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Pentominoes packing a 3x20 rectangle

Gengulphus
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Pentominoes packing a 3x20 rectangle

#385643

Postby Gengulphus » February 11th, 2021, 12:24 pm

A polyomino is a 2-dimensional shape consisting of a number of equal-sized squares joined together along their edges. Regarding reflections and rotations of a polyomino as being the same polyomino, there is one monomino (which is just a single square), one domino (the familiar two squares joined together, and the origin of all these names), two trominoes (the three squares can either be joined together in a line, or with a right angle in the middle), five tetrominoes (which I won't bother to list here) and twelve pentominoes:

+---+           +---+---+       +---+               +---+---+   +---+       +---+---+---+
| O | | P P | | Q | | R R | | S | | T T T |
+ + + + + + +---+ +---+ + +---+ +---+ +---+
| O | | P P | | Q | | R R | | S S | | T |
+ + + +---+ + + +---+ + +---+ + + +
| O | | P | | Q | | R | | S | | T |
+ + +---+ + +---+ +---+ + + +---+
| O | | Q Q | | S |
+ + +---+---+ +---+
| O |
+---+

+---+ +---+ +---+ +---+ +---+ +---+ +---+---+
| U | | U | | V | | W | | X | | Y | | Z Z |
+ +---+ + + + + +---+ +---+ +---+ +---+ + +---+ +
| U U U | | V | | W W | | X X X | | Y Y | | Z |
+---+---+---+ + +---+---+ +---+ +---+ +---+ +---+ +---+ + + +---+
| V V V | | W W | | X | | Y | | Z Z |
+---+---+---+ +---+---+ +---+ + + +---+---+
| Y |
+---+

I've labelled each of them with a letter from O to Z to allow them to be quickly referred to as e.g. "the R pentomino". I got this particular labelling from Conway, and its main advantage is that by remembering that it uses the last twelve letters of the alphabet and that the shapes of the letters give hints about the shapes of the pentominos, it allows you to remember the list of the twelve pentominos reasonably easily. Its disadvantage is that some of the hints aren't very good - in particular, O (think of a very tall and thin letter O) and Q (think of an O with a tail), and it isn't exactly obvious which of S and Z is which. (So some people prefer to think of the O and Q pentominoes as instead being more obviously named as the I and L pentominoes, for example: the set of letters I,L,P,R-Z is less memorable than O-Z, but the hints about the pentomino shapes are better.)

Anyway, the puzzles of tiling rectangles with the twelve pentominoes are fairly well-known. As the twelve pentominoes are made up of five squares each, they cover a total of sixty squares, and so can potentially tile 3x20, 4x15, 5x12 and 6x10 rectangles. And in fact, all four of those rectangles can be tiled, most of them in quite a lot of ways - the 6x10 rectangle can be tiled in 2339 different ways, the 5x12 rectangle in 1010 different ways and the 4x15 rectangle in 368 different ways (not counting rotations and reflections of solutions as different). That basically puts the puzzle of finding all the tilings of those rectangles beyond the reach of sensible human endeavours - those numbers have all been found by computer searches.

But there are just two possible tilings of the 3x20 rectangle by the twelve pentominoes, and the puzzle of finding both of them can be solved by humans without the aid of computers. What I'm wondering is what the best line of reasoning is that leads to the two solutions - "best" essentially meaning eliminating as many cases as possible in a 'wholesale' way, to avoid doing a massive search of lots of different possibilities. As an example of a step in such a line of reasoning, consider how much of the middle row of the 3x20 rectangle each of the P-Z pentominoes must cover: the X pentomino must cover three of its squares, the P, R, S and W pentominoes must each cover at least two of its squares and the Q, T, U, V, Y and Z pentominoes must each cover at least one of its squares. So between them, they must cover at least 17 of the 20 squares of the middle row, and so the O pentomino cannot run along the middle row, but must run along either the top row or the bottom row (and we might as well assume that it's the top row, since solutions with it on the top row and solutions with it on the bottom row are mirror images of each other).

Note that what I am after is the reasoning. Just posting the solutions without reasoning gets no credit! Posting interesting lines of reasoning that pin down some aspects of the solutions gets rather more, even if they don't fully solve the problem.

Gengulphus

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Re: Pentominoes packing a 3x20 rectangle

#385863

Postby 9873210 » February 11th, 2021, 10:56 pm

r,t,x,z must cross all three rows, so there must be a multiple of 5 squares between them. If r and t are vertical (in the orientation shown) then the centers of these must be in columns 3,8,13 and 18. Even if z is sideways this leaves no room for o. So at least one of r or t is sideways. This increases the count in the center row to at least 18. That leaves no room for the long arms of q or y. So the long arms of both are either on the top or the bottom row.

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Re: Pentominoes packing a 3x20 rectangle

#385872

Postby 9873210 » February 12th, 2021, 12:09 am

9873210 wrote:r,t,x,z must cross all three rows, so there must be a multiple of 5 squares between them. If r and t are vertical (in the orientation shown) then the centers of these must be in columns 3,8,13 and 18. Even if z is sideways this leaves no room for o. So at least one of r or t is sideways. This increases the count in the center row to at least 18. That leaves no room for the long arms of q or y. So the long arms of both are either on the top or the bottom row.

oops.

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Re: Pentominoes packing a 3x20 rectangle

#385877

Postby 9873210 » February 12th, 2021, 12:40 am

r,t,x,z must cross all three rows, so there must be a multiple of 5 squares between them and between them and each end. If r and t are vertical (in the orientation shown) then the centers of these must be in columns 3,8,13 and 18. To fit the o the z must be sideways. That fills 19 of the squares of the middle row. If one of r and t are sideways that fills at least 18 squares on the second row. So at least 18 squares on the second row are filled. That leaves no room for the long arms of q or y in the middle row. So the long arms of both are either on the top or the bottom row.

