Re: Re: programming competition

*To*: mathgroup at smc.vnet.net*Subject*: [mg4915] Re: [mg4866] Re: programming competition*From*: "w.meeussen" <w.meeussen.vdmcc at vandemoortele.be>*Date*: Fri, 4 Oct 1996 00:17:48 -0400*Sender*: owner-wri-mathgroup at wolfram.com

At 22:42 26-09-96 -0400, you wrote: >Xah wrote: >>Problem: >>Suppose you have a ordered list {p1, p2, ... , pn} where points has the form >>{x,y}. You want to modify the list so any neighboring points will have a >distance >>less or equal to a given value maxLength. You do this by adding points in >between >>points. For example, suppose p3 and p4 has length greater than maxLength. >Your new >>list should then be >>{p1, p2, p3, newP1, newP2,..., newPm, p4, p5, ... , pn} where >>newP1, ...newPm lies on the line p3 p4. > >This was Xah's list of points: > >In[26]:= > li = Join[ {{0,0},{1,1}}, Table[{i,1}, {i,1,2,1/5}], {{2,1}, {3,3}}]; > >Xah gave linearInterpolate[] as a solution. > (****************** portion deleted ************************) > >In[34]:= > linearInterpolate2[ li, .3] >Out[34]= > {{0.2, 0.2}, {0.4, 0.4}, {0.6, 0.6}, {0.8, 0.8}, {0, 0}, {1, 1}, > > 6 7 8 9 > {1, 1}, {-, 1}, {-, 1}, {-, 1}, {-, 1}, {2, 1}, {2.125, 1.25}, > 5 5 5 5 > > {2.25, 1.5}, {2.375, 1.75}, {2.5, 2.}, {2.625, 2.25}, > > {2.75, 2.5}, {2.875, 2.75}, {2, 1}, {3, 3}} (****************** portion deleted ************************) > >This is better, but it still doesn't give the answer in the form Xah >specified; the new points are not positioned in the list between the points >they're being interpolated between, i. e. they are not >{p1, p2, p3, newP1, newP2,..., newPm, p4, p5, ... , pn}. > >With Lichtblau's and Hayes' admonitions to use procedural programming >to solve procedural problems ringing freshly in my ears, I offer the >following: > (****************** portion deleted ************************) >In[70]:= > insertPoints[li, .3] >Out[70]= > {{0, 0}, {0.212132, 0.212132}, {0.424264, 0.424264}, > > {0.636396, 0.636396}, {0.848528, 0.848528}, {1., 1.}, {1., 1.}, > > {1.2, 1.}, {1.4, 1.}, {1.6, 1.}, {1.8, 1.}, {2., 1.}, {2., 1.}, > > {2.13416, 1.26833}, {2.26833, 1.53666}, {2.40249, 1.80498}, > > {2.53666, 2.07331}, {2.67082, 2.34164}, {2.80498, 2.60997}, > > {2.93915, 2.8783}, {3., 3.}} > >Xah's description of the problem didn't say anything about having the >points evenly spaced, so they're not. But it would be simple to modify >one line of code so they were evenly spaced. > >As Lichtblau points out, repeated copying of lists makes the problem >O(n^2). > >In[137]:= > pointsList1 = Table[{Random[], Random[]}, {100}]; > pointsList2 = Table[{Random[], Random[]}, {1000}]; > >In[139]:= > linearInterpolate2[ pointsList1, .3]; // Timing > linearInterpolate2[ pointsList2, .3]; // Timing >Out[139]= > {3.8 Second, Null} >Out[140]= > {15.75 Second, Null} > >Avoiding copying of lists keeps the problem to O(n) > >In[143]:= > insertPoints[pointsList1, .3]; // Timing > insertPoints[pointsList2, .3]; // Timing >Out[143]= > {2.4 Second, Null} >Out[144]= > {4.25 Second, Null} > >I periodically need to do fairly complicated operations on >small lists. Using the list manipulation functions may make the >problem O(n^2), but for small lists this doesn't really matter. >Using Append[], Rest[], and similar functions simplifies the >code and makes it easier to develop and debug, compared to >using a procedural approach and trying to write the fastest >code possible. In Xah's and Richard Gaylord's (actually, Don >Piele's) problems it was easier to develop a solution using >a