Puzzle Description:

whacker.

#1: Megan Kennedy (MCGal) on Aug 3, 2012

How are you supposed to solve with clues blotted out

Made a lucky guess and the rest fell into place.

I had to guess as to whether I needed to add to the blots or not. After I guessed, then it all worked out.

You don't guess, you count. The total black row and column clues must be equal. Try again.

got it with no guessing. nice use of the blots that i haven't seen before!

no guessing,good one brian :)

it's do-able...very little blots too!

I solved this puzzle many times with the blots in different places to make it non-trivial and in the process figured out some blot logic.

I saved the blots for last. Instead, I looked for a place to put the 1s and 3s, because the clues indicated a diagonal with a bend near the top.

Found to be solvable with moderate lookahead by gator.

Great use of combining the blots with summing logic. :)

I'm way out of practice on solving, but I was able to do this with a bit of summing. The bottom four rows were pretty easy to solve, and that took care of all the green blots. So we just had the two black blots left. Adding the black numbers that hadn't been solved yet, got me 34 for the row clues and 30 for the column clues. That meant that there had to be exactly four more cells covered by the two blots. Knowing that there had to be at least one more black cell covered by the blotted blocks told me that at least one of R11C15 or R11C16 had to be black, which fact let me solve row 10 and accounted for two of the four missing blacks. By the same reasoning, we know at least one of R10C15 or R10C16 must be black, so R10C14 and R10C15 must be black. We now have only one more black cell to place in the blotted columns and only two places it can be: either in R9C15 or R10C16. One of these must be black, the other must be white, or the sums won't work out right. The case of R9C15 being black and R10C16 being white leads very quickly to a contradiction, so it must be the other way around. The rest of the puzzle solves very easily.

Yes, I suspect that this is going to be a standard logic trick for blotted puzzles. The trivial case is when only one clue number is blotted, in which case you can easily calculate it's value from all the other clue numbers (webpbn will actually discourage people from posting puzzles with blots as stupid as that.) But whenever the blots are just in the rows, or just in the columns, you always calculate how many cells must be in them all together, and do some reasoning based on that. If there are blots in both the row and column clues, then it'll be trickier ... but not completely useless. You could figure out that there is exactly one more cell among all the column blots than there are among the row blots...which would prove that at least one of the column blots is at least a 2, a fact that might well prove useful.

So, if there weren't easier ways to solve the green part of this puzzle, you add up the green row clues (9) and the green column clues (23) and know that the blot in row 15 must be at least a 14, and that fact alone would let you mark the middle 8 columns of that row as green.

So I think summing, which is rarely useful on regular puzzles, might be quite an important technique on blotted puzzles.

Anyway, a very thought provoking puzzle. Thanks Brian.

I think the image is great.

Jan. That is exactly how I did the black as well.
You are welcome.

Thank you Tara.

Excellent solve! Loved it! Thanks Brain. It could be an entry!

Thank you. I had not even looked at this week's theme.

While I think it would be a good contender, I'm not sure it is eligible. It was created prior to the end of the last WCP. Can we just rename old puzzles and them be in contention?

Me pointing this out is a bit selfish, because I would really like to see another Brian blot-puzzle!

I'll keep them coming. I like the new twist.

This was actually the first time I've used summing to solve a puzzle, so it's a good thing Jan gave a little walkthrough there. Still, now I've been forced to use it once, I'll hopefully remember it in future.

There is an interesting way to solve obscured clues. If you count the black digits on top (NOT the squares), they add up to 36. On the side, they total 46. 36 cells based on the top will be distributed on the left numbers as well; this means 10 cells in the square clues MUST be true!

Knowing this, I had nine of the cells allotted and knew that only one of two cells above could be true, making the rest of the columns false. I continue solving from R10 (the 3) but didn't get very far (only filling a pair of cells and knocking out the rest of the rows/columns).

Here is where a little look-ahead is required. One square currently has five cells, the other four. If you fill the sixth (R9C15), you effectively force the fifth cell in the other (R10C16) to be true, and there is no solution. Therefore, R9C15 is false and the squares are completed. The puzzle easily solves from here.

Easy except I had to guess to get started in the upper left corner.

I thought I was clever figuring out summing the blots so the horizontal black numbers equalled the vertical items, but it looks like a bunch of people beat me by several years!

It is interesting to see this sort of different logic. Jan's comment on how to get the greens also usefully summarizes some stuff I had bouncing in my head but couldn't quite describe.

So educated guesses knock this out quickly, but you may miss some neat logical details.

I second the other Andrew's comment above.

I let this puzzle simmer in my partially-solved bank for about... 7 years. Came back to it a month ago and thought for a while. Finally looked at it from a new angle today and solved the same way.

I felt pretty bright. Glad everyone else discussed it 8 years ago.

Wait, are you two different people? I did not know that! Come to think of it, you do have different spellings of your last names. I honestly have no idea whether you are the same person or different people. How did I miss that?!

Goto next topic

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