Re: Testing a random integer generator
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- Subject: [mg130395] Re: Testing a random integer generator
- From: Peter Pein <petsie at dordos.net>
- Date: Mon, 8 Apr 2013 00:06:16 -0400 (EDT)
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Am 07.04.2013 03:55, schrieb Clif McInnis: > I am trying to test the random integer generator in a program to make sure > that it gives the type of output that I am looking for. I expect it to generate (p<>q)<10 slightly less than 1/8 of the time and (p=q)<10 slightly more than 1/8 of the time, 9<(p<>q)<20 slightly less than 3/8 of the time, and 9<(p=q)<20 slightly more than 3/8 of the time. Regardless of my expectations I would like to see how the numbers actually come out, and to that end was attempting to test the test the code, but I can not get it to generate (print) any numbers. I am sure that I have made an error somewhere in the code, but I can not find it and am not getting any error messages. Thank you for any and all help. > > Clif McInnis > > > Initialization :> ({x = 0, p = 0, q = 0}) This initializes nothing because of RuleDelayed ( :> ), but would be unusual too if Rule ( -> ) had been used. > While[x < 100, { As x is not used inside the loop, a Do[ ... , {100}] is sufficient. If you want to use the generated pairs, the use of Table is the "natural" way to generate them in Mathematica. > Clear[p, q], no need to clear, because the values are set in the next statement. A Clear deletes the zero value from p too; it does not set the values of p and q to 0 -- be careful! > q = RandomInteger[{1, 4}], If[q < 4, > {p = RandomInteger[{10, 20}], > q = RandomInteger[], > If[q == 1, q = p, q = RandomInteger[{10, 20}]]}, {p = RandomInteger[{1, 9}], > q = RandomInteger[], > If[q == 1, q = p, q = RandomInteger[{1, 9}]]}], Print[p], > Print[q]}; x++] > randomPairs[number_] := Table[ Block[{integerrange = If[RandomInteger[3] == 0, {1, 9}, {10, 20}] }, If[RandomInteger[] == 0, ConstantArray, (*else*) Table][RandomInteger[integerrange], {2}] ], {number}] To see if this does what you want, generate a few pairs SeedRandom[1]; (* for reproducable rsults only *) data = randomPairs[1000]; how many have got the same element twice? Count[data, {p_, p_}] --> 566 and how many have got the smaller values? Count[data, {p_, _} /; p < 10] --> 255 It seems to me that it works :) Take a look at e.g.: Histogram3D[Flatten[GatherBy[data, {SameQ @@ # &, #[[1]] < 10 &}], 1], ChartStyle -> {Darker[Green, .5], Darker[Red, .5], Green, Red}] Peter