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#1: Aldege Cholette (aldege) on Jun 30, 2012 [SPOILER]

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nice pic, pal

Thx Tom buddy.:)

Nice.

After line logic ran out, I used 2-way logic in the lower right corner

Next time LL failed, I used 2-way/stair step logic in the lower left

The 3rd time LL failed, I used a combination of 2-way and visual logic. So I had to guess a bit around his ear and beard. :(

Thx Kristen.:)

After initial line logic, I used edge logic on the 5 clue in row 12 to make R12C16-R12C17 and R12C23 dots. More line logic.

Next I used edge logic on the 13 clue in column 7 to make R3C7 a dot. More line logic.

Next I used internal edge logic on the 8 clue in column 8 to make R17C8 a dot. More line logic.

I also made R30C27 a dot, but I didn't see where this helped. I'm stuck at this point. Any help?

Did you see him in concert, or does he live nearby?

R30C22 in that last step right gator?

I'm stuck at the same place, Gator. I keep coming back to it, but I'm not finding anything.

i got one black further than you guys but that was it

the bottom "6" clues in c14-15 (of which you already have 4 of each) cant both go up into r23 so at least one goes down into r29 (you can make r29c15 black)

it doesnt help much though

HELP!

i actually got further but had to leave computer for a while. still stuck though, but here are my steps beyond gator's:

whether C14R23 or C14R29 is black, C15R29 is black

since either C7R24 or C9R24 are black, the black in the row above is part of the "2", dotting C3-6R23

whether C19R14 or C19R17 is black, C20R14 or C20R17 will be black (respectively)...either way C20R14 is black.

whether C18R23 or C18R30 is black, C17R30 is black.

whether C12R28, C12R29 or C12R30 is black, it will form a black to it's left, so the black in C11R26 must be a 1, so dot C11R27.

group IEL(?): if the 2-2-4 in R28 stack left, the dots formed in R30 push the 3 in R30 to include C23R30 which violates the original premise...so C23R29 is black. (i didn't explain that very well)

we know that whether C10R16 or C11R16 are black it will force the 2 in R17 to one side or the other...either way, C11-12R18 are dots.

here are the rest of my steps (it is up to someone else whether any of my steps were enough look-ahead to be guessing):

there are three places (R14, R17 and R19) for the two 1's in C12 to go. no matter which two are black, C11R14 and C11R19 are dots. ll.

now, whether C12R17 or C12R19 is black (and therefore C9R17 or C9R19 is a dot), C9R17 is a dot.

the first two 2's in R30 can't pack into C3-7 because it would violate R29, so C11R30 is black.

the 3-2 at the bottom of C9 can't pack into R19-24 because it would violate C7, so C9R27 is black.

whether the unknown 2 in C10 is high or low, C9R19 is a dot.

ll to end

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in your "group IEL(?)" step i think you meant r29 instead of r28 (and that series of moves itself makes perfect sense) but even so i think that is clearly more than a simple 2-step look ahead

and then one part of your next step confused me though: i agree that either c10r16 or c11r16 must be black and it does indeed force the 2 clue in r17 to the opposite side (if c10r16 is black, then the 2 clue below it is in c11-12 OR if c11r16 is black, then the 2 clue below it is in c9-10)
all of this is fine
furthermore no matter which one is black (of c10r16 or c11r16) it is true that c11r18 is always a dot
that is fine too
if c10r16 is the black, then c12r18 is a dot as well, but based on what we have so far, if c11r16 is the one that is black, why does c12r18 have to be a dot as well (without looking several steps ahead)?

assuming that the above step somehow is logically acceptable, lets move on to your next step (the first step of comment 13): shouldnt it be the 3 places that the two "1" clues in c12 can go are r14, r17, and "r19-20" (not just r19)? at least i dont know how you have something already marked in c12r20 but the logic is still the same to make c11r14 and c11r19 both dots

the rest of your steps are fine but i am still thinking that the two steps of yours that i addressed in my first two paragraphs are more than a simple 2-move look ahead
but maybe not

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Ok, I followed Gator's hints, and then started David's. The group IEL step seems to be missing some explanation, but the conclusion is valid:
If R29 C15-17 is the 3, R30 C16-17 will also need to be part of the 3 in R30. There is exactly enough room for the remaining clues in R29, but if they are placed that way, there isn't enough room left for the first 2 in R30. Therefore R29C23 must be black.

To address bugaboo's confusion about David's next step: R18 only has a single 1 clue left. If the 2 in R17 is placed in C9-10, it triggers the 3 in C9, which accounts for the 1 in R18. That's why R18C12 can be safely marked white.

I'm going to diverge from David's hints here. This step covers the missing mark in R12C20 that bugaboo pointed out, but it also seems to make the 2 steps about the 1s in C12 irrelevant:

C28: If the 7 goes down to R28, C10 is made invalid, so R21C12 is black. LL.

The rest of David's guide is valid, starting here: "the first two 2's in R30 can't pack into C3-7 because it would violate R29, so C11R30 is black."

Found to be solvable with moderate lookahead by wombatilim.

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