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Posts: 136
 
Location: India | TiiT - 2020-04-14 12:58 AM
It would be nice if only author answers the questions. In that way we can be sure that the answers are correct and it doesn't get messy.
Thanks Riad for the answers!
Your quick retort was rude and unnecessary, if not ridiculous. On the contrary of what you think, and if you have not been following, the participants are NOT the source of mess. Instead they are the ones who are doing the admins' job.
Everyone is confused with many of the puzzles. Even after all that tedious messaging back and forth, look at the last question here - Chaoticiak has still got the rules of puzzle 10 wrong (or I got it wrong). I am also not sure yet as the 'suggested' but not confirmed second version appears to be a trivial puzzle. You can imagine how shabbily all of this was managed. The earliest queries regarding Puzzle 10, for example, came from me and they were selectively taken down for no reason!
The random change in Puzzle 11 is unacceptable just 2 days shy of the deadline. It would be unfair as I had not considered corner touch until now. |
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Posts: 136
Location: India | Riad Khanmagomedov - 2020-04-13 9:05 PM
anurag - 2020-04-12 11:43 PM
Puzzle 11: If the solution has multiple occurences of a card, but only one that does not share borders with other cards, is it acceptable?
Cards may not overlap.
I do not understand how that answers the question. If language is a problem, the admin should own the responsibility of a mediator. |
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Posts: 136
Location: India | chaotic_iak - 2020-04-14 9:38 AM
Considering the "white streets may not form 2x2 area" rule in 10 is actually not part of the Town ruleset, but a requirement for the author (that we must form a unique Town puzzle, which happens to have no 2x2 white area)
I think I understand your third version. Is it different from both versions described here previously? I think it is. One can take the liberty to innovate, given that the instructions were woefully inadequate. |
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Location: India | Puzzle 8: Hex maze
Answer format is total number of obtuse angles of the path. Do I have to count (the number of obtuse angles) both sides of each turn of the path? |
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Location: United States | Swagatam - 2020-04-13 1:58 PM
Puzzle 8: Hex maze
Answer format is total number of obtuse angles of the path. Do I have to count (the number of obtuse angles) both sides of each turn of the path?
An obtuse angle measures strictly between (and not including) 90 and 180 degrees, so even if you did count both sides, you should arrive at the same answer. |
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Posts: 139
Location: Estonia | TiiT - 2020-04-14 9:58 PM
It would be nice if only author answers the questions. In that way we can be sure that the answers are correct and it doesn't get messy.
Thanks Riad for the answers!
Sorry about my expression. I agree that solvers make an important part in here.
So I put it in this way.
Maybe it would be nice if only those solvers are responsing who understand the rules (in addition to authors responses). It doesn't make sense if the same person answers the questions and at the same time repeats that he doesn't understand anything. |
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Posts: 136
Location: India | Puzzle 10: As it stands now, any clarity remains elusive. There are at least three versions that the contestants could innovate. Two that are straightforward, and a third that asks me to create a town that was a clouds by definition and clues, but is actually somehow a town by solution. So, I guess it would be preferable to accept a solution based on any of the three versions as the participant likes. |
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Posts: 136
Location: India | Apparently, either puzzle 10, or the whole contest, is scrapped. |
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Posts: 187
Location: New Zealand | Sorry Anurag, you are filling this thread with complaints. You've posted 30 times already: take a breath, go for a walk.
I think Riad does a wonderful job every year, creating a contest for our enjoyment, even if there are small language difficulties.
Ivan's comment agrees with Riad's version, there is no ambiguity any more.
Let me repeat the instructions in my own words, which might help? |
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Posts: 187
Location: New Zealand | chaotic_iak had a good question: is "Borders of cards cannot have common segments" used when finding the unique contours, or a condition that the unique solution must have?
I assume the latter, see below.
These are not full rules, just helping to explain the official rules.
10.
Construct a normal Clouds/Radar/Regenwolken/Town puzzle. (e.g. https://www.janko.at/Raetsel/Regenwolken/index.htm)
It must have a unique solution.
That unique solution cannot have any 2x2 white areas.
Maximize N, the number of white words if this was a crossword grid.
11.
Place diamonds into a grid (no larger than 16x13).
There must be exactly one possible way to place the 9 transparent cards to match all of the diamonds.
That unique solution cannot have two cards sharing an edge, but they can overlap or touch at corners.
