User sleepless in beantown - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T09:13:47Z http://mathoverflow.net/feeds/user/8676 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51169/flux-through-a-mobius-strip/51183#51183 Answer by sleepless in beantown for Flux through a Mobius strip sleepless in beantown 2011-01-05T06:42:45Z 2011-01-09T03:04:32Z <p>The Mobius resistor concept seems like bunk/hokum because there is only one side to the resistor: both leads connecting to such a resistor would be connected to the <strong>same side</strong> and would theoretically have $0$ (zero) resistance for circuit-modeling purposes. </p> <p>The three answers prior to my comment here by Douglas Zare, David Speyer, and Scott Carnahan should all be sufficient to convince you that since you can't orient this surface, you cannot define a positive/negative orientation for the direction of flow through this surface and thus cannot define a flux through a mobius strip.</p> http://mathoverflow.net/questions/50779/floating-polyhedra-with-fair-equilibria/50787#50787 Answer by sleepless in beantown for Floating polyhedra with fair equilibria sleepless in beantown 2010-12-31T04:51:41Z 2010-12-31T04:51:41Z <p>Quick thought: make a four-sided pyramid of ice, square base of edge length $x$ and have the height be less than $x$. This type of ice-cube or pyramidal ice-berg has the majority of its buoyancy at the square face, thus the subset of size $1$ of {flat square base} is more likely to be the "face up" face in a container. This could probably be tested better with a plastic homogeneous pyramid rather than with an ice-cube pyramid which would melt during the experiment and change the results over time.</p> <p>There's probably a decent ratio $a$ such that if the height of the pyramid is less than $ax$ that the pyramidal-square base is significantly biased to be the "face up" floating face.</p> http://mathoverflow.net/questions/50763/what-is-the-dimensionality-of-a-tree/50777#50777 Answer by sleepless in beantown for what is the dimensionality of a tree? sleepless in beantown 2010-12-31T01:28:59Z 2010-12-31T01:28:59Z <p>It makes more sense to speak of the dimensionality of an <strong>embedding</strong> of a tree or graph-structure into the $n$-dimensional space $\mathbb{R}^n$, rather than to speak of the dimensionality of a tree or a graph-structure independently of an embedding into space.</p> <p>Even then, it's a different question talking about the Hausdorff dimension of the $n$-dimensional embedding. Does that help clarify what question you would really like to ask? What kind of tree structures are you dealing with? Natural objects represented as trees? Random structures with random but known branching patterns that can be represented as parametrized structures (branching factors or probabilites, outdegree of vertices, etc.)? Infinite structures with some known properties which you would like to model? Is it a fractal structure with scale-free parameters that you need to understand better?</p> <p>Gerald Edgar's reference is good, but I think that you need to specifically think about the geometric embedding of your tree structure into $\mathbb{R}^n$ or into $\mathbb{Z}^n$ before you venture further with your question. Is there a physical reason for $n$ to be a particular value? Is there a branching factor that will cause the fractal dimension to be much smaller than $n$?</p> http://mathoverflow.net/questions/50711/how-long-is-the-longest-path-in-the-game-tree-of-chess/50727#50727 Answer by sleepless in beantown for How long is the longest path in the game tree of chess? sleepless in beantown 2010-12-30T13:56:20Z 2010-12-30T14:25:01Z <p>A long comment, too long and painstakingly difficult to keep re-editing in the comment boxes:</p> <p>If you don't <strong>require</strong> a draw to be declared, there are multiple scenarios in which king vs. king or (king+queen) vs. (king+queen) can play on infinitely; in that case, the game tree of chess is unbounded. There must be a strict rule for when to prune a branch in the game tree. @Didier-Piau, the upper-bound concept as posited by the poster of this question appears to have 3 mistakes in it. </p> <p>It may be the concept of {white pawn, white other, black pawn, black other, empty}$^{64}$, which has a set size of $5^{64}$. </p> <ul> <li><p>This makes the mistake of lumping all of the pieces into $4$ categories. Even if you define the pieces to be {Black, White} $\times$ {Pawn, Queen, King, Rook, Knight, Bishop}, and allow for an empty space, then $13^{64}$ would be a better (but still grossly overlarge) upper-bound on the number of chess board configurations as it included multiple implausible configurations with an impossible count of pieces. A better guess might be the combinatorial (64 choose 32) + (64 choose 31) + ... (64 choose 1), and that can be pruned in many ways such as if the last board position has only one piece in it, then that last piece could only be the winning side's king, etc.</p></li> <li><p>It makes the mistake of conflating the number of possible positions or "boards" of a chess game with the number of paths through these possible boards; this is equivalent to the error of confusing the number of vertices in a directed graph with the number of paths leading out from a starting vertex. </p></li> <li><p>And it makes the error of not being rigorously defined: for example, defining the tree correctly, as the tree starts out from a fixed board position.</p></li> </ul> http://mathoverflow.net/questions/50624/happy-new-prime-year/50626#50626 Answer by sleepless in beantown for Happy New Prime Year! sleepless in beantown 2010-12-29T04:31:22Z 2010-12-29T20:18:18Z <p>You missed the excitement of 1998, with $1999=1999, 2000=2^4 \cdot 5^3, 2001=3 \cdot 23 \cdot 29,$ and $2002=2 \cdot 7 \cdot 11 \cdot 13$</p> <p>A quick set of thoughts, too long to fit in a comment:</p> <p>this requires finding a "prime gap" of length $k-1$, since $s(n+1)=1$ means <strike>that $n+1$ is prime</strike> that $n+1$ is either prime or a power of a prime, but the next $k-1$ digits are composite since s(n+x)>1 for $2 \le x \le k$. This also means that $s(n+2)=2$ only because $s(n+2)$ is even, thus $2$ is one of the factors and implies that $(n+2)/2$ is prime (or that $(n+2)/2^j$ is prime for some $j \in \mathbb{Z}$), since $n+2$ only has two factors and one of them is $2$ (or $2^j$). </p> <p>For $s(n+k)$ to have $k$ distinct prime factors means that it has to be at minimum a product of the first $k$ prime numbers, while it definitely has to be a multiple of the product of $k$ prime numbers. So the two key restrictions are that s(n+k) is $k$-composite (has $k$ prime factors) and that both (n+1) and (n+2)/2 are prime numbers.</p> <p>Hmm, I thought something about the fact that either s(n+2) or s(n+4) would be divisible by $4$ while the other would be divisible by $2$ but not by $4$ would play some role in this.</p> <p>Here are some quick results from running "bash", "factor", and "sed" and "awk" at the command line:</p> <p>If you want an ascending run of 1,2, and 3 prime factors, the smallest example starts at $n=63$, with $64=2^6, 65=5 \cdot 13, 66=2 \cdot 3 \cdot 11$</p> <p>If you want an ascending run of 1,2,3, and 4 prime factors, we already missed the exciting years of $n=1866$ and $n=1998$</p> <pre><code>1867: 1867 1868: 2 2 467 1869: 3 7 89 1870: 2 5 11 17 1999: 1999 2000: 2 2 2 2 5 5 5 2001: 3 23 29 2002: 2 7 11 13 </code></pre> <p>And the next few years with ascending runs of 1,2,3, and 4 prime factors will start after the years 3216, 4056, and 4176 with 3217, 4057, and 4177 as prime years. Unfortunately, these computational results are not giving me the germ of any shortcut or understanding. There are also some descending sequences in terms of the number of prime factors, and their placement also does not help.</p> <p>If you want an ascending run of 1,2,3,4, and 5 prime factors, we have to wait almost half-a-million years to get to the exciting years of $n=491850$ and $n=521880$ for $k=5$</p> <pre><code>491851: 491851 491852: 2 2 122963 491853: 3 19 8629 491854: 2 11 79 283 491855: 5 7 13 23 47 521881: 521881 521882: 2 260941 521883: 3 3 3 3 17 379 521884: 2 2 11 29 409 521885: 5 7 13 31 37 </code></pre> <p>Now with four numbers computed and found, I searched the OEIS and found the corresponding sequence. Since the Online Encyclopedia already has this sequence, I'm hanging up my computational hat and heading off to work. :)</p> <blockquote> <p><a href="http://oeis.org/A086560" rel="nofollow">http://oeis.org/A086560</a></p> <p>Start of first run of n successive numbers in which i-th number has exactly i distinct prime divisors for i = 1..n</p> <pre><code>2, 5, 64, 1867, 491851, 17681491, 35565206671 </code></pre> <p>J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 64, p. 23, Ellipses, Paris 2008.</p> </blockquote> http://mathoverflow.net/questions/50522/obtaining-all-vectors-of-given-length-and-with-with-1-entries-from-a-given-o/50623#50623 Answer by sleepless in beantown for obtaining all vectors of given length and with with $+-1$ entries from a given one sleepless in beantown 2010-12-29T04:13:22Z 2010-12-29T04:13:22Z <p>Simply represent each vector as a binary integer:</p> <p>$v_0 = "0 0 0 ... 