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October 12th, 2011, 06:02 PM  #1 
Joined: Aug 2009 From: Copenhagen, Denmark Posts: 45 Thanks: 0  Mathematica: Applying rules to random integer generation
Hello I have a problem in Mathematica. In the range 130 I will like Mathematica to choose five almost random integers. I can do so by writing: RandomSample[Range[30], 5] The thing is how to apply some rules, restricting the random command. Like this: 1) At most 2 integers =<12 2) At least one integer >=25 3) Each of the integers 3,4 and 7 can't be a divisor for more than 3 of the random chosen integers (so e.g. 3121530 or 7212814 is not possible) 4) At most one "pair of integers" (if both e.g. 3 and 4 are chosen, I call it a "pair"). Anyone with suggestions on how to solve this in Mathematica language? 
October 12th, 2011, 08:11 PM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 15,643 Thanks: 808 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Mathematica: Applying rules to random integer generation
Write a function that checks if the conditions are met, then define a function that stores a random number in a variable and checks if the conditions are met; if so return the variable's value, if not recurse.

October 13th, 2011, 05:42 PM  #3  
Joined: Aug 2009 From: Copenhagen, Denmark Posts: 45 Thanks: 0  Re: Mathematica: Applying rules to random integer generation Quote:
chosen = Sort[RandomSample[Range[36], 7], Less]; less12 = Count[Select[chosen, # <= 12 &], _Integer]; great31 = Count[Select[chosen, # > 31 &], _Integer]; divisor = Count[Select[chosen, Divisible[#, {3, 8}] &], _Integer]; test = If[less12 == 2, If[great31 == 1, If[divisor <= 5, chosen, False], False], False] If the random generation is False, then the elements in the list either contains more than two integers <12, no integers >31, or more than five elements with 38 as a divisable integer. If every condition is met, I get the list generated in "chosen". The thing is  I now want to loop the procedure, so I don't have to run it manually a lot of times to get anything other than False. I have thought about the while command. It will have to be something like: 1) Check if "test" is true 2) If yes > print list. If no> make a new random sample and repeat the test. Any input on this?  
October 13th, 2011, 09:41 PM  #4  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 15,643 Thanks: 808 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Mathematica: Applying rules to random integer generation Quote:
 
October 14th, 2011, 05:28 AM  #5  
Joined: Aug 2009 From: Copenhagen, Denmark Posts: 45 Thanks: 0  Re: Mathematica: Applying rules to random integer generation Quote:
Code: list = Sort[RandomSample[Range[36], 7]]; f[x_] := If[Count[Select[list, # <= 12 &], _Integer] == 2, If[Count[Select[list, # > 31 &], _Integer] == 1, If[Count[Select[list, Divisible[#, {3, 8}] &], _Integer] <= 5, list, False], False], False] While[f[x] == False, list = Sort[RandomSample[Range[36], 7], Less]; If[list == f[x], a]] f[x] Now the last challenge: This procedure only produces one result. How to make the first 5 results appear at one time?  
October 14th, 2011, 05:51 AM  #6  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 15,643 Thanks: 808 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Mathematica: Applying rules to random integer generation Quote:
Quote:
 
October 14th, 2011, 06:02 AM  #7  
Joined: Aug 2009 From: Copenhagen, Denmark Posts: 45 Thanks: 0  Re: Mathematica: Applying rules to random integer generation Quote:
 
October 14th, 2011, 06:55 AM  #8 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 15,643 Thanks: 808 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Mathematica: Applying rules to random integer generation
This is the most basic possible use of the function. If you want a 5 random numbers as generated by g[], you just write List[g[], 5].

October 14th, 2011, 07:10 AM  #9  
Joined: Aug 2009 From: Copenhagen, Denmark Posts: 45 Thanks: 0  Re: Mathematica: Applying rules to random integer generation Quote:
Code: list = RandomSample[Range[36], 7]; f := If[1 <= Count[Select[list, # <= 12 &], _Integer] <= 2, If[Count[Select[list, # > 31 &], _Integer] >= 1, If[Count[Select[list, Divisible[#, {2, 7}] &], _Integer] <= 5, list, False], False], False]; While[f == False, list = RandomSample[Range[36], 7]; If[list == f, f]] f Code: {1, 7, 15, 16, 25, 28, 34} Code: Table[Sort[f], {i, 5}] Code: {1, 7, 15, 16, 25, 28, 34}, {1, 7, 15, 16, 25, 28, 34}, {1, 7, 15, 16, 25, 28, 34}, {1, 7, 15, 16, 25, 28, 34}, {1, 7, 15, 16, 25, 28, 34}  
October 14th, 2011, 09:08 AM  #10 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 15,643 Thanks: 808 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Mathematica: Applying rules to random integer generation
You need tro wrap everything that generates the random numbers into a single function and run that repeatedly. If you generate it with one call and simply test it it will be the same each time.


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applying, generation, integer, mathematica, random, rules 
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