Puzzle Description:

Antique Key

#1: Gator (Gator) on Jul 14, 2009 [HINT]

This is NOT a beginner puzzle. Attempting to solve this puzzle may cause head pain and/or brain damage. By trying to solve this puzzle, you agree that the designer will not be held liable.

This puzzle requires multi-step contradictions to complete. You will need to focus on where the 7 clues can and cannot go in conjunction with the 4 clues along the bottom and right.

Good luck!

Definitely a challenge! I used edge logic on the right side to help get started. After that, standard LBL techniques worked. No guessing required.

I used edge logic on the 4's in the far R column and the bottom row. I got down to 8 available spaces for each of those 4's (I eliminated the first 7 spaces of R14, and the first 6 plus the last square of C15) which didn't help that much since I then seemed to get stuck. I was able to do trial and error, and luckily my first guess solved it. Suggestions on what do do at this point? No other edge logic worked.

Consider the 4 in the bottom (non-blank) row. As Adam says, edge logic allows it to be in any of the last eight columns, which doesn't help very much. However, notice that wherever it is among those eight columns, it has to cross at least one number greater that is 4 or greater. (The six in column 9, the 7 in column 12, and the 4's in columns 13 and 15). That means that the 7 in row 11 must cross at least one of those columns. That means it cannot start in column 1 or two, so you can place dots in R11C1 and R11C2. Similar logic along the right edge let you put a dot in R1C12.

Once you have that, you can set a bunch more cells in row 8 and column 9.

And then I'm stuck.

Or maybe I'm not.

Consider again the seven in row 11. Suppose the cell in column 4 (with the clue 4) is black. Then the cell R10C4 must also be black.

If those two cells are black, the both the cells R10C20 and R11C20 must be white. And this cannot be because column 20 would be unsolvable.. Ergo, cell R11C4 must be white.

Similar things can be done with the other 7.

Still not solved though.

From whence one was after my last hint, you can do some line solving and wind up here:

Best thing I could think of here was to consider what happens if we put a dot at the junction of the two sevens, at R11C12. We consider the effect of that on row 19. It's easy to see that we get two dots in that row, at column 13 and column 12. And this is an unbearable thing, as those two dots eliminate any possibility of placing the 4 in row 19. So R11C12 cannot be a dot and must be black.

Line solving does the rest.

So I guess this is logically solvable. All of those things were hard to find, but not too hard to see once you found them. But that's pretty danged hard going.

Found to be logically solvable by jan.

Assuming you filled in R11C10, R11C11, R9C12, and R10C12 after blanking out R11C4 and R3C12, look at row 10 (and likewise column 11). Either R10C12 is a part of the 4 clue or the 1 clue. Regardless, R10C13 has to be blank (likewise R12C12). Also more blanks can be filled in row 10 (columns 1 - 5) and column 11 (rows 1 - 4). More blanks can be filled in after that.

Next consider column 12. The 4 clue can either go in rows 11 - 14 or rows 6 - 7 (rows 5 - 9 actually, but 6-7 would always be filled in). If it is placed in 6-7, R7C15 would be blank, but R11C15 would have to be filled in which will cause a contradiction for row 11 (R11C15 causes R11C13 to be filled in). So the 4 clue must go in rows 11-14.

It should play out normally from there.

There may be a different way to get to a solution, but that is how I got there. :)

EDIT: That confirms there are different ways to the solution. :) It seems we were detailing it out at the same time.

Gator evidentally didn't see my responses #6 & #7 before posting #8. So #8 partly describes how to get to the picture in #6 and partly tells how to continue after that.

#4, #5, and #6 - the hints seem to indicate a larger picture than it is...

I hate you Gator! I had to look at Jan's hints. I read through #5 to get me started and then found I had a bunch more than Jan did when I looked at his #6.

Another Zelda reference! :) Still this is one of those puzzles that I really couldn't care less what the actual picture is. Just so long as I solved it I'm happy. :P

I have to make another comment.

I snarled and growled as soon as I saw the clues. The numbers were way to small to fit in the rows and columns. It's like my arms may as well be too short to reach the keyboard! The feeling of defeat crossed me instantly with this one. It didn't even lure me in to 51% solved before I knew I couldn't solve it on my own. Grrrr.....

Yeah, I tried to give a warning on this in the first comment. The clues are the first indication that the puzzle will be a struggle. It's ok if you hate me - I can take it. :(

sometimes "Edge Logic" is too much like guessing...like in this puzzle...when you have to use it in several directions at once making for several tries before finding the correct combination
it helped that i had a pretty good idea what the image might be from the title

I used a little edge logic, and some trial and error on the bottom - solved it in about 2 minutes. Has it been revised?

Jan, I resolved this using your hints. And of course it worked. But my question is this: Isn't that sort of "trial and error" logic? When I solved this the first time, I had to make an assumption that a black or a dot went in some given space, to see what would happen and prove that it could or couldn't go there. Is there any way to "logically" know that a certain cell can't have a black or a dot in it?

I still am doubting the true logical solvability of this one.