Minimize the number of diamonds.
12.
Place some or all of the blue pieces in a 7x7 grid such that they don't touch each other, not even diagonally.
Choose some or all of the black ships to be placed.
N is the number of solutions for placing your black ships.
We don't actually define the black ships in the answer format, but they are found by subtracting the number of blue pieces from the ABCD totals. |
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Posts: 11
Location: Indonesia | For puzzle 11, I have a feeling that sharing an edge is a part of Diamond ruleset instead. This is based on Riad's September 2012 contest if the puzzle is indeed reused: https://logicmastersindia.com/lmitests/dl.asp?attachmentid=269&view=... (puzzle 1).
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Posts: 136
Location: India | kiwijam - 2020-04-15 5:28 PM
chaotic_iak had a good question: is "Borders of cards cannot have common segments" used when finding the unique contours, or a condition that the unique solution must have?
I assume the latter, see below.
These are not full rules, just helping to explain the official rules.
10.
Construct a normal Clouds/Radar/Regenwolken/Town puzzle. (e.g. https://www.janko.at/Raetsel/Regenwolken/index.htm)
It must have a unique solution.
That unique solution cannot have any 2x2 white areas.
Maximize N, the number of white words if this was a crossword grid.
11.
Place diamonds into a grid (no larger than 16x13 ).
There must be exactly one possible way to place the 9 transparent cards to match all of the diamonds.
That unique solution cannot have two cards sharing an edge, but they can overlap or touch at corners.
Minimize the number of diamonds.
12.
Place some or all of the blue pieces in a 7x7 grid such that they don't touch each other, not even diagonally.
Choose some or all of the black ships to be placed.
N is the number of solutions for placing your black ships.
We don't actually define the black ships in the answer format, but they are found by subtracting the number of blue pieces from the ABCD totals.
Funny. |
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Posts: 45
Location: India | Is this possible that if I put all the numbers outside the grid (which indicate the number of shaded cells) it may lead to multiple solutions? [For Town puzzle or any of this kind of puzzle] |
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Location: Spain | Congratulations for the contest !!
Puzzle 9:
I don't understand very well: "At the same time,starting from theb lack circle, draw a loop along the lines of the grid,passing through all rows and columns."
Can someone explain that to me in more detail?
Thanks.
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Location: Spain | Ok, I have seen that it can be solved without understanding it very well... |
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Location: Japan | kiwijam - 2020-04-15 8:28 PM
chaotic_iak had a good question: is "Borders of cards cannot have common segments" used when finding the unique contours, or a condition that the unique solution must have?
I assume the latter, see below.
These are not full rules, just helping to explain the official rules.
10.
Construct a normal Clouds/Radar/Regenwolken/Town puzzle. (e.g. https://www.janko.at/Raetsel/Regenwolken/index.htm)
It must have a unique solution.
That unique solution cannot have any 2x2 white areas.
Maximize N, the number of white words if this was a crossword grid.
11.
Place diamonds into a grid (no larger than 16x13 ).
There must be exactly one possible way to place the 9 transparent cards to match all of the diamonds.
That unique solution cannot have two cards sharing an edge, but they can overlap or touch at corners.
Minimize the number of diamonds.
12.
Place some or all of the blue pieces in a 7x7 grid such that they don't touch each other, not even diagonally.
Choose some or all of the black ships to be placed.
N is the number of solutions for placing your black ships.
We don't actually define the black ships in the answer format, but they are found by subtracting the number of blue pieces from the ABCD totals.
Thanks kiwijam for summarizing rules, but I don't agree with the interpretation of the rule 11. I agree with athin, the Diamonds rule set includes the border rule.
It becomes a problem in the following case.
(use 2,3,4 cards)
~~~~~~~
D~D~D~~
~~~~~~~
~~~~~~~
~~~~~~~
~~D~~~D
~~~~~~~
~~D~~~D
The position of the card 2 is unique or not?
I imagine each participant has different interpretations. It's a pity that the winner will change depending on the Riad's final judge after the contest....
I hope Riad will answer that question as soon as possible not to ruin this wonderful contest.
Edited by panista 2020-04-16 4:55 PM
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Location: Germany | yes |
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Moscow Puzzle Cup 2016 Author
Posts: 234
Location: Russia | panista - 2020-04-16 4:53 PM
kiwijam - 2020-04-15 8:28 PM
chaotic_iak had a good question: is "Borders of cards cannot have common segments" used when finding the unique contours, or a condition that the unique solution must have?
