German Puzzle championship
German Puzzle championship
With very short notice, the online qualification of the German Puzzle championship will take place next weekend.
The duration of the test will be of 150 minutes (apparently with no extension), and will be running from April 26th at 12 until the 29th at 12 noon (German time).
At the moment the contest page is not yet activated. The instruction booklet (only in German for the moment) can be downloaded from this link
http://www.logicmasters.de/dokumente/L ... tungen.pdf
Submission as usual, wrong submission will be penalized.
stefano
The duration of the test will be of 150 minutes (apparently with no extension), and will be running from April 26th at 12 until the 29th at 12 noon (German time).
At the moment the contest page is not yet activated. The instruction booklet (only in German for the moment) can be downloaded from this link
http://www.logicmasters.de/dokumente/L ... tungen.pdf
Submission as usual, wrong submission will be penalized.
stefano
Re: German Puzzle championship
The contest page with more detail (in english) is now available: http://www.logicmasters.de/LM/qualifik ... ?chlang=ensf2l wrote:With very short notice, the online qualification of the German Puzzle championship will take place next weekend. [...]
Re: German Puzzle championship
Just to make sure  the instructions are also available in English.
Re: German Puzzle championship
If anyone was considering entering this, and was 'bewildered' by the hardest puzzle (a new hexagonal Masyu variant), I've created a practise puzzle at http://www.logicmasters.de/Raetselport ... ?id=0001PB.
At least it might give an indication whether you should invest time solving the real one. I'm happy to give any tips on solving it if wanted.
At least it might give an indication whether you should invest time solving the real one. I'm happy to give any tips on solving it if wanted.

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Re: German Puzzle championship
Nice one James! BL 1/3 need some intuition to connect, but the rest was nicely logical.
Considering doing the contest just for this one puzzle now
Considering doing the contest just for this one puzzle now
Re: German Puzzle championship
This is not exactly same puzzle, but I think it can be useful:
Hex Masyu by Dave Tuller, USPC 2010: http://wpc.puzzles.com/uspc10/
Hex Masyu by Dave Tuller, USPC 2010: http://wpc.puzzles.com/uspc10/
Re: German Puzzle championship
Felt a bit weird solving, but I think it builds on most of the obvious problems with hex masyu and leaves you with the main 120 degree constraint. I haven't thought about the logic entirely rigorously, but I enjoyed how you could feel out a solution. In that respect it's a lot like numberlink. If it feels good then you're almost certainly on the right track. Nice puzzle James!kiwijam wrote:If anyone was considering entering this, and was 'bewildered' by the hardest puzzle (a new hexagonal Masyu variant), I've created a practise puzzle at http://www.logicmasters.de/Raetselport ... ?id=0001PB.
At least it might give an indication whether you should invest time solving the real one. I'm happy to give any tips on solving it if wanted.
Re: German Puzzle championship
Actually I had no idea how to create a hex grid, so I found Dave's puzzle first, deleted his clues and pasted my own ones in. So that's why they might look similar...yureklis wrote:Hex Masyu by Dave Tuller, USPC 2010: http://wpc.puzzles.com/uspc10/
Re: German Puzzle championship
Yes, most of the puzzle only requires 'shortterm' deductions, as I was trying to demonstrate how the basic logic works.PuzzleScot wrote:Nice one James! BL 1/3 need some intuition to connect, but the rest was nicely logical.
Considering doing the contest just for this one puzzle now
But I intended there to be two options for that black pair in the bottomleft, and if the solver follows one of the paths they soon reach a contradiction down by the asterisk.

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Re: German Puzzle championship
I used the 'uniqueness' rule to deduce the direction of that pair. (One way would leave the very left loopback having 2 valid routes out).kiwijam wrote:I intended there to be two options for that black pair in the bottomleft, and if the solver follows one of the paths they soon reach a contradiction down by the asterisk.
Uniqueness came in handy in 1 or 2 other places too  eg, on one of the white pairs in the bottom right.
Re: German Puzzle championship
Well that was fun  first online competition for ages, and feeling the rustiness.
