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[math-fun] perfect packings versus tilings
by James Propp 21 Feb '06

21 Feb '06

>PS, I like Klarner's distinction: perfect packing -> tiling. Can someone say what Klarner's distinction is? (I have used the two phrases interchangeably in my own work.) Jim Propp

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Re: [math-fun] AB=B.AA; Gosper's packing problem
by dasimov＠earthlink.net 21 Feb '06

21 Feb '06

Rich writes: << Gosper's packing puzzle is to fit as many 4x4x1 bricks as possible into a 7x7x7 cube. I'm assuming an "orthogonal, registered" fit, with each brick oriented parallel to a cube face, and placed at integral subcube boundaries. The answer is 18 bricks. [Here is a packing of 18.] My proof is the same as Michael Reid's: Consider the (orthogonal) lines going through the center of the 7x7x7 cube. Take the union, remove the center cell. Resulting collection of subcubes has 18 cells. Any 4x4x1 brick must occupy at least one of these cells. Note that the simple volume constraint gives an upper bound of 7^3/4^2 = 343/16 = 21.4375. >> Cute proof. As usual I'm interested in the torus version of this cube puzzle. QUESTION: Suppose we identify the opposite faces of this 7x7x7 cube, to get a 7x7x7 three-torus T. How many registered* 4x4x1 bricks will fit ? (Of course from the cube solution & the volume constraint, the answer lies in {18, 19, 20, 21}. And the proof of Michael & Rich for the cube no longer works for the torus.) --Dan ____________________________________________ * To register your bricks, just send me a check for $50.

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P.S. Re: [math-fun] Fractals and (real) snowflakes
by dasimov＠earthlink.net 20 Feb '06

20 Feb '06

3. I also wonder: What is this "fractal analysis"; Anyway? I'd expect someone trying to find a "fingerprint" of Pollock's word to use Fourier analysis: taking Fourier transforms of his known works and comparing these with the ones in question. But "fractal analysis" of paintings ???? Sounds kind of loosey-goosey to me. --dan

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[math-fun] Fractals and (real) snowflakes
by dasimov＠earthlink.net 17 Feb '06

17 Feb '06

Last week the NY Times ran an article about how some physics professor was analyzing allegedly genuine Jackson Pollock paintings by using "fractal geometry". One line reads: << In previous years Dr. Taylor examined 14 indisputably authentic Pollock paintings by using what is known as fractal geometry, or looking for patterns that recur on finer and finer magnifications, like those in snowflakes. >> This is the first that I heard that snowflakes have patterns that recur on finer and finer magnifications. Needless to say, many interesting fractals can be constructed from recursive algorithms that do in fact end up with identical patterns on finer and finer scales. But I hadn't heard this is a necessary ingredient of a fractal shape. (Technically, there seems to be no fixed definition of a fractal. Everyone agrees that a subset of say the plane having fractional Hausdorff dimension is a fractal. But there are also subsets that have Hausdorff dimension = 1 that most agree should also be called fractals. (E.g., starting with one solid square, divide each (remaining) square into a 2x2 pattern of half-size squares, and discard the NW and SE ones. Then divide each (remaining) square each into a 2x2 pattern, but this time discard the SW and NE ones for each 2x2 pattern. Iterate indefiniitely. The limiting set has Hausdorff dim = 1.) 1, Is This True About (real) Snowflakes??? 2, Does Every Fractal have (in some sense) Repeating Patterns at Finer & Finer Scales? --Dan

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Re: [math-fun] AB=B.AA
by Christian G.Bower 15 Feb '06