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Re: Pentominoes packing a 3x20 rectangle

#385879

Postby 9873210 » February 12th, 2021, 1:08 am

Are we allowed to use "there are just two possible tilings of the 3x20 rectangle" as a given? There are a few deductions that can be made from that. It's surprising how often that knowing there is a unique (or in this case semi-unique) solution gives the answer away.

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Re: Pentominoes packing a 3x20 rectangle

#385904

Postby Gengulphus » February 12th, 2021, 9:03 am

9873210 wrote:Are we allowed to use "there are just two possible tilings of the 3x20 rectangle" as a given? There are a few deductions that can be made from that. It's surprising how often that knowing there is a unique (or in this case semi-unique) solution gives the answer away.

Good question! And yes, I can see at least one way in which treating there being only two possible tilings as a given would help, by allowing deductions about where the P, T, U and V pentominoes can be placed...

But I didn't intend it to be treated as a given - it was just meant as an indication of why (unlike the other rectangles) it might be sensibly solvable by a human.

Incidentally, I was prompted to post this puzzle by a friend of mine mentioning a fairly classic science fiction book in which the 'viewpoint' character was impressed by his friend managing to find all the solutions by hand, which he regarded as especially impressive given that there are only two solutions. Whereas actually finding all the solutions of any of the other three rectangles by hand would be hugely more impressive! So for those whose interests lie elsewhere, a 'pub quiz' type question: name the book and its author.

Gengulphus

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Re: Pentominoes packing a 3x20 rectangle

#385926

Postby Gengulphus » February 12th, 2021, 10:02 am

9873210 wrote:r,t,x,z must cross all three rows, so there must be a multiple of 5 squares between them and between them and each end. If r and t are vertical (in the orientation shown) then the centers of these must be in columns 3,8,13 and 18. To fit the o the z must be sideways. ...

That step seems far from obvious to me. It's possible for instance to put the R vertically centred in column 3 and the T vertically centred in column 8 and fill in the small gaps around them as follows:

+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| U U | R R | W W | T T T | | | | | | | | | | | |
+ +---+ +---+ +---+---+ +---+---+---+---+---+---+---+---+---+---+---+---+
| U | R R | W W | P P | T | | | | | | | | | | | | |
+ +---+ + +---+ + +---+---+---+---+---+---+---+---+---+---+---+---+
| U U | R | W | P P P | T | | | | | | | | | | | | |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+

and there's still plenty of space to put a vertically-oriented Z and an O into the unused squares with each resulting chunk of unused squares containing a multiple of 5 squares. I'm not saying that any of those ways of placing them leads to a solution - they don't! - but it requires a good deal of case-by-case analysis to show that for each of them, at least one of the chunks of unused squares cannot be tiled without re-using a pentomino. And that's only one of the cases of placing the vertically-oriented R and T pentominoes, and there are essentially 12 such cases. (The R can be placed centred on any of columns 3, 8, 13 and 18, and then the T can be placed centred on any one of the three columns not used by the R and either the 'right way up' or 'upside down'. The R can also be placed the same way up as shown in the above diagram, reflected top-bottom, reflected left-right or rotated by 180 degrees, giving an apparent extra factor of 4, but since we're treating rotations and reflections of entire solutions as the same, we can fix on one of those orientations of the R.)

So I'm not saying that your argument is wrong - but it does seem either to be skipping over a good deal of case-by-case analysis, which is exactly the sort of thing I want to minimise, or to be failing to mention a general reason why the O and a vertically-oriented Z are incompatible with each other, no matter how the vertically-oriented R and T are positioned.

Gengulphus

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Re: Pentominoes packing a 3x20 rectangle

#386059

Postby SteelCamel » February 12th, 2021, 2:07 pm

987310 mentioned the X as well. Placing the X and the vertical Z gives you:

+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| U U | R R | W W | T T T | | | Z Z | | | | | X | | |
+ +---+ +---+ +---+---+ +---+---+---+---+ +---+---+---+---+ +---+---+
| U | R R | W W | P P | T | | | | | Z | | | | X X X | |
+ +---+ + +---+ + +---+---+---+---+ +---+---+---+---+ +---+---+
| U U | R | W | P P P | T | | | | | Z Z | | | | X | | |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+


There are no blocks of five squares in a row to place the O. Swapping over the X and Z doesn't help.

A shorter answer is that the O must be horizontal, which means it must cross one of columns 3,8,13 and 18 and occupy one square in one of those columns. But R, T, Z and X must be in those columns, and vertical R, vertical T, vertical Z and X occupy all three rows in their centre column. So it's not possible to place vR, vT, vZ, X and O since between them they occupy thirteen of the twelve squares in columns 3,8,13 and 18. At least one of R, T or Z must be horizontal.

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Re: Pentominoes packing a 3x20 rectangle

#386112

Postby Gengulphus » February 12th, 2021, 4:19 pm

SteelCamel wrote:987310 mentioned the X as well. ...

A shorter answer is that the O must be horizontal, which means it must cross one of columns 3,8,13 and 18 and occupy one square in one of those columns. But R, T, Z and X must be in those columns, and vertical R, vertical T, vertical Z and X occupy all three rows in their centre column. So it's not possible to place vR, vT, vZ, X and O since between them they occupy thirteen of the twelve squares in columns 3,8,13 and 18. At least one of R, T or Z must be horizontal.

Thanks! I misread "the centres of these" as referring back to "r and t" in "If r and t are vertical"; as you indicate, the argument makes more sense if one sees it as referring back to the somewhat earlier "r,t,x,z" in "r,t,x,z must cross all three rows".

