[LMI] Puzzle Marathon  1st to 10th March

 Posts: 129
 Joined: Fri 09 Jul, 2010 12:36 pm
[LMI] Puzzle Marathon  1st to 10th March
Link : http://logicmastersindia.com/2013/03P/
• 12 large puzzles
• 12 different authors
• solve each puzzle separately
• dates : 1st  10th March
• 12 large puzzles
• 12 different authors
• solve each puzzle separately
• dates : 1st  10th March

 Posts: 129
 Joined: Fri 09 Jul, 2010 12:36 pm
Re: [LMI] Puzzle Marathon  1st to 10th March
!!! Last 3 days for Puzzle Marathon !!!

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Re: [LMI] Puzzle Marathon  1st to 10th March
Well, what a lovely set of puzzles!
I didn't have time to compete seriously (or at all, really), but am very impressed by the solve times of the top people!
Nice to see another Neil/James duel. I see Neil doesn't like playing 2nd fiddle to anyone and upped his game a little! James  your move
I didn't have time to compete seriously (or at all, really), but am very impressed by the solve times of the top people!
Nice to see another Neil/James duel. I see Neil doesn't like playing 2nd fiddle to anyone and upped his game a little! James  your move
Re: [LMI] Puzzle Marathon  1st to 10th March
I finished all 12 puzzles on the first weekend of the competition, then checked the leaderboard about 3 times a day for the past week to see how many people had overtaken me! To my surprise, not many did pretty happy with that performance!PuzzleScot wrote:Nice to see another Neil/James duel. I see Neil doesn't like playing 2nd fiddle to anyone and upped his game a little! James  your move
All 12 puzzles solved really smoothly, so if anyone hasn't had a chance to attempt them yet, I recommend having a go now that the time pressure is off! Happy to give hints if anyone needs them.
Re: [LMI] Puzzle Marathon  1st to 10th March
I did a few  took a while to get the Skyscrapers to work  but not enough time to attempt it seriously though... Great competition
Re: [LMI] Puzzle Marathon  1st to 10th March
Not in the least smoothly for me, in some cases  I broke three of them repeatedly, which on Sunday was getting annoying (especially as I was out all day  although for once the violas had some tacet movements in which I made some progress).
I enjoyed the others though, and might come back to the broken one.
I enjoyed the others though, and might come back to the broken one.
Re: [LMI] Puzzle Marathon  1st to 10th March
TeaserPuzzleScot wrote: James  your move
Not a great week for me, never really got into them this time, but well done Neil for leading the way.
Re: [LMI] Puzzle Marathon  1st to 10th March
I was asked for some hints for the Star Battle. Here's my entire writeup, which I'll post publicly in case it helps anyone else.
I can’t remember if this is exactly how I did it, so I may have made some assumptions below that don’t quite hold. If you spot any, let me know and I’ll see if I can prove them! It may not be too coherent either, as I’ll write things as I think of them, because that’s how I solve! So apologies again. I hope it’s useful!
Label rows 117, cols are AQ, segments will be referred to by their top left cell.
There are 3 segments entirely within cols O,P,Q, so no other stars can be in those cols.
In the 3x3 areas, there must be at least 1 star on each of the top and bottom rows. This means row 4 cols HM, row 8 cols HM, row 10 cols EJ and row 14 cols EJ are empty.
H57, 1113 & 1517 must all contain at least one star. Hence they each contain exactly one, and the rest of the col is empty. G6,I6,G12,I12,G16,I16 are therefore empty.
Since for any 3x3 box, the outer edges must each contain at least one star, there cannot be any stars in cells adjacent to all 3. Eg J6 is empty, since it touches K57. Similarly, K6 is empty, N6, D12,K12,H12,L6,F12 too.
If E12 were a star, G11 & G13 would have to be, blocking the 3x3 next to it. So E12,J12,H12,M12 are blank. So for rows 5&7, all stars must be in the 3x3 boxes, same for 11&13.
Segment O4 is now divided into 3 2x1 boxes, each of which must have a star. So P9&Q9 are blank. So Segment O8 has all 3 stars in col O, rest of the col is blank so P4 & P6 are stars. O9 is empty, as is M9, else there’s not enough space in segment I2.
Consider segment I14. I16 & G17 are empty, so F17 & H17 are stars. Looking at segment E14, starts must be E15,G15,D17, and I15 must then be a star too. Rows 15&17 are done.
Segment M14 must have starts in M14,L16,N16. Segment D7 has D8,D10 and (A or B)16.
Segment E11 has (F or G)11, E13,G13. Segment H11 has H11,J11 and (I or J)13 thus F11 is a star.
