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Caravel Forum : Caravel Boards : General : Invariants of Tar Cutting (Or, Why Can't I Remove That Bloody Tar Blob)
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coppro
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icon Re: Invariants of Tar Cutting (+1)  
The other issues is that the shape:
XX
XXX
 XX

is completely cuttable, yet its interior corners are in the shape:
X
 X

This shape is cuttable, but only when there is no L-tetromino containing it.

[Last edited by coppro at 05-17-2014 05:42 PM]
05-17-2014 at 05:42 PM
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icon Re: Invariants of Tar Cutting (0)  
Good point. I should try to avoid threads like this before I've had any coffee, I'm bound to say something dumb. So it's definitely true that all cuttable gel blobs can be tiled with L-trominos, but the converse is not true. A cuttable L-tromino of enclosed corners needs to have a cuttable gel corner "inside" it, like this:
CX
XX

(X = enclosed corner, C = cuttable cell)

Removing cells will of course expose other cells and change their cuttability, which makes this a very interesting problem. For a cell of gel to be cuttable, it must have 4 orthogonally adjacent cells, this is probably more useful for analysis than concave corners.

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[Last edited by Pinnacle at 05-17-2014 06:57 PM : I have no idea what I'm talking about.]
05-17-2014 at 05:46 PM
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Kallor
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icon Re: Invariants of Tar Cutting (0)  
I think we have seen these 2 counterexamples in this thread: 3x4 tile gel blob, an uncuttable shape such that the interior corners consists of L-trominoes only, and the two 2x2 tile blobs intersecting at on tile, a cuttable shape with only two interior corners that of course don't consist of L-trominoes.
So neither the claim nor the converse is true.

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05-17-2014 at 06:30 PM
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Kallor
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icon Re: Invariants of Tar Cutting (0)  
Oh by the way, I don't have my DROD here so I can't test this (can be tested with TSS elements at least) but is any gel tile with all 4 orthogonal neighbours cuttable or does it need the 1 or 2 diagonal non-neighbours as well? I other words: does Pinnacle's condition completely descript the cutting rules?

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05-17-2014 at 06:39 PM
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icon Re: Invariants of Tar Cutting (0)  
I am not talking about tiling gel blobs with


XX
X

and 

X
 X


I am talking about the enclosed corners.

And - of course - this is not a sufficient condition

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05-17-2014 at 06:44 PM
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icon Re: Invariants of Tar Cutting (+1)  
Kargaros wrote:
I believe I have convinced myself:

[...]

The secret lies in the inequality h(w) - h(v) <= d(v,w) (which Thurston sadly wrote with >= instead of <=).

Nicely done, Kargaros.

It occured to me that there's a slightly different way to approach the construction of the height function which avoids the iterative process. Having defined h on the set of boundary vertices B, we can define a function H on all vertices of R by setting H(v) = min{ h(b) + d(b,v) | b an element of B }. (I think this is what Thurston meant to write on page 767.) Then if there is an edge in R from v to w, we have d(b,w) <= d(b,v) + 1 for all b, so H(w) <= H(v) + 1. Similarly, there is a sequence of three edges in R going from w to v (just continue around one of the neighboring squares), so we also have H(v) <= H(w) + 3. Then -3 <= H(w) - H(v) <= 1, so H(w) - H(v) is either 1 or -3 by the modulo 4 congruence. We can then apply your analysis to see that each square has exactly one edge with a change of -3.

In this formulation, the tricky part is that when v is a boundary vertex, we can only be sure that H(v) <= h(v). What we actually want is to have H(v) = h(v). For this to hold, we should have h(v) <= h(w) + d(w,v) for any boundary vertex w, which is exactly the crucial inequality.

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05-17-2014 at 06:52 PM
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Pinnacle wrote:
Does gel actually reduce to L-tromino tiling on the enclosed corners? From a bit of experimentation, it looks like it does,
Definitely not.

and we know that Golomb's proof applies to gel.
I have seen this somewhere in the Giant Jewel hold. as anybody seen a proof?

The most basic cuttable tar shape fits a single domino, and the most basic cuttable gel shape fits a single L-tromino.

Unfortunately it is not that simple. A tar configuration is cuttable if and only if the set of enclosed corners is tileable with dominoes. For gel the situation is not that simple.

adS

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05-17-2014 at 07:00 PM
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tempestadept
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icon Re: Invariants of Tar Cutting (+1)  
adS wrote:
and we know that Golomb's proof applies to gel.
I have seen this somewhere in the Giant Jewel hold. as anybody seen a proof?
Which proof do you mean?
Tileability of a 2^n \times 2^n square with a single unit square removed is easily proven by induction. The fact that this tiling has a cutting sequence can be seen from its construction.