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Re: Pentominoes packing a 3x20 rectangle

#386145

Postby Gengulphus » February 12th, 2021, 6:02 pm

A tightening up of 9873210's result:

Regard the 3x20 rectangle as four 3x5 rectangles placed end to end:

+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| : : : |
+ + + + + + + + + + + + + + + + + + + + +
| : : : |
+ + + + + + + + + + + + + + + + + + + + +
| : : : |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+

There are six pentominoes that span three rows and three columns (R,T, V, W, X and Z) and which must therefore cross from the top row to the bottom row no matter how they're placed. Each of them must divide the rectangle into the chunk of squares to their left and the chunk of squares to their right, with each of those chunks containing a multiple of 5 squares, or be nestled into the left or right end of the 3x20 rectangle to leave no squares between it and the end, which is only possible for T and V (and is essentially just the case that one of the left chunk or the right chunk contains no squares and the other 55 squares, which is a degenerate case of the multiple-of-5 requirement). Of the six pentominoes, V, W and X have essentially only one possible orientation, and the multiple-of-5 requirement forces V and X to be in one of the 3x5 rectangles and W to be partly in each of two consecutive 3x5 rectangles, as illustrated below (or a rotation or reflection):

+---+---+---+---+---+   +---+---+---+---+---+   +---+---+---+---+---+---+---+---+---+---+
| V | | | | X | | | | W | : |
+ + + + + + + +---+ +---+ + + + + + +---+ + + + + +
| V | | | | X X X | | | | W W | |
+ +---+---+ + + + +---+ +---+ + + + + +---+ +---+ + + + +
| V V V | | | | X | | | | W : W | |
+---+---+---+---+---+ +---+---+---+---+---+ +---+---+---+---+---+---+---+---+---+---+

Similarly, T will be in a 3x5 rectangle if vertical (left diagram) or horizontal (right diagram):

+---+---+---+---+---+   +---+---+---+---+---+
| | T T T | | | T | |
+ +---+ +---+ + + +---+---+ + +
| | T | | | T T T | |
+ + + + + + + +---+---+ + +
| | T | | | T | |
+---+---+---+---+---+ +---+---+---+---+---+

and Z will be in a 3x5 rectangle if vertical (left diagram) or horizontal (right diagram):

+---+---+---+---+---+   +---+---+---+---+---+
| | Z Z | | | | Z | |
+ +---+ + + + + + +---+---+ +
| | Z | | | | Z Z Z | |
+ + + +---+ + + +---+---+ + +
| | Z Z | | | | Z | |
+---+---+---+---+---+ +---+---+---+---+---+

and R will be in a 3x5 rectangle if vertical (left diagram) or partly in each of two consecutive 3x5 rectangles if horizontal (right diagram):

+---+---+---+---+---+   +---+---+---+---+---+---+---+---+---+---+
| | R R | | | | R | : |
+ +---+ +---+ + + + + + +---+---+ + + + +
| | R R | | | | R R R | |
+ + + +---+ + + + + +---+ +---+ + + + +
| | R | | | | R | |
+---+---+---+---+---+ +---+---+---+---+---+---+---+---+---+---+

Now observe that T, X, Z and vertical R all necessarily occupy the central square of their 3x5 rectangles, so must be in different 3x5 rectangles. So if R is vertical, then R,T, X and Z must occupy all four 3x5 rectangles, and so V must share a rectangle with one of them.

But any orientation of V overlaps with vertical R, vertical T, X or vertical Z, and two of its orientations overlap with horizontal T while the other two are incompatible with it because between them, they leave an isolated uncovered square. So if both R and Z are vertical, V cannot be placed - i.e. at least one of R and Z must be horizontal.

That analysis suggests splitting into two cases to me:

A) R is vertical and Z is horizontal, in which case R, T, X and Z occupy the four 3x5 rectangles and V must share Z's 3x5 rectangle, while W must be fitted in between two of those rectangles.

B) R is horizontal, in which case T, X and Z occupy three of the four 3x5 rectangles, and R, W and V must all be fitted between them.

Of course, splitting into cases is undesirable, but at least there are only two of them, and furthermore, case A) looks quite tightly constrained to me, so might be quite quickly eliminated...

Gengulphus

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Re: Pentominoes packing a 3x20 rectangle

#386289

Postby 9873210 » February 13th, 2021, 9:00 am

Gengulphus wrote:A tightening up of 9873210's result:

That analysis suggests splitting into two cases to me:

A) R is vertical and Z is horizontal, in which case R, T, X and Z occupy the four 3x5 rectangles and V must share Z's 3x5 rectangle, while W must be fitted in between two of those rectangles.

B) R is horizontal, in which case T, X and Z occupy three of the four 3x5 rectangles, and R, W and V must all be fitted between them.

Of course, splitting into cases is undesirable, but at least there are only two of them, and furthermore, case A) looks quite tightly constrained to me, so might be quite quickly eliminated...

Gengulphus


A) R is vertical and Z is horizontal, in which case R, T, X and Z occupy the four 3x5 rectangles and V must share Z's 3x5 rectangle, while W must be fitted in between two of those rectangles.

We need to fit the following togeather
+---+---+---+---+---+   
| | R R | |
+ +---+ +---+ +
| | R R | |
+ + + +---+ +
| | R | |
+---+---+---+---+---+

+---+---+---+---+---+
| V | Z | |
+ + +---+---+ +
| V | Z Z Z | |
+ +---+---+ + +
| V V V | Z | |
+---+---+---+---+---+

+---+---+---+---+---+
| | X | |
+ +---+ +---+ +
| | X X X | |
+ +---+ +---+ +
| | X | |
+---+---+---+---+---+

+---+---+---+---+---+ +---+---+---+---+---+
| | T T T | | | T | |
+ +---+ +---+ + + +---+---+ + +
| | T | | or| T T T | |
+ + + + + + + +---+---+ + +
| | T | | | T | |
+---+---+---+---+---+ +---+---+---+---+---+

+---+---+---+---+---+---+---+---+---+---+
| | W | : |
+ + + + +---+ + + + + +
| | W W | |
+ + + +---+ +---+ + + + +
| | W : W | |
+---+---+---+---+---+---+---+---+---+---+

(I will treat the orientation of W as fixed, everything else can rotate/reflect)

================================================================================
To the left of the W we must have an vR or vT (hZ overlaps; X and hT leave holes)
+---+---+---+---+---+---+---+---+---+---+
| | T | W | : |
+ + + + +---+ + + + + +
| | T | W W | |
+ +---+ +---+ +---+ + + + +
| | T T T | W : W | |
+---+---+---+---+---+---+---+---+---+---+

+---+---+---+---+---+---+---+---+---+---+
| | R | W | : |
+ +---+ + +---+ + + + + +
| | R R | W W | |
+ +---+ +---+ +---+ + + + +
| | R R | W : W | |
+---+---+---+---+---+---+---+---+---+---+

================================================================================

On the right of the W we cannot have an X, or a vR in one orientation (these leave holes)
So one of these 4:

+---+---+---+---+---+---+---+---+---+---+
| | W | P : P P | R | |
+ + + + +---+ + + +---+ +
| | W W | P P | R R | |
+ + + +---+ +---+---+ +---+ +
| | W : W | R R | |
+---+---+---+---+---+---+---+---+---+---+

Inverting the R requires 2 Ws.

