June 2013 - mathfun@mailman.xmission.com

Re: [math-fun] Atrocious Mma 8&9, WolframAlpha bug:
by Bill Gosper 26 Jun '13

26 Jun '13

JimP>If you make a trip through the six sectors of Z^2 that surround n=k=0, extrapolating analytically (in the discrete sense of the term), you'll find that the binomial coefficient function changes sign. So there's a sense in which (n choose k) is best thought of as living not in Z^2 but a double-cover thereof, with a branch point at (0 choose 0). I have been puzzled by this for a long time. Jim Propp For simplicity I was avoiding Z^2 and cowering in R^2, but I can't for the life of me imagine a branch point at (0 choose 0). Can you expound this sign change? On Tue, Jun 25, 2013 at 3:59 PM, Fred lunnon <fred.lunnon(a)gmail.com> wrote: There does seem to be a genuine ambiguity involved in extending binomial coefficients to negative arguments: if they are to be Taylor coefficients in the expansion of (1+x)^n , then plainly it follows that n_C_k = 0 for k < 0 . But if you define them (less naïvely) via n_C_k = 1/( (n+1)*Beta[m+1, n-m+1] ) it is evident from (say) the plot at http://mathworld.wolfram.com/BetaFunction.html that the "radiation warning" and bilateral symmetry are indeed correct. Then it's not the symmetry which breaks down: Binomial[n, m] - Binomial[n, n-m] = 0 , for all integer n,m but the recursion: Binomial[n, m] - Binomial[n-1, m] - Binomial[n-1, m-1] = 0 _except_ at m = n = 0, where it equals -1 . I haven't investigated the situation for non-integer arguments. But below Bill --- along with the CAS's --- seems to be trying to have his cake and eat it [which might help explain why he's having trouble plotting it. Mind the carpet!] Either both triangle and surface are symmetric, or neither is; and either you define 'em via Taylor series, or via Gamma functions --- but not both at once. Fred Lunnon Sorry to bifurcate on your carpet (or a least leave crumbs), but I *can* have my cake and eat it too. (n choose m) := Limit[Limit[x!/(y!*(x-y)!),y->m],x->n] The surface is clearly symmetrical under y -> x-y. The asymmetry comes from the order of the limitry. (x choose y) "goes vertical" at negative integer x, integer y, and can take on any value depending on the direction of approach. This is why it's so hard to intelligibly plot. One way to think of the negative x realm is adjacent, diagonal strips of curly-edged lasagna, uncooked. Except the curls have infinite amplitude at the edges. Now, slide alternate strips half a ruffle, and "glue" each upward-going vertical edge to the downward- going vertical edge of the adjacent strip. Now that's a surface. --rwg Noninteger arguments are totally piece of cake. On 6/25/13, Bill Gosper <billgosper(a)gmail.com> wrote: In[509]:= Table[Binomial[-1, k], {k, -3, 3}] Out[509]= {1, -1, 1, 1, -1, 1, -1} But In[550]:= Table[Binomial[m1, k], {k, -3, 3}] Out[550]= {0, 0, 0, 1, m1, 1/2 (-1 + m1) m1, 1/6 (-2 + m1) (-1 + m1) m1} In[551]:= % /. m1 -> -1 Out[551]= {0, 0, 0, 1, -1, 1, -1} Macsyma, since 40 years, (c1) makelist(binom(-1,k),k,-3,3) (d1) [0, 0, 0, 1, - 1, 1, - 1] I expect as much from Maple. Mma Doc: "As a bivariate function, Binomial is not continuous in both variables at negative integers: In[1]:= Binomial[-3, -5] Out[1]= 0 In[2]:= Binomial[-3.0000001, -5.0000001] Out[2]= 6. " But in inconvenient fact, In[552]:= Binomial[-3, -5] Out[552]= 6 I don't see how this eluded their test suite unless the latter is corrupted. Mma Doc: "In general, (n m) is defined by \[CapitalGamma] (n+1)/(\[CapitalGamma](m+1)\[CapitalGamma](n-m+1)) or suitable limits of this." For the record, these limits are Limit[Limit[x!/(y!*(x-y)!),y->m],x->n] This is how the bilateral symmetry of Pascal's triangle breaks down, even though the surface x!/(y!*(x-y)!) (appropriately skewed) is bilaterally symmetric. (I have repeatedly failed to make a convincing plot of this.) For anyone still doubting the symmetry breakdown for negative integer n: The recurrence that generates each row of Pascal's Triangle from the previous would place a "2" at the apex (0,0), if Out[509] were correct! --rwg Here is an almost correct technique for printing number triangles, of which Julian must repeatedly remind me: pt[n_Integer, from_Integer: 0] := TableForm[Table[If[EvenQ[i + j], "", Binomial[-1 + i, -Floor[n/4] + (1 + i + j)/2] /. 0 -> ""], {i, from, n}, {j, -Ceiling[3*n/4], n + 1}]] 10 -4 1 1 -4 6 -3 1 1 -3 6 -4 3 -2 1 1 -2 3 -1 1 -1 1 1 -1 1 -1 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 (Assuming GMail hasn't murdered it,) the upper left wedge is the spurious subject of this message. Pascal's Triangle is really Pascal's biwedge. NOT Pascal's Radiation Warning. _______________________________________________