[Last edited by tempestadept at 05-17-2014 07:06 PM]
05-17-2014 at 07:06 PM
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Kargaros
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icon Re: Invariants of Tar Cutting (+2)  
Another thing I realized while working to understand Bill Thurston's domino tiling/tar cutting algorithm:
We know that every tar blob has an associated configuration of enclosed corners, and that the tar blob can be reconstructed from the configuration of corners. However, not every corner configuration is actually realized by a tar blob.

If R is a configuration of enclosed corners in a tar blob, then R satisfies the following: If either of the two patterns of X's below (or rotations hereof) appear in R, then the middle O must be in R as well.
 X     X
XO     O
  X    X
From what I could work out, this property is also enough to ensure that a configuration R is realized by a tar blob.

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05-17-2014 at 07:24 PM
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tempestadept
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Kargaros wrote:
If R is a configuration of enclosed corners in a tar blob, then R satisfies the following: If either of the two patterns of X's below (or rotations hereof) appear in R, then the middle O must be in R as well.
 X     X
XO     O
  X    X

From what I could work out, this property is also enough to ensure that a configuration R is realized by a tar blob.
Nice observation (and indeed provable).

[Last edited by tempestadept at 05-17-2014 07:46 PM]
05-17-2014 at 07:46 PM
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tempestadept wrote:
adS wrote:
Tileability of a 2^n \times 2^n square with a single unit square removed is easily proven by induction. The fact that this tiling has a cutting sequence can be seen from its construction.

I know Golomb's proof - I even have the book :)

What does "Tileability " mean here? A 4x4 gel blob with a corner removed is clearly not cuttable.

Does it mean that a gel blob whose enclosed corners are a 2^nx2^n square with a single corner removed is cuttable?

Please explain.

adS


adS

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05-17-2014 at 08:12 PM
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tempestadept
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icon Re: Invariants of Tar Cutting (0)  
I mean not the shape of the blob itself, but the enclosed corners configuration.
Tileability means existence of tiling with L-triominoes.

[Last edited by tempestadept at 05-17-2014 08:19 PM]
05-17-2014 at 08:18 PM
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icon Re: Invariants of Tar Cutting (+1)  
tempestadept wrote:
I mean not the shape of the blob itself, but the enclosed corners configuration.
Tileability means existence of tiling with L-triominoes.

Hm, according to Kargaros' remarks the only enclosed corner configuration could be a square with a corner missing. You can't have a square with an interior corner missing as. So the only possible cases are


...

Right?

adS




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[Last edited by adS at 05-17-2014 08:51 PM]
05-17-2014 at 08:51 PM
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tempestadept
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icon Re: Invariants of Tar Cutting (+1)  
If you remove an inside square of the blob, this corresponds to removing 2x2 enclosed corners. If that removed block is well positioned on a step-2 grid (it should either coincide with one of the grid cells or consist of corners of 4 different cells) then the configuration would be tileable and cuttable (by a mimic).
If the text isn't clear I can illustrate it with some figures tomorrow.
05-17-2014 at 09:04 PM
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tempestadept wrote:
If you remove an inside square of the blob, this corresponds to removing 2x2 enclosed corners. If that removed block is well positioned on a step-2 grid (it should either coincide with one of the grid cells or consist of corners of 4 different cells) then the configuration would be tileable and cuttable (by a mimic).
If the text isn't clear I can illustrate it with some figures tomorrow.

If you remove 2x2 enclosed corners Golombs's "theorem" doesn't help at all.

Edit: Hm, you are right. The 2x2 can be tiled with a tromino and a hole.

So probably we may claim: a (2^n+1)x(2^n+1)gel blob with a single missing square is cuttable.

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[Last edited by adS at 05-17-2014 09:26 PM]
05-17-2014 at 09:13 PM
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tempestadept
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adS wrote:
tempestadept wrote:
If you remove an inside square of the blob, this corresponds to removing 2x2 enclosed corners. If that removed block is well positioned on a step-2 grid (it should either coincide with one of the grid cells or consist of corners of 4 different cells) then the configuration would be tileable and cuttable (by a mimic).
If the text isn't clear I can illustrate it with some figures tomorrow.

If you remove 2x2 enclosed corners Golombs's "theorem" doesn't help at all.
A 2x2 can be thought as an L-tile + unit square. If this tile is included in Golomb's tiling for this unit square missing, then everything works.