+---+---+---+---+---+---+---+---+---+---+
| | W | Y : Y Y Y | Z | V |
| + + + +---+ +---+---+ + +
| | W W | Y | Z Z Z | V |
+ + + +---+ +---+ +---+---+ +
| | W : W | Z | V V V |
+---+---+---+---+---+---+---+---+---+---+

(Inverting the ZV leaves holes, Y is forced)
We will discard this below (because the O cannot be place)

+---+---+---+---+---+---+---+---+---+---+
| | W | P : P P | T | |
+ + + + +---+ + + + + +
| | W W | P P | T | |
+ + + +---+ +---+---+ +---+ +
| | W : W | T T T | |
+---+---+---+---+---+---+---+---+---+---+

(inverting the T requires 2 Ws, P is forced)

+---+---+---+---+---+---+---+---+---+---+
| | W | O : O O O O | T |
+ + + + +---+---+---+---+---+ +
| | W W | S S | T T T |
+ + + +---+ +---+ +---+---+ +
| | W : W | S S S | T |
+---+---+---+---+---+---+---+---+---+---+

(S and O are forced)

================================================================================
+---+---+---+---+---+---+---+---+---+---+
| | W | Y : Y Y Y | Z | V |
| + + + +---+ +---+---+ + +
| | W W | Y | Z Z Z | V |
+ + + +---+ +---+ +---+---+ +
| | W : W | Z | V V V |
+---+---+---+---+---+---+---+---+---+---+

cannot have a vT,vR and X because that would prevent placing to O
+---+---+---+---+---+---+---+---+---+---+
| | R | W | Y : Y Y Y | Z | V |
| +---+ + +---+ +---+---+ + +
| | R R | W W | Y | Z Z Z | V |
+ +---+ +---+ +---+ +---+---+ +
| | R R | W : W | Z | V V V |
+---+---+---+---+---+---+---+---+---+---+

So we must have hT vR and X. We can place the vR.
Note that the open side of the vR looks like the side of the X.
If we put the hT next to the X we cannot fit the O,
If we put the hT next to the V we also cannot fit the O.
So this configuration cannot be part of the solution.
================================================================================
The remaining three possibilities for the right of the W contain a T or an R,
This forces the left of the W (which must be an R or a T)
+---+---+---+---+---+---+---+---+---+---+
| | T | W | P : P P | R | |
+ + + + +---+ + + +---+ +
| | T | W W | P P | R R | |
+ +---+ +---+ +---+---+ +---+ +
| | T T T | W : W | R R | |
+---+---+---+---+---+---+---+---+---+---+

+---+---+---+---+---+---+---+---+---+---+
| | R | W | P : P P | T | |
+ +---+ + +---+ + + + + +
| | R R | W W | P P | T | |
+ +---+ +---+ +---+---+ +---+ +
| | R R | W : W | T T T | |
+---+---+---+---+---+---+---+---+---+---+

The above are mirror images of each other and use the same tiles.
We need only consider the second and all conclusions follow for the first by symmetry.

+---+---+---+---+---+---+---+---+---+---+
| | R | W | O : O O O O | T |
+ +---+ + +---+---+---+---+---+ +
| | R R | W W | S S | T T T |
+ +---+ +---+ +---+ +---+---+ +
| | R R | W : W | S S S | T |
+---+---+---+---+---+---+---+---+---+---+

================================================================================
The two cases remaining use the R and the T, so we are left to place the hZV block and the X.

Also both cases remaining R in the same orientation on the end.
This cannot be next to the X because the only way to fill the resulting hole is with 2 Us, 2 Ys or 2 Qs.
It cannot face the V side of the VZ block because that would require a U and leave the X on the end requiring a second U.
It cannot face the Z side of the VZ block because that would leave the X to require 2 Us
(Can somebody clean up the above using a counting argument instead of an enumeration).
The only remaining possibility is to cap the R with a U and make it the end of the pattern. But this rules out the
second pattern because it leaves the X facing the hT or the other end requiring a second U.

So we are down one remaining possibility (plus the symmetry) with the hZV and X to be placed:
+---+---+---+---+---+---+---+---+---+---+
| U U | R | W | P : P P | T | :
+ +---+ + +---+ + + + + +
| U | R R | W W | P P | T | :
+ +---+ +---+ +---+---+ +---+ +
| U U | R R | W : W | T T T | :
+---+---+---+---+---+---+---+---+---+---+


Since the U is taken we must place the X next and the V on the end.
Note that the ZV block can be inverted but the symmetry of the X means we don't
need to consider both cases.

+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| U U | R | W | P : P P | T | : | X | : | Z | V V V |
+ +---+ + +---+ + + + + + +---+ +---+ + + +---+---+ +
| U | R R | W W | P P | T | : | X X X | : | Z Z Z | V |
+ +---+ +---+ +---+---+ +---+ + +---+ +---+ + +---+---+ + +
| U U | R R | W : W | T T T | : | X | : | Z | V |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+

Placing the O leaves a P shaped hole, but the P is already used

So case A does not work.