0 0 0" = b_{n-1} b_{n-2}... b_2 b_1 b_0 = 0$</p> <p>so the binary digit sequence representing $i$ is a decimal number $d$</p> <p>$$d = \sum_{j=0}^{j=n-1} b_j \cdot 2^{j}$$</p> <p>Given such a binary representation, you can transform the binary digits into the coefficients by mapping $0 \to -1$ and $1 \to 1$</p> <p>$b_j = 0 \to r_j = (-1)$ and</p> <p>$b_j = 1 \to r_j = (+1)$</p> <p>Then, given a $v_j$ as a binary digit, generate $v_{j+1}$ by adding $1$ to the binary representation.</p> <p>There is no shortcut around having to test each of the $2^n$ possibilities, so this does not make your overall calculations faster. It just makes it easier to generate the binary digits.</p> <p>A simpler way is to iterate your index as an integer (call it $z$) from $0$ to $n-1$ and generate the binary representation of each $z$. There are many ways to do that quickly which you can find with simple searches on the internet or by asking at stackexchange.com</p> http://mathoverflow.net/questions/50294/point-in-polygon-algorithm-from-the-viewpoint-of-a-robot/50299#50299 Answer by sleepless in beantown for Point in Polygon algorithm from the viewpoint of a robot sleepless in beantown 2010-12-24T19:04:03Z 2010-12-24T20:26:39Z <p>Lookj up "level sets".</p> <p>The wikipedia page "Point in Polygon" talks about algorithms that can be used when the polygon's coordinates are known. The proposed solution of the length of a path following the fence along the fence (call it $d_0$) and a path following along the fence but maintaining a constant distance of $x=1$ meter (call it $d_1$) will work for a robot that can do the tasks you're asking of it. However, the answer of the difference in length being $2 \pi \sim 6.28$ meters would only apply if the fence is perfectly circular.</p> <p>Given a map or diagram of the fence, generate multiple contours or level sets of points which are a constant distance from the fence. You'll end up with something that looks like a contour map or topographical map that the U.S. Geological surveys generates. Notice that for each distance $x$ (up to a certain limiting value), the level sets for $d_x$ may contain points inside the fence as well as outside the fence. Once $x$ is greater than the radius of the circle, the level sets for $d_x$ such that $x \gt r$ will only contain points outside the circle. For fences with concavities (like a pinched figure 8), the inner level set may break up into multiple non-connected paths.</p> <p>If the fence is square, width edge length $2r$, then the close-fence contour $d_0$ will be $4 \times 2r = 8r$, whereas the 1-meter level set will be </p> <ul> <li><p>$4 \times (2r) + 4 \times (\frac{1}{2} \pi) = 8r + 2 \pi$ if the robot's path is 1-meter outside the fence (which consists of the edges translated outward a distance of 1-meter, and of quarter-circle arcs at each of the corner, as correctly pointed out by Mark Bennet's comment below).</p></li> <li><p>$4 \times (2r-2) = 8r - 4$ if the robot path is 1-meter inside the fence.</p></li> </ul> <p>Thus for square, non-convex, and pretty much any noncircular fence, the level-set path one meter of the fence will not be $2 \pi \sim 6.28$ meters different from the level-set path of distance $0$ from the fence.</p> <p>The generalization, however, will still apply that the level set path of distance $x$ away from the fence will be smaller ($d_x \lt d_0$) if the robot follows the level set path within the fence, vs. larger if the robot follows the level set path outside the fence ($d_x \gt d_0$).</p> http://mathoverflow.net/questions/49862/probability-theory-and-measuring-the-true-strength-of-chessplayers/49863#49863 Answer by sleepless in beantown for Probability theory and measuring the true strength of chessplayers sleepless in beantown 2010-12-19T09:28:37Z 2010-12-19T09:28:37Z <p>David, your question makes the assumption that players will stochastically pick a move in the current possible set of branches, and does not say anything about the current depth of the tree. I believe that for certain states, particularly those labeled "end-games", it is possible for an astute player (or an experienced player) (the set of astute and experienced players are not equal) to have a higher $W$ percentage against an equally skilled or worse opponent.</p> <p>Thus I believe that $W$ and $D$ are not just functions of one player $P_1$, but also of - the opponent $P_2$ and of - the current-depth of the game tree (= the number of moves played thus far), and - the current-state (global and local) of the game board.</p> <p>$P_1$'s $W$ may change for a different opponent and for certain opening sequences or end-games with which they are familiar.</p> <p>Now if you allow the assumption that you do not have an oracle evaluation function, but do have the win and draw percentages for two players, $P_1=(W_1, D_1)$ and $P_2=(W_2,D_2)$, your question in the second paragraph asks if that is sufficient to allow for calculating the probabilities of one player dominating over the other. I do not believe that there is a way to calculate this, as the $W$ and $D$ ratios are going to have to be calculated as a measure over all possible game board states, and the finite sampling of win and draw ratios for a finite number of games and board positions will not be sufficient to allow for such an extrapolation to be made.</p> <p>Back to my first ruminations: the $W$ and $D$ will depend on the depth of the game tree and the relative experience of the players. If a player picks a bad move but has better experience, she could still recover and win at a later point in the game. If a player picks a bad move but does not have much experience, they are less likely to be able to recover and get to a position of advantage. </p> <p>An experienced player recognizing classic openings may play by rote for a few moves, or may in fact feint and play slightly askew to see how her opponent responds. This type of psychological repertoire and skill cannot be encoded and captured in a two parameter model, and is also why I think $W$ and $D$ ratios are not just a function of the player $P_1$ but also of the opponent.</p> http://mathoverflow.net/questions/49671/weighted-distribution/49693#49693 Answer by sleepless in beantown for Weighted Distribution sleepless in beantown 2010-12-17T00:00:24Z 2010-12-17T00:00:24Z <p>This is a basic probability and distribution question, I believe, and I hope it is not homework. Based on Yemon Choi's comments, I'm posting this answer:</p> <p>For the sake of example, I will relabel your fruit</p> <ul> <li>$a$ = apple</li> <li>$b$ = banana</li> <li>$c$ = cherry</li> </ul> <p>and I will relabel your scoring system</p> <ul> <li>old $1$ = developing = $-1$</li> <li>old $2$ = performing = $0$</li> <li>old $3$ = leading = $+1$</li> </ul> <p>Presume that you've set the scoring correctly so that the distribution of $a$ is 15% = | (-1) | , 70% = | (0) | , and 15% = | (+1) |, of $b$ is 15% = | (-1) | , 70% = | (0) | , and 15% = | (+1) }, and of $c$ is 15% = | (-1) | , 70% = | (0) | , and 15% = | (+1) |.</p> <p>There are now 27 categories for the ordered triplet $(a,b,c)$; each of these 27 categories is $c_i \in$ {$-1, 0, 1$}$^3$. Calculate the distributions of each category. Draw as a histogram. Select the score ($85$th percentile) such that $15$% of your group exceeds it as the leading category. Select the score ($15$th percentile) such that 70%+15% of your group exceeds it as the lower-most value of your performing category, and use the score of the leading category as the upper-most boundary of your performing category. Use the $0$th to $15$th percentile as your bounds for your developing category.</p> <p>Rescale the scores to {1, 2, 3} from {-1, 0, 1} which I used in this example to get the answer which you want. If this really is not homework, you won't have a problem doing that. If this is homework, you really should have asked somewhere else.</p> http://mathoverflow.net/questions/49577/voronoi-polygons-of-general-point-patterns-forced-to-honeycomb/49615#49615 Answer by sleepless in beantown for Voronoi polygons of general point patterns forced to honeycomb? sleepless in beantown 2010-12-16T08:28:13Z 2010-12-16T08:28:13Z <p>If the Voronoi tesselation of a set of points is a honeycomb pattern, then the points are all located at the center of each of the hexagons of the honeycomb, except for the edges. This is only true for a simple Euclidean geometry, of which "London" is a small enough region (steradian/solid angle of the Earth's spherical surface) that a patch of the Euclidean 2-d plane is a reasonable approximation.</p> <p>So the result you are looking for is going to be a "hexagonal grid", similar to what you would see in board games that have playing fields divided into hexagons. If you limit yourself to small patches of the Earth, say London, or Paris, rather than trying to do it for all of the Earth at once, then a regular hexagonal grid will be fine. </p> <p>Once you have a regular hexagonal grid, you will have to choose the scale you would like to use (the edge length of the hexagons, or the area encompassed by each of the hexagons). Notice that on the board games, they colour the countries by the hexagon, and approximate the "border" as boundaries between the hexagons, rather than allowing a hexagon to straddle two different countries.