Discrrete mathematics for computer science counting

Counting First Principles 3/22/19 Counting in Gambling What fraction of poker hands are “a pair of Jacks?” (probability of a pair of Jacks) 3/22/19 Counting in Algorithms # ops to update a data structure (# comparisons needed to sort n items) # steps in a computation (# n multiplies to compute d ) 3/22/19 Counting in Cryptography # possible passwords # possible keys 3/22/19 lec 10W.4 Sum Rule A B If sets A and B are disjoint, then |A ∪ B| = |A| + |B| 3/22/19 The Sum Rule • Class has 17 women, 25 men so total enrollment = 17 + 25 = 42 • 26 lower case letters, 26 upper case letters, and 10 digits, so # characters = 26+26+10 = 62 3/22/19 lec 10W.6 The Product Rule If there are men and women, there are 4⋅3 = 12 different man/woman couples 3/22/19 Product Rule If |A| = m and |B| = n, then |A × B| = m⋅n A = {a, b, c, d}, B = {1, 2, 3} A × B = {(a,1),(a,2),(a,3), (b,1),(b,2),(b,3), (c,1),(c,2),(c,3), (d,1),(d,2),(d,3) } 3/22/19 Product Rule: Counting Strings # length-4 binary strings = |B × B × B × B| = |B | where B ::= {0,1} =2·2·2·2=2 3/22/19 Product Rule: Counting Strings # length n strings from an alphabet of size m is m 3/22/19 n 10 Example: Counting Passwords Password conditions: • • • • 3/22/19 characters are digits & letters between & characters long starts with a letter case sensitive 11 Counting Passwords L ::= {a,b,…,z,A,B,…,Z} D ::= {0,1,……,9} Pn ::= length n words starting w/letter = L × (L ∪ D) 3/22/19 n-1 12 Counting Passwords |L × (L ∪ D) n-1 = |L|·|(L ∪ D)| | n-1 = |L|·(|L| + |D|) = 52·(52+10) 3/22/19 n-1 n-1 13 Counting Passwords set of passwords: P ::= P6∪ P7∪P8 |P| = |P6|+|P7|+|P8| = 52·(62 +62 +62 ) 14 ≈ 19·10 3/22/19 14 # 4-digit nums w/ ≥ one cases by 1st occurrence of 7: x: any digit o: any digit ≠ 7xxx or o7xx or oo7x or ooo7 • 103 + 9·10 3/22/19 = 3439 + ·10 + 15 at least one 7: another way |4-digit nums w/ ≥ one 7| = |4-digit nums| − |those w/ no 7| = 3/22/19 104 − 94 = 3439 16 Cracking Passwords by Exhaustive Search • Suppose a password has to be a string of letters and digits beginning with a letter • • So there are 52∙62n-1 passwords of length n • How big does n have to be for it to take more than a day to try them all? Suppose a computer can try a password every billionth of a second (one nanosecond per try) 3/22/19 17 More than a day: • • • • n-1 52∙62 > number of nanoseconds in a day = 24∙60∙60∙10 n-1 12 62 > 1.66∙10 n-1 > log (1.66∙10 12 )/log 62 ≈ 6.8 So if passwords are at least characters long it would take more than a day to try them all by exhaustive search 3/22/19 18 FINIS 3/22/19 19 .. .Counting in Gambling What fraction of poker hands are “a pair of Jacks?” (probability of a pair of Jacks) 3/22/19 Counting in Algorithms # ops to update... (d,1),(d,2),(d,3) } 3/22/19 Product Rule: Counting Strings # length-4 binary strings = |B × B × B × B| = |B | where B ::= {0,1} =2·2·2·2=2 3/22/19 Product Rule: Counting Strings # length n strings... 3/22/19 n 10 Example: Counting Passwords Password conditions: • • • • 3/22/19 characters are digits & letters between & characters long starts with a letter case sensitive 11 Counting Passwords