I assume the latter, see below.
These are not full rules, just helping to explain the official rules.
10.
Construct a normal Clouds/Radar/Regenwolken/Town puzzle. (e.g. https://www.janko.at/Raetsel/Regenwolken/index.htm)
It must have a unique solution.
That unique solution cannot have any 2x2 white areas.
Maximize N, the number of white words if this was a crossword grid.
11.
Place diamonds into a grid (no larger than 16x13 ).
There must be exactly one possible way to place the 9 transparent cards to match all of the diamonds.
That unique solution cannot have two cards sharing an edge, but they can overlap or touch at corners.
Minimize the number of diamonds.
12.
Place some or all of the blue pieces in a 7x7 grid such that they don't touch each other, not even diagonally.
Choose some or all of the black ships to be placed.
N is the number of solutions for placing your black ships.
We don't actually define the black ships in the answer format, but they are found by subtracting the number of blue pieces from the ABCD totals.
Thanks kiwijam for summarizing rules, but I don't agree with the interpretation of the rule 11. I agree with athin, the Diamonds rule set includes the border rule.
It becomes a problem in the following case.
(use 2,3,4 cards )
~~~~~~~
D~D~D~~
~~~~~~~
~~~~~~~
~~~~~~~
~~D~~~D
~~~~~~~
~~D~~~D
The position of the card 2 is unique or not?
I imagine each participant has different interpretations. It's a pity that the winner will change depending on the Riad's final judge after the contest....
I hope Riad will answer that question as soon as possible not to ruin this wonderful contest.
I see a unique position for 2. This is a card with diamonds P2C3 and P6C3. What other options are possible? |
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Posts: 187
Location: New Zealand | Oh dear. My interpretation was wrong then, that I posted a day ago.
According to my understanding of the rules, this diagram fails the sentence "All transparent cards from 2 to 4 are put on the table so that it is possible to uniquely restore their contours".
Because there are three places the 2-card might fit.
This is the opposite approach to puzzle 10, where the extra rule (no white 2x2) applies to the unique solution, not to the solving process.
Unfortunately it is 1:30am here, and I don't have time to start again... |
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Moscow Puzzle Cup 2016 Author
Posts: 234
Location: Russia | kiwijam - 2020-04-16 7:02 PM
Oh dear. My interpretation was wrong then, that I posted a day ago.
According to my understanding of the rules, this diagram fails the sentence "All transparent cards from 2 to 4 are put on the table so that it is possible to uniquely restore their contours".
Because there are three places the 2-card might fit.
This is the opposite approach to puzzle 10, where the extra rule (no white 2x2) applies to the unique solution, not to the solving process.
Unfortunately it is 1:30am here, and I don't have time to start again...
We can extend the competition for a day.
I want to say about the puzzle 10. Its text changed the day before the start. My friends from India didn't like my translation using a translator. There was a misunderstanding. As a result, the text has changed a lot.
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Moscow Puzzle Cup 2016 Author
Posts: 234
Location: Russia | kiwijam - 2020-04-16 7:02 PM
Oh dear. My interpretation was wrong then, that I posted a day ago.
According to my understanding of the rules, this diagram fails the sentence "All transparent cards from 2 to 4 are put on the table so that it is possible to uniquely restore their contours".
Because there are three places the 2-card might fit.
This is the opposite approach to puzzle 10, where the extra rule (no white 2x2) applies to the unique solution, not to the solving process.
Unfortunately it is 1:30am here, and I don't have time to start again...
My text using the translator looked like this:
Place dark rectangles in the 12 x 10 grid, the lengths of the sides of which are at least 2 cells. They do not touch each other, not even diagonally. The number outside the grid shows the number of dark cells in the corresponding row. The puzzle must have a unique solution in your placement of the numbers. In this case, the white cells must form streets with a width of one cell, i.e. the white area of the streets cannot contain fragments of 2 x 2 cells. Maximize the number of streets. Among solutions with equal result the best will be the one in which the sum of the given numbers is less. |
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Posts: 136
Location: India | Riad Khanmagomedov - 2020-04-16 7:21 PM
My text using the translator looked like this:
Place dark rectangles in the 12 x 10 grid, the lengths of the sides of which are at least 2 cells. They do not touch each other, not even diagonally. The number outside the grid shows the number of dark cells in the corresponding row. The puzzle must have a unique solution in your placement of the numbers. In this case, the white cells must form streets with a width of one cell, i.e. the white area of the streets cannot contain fragments of 2 x 2 cells. Maximize the number of streets. Among solutions with equal result the best will be the one in which the sum of the given numbers is less.