</p> <p>You will always have the problem of deciding between very small hexagons, with many partitions and small populations but better approximations of the boundaries of the post-codes, and larger hexagons, with fewer partitions to deal with each containing larger populations but which will resolve to a poorer approximation of the boundaries of the post-codes. The scale which you decide to use will probably depend upon the nature of the similarity of the members of the group clusters. You may be introducing an artificial group of boundaries which you may not need to.</p> <p>You may also want to look at <a href="http://en.wikipedia.org/wiki/K-means_clustering" rel="nofollow">k-means clustering</a> as a way of generating a clustering of these data points. It will not give you a regular hexagonal tesselation, but will give you a reasonable approximation of where clusters are located in your data-set. One problem with k-means, fuzzy-c-means, (and similar clustering techniques) is that the centroids are defined by the location of the members of the clusters, and the final segmentation which is reached by the iterative algorithm is strongly dependent upon the choise of the initial points of the clusters, and on the number of clusters ($k$) selected. Selecting the proper number of clusters is essential. </p> <p>This is usually a big problem for particular data-sets, however, in your case, your desire to use the "city centers" or "post-code centers" as the centers may be an advantage for you in the use of k-means. Set the number of clusters to be the number of post-codes which you wish to use. Set the initial centers of those clusters to be at the "geographic center" of those post-codes. Then let the k-means clustering algorithm run on your data set. The algorithm may coalesce multiple nearby clusters into single clusters. </p> <p>The end-result of k-means will <strong>not</strong> be a regular hexagon tesselation. It will, however, give you a good pointer as to which local areas can be categorized together and considered as a single cluster. You can then decrease (or increase) $k$ and rerun the algorithm and allow it to refine the clustering.</p> <hr> <p>If you are dead-set on having a regular hexagonal tesselation, perhaps use the k-means to see how far apart the "main" clusters are geographically. Then generate a regular "hexagonal grid" on that scale, and then run a few different simulations where you translate and rotate the regular hexagonal grid to see if there is a particular orientation and scale which best captures the features you wish to ascertain or convey (depending on if you're trying to understand something, or convince someone else of something :) ). </p> <p>There is no getting around preliminary exploratory data analysis of experimental or phenomenological data sets. Best wishes.</p> http://mathoverflow.net/questions/49384/tools-for-long-distance-collaboration/49396#49396 Answer by sleepless in beantown for Tools for long-distance collaboration sleepless in beantown 2010-12-14T15:37:26Z 2010-12-14T15:37:26Z <p>What needs to exist is a system that's akin to this site's functionality that could be used on a personal or university server and allow multiple people to contribute (via password-protected entry to the web-site) together to a notebook page which contains $\LaTeX$ markup and does it in a clean fashion. </p> <p>Perhaps the newer incarnation or instantiation of MO being tested on alpha.mathoverflow.net would allow for something like "private question pages" which are invitation only and could be used as an adjunct for white-board like functionality while the participants also use a telephone or skype or any other tools for instant collaboration. This technique would also allow for asynchronous updating by the collaborators if they happen to be living/working in different time zones.</p> http://mathoverflow.net/questions/49197/what-are-some-examples-of-journals-that-will-accept-undergraduate-student-researc/49203#49203 Answer by sleepless in beantown for What are some examples of journals that will accept undergraduate student research? sleepless in beantown 2010-12-13T02:38:38Z 2010-12-13T02:38:38Z <p>Electronic Journal of Combinatorics might be a good candidate considering the topic. My contention is that if the paper has a real result which is mathematical and combinatoric in nature, and is well-written, then the education-level of the author(s) should not play a role in whether the paper is appropriate or not for a particular journal.</p> <p>Whether or not one or more of the authors is an undergraduate still working on their studies or already holds a Ph.D. or teaching position should not be a factor. However, the best person to tell you about the most appropriate forum in which to attempt to publish your findings is <strong>your mentor</strong>, the professor sponsoring or advising you as you do this research. It is your advisor's job to advise about something like this, and they will have the best and most appropriate answer for you. People who don't know the details of your work, as the rest of these readers/commentators on Mathoverflow and slashdot are, cannot give you an informed answer. You should look closer to home, and ask your teachers and your undergraduate advisors, or an appropriate mathematician in your local mathematics department.</p> http://mathoverflow.net/questions/49082/individual-mathematical-objects-whose-study-amounts-to-a-subdiscipline/49123#49123 Answer by sleepless in beantown for Individual mathematical objects whose study amounts to a (sub)discipline? sleepless in beantown 2010-12-12T06:09:52Z 2010-12-12T07:00:04Z <p><a href="http://en.wikipedia.org/wiki/Conways_Game_of_Life" rel="nofollow">Conway's Game of Life</a> in 2-dimensions, as my exemplar instance in the class of (what used to be my overly general answer of...) <strong>Automata</strong>: deterministic finite state machines and nondeterministic and probabilistic automata and the theory behind them leading to things like acceptors of regular languages and the concepts of simulation, computational equivalence and computability as in Turing machines and "Turing equivalent", and the concept of "power of computing", computational complexity and complexity classes, bisimulation (and the equivalent computing power of single-tape vs. multi-tape and other classes of Turing machines, and the equivalent computing power of systems which can simulate other systems). </p> http://mathoverflow.net/questions/49082/individual-mathematical-objects-whose-study-amounts-to-a-subdiscipline/49121#49121 Answer by sleepless in beantown for Individual mathematical objects whose study amounts to a (sub)discipline? sleepless in beantown 2010-12-12T06:02:48Z 2010-12-12T06:02:48Z <p>Knots. Quandles and Racks.</p> http://mathoverflow.net/questions/48732/can-we-alter-the-axioms-of-euclidean-space-to-have-mathbbq3-as-a-unique-mod/48919#48919 Answer by sleepless in beantown for Can we alter the axioms of Euclidean space to have $\mathbb{Q}^3$ as a unique model? sleepless in beantown 2010-12-10T13:15:41Z 2010-12-10T19:07:11Z <p>As Colin Tan said, "[using] only countably many symbols, then there will always be a countable models." Whereas the field $\mathbb{Q}$ of rational numbers uses a finite-number of symbols for an uncountable number of rational numbers. I misunderstood and misapplied a concept.</p> <p>The field generated by rational numbers is quite different from the "approximation space" rendered by using a finite number of bits interpreted as a floating-point number low-precision approximation to real numbers. I'm editing my answer to point out my misunderstanding. @Hans-Stricker, I've fixed my error by pointing it out, but leaving it up (below the ruled line) so that some other bit-flipper like me will see why {0,1}$^n \times${0,1}$^n$ is not equivalent to $\mathbb{Q}$</p> <p>below this is my original (erroneous) answer</p> <hr> <p>Similarly, every numerical simulation in physics (or chemistry, biology, physiology, or medicine) always has to use <strong>finite precision</strong> representation of values, such that there is a limit to the largest and smallest integer represented by a fixed number of bits, and such that there is a limited amount of "floating-point-precision" available in dividing the bits of a floating-point representation of a real number into a set number of bits for the mantissa and a set number of bits for the exponent.</p> <p>For example, assuming that $d=64$-bits are used to represent "real numbers" as floating point numbers in computations, $m=48$ bits may be allocated to the mantissa, allowing the numerator to be $2^{48}$ yielding approximately $14$ digits of base-ten specificity to the numerator; this leaves $d-m=16$ bits to the exponent which may be signed (+/-) yielding a range of -32768 to +32767.</p> <p>In this case, the floating point number is in the range $n\times 2^{d-m}$, where $-(2^{47} \le n \le +(2^{47}-1)$, and ${-32768} \le d \le {32768}$.