Puzzle Description:

#1: Brian Bellis (mootpoint) on Nov 2, 2013 [HINT]

This was really HARD to start. When I encounter something I have to guess. I start at the far left and work until I find a conflict. I knew this was going to be a long time until I got to a conflict. Luckily, I never did.

It was too easy ... especially for me. At first sight I recognized NOT TO USE any line logic, but must to draw to four corners. This is a main step of my special method. And after that I could solve it less than 2-3 minutes :)

Special method=guess?

Nope. Not guess. We use in this method simple conjectural variates. BUT! It's very easy! Combined with "edge" logic, it can be seemed at first sight. (No need for any counting) :)
When you use "line logic", you have to guess. So: DON'T USE LL in this case.

This puzzle is a real pilot project for it. There must be dots in all corners! After that this is an 0* difficulty :)

This takes some tricky edge logic, but it can be done.

Look at column 1. The clues are 5,5. The clues in c2 are also 5,5.

This tells us something right away. the clues in c2 must be right next to the clues in c1 *unless* there are enough 1s in the row clues to "hide" one or both of the 5s from r2.

Now look at the row clues - there's only one 1 that can apply to c1 - column ten's.

No matter where we put one of the c1 5s, the other one cannot cross the 1 or it would make a total of three blocks in c2 where there's only two clues.


OK, now let's start trying to place the top 5. Remember its range is only r1-14 because the bottom 5 must fit, too.

So assuming the bottom 5 is all the way down, will the top 5 fit in c10-14? No- the 1 in r10 kills that as it would force the 5 in c2 to run into the other 5 in c2.

It also cannot end in rows 11, 12, or 13 because that would make too many blocks in c2 (because of crossing the 1)

So now our range for the top 5 is shortened to r1-10.

However, deep edge logic - comparing it to c3 - means that we cannot go below the 2 in r6 or there would be three blocks in c3. (One due to the bottom 5 and two due to the top 5)

So the range of the top 5 in c1 is r1-6 and c1r2-5 can be marked black.

Then Line logic.


Now we can do edge logic to try places for the bottom 5.
Its current range is r7-12 & r15-20.

Anywhere we put it in r7-12 will cross the 1 in r10 and create two new blocks in r2 (where there's only one clue available), so c1r7-12 is white.

And the rest solves with line logic.


Hard to start, but moderate lookahead.

Found to be solvable with moderate lookahead by infrapinklizzard.

Definitely moderate lookahead. I won't detail a method, as that's been done, but once you get two of the corners filled in, it works by line logic.

it's always funny to me when people make an assumption and, because it worked out, think it's logic and not guessing. yes...we can sort of "see" what will probably happen, but that doesn't mean that it MUST happen, which makes it "guessing". i can make a puzzle VERY very similar to this one in clue formation that does not have black squares in the corners. When you've done a bunch of these and can think ahead a bit you do start to see the most likely formations, but if you can't step through the squares and clues with a logical explanation, it's just the educated guessing of an intelligent mind.

fun, difficult puzzle!

I tried to figure where the left of the 2 5's started in row 20. If it starts from column 8/9 then C17, 10, 12 are black and 11 is a dot, a contradiction. If 6 or 7, you get a contradiction in column 16. But 9 is the farthest right it can start without crowding the other 5.

So you have a black square at R20C5.

Then line logic works. It's neat to see it unfold--I thought I might have to use more edge logic.

Goto next topic

You must register and log in to be able to participate in this discussion.