CountingWarmup Strings of bits • • • • • • How many have a run of at least 1s? Let S={0,1}7 Let R⊆S be the strings with a run of 1s R={0,1}*111{0,1}*∩S To count R, split it into disjoint subsets and count the subsets Break R into subsets according to the position of the FIRST block of three 1s • A string in R could be of the form 111xxxx, 0111xxx, x0111xx, xx0111x, or abc0111 where x is or 1, and abc is any string of bits except 111 • 16+8+8+8+7=47 • Sum rule: If R=R1∪…∪Rn and the Ri are pairwise disjoint, then |R|=|R1|+…+|Rn| Strings of bits • How many members of S have a run of exactly 1s but not longer? • Easiest to calculate how many have a run of at least four 1s and take the difference: 1111xxx, 01111xx, x01111x, xx01111 so 20 total • 47-20 = 27 • General rule: If A⊆B then |B-A|=|B|-|A| Strings of bits • How many members of S have a run of at least 1s or at least 0s? • # with 1s + # with 0s - # with both (so they will not get counted twice) • General rule: |A∪B|=|A|+|B|-|A∩B| • Strings with both a run of 1s and 0s = 000111x, 111000x, 00001111, 1111000, 1000111, 0111000: possibilities • 47+47-8=86 ... subsets according to the position of the FIRST block of three 1s • A string in R could be of the form 111xxxx, 0111xxx, x0111xx, xx0111x, or abc0111 where x is or 1, and abc is any string of bits