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Heavily inspired by ABDEC.

For each level below (40 total), connect the squares. Each level has a unique solution. New mechanics are introduced, but will remain backwards-compatible (that is, previous levels still have the same unique solution and no new solutions are introduced). You have to deduce the objective. Since puzzles are meant to be solved in order, the puzzles have been spoilerized. Partial Solutions are Welcome.

Puzzle 1, Puzzle 2, Puzzle 3, Puzzle 4, Puzzle 5, Puzzle 6, Puzzle 7, Puzzle 8, Puzzle 9, Puzzle 10, Puzzle 11, Puzzle 12, Puzzle 13, Puzzle 14, Puzzle 15, Puzzle 16, Puzzle 17, Puzzle 18, Puzzle 19, Puzzle 20, Puzzle 21, Puzzle 22, Puzzle 23, Puzzle 24, Puzzle 25, Puzzle 26, Puzzle 27, Puzzle 28, Puzzle 29, Puzzle 30, Puzzle 31, Puzzle 32, Puzzle 33, Puzzle 34, Puzzle 35, Puzzle 36, Puzzle 37, Puzzle 38, Puzzle 39, and Puzzle 40.

1:

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2:

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3:

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4:

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5:

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6:

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7:

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8:

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9:

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10:

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11:

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12:

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13:

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14:

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15:

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16:

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17:

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18:

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19:

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20:

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21:

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22:

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23:

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24:

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25:

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26:

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27:

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28:

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29:

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30:

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31:

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32:

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33:

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34:

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35:

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36:

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37:

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38:

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39:

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40:

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  • 2
    $\begingroup$ The images in the spoiler blocks are the same as the image links in the main text, right? $\endgroup$
    – justhalf
    Commented Feb 13 at 13:02
  • $\begingroup$ @justhalf yeppy $\endgroup$
    – Sny
    Commented Feb 13 at 13:20
  • 7
    $\begingroup$ Probably this puzzle is a bit too space-hungry. Maybe StackExchange is not the best place for this kind of long series puzzle. $\endgroup$
    – justhalf
    Commented Feb 13 at 14:23
  • $\begingroup$ sadgy sadgy sadge $\endgroup$
    – Sny
    Commented Feb 13 at 14:52
  • $\begingroup$ Cool puzzle! Could make a nice game from this. $\endgroup$ Commented Feb 14 at 1:44

2 Answers 2

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Full solution (unsure about 32, and probably 39 and 40).

1.

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The objective is to connect the cells in square formation. Here it's simple.

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Here we learn that we may need to make multiple squares

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Here we learn that we might need to figure out the partition in a more complex way.

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Here we learn that the squares might not be aligned

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Now things are getting more interesting. Now we learn the squares can be scaled, and they can intersect.

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Here we learn the squares may overlap in edges, not just intersect.

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Here we learn the shape may be other than square. We have triangle here.

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Here we learn the shape can be rotated.

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Here we learn the shapes don't have all be of same rotation or scale.

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Nothing new, probably we learn that we can do cross boundary with triangle shape, just like we did with the squares.

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Here we learn there could be combination of shapes.

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Another combination.

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Another combination.

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Here we learn the rotation can be other than multiples of 90 degrees.

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Nothing new.

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Nothing new.

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Nothing new.

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Here we see the rotation can be quite tricky.

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Nothing new.

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Another shape, with different nodes. Here we learn the node can be anywhere in the shape, not just vertices.

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Here we see two new shapes with new placements of the nodes. The triangle even only have two nodes here.

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Other unusual node placements

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Here we learn that not all shapes are required to be present. So the shapes apparently are "allowed shapes". For uniqueness discussion, see 24.

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The number of squares is odd, so we need one 3-node triangle, and two 2-node triangles.
This is interesting. Given that the solution is unique, we can learn that even without nodes, the shapes' vertices should be grid aligned (otherwise, 23 would have two solutions, using 3 of the 2-node triangles). So the configuration below is not valid:
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Again, the grid-aligned vertices ensure uniqueness of the solution.

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Nothing new.

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Here we see new colored shape (red). Given that the square appears twice, while previously we can use each shape multiple times anyway, this probably mean that the red shapes must appear exactly once. We'll perhaps learn more in later puzzles.

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Nothing new. Our hypothesis that red shapes must be used exactly once still holds.

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Here we learn new rule again, the shape has solid edge, which probably means center of cells that the shape touches become part of this shape. The color is red, so it's used exactly once.
And we learn that the shape can extend the image border.

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Another out-of-the-box solution. The square vertices are still aligned to cell centers (sorry I can't extend the image)

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Another out-of-the-box solution. The requirement for the edges to touch cell edges to count means the solution is unique.

This one is interesting.
Since this puzzle is symmetric left-right, the only possible way to have a unique solution would be to have something that is also symmetric left-right.
For example, this solution, if valid, is not unique since we can reflect it left-right:
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So we need to make that solution invalid, but I can't find a rule that make that invalid. I found a symmetric left-right solution, as follows:
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which is probably the intended solution for this.
The following is probably invalid due to some cells covered by two shapes (since the two triangles pass the center of cells at the bottom):
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There is also another left-right symmetric possible solution which I can't eliminate:
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Our first non-polygon shape. Not sure what restrictions for the location of the circle.

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One circle, and another big circle. I only drew the bottom right quarter-circle of the big one here.

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The red triangle with 3 nodes must exist, so the other must also be a 3-node triangle. As adj138 pointed out in comments, though, this is not unique either. Another solution is to cover the top-left, bottom-left, and bottom-right cells with one triangle, and the leftover cells with another.

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Our first non-2D shape.

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Solid edge covers everything it passes.

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Red shape must cover at least 1 cell (since it appears exactly once), so it must cover exactly 3 cells out of 6, so the other 3 can be covered by the yellow triangle.

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Red circle must exists, so it must cover the outer cells, the leftover 2 cells by the yellow edge. There must be some other rules that ensure uniqueness here which I can't figure out. Otherwise, the red circle can just cover 2 cells, and then the leftover 4 cells be covered by two yellow edges.

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Probably this is the intended solution, interpreting the red filled circle as a single dot. Although I'm not sure how to exclude other solutions, like this one:
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So what's left is probably some missing additional rules to ensure uniqueness in 39, and a few others.

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  • $\begingroup$ #32 and #39 are my mistake, sorry $\endgroup$
    – Sny
    Commented Feb 13 at 14:52
  • $\begingroup$ For #39 you can make the edge red. For #32, need to change the cell placements, unfortunately. $\endgroup$
    – justhalf
    Commented Feb 13 at 15:27
  • $\begingroup$ * #32 and #40 are my mistake, but #39 is because the center of the circle must be a lattice point $\endgroup$
    – Sny
    Commented Feb 13 at 15:29
  • $\begingroup$ Oh yah... forgot circle can just cover four points. But this series probly won't continue anyways... $\endgroup$
    – Sny
    Commented Feb 13 at 15:30
  • 1
    $\begingroup$ @ajd138 thanks, updated $\endgroup$
    – justhalf
    Commented Feb 14 at 3:51
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Intended solutions:

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probly will make a big image instead of several small ones next time ...

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