</p> <p>If the total number of bits is $d$, the number of bits allocated to the exponent, $m$, may be decreased while simultaneously increasing the number of bits, $d-m$, allocated to the mantissa, increasing the "precision" of the numerator while decreasing the range over $\mathbb{R}$ spanned by this particular approximating set of {0,1}$^m \times${0,1}$^{d-m}$ (which is <strong>not</strong> equivalent to $\mathbb{Q}$, as I erroneously stated originally)</p> <p><strike>Thus every numerical simulation is already, in a way, based on $\mathbb{Q}^d$ when models of $d$-dimensional systems are created and iterated using Euler or Runge-Kutta of whatever order. </strike></p> http://mathoverflow.net/questions/48913/uneven-spaced-time-series/48923#48923 Answer by sleepless in beantown for uneven spaced time series sleepless in beantown 2010-12-10T13:56:50Z 2010-12-10T13:56:50Z <p>Fourier transforms depend upon the fact that the modeled signal are going to be infinite in time-span and time-extent. While it is possible to get a very good example of a time-limited signal by using a finite set of Fourier coefficients, the finite-fourier-coefficient-approximation always ends up with "ringing artefacts" at any high-frequency edges beyond the bandwidth-limited approximation. These artifacts arise from the fact that Fourier decomposition using the "infinite-time-extent" sine-wave as its base-component.</p> <p>This type of problem in representing "limited-time-span" signals is what led to the concepts of "wavelets" and wavelet-transforms, using such limited-time-span base components such as the Haar wavelet. This is a slightly different problem from having non-equally-spaced-in-time samples extracted from a time series, but even then in these cases, there is the assumption that the underlying time series is continuous over time or is composed of the superposition of multiple discrete events occuring as Bernoulli or Poisson processes over time with some convolution of the discrete events by a smoothing factor (volcano eruption or geyser spouting, with the effluent "smoothed out" by prevailing winds or water currents).</p> http://mathoverflow.net/questions/48882/weighted-polytope/48915#48915 Answer by sleepless in beantown for Weighted Polytope sleepless in beantown 2010-12-10T12:58:00Z 2010-12-10T12:58:00Z <p>If you are restricting the weighting of the edges (it could just as equivalently be, in this case, weighting of vertices which would yield the same result) to be $\pm 1$, then you've got the weighting of a face being equivalent to the parity of the number of $-1$ weighted edges.</p> <p>Every $1$ edge contributes no change to the weighting of a face, while every $-1$ edge flips the parity of the weight of the face, thus the weight of a face is $+1$ if there is parity $0$ (an even number of $-1$ weighted edges) and the weight of a face if $-1$ if there is parity $1$ (an odd number of $-1$ weighted edges) surrounding the face.</p> <p>A similar simplification can be seen for the $j+1$-face: the weight of a $j+1$-face is the parity of the number of $-1$ weighted faces which surround it.</p> <p>Of course, the obvious physical weightings of polytopes exist: where the weights represent </p> <ul> <li><p>lengths as distances along a path, such as traveling salesman problems, wiring length in a communication network or the network topology of a connected grouping of computers (hypercubes such as the Connection Machine, for example)</p></li> <li><p>time delays in network signal propagation (either as maximum time delays in order to calculate the maximal propagation delay times, or as average time delays) representing amount of time for a signal to pass through or the congestion of a network,</p></li> <li><p>strength of connectivity or affinity, e.g. bond-strength in chemical structures as in "single-bond", "double-bond", or "triple-bond" for covalent bonds, or energy-of-bond representing the amount of energy required to break a chemical covalent bond (or other type of band, such as Van der Waals, weak hydrogen bonds, etc.)</p></li> <li><p>strength of known linkage, as in similarity of sequence for molecular structures made of replicated subunits for DNA or RNA or proteins, or as in interactions between biochemical pathways (some of which form multiple interacting cycles, which can be represented as multidimensional graph structures).</p></li> </ul> <p>If you look at your +1, -1 weighting of edges, you'll see that the same results would be obtained if the weighting were applied to the vertices, as each face is an $n$-gon polygon, and every (simple non-self-intersecting planar) polygon has the same number of edges and vertices. Is there any particular reason that you are applying the weighting as an attribute to the edges, rather than to the vertices? What kinds of systems are you looking at?</p> http://mathoverflow.net/questions/48642/parametrization-of-the-boundary-of-the-mandelbrot-set/48655#48655 Answer by sleepless in beantown for Parametrization of the boundary of the Mandelbrot set sleepless in beantown 2010-12-08T16:44:47Z 2010-12-08T16:44:47Z <p>My conjecture would be that such a parametrization would not work. Try something similar for a simpler (in certain viewpoints) structure such as a Koch snowflake. Would your approach to parametrization allow you to generate a function based on $n$, the number of recursive iterations used to generate the snowflake to a certain depth? I would think not. You might be able to, at least for the Koch curve, parametrize the "rubber band" hull around it, but that would be trivial for most recursively defined objects.</p> http://mathoverflow.net/questions/48379/social-reading-platform-for-math-or-latex-texts/48390#48390 Answer by sleepless in beantown for Social Reading Platform for Math or LaTeX texts sleepless in beantown 2010-12-05T20:43:54Z 2010-12-05T20:43:54Z <p>Your "social reading platform" looks like what HTML and the WWW=World-Wide-Web was supposed to be when Tim Berners-Lee first set up a web-server and web-browser platform applications at CERN on a NeXT machine using Objective-C (I think he programmed it in objective C). Researchers were supposed to have their web pages listing and highligthing their research with hyperlinks pointing to the publications and datasets. If you look at the majority of academic webpages, the pulication and research interests <em>are</em> listed in that way. It's just that the majority of the internet world has gone into <strong>walled gardens</strong> such as the social media pages, with the cost of entry usually being the loss of any privacy or control over what can be done with your user-provided content. Look at the issues discussed on the <a href="http://meta.mathoverflow.net" rel="nofollow">meta</a> website here at mathoverflow about why there hasn't been an upgrade to the StackExchange 2.0 software.</p> <p><a href="http://en.wikipedia.org/wiki/Wiki" rel="nofollow">Wiki</a> pages (not just that encyclopedic site that everyone uses, but a wiki page and wiki server which you can set up for yourself) allow for multiple users to modify a text using html or internal markup language. The requirement that $\LaTeX$ be usable in the markup language may require the use of MathML, or MathJax, or the jsMath package.</p> <p>I think the correct answer is most likely an internal wiki server, with password-accessed accounts for modifying the wiki-pages. The problem is going to lie in placing a full copy of possibly copy-righted material, particularly in the case of wanting to do an "annotated version" of a research paper, or of a book chapter. If the author of a particular paper or book chapter, or the full book itself, wanted to do the experiment and set up their own wiki for the paper or book, and allow either free-for-all access or password-required gateway granted access to allow modifications and annotations, I would be very interested in taking part in that collaborative effort.</p> http://mathoverflow.net/questions/47636/are-there-moves-between-reidemeister-moves/47667#47667 Answer by sleepless in beantown for Are there moves between Reidemeister moves? sleepless in beantown 2010-11-29T12:13:42Z 2010-11-29T12:13:42Z <p>Take a look at page 180 of <em>Low dimensional topology</em> by Tomasz Mrowka, Peter Steven Ozsvát, (following Ben Webster's comment about <strong>Movie Moves</strong> elucidated by Baez and Langford and their 30 basic movie moves, and by Carter and Saito who describe a 31st basic movie move.) A movie move is a sequence of frames of a braid (or subregion of a knot, I suppose). </p> <p>Carter and Saito have a theorem that</p> <blockquote> <p>two movies represent the same tangle cobordism iff they can be related by a sequence of <em>movie moves</em></p> </blockquote> <p>If you take the subset of Movie Moves where each movie is a composition of a sequence of Reidemeister moves, it seems like that would be similar or equivalent to what you are calling "Higher Reidemeister moves." Am I understanding you correctly?</p> <p>I would point you out to the appropriate page on that wiki o' info, but "Movie moves" does not even show up on their search page.