Also edit: only some squares leave the blob cuttable. Removing non-corner square on the border removes 2 enclosed corners and therefore breaks divisibility by 3. Inside squares are more tricky. I'll elaborate a bit later.

[Last edited by tempestadept at 05-17-2014 09:34 PM]
05-17-2014 at 09:23 PM
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Is it true that if a glob of gel is cuttable, a glob of gel which entirely encloses the original (that is, as if it grew with no barriers) is also cuttable? It feels to me as if it is. If so, we do the inverse (remove any part immediately next to an open tile), first labelling any interior corners (cuttable spaces) and examine what remains:

1) If any tile that previously had gel on it cannot trace a path through tiles with gel remaining to a previously interior corner, it is not cuttable.
2) If there is a section that looks like this:
xxx
GGG
xxx

The gel is not cuttable.
3) If there is any stable gel that remains, repeat for each section of stable gel.

As far as my research goes, this seems to work.

EDIT: Upon second thought, this might be wrong, but I don't know.

[Last edited by Someone Else at 05-17-2014 09:46 PM]
05-17-2014 at 09:40 PM
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adS wrote:
So probably we may claim: a (2^n+1)x(2^n+1)gel blob with a single missing square is cuttable.

This is too general. Take a 9x9 blob and remove the square 3 steps east and 2 steps south of the NW corner square. In this situation, you must stab NW from the removed square in order to remove the enclosed corner 2 steps east of the northwesternmost enclosed corner. After doing that, it's impossible to remove the northwesternmost square of the blob.

I think you'll want to require the removed square to have the same color as the corner squares. Also, it feels like there's going to be a bunch of special cases having to do with squares near the edge of the blob.

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05-17-2014 at 09:51 PM
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Watcher wrote:
adS wrote:
I think you'll want to require the removed square to have the same color as the corner squares. Also, it feels like there's going to be a bunch of special cases having to do with squares near the edge of the blob.

Yes, I noticed this myself a few minutes ago :)

The removed square must have the same color as the corner squares since cutting a square always leaves cuttable squares of the same color only.

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05-17-2014 at 10:19 PM
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icon Re: Invariants of Tar Cutting (+1)  
I have experimented a bit in the editor.

A 7x7 square with a corner removed is cuttable.

Of course, to cover the enclosed corners you must use (at least one)

x
 x


shapes since the number of enclosed corners is not even divisible by three.

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05-17-2014 at 10:42 PM
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icon Re: Invariants of Tar Cutting (+10)  
Someone Else wrote: Is it true that if a glob of gel is cuttable, a glob of gel which entirely encloses the original (that is, as if it grew with no barriers) is also cuttable? It feels to me as if it is.

EDIT: Upon second thought, this might be wrong, but I don't know.
How about the following blob of Gel?
XX XX
XXXXX
XXXXX
This blob is completely removable, but if you let it grow once, it cannot be cut at all.

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[Last edited by Kargaros at 05-18-2014 03:28 AM]
05-17-2014 at 11:22 PM
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icon Re: Invariants of Tar Cutting (+1)  
As for the Golomb's method, I can prove the following:

Theorem. If the corners of the original square have coordinates (0,0) and (2^n, 2^n) and the removed square has coordinates (x,y), where maximum powers of 2 that divide x and y are equal, then the blob is cuttable. Here for convevience we can think of 0 being divisible by at most 2^n.

This means that having the same color as corners is only a necessary condition. As for the other (x,y) pairs, uncuttability might be provable in general (because we have only one "source" of cuttability), but I'm not sure yet.

[Last edited by tempestadept at 05-21-2014 10:11 PM]
05-18-2014 at 02:28 AM
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adS wrote:
If you remove 2x2 enclosed corners Golombs's "theorem" doesn't help at all.
And here I thought that Golomb's Theorem was: When you cut gel, the gel babies will move like rock golombs.

Seriously, this is the best conversation I've seen online anywhere in ages. Even if I don't understand a word (or strange Greek-like symbol) of it.
05-18-2014 at 04:25 PM
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efficiency(Sigma monsters) = swings + moves + backtracking

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05-18-2014 at 11:15 PM
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icon Re: Invariants of Tar Cutting (+1)  
I was still curious about these things and did some googling today. While destroying gel is not the same thing as finding a L-tromino tiling of a planar region
(for various reasons such as
1. the L-tromino of enclosed corners is not the only permitted tile, also two enclosed corners diagonally from each other is allowed,
2. it's not enough for the tiling to exist, we need to be able to remove permitted tiles one by one so that each removed L-tromino has emptiness on its inward corner and each removed tile of 2 diagonal enclosed corners has emptiness at its both inward corners,
3. there sometimes is safety concerns with the gel babies
),
this article by Moore and Robson answers some questions that could have been asked on this thread. They show (by reducing the question to certain special case of Boolean satisfiability problem) that the problem of determining whether a given planar region has a L-tromino tiling is NP-hard. That is, it is very unlikely that there exists a polynomial time algorithm like Thurston's (linear time?) algorithm for the domino tiling problem.