15 Feb '06

I think I have a formula for the labeled (tabtot) count: For a partition of n=1*a_1+2*a_2+3*a_3+...+j*a_j into k parts k=a_1+a_2+...+a_j we have PROD {i=1 to j} j^a_j * n^(k*(k-1)/2) * n! / PROD {i=1 to j} ((j!)^a_j * a_j!) To get a_labeled(n) sum these products for each partition of n. I get: 1: Total 1 Idem 0 1 2: Total 4 Idem 0 2 2 3: Total 48 Idem 0 3 18 27 4: Total 4964 Idem 0 4 96 768 4096 5: Total 10089780 Idem 0 5 400 11250 312500 9765625 6: Total 472010285382 Idem 0 6 1440 116640 11197440 1813985280 470184984576 7: Total 558745410230718214 Idem 0 7 4704 972405 263533760 148299505725 199397583417606 558545864083284007 The unlabeled (dist) version I'm sure can be done through Polya enumeration, but I doubt I have the patience to work out the formula. I'm wondering about the associative version of this. If we added associativity to the requirements, I suspect what we get could be described as a meet semilattice of pointed sets, but I will need to do more investigation to verify that. Certainly it would be dominated by the commutative idempotent semigroups which are just meet semilattices. Christian ------ Original Message ------ Received: Tue, 14 Feb 2006 12:49:42 PM PST From: "Schroeppel, Richard" <rschroe(a)sandia.gov> To: math-fun(a)mailman.xmission.comCc: rcs(a)cs.arizona.edu Subject: [math-fun] AB=B.AA; Gosper's packing problem > Good work on the AB=B.AA problem! > This might even lead to a formula for the number of tables of each size. > My data for N=2,3,4,5 is > 2: > dist 2 tabtot 4 > idemspec 0 1 1 > isospec 1.2 > 3: > dist 11 tabtot 48 > idemspec 0 1 3 7 > isospec 1.6 2.3 3 6 > 4: > dist 234 tabtot 4964 > idemspec 0 1 5 36 192 > isospec 1.185 2.38 3.6 6.5 > 5: > dist 85196 tabtot 10089780 > idemspec 0 1 7 108 2725 82355 > isospec 1.83035 2.2023 3.57 4.34 5.6 24.3 6.36 8 20 > > Dist is the number of distinct, non-isomorphic tables; Tabtot is the > number of tables with isomorphic tables counted. Idemspec is the number > of (non-isomorphic) tables with each number of idempotents -- the "all > elements are idempotent" class seems dominant. Isospec reports on the > number of tbales with each (size of) automorphism group: For the 85196 > tables with five elements, 83035 have no automorphisms (group size =1, self), > while three tables have a size-24 group, and there's one table each for > group sizes 8 and 20. > > The observations that squares (AA) are idempotent, and that an element and its > square have identical rows and columns, constrains the table structure quite > a bit. >

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[math-fun] torus packing
by Michael Reid 15 Feb '06

15 Feb '06

> Can anybody find a brick-packing problem where you can fit more bricks > into the (3D) torus than you can into the cube? Bricks are AxBxC, not > arbitrary polycubes. How about the 2D version? twenty-five 1x6 rectangles can (perfectly) pack a 10x15 torus, whereas only twenty-four fit into the corresponding non-toroidal rectangle. i believe this example is given by golomb in his book, "polyominoes", but i don't have it at my fingertips at the moment. mike

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Re: [math-fun] AB=B.AA; Gosper's packing problem
by dasimov＠earthlink.net 15 Feb '06

15 Feb '06

Franklin writes: << It's still 18. Note first that each 7x7x1 plane can contain at most one entire brick, so in order to fit more than 18 bricks, one of the directions must have one entire brick in each plane; call these the base set. Looking at each of the other dimensions, there are 28 1x4 cross-sections of the base set distributed amongst the 7 planes. Since no plane can have more than 7 of these (one from each brick in the base set), there must be at least two planes with at least 4 such cross-sections (5*3+6+7=28). A little fiddling will show that it is impossible to fit a brick and 4 parallel cross-sections into a single plane (they have to be parallel because they come from bricks aligned in the same dimension). So each of the other dimensions can have at most 5 bricks, for 17 total, contrary to our assumption. >> That's pretty subtle reasoning --nice! Do you also know the story for how many NxNx1's can fit in a (2N-1)-cube, as you suggested? (Or torus?) Is the question of how many registered KxLxM's can fit in a PxQxR cube (or torus) solved in general? --Dan