</p> http://mathoverflow.net/questions/47471/classification-of-surfaces-composed-of-circles/47491#47491 Answer by sleepless in beantown for Classification of surfaces composed of circles sleepless in beantown 2010-11-27T08:23:31Z 2010-11-27T08:31:59Z <p>If you had a torus in $\mathbb{R}^3$ created by sweeping a circle of radius $1$ perpendicular to the path defined by the circle $x^2+y^2=1$ (or equivalently the parametrized version of the circle as $x=sin(\theta), y=cos(\theta), 0 \le \theta \le 2\pi$), then the point $(0,0,0)$ is contained in an infinite number of circles.</p> <p>I believe you are asking about the outer hull of a swept surface defined by sweeping an outline/path object along another outline or path object. As Michael Hardy pointed out, there is more than one way to specify a torus as the sweep of a circle. In fact, there are four ways with the Villarceau circles. Sometimes, specifying one of the non-obvious circles is a better way to solve certain problems (see <a href="http://mathoverflow.net/questions/37295/advanced-view-of-the-napkin-ring-problem/" rel="nofollow">http://mathoverflow.net/questions/37295/advanced-view-of-the-napkin-ring-problem/</a> about the volume of napkin rings)</p> <p>There's a great image of the non-obvious Villarceau circles at Benoît Kloeckner's web site:</p> <p><a href="http://www-fourier.ujf-grenoble.fr/~bkloeckn/images/villarceau.png" rel="nofollow">http://www-fourier.ujf-grenoble.fr/~bkloeckn/images/villarceau.png</a></p> <p>If the radius of curvature of the sweep-path becomes smaller than the radius of the circle being swept, then you get strangeness unless your meshing algorithm remembers to remove the extraneous mesh-points that are within the swept-volume's outer hull.</p> <p>I am not sure I'm completely following your differentiation between <em>circled</em> vs. <em>hoop</em> surfaces, unless you mean that in a circled surface the swept circle is maintained perpendicular to the sweep path, whereas in a hoop surface, it is not. Is that what you meant?</p> <p>It's also possible to define a sweep surface where the radius of the swept circle also varies parametrically along the sweep path, and if you do that in two-dimensions, you can emulate the behaviour of an ink-pen or a marker where increasing the pressure (or slowing the speed of the pen) leads to a thicker weight line being drawn at certain regions of the "writing path".</p> http://mathoverflow.net/questions/47181/can-we-uniquely-define-a-graph-to-have-the-topology-of-a-polytope-via-proper-edge/47464#47464 Answer by sleepless in beantown for Can we uniquely define a graph to have the topology of a polytope via proper edge length selection? sleepless in beantown 2010-11-26T21:45:52Z 2010-11-26T22:40:15Z <p>You are asking about the <strong>embedding of a graph structure into 3-space</strong> $\mathbb{R}^3$. A graph structure by itself does not specify its embedding into $n$-space. In chemistry, these two different chiral instances of (tetrahedral) molecules below would be called stereo-isomers or enantiomers of each other.</p> <p>In mechanical engineering, you'd be talking about building trusses and support structures, and a lot is known about the fact that quadrilaterals do not define a rigid structure. Quadrilaterals are easily sheared within a plane, and are not restricted to being coplanar, whereas triangular faces are at least limited to being coplanar.</p> <p>Also, the presence of these constraints (on edge length and vertex-edge connectivity) also does not mean that it would be impossible to build partial structures that meet the specified partial constraints but which cannot be built upon to complete the structure. In other words, a "naive constructor" cold generate a partial assembly which is a configuration which is impossible to continue onto a final desired construction. There could be <strong>dead-end</strong> partial constructions which could not be completed. This type of problem could partially be avoided by also imposing a temporal constraint, or a sequence constraint, e.g. first add this, then add that. </p> <p>However, there are chirality issues in play which cannot be avoided.</p> <p>If the "vertices" do not impose restrictions on relative angles, then there are no additional contraints beyond edge-length, and the graph-structure and edge lengths will not usually define a single embedding in 3-space, relative to transformations such as translation and rotation.</p> <p>If by topology, you do not also mean chirality, you may be correct. If you allow chirality differences to mean something, then there is a simple counterexample in the tetrahedron.</p> <p>Let this tetrahedron $T_1$ in $\mathbb{R}^3$ be defined with a base triangle $ABC$ with the points $A=(0,0,0), B=(0,1,0), C=(1,0,0)$ and the top of the tetrahedron at $D=(0,0,1)$. Let the edge lengths of the skeleton of this polytope be defined based on this baseline instantiation in 3-space, $|AB|=1, |AC|=1, |BC|=\sqrt{2}, |AD|=1, |BD|=\sqrt{2}, |CD|=\sqrt{2}$.</p> <p>Now note that if $D$ is instead placed at $D_2=(0,0,-1)$, that the this alternate tetrahedron (let's call it $T_2=ABCD_2$), has the same edge lengths as $T_1$, but has the mirror chirality. If we labeled the vertices with $A,B,C,D$, it is not possible to rotate and translate $T_1$ into $T_2$, whereas it is possible to turn $T_1$ inside-out and transform it into $T_2$.</p> <p>If you don't have all triangular faces, e.g. you use the edge lengths of a cube as the only constraints on a skeleton of a cube, you'll quickly see the problem that engineers found in constructing trusses with square faces: parallelograms are not necessarily "rigid" and can be sheared easily and still maintain the correct edge-lengths between vertices. Thus it's not possible to build a rigid skelton with only square faces.</p> <p>Thus, it depends on the axiomatic construction of your objects:</p> <p>if you disallow disassembly and reconstruction, then the tetrahedra $T_1$ and $T_2$ are separate chiral mirror-images of each other. If you allow for disassembly and reconstruction, then $T_1$ and $T_2$ have the same topology. If you also define "topologically equivalent" to allow for elastic stretching (at least for transforming from one 3-d realization to another, then back to being solid and rigid while in a specific 3-d realization), then $T_1$ can be transformed into $T_2$ by pushing the vertex $D$ through the center of the face $ABC$ and onto the other side. If the faces actually have a physical planar object defining that face (like a kite has its tissue paper), then this sort of transform is disallowed and the mirror image tetrahedra $T_1$ and $T_2$ are different.</p> <p>You can also visualize this by allowing the edges to be made of elastic springy rods with spring constants $k_i$. If the $k$'s are very large, then the springs are very stiff and the inversion will be impossible; if the $k$'s are small, the springs have a lot of give and it's easily possible to change between the two mirror-image configurations.</p> http://mathoverflow.net/questions/47395/universities-not-listed-on-mathjobs/47398#47398 Answer by sleepless in beantown for Universities not listed on mathjobs sleepless in beantown 2010-11-26T04:55:55Z 2010-11-26T04:55:55Z <p>Jane, I don't think you hurt your chances by asking if they might have a position available. If you end up going out to one institute for an interview, call other nearby colleges and see if you could drop by and sit in on a seminar and talk with the chair if at all possible. Definitely talk with the faculty there; some of them might even have suggestions as to other facilities who may be in need of someone, soon if not at that particular time. </p> <p>When you speak with the department chair's secretary/office manager, make sure to find out if there are any colloquia scheduled around the time that you would be visiting that part of the country. See if the department chair might be willing to meet with you, even if they don't have a job available there at the moment, so that you may introduce yourself as someone who is going to be part of the mathematics community in that geographic region since you'll probably land a job somewhere nearby. It never hurts to ask for advice; and if you want a job, it never hurts to let people know that you're looking. You may also learn a bit from talking with recent hires about the local mathematics community, and where spots might be opeining up nearby.</p> http://mathoverflow.net/questions/47188/non-mathematician-submitting-to-top-maths-journal/47190#47190 Answer by sleepless in beantown for Non-mathematician submitting to top maths journal? sleepless in beantown 2010-11-24T03:33:20Z 2010-11-24T03:33:20Z <p>What do you have to lose by submitting an article for publication? You'll have an even better record/credentialing/verification of the work you've put into it by being published in a Journal of good reputation. In the worst case, you will get a rejection letter, perhaps with a good explanation of why they are rejecting the article. The in-between case is pretty good too: you'll receive a referee report which may criticize your approach, suggest particular points to be polished and corrected, perhaps suggest a different approach to take, perhaps suggest that some of this work has been done by others who you shoud read and study or perhaps refer to in your work.