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Re: Pentominoes packing a 3x20 rectangle

#386304

Postby Gengulphus » February 13th, 2021, 9:34 am

9873210 wrote:
Gengulphus wrote:A tightening up of 9873210's result:

That analysis suggests splitting into two cases to me:

A) R is vertical and Z is horizontal, in which case R, T, X and Z occupy the four 3x5 rectangles and V must share Z's 3x5 rectangle, while W must be fitted in between two of those rectangles.

B) R is horizontal, in which case T, X and Z occupy three of the four 3x5 rectangles, and R, W and V must all be fitted between them.

Of course, splitting into cases is undesirable, but at least there are only two of them, and furthermore, case A) looks quite tightly constrained to me, so might be quite quickly eliminated...

Gengulphus


A) R is vertical and Z is horizontal, in which case R, T, X and Z occupy the four 3x5 rectangles and V must share Z's 3x5 rectangle, while W must be fitted in between two of those rectangles.

We need to fit the following togeather
+---+---+---+---+---+   
| | R R | |
+ +---+ +---+ +
| | R R | |
+ + + +---+ +
| | R | |
+---+---+---+---+---+

+---+---+---+---+---+
| V | Z | |
+ + +---+---+ +
| V | Z Z Z | |
+ +---+---+ + +
| V V V | Z | |
+---+---+---+---+---+

+---+---+---+---+---+
| | X | |
+ +---+ +---+ +
| | X X X | |
+ +---+ +---+ +
| | X | |
+---+---+---+---+---+

+---+---+---+---+---+ +---+---+---+---+---+
| | T T T | | | T | |
+ +---+ +---+ + + +---+---+ + +
| | T | | or| T T T | |
+ + + + + + + +---+---+ + +
| | T | | | T | |
+---+---+---+---+---+ +---+---+---+---+---+

+---+---+---+---+---+---+---+---+---+---+
| | W | : |
+ + + + +---+ + + + + +
| | W W | |
+ + + +---+ +---+ + + + +
| | W : W | |
+---+---+---+---+---+---+---+---+---+---+

(I will treat the orientation of W as fixed, everything else can rotate/reflect)
...
So case A does not work.

Yes, I think your argument works. But it can be simplified in at least one way, namely by applying the middle-row argument: the 'horizontal T' possibility can be eliminated, because if we use it, then those six pentominoes occupy 2+4+3+3+2 = 14 middle-row squares. The remaining pentominoes require no middle-row squares for the O, at least 1 for each of the Q, U and Y and at least 2 for each of the P and S, for a total of at least another 7 middle-row squares, or at least 21 middle-row squares in total, when there are only 20.

I may be on the track of a consequential simplification, but haven't worked it out fully yet...

Gengulphus

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Re: Pentominoes packing a 3x20 rectangle

#386346

Postby 9873210 » February 13th, 2021, 11:21 am

Gengulphus wrote:
9873210 wrote:
Gengulphus wrote:A tightening up of 9873210's result:
...
So case A does not work.

Yes, I think your argument works. But it can be simplified in at least one way, namely by applying the middle-row argument: the 'horizontal T' possibility can be eliminated, because if we use it, then those six pentominoes occupy 2+4+3+3+2 = 14 middle-row squares. The remaining pentominoes require no middle-row squares for the O, at least 1 for each of the Q, U and Y and at least 2 for each of the P and S, for a total of at least another 7 middle-row squares, or at least 21 middle-row squares in total, when there are only 20.

I may be on the track of a consequential simplification, but haven't worked it out fully yet...

Gengulphus

Considerable simplification.
The vR,vT and X fill the center column of 3 of the 3x5blocks. The O is also forced to be above the Z, with the P below the O (although the orientation of the P is not known. This fills the central column of the remaining 3x5block.

The central columns being filled gives us the following, where x means is used and the dots might not be used.

,,x,,,,x,,,,x,,,,x,,
,,x,,,,x,,,,x,,,,x,,
,,x,,,,x,,,,x,,,,x,,

One of the 3x4 empty blocks is filled by the P,O,Z and a single square from an adjacent tile.
One of the others contains the W which spans all 3 rows.
So the only empty space that is 4 wide is at most a single 4x3 block which must contain the Q,Y and S. And 15 into 12 (actually 10) won't go.

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Re: Pentominoes packing a 3x20 rectangle

#386359

Postby Gengulphus » February 13th, 2021, 11:48 am

Gengulphus wrote:Yes, I think your argument works. But it can be simplified in at least one way, namely by applying the middle-row argument: the 'horizontal T' possibility can be eliminated, because if we use it, then those six pentominoes occupy 2+4+3+3+2 = 14 middle-row squares. The remaining pentominoes require no middle-row squares for the O, at least 1 for each of the Q, U and Y and at least 2 for each of the P and S, for a total of at least another 7 middle-row squares, or at least 21 middle-row squares in total, when there are only 20.

I may be on the track of a consequential simplification, but haven't worked it out fully yet...

I've now done that, and it's worked out quite a lot shorter, and I think simpler. Having eliminated the possibility of the T being horizontal as above, we need to fit the following four 3x5 rectangles together (I won't bother looking at the W in this argument):

+---+---+---+---+---+      +---+---+---+---+---+
| | R R | | | V | Z | |
+ +---+ +---+ + + + +---+---+ +
| | R R | | | V | Z Z Z | |
+ + + +---+ + + +---+---+ + +
| | R | | | V V V | Z | |
+---+---+---+---+---+ +---+---+---+---+---+

+---+---+---+---+---+ +---+---+---+---+---+
| | X | | | | T T T | |
+ +---+ +---+ + + +---+ +---+ +
| | X X X | | | | T | |
+ +---+ +---+ + + + + + + +
| | X | | | | T | |
+---+---+---+---+---+ +---+---+---+---+---+

in which I'll treat the Z+V rectangle's orientation as fixed, while the other three rectangles can be rotated or reflected.

Where can the Z+V rectangle go? It cannot be the rightmost of the four rectangles, as that would force the empty space at its right end to be filled with a second V. So the rightmost rectangle must be one of the others, which in each case forces the empty space at its right end to be filled with a U or a P.

