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


Greener Grasses is apparently a new genre of grid puzzle. Pitched by Joseph Howard and executed by a friend of mine, Ivan Koswara a.k.a chaotic_iak. This was first appeared in 24-Hour Puzzle Championship 2018 and lately revived in Ivan's 25 Years in Logic Masters India last month (November 2020).

When I first trying to solve this genre, it took me a long time until I grasped the idea and then hooked to explore more of this genre. Stated by Ivan in the 25 Years, the number of published Greener Grasses puzzles apparently can be counted on one hand. So here I am, trying to pick this one up as I do believe the genre has a great potential~

Also, as this is the entree on the genre here, I write this in a "classic" way (with... a supposedly-not-too-hard difficulty...) I personally recommend you to check the puzzles in the 25 Years to see how neat this genre can be! :)

Nb. Uhh.. Look at me getting invested in it by writing a cute puzzle as the example... >w<
Nb2. Ok, I asked Ivan to playtest this... And it looks like this is harder than I expected... Oh well... ><


Rules of Greener Grasses:

  1. Divide the white cells into regions of the indicated size (top-right of the grid.)
  2. Whenever two regions share a side, each region has something the other does not: a symbol that the other region has less of. (For example, a region containing AA and AB may share a side: the AA region has more A's than the other, and the AB region has more B's than the other. But AB and B may not share a side; the B region doesn't have anything the AB region doesn't have.)

enter image description here


enter image description here

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  • $\begingroup$ In the example provided, the NE and ES pieces seem to break the rules (they have the same amount of E's). What did I miss here? :0 $\endgroup$ – oAlt Dec 5 '20 at 11:23
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    $\begingroup$ @oAlt NE has something ES doesn't have, which is N. ES has something NE doesn't have, which is S. $\endgroup$ – athin Dec 5 '20 at 11:25
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I'm using the penpa editor so the trees are gone unfortunately...

Step 1:

Starting at the bottom left, we can make some observations. If we were to form a 3-cell group with D in R10C2 using R10C1 and R10C3 (solo D), then the C in R9C1 would be forced to join up with either the A in R7C1 or the A in R8C1. In either case, it would be forced to share an edge with a 3-cell group containing "AC" or a single 'A'. So, this runs into a contradiction.

If we were to form a 3-cell group with D without using the cell R10C1, then a "DC" group would be forced to align with a single "C" group, which is again a contradiction. Therefore, the D in R10C2 must join up with the C in R9C1. Avoiding contradictions, we get here:

GG_1

Step 2:

Then, notice the two "A"s in the lower left must join up together. Otherwise, we get groups of 3 sharing an edge where both groups have the same amount of As. For similar reasons, the cell R8C3 must include both C and B.

GG_2

Step 3:

The cell R10C5 must include 'B'. Otherwise, it would be forced to group up with a 'C' and it already shares an edge with 2 Cs. After joining, the B cannot extend upwards and neither can it join with the C next to it (Otherwise, the C in R10C7 would share an edge with it). Therefore, the last cell of that group must be R10C6.

After that, the cell R10C8 cannot join up with the D in R9C8. Otherwise, the cell in R8C8 would share an edge with same letters. Therefore, it must join up with A in R10C9.

With this, we can finally make some knock-on deductions and solve the bottom part of the grid.

GG_3

Step 4:

Next, we look at the mid-left. The Bs in R4C1 and R5C2 cannot join together (otherwise, the C in R6C2 runs into a contradiction. If either of the Bs were to include R5C1 as part of their group, we run into a contradiction. Therefore, that cell must join up with C. This forces the B in R4C1 to join up with the B in R3C2.

GG_4

Now, the B in R5C2 cannot join with D (the A runs into a contradiction) and cannot join with cell R6C3. Therefore, it must go up. The corresponding chain deductions lead to:

GG_5

Step 5:

Now, the cell in R6C5 cannot go up. It would either be forced to join with C (a contradiction) or be an empty group. Therefore, it has to join with the B in R6C7. This means the cell R5C5 must be connected to the C in R4C5

After that, the cell R5C6 cannot be a part of the C group (otherwise the A in cell R4C6 would be forced to join up with either R3C5 or the C in R3C7. Since the A cannot be a part of the C (the D in R5C7 runs into a contradiction), the C must join up with R3C5.

GG_6

Step 6:

Next, if the C in R7C8 was to join up with R5C8, then the B in R5C9 would be forced to join up with the C in R4C10 and it would share an edge with a group with single 'C'. Therefore, C in R7C8 must join up with the D in R6C9.

GG_7

Following that, if the B in R5C9 were to join up with D in R5C7, then we run into a contradiction because there would be two groups with "CD" in them sharing an edge, like so:

GG_8_wrong Therefore, the B must instead join up with the D in R4C8. From there, the D in R5C7 is forced to link up with the A in R4C6. The cell in R3C6 must belong to the "DB" group as the C in R3C7 cannot touch it. From the knock-on deductions, we can finish the grid:

enter image description here

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    $\begingroup$ actually, do you know that I use penpa-edit too to put the trees? :) anyway this is a great-solving, well done! $\endgroup$ – athin Dec 6 '20 at 3:45
  • $\begingroup$ @athin Oh yeah, now I saw it under the "special 1" selection after your comment. $\endgroup$ – Alaiko Dec 6 '20 at 6:45

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