BETWEEN POLYOMINOES, then (5) BUILD A DOMINOES comes next.

One suggestion maybe: For an answer format involving the number of turns, especially (8) BUILD A HEXMAZE, it is easy to miscount the number. I have to recount like 5 times until I'm sure that the answer is 101 instead of 100 (3 times I counted 100, and only 2 times I counted 101.) Maybe the answer format can be simpler and easier like the content/wall of some rows, or the total number of turns (not just obtuse one.)

Edited by athin 2020-04-20 5:09 AM

Congratulations to Lukasz Kalinowski, Hugo van Rooijen and Tomoya Kimura for taking the top three places.

Thanks to the 112 participants from 31 countries. The participation was much higher compared to 2019. 36 participants completed the first 9 puzzles, and 24 participants scored points in the Optimzers. The median score of the test was 47. Italy had 14 participants, India 12, USA and France had 11 participants each.

Puzzle #1 (Battleships with Losses) was solved by the highest number of participants (96). Puzzles #8 (Build a Hex Maze) and #9 (Chain between Polyominoes) were each solved by 62 participants, the least among the first 9 puzzles.

Thanks Riad for the engaging puzzles!

And, thanks to all participants for healthy discussions, suggestions and constructive criticisms!