procedural approach, but it is often easier to use an approach >that results in slower code. What I ask from Mathematica is that >the code be as clear and self-documenting as possible, the >development process be as short and painless as possible, and >the result be correct. If speed were that important to me >I'd be writing in a compiled language. The problem with many >blocks of code is not that they're slow, but that >they're confusing. This makes it more difficult to debug >and be sure of getting a correct result. Overall, I think >top-down design and attention to formatting are more important >that speed. Because of it's clarity, I liked Paul Abbott's pattern >matching approach to the Gaylord-Piele problem. It worked fine >on short lists and you could tell at a glance what the code was >doing. > >-- >Bob Hall | "Know thyself? Absurd direction! >rhall2 at gl.umbc.edu | Bubbles bear no introspection." -Khushhal Khan Khatak > > > > Bob, it seems to be a kind of problem suitable to a rule-based attack : for a simplified one-dimensional case : test=Table[n+Random[Real,{1,10}],{n,0,1000,10}] a rule "rule1" is defined that shows the inserted numbers as lists : rule1={a___,b_/;NumberQ[b],c_/;NumberQ[c],d___}/;Abs[b-c]>dif:> {a,Range[b,c-dif,dif],c,d} Timing[test//.rule1] {2.242 Second, {{6.00759, 9.00759, 12.0076}, {15.7191}, ... etc ... {963.993, 966.993}, {972.051, 975.051, 978.051}, {982.462, 985.462, 988.462}, {993.766, 996.766, 999.766, 1002.77, 1005.77}, 1009.67}} a simple Flatten[] gets rid of the extra list-envelopments. So far, still lean & mean, don't you think? Now, in two dimensions, why not use complex numbers, and keep the same strategy ? testc=Table[{n+Random[Real,{1,10}],n+Random[Real,{1,10}]},{n,0,1000,10}] rule2={a___, b_/;NumberQ[b], c_/;NumberQ[c], d___} /; Abs[b-c]>dif :> {a, Range[b,c-(c-b) dif/Abs[c-b], (c-b) dif/Abs[c-b]], c, d} we only need to switch between {x,y} and x+Iy and back : Timing[{Re[#],Im[#]}&/@Flatten[Apply[(#1+I #2) &, testc,1]//.rule2]] gives: {5.013 Second, {{6.67336, 4.32093}, {8.47046, 6.72311}, {10.2675, 9.12529}, ... etc ... {917.71, 920.338}, {921.006, 924.902}, {923.581, 926.442}, {926.156, 927.982}, {928.73, 929.522}, {931.97, 931.459}, {933.657, 933.94}, {935.344, 936.421}, {937.031, 938.901}, {938.718, 941.382}, {940.406, 943.863}, {943.143, 947.888}, {945.831, 949.221}, {948.518, 950.554}, {951.281, 951.925}, {953.626, 953.796}, {955.972, 955.667}, {958.317, 957.537}, {960.662, 959.408}, {963.008, 961.279}, {965.353, 963.149}, {968.588, 965.73}, {969.662, 968.532}, {970.735, 971.333}, {971.808, 974.135}, {973.642, 978.922}, {976.562, 979.608}, {979.483, 980.293}, {982.403, 980.979}, {985.324, 981.665}, {989.648, 982.68}, {990.212, 985.627}, {990.775, 988.574}, {991.339, 991.52}, {991.902, 994.467}, {992.545, 997.828}, {995.042, 999.49}, {997.54, 1001.15}, {1000.04, 1002.81}, {1002.92, 1004.73}}} I admit that it took me some time to get the Range[...] arguments right (needs rule delayed ":>", not ReplaceAll "->" to avoid unpleasant and unhelpful red remarks about the arguments not being Integers (silly, they need't be Integers at all). Also, I fell into the infinite regression twice (because of //. ). All in all, the ReplaceRepeated just scans over the list once, because the replacement generates no numbers, but lists. Final remark : it would be more elegant if a condition like b_Number were allowed, in stead of b_/;NumberQ[b] I lost quite some time on that. Wouter. ==== [MESSAGE SEPARATOR] ====

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