Japanese crossword «Roots»
| Size: 45x35 | Picture: | Difficulty: | Added: | 03.12.13 | Author: Irina-belko61 |
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show: 18 🗨
If this is not difficult enough please tell me how to continue. I´m totally stuck here.
https://imgbox.com/hcnqbdaR
replyhttps://imgbox.com/hcnqbdaR
show: 17 🗨
<spoiler>
Hint: Look at right up side. If u start 5 from begining, the left 4 is gonna block there ;)
</spoiler>
replyHint: Look at right up side. If u start 5 from begining, the left 4 is gonna block there ;)
</spoiler>
show: 16 🗨
There's neither rhyme nor reason to your hint. It's gibberish.
replyshow: 14 🗨
show: 13 🗨
Your genius won't help me any further. How was this helpful?
The hint still makes no sense to me at all. So please explain.
replyThe hint still makes no sense to me at all. So please explain.
show: 12 🗨
Well, this reply is a bit late, but maybe you or someone else is still interested in the solution of this puzzle. :D
You probably have to use edge logic in order to solve the puzzle.
Putting the 5 in the upper right corner for example results in this
( https://imgur.com/a/hvAlaNx ) contradiction and therefore you have to put an X in C45/R1. You can cross out some more squares in this column using edge logic and-more importantly- you can also use edge logic in row 2,5 and 6 to find even more Xs in the upper right corner. ( https://imgur.com/a/hDcXoU8 )
I hope this helps. :)
replyYou probably have to use edge logic in order to solve the puzzle.
Putting the 5 in the upper right corner for example results in this
( https://imgur.com/a/hvAlaNx ) contradiction and therefore you have to put an X in C45/R1. You can cross out some more squares in this column using edge logic and-more importantly- you can also use edge logic in row 2,5 and 6 to find even more Xs in the upper right corner. ( https://imgur.com/a/hDcXoU8 )
I hope this helps. :)
show: 11 🗨
I wasn't aware of edge logic. Thank you for clearing that up. It wasn't gibberish after all for which statement I apologize.
However, it is solvable without using edge logic. I found the 'traditional' solution in row 20 which was really hard to deduce. I've never had to use edge logic before and as it turns out, not even now.
I have doubts edge logic is a legit way to solve nonograms.
replyHowever, it is solvable without using edge logic. I found the 'traditional' solution in row 20 which was really hard to deduce. I've never had to use edge logic before and as it turns out, not even now.
I have doubts edge logic is a legit way to solve nonograms.
show: 10 🗨
Oh nice, I like your solution! It took me some time to see the two squares that have to be filled in in this row, really clever! :)
Well, tbh I don't mind using edge logic whenever I'm stuck on a puzzle. It's a logical way after all to cross out some squares and I think edge logic is even required to solve some of the puzzles in Mario's Picross 2 for the Gameboy and Mario's Super Picross for the SNES.
But yeah, apparently edge logic isn't required to solve the puzzles on this site or the ones from the newer Jupiter games for Nintendo Switch, so I also understand why some people refuse to use edge logic. :)
replyWell, tbh I don't mind using edge logic whenever I'm stuck on a puzzle. It's a logical way after all to cross out some squares and I think edge logic is even required to solve some of the puzzles in Mario's Picross 2 for the Gameboy and Mario's Super Picross for the SNES.
But yeah, apparently edge logic isn't required to solve the puzzles on this site or the ones from the newer Jupiter games for Nintendo Switch, so I also understand why some people refuse to use edge logic. :)
It definitely is legit, even if you don't like using it. I used it in this puzzle and I use it in many puzzles if I get stuck
replyshow: 2 🗨
Specifically, I used it to get out of being stuck in the same spot you were :)
replyThe goal of solving a difficult puzzle is not taking an alternate route to solving it the easy way. You may have solved it, but you skipped the hard part.
Of course, everyone is free to solve a puzzle their own way. There are no strict rules, but using edge logic is not the way as was originally intended. This is evident by the fact that if you take a smart hint, it will never point to edge logic.
Edge logic isn't guesswork, okay. But if you use it, you cheated. That's how I see it.
replyOf course, everyone is free to solve a puzzle their own way. There are no strict rules, but using edge logic is not the way as was originally intended. This is evident by the fact that if you take a smart hint, it will never point to edge logic.
Edge logic isn't guesswork, okay. But if you use it, you cheated. That's how I see it.
I found the 'traditional' solution in row 20could you please describe this solution?
show: 5 🗨
The solution is not only hard to understand, but also difficult to explain.
I will give it a go though. The solution lies in the last four digits of row 20.
For readability let's call r20c25 = A, r20c27 = B, r20c29 = C, r20c31/c32 = D
1) Cell A must be part of the 11. Duh, obviously :)
2) If D is part of the 5, it's impossible for the 11 to end up to the right next of cell A, for the next 4 cells (i.e. up to and including cell C).
3) If D is NOT part of the 5, then D becomes part of the 11, as a result.
Thus, all between A and D is blacked out (so, that includes cells B and C).
4) If the 11 is fully filled in to the left side of A, then D must be part of the 5. Consequence: B and C have to be filled in (i.e. 11-1-1-5 filled in contiguously).
Conclusion: Considering previous arguments, B and C must be completed.
It's complicated, I know. It took me considerable time to get my head around this incredibly hard problem. But it was fun working on it nonetheless, and it still is.
Hopefully this will help.
replyI will give it a go though. The solution lies in the last four digits of row 20.
For readability let's call r20c25 = A, r20c27 = B, r20c29 = C, r20c31/c32 = D
1) Cell A must be part of the 11. Duh, obviously :)
2) If D is part of the 5, it's impossible for the 11 to end up to the right next of cell A, for the next 4 cells (i.e. up to and including cell C).
3) If D is NOT part of the 5, then D becomes part of the 11, as a result.
Thus, all between A and D is blacked out (so, that includes cells B and C).
4) If the 11 is fully filled in to the left side of A, then D must be part of the 5. Consequence: B and C have to be filled in (i.e. 11-1-1-5 filled in contiguously).
Conclusion: Considering previous arguments, B and C must be completed.
It's complicated, I know. It took me considerable time to get my head around this incredibly hard problem. But it was fun working on it nonetheless, and it still is.
Hopefully this will help.
show: 4 🗨
Thank you for your explanation. But I don't understand why you claim that this is a "traditional" way that was "intended" and the edge logic is cheating.
replyshow: 3 🗨
I don't claim anything. It's just how I see it, i.e.my opinion, which I clearly stated. Furthermore, I can't think of anything to make it more clear than I already did, except that anyone who claims this 4.5-star puzzle is easy, used edge logic to solve it.
replyshow: 2 🗨
Very difficult, it took me several hours! Wow! Thanks for the challenge, I enjoyed it.
replyExcellent puzzle! Very challenging to get started but once I got a foothold, smooth sailing, and lovely picture!
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