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Mathematics for Computer Science Pigeonhole Principle MIT 6.042J/18.062J Mapping Rule: total injection from A to B implies |A| |B|. Generalized Counting Rules Albert R Meyer, April 12, 2010 If |A| > |B| , then no total injection from A to B. Albert R Meyer, lec 10M.1 Pigeonhole Principle April 12, 2010 lec 10M.2 Pigeonhole Principle If more pigeons then some hole must have two pigeons! than pigeonholes, Albert R Meyer, April 12, 2010 example: 5 Card Draw set of 5 cards: must have 2 J 10 J 66 10 22 lec 10M.4 5 cards (pigeons) JJ with the same suit. April 12, 2010 April 12, 2010 5 Card Draw Image by MIT OpenCourseWare. Albert R Meyer, Albert R Meyer, lec 10M.3 lec 10M.5 4 suits (holes) Albert R Meyer, April 12, 2010 lec 10M.6 Pigeon clip art © source unknown. All rights reserved. This content is excluded from our Creative Commons license. For more information, see http://ocw.mit.edu/fairuse 1 10 Card Draw 10 Card Draw 10 cards: how many have the same suit? NO! < 3 cards in every hole? Albert R Meyer, April 12, 2010 lec 10M.7 10 Card Draw Albert R Meyer, April 12, 2010 lec 10M.8 Generalized Pigeonhole Principle If n pigeons and h holes, then some hole has # cards with same suit 10 =3 4 pigeons. “ceiling,” means round up Albert R Meyer, April 12, 2010 lec 10M.9 Albert R Meyer, April 12, 2010 Generalized Product Rule Generalized Product Rule # lineups of 5 students in 6.042? let S::= 6.042 students |S| = 91 so |lineups of 5 students| =NO! 915 ? lineups have no repeats: |seqs in S5 with no repeats| ? |seqs in S5 with no repeats| Albert R Meyer, April 12, 2010 lec 10M.11 lec 10M.10 91 choices for 1st student, 90 choices for 2nd student, 89 choices for 3rd student, 88 choices for 4th student, 87 choices for 5th student 91! = 9190898887 = 86! Albert R Meyer, April 12, 2010 lec 10M.12 2 Generalized Product Rule Division Rule Q a set of length-k sequences #6.042 students = if n1 possible 1st elements, n2 possible 2nd elements (for each first entry), n3 possible 3rd elements (for each 1st & 2nd entry,…) then, |Q| = n1n2nk Albert R Meyer, April 12, 2010 lec 10M.13 Division Rule Let A::= permutations of {1,2,…,13} B::= size 4 subsets |A| = k|B| (generalized Bijection Rule) April 12, 2010 lec 10M.15 Counting Subsets map a1 a2 a3 a4 a5…a12 a13 A to {a1, a2, a3, a4} B Albert R Meyer, April 12, 2010 lec 10M.16 Counting Subsets a1 a3 a2 a4 a5…a12 a13 also maps to {a1, a2, a3, a4} 13! = |A| = (4!9!)|B| so # of size 4 subsets is so does a1 a3 a2 a4 a13 … a12 a5 4! perms lec 10M.14 How many size 4 subsets of {1,2,…,13}? k-to-1, then Albert R Meyer, April 12, 2010 Counting Subsets if function from A to B is Albert R Meyer, 9! perms all map to same set 4!9!-to-1 Albert R Meyer, April 12, 2010 lec 10M.17 Albert R Meyer, April 12, 2010 lec 10M.18 3 Counting Subsets counting 2-pair poker hands # m element subsets of an n element set is Albert R Meyer, April 12, 2010 a 2-pair hand has •2 cards of some rank •2 cards of a second rank •1 card of still a third rank lec 10M.19 counting 2-pair poker hands K, K, A, A, 3 April 12, 2010 lec 10M.22 counting 2-pair poker hands then choose: • 1st pair suits • 2nd pair suits • last card suit Albert R Meyer, lec 10M.21 counting 2-pair poker hands Albert R Meyer, April 12, 2010 lec 10M.23 counting 2-pair poker hands successively choosing: K, A, 3, {, }, {, }, specifies 2-pair hand: 4 2 4 2 sets of 2 suits sets of 2 suits (4 suits) April 12, 2010 April 12, 2010 to count, choose: • 1st pair rank (13 ranks) (12 ranks left) • 2nd pair rank • last card rank (11 ranks left) a 2-pair hand: Albert R Meyer, Albert R Meyer, K, K, A, A, 3 lec 10M.24 Albert R Meyer, April 12, 2010 lec 10M.25 4 counting 2-pair poker hands so # 2-pair hands is this method counts 6-tuples [1st card ranks][2nd card ranks] [last card rank] [1st card suits] [2nd card suits] [last card suit] correctly 4 4 13 12 11 4 2 2 NO! Albert R Meyer, April 12, 2010 counting 2-pair poker hands lec 10M.26 counting 2-pair poker hands but the correspondence to 2-pair hands is not a bijection: Albert R Meyer, April 12, 2010 lec 10M.27 counting 2-pair poker hands the bug to count, choose: • 1st pair rank (13 ranks) (12 ranks left) • 2nd pair rank • last card rank (11 ranks left) (K, A, 3, {, }, {, }, ) K, K, A, A, 3 (A, K, 3, {, }, {, } , ) Albert R Meyer, April 12, 2010 lec 10M.28 counting 2-pair poker hands the bug to count, choose: • 1st pair rank (13 ranks) (12 ranks left) • 2nd pair rank • last card rank (11 ranks left) either pair might be 1st Albert R Meyer, April 12, 2010 lec 10M.31 Albert R Meyer, April 12, 2010 lec 10M.30 counting 2-pair poker hands map from 6-tuples (K, A, 3, {, }, {, }, ) to 2-pair hands K, K, A, A, 3 is 2-to-1 Albert R Meyer, April 12, 2010 lec 10M.32 5 counting 2-pair poker hands so # 2-pair hands is so # 2-pair hands is really 4 4 13 12 11 4 2 2 4 4 13 12 11 4 2 2 2 1 NO! Albert R Meyer, April 12, 2010 counting 2-pair poker hands lec 10M.33 Albert R Meyer, April 12, 2010 lec 10M.34 Team Problems Problems 14 Albert R Meyer, April 12, 2010 lec 10M.35 6 MIT OpenCourseWare http://ocw.mit.edu 6.042J / 18.062J Mathematics for Computer Science Spring 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.