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[math-fun] Atrocious Mma 8&9, WolframAlpha bug:
by Bill Gosper 25 Jun '13

25 Jun '13

In[509]:= Table[Binomial[-1, k], {k, -3, 3}] Out[509]= {1, -1, 1, 1, -1, 1, -1} But In[550]:= Table[Binomial[m1, k], {k, -3, 3}] Out[550]= {0, 0, 0, 1, m1, 1/2 (-1 + m1) m1, 1/6 (-2 + m1) (-1 + m1) m1} In[551]:= % /. m1 -> -1 Out[551]= {0, 0, 0, 1, -1, 1, -1} Macsyma, since 40 years, (c1) makelist(binom(-1,k),k,-3,3) (d1) [0, 0, 0, 1, - 1, 1, - 1] I expect as much from Maple. Mma Doc: "As a bivariate function, Binomial is not continuous in both variables at negative integers: In[1]:= Binomial[-3, -5] Out[1]= 0 In[2]:= Binomial[-3.0000001, -5.0000001] Out[2]= 6. " But in inconvenient fact, In[552]:= Binomial[-3, -5] Out[552]= 6 I don't see how this eluded their test suite unless the latter is corrupted. Mma Doc: "In general, (n m) is defined by \[CapitalGamma] (n+1)/(\[CapitalGamma](m+1)\[CapitalGamma](n-m+1)) or suitable limits of this." For the record, these limits are Limit[Limit[x!/(y!*(x-y)!),y->m],x->n] This is how the bilateral symmetry of Pascal's triangle breaks down, even though the surface x!/(y!*(x-y)!) (appropriately skewed) is bilaterally symmetric. (I have repeatedly failed to make a convincing plot of this.) For anyone still doubting the symmetry breakdown for negative integer n: The recurrence that generates each row of Pascal's Triangle from the previous would place a "2" at the apex (0,0), if Out[509] were correct! --rwg Here is an almost correct technique for printing number triangles, of which Julian must repeatedly remind me: pt[n_Integer, from_Integer: 0] := TableForm[Table[If[EvenQ[i + j], "", Binomial[-1 + i, -Floor[n/4] + (1 + i + j)/2] /. 0 -> ""], {i, from, n}, {j, -Ceiling[3*n/4], n + 1}]] 10 -4 1 1 -4 6 -3 1 1 -3 6 -4 3 -2 1 1 -2 3 -1 1 -1 1 1 -1 1 -1 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 (Assuming GMail hasn't murdered it,) the upper left wedge is the spurious subject of this message. Pascal's Triangle is really Pascal's biwedge. NOT Pascal's Radiation Warning.

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[math-fun] 217-step sliding block puzzle
by Adam P. Goucher 25 Jun '13

25 Jun '13

(Hopefully this will help to provide the much-needed break from the inundation of math-fun messages entitled `yacht question'!) Subject to the constraint that pieces are only distinguished by shape, there is a sliding block puzzle in a 4-by-4 box that requires 217 moves (in the `steps' metric) to solve: http://cp4space.wordpress.com/2013/06/25/cipher-33-puzzling-ii/ The state-space of the puzzle is a graph with 9591 vertices and a diameter of 217: http://cp4space.files.wordpress.com/2013/06/sbp-graph.png Is there a harder (i.e. requiring more moves) sliding block puzzle in a 4-by-4 box? This is the most difficult pair of positions reachable from the configuration provided by Neil Bickford as the hardest `simple^3 sliding block puzzle'. Sincerely, Adam P. Goucher

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Re: [math-fun] yacht question
by Guy Haworth 25 Jun '13

25 Jun '13

In general, the bow and the stern of a yacht do not travel the same distance. Counterexample: a yacht moored at the bow. The bow describes an arc with a smaller radius than the stern does. 'Perturbation theory': attach a long, horizontal bowsprit-like stick to the bow which makes no difference to the Centre of Lift or the Centre of Gravity. This makes no difference to the locus of the stern but exaggerates the motion of what was the bow and is therefore greater. 'Three dimensions': The bow seems to bob up and down much more than the stern, and presumably has a longer locus measured in 3D Guy Haworth

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[math-fun] 217-step sliding block puzzle
by Bill Gosper 25 Jun '13

25 Jun '13

Neil's off the grid for a week, but the url for his simple^3 version is http://nbickford.wordpress.com/category/sliding-block-puzzles/ --rwg We should bug him to post his 5x4 goodies. AdamG> (Hopefully this will help to provide the much-needed break from the inundation of math-fun messages entitled `yacht question'!) Subject to the constraint that pieces are only distinguished by shape, there is a sliding block puzzle in a 4-by-4 box that requires 217 moves (in the `steps' metric) to solve: http://cp4space.wordpress.com/2013/06/25/cipher-33-puzzling-ii/ The state-space of the puzzle is a graph with 9591 vertices and a diameter of 217: http://cp4space.files.wordpress.com/2013/06/sbp-graph.png Is there a harder (i.e. requiring more moves) sliding block puzzle in a 4-by-4 box? This is the most difficult pair of positions reachable from the configuration provided by Neil Bickford as the hardest `simple^3 sliding block puzzle'. Sincerely, Adam P. Goucher