The Z+V rectangle also cannot be the second from the left, since then both ends of the leftmost rectangle would be forced to be filled with a U or a P, and there aren't enough Us and Ps to go around! And similarly, it cannot be the third from the left, because the empty spaces at the left end of the leftmost rectangle and the right end of the second-from-left rectangle would each have to be filled with a U or a P.

So the Z+V rectangle has to be the leftmost. Now look at the 10-square empty spaces between the four rectangles. The one between the Z+V rectangle and the second-from-left rectangle must be of one of the shapes (in which asterisks indicate squares occupied by one of R, X and T):

+---+---+---+---+---+---+---+---+...      +---+---+---+---+---+---+---+---+...
| V | Z | | * * ... | V | Z | | * ...
+ + +---+---+ + +---+ +... + + +---+---+ + +---+ +...
| V | Z Z Z | | * ... | V | Z Z Z | | * * ...
+ +---+---+ + + + + +... + +---+---+ + + +---+ +...
| V V V | Z | | * ... | V V V | Z | | * ...
+---+---+---+---+---+---+---+---+... +---+---+---+---+---+---+---+---+...

+---+---+---+---+---+---+---+---+...
| V | Z | | * ...
+ + +---+---+ + + + +...
| V | Z Z Z | | * ...
+ +---+---+ + + +---+ +...
| V V V | Z | | * * ...
+---+---+---+---+---+---+---+---+...

and the other two must each be of one of the shapes:

...+---+---+---+---+---+---+...      ...+---+---+---+---+---+---+...
... * | | * ... ... * | | * * ...
...+ +---+ + +---+ +... ...+ +---+ + +---+ +...
... * * | | * * ... ... * * | | * ...
...+ +---+ + +---+ +... ...+ +---+ + + + +...
... * | | * ... ... * | | * ...
...+---+---+---+---+---+---+... ...+---+---+---+---+---+---+...

...+---+---+---+---+---+---+... ...+---+---+---+---+---+---+...
... * * | | * * ... ... * * | | * ...
...+ +---+ + +---+ +... ...+ +---+ + + + +...
... * | | * ... ... * | | * ...
...+ + + + + + +... ...+ + + + +---+ +...
... * | | * ... ... * | | * * ...
...+---+---+---+---+---+---+... ...+---+---+---+---+---+---+...

Now, the O can only fit into two of the three possibilities for the first of those empty spaces, and whichever it is placed in, the other pentomino in that empty space is forced to be the P.

That done, the Q can only fit into three of the four possibilities for one of the other empty spaces, and whichever way it is fitted into that space, either the remainder of that space is broken up into unfillable holes or the other pentomino in that space has to be a P.

So any attempt to complete case A fails because it contains two Vs, or it contains three pentominoes that are either U or P, or it doesn't contain at least one of O and Q, or it contains unfillable holes, or it contains two Ps.

Gengulphus

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Re: Pentominoes packing a 3x20 rectangle

#386365

Postby 9873210 » February 13th, 2021, 12:00 pm

So far we have:

1) O is in the top or bottom row.
2) The long arms of the Q and Y are in the top or bottom row.
3) R is horizontal
4) At least one of T and S is vertical.

This nails down 18 of the center row squares. If either T or S is horizontal we have a complete count and know something about the orientation of the remainder P S and Q. Conversely if P S or Q is not minimal in the center row then both T and S are vertical.
Last edited by 9873210 on February 13th, 2021, 12:14 pm, edited 1 time in total.

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Re: Pentominoes packing a 3x20 rectangle

#386369

Postby Gengulphus » February 13th, 2021, 12:03 pm

9873210 wrote:Considerable simplification.
The vR,vT and X fill the center column of 3 of the 3x5blocks. The O is also forced to be above the Z, with the P below the O (although the orientation of the P is not known. This fills the central column of the remaining 3x5block.

The central columns being filled gives us the following, where x means is used and the dots might not be used.

,,x,,,,x,,,,x,,,,x,,
,,x,,,,x,,,,x,,,,x,,
,,x,,,,x,,,,x,,,,x,,

One of the 3x4 empty blocks is filled by the P,O,Z and a single square from an adjacent tile.
One of the others contains the W which spans all 3 rows.
So the only empty space that is 4 wide is at most a single 4x3 block which must contain the Q,Y and S. And 15 into 12 (actually 10) won't go.

Quite a bit there in common with my less-succinctly expressed simplification, but bringing in the O at an earlier stage than I did and the use of the W are both nice, making your simplification even simpler than mine (e.g. it eliminates my 4-way case split of the possible shapes of the two empty spaces between the R, T and X pentominoes). I'm leaving my post as it is, though: it may contain ideas that are useful in solving case B.

Gengulphus

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Re: Pentominoes packing a 3x20 rectangle

#386485

Postby 9873210 » February 13th, 2021, 7:10 pm

Again look at the 3x5 boxes. Label them A,B,C,D.
There are 3 internal boundaries. Label them A:B B:C and C:D
Label the two ends |A and D|
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| : : : |
+ + + + + + + + + + + + + + + + + + + + +
|A A A:B B B:C C C:D D D|
+ + + + + + + + + + + + + + + + + + + + +
| : : : |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+

Now hR and W must cross boundaries.
They do not nest, so they must cross different boundaries.
That accounts for at least one edge of each of B and C.
X does not nest with any of the four possible configurations of W and hR.
Thus X in either A or D. Wlog (Without Loss of Generality) put it in A.
U must be vertical against |A. And B:C and C:D are crossed by hR and W (in either order).
V cannot be against A:B because in one direction it overlaps X and in the other it requires a second U.
So L is against D| and wlog we can put the long arm at the top.
(The two wlogs have used the fact that rotations and reflections are considered
equivalent, they have also used up all the symmetry so we can't wlog again.)


+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| U U | X | : : : | V V V |
+ +---+ +---+ + + + + + + + + + + + + +---+---+ +
|AU | X X X | A:B B:C C:D | VD|
+ +---+ +---+ + + + + + + + + + + + + + + + +
| U U | X | : : : | V |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+


T must be vertical because hT cannot abut A:B (Overlap X or use a second U) or B:C or C:D (overlap W or hR)
P must be horizontal for the same reason.

