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IIT MATH 149 - Randomization

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RandomizationPatrick Dale McCray, 31 January 2008Revised, 27 January 2009Application for second recitation, Spring, 2011. "RandomizationSecond_v5.nb"<< DiscreteMath`Combinatorica`Random K subsetGenerate a list of 9 students from 149-01, 149-02 classess.Random@Integer, 810000000, 99999999<D84140267First time generated the eight digit integer 61068477.Second time used the seed 26373360.Third time used the seed 64665653. (??)Fourth time used the seed 56864032.Fifth time used the seed 62783836Sixth time used the seed 45371279SeedRandom@61068477DSeedRandom@26373360DSeedRandom@56864032DSeedRandom@62783836DSeedRandom@45371279DSeedRandom@84140267DClass14901 = 8 1, 3, 4, 5, 7, 9 <81, 3, 4, 5, 7, 9<For recitation 2 we need to pick 3 people from 149 01. These six people spoke in the first recitation.R02 = RandomKSubset@Class14901, 3D83, 4, 7<Recitate02 = 8 2, 3, 4, 6, 7, 8<82, 3, 4, 6, 7, 8<These six people will speak in the second recitation.Recitate03 = 81, 2, 5, 6, 8, 9<81, 2, 5, 6, 8, 9<These six people will speak in the third recitation. Then they will speak in the same groups for the next threerecitations.For the second recitation there are a maximum of ten problems. And for Math 149 01, only six people will bespeaking.TenProblems = 81, 2, 3, 4, 5, 6, 7, 8, 9, 10<81, 2, 3, 4, 5, 6, 7, 8, 9, 10<Problems14901Rec2 = RandomKSubset@TenProblems, 6D81, 2, 5, 6, 8, 9<And now we need randompermutations, on six and on ten objects.8RandomPermutation@6D, RandomPermutation@10D<885, 3, 2, 1, 6, 4<, 82,10,5,7,6,9,1,3,4,8<<R02 = RandomKSubset@Class14902, 9D82, 5, 12, 13, 19, 20, 22, 23, 24<27 + R02829, 32, 39, 40, 46, 47, 49, 50, 51<8RandomPermutation@9D, RandomPermutation@9D<881, 6, 5, 3, 2, 9, 8, 4, 7<, 84, 1, 6, 2, 3, 5, 9, 7, 8<<2 RandomizationSecond_v5.nbFor recitation 3 we need to pick, for 149001, those 5 people who have not presented twice.For 149002, those 7 people who have not yet presented.8RandomPermutation@5D, RandomPermutation@7D<881, 2, 5, 3, 4<, 87, 6, 4, 2, 3, 1, 5<<For recitation 4 we need to assign problems to 5 people for 149001.For 149002, we need to assign problems to 9 people.Since we do not want to assign problem 6, the numbers from 1 to 10will be permuted, but problem 6 will not be assigned.8RandomPermutation@5D, RandomPermutation@10D<885, 4, 2, 1, 3<, 810,4,9,5,7,8,2,6,3,1<<For recitation 5 we need to assign problems to 5 people for 149001.For 149002 we need to assign problems to 8 people.ProblemsR5 := 81, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12<Reject problem 7 if selected, otherwise take first 5, for 149001:8RandomKSubset@ProblemsR5, 6D, RandomKSubset@ProblemsR5, 9D<884, 5, 7, 9, 11, 12<, 81, 2, 3, 5, 6, 8, 9, 10, 12<<So, the subsets for 149001, 149002 are 883, 6, 8, 10, 11<, 81, 3, 4, 6, 8, 9, 10, 12<<883, 6, 8, 10, 11<, 81, 3, 4, 6, 8, 9, 10, 12<<8RandomPermutation@5D, RandomPermutation@8D<881, 5, 3, 2, 4<, 87, 6, 3, 2, 1, 8, 4, 5<<RandomizationSecond_v5.nb


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IIT MATH 149 - Randomization

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