Col N has 1 star already, and segment I must have at least 2 more in that col hence rest of col is empty. Also, 3rd star in segment I2 is I2 or I3, thus J2 and J3 are empty.
Look at cols HJ. There are 4 stars already. Segment H5 has 3 more within those cols, and (I2 or I3) and (I13 or J13) are stars, so rest of those 3 cols are empty.
H1 is now a 3x3 box, so L2 is empty.
Segment I2 has exactly 2 stars in col N. If these are N8 & N10, segment O8 is blocked, hence N4 is a star. Then segment K5 has stars in K7,M7 and (K or L)5. So segment H5 has stars in H5,J5 and (H or I)7. Then L5 must be a star, and col H is filled, so H7 is empty and I7 is a star.
So one of N810 is a star. Thus O8 & O10 can’t both be stars, so O12 & O14 are. For segment Q9, stars must be in Q12, Q14 and (P or Q)10. Row 14 is done.
For segment H1, K2 & L1 must be empty, thus K1 is a star. Segment B1 must have at least 2 stars in row 1. It can have at most 2, thus it must have exactly 2. So 3rd star is F2 or F3, so E2,E3,G2,G3 are empty. So rest of row is empty, and rest of row 1 is empty. So M2 is a star, and then K3 must be too. M & K are done.
N9 can’t be a star, it would block row O. Row 9 must have stars in (A or B), (F or G) and L. So 8A,8B,10A,10B,8F,8G are blank. Rows 8 & 10 can then be finished off, and col L is done.
Since F2 or F3 is a star, rest of col is empty, so G9 is a star. Col G is done. 3rd star in J must be J13. D is done. For segment B1, stars must be in E1, and hence F3. E is done. Row 6 can be finished. B3 must be empty, else it blocks segment A1, so C3 is 3rd star in that segment. Thus segment A1 has stars in A2 & A4. So C1 is a star. A & C are done, so B can be filled in, and the final star must be in I2.
Hopefully that’s helped, let me know if any of it is not completely clear!
I can’t remember if this is exactly how I did it, so I may have made some assumptions below that don’t quite hold. If you spot any, let me know and I’ll see if I can prove them! It may not be too coherent either, as I’ll write things as I think of them, because that’s how I solve! So apologies again. I hope it’s useful!
Label rows 117, cols are AQ, segments will be referred to by their top left cell.
There are 3 segments entirely within cols O,P,Q, so no other stars can be in those cols.
In the 3x3 areas, there must be at least 1 star on each of the top and bottom rows. This means row 4 cols HM, row 8 cols HM, row 10 cols EJ and row 14 cols EJ are empty.
H57, 1113 & 1517 must all contain at least one star. Hence they each contain exactly one, and the rest of the col is empty. G6,I6,G12,I12,G16,I16 are therefore empty.
Since for any 3x3 box, the outer edges must each contain at least one star, there cannot be any stars in cells adjacent to all 3. Eg J6 is empty, since it touches K57. Similarly, K6 is empty, N6, D12,K12,H12,L6,F12 too.
If E12 were a star, G11 & G13 would have to be, blocking the 3x3 next to it. So E12,J12,H12,M12 are blank. So for rows 5&7, all stars must be in the 3x3 boxes, same for 11&13.
Segment O4 is now divided into 3 2x1 boxes, each of which must have a star. So P9&Q9 are blank. So Segment O8 has all 3 stars in col O, rest of the col is blank so P4 & P6 are stars. O9 is empty, as is M9, else there’s not enough space in segment I2.
Consider segment I14. I16 & G17 are empty, so F17 & H17 are stars. Looking at segment E14, starts must be E15,G15,D17, and I15 must then be a star too. Rows 15&17 are done.
Segment M14 must have starts in M14,L16,N16. Segment D7 has D8,D10 and (A or B)16.
Segment E11 has (F or G)11, E13,G13. Segment H11 has H11,J11 and (I or J)13 thus F11 is a star.
Col N has 1 star already, and segment I must have at least 2 more in that col hence rest of col is empty. Also, 3rd star in segment I2 is I2 or I3, thus J2 and J3 are empty.
Look at cols HJ. There are 4 stars already. Segment H5 has 3 more within those cols, and (I2 or I3) and (I13 or J13) are stars, so rest of those 3 cols are empty.
H1 is now a 3x3 box, so L2 is empty.