Re: German Puzzle championship
Same here. Thanks very much to those who set it.
No specifics, but I spent 41 minutes on a puzzle, reached a contradiction, left it, came back later, had another go, spent another 40 minutes and reached the same contradiction.
It's my own fault, but... my performance was not what I was hoping for.
No specifics, but I spent 41 minutes on a puzzle, reached a contradiction, left it, came back later, had another go, spent another 40 minutes and reached the same contradiction.
It's my own fault, but... my performance was not what I was hoping for.
Re: German Puzzle championship
Snap!drsteve wrote:Well that was fun  first online competition for ages, and feeling the rustiness.
Two or three that I spent quite a bit of time on and found errors in my solving.
Thanks to all concerned.
Re: German Puzzle championship
The results are out.
Re: German Puzzle championship
Good scores, gang!
I dreamt last night that someone here (not saying who, but they took the test) provided a walkthrough of the KropkiKakuro, which is the one that I spent 40+ minutes on, twice, without a solution. I'm happy enough with my solution to the top three rows (though without having seen the right answers, I can't be sure whether or not they're correct) and kept coming to the same result for the fourth shorter row, which may be dubious. I kept getting a contradiction in the lowerlelft corner, though. If someone could step through at least that bit of it then I would be most appreciative.
Also, startingpoints for Transporter and the 15s puzzle would also be gratefully received. I thought Transporter might be a bit like an Arukone except you connected the lines to each other, but that wasn't close to being the way to crack it, and I tried some StarBattlelike 2x2 box thinking to the 15s puzzle, but couldn't really place more than 23 digits for sure as a result.
I dreamt last night that someone here (not saying who, but they took the test) provided a walkthrough of the KropkiKakuro, which is the one that I spent 40+ minutes on, twice, without a solution. I'm happy enough with my solution to the top three rows (though without having seen the right answers, I can't be sure whether or not they're correct) and kept coming to the same result for the fourth shorter row, which may be dubious. I kept getting a contradiction in the lowerlelft corner, though. If someone could step through at least that bit of it then I would be most appreciative.
Also, startingpoints for Transporter and the 15s puzzle would also be gratefully received. I thought Transporter might be a bit like an Arukone except you connected the lines to each other, but that wasn't close to being the way to crack it, and I tried some StarBattlelike 2x2 box thinking to the 15s puzzle, but couldn't really place more than 23 digits for sure as a result.
Re: German Puzzle championship
Oooh yeah some good results. Sadly I couldn't find time to have a go during the weekend, but I might give it a go over the next couple of evenings. Congrats on the great scores everyone!
Re: German Puzzle championship
Nice to see that a few of us would still be in with a shot of getting into the German squad. Y'know, if we were German and everything...
Pretty pleased with my performance. Guards I can't see how to solve at all if anyone has a logical walkthrough it would be much appreciated. As for Battleships, I placed the 3 long ones quickly, but even after the test it took ages to get the rest sorted, I just had a brainfreeze with the 2s. But happy with the rest.
Transporter I didn't solve completely logically, but I can give my rough thought process. First, I connected the G to the G round the outside, using numberlink logic/ uniqueness, and the same with the A and A. D to D could then be done too. Mentally connecting E to E with a straight line divided the rest of the grid into a North and South. In the North, the C to C path is almost forced, bearing in mind that 2 line segments must emanate from the C Cube, and the A Cube. Joining C circle to the A cube seemed sensible next, because doing anything else would cross the E to E 'line'. Following numberlink logic on the bottom half, the rest should fall almost immediately. Hope that helps; if you want a purely logical solution, I can't help!
By 15s puzzle, I assume you mean Capsules. I think I did solve that logically, so I'll see if I can remember my thought process. It took longer than the Transporter though, despite being worth fewer points.
There's only one place for a 1 in the top Ipentomino. There are 2 places for a 2 in the bottom N. Either way, the 2 in the P must be in the top row of 3. So the 2 in the I can't be next to the 4. Also, if the 2 in the I is the leftmost, there's no room for a 2 in the topleft W. So there's a 2 in row 1 col 4, and therefore in row 2 col 6.