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[math-fun] 217-step sliding block puzzle
by Bill Gosper 25 Jun '13

25 Jun '13

Neil's off the grid for a week, but that looks like the same set of pieces in his simple^3 132 mover at http://nbickford.wordpress.com/category/sliding-block-puzzles/ --rwg AdamG> (Hopefully this will help to provide the much-needed break from the inundation of math-fun messages entitled `yacht question'!) Subject to the constraint that pieces are only distinguished by shape, there is a sliding block puzzle in a 4-by-4 box that requires 217 moves (in the `steps' metric) to solve: http://cp4space.wordpress.com/2013/06/25/cipher-33-puzzling-ii/ The state-space of the puzzle is a graph with 9591 vertices and a diameter of 217: http://cp4space.files.wordpress.com/2013/06/sbp-graph.png Is there a harder (i.e. requiring more moves) sliding block puzzle in a 4-by-4 box? This is the most difficult pair of positions reachable from the configuration provided by Neil Bickford as the hardest `simple^3 sliding block puzzle'. Sincerely, Adam P. Goucher

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Re: [math-fun] yacht question
by Henry Baker 25 Jun '13

25 Jun '13

The front wheel of a bicycle travels more than the rear wheel, because it is the steering wheel. (The rear wheel track always points at the direction of tangent contact of the front wheel, more-or-less.) I would imagine that since the rudder in a (prototypical) yacht is in the rear, that the rear end of a yacht travels further than the front end. On submarines & some very large ships, they have front rudder-like steering apparatuses, as well, so perhaps they are more balanced. The real question is why boats stear from the rear (other than the fact that it is more comfortable for the captain to be there (not crashing up & down in the front), and perhaps it is the case that even with more modern technology, no one bothered to re-think where the best place to put the rudder would be. A little-known 'turning point' in history took place right around the turn of the century 1900, when the rudder wheels started turning in the direction of the turn, instead of the reverse, which had been standard up until then. This change may have been done to make it consistent with those new-fangled automobiles. Many movies get this wheel direction/turning direction relationship of these older ships wrong. http://en.wikipedia.org/wiki/Ship's_wheel At 01:55 PM 6/24/2013, James Propp wrote: >Do the front and back of a yacht travel the same distance? > >(Let's assume that the earth is flat for purposes of this problem.) > >Here's a question that I think is equivalent: If a yacht travels in a >circle, do the front and back ends of the yacht travel on circles of the >same radius? > >Jim Propp

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Re: [math-fun] yacht question
by Henry Baker 24 Jun '13

24 Jun '13

But these rear rudders were ubiquitous even on sailing ships: https://en.wikipedia.org/wiki/Rudder Here's something I just learned from Wikipedia: "Rudder post and mast placement defines the difference between a ketch and a yawl, as these two-masted vessels are similar. Yawls are defined as having the mizzen mast abaft (i.e. "aft of") the rudder post; ketches are defined as having the mizzen mast forward of the rudder post." At 04:13 PM 6/24/2013, Fred lunnon wrote: >On a small motor vessel driven by a single screw propeller, a considerable amount of >steering force is generated by the action of the propeller stream on the rudder, which >therefore has to be in line with and adjacent to the propeller.

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[math-fun] yacht question
by James Propp 24 Jun '13

24 Jun '13

Do the front and back of a yacht travel the same distance? (Let's assume that the earth is flat for purposes of this problem.) Here's a question that I think is equivalent: If a yacht travels in a circle, do the front and back ends of the yacht travel on circles of the same radius? Jim Propp

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[math-fun] refining the notion of density
by James Propp 24 Jun '13

24 Jun '13

Dan Asimov's remark "Finally, the optimality of a sphere packing is determined by the limit of its density in a ball B(R) of radius R, as R -> oo, when this limit exists, which is independent of the center of the ball. This means that any modification of a packing in a bounded region will have no effect on the limiting density." reminds me of a research program currently on my back-burner, awaiting some ideas to make it fully workable. It's a theory in which density is measured not by real numbers but by a richer non-Archimedean number system that allows one to simultaneously consider regular packings, line defects (which are infinitesimal compared to regular packings), and point defects (which are infinitesimal compared to line defects). In this theory, modifying a packing in a bounded region, by removing n spheres and replacing them by n' spheres, changes the "refined density" by an infinitesimal but non-zero amount proportional to n-n'. I can prove some of the basic lemmas in this theory, but what I really want to do in the first article on the theory is to show that, relative to my refined notion of density, the only densest packings of the plane by unit disks are the ordinary densest lattice packings. Maybe one of the existing proofs of Fejes Toth's theorem can be modified to work with refined density, but I don't know how to do this yet. Anyone interested in more details? Jim Propp

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