You'd think it would be easy at this point, but apparently one more trick is required.

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Re: Pentominoes packing a 3x20 rectangle

#386506

Postby Gengulphus » February 13th, 2021, 9:56 pm

9873210 wrote:
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| U U | X | : : : | V V V |
+ +---+ +---+ + + + + + + + + + + + + +---+---+ +
|AU | X X X | A:B B:C C:D | VD|
+ +---+ +---+ + + + + + + + + + + + + + + + +
| U U | X | : : : | V |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+


T must be vertical because hT cannot abut A:B (Overlap X or use a second U) or B:C or C:D (overlap W or hR)
P must be horizontal for the same reason.

You'd think it would be easy at this point, but apparently one more trick is required.

Suppose Z is vertical. Then the vertical T and vertical Z must be placed centrally in boxes B and C (in one order or the other) and the empty space between them must be of one of the following forms (with the asterisks on one side being T and on the other side being Z):

...+---+---+---+---+---+---+...      ...+---+---+---+---+---+---+...
... * * | : | * * ... ... * * | : | * ...
...+ +---+ + +---+ +... ...+ +---+ + + + +...
... * | B:C | * ... ... * | B:C | * ...
...+ + + + + + +... ...+ + + + +---+ +...
... * | : | * ... ... * | : | * * ...
...+---+---+---+---+---+---+... ...+---+---+---+---+---+---+...

Either W or horizontal R must be in that empty space, and it's easy to see that horizontal R fragments either possibility into unfillable holes no matter how it's placed. W can be fitted into either of them without such fragmentation, but the only ways to do it in the right-hand space require a second W, while the only ways to do it in the left-hand space require P to accompany the W. So the empty space must be of the left-hand form and filled with P and W.

Now look at the empty space between A and B. It must be of the following form (where again asterisks are T or Z):

+---+---+---+---+---+---+---+---+...
| U U | X | : | * ...
+ +---+ +---+ + + + +...
| U | X X X | A:B | * ...
+ +---+ +---+ + +---+ +...
| U U | X | : | * * ...
+---+---+---+---+---+---+---+---+...

We need to fill that space using two pentominoes. The first square on the top row of the space and the first square on the bottom row of the space must be in different pentominoes, since otherwise they would both have to be in a second U. The first square on the middle row of the space therefore has to be in the same pentomino as one of them: depending which one it is, it's easy to see that the division of the space must be one of the following two:

+---+---+---+---+      +---+---+---+---+
| Y Y Y Y | | Q Q Q Q |
+---+ +---+---+ +---+---+---+ +
| Y | S S | | P P | Q |
+---+---+ +---+ +---+ + +---+
| S S S | | P P P |
+---+---+---+ +---+---+---+

The second of those would duplicate P, so it must be the first of them.

So far, we've used the P, S, T, U, V, W, X, Y and Z pentominos, leaving O, Q and R to fill the remaining empty space surrounding C:D. However, that space always has 6 squares on the middle row (2 in box C, 4 in box D), O doesn't occupy any of them, Q occupies 1 of them, and so R must occupy all of the remaining 5 - which is of course impossible.

So Z must be horizontal.

Gengulphus

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Re: Pentominoes packing a 3x20 rectangle

#386514

Postby 9873210 » February 13th, 2021, 10:41 pm

Gengulphus wrote:
...

So Z must be horizontal.

Gengulphus

You can shorten that slightly.

If Z is vertical the the count on the middle row requires both P and S to have 3 squares on the middle row. (The counts for everything but Z.P and S are already known) Thus you need to consider only one of the A:B or B:C gaps since either one forces the P or S have the wrong orientation.

SInce Z is horizontal we can deduce that S and P each have 2 squares on the middle row.

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Re: Pentominoes packing a 3x20 rectangle

#386545

Postby 9873210 » February 14th, 2021, 6:42 am

I guess this is the part where you just have to slog through it
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| U U | X | : : : | V V V |
+ +---+ +---+ + + + + + + + + + + + + +---+---+ +
|AU | X X X | A:B B:C C:D j j i | VD|
+ +---+ +---+ + + + + + + + + + + + + + + + +
| U U | X | : : : b b b a | V |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
consier squares a and i.
If a and i are different then we need to extend i through j and a through b.
This would mean that i is part of S,Y,Q or O. Bt we we know that the long sides
of all of these are not in the middle row. So i and a are part of the same tile.
an exhaustive search leaves me with P,Q and Z
============================================================================================
Case P.
The count in the middle row tells us which way up P is.
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| U U | X | : : : | V V V |
+ +---+ +---+ + + + + + + + + + + + + +---+---+ +
|AU | X X X | A:B 1B:C C:D | p p | VD|
+ +---+ +---+ + + + + + + + + + + + +---+ + + +
| U U | X | : : : | p p p | V |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
Z and T go in boxes B and C
Z is horizontal. If Z goes in B the shape between X and Z will be a mirror image
for either orientation. It can only filled with UV or PO, but P and U are already
in use.
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| U U | X | : | Z | : : | V V V |
+ +---+ +---+ + +---+---+ + + + + + + + + +---+---+ +
|AU | X X X | A:B | Z Z Z | 1B:C C:D | p p | VD|
+ +---+ +---+ + + +---+---+ + + + + + + +---+ + + +
| U U | X | : | Z | : : | p p p | V |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
So T goes in B and Z in C
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| U U | X | : T a a : a a ab b b : b b | V V V |
+ +---+ +---+ + + + + + + + +---+ + + + +---+---+ +
|AU | X X X | A:B | T | 1B:C | Z Z Z | C:D | P P | VD|
+ +---+ +---+ + + + + + + + +---+ + + +---+ + + +
| U U | X | : T c c : c c c : | P P P | V |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
O must be in one of the positions aaaaa,bbbbb or ccccc. But any one of these
interferes with placing W and R on the B:C and C:D boundaries.
So P is ruled out.
========================================================================================
Case Q