If the right arm of the X is a 1, you can't get a 1 in the right L. So the 1 in the X is the bottom arm.
There can't be a 2 on the left arm of the T, because of the N as previously mentioned. So the 2 is row 5 col 7 or 8. But if it's col 8, there's no room for a 2 in the F. So it's col 7.
If row 4 col 8 is a 4, there's no room for 4 in the T. So it's a 3, and so is row 5 col 6. The 1 and 4 can also be placed in the T.
Row 4 col 6 can only be a 1; the cell above must then be a 3, and the 5 and 4 can also be placed in the P. The I can now be finished, and then a 5 can be placed in the W in row 3 col 3, and another 5 in the Z in row 5 col 2.
The 2 in the N must be in row 4 col 5, and the 3 and 5 can then be placed in the N too.
There's only 1 place for a 3 in the Z, and the 1 and 4 can then be placed. The 4 can now be placed in the leftmost Y, and then in the W.
The 5 can be placed in the Y, and the 1s and 2s in the Y and W can be finished off.
On the right side now: Row 2 col 7 only has one option, a 5.
If the right arm of the X was a 4, row 4 col 10 would also have to be a 4, and there'd be no room for a 4 in the bottom F. So the right arm of the X is a 3, the top arm is the 4.
Row 1 col 9 must be a 1, and there's now only one option for the 3 in the L. Row 3 col 9 has only one option, a 4. And finally the L can be finished off, 4 above 2, and the F can be finished, 5 above 2.
Hope my notation is clear!
Pretty pleased with my performance. Guards I can't see how to solve at all if anyone has a logical walkthrough it would be much appreciated. As for Battleships, I placed the 3 long ones quickly, but even after the test it took ages to get the rest sorted, I just had a brainfreeze with the 2s. But happy with the rest.
Transporter I didn't solve completely logically, but I can give my rough thought process. First, I connected the G to the G round the outside, using numberlink logic/ uniqueness, and the same with the A and A. D to D could then be done too. Mentally connecting E to E with a straight line divided the rest of the grid into a North and South. In the North, the C to C path is almost forced, bearing in mind that 2 line segments must emanate from the C Cube, and the A Cube. Joining C circle to the A cube seemed sensible next, because doing anything else would cross the E to E 'line'. Following numberlink logic on the bottom half, the rest should fall almost immediately. Hope that helps; if you want a purely logical solution, I can't help!
By 15s puzzle, I assume you mean Capsules. I think I did solve that logically, so I'll see if I can remember my thought process. It took longer than the Transporter though, despite being worth fewer points.
There's only one place for a 1 in the top Ipentomino. There are 2 places for a 2 in the bottom N. Either way, the 2 in the P must be in the top row of 3. So the 2 in the I can't be next to the 4. Also, if the 2 in the I is the leftmost, there's no room for a 2 in the topleft W. So there's a 2 in row 1 col 4, and therefore in row 2 col 6.
If the right arm of the X is a 1, you can't get a 1 in the right L. So the 1 in the X is the bottom arm.
There can't be a 2 on the left arm of the T, because of the N as previously mentioned. So the 2 is row 5 col 7 or 8. But if it's col 8, there's no room for a 2 in the F. So it's col 7.
If row 4 col 8 is a 4, there's no room for 4 in the T. So it's a 3, and so is row 5 col 6. The 1 and 4 can also be placed in the T.
Row 4 col 6 can only be a 1; the cell above must then be a 3, and the 5 and 4 can also be placed in the P. The I can now be finished, and then a 5 can be placed in the W in row 3 col 3, and another 5 in the Z in row 5 col 2.
The 2 in the N must be in row 4 col 5, and the 3 and 5 can then be placed in the N too.
There's only 1 place for a 3 in the Z, and the 1 and 4 can then be placed. The 4 can now be placed in the leftmost Y, and then in the W.
The 5 can be placed in the Y, and the 1s and 2s in the Y and W can be finished off.
On the right side now: Row 2 col 7 only has one option, a 5.