Why is this discussion still happening? I thought it was "no more ambiguous" or even unambiguous from the beginning and I was posting countless questions forn nothing.
kiwijam - 2020-04-16 7:02 PM
Oh dear. My interpretation was wrong then, that I posted a day ago.
According to my understanding of the rules, this diagram fails the sentence "All transparent cards from 2 to 4 are put on the table so that it is possible to uniquely restore their contours".
Because there are three places the 2-card might fit.
This is the opposite approach to puzzle 10, where the extra rule (no white 2x2) applies to the unique solution, not to the solving process.
Unfortunately it is 1:30am here, and I don't have time to start again...
I thought kiwijam cannot be wrong ;)
Edited by anurag 2020-04-16 7:31 PM
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Location: India | 24 hours Extension
The contest duration is extended by 24 hours.
Answers can now be submitted/modified till 23:59 Moscow time on 17th April (20:59 GMT 17-04-2020) .
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Posts: 136
Location: India | Riad Khanmagomedov - 2020-04-16 7:21 PM
kiwijam - 2020-04-16 7:02 PM
Oh dear. My interpretation was wrong then, that I posted a day ago.
According to my understanding of the rules, this diagram fails the sentence "All transparent cards from 2 to 4 are put on the table so that it is possible to uniquely restore their contours".
Because there are three places the 2-card might fit.
This is the opposite approach to puzzle 10, where the extra rule (no white 2x2) applies to the unique solution, not to the solving process.
Unfortunately it is 1:30am here, and I don't have time to start again...
My text using the translator looked like this:
Place dark rectangles in the 12 x 10 grid, the lengths of the sides of which are at least 2 cells. They do not touch each other, not even diagonally. The number outside the grid shows the number of dark cells in the corresponding row. The puzzle must have a unique solution in your placement of the numbers. In this case, the white cells must form streets with a width of one cell, i.e. the white area of the streets cannot contain fragments of 2 x 2 cells. Maximize the number of streets. Among solutions with equal result the best will be the one in which the sum of the given numbers is less.
Riad, the translation from your original text sounds much better than what the admins reduced it to. Not surprising, given that "You must submit before ..." has been there for years now in bold font on the submission pages. |
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Moscow Puzzle Cup 2016 Author
Posts: 234
Location: Russia | anurag - 2020-04-16 7:36 PM
Riad Khanmagomedov - 2020-04-16 7:21 PM
kiwijam - 2020-04-16 7:02 PM
Oh dear. My interpretation was wrong then, that I posted a day ago.
According to my understanding of the rules, this diagram fails the sentence "All transparent cards from 2 to 4 are put on the table so that it is possible to uniquely restore their contours".
Because there are three places the 2-card might fit.
This is the opposite approach to puzzle 10, where the extra rule (no white 2x2) applies to the unique solution, not to the solving process.
Unfortunately it is 1:30am here, and I don't have time to start again...
My text using the translator looked like this:
Place dark rectangles in the 12 x 10 grid, the lengths of the sides of which are at least 2 cells. They do not touch each other, not even diagonally. The number outside the grid shows the number of dark cells in the corresponding row. The puzzle must have a unique solution in your placement of the numbers. In this case, the white cells must form streets with a width of one cell, i.e. the white area of the streets cannot contain fragments of 2 x 2 cells. Maximize the number of streets. Among solutions with equal result the best will be the one in which the sum of the given numbers is less.
Riad, the translation from your original text sounds much better than what the admins reduced it to. Not surprising, given that "You must submit before ..." has been there for years now in bold font on the submission pages.
Anurag, the second version also appeared through my efforts. We were looking for a more understandable wording, but as time has shown, we have not clarified the issue. I am very grateful to the administrators for their work. They have supported me for many years. Please note that the entire serious workload of the organizers is based on enthusiasm. |
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WSPC 2017 • Schedule • Participation • Location • WSC • WPC • Contact •Search •Recent
Riad Khanmagomedov's April Contest — 8th to 18th April 2020