I don't /really/ hate you Gator. Just in a small way because I don't like to be beat by a puzzle. :P

I did what Robyn did, started with edge logic for the 4 on the right. Two of those pixels would have to be in Rows 11 and 12. From there, it fills in nicely, with some thinking ahead in a few places.

This was a fun challenge and very nice image. I don't hate you, Gator. On the contrary, I'm getting addicted to your puzzles. Keep 'em coming! :-)

I also started with edge logic on the 4 at the bottom and the right. Like Adam, I came up with only 8 spots in the column and row where the 4 could fit. Jan seems to jump to a conclusion, from there, that the 7s must cross each other. I can't get to that conclusion.

For me, I started with the 4 in the last column and tried its fit. Fortunately, I hit the correct one on the second try. To me, that's trial and error (and I suspect to everyone else). I don't find that logically solveable but rather to require guessing.

But I'm probably missing something.

#4 starts the logic progression, although it should say "(The six in column 9, the 7 in column 12, and the 4's in columns 13 and 15)". This guarantees that R11C9, R11C12, R11C13, or R11C15 will be black. Since one of those has to be filled in, then it would be impossible for R11C1 and R11C2 to also be filled in. So R11C1 and R11C2 are dots. This allows R11C9 to be black. Because of the 6 clue in column 9, R9C9 and R10C9 are also black, and R1C9 through R5C9 are dots.

The same logic can be applied to column 12.

I'll continue this later as I am now out of time.

Oops, I fixed the wrong column numbers in my hint in #4, which gator pointed out.

I think these "proofs by contradiction" are short enough to be worked out in a person's head, so I think they don't count as "guessing".

I think edge logic only gets you this far:

The fours could be any place in the remaining 8 squares without contradicting the "1 2 1" clues or any other clue further inward.

Jan, in your drawing in #22, you have a dot in R7C15 that shouldn't be there.

Yup, I did. Fixed now. Can't do anything right this AM.

This puzzle is the perfect exemple of a small puzzle being able to be fun and tough to solve and in the end give you a great image

this one is hall of fame material as far as I'm concern

I agree with Sylvain!

Holy cow this was an awesome puzzle. Had to put it away and come back to it after awhile, but I solved it tonight without any hints and thoroughly enjoyed it. Another Gator masterpiece!

Thanks Matt!

Used trial-and-error. Got it wrong the 1st time; cleared the puzzle and got it right the 2nd time. Not that big of a deal with a 15x15 puzzle.
I've probably solved over 2,000 of these puzzles (between Games magazine, www.griddlers.net, and of course this website). This is definitely one of the tougher 15x15 puzzles out there.

wow
hundred percent agree with comment 25

how did I miss this jewel, thx Gator

Your welcome :)

I decided to make a more comprehensive hint for this puzzle since the above is in many posts.

Edge logic in row 14 will make R14C1-7 dots. Edge logic in column 15 will make R1C15-R6C15 dots.

Next consider that the 7 clue in row 11 is all the way to the left (or that R11C9-15 are dots). This makes it so that R14C9 and R14C12 are also dots making row 14 invalid. So the 7 clue will have to move more to the right so that row 14 is not invalid. From this we can dot R11C1-2. Now we get a little line logic.

Next do the same logic with the 7 clue in column 12. If R8C12-R14C12 are all dots, then R8C15 and R11C15 would be dots making column 15 invalid. So the 7 clue will have to move down. This lets us dot R1C12. A little more line logic.

Next look back at the 7 clue in row 11. If it extends to R11C4, then R10C4 would have to be black. This would cause both R10C15 and R11C15 to be dots making column 15 invalid. So R11C3 and R11C4 must be dots. More line logic. This gets us to the picture in post #6 above.

Look at the 4 clue in column 12 next. It can go on either the left or right side of the dot at R12C11. If it goes on the left, then R12C7-8 would be black. This causes R14C8 and R14C12 (2 move lookahead) to be dots making row 14 invalid. So the 4 clue must go on the right side. The rest solves with line logic.

I hope that makes it a bit more clear.

I feel like announcing I managed to solve this puzzle, haha. I think the last lookahead step I did was something more complicated than it needed to be, but I refuse to look at hints when I'm solving. It was...fun? I already had a headache when I started it, but this didn't make it any worse. :P

I feel like I just climbed Mt. Everest, but the puzzle was so small... Maybe I climbed a hill that was just unusually hard to ascend. Maybe it was super rocky terrain and slippery because it just rained. And it was cold out so it felt like the temperatures at a mountain peak...

An apt metaphor, I hope, for this tricky but conquerable puzzle.

This is one of the hardest 2 color 15x15 puzzles in the database. :) I'm glad you were able to conquer it!

nice image. I did have to guess twice.

After the initial edge logic, look at the four in R14. If it is in C13-C16, then R12C13 is a dot, R12C14 is black, therefore R12C17 is black. If R14C13 is a dot, then R14C17 iis black, and again, R12C17 is black. Once you have that, the rest follows.

Goto next topic

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