If the right arm of the X was a 4, row 4 col 10 would also have to be a 4, and there'd be no room for a 4 in the bottom F. So the right arm of the X is a 3, the top arm is the 4.
Row 1 col 9 must be a 1, and there's now only one option for the 3 in the L. Row 3 col 9 has only one option, a 4. And finally the L can be finished off, 4 above 2, and the F can be finished, 5 above 2.
Hope my notation is clear!
Re: German Puzzle championship
Can't say that I solved guards logically at all. I basically put one in where it looked like it was going to cover loads of rooms, then did the same with the others. Took me about three attempts of tweaking to get it right. On successful attempt, I think I started with D1 and G4 then the others seemed to fall in.

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Re: German Puzzle championship
Does anyone have the password? Can't see any way of obtaining it.
Thank you.
Thank you.
Re: German Puzzle championship
http://forum.logicmasters.de/showthrea ... 3#pid19913david mcneill wrote:Does anyone have the password? Can't see any way of obtaining it.

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Re: German Puzzle championship
Thanks Berni.
Re: German Puzzle championship
Thank you, Neil; that was clear, and I appreciate your time and effort on the walkthrough  both notation and reasoning were clear in context.
I got the first sentence in practice, but that was the first bit I didn't get. Note to self: there's more of these at Rohan Rao's blog for practice.Nilz wrote:There are 2 places for a 2 in the bottom N.
Re: German Puzzle championship
You're welcome; I'm happy to help! With things like this where there are no obvious placements I just find that placing a number on the border between two cells, if you know it goes in one of the two, works wonders.dickoon wrote:Thank you, Neil; that was clear, and I appreciate your time and effort on the walkthrough  both notation and reasoning were clear in context.
I got the first sentence in practice, but that was the first bit I didn't get. Note to self: there's more of these at Rohan Rao's blog for practice.Nilz wrote:There are 2 places for a 2 in the bottom N.
Re: German Puzzle championship
By request: A walkthrough of the KropkiKakuro.
Obviously with no dots, this is sort like "nonconsecutive" only more so.
This tends to involve thinking 23 steps ahead, so apologies if any of this seems convoluted.
Step 1:
Consider the top left "8" row. if it were a "125", the 5 would have to be in the middle to separate the 1 and 2.
But if there's a 5 in the middle, the "10" going down from there is impossible (523 has 23, 541 has 54, 514 has 1 in 2nd row)
So the 8 is a 134 with the 1 in the middle as a spacer.
Further, the only way to complete the "10" down from the 1 is 172. (the only allowable 2cell 9's are 18 and 27, etc).
If there were a 4 in the top left cell, the "16" going down is impossible (457 has 45, 475 conflicts with the 7, 439 has 43, 493 rubs 3 against 2).
So we have a 3 in the top left, which must complete as 358, since 367 doesn't work and there would be no place to put a 4 in that column.
So, thus far as Step 1, we have:
3 1 4 _
5 7
8 2
Step 2:
Let's look at the 2nd row. We know the remaining numbers are 34689. All of those except the 9 have
a conflict for the slot next to the 7, so that's a 9.
The 4 has conflicts placed next to the 3 or the 8, so it must be on the end of the remaining string of 4 cells.
The order must be 4 6 8 3 or 3 8 6 4. If there were a 3 on the right, we'd have trouble with the 20 in the last column.
A 3 would force it to finish with an 8 or 9, either of which causes issues with the 11 in the top right row. (128 is not ok here).
Thus the 2nd row is 5 7 9 3 8 6 4
We fill in the 7 and 9 in the last column, and the 1 and 3 to finish the "11" (3 can't be above the 6".
3 1 4 _ 3 1 7
5 7 9 3 8 6 4
8 2 . . . . 9
Step 3:
In the 3rd row, there is no place we could put a 3. Since we are missing 2 things that add to 8, we are missing 3,5.
The remaining numbers are 1467.
We have to start with 6 or 7 next to the 2. Then a 1 (can't be the other of the 6/7, and the 3 above conflicts with a 4)
We have to have a 6 under the 8, since 4 and 7 conflict with 8, so that resolves the sequence as 7 1 6 4.
