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This is an entry for Fortnightly Topic Challenge #44: Introduce a new grid deduction genre to the community.


Here is a standard Cross the Streams puzzle. The genre is invented by Grant Fikes which combines Nonogram and wildcard clues.

Rules of Cross the Streams:

  1. Shade some empty cells black to create a single group of black cells that are all connected to each other through their edges. No 2x2 cell area within the grid contains all black cells.
  2. Numbers to the left/top of the grid represent the groups of consecutive black cells which are in that row/column in order, either from left to right or from top to bottom. (For example, a clue of "3" means the row or column has three consecutive black cells, and a clue of "3 1" means that the row or column has a group of three consecutive black cells followed by a single black cell, separated by at least one white cell.)
  3. A question mark (?) represents a group of consecutive black cells whose size is unknown; an asterisk (*) represents any number of unknown groups of black cells, including none at all.

enter image description here


enter image description here

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    $\begingroup$ Is your mini-example solvable as it stands? The 4th column appears to be missing a value...? $\endgroup$ – Stiv Dec 7 '20 at 9:46
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    $\begingroup$ @Stiv oops you're right, I made a wrong deduction, I'll fix but it's the 6th column: it should be "2?1". The 4th column is still "*". The asterisk basically can represent anything. $\endgroup$ – athin Dec 7 '20 at 10:08
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    $\begingroup$ Ah, I see - well I was accidentally helpful then! Hadn't grasped that an asterisk could represent "any number of unknown groups of black cells"... Thanks for clarifying :) $\endgroup$ – Stiv Dec 7 '20 at 10:19
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    $\begingroup$ And ah, someone downvoted this haha, not a big deal but I'm actually curious of the reason >< $\endgroup$ – athin Dec 8 '20 at 1:06
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    $\begingroup$ Nice new puzzle athin! $\endgroup$ – justhalf Dec 8 '20 at 8:19
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The complete grid:

enter image description here

Reasoning:

In row 9 we can fill in two blocks of two just by simple counting, since the row must be at least "3 3 1". In the upper right corner, if we assume R2C9 is shaded, then this forces all of R2C8-9 and R3C8-9 to be shaded, contradicting the no 2x2 rule. So R2C9 is unshaded, forcing the squares above and to its right to also be unshaded, and then counting forces R2C6-7 to be shaded. The grid thus far:

enter image description here

The quicker-picker-upper (added later):

I originally had a longer contradiction argument to exclude the possibility that R2C8 is unshaded, but that's because I forgot the connectivity rule at first, and so did not immediately exclude the possibility that R1C10 could be shaded. With that correct deduction, simple counting shows that R6-7C10 need to be shaded for the 3-block in column 10, which forces R4-5C9 to be shaded for the 3-block in column 9, which forces R2-R3C8 to be shaded for the 3-block in column 8. This leads into the rest of the solution reasonably well, since I focused on the left side next, then came back to the right.

The original Long-Developing Contradiction:

By way of contradiction, assume R2C8 is not shaded. Thus gives us the 3 block in row 2 and column 8. Exactly one of R3C5 or R3C6 must be unshaded; were both unshaded, the two 3-blocks in these columns would have to be side-by-side, creating multiple 2x2 shaded blocks. If R3C5 is unshaded, then R4-6C5 and R8-10C5 must be the 3-blocks in C5, which only leaves room for one 3-block in C6. So R3C5 must be shaded and R3C6 unshaded. This forces the location of the 3-blocks in C6, which leaves only one location for the bottom 3-block in C5. Some additional simple deductions leave us with:

enter image description here

Focus now on C9 and C10. The 3-block in C9 must contain R6-7C9, which forces R3-4C9 to be unshaded. But then R4C10 cannot be shaded, since that would force all of R3-4C4-5 to be shaded. Thus the 3-block in C10 must also contain R6-7C10, a final contradiction.

Moving forward:

All of that simply shows that R2C8 must be shaded, but this shows that R3C8 is shaded, and that R2C5 is unshaded, which forces the two 3-blocks below it, of which we can place 2 blocks of each. But one of these forces R8C6 to be unshaded, which forces the 3-blocks in C6. These placements also force the positions of the 3-blocks in R9. The grid thus far:

enter image description here

In row 3, the 3-block cannot start before column 3, due to the ? before the 3, so it must be C4-6. In row 4, we need two blocks right of the 3-block, so the 3-block must be in C1-5, forcing R4C3 to be shaded. This forces R1C3 to be unshaded, since the initial 3-block in C3 must contain R4C3. Similar logic in R6 shows that R6C2-3 are both shaded. Together, these force the 3-block in column 3, which then forces R2C4 to be shaded. In column 4, R5C4 must be unshaded, since it would create a 4-block, leaving no room for a 3- and a smaller-block to the right. This indeed forces the 3-block in row 5 to be C5-7. Also in column 7, the 3-block must go between rows 7-10, forcing R8C7 to be shaded. The grid thus far:

enter image description here

Finishing up the left side:

In row 4, the 3-block must be in the first 3 columns, which forces R1C1 to be unshaded. In addition, the 3-block in the second column has to be R2-4. The only other place it could be is R8-10, but if those blocks are all shaded, then connectivity forces R7C2 to be shaded as well. This then forces the 3-block in column 1 to be R4-6. This then forces R6C4 to be unshaded, since there is nowhere else for the 3-block in R6 to go. Connectivity forces additional squares in column 2 in R7-8. After ensuring we get no shaded 2x2, connectivity again forces us to bridge across column 4 in row 10, from C3-C5. Finally, R10C1 must be shaded in order to get four distinct groups in R10. The grid thus far:

enter image description here

Finishing up:

The 3-block in columns 4 and 5 are now forced, as is the 3-block in row 8. The latter forces column 10's 3-block to lie between R3 and R7, so R5C10 is definitely shaded. There are thus only two places C9's 3-block can go: either R3-5 or R6-8. But note: R9C9 cannot be unshaded! If it were, then the shaded blocks in R10C7-10, of which there must be at least 2, must be connect out through R10C7 in a single block, but there must be at least two blocks there. So the 3-block in C9 must be R3-R5. The same connectivity and two blocks consideration in the lower right corner force R10C7 to be shaded: otherwise all shaded blocks would have to escape through column 9. Connectivity forces R7C9 to be shaded. The rest falls out with simple deduction.

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  • $\begingroup$ Oh well, the "long-developing contradiction" isn't quite the intended. Actually, what you left at the end (the last 3 columns) are the key for the breaking-in to avoid that long contradiction. Try to find how can you place the threes on the last 3 columns after you unshade the upper-right. $\endgroup$ – athin Dec 8 '20 at 1:03
  • $\begingroup$ @athin I forgot about the connectivity rule until near the end, so I didn't exclude the upper right corner until way too late. Let me have another quick look. $\endgroup$ – Jeremy Dover Dec 8 '20 at 3:24
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    $\begingroup$ @athin Fixed up the answer a little bit. Wish I'd seen that sooner...I'd have an hour of my life back :-) Great puzzle by the way...thoroughly enjoyed it! $\endgroup$ – Jeremy Dover Dec 8 '20 at 3:33
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    $\begingroup$ Thanks a lot for the fix and the compliment! You did a great job, very well done! :) $\endgroup$ – athin Dec 8 '20 at 4:00

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