</p> <p>If you're lucky enough to be asked to rewrite and resubmit for consideration, you're possibly on your way to being published. If not, you'll at least have made some progress and educated yourself about the academic publication culture, and will be more prepared for the next article which you prepare for submission.</p> <p>May I recommend that you find out what your activation energy level is, exceed it, and go ahead and clean up your paper and submit it for publication. Best wishes and good luck. <em>Go for it!</em></p> <p>Would you mind sharing the arxiv link to your work?</p> http://mathoverflow.net/questions/46684/fair-but-irregular-polyhedral-dice/46774#46774 Answer by sleepless in beantown for Fair but irregular polyhedral dice sleepless in beantown 2010-11-20T20:38:52Z 2010-11-23T06:25:40Z <p>I just noticed Joseph's comment in his question about Markov chains. My observations about the correctness of trying to use Markov chains to describe the rolling of a die, fair or unfair:</p> <p>If by <em>state</em> in the Markov chains, you mean just the "face" it is currently on or the "face" which is lower-most in attitude at a particular point in time, then it is inappropriate to use Markov chains because the likelihood of transitioning from die face $F_i$ to die face $F_j$ is not purely dependent upon the current state. If $F_j$ and $F_k$ are two "faces" adjacent to face $F_i$, then the likelihoods of transitioning $F_i \to F_j$ vs. $F_i \to F_k$ is not just dependent on the "current state" being $F_i$, but also dependent upon the velocity, position, and orientation of the die. The "faces" are necessary but <strong>not sufficient</strong> to encode state in such a way for Markov chains to be applicable: that the Bayesian requirement that "current state" at time $t$ is all that is needed to be known in order to be able to predict the likelihood of the state at time $t+1$ (if you talk about discrete time) or time $t+\varepsilon$ if you talk about continous time.</p> <p>If by "state", you try to get around this factor that only current state be considered and not the history of how you came to currently be in that state, then you could try to add the vectors of position, velocity, and orientation as extra "states", which is valid in numerical simulation, because ultimately all reals are still encoded into limited precision "floating point" representations of reals. However, the transition table would be huge if you allowed even for 16-bit floating point representation.</p> <p>I do not think that history-less "Markov chains" can be applied in this situation.</p> <p>older answer components below</p> <hr> <p>To answer Benoît Kloeckner's comment that &lt;&lt;[then] the solid angle under which each face is seen from the center of gravity alone would determine the probability of the dice landing on that face. But to determine all polyhedra for which this solid angle is constant is already a nice problem.>></p> <p>I don't believe that having similar solid angles is sufficient to determine the equal probabilities of the die landing on faces with similar solid angles.</p> <p>Here is a construction for 2-d die (which can easily be converted into prismatic die, disregard if the die lands on "top" or "bottom" face, and look at the relative probabilities of landing on the prismatic faces)</p> <p>Using polar coordinates $(r,\theta)$ , let's define a fair hexagonal die's profile as the closed path determined by the six vertices at </p> <p>$(1,\frac{\pi}{3}), (1,\frac{2\pi}{3}), (1,\pi), (1,\frac{4\pi}{3}), (1,\frac{5\pi}{3}), (1,{2\pi})$</p> <p>Now let us define an unfair hexagonal die's profile as the path defined by the polar coordinates</p> <p>$(1,\frac{\pi}{3}), (1,\frac{2\pi}{3}), (100,\pi), (1,\frac{4\pi}{3}), (1,\frac{5\pi}{3}), (100,{2\pi})$</p> <p>Now this die's center of mass (center of gravity) remains at $(0,0)$ since the material the die is composed of has uniformly homogeneous density. This unfair die also has each prismatic face subtending equal solid angles (and equal angles of $\pi/3$ for each edge in the $2$-dimensional case), however the this unfair die is highly biased towards landing on two faces to the detriment of the other four faces probabilities.</p> <p>Thus Benoît Kloeckner's conjecture that</p> <p>"the solid angle under which each face is seen from the center of gravity alone would determine the probability of the dice landing on that face"</p> <p>is incorrect.</p> <p>In fact, using this polar coordinate approach, it can be seen that using <em>any</em> three radii greater than $0$ in length yields a rotationally symmetric die profile with <em>equiangular faces</em> (<strong>edges which subtend equal angles in $2$-d, prismatic faces which subtend equal steradians of solid angle in $3$-d</strong>) and with center of mass still at $(0,0)$:</p> <p>$(r_1,\pi/3), (r_2,2\pi/3), (r_3, \pi), (r_1,4\pi/3), (r_2,5\pi/3), (r_3,2\pi)$</p> <p>but very few of these would be fair. Particularly the non-convex profiles, which also are equiangular, but make it possible to land on pairs of vertices/edges without landing on a specific face</p> http://mathoverflow.net/questions/46793/how-many-winning-configurations-can-you-have-in-a-nxn-tic-tac-toe-game-where-play/46800#46800 Answer by sleepless in beantown for How many winning configurations can you have in a nxn Tic-Tac-Toe game where players win if a they get n/2 in either a row or column, consecutively. sleepless in beantown 2010-11-21T05:13:49Z 2010-11-22T03:35:42Z <p>The answer depends upon, of course, if the players make their moves intelligently or stupidly. In either case, build an alpha-beta tree of move sequences and search either depth-first or breadth-first, stopping and backtracking as winning configurations are found. However, there are some rigorous points you have not defined about this question, including whether perhaps it came to you as a homework problem. Hoping that this is not the case, here's a take on it:</p> <p>Yes, if $n=2$, then the first player to move obviously wins by picking any one of the four open spots, and the game is over. If $n=3$ (which you don't define for odd $n$, but lets say for odd $n$, $(n+1)/2$ spaces wins, then the first player to move wins automatically also, because no matter which space they take in the $3 \times 3$ grid, the next player can only block one space, and in the 2nd round, first player easily wins.</p> <p>For $n=4$, the first player wins, with the same strategy as shown for $n=3$. $A$ picks any square in the $4 \times 4$ grid, $B$ picks any other square but can only block $A$ in <strong>one</strong> of the two dimensions in which this grid lies, therefore on the first step of the second round, $A$ wins by picking a square adjacent to his first square.</p> <p>Are you sure that you've thought this out for the smaller grid games?</p> <hr> <p>So for the $4 \times 4$ configuration, assuming that $A$ plays intelligently, there are </p> <p>$4 \times 4 = 16$ first moves for $A$</p> <p>$16 - 1 = 15$ first moves for $B$, if $B$ plays stupidly, or either $2, 3,$ or $4$ moves if $B$ plays intelligently</p> <p>$4 \times 1 + 8 \times 2 + 4 \times 3 = 4+16+12=32$ second moves for $A$ which win the game and end the game ($A$ could have picked one of the $4$ corners, $B$ blocks one way leaving $1$ way for $A$ to play, if $A$ picks one of the non-corner edge pieces (8 ways to do that), then $B$ blocks one way leaving two ways for $A$'s second winning move, and if $A$'s first move is a center piece, $B$ can block one of the $4$ adjacent squares, leaving $3$ adjacent squares for $A$ to pick and win.</p> <p>So if $A$ and $B$ both play intelligently, the sum of these are the possible games for a $4 \times 4$ grid:</p> <p>$4 \cdot 2 \cdot 1 + 8 \cdot 3 \cdot 2 + 4 \cdot 4 \cdot 3 = 8 + 48 + 48 = 104$ </p> <p>If $A$ and $B$ do not play intelligently, then the number of possible games is much larger, and the number of "winning configurations" or "end positions" is much larger.</p> http://mathoverflow.net/questions/46684/fair-but-irregular-polyhedral-dice/46707#46707 Answer by sleepless in beantown for Fair but irregular polyhedral dice sleepless in beantown 2010-11-20T03:25:23Z 2010-11-20T03:25:23Z <p>Here's an argument for an irregular octahedral-prism die:</p> <blockquote> <p>Have two faces be regular octagons with edge-length $1$</p> <p>Have the prism height be $h$</p> <p>Call the $8$ rectangular prismatic faces with labels "1" through "8"</p> <p>Call the octahedral faces "9" and "10" or "top" and "bottom"</p> </blockquote> <p>It may require physical simulation to find the critical heights I am defining below, and it will obviously be a function of the material used to make the die , the surface onto which it is rolled, and of the frictional coefficients $\mu_{die}$ and $\mu_{felt}$, and the nature of the rolling mechanism used to toss the die onto the felt (most likely the tumbling action in 6-dimensional state-space describing the initial velocity and rotational movement of the die and the height at which it is released, etc.)</p> <p>Because of symmetry, we can say</p> <p>$p_{top}=p_{bottom}=x,$ </p> <p>$p_{i\in(1,8)}=y$,</p> <p>$2x+8y=1$,</p> <p>$0 \le x \le 1 \textrm{ and } 0 \le y \le 1$</p> <p>There will be a critical height $h_{min}$, where for $h \lt h_{min}$, the bias for "top" and "bottom" will be greater than for the prismatic faces, and $x \gt y$.