3 1 4 _ 3 1 7
5 7 9 3 8 6 4
8 2 7 1 6 4 9
Step 4:
Let's look at the 6th column. It is missing two things that add to 9. We already have 1,4,6 so missing pair is 27.
But the only allowable pairs for that 9 at the bottom right are 18 or 27, so it's 81 (due to 1 directly above in 6th col).
The other entries in that column are 935 in some order. The 9 must be below the 4 since 3, 5 conflict.
We note the (35) pairs for later. The remaining 13 in that bottom right column must
be a 5 and 8, since 67 is a conflict, and there is no place to put a 4 (it conflicts with both the 3 and the 5 next to it),
so 49 is out.
3 1 4 _ 3 1 7
5 7 9 3 8 6 4
8 2 7 1 6 4 9
. . . . . 9
. . . . . .35 58
. . . . . .35 58
. . . . . .8 1
(breaking here to start a new message as this is getting long)
Obviously with no dots, this is sort like "nonconsecutive" only more so.
This tends to involve thinking 23 steps ahead, so apologies if any of this seems convoluted.
Step 1:
Consider the top left "8" row. if it were a "125", the 5 would have to be in the middle to separate the 1 and 2.
But if there's a 5 in the middle, the "10" going down from there is impossible (523 has 23, 541 has 54, 514 has 1 in 2nd row)
So the 8 is a 134 with the 1 in the middle as a spacer.
Further, the only way to complete the "10" down from the 1 is 172. (the only allowable 2cell 9's are 18 and 27, etc).
If there were a 4 in the top left cell, the "16" going down is impossible (457 has 45, 475 conflicts with the 7, 439 has 43, 493 rubs 3 against 2).
So we have a 3 in the top left, which must complete as 358, since 367 doesn't work and there would be no place to put a 4 in that column.
So, thus far as Step 1, we have:
3 1 4 _
5 7
8 2
Step 2:
Let's look at the 2nd row. We know the remaining numbers are 34689. All of those except the 9 have
a conflict for the slot next to the 7, so that's a 9.
The 4 has conflicts placed next to the 3 or the 8, so it must be on the end of the remaining string of 4 cells.
The order must be 4 6 8 3 or 3 8 6 4. If there were a 3 on the right, we'd have trouble with the 20 in the last column.
A 3 would force it to finish with an 8 or 9, either of which causes issues with the 11 in the top right row. (128 is not ok here).
Thus the 2nd row is 5 7 9 3 8 6 4
We fill in the 7 and 9 in the last column, and the 1 and 3 to finish the "11" (3 can't be above the 6".
3 1 4 _ 3 1 7
5 7 9 3 8 6 4
8 2 . . . . 9
Step 3:
In the 3rd row, there is no place we could put a 3. Since we are missing 2 things that add to 8, we are missing 3,5.
The remaining numbers are 1467.
We have to start with 6 or 7 next to the 2. Then a 1 (can't be the other of the 6/7, and the 3 above conflicts with a 4)
We have to have a 6 under the 8, since 4 and 7 conflict with 8, so that resolves the sequence as 7 1 6 4.
3 1 4 _ 3 1 7
5 7 9 3 8 6 4
8 2 7 1 6 4 9
Step 4:
Let's look at the 6th column. It is missing two things that add to 9. We already have 1,4,6 so missing pair is 27.
But the only allowable pairs for that 9 at the bottom right are 18 or 27, so it's 81 (due to 1 directly above in 6th col).
The other entries in that column are 935 in some order. The 9 must be below the 4 since 3, 5 conflict.
We note the (35) pairs for later. The remaining 13 in that bottom right column must
be a 5 and 8, since 67 is a conflict, and there is no place to put a 4 (it conflicts with both the 3 and the 5 next to it),
so 49 is out.
3 1 4 _ 3 1 7
5 7 9 3 8 6 4
8 2 7 1 6 4 9
. . . . . 9
. . . . . .35 58
. . . . . .35 58
. . . . . .8 1
(breaking here to start a new message as this is getting long)