Puzzle Description:

"Isn't it romantic sitting here by that old flag, your arm around me, looking at the starry sky, with the full earth up there?"

#1: Aldege Cholette (aldege) on Nov 4, 2011

A truly wonderful puzzle larry,her head on his shoulder is fantastic,one of the best puzzles i've ever seen,you've commented on my B&W's,well ditto to you larry,this is fantastic,thx.:)

After the initial line logic, I did edge logic on row 33 to make R33C12 and R33C28 dots. More line logic.

Next I did edge logic on column 29 to make R23C29-R25C29 dots. More line logic.

Next I did edge logic on column 27 to make R33C27 a dot, then R29C27 a dot. More line logic.

Staying on column 27, the 3 clue can either go in R24C27-R26C27 or R30C27-R32C27 (two-way logic). This will make either R25C26 or R31C26 black. This lets us dot most of the rest of column 26. More line logic.

Next I did two-way logic on R30C20. To complete the 2 clue in column 20, either R29C20 or R31C20 has to be black. This makes either R29C21 or R31C21 black. But if R31C21 is black, then R29C21 has to be black. So in either case R29C21 is black. More line logic.

And at this point I am stuck. Anyone have another move?

Is this a WCP entry? You should change the title!

what's funny is that i did the same logic in the same order and am stuck in the same place...still trying, but my lunch break is over right now...will hit again later.

here's a big lookahead move that breaks it loose enough to LL to the end:

Imagine the 5 (R25) being in the left gap. the dots down C16 will lock the 9 in R30 to the right, moving the 4s in C13 and C15 up one bay violating R26, so the 5 (R25) goes right.

separate from that i found a couple of dots that didn't help me much, but you can add them if you didn't have them already:

- two way the 1 in R13 to dot C15R14

- three way the 1 in R14 to dot C16R15

i'm going to look a little more for break-through move that doesn't take as much lookahead...

i was going to say ditto the logic so far but i guess i have to say a double ditto since dbouldin already did it

(i had the two-way logic dot from the 1 clue in r13 but not the three-way logic dot from the 1 clue in r14 - that was a nice find by the way even though it didnt help)

i can not see anything else "logical" without looking too many moves ahead

Pretty cool image, Larry. Too bad it requires guessing.

I love this image Larrry, funny and romantic, sweet! :)

So I've been spending some time with the 4's in the middle of C13,14,15 and how they can be arranged in the three vertical gaps. Here are the basic ground rules:

Top: No more than one of the 4s (R19)

Middle: No more than two of the 4s and if there are two, then they are in adjacent columns (R26) AND if the center is in the middle, then at least one of the other two will be at the bottom because the 1's under it would trigger something (R30 & R31).

Bottom: The middle one can't be there because it has 1's under it AND if the left one is there, so is the right one (R33)

Using those, here are the only possible combinations:
Left-Top......Center-Middle...Right-Bottom
Left-Middle...Center-Top......Right-Bottom
Left-Middle...Center-Middle...Right-Bottom
Left-Bottom...Center-Top......Right-Bottom
Left-Bottom...Center-Middle...Right-Bottom

So, the right one (C15) HAS to be at the bottom.

There is probably a more straight-forward way of wording this without the chart and such, but my brain is too fried to figure it out right now. Also, I don't know how this is looked at from a "solvability" stand-point, but as luck would have it I don't really care ;)

takes some time but work through options for the 9 in row 11 and the 10 in row 8 - then its logical - no guessing required in the end.

Found to be logically solvable by gator.

How about this? R30C15-R33C15 will either be all black or all dots. Let's assume they are all dots. Looking one move ahead (just concentrating on the cells that will show a contradiction), R30C23, R30C24, R33C23, and R33C24 will be black. Looking a second move ahead, R31C23 would be black and R31C24 would be a dot. But as R31C22 is already a dot and there cannot be another 1 clue in row 31 (there are only 2 clues left), then row 31 would be invalid. SO R30C15-R33C15 have to be black.

The rest solved with line logic.

i'll take your word for it...my head starts hurting just looking at the solution ;)

Could not see it at first. Then I leaned back and this wonderful image appeared. Tough solve and I did have to guess twice.

High praise from me. Awesome puzzle. Thanks

I loved it all the way to the moon and back!

Goto next topic

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