</p> <p>There will also be a critical height $h_{max}$ where because the long axis is substantially longer than the flat octahedral faces, the momentum will be such as to cause a die landing on an octahedral face "top" or "bottom" to continue to move and fall/topple onto a rectangular prismatic face, with each of the 8 prismatic faces equally likely.</p> <p>Somewhere between the critical maximal and minimal heights will be a fair height, $h_{fair}$, where this non-regular die will roll fairly onto any of the ten faces with $$p=\frac{1}{10}=0.1$$</p> <p>Now this is a rotationally symmetric die with 8-rotations around the long axis and 2 orientations of the long axis, so it's not the "assymetric die" you've asked for, but it could be the beginning for a similar construction of an asymmetric die.</p> http://mathoverflow.net/questions/45920/shortest-grid-graph-paths-with-random-diagonal-shortcuts/45930#45930 Answer by sleepless in beantown for Shortest grid-graph paths with random diagonal shortcuts sleepless in beantown 2010-11-13T15:40:24Z 2010-11-13T16:01:10Z <p>For an $n \times n$ grid, the probability of finding a path of length $n\sqrt{2}$ is $1/2^n = 2^{-n}$.</p> <p>For a grid of size $(n,0)$ or $(0,n)$, the expected path length is $n$ with probability $p=1$. Let's call the expected path length $L(x,y)$</p> <p>$$L(n,0)=L(0,n)=1 \cdot n = n$$</p> <p>For a grid of size $(n,1)$ or $(1,n)$, the expected path length is $n+1$ if all possible diagonals face the incorrect way, and $n+\sqrt{2}$ if there exists at least one-diagonal facing the correct way to create a short-cut:</p> <p>$$L(n,1)=L(1,n)=\binom{n}{1} \cdot \frac{1}{2^n} \cdot n + (1 - \binom{n}{1} \cdot \frac{1}{2^n}) \cdot (n+\sqrt{2} )$$</p> <p>$$L(n,1)=L(1,n)= n + (1 - \binom{n}{1} \cdot \frac{1}{2^n}) \cdot \sqrt{2}$$</p> <p>$$L(n,1)=L(1,n)= n + (\frac{2^n -1}{2^n}) \cdot \sqrt{2}$$</p> <p>For a grid size $(n,2)$ or $(2,n)$, the expected shortest path length is $n+2$ if in all of the locations, there are no correct facing diagonals; $n+1+\sqrt{2}$ if the short-cut diagonal only occurs in the last square (top-most); or length $n+2\sqrt{2}$, if there is a short-cut diagonal in one of the first column's $n-1$ lower squares, and a short-cut diagonal in one of the second column's upper squares after the lower square. </p> <p>This could probably be written as a recursive formula to see what the limit yields, as $L(0,0)=0$, $L(1,0)=L(0,1)=1$, $L(1,1)=\sqrt(2)\cdot\frac{1}{2}+2\cdot\frac{1}{2}=1+\frac{\sqrt{2}}{2}$, </p> http://mathoverflow.net/questions/45802/undergraduate-math-research/45810#45810 Answer by sleepless in beantown for Undergraduate math research sleepless in beantown 2010-11-12T10:32:02Z 2010-11-13T05:56:13Z <p>Since you are a student who's already interested in going on to graduate school and is specifically asking about finding a topic to study at your undergraduate level program at McNair, please <em>disregard the negative nattering nabobs</em> whose answers and comments suggest that undergraduates have no place or business in trying to perform research, whether it's research as defined for all scientists or the "research experience" that is put together for undergraduates and for advanced high-school students. Undergraduates can definitely perform research, or even benefit from going through a structured and well-administered "research experience".</p> <p>I agree with Peter Shor about finding a mentor, or multiple mentors, as soon as possible. There's no reason you have to be limited to getting advice from just one professor or teacher.</p> <p>I agree with Ben Webster, specifically about speaking with professors in order to get a reasonable idea about the level of work that would be needed for you to perform useful research at an undergraduate level. A few other suggestions come to mind:</p> <ul> <li><p>if you are at an institution that offers Masters and Ph.D. level degrees in mathematics, then your institution's library should have multiple <strong>research journals in hard-copy</strong>. I have found that it is much easier to go to the stacks in the library and browse through one or two year's worth of Tables of Contents and Abstracts in one journal in an afternoon or evening. This will familiarize you with the types of research papers being published currently, and make you aware of what "quanta" of research is enough to be a single research article.</p></li> <li><p>make sure to attend Seminars, Colloquia, and (if your school's graduate students have one) any graduate research seminar courses that you can find time for. This will allow you to become more familiar with various subtopics within the topics of your interests, and to see what the current areas of interest are for local and visiting faculty members. </p></li> <li><p>Colloquia are great as they often start by including a brief history of the topic by an expert in that field. </p></li> <li><p>Seminars are great because they allow students to see the social aspect of math, including the give-and-take and the critical comments and requests for more detail and explanation, even by tenured faculty who don't follow a speaker's thought processes. </p></li> <li><p>Graduate student seminar presentations are great because a student observes how graduate students can falter during presentations, how they are quizzed/coached/criticized/mentored/assisted by faculty during their presentations.</p></li> <li><p>I'll admit that I'm not sure attending dissertation defenses would be of any serious benefit to the undergraduate student, other than observing the interaction level (animosity level?) between faculty and graduate students.</p></li> <li><p>absolutely make sure to schedule some time to meet with mathematics professors who specialize in the fields of your interest, and communicate your desire to do research while you are an undergraduate, and communicate your desire to go on to graduate studies in mathematics.</p></li> <li><p>look on the internet and search for undergraduate opportunities for research in mathematics. I guarantee you will find quite a number of web sites that can give you more information. MIT has an undergraduate research opportunity program that many of their students take advantage of. Your institution may have professors who can speak with you and give you advice. </p></li> </ul> <p>Also, make sure to speak with more than one professor, and <strong>do not take any single person's advice as being the final word.</strong> Mathematicians are human beings too, and subject to the foibles and inclinations and disinclinations that all human beings have. If you run into disgruntled and critical individuals, do not let that dissuade you from going on into mathematics or decrease your desires. If you run into overly optimistic individuals who praise you too much and are too eager to take you on to do "scut work" computer programming, thank them for their time and let them know you'll come back to speak with them after you've spoken with other professors and weighed your options. Don't turn anyone down immediately. Always be polite in speaking with professors and teachers. Ask them how they chose their topics for their degrees, and you'll learn a lot.</p> http://mathoverflow.net/questions/22522/how-is-a-permutation-taken-as-an-equivalent-of-a-hash-function-in-minwise-indepen/45813#45813 Answer by sleepless in beantown for How is a permutation taken as an equivalent of a hash function in MinWise independent permutations? sleepless in beantown 2010-11-12T11:12:41Z 2010-11-12T11:12:41Z <p>A permutation is the equivalent of an unbiased "good" hash function because it has even distribution. Since a permutation maps (1,2,...,n) $\to$ permuted-list-of(1,2,..., n), if the original set $A$ is an unbiased random selection of the range $1$ to $n$, then the likelihood that the $B$="permutation-hash" of $A$ will contain the minimum-value of $1$ to $n$ is the size of the set $A$ divided by $n$.</p> <p>If the hash function is equitable and well-distributed, then the range of input values will be "hashed" into and map into the output range in an evenly distributed fashion. If the domain is a group of 256 characters, and the hash is a function which outputs 8 characters, then you want the domain to be partitioned into as closely-equal-sized subsets based on the what the hash of the domain words are.</p> <p>A hash normally reduces the size of the object being hashed, thus because of the pigeonhole priniciple, at least one element of the range of the hash must map to more than one item in the domain. An equitable hash attempts to minimize the variance in the size of the bins which the domain is partitioned into. An inequitable hash leads to some of the bins being very different in size from the other bins. Thus a hash is usually a <strong>one-way</strong> mapping: given the hash of a message, it is not possible to determine what the original message consisted of, even though it may be possible to deduce what partition the message is part of (the partition whose hash digest is ...)