# Create a function that takes two parameters and returns one output?

Posted 1 year ago
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 Hello Wolfram Community,I have a function that takes two parameters and is essentially supposed to return one of two things: "line Crossed" or "no line Crossed". However with this current function I am getting Null as my output. Why is this and how can I achieve my desired output. Here is the code: needleCross[length_, distance_] := { Block[ {midPoint = RandomReal[distance], rotationAngle = RandomReal[{-Pi/2, Pi/2}]}, xCoordLinePoints = {midPoint - (Cos[rotationAngle]*length/2), midPoint + (Cos[rotationAngle]*length/2)}]; If[xCoordLinePoints[[1]] <= 0 || xCoordLinePoints[[2]] >= distance, Print["Line Crossed"], Print["No Line Crossed"]]} any advice would help!Thanks
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Posted 1 year ago
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Posted 1 year ago
 Hi,when I run it it does print out the result and gives the output Null.If you only want the output you might want to use: needleCross[length_, distance_] := {Block[{midPoint = RandomReal[distance], rotationAngle = RandomReal[{-Pi/2, Pi/2}]}, xCoordLinePoints = {midPoint - (Cos[rotationAngle]*length/2), midPoint + (Cos[rotationAngle]*length/2)}]; If[xCoordLinePoints[[1]] <= 0 || xCoordLinePoints[[2]] >= distance, "Line Crossed", "No Line Crossed"]}[[1]]; So now you can simulate Buffon's needle experiment to estimate Pi. We know that $\pi \approx \frac{2 \cdot length}{distance \cdot P}$. P is the probability to hit a line. So lets get an estimate for P.If your function is correct then M = 1000000; P = N[(Count[Table[needleCross[1, 1.2], {M}], "Line Crossed"]/M), 10] should give you an estimate of the probability that a needle crosses a line. Then an estimate of Pi should be 2./(1.2*P) which in my case gives 3.13967, 3.1468, 3.13999, 3.13775, 3.14344 for a couple of different runs.If we assume someone really keen on the experiment would run it 400 times on 8000 needles. Then we would get: estimatesP = Monitor[Table[M = 8000; P = N[(Count[Table[needleCross[1, 1], {M}], "Line Crossed"]/M), 10], {i, 1, 400}], i]; piestimates = 2./(1*#) & /@ estimatesP; dist = SmoothKernelDistribution[piestimates]; Plot[ PDF[dist, x], {x, 3.03, 3.28}, Epilog -> Line[{{Mean[dist], 0}, {Mean[dist], 20}}], Filling -> Bottom, PlotTheme -> "Marketing", LabelStyle -> Directive[Bold, Medium], ImageSize -> Large] The expected value is: Mean[SmoothKernelDistribution[piestimates]] or equivalently Mean[piestimates] which equals 3.14029.If we assume that the values are Gaussian distributed we can also represent that: Plot[PDF[EstimatedDistribution[piestimates, NormalDistribution[\[Mu], \[Sigma]]], x], {x, 3.03, 3.28}, Epilog -> Line[{{Mean[dist], 0}, {Mean[dist], 20}}], Filling -> Bottom, PlotTheme -> "Marketing", LabelStyle -> Directive[Bold, Medium], ImageSize -> Large] Cheers,Marco
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