Here is one Fillomino puzzle:

Remove all fours, and it turns into a Slitherlink:

Remove all threes, too, and it turns into a Tapa:

Remove all twos as well and it turns into a Simple Loop (with the ones acting as black squares):

Here is one Fillomino puzzle:

Remove all fours, and it turns into a Slitherlink:

Remove all threes, too, and it turns into a Tapa:

Remove all twos as well and it turns into a Simple Loop (with the ones acting as black squares):

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This puzzle solves as an All Ones Slitherlink, an All Twos Slitherlink, and an All Threes Slitherlink.

This puzzle solves both as a regular Slitherlink and a Liar Slitherlink.

Here is one innocent-looking 5 by 5 Slitherlink puzzle:

Inflate it to twice its original size, though, and it turns into a full-blown 10 by 10 Liar Slitherlink:

Here is one two-step incremental Slitherlink puzzle. With ordinary addition, you very probably couldn’t do any better in terms of the number of steps.

What about addition modulo four, though? This means that, on each step, all zeros become ones, all ones become twos, all twos become threes, and all threes are set back to zeros. This way, you might just be able to make an infinite number of steps! (With the same four puzzles repeating over and over, of course.)

When you state the task like this, it isn’t difficult to achieve at all:

So, let us further require that all four puzzles be (preferably, very) different. Is *that *possible? Yes, it is:

And here are the successive increments. First step:

Second step:

Third step:

Here is one Slitherlink puzzle.

Increase all clues by one, and a new Slitherlink puzzle appears:

Increase all clues by one once again, and you have yet another puzzle: