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Fw: Fw: [math-fun] Fractile application: antennas
by Thane Plambeck 23 Nov '03

23 Nov '03

Here's an opinion from an (anonymous) friend on the antennas > > On Fri, Nov 21, Thane Plambeck writes ... > > Thane> Any comments... > > You're in trouble because I'm not really an antenna guy. Of course, > I do have an opinion ... > > There are a few key parameters of an antenna that determine its performance > (I'm going to describe them in terms of reception, but all of the > statements have a corresponding transmission formulation). The first > parameter is effective aperture (units of m^2). Effective aperture is not > the same as the physical area of an antenna, but it's very closely related. > The antenna's job is to collect RF energy in the form of a flux density > (Watts/m^2) from space. Physically small antennas have small effective > apertures; tiny antennas can't collect as much energy as big antennas > regardless of whether they're made from clever space-filling curves or coat > hangers. The second two parameters are radiation and circuit impedance, > and respectively describe how efficiently antennas couple RF energy into > free space and between their terminals and some other circuit. These are > quantities that are functions of antenna geometry and materials. If > fractal designs have any fundamental efficiency benefits to offer, it is in > the area of antenna geometry. > > One issue that antenna manufacturers face is that the coupling efficiency > and directivity (radiation pattern) of a particular antenna design is a > function of wavelength (among other things). For relatively broadband > applications, e.g., for UHF televsion which operates at frequencies from > 470-890 MHz in the US, special geometries are used to provide efficiency > across a broad range of operating frequencies. These antennas have > self-similar structure resulting in efficient reception across a wide range > of frequencies; the most familiar example is a log-periodic antenna which > you would recognize by sight if not by name (see, e.g., > http://www.ramayes.com/Log_Periodic_Antennas.htm) The self-similar > structure of various fractal curves might also have good properties in this > regard, but I can't immediately think of any physical reason why they would > have exceptional performance as compared to the alternatives. > > Another fractal antenna claim is that by "folding" the array geometry, an > antenna can be made small (retain its effective aperture) while > maintaining performance. I've been approached by other people working with > folded antennas, e.g., http://www.skycross.com, and the problem always > seems to be that these types of antennas are inefficient in coupling energy > to a transmitter or receiver. > > Finally, regarding the claims of enabling fast algorithms for beamforming, > there are probably some clever recursive algorithms that one can develop > for an array with fractal geometry. The real question is whether the > algorithms are less computationally intensive than standard techniques > (which typically require only the solution of a linear system of equations > of order equal to the number of elements in the array) for arrays of > realistic size. > > So, I'm suspicious. Without giving the subject the thought that it no > doubt deserves, my intuition is that there are some nice > mathematical/algorithmic properties of arrays with fractal geometry but > that the arrays being proposed by these people have no significant > performance advantage relative to the other self-similar and folded designs > that are well known in the industry. > >

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[math-fun] Fractile application: antennas
by Richard Schroeppel 21 Nov '03

21 Nov '03

Source: Penn State Date Posted: 2003-10-20 http://www.sciencedaily.com/releases/2003/10/031020054315.htm FRACTAL-SHAPED TILES DEVELOPED FOR NEW BROADBAND ANTENNA CLASS Penn State engineers have developed innovative design methods for a new class of antennas composed of an array of fractal-shaped tiles that offer anywhere from a 4:1 to 8:1 improvement in bandwidth compared to their conventional counterparts. Many natural objects, such as tree branches and their root systems, peaks and valleys in a landscape and rivers and their tributaries are versions of mathematical fractals which appear pleasingly irregular to the eye but are actually made of self-similar, repeated units. The new broadband antennas are composed of irregular but self-similar, repeated fractal-shaped unit tiles or "fractiles" which cover an entire plane without any gaps or overlaps. The outer boundary contour of an array built of fractiles follows a fractal distribution. Dr. Douglas H. Werner, professor of electrical engineering and senior scientist in Penn State's Applied Research Laboratory, will describe the new antennas and their generation at the 2003 IEEE AP-S Topical Conference on Wireless Communication Technology, Oct. 16, in Honolulu, Hawaii. His paper is "A New Design Methodology for Modular Broadband Arrays Based on Fractal Tilings." His co-authors are Waroth Kuhirun, graduate student, and Dr. Pingjuan Werner, associate professor of electrical engineering. While fractal concepts have been used previously in antenna design, Werner and his research team are the first to introduce a design approach for broadband phased array antenna systems that combines aspects of tiling theory with fractal geometry. Once the specific fractile array has been designed, the Penn State team exploits the fact that fractal arrays are generated recursively or via successive stages of growth starting from a simple initial unit, to develop fast recursive algorithms for calculating radiation patterns. Using the recursive property, they have also developed rapid algorithms for adaptive beam forming, especially for arrays with multiple stages of growth that contain a relatively large number of elements. Werner says, "The availability of fast beam forming algorithms is especially advantageous for designing smart antenna systems." The Penn State team has also shown that a fractile array made of unit tiles based on the Peano-Gosper curve, for example, offers performance advantages over a similar-sized array with conventional square boundaries. The Peano-Gosper fractile array produces no grating lobes over a much wider frequency band than conventional periodic planar square arrays. Werner explains that "Grating lobes are sidelobes with the same intensity as the mainbeam. They are undesirable because they take energy away from the main beam and focus it in unintended directions, causing a reduction in the gain of an antenna array." The University is patenting the team's approach to Peano-Gosper and related fractile arrays. The team has also been awarded a grant through the Applied Research Laboratory to build and test a prototype.

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[math-fun] CA Whites may be nastier than NH Whites
by Eugene Salamin 21 Nov '03

21 Nov '03

An automated weather station has been operating since September on 14250 ft White Mountain Summit, the high point of the White Mountain Range, which is the next range east of the California Sierras. Already it has recorded a wind gust of 154 MPH. I expect it won't be long before they break the current record of 231 MPH held by New Hampshire's Mt. Washington. This can be a very pleasant place to visit in the summer: most scenic drive in California, 4700 year old bristlecone pines, 7.5 mi 3000 ft hike to summit, early July wildflowers. It can also be nasty. On a backpack to Pellisier Flats, I woke up in a well below freezing whiteout at 13000 ft and had to navigate out by compass. http://www.wrcc.dri.edu/weather/wmtn.html http://www.wmrs.edu/ http://www.sonic.net/bristlecone/ Gene __________________________________ Do you Yahoo!? Free Pop-Up Blocker - Get it now http://companion.yahoo.com/

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[math-fun] Method for magic cubes
by Christian Boyer 18 Nov '03

18 Nov '03

Some words about the method used, by Walter and me, for our perfect magic cube of order 5. We have first computed a large amount of auxiliary cubes of order 3. These auxiliary cubes are central "symmetrical", meaning that all their 13 alignments of 3 numbers going through the center have x+63+y = 189. These auxiliary cubes have some other partial magic characteristics: 29/49 alignments (including 14/18 diagonals and 4/4 triagonals) have already the same sum = 189. You can check these characteristics, looking at the central cube of order 3 included in our magic cube of order 5. And after that, using these auxiliary cubes, we have tried to fill always by computer the 5^3 - 3^3 = 98 missing numbers, mainly using complementary numbers x+189+y = 315. That's why you can seen a lot of various symmetries in our cube. Running several computers during several weeks before to find the first solution. With this method, it is very easy to find cubes of order 5 with 26 magic diagonals. And more than 1500 cubes with 28 magic diagonals were found before to find our first cube with 30 magic diagonals. We used more than 80,000 different auxiliary cubes of order 3 before to find the first solution. The method used previously by Walter for his cube of order 6 is similar, using a central auxiliary cube of order 4. Christian Boyer. www.multimagie.com/indexengl.htm > -----Message d'origine----- > De: Richard Schroeppel [mailto:rcs@CS.Arizona.EDU] > Date: lundi 17 novembre 2003 21:24 > À: cboyer(a)club-internet.fr > Objet: magic cubes > > Feel free to send more details about the construction/search methods for > the 5^3 and 6^3 to the math-fun list. > > Rich Schroeppel rcs(a)cs.arizona.edu >

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[math-fun] Re: powers of PseudoAntisymmetric (-1,0,1)- Matrices
by Dan Hoey 17 Nov '03

17 Nov '03

"wouter meeussen" <wouter.meeussen(a)pandora.be> writes: > One counter-example to spoil things a bit: What is this a counterexample of? > for n=7, a small sample of (-1,0,1)-matrices whose powers are > 'same', shows that not all are pseudo-Antisymmetric (= AntiSymmetric > + Diagonal) What about [ 0 1 ] [ 1 0 ] ? Isn't that a non-PA (-1,0,1)-matrix whose powers are PA (-1,0,1)-matrices? But the (powerlength = divisor of 12) remains valid. That's not true of the permutation matrix [ 0 1 0 0 0 ] [ 0 0 1 0 0 ] [ 0 0 0 1 0 ] [ 0 0 0 0 1 ] [ 1 0 0 0 0 ]. Earlier, you wrote > > Consider > > the (-1,0,1)-matrices T with properties : Det[T] not zero > > (invertible), all powers T^k are also invertible (-1,0,1) matrices. > > Properties: > > powerlength of T divides 12, > > Det[t] is 1 or -1, > > T is pseudoAntisymmetric, What are you asserting or conjecturing about those matrices and those properties? By the way, a more standard term for "powerlength" would be "order", or "multiplicative order" if you use the word "order" for some other property. Certainly any matrix of finite multiplicative order must have unit determinant, but I can't see any other likely conjecture. Dan

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[math-fun] Re: powers of PseudoAntisymmetric (-1,0,1)- Matrices
by wouter meeussen 15 Nov '03

15 Nov '03

One counter-example to spoil things a bit: for n=7, a small sample of (-1,0,1)-matrices whose powers are 'same', shows that not all are pseudo-Antisymmetric (= AntiSymmetric + Diagonal) try: { {-1, 0, 0, 1, 0, 0, 0}, { 0, -1, 0, 0, 0, 0, 0}, { 0, 0, 0, 0, 0, 1, -1}, {-1, 0, 0, 0, 0, 0, 0}, { 0, 0, 0, 0, 1, 0, 0}, { 0, 0, -1, 0, 0, -1, 1}, { 0, 0, -1, 0, 0, -1, 0} } so my program might have missed some for n=4 or n=5, since I examined only the pseudo-antisymmetric ones. But the (powerlength = divisor of 12) remains valid. When interpreted as adjacency-matrices of (oriented?) graphs, do their matrix-powers have a 'graph-theoretical meaning'? Wouter. ----- Original Message ----- From: "wouter meeussen" <wouter.meeussen(a)pandora.be> To: "math-fun" <math-fun(a)mailman.xmission.com>; "Seqfan (E-mail)" <seqfan(a)ext.jussieu.fr> Cc: "Marc LeBrun" <mlb(a)fxpt.com> Sent: Saturday, November 08, 2003 7:27 PM Subject: moRe: powers of PseudoAntisymmetric (-1,0,1)- Matrices > rehash of thread 15/08/2003, with some news. > OEIS?Anum=A072148 Sequence: 2,14,92,796,7672 > > Definitions: > pseudoAntisymmetric : T(i,j)= -T(j,i) for j<i , so T = diagonal+Antisymmetric. > (my definition, forgive..) > > powerlength: minimal p>0 so that T^p = Identity > > > > Consider > the (-1,0,1)-matrices T with properties : Det[T] not zero (invertible), > all powers T^k are also invertible (-1,0,1) matrices. > > > Properties: > powerlength of T divides 12, > Det[t] is 1 or -1, > T is pseudoAntisymmetric, > > the powers T^k need not be all pseudoAntisymmetric: > > for 4x4 matrices, > all those with 8 non-zero elements have > powerlength 4, and their powers 2 and 3 are not pseudoAntisymmetric; > > for the 5x5 matrices, > all those with 9 non-zero elements have > powerlength 4, and their powers 2 and 3 are not pseudoAntisymmetric; > all those with 10 non-zero elements have > powerlength 12, and their powers 2,3,6,7,10 and 11 are not pseudoAntisymmetric; > > There is a system in this madness, > but this margin is too small... > > > W. > > (I owe Marc LeBrun <mlb(a)fxpt.com> for help, > partial insight & lots inspiration, thanx Marc) > > > I put the 796 4-by4 and the 7672 5-by-5 on > http://users.pandora.be/Wouter.Meeussen/pseudoAntisymmMatrixPowers_4.txt > http://users.pandora.be/Wouter.Meeussen/pseudoAntisymmMatrixPowers_5.txt > > > >

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[math-fun] Magic torus of maximum magicity
by asimovd＠aol.com 14 Nov '03

14 Nov '03

Rich announces: << Congratulations to Christian Boyer and Walter Trump, who have found a magic cube of size 5! >> That's really cool! (I hadn't even knownrealized this question was unsolved.) I can't help wondering about whether there is such a thing as a "magic torus" T with maximum magicity. Meaning: Viewing T as the group (Z_k)^n, consider all "affine lines" of the form L = { v + j*w | v,w are in T, and 0 <= j < k } such that card(L) = k. QUESTION: For which (k,n) does there exist an assignment of 1,2,...,k^n to the points of (Z_k)^n such that *every* affine line has the same sum? Or less generally, what is known about this question? --Dan

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[math-fun] magic cube, size 5
by Richard Schroeppel 14 Nov '03

14 Nov '03

Congratulations to Christian Boyer and Walter Trump, who have found a magic cube of size 5! They sent me a .gif picture of the cube. I've transcribed it below. K=315. I spot checked a couple of rows but haven't done a full check of all 109 lines. [109 = (7^3-5^3)/2.] There is some amount of symmetry, opposite cells adding to 126, but it's irregular. Sometimes the symmetry is reflection through a center plane rather than through the center cell. Hopefully we'll hear more about the technique for finding the cube soon. I think this leaves size 6 as the only magic cube size of undetermined existence. Gardner's columns c. 1974 had a magic cube of size 7, and I think 8+ had already been done. Size 7 is a straightforward generalization of the stepping method for the magic 3^2 and 5^2. I proved there's no magic 4^3 (published in MGs column), and the nonex. of 3^3 was already well known. Curiously, the magic 6^2 is harder to construct than squares of size 4,5,7,8. 4K+2 seems to be a hard size. I have some simple results on magic 6^3 properties: various groups of cells must have the expected average value. (Make a semihemidemi-magic 3^3 by folding the 6^3 along the three center planes, adding the cells. Various rows & agonals must have the average value, and algebra gives a little bit more.) The nonexistence of magic 5^4 follows from each 5^3 having the same required center value. There should be enough line types to prove that a magic 7^4 must have center 1201, but I haven't done the algebra. I've tried to show the impossibility of the magic 6^4 but have only partial results, which don't lean either way. This magic 5 cube supports my conjecture that existence of a magic object (of size >1) is equivalent to the corresponding tictactoe game being a draw. (Since 4^3 TTT is a long battle for the win, 5^3 seems certain to be a draw.) Let the analysis begin! Rich rcs(a)cs.arizona.edu ----- Walter Trump & Christian Boyer Magic 5^3 (<= Nov 14, 2003) 25 16 80 104 90 115 98 4 1 97 42 111 85 2 75 66 72 27 102 48 67 18 119 106 5 91 77 71 6 70 52 64 117 69 13 30 118 21 123 23 26 39 92 44 114 116 17 14 73 95 47 61 45 76 86 107 43 38 33 94 89 68 63 58 37 32 93 88 83 19 40 50 81 65 79 31 53 112 109 10 12 82 34 87 100 103 3 105 8 96 113 57 9 62 74 56 120 55 49 35 121 108 7 20 59 29 28 122 125 11 51 15 41 124 84 78 54 99 24 60 36 110 46 22 101 ----- From: "Christian Boyer" <cboyer(a)club-internet.fr> To: <rcs(a)CS.Arizona.EDU> Cc: "Walter Trump" <TrumpNbg(a)aol.com> Subject: "The center of an order 5 magic cube must be 63" Date: Fri, 14 Nov 2003 18:05:49 +0100 Dear Richard, You will probably be happy by these news. As reported by Martin Gardner, and as reported at http://www.inwap.com/pdp10/hbaker/hakmem/number.html#item50, you have proved in 1972 that a perfect magic cube of order 5, if it exits, has a central value = 63. But nobody had found such a cube. We, my German friend Walter Trump and me, have computed this week the first perfect cube of order 5. 109 ways to sum 315. Of course, as you proved, its central value is 63: look at the attached file! Best regards. Christian Boyer. www.multimagie.com/indexengl.htm

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[math-fun] 15 puzzle - one of rich's nits
by Chris Landauer 14 Nov '03

14 Nov '03

hihi, all - rich's first nit about the set of reachable positions: indeed each move is a transposition in S16, but it is easy to see that the only way for the blank space to get back to its original position requires an even number of them, so all of those permutations are even, and my previous note was about a proof (i think) that all even permutations are possible, if the outline is thick enough (as the square is) then the set of all positions is A15 in S16 fixing 16 with the added transpositions making a transitive group on all 16 positions, with overall size therefore (15!/2) x 16, though it is not A16 (as rich said) the key to the proof that you can get all of A15 is to show that in any 2x2 subsquare, with any one of the four unit squares fixed, you can get the 3-cycle for the other three unit squares (almost): pick a 2x2 subsquare, and one of the unit squares in it that is not a corner of the entire square - then it is easy to move the blank spot to a position adjacent to the fixed unit square in that 2x2 subsquare without disturbing any of the unit squares in it - move the fixed one out into the blank spot, cycle the other three unit squares, move the fixed one back, and reverse the moves to get the blank spot back to position 16 this is a lot of 3-cycles, i think easily enough of them to generate all of A15 (i do remember having a real argument for that, but i can't remember what it was) more later, cal

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[math-fun] 15 puzzle - a few nits
by Richard Schroeppel 13 Nov '03

13 Nov '03

A few minor nits on the 15 puzzle. I've always seen the puzzle discussed as "half the arrangements are reachable", and simply assumed it was the even permutations of the sliding pieces that were reachable. I was trying to write down a mapping from the (nonnegative) integers < 16!/2 to the reachable positions and ran into trouble. The mapping to even permutations of [1,16] is easy enough, but a tiny problem appears. [I'm calling the blank space 16 here.] If the blank is in the last position (lower right corner, assuming standard reading order for the cells), then the piece to the left can be moved into the blank cell. This is a swap, which switches an even permutation to an odd one and vice versa. The truth appears to be that indeed, exactly half the positions are reachable, but the description is more complicated. Suppose we color the puzzle cell positions with a checkerboard of black and white squares. The a position is achievable iff it's an even permutation and the blank is at a black square, or if it's an odd permutation and the blank is at a white square. Still 50%, but the mapping rule for the integers will need a little fixup. --- Second nit. The "max solution is 80" paper doesn't tell us the graph diameter, but only that all nodes are within distance 80 of the starting position. (Along with giving 2 of the furthest- from-start positions.) I think we can say that the diameter is at most 82, since a position with the blank in a central or edge position can be morphed into the initial position in one or two moves (plus a relabeling of the tiles). Redoing the search with an initial position where the blank is on the edge and getting a final answer of 79 would prove diameter 80. --- The group of positions for Rubik's is much bigger than 16! But solving individual positions in the rigorous minimum number of moves is feasible now, and has been for a while, since you can work from both Start and Goal towards the middle, and there are only around 10^10 middle positions. If we were lucky, a similar situation might apply to the 15 puzzle, with only about sqrt(16!/2) ~= 3 million positions being reachable in 40 moves. I have no idea if this is true. --- 10^13 positions can be represented as a bit table (reachable in <=K steps, or maybe exactly K steps). That's only 1.25 TB. A lot of storage, but 100GB disks are cheaply available. --- Wechsler's solution algorithm (his note 4, with three lists) can be simplified slightly for the 15 puzzle, since each move changes the color of the "blank square". The reachable positions can be subdivided exactly into two equinumerous classses, those with the blank on a black square (which are reachable in an even number of moves), and those with the blank on a white square (which are an odd distance from Start). When discarding positions reachable from the Current List, only the Previous and Next Lists need be examined; they are of opposite color to the Current List. A good encoding of positions can give considerable locality for the 1-neighbor operator, so the implied sorting step for uniquing the 1-neighbors of the Current List is tractable. --- During the 15-puzzle craze (1880?) there were published problems with prizes and best-found solutions. I wonder if anyone has compared these answers to "truth". Rich rcs(a)cs.arizona.edu

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