</p> <p>For example, the hash function EVEN-ODD, equivalently called REMAINDER-MODULO-2, is easily calculated by looking at the last (lowermost) bit of the message. If the range is a subset of $\mathbb{Z}$, e.g. the numbers $1...n$, then EVEN-ODD is an equitable hash, as the size of the partitions will differ by at most $1$.</p> <p>A permutation is equivalent to performing a hash which does not reduce the size of the domain as it maps into the range, it performs a bijection. As such, the size of each partition is $1$, the variance of the sizes of the partitions is $0$, and these partitions are as equitably sized as possible.</p> http://mathoverflow.net/questions/53601/which-platonic-solids-can-form-a-topological-torus/53658#53658 Comment by sleepless in beantown sleepless in beantown 2011-02-01T02:20:51Z 2011-02-01T02:20:51Z @Tracy-Hall, my eyes must be missing the reference to the 1972 paper on this page. What/where exactly is the paper you're referring to? Thanks for the Martin Garner reference. http://mathoverflow.net/questions/51169/flux-through-a-mobius-strip/51177#51177 Comment by sleepless in beantown sleepless in beantown 2011-01-05T06:47:34Z 2011-01-05T06:47:34Z @Scott-Carnahan, the Mobius resistor concept missing the fact that the only resistance being provided between two leads connected to it is the internal resistance of the conductor. In electrical circuit modeling, you take the resistance between two components connected to the same lead to be $0$ (zero). There's no need to worry about &quot;self-inductance&quot; with a mobius resistor because you're also connecting to the same side of a capacitor. It might act a little bit like an antenna perhaps, but not as a resistor in any useful sense of the word. A mobius resistor is a <b>short circuit</b>. http://mathoverflow.net/questions/51176/autocorrelation-of-a-1-valued-random-process-with-certain-statistics Comment by sleepless in beantown sleepless in beantown 2011-01-05T06:28:08Z 2011-01-05T06:28:08Z Why do you make the statement or the claim that the switches are not independent here? Obviously, switches between $+1$ alternate with $-1$, so in one sense of sequence alone with disregard to the time between the sign changes there is a predictable component: $+1$ follows $-1$ follows $+1$... However, the time interval between the sign changes is still a random variable, isn't it? Also, the $\LaTeX$ command you want for $\pm$ is \pm, not \plusminus http://mathoverflow.net/questions/50779/floating-polyhedra-with-fair-equilibria/50787#50787 Comment by sleepless in beantown sleepless in beantown 2010-12-31T05:00:36Z 2010-12-31T05:00:36Z Prismatic die with a regular polygon bases would also exhibit the same behaviour with each prismatic face equally likely to float &quot;up&quot; with the $2$ flat polygon-base faces less likely to float &quot;up&quot; as long as the height of the prismatic die is larger than $ar$ where $r$ is the &quot;radius&quot; of the polygonal base. Extreme example 1: wooden nickel is going to float face up or face down, not very likely to float on its edge. Extreme example 2: hexagonal wooden pencil dowel (without metal/rubber eraser nub) is going to float with one of its prismatic faces &quot;up&quot; rather than one of the end faces. http://mathoverflow.net/questions/50779/floating-polyhedra-with-fair-equilibria/50787#50787 Comment by sleepless in beantown sleepless in beantown 2010-12-31T04:55:24Z 2010-12-31T04:55:24Z If the height is greater than $ax$, than the triangular sloped faces of the pyramid are all equally likely to be the &quot;face up&quot; stable face, with the &quot;base face&quot; unlikely to be the &quot;face up&quot; face. http://mathoverflow.net/questions/50779/floating-polyhedra-with-fair-equilibria/50787#50787 Comment by sleepless in beantown sleepless in beantown 2010-12-31T04:53:43Z 2010-12-31T04:53:43Z This square-based (or more-sided regular polygon base pyramid) would also have a second equilibrium point with the pointy pyramidal tip facing up, but that equilibrium would be an <b>unstable equilibrium</b> whereas the equilibrium with the pyramidal base facing upwards is a <b>stable equilibrium</b>. http://mathoverflow.net/questions/50779/floating-polyhedra-with-fair-equilibria Comment by sleepless in beantown sleepless in beantown 2010-12-31T04:47:48Z 2010-12-31T04:47:48Z So you're asking if there are preferred orientations of floaty objects like ice-cubes and ice-bergs, particularly if they're shaped into polyhedra? http://mathoverflow.net/questions/50624/happy-new-prime-year/50752#50752 Comment by sleepless in beantown sleepless in beantown 2010-12-31T01:39:29Z 2010-12-31T01:39:29Z @Ricardo and @Wadim-Zudilin, yes the numerologists will see a way to make every year interesting, and the paradox of the first un-interesting year becoming interesting because of the fact that it <b>is</b> the first un-interesting year. ;&gt;) Also, see Mayan, Hebrew, Chinese, Indian, and many other cultural artefacts for different origins and phase-shifts for yearly and weekly and monthly calendar objects. It's amazing how human beings have dissected the night sky and the numbers we pluck out of the sky and thin air. http://mathoverflow.net/questions/50711/how-long-is-the-longest-path-in-the-game-tree-of-chess/50727#50727 Comment by sleepless in beantown sleepless in beantown 2010-12-30T14:23:45Z 2010-12-30T14:23:45Z @Didier-Piau, I'm only guessing as to the original poster's intent or thoughts, but considering the confusion in the question between paths on a directed graph and the number of vertices, I surmised that the mistake came from something like that. Thanks for pointing out the specific branch-pruning rule; I like Joel's comment about how it's possible for two colluding (or clueless) players to keep the game going longer pointlessly. http://mathoverflow.net/questions/50718/how-do-i-tweak-a-linear-regression-equation-to-minimize-variance-rather-than-sd Comment by sleepless in beantown sleepless in beantown 2010-12-30T14:16:23Z 2010-12-30T14:16:23Z I believe you're asking in the wrong forum for this type of question. You may find the error in your code's mathematics by possibly asking on math.stackexchange.com, or you may find the error in your mathematic's program code by asking on stackoverflow.com; in either case, please look at the FAQ's and note that what you are asking is not a mathematical research level question. <b>You should either pay a consultant to correct your code for you if you are at work, or you should do your own homework if you are at school.</b> http://mathoverflow.net/questions/50711/how-long-is-the-longest-path-in-the-game-tree-of-chess/50727#50727 Comment by sleepless in beantown sleepless in beantown 2010-12-30T14:10:22Z 2010-12-30T14:10:22Z I meant to say &quot;the total number of paths which being at a fixed starting vertex&quot; which can be much much larger than the number of vertices, not &quot;the number of paths leading out from a starting vertex&quot; which sounds like it is limited to &quot;the outdegree of the starting vertex&quot;. Alas, poor rigour in my own statements. http://mathoverflow.net/questions/50711/how-long-is-the-longest-path-in-the-game-tree-of-chess Comment by sleepless in beantown sleepless in beantown 2010-12-30T14:05:24Z 2010-12-30T14:05:24Z Why the down-votes without comments or explanations? It's an interesting question, even if it is poorly formed and not rigorously defined, and does not explain the question asker's belief and reasoning behind why he posits that $5^{64}$ is the upper bound on &quot;all configurations&quot;? It's easy to think of reasons to criticize the question; so I find it sad that people are down-voting without taking the effort to leave a comment. And why not leave feedback allowing the original poster to edit and improve the question? There's no reason to pile on down-votes indiscriminately http://mathoverflow.net/questions/50711/how-long-is-the-longest-path-in-the-game-tree-of-chess Comment by sleepless in beantown sleepless in beantown 2010-12-30T13:44:51Z 2010-12-30T13:44:51Z If you don't <b>require</b> a draw to be declared, there are multiple scenarios in which king vs. king or (king+queen) vs. (king+queen) can play on infinitely; in that case, the game tree of chess is unbounded. There must be a strict rule for when to prune a branch in the game tree. @Didier-Piau, the upper-bound concept appears to have 3 mistakes in it. It may be the concept of {white pawn, white other, black pawn, black other, empty}$^64, which has a set size of$5^64$http://mathoverflow.net/questions/50624/happy-new-prime-year/50626#50626 Comment by sleepless in beantown sleepless in beantown 2010-12-29T19:41:30Z 2010-12-29T19:41:30Z @A-Rex asaurus, the$n+1$being prime was my original conjecture which I did not edit out when I found the first counter example for$k=3$,$n=63$, with$n+1=64=2^6$,$n+2=65=5\cdot13$,$n+3=66=2\cdot3\cdot11$. So$n+1$only has to be prime (or a power of a prime) if$n$is even; if$n$is odd,$n+1$can be a power of an even prime thus it must be a power of$2\$. http://mathoverflow.net/questions/50624/happy-new-prime-year/50626#50626 Comment by sleepless in beantown sleepless in beantown 2010-12-29T08:05:27Z 2010-12-29T08:05:27Z @Gerhard-Paseman: Perlify it into a one-liner, and I'll see your code and raise it to APL! Or maybe get out the punch-cards and do some FORTRAN-WATFOR.