-Kallor

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[Last edited by Kallor at 05-24-2014 04:26 PM]
05-24-2014 at 04:25 PM
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Kallor wrote:
I was still curious about these things and did some googling today. While destroying gel is not the same thing as finding a L-tromino tiling of a planar region
(for various reasons such as
1. the L-tromino of enclosed corners is not the only permitted tile, also two enclosed corners diagonally from each other is allowed,
as I have said several times.

2. it's not enough for the tiling to exist, we need to be able to remove permitted tiles one by one so that each removed L-tromino has emptiness on its inward corner and each removed tile of 2 diagonal enclosed corners has emptiness at its both inward corners,
This is probably much harder to decide than tileability.

3. there sometimes is safety concerns with the gel babies
Well, sometimes we want to know if a mimic can cut the blob :)
Other blobs may require to place a mimic or a clone in a hole in the blob. There is a really hard room in advanced concepts (2?) that is only solvable by placing a mimic in one of several holes (of tar).



),
this article by Moore and Robson answers some questions that could have been asked on this thread. They show (by reducing the question to certain special case of Boolean satisfiability problem) that the problem of determining whether a given planar region has a L-tromino tiling is NP-hard. That is, it is very unlikely that there exists a polynomial time algorithm like Thurston's (linear time?) algorithm for the domino tiling problem.
-Kallor
I have seen the article before - Google. Unfortunately I do not really understand it.

Yes, Thurston's algorithm is linear and there another one for tar blobs (domino tileability) with holes which is polynomial.

I agree that there is little hope for gel.

adS


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05-24-2014 at 04:43 PM
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icon Re: Invariants of Tar Cutting (+2)  
I doodled some stuff last night and came to the conclusion that the proof of Moore and Robson can easily be modified to concern gel. That is when we assume we have a lone swordsman with access to all gel borders (via tunnels or clone potions or whatever) and no worries about the gel babies the problem of deciding whether the gel can be destroyed is NP-hard (with respect to the size of the gel blob).
The reduction of Moore and Robson to 1-in-3-SAT fails in Figures 9 and 10 because of the possibility to remove blobs with 2 eclosed corners diagonally (with 2 empty corners), but new constructions without this problem were easy enough to come up with (attached picture).

In fact, unlike with L-trominoes only (it seems), it was possible to make a NOT-gate too (also in the attached picture). Unfortunately, I could not invent any other logical gates to be able to use the nicer proof of reducting the problem to Boolean satisfiability directly, like Moore and Robson did with L-trominoes and 2x2-blocks. EDIT: Silly me, New Figure 10 is in fact an OR-gate if we label two wires as input wires and the third one as an output wire and take into account that the direction of "current" changes. EDIT OF EDIT: No it's not, there is no output for two "true":s. New figure 9 is a wire splitter, so we can in fact use the easier proof of reducing to Boolean sitisfiablity in the case of gel. Appartently the problem with L-trominoes only was not the logical gates, but the variable node, which can easily be made with gel, like this:

.......
....*..
...***.
.***...
**.....
.......


As for the case of domino tilings, the link earlier assumed that the number of holes is bounded. Moore and Robson remark that the number domino tilings can be reduced to the determinant of some "modified adjancency matrix" ([18],[19]), thus making solvability a polynomial time problem (but maybe not n*log(n) as in your (adS's) link with the bounded number of holes perhaps? EDIT: detecting non-zero determinant has complexity O(n^3) so the earlier link beats the matrix point of view.).

-Kallor

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[Last edited by Kallor at 05-26-2014 07:24 AM]
05-25-2014 at 09:10 AM
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Erm, but for the purpose of players everything can be decided in O(1), since we have a board of constant size, right?

adS

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05-25-2014 at 09:59 AM
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I'd guess so :)

-Kallor

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05-25-2014 at 11:16 AM
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icon Re: Invariants of Tar Cutting (0)  
Has anyone solved the following problem?

How many tarstuff shapes are possible for n enclosed corners?

e.g.
n=1 : 1
n=2 : 2 without rotation, 6 including rotation

Obviously it is the unique shapes, without rotation, that is more interesting.

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