8
$\begingroup$

Four Cells is an area dividing puzzle. You may see an introduction to this puzzle here: Four Cells: An Introduction


Rules of Four Cells:

  • The grid is to be divided along the grid lines into areas containing exactly four cells.
  • A number in a cell indicates how many of its four sides are segments of area boundaries. Note that this also includes the border of the grid.
  • Line segments of area boundaries should not be left dangling.
  • An area may contain multiple number cells (including none).

An example is shown in the link above if further clarification is needed.


Shown below is the actual puzzle to solve:

Four_Cells_2

Good luck and have fun!

P.S. You may be wondering, "What's the big deal about not having any 3s?" Nothing. I just needed something for the title.

$\endgroup$
8
$\begingroup$

The Top

The 1 at the top is already fulfilled, and we can make progress with the two 2’s, also at the top.

step1

If the lower end of the top right 2 goes down, it would make the 2 below it have three borders instead of just two. Therefore, the top right 2 must form a square instead.

step2

Same logic can be applied to the downward extension of the 2 at Row 1 Column 5 and the 2 at Row 3 Column 8.

step3

The Middle Left and the Two Down Under

Let us also turn our attention to the 1 in the middle left. If its border were anywhere other than to its right, the 2 would always be forced to have three borders. Therefore, the border of the 1 must be to its right.

step4

The square to the top left of the same 1 must be obtained by the 2 in the top left; otherwise, it will belong to a region that does not have exactly four squares. And there will only be one way that this is fulfilled:

enter image description here

The square at Row 3 Column 3 cannot extend rightward, as it would force the 2 to its right to have three borders. Therefore it must extend downward.

step6

Note than the 2 at the bottom must extend fully horizontally. Otherwise, either a group of less than four cells would be trapped, or the 2 would obtain three borders.

step7

The Bottom Right, and Finishing Off the Twos

Now the unsolved 1 is limited as to how it can extend: two configurations of the area where the 1 will be yield a square trapped to the bottom left of the 1. Therefore, we are sure that the 1 will extend upward and rightward. It also cannot include the 2 to its right, because of the same problem that arose in previous steps: the 2 will have three borders.

step8

And quickly we see that the 1 must extend downward, or else it would trap a few squares below it.

step9

Here it took me a bit of time to figure out that the 2 at the top cannot form a square, or the 2 at Row 3 Column 4 would be forced into an area of six squares.

step10

Therefore it must form an S-shape instead.

step11

Finally, the 2 at Row 3 Column 6 cannot extend downward, or else it would trap seven squares to the right. Therefore, it must extend rightward. The puzzle is completed after this.
step12

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.