Again the center row count tells us which way up Q is
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| U U | X | : : : | V V V |
+ +---+ +---+ + + + + + + + + + + + + +---+---+ +
|AU | X X X | A:B B:C C:D | Q | VD|
+ +---+ +---+ + + + + + + + + + + +---+---+---+ + +
| U U | X | : : | Q Q Q Q | V |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
Putting hR or W on the C:D boundary forces R and S.
T and Z in the centers of B and C force T.
Which forces W on the B:C boundary
Which gives the orientation of Z, and forces Y
O and P are trivial.
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| U U | X | P P : P | Z | Y Y Y : Y | W | T | R | S : S S | V V V |
+ +---+ +---+ + + +---+---+ +---+ + + +---+---+ +---+---+ +
|AU | X X X | PA:BP | Z Z Z | YB|CW W | T | R RC:DR | S S | q | VD|
| +---+ +---+---+---+---+---+ +---+ +---+ +---+ +---+---+---+ + +
| U U | X | O O : O O O | Z | W : W | T T T | R | q q q q | V |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
Ding Ding we have a winner!!!!!!!!

========================================================================================
case Z

The Z forces W on the C:D boundary
Which forces Y
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| U U | X | : : | W : W | Z | V V V |
+ +---+ +---+ + + + + + + + + +---+ +---+ +---+---+ +
|AU | X X X | A:B B:C | W WC|DY | Z Z Z | VD|
+ +---+ +---+ + + + + + + + + + +---+ +---+---+ + +
| U U | X | : : | W | Y : Y Y Y | Z | V |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+

Case Z/T in box B
T goes in box B or C. Try Box B
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| U U | X | : T a a : a a | W : W | Z | V V V |
+ +---+ +---+ + + + + + + + + +---+ +---+ +---+---+ +
|AU | X X X | A:B | T | B:C | W WC|DY | Z Z Z | VD|
+ +---+ +---+ + + + + + + + + + +---+ +---+---+ + +
| U U | X | : T b b : b b | W | Y : Y Y Y | Z | V |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
The hR on the B:C boundary uses exactly one of the four a squares and exactly
one of the four b squares. The following is the only option that allows O to be
placed
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| U U | X | : | T | R | O : O O O O | W : W | Z | V V V |
+ +---+ +---+ + + + + +---+---+---+---+---+ +---+ +---+---+ +
|AU | X X X | A:B | T | R RB:CR | | W WC|DY | Z Z Z | VD|
+ +---+ +---+ + +---+ +---+ +---+ + + +---+ +---+---+ + +
| U U | X | : | T T T | R | | W | Y : Y Y Y | Z | V |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
Q is forced but that leaves two Ps and no S so no solution here.

Case Z/in box C
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| U U | X | : : | T T T | W : W | Z | V V V |
+ +---+ +---+ + + + + + + +---+ +---+ +---+ +---+---+ +
|AU | X X X | A:B B:C | T | W WC|DY | Z Z Z | VD|
+ +---+ +---+ + + + + + + + + + +---+ +---+---+ + +
| U U | X | : : | T | W | Y : Y Y Y | Z | V |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
hR on B:C is forced
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| U U | X | : | R | T T T | W : W | Z | V V V |
+ +---+ +---+ + + + + +---+ +---+ +---+ +---+ +---+---+ +
|AU | X X X | A:B | RB:CR R | T | W WC|DY | Z Z Z | VD|
+ +---+ +---+ + + + + +---+---+ + + +---+ +---+---+ + +
| U U | X | : : | R | T | W | Y : Y Y Y | Z | V |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
O P Q S are left. These fill 5,3,4 and 3 squares in the top and bottom rows.
We need sets that add up to 7 and 8 so Q is in the top row and O is in the bottom row.
Try two the possible positons for Q
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| U U | X | Q Q : Q Q | | R | T T T | W : W | Z | V V V |
+ +---+ +---+---+---+ + + +---+ +---+ +---+ +---+ +---+---+ +
|AU | X X X | SA:BS | Q | | RB:CR R | T | W WC|DY | Z Z Z | VD|
+ +---+ +---+ + +---+ + +---+---+ + + +---+ +---+---+ + +
| U U | X | : S S S : | R | T | W | Y : Y Y Y | Z | V |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
This forces S to a place that rules out O.
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| U U | X | P P : P | Q Q Q Q | R | T T T | W : W | Z | V V V |
+ +---+ +---+ + + +---+---+---+ +---+ +---+ +---+ +---+---+ +
|AU | X X X | PA:BP | Q | S S | RB:CR R | T | W WC|DY | Z Z Z | VD|
+ +---+ +---+---+---+ +---+ +---+---+ + + +---+ +---+---+ + +
| U U | X | O O : O O O | S S : S | R | T | W | Y : Y Y Y | Z | V |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
Q forces P forces S forces O and we have another winner.

So we have exactly two unique solutions (and the reflections and rotations).
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| U U | X | P P : P | Q Q Q Q | R | T T T | W : W | Z | V V V |
+ +---+ +---+ + + +---+---+---+ +---+ +---+ +---+ +---+---+ +
|AU | X X X | PA:BP | Q | S S | RB:CR R | T | W WC|DY | Z Z Z | VD|
+ +---+ +---+---+---+---+---+ +---+---+ + + +---+ +---+---+ + +
| U U | X | O O : O O O | S S : S | R | T | W | Y : Y Y Y | Z | V |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| U U | X | P P : P | Z | Y Y Y : Y | W | T | R | S : S S | V V V |
+ +---+ +---+ + + +---+---+ +---+ + + +---+---+ +---+---+ +
|AU | X X X | PA:BP | Z Z Z | YB|CW W | T | R RC:DR | S S | q | VD|
| +---+ +---+---+---+---+---+ +---+ +---+ +---+ +---+---+---+ + +
| U U | X | O O : O O O | Z | W : W | T T T | R | q q q q | V |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+

N


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