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[math-fun] magma
by Christian G.Bower 14 Feb '06

14 Feb '06

So magma is a name used for that. I was introduced to it as "groupoid" in the 1968 book _Introduction to Algebraic Structures_ by Azriel Rosenfeld. The EIS has called them groupoids since before I knew about it. (The wikipedia article mentions the name also.) Then there's that annoying fact that "groupoid" also refers to group-like categories, so maybe a new name is needed to avoid confusion. Then, too often, the things that should avoid confusion only create more. Regarding Rich's counts, since commutivity + idempotent ==> Rich's condition, the all idempotent case is given by A030257. Christian ------ Original Message ------ Received: Tue, 14 Feb 2006 04:54:40 AM PST From: dasimov(a)earthlink.net To: math-fun <math-fun(a)mailman.xmission.com> Subject: Re: [math-fun] AB = B.AA seems to imply commutativity? > Rich writes: > > << > I've been exploring the binary operator *, which satisifies the > rule A*B = B*(A*A). I like to write it in dot notation as AB = B.AA . [snip] > > I just learned the word "magma" for a set S together with a binary operation *. > Call this magma (S,*). > ...

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[math-fun] how many 4x4x1 blocks
by R. William Gosper 14 Feb '06

14 Feb '06

fit in a 7x7x7? You might be surprised, especially if you have the physobs! (I.e., nice puzzle.) My guess: ceiling(pi^2/Euler's constant). --rwg

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[math-fun] AB=B.AA; Gosper's packing problem
by Schroeppel, Richard 14 Feb '06

14 Feb '06

Good work on the AB=B.AA problem! This might even lead to a formula for the number of tables of each size. My data for N=2,3,4,5 is 2: dist 2 tabtot 4 idemspec 0 1 1 isospec 1.2 3: dist 11 tabtot 48 idemspec 0 1 3 7 isospec 1.6 2.3 3 6 4: dist 234 tabtot 4964 idemspec 0 1 5 36 192 isospec 1.185 2.38 3.6 6.5 5: dist 85196 tabtot 10089780 idemspec 0 1 7 108 2725 82355 isospec 1.83035 2.2023 3.57 4.34 5.6 24.3 6.36 8 20 Dist is the number of distinct, non-isomorphic tables; Tabtot is the number of tables with isomorphic tables counted. Idemspec is the number of (non-isomorphic) tables with each number of idempotents -- the "all elements are idempotent" class seems dominant. Isospec reports on the number of tbales with each (size of) automorphism group: For the 85196 tables with five elements, 83035 have no automorphisms (group size =1, self), while three tables have a size-24 group, and there's one table each for group sizes 8 and 20. The observations that squares (AA) are idempotent, and that an element and its square have identical rows and columns, constrains the table structure quite a bit. --------- Gosper's packing puzzle is to fit as many 4x4x1 bricks as possible into a 7x7x7 cube. I'm assuming an "orthogonal, registered" fit, with each brick oriented parallel to a cube face, and placed at integral subcube boundaries. The answer is 18 bricks. A view DOWN on the bottom half of the packing is AAAABBB AAAABBB AAAABBB CCCCBBB CCCC--- CCCC--- CCCC--- where the As and Bs are sideways stacks of 3 bricks each, and the Cs are a vertical stack of 3 bricks. My proof is the same as Michael Reid's: Consider the (orthogonal) lines going through the center of the 7x7x7 cube. Take the union, remove the center cell. Resulting collection of subcubes has 18 cells. Any 4x4x1 brick must occupy at least one of these cells. Note that the simple volume constraint gives an upper bound of 7^3/4^2 = 343/16 = 21.4375. Rich

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