Segment I2 has exactly 2 stars in col N. If these are N8 & N10, segment O8 is blocked, hence N4 is a star. Then segment K5 has stars in K7,M7 and (K or L)5. So segment H5 has stars in H5,J5 and (H or I)7. Then L5 must be a star, and col H is filled, so H7 is empty and I7 is a star.
So one of N810 is a star. Thus O8 & O10 can’t both be stars, so O12 & O14 are. For segment Q9, stars must be in Q12, Q14 and (P or Q)10. Row 14 is done.
For segment H1, K2 & L1 must be empty, thus K1 is a star. Segment B1 must have at least 2 stars in row 1. It can have at most 2, thus it must have exactly 2. So 3rd star is F2 or F3, so E2,E3,G2,G3 are empty. So rest of row is empty, and rest of row 1 is empty. So M2 is a star, and then K3 must be too. M & K are done.
N9 can’t be a star, it would block row O. Row 9 must have stars in (A or B), (F or G) and L. So 8A,8B,10A,10B,8F,8G are blank. Rows 8 & 10 can then be finished off, and col L is done.
Since F2 or F3 is a star, rest of col is empty, so G9 is a star. Col G is done. 3rd star in J must be J13. D is done. For segment B1, stars must be in E1, and hence F3. E is done. Row 6 can be finished. B3 must be empty, else it blocks segment A1, so C3 is 3rd star in that segment. Thus segment A1 has stars in A2 & A4. So C1 is a star. A & C are done, so B can be filled in, and the final star must be in I2.
Hopefully that’s helped, let me know if any of it is not completely clear!
Re: [LMI] Puzzle Marathon  1st to 10th March
Thanks Neil. For you others, this 'marathon' puzzle only took Neil 11:52 minutes to finish, the fastest of the twelve.Nilz wrote: Label rows 117, cols are AQ, segments will be referred to by their top left cell.
There are 3 segments entirely within cols O,P,Q, so no other stars can be in those cols.
In the 3x3 areas, there must be at least 1 star on each of the top and bottom rows. This means row 4 cols HM, row 8 cols HM, row 10 cols EJ and row 14 cols EJ are empty.
H57, 1113 & 1517 must all contain at least one star. Hence they each contain exactly one, and the rest of the col is empty. G6,I6,G12,I12,G16,I16 are therefore empty.
Since for any 3x3 box, the outer edges must each contain at least one star, there cannot be any stars in cells adjacent to all 3. Eg J6 is empty, since it touches K57. Similarly, K6 is empty, N6, D12,K12,H12,L6,F12 too.
If E12 were a star, G11 & G13 would have to be, blocking the 3x3 next to it. So E12,J12,H12,M12 are blank. So for rows 5&7, all stars must be in the 3x3 boxes, same for 11&13.
An alternative way to get started: Two adjacent 3x3 squares mean you need 6 stars to fit into a 6x3 rectangle. This is only possible with 3 in the top row, none in the middle, 3 in the bottom. This can be proved by imagining 2x2 tiles covering the shape.
Re: [LMI] Puzzle Marathon  1st to 10th March
Yes, I remembered getting to the stage where none of the middle row could be stars, but couldn't quite remember exactly how I got there. Taking those 3 cells one at a time and seeing what would happen if a star was there, should hopefully convince any skeptics that that bit of logic does hold, and all of the middle row must be empty.kiwijam wrote:Thanks Neil. For you others, this 'marathon' puzzle only took Neil 11:52 minutes to finish, the fastest of the twelve.Nilz wrote: Label rows 117, cols are AQ, segments will be referred to by their top left cell.
There are 3 segments entirely within cols O,P,Q, so no other stars can be in those cols.
In the 3x3 areas, there must be at least 1 star on each of the top and bottom rows. This means row 4 cols HM, row 8 cols HM, row 10 cols EJ and row 14 cols EJ are empty.
H57, 1113 & 1517 must all contain at least one star. Hence they each contain exactly one, and the rest of the col is empty. G6,I6,G12,I12,G16,I16 are therefore empty.
Since for any 3x3 box, the outer edges must each contain at least one star, there cannot be any stars in cells adjacent to all 3. Eg J6 is empty, since it touches K57. Similarly, K6 is empty, N6, D12,K12,H12,L6,F12 too.
If E12 were a star, G11 & G13 would have to be, blocking the 3x3 next to it. So E12,J12,H12,M12 are blank. So for rows 5&7, all stars must be in the 3x3 boxes, same for 11&13.
An alternative way to get started: Two adjacent 3x3 squares mean you need 6 stars to fit into a 6x3 rectangle. This is only possible with 3 in the top row, none in the middle, 3 in the bottom. This can be proved by imagining 2x2 tiles covering the shape.