Transcript warmup

Counting Warmup
Strings of 7 bits
•
•
•
•
•
•
How many have a run of at least 3 1s?
Let S={0,1}7
Let R⊆S be the strings with a run of 3 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 0 or 1, and abc is any string of
3 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 7 bits
• How many members of S have a run of
exactly 3 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 7 bits
• How many members of S have a run of at
least 3 1s or at least 3 0s?
• # with 3 1s + # with 3 0s - # with both (so
they will not get counted twice)
• General rule: |A∪B|=|A|+|B|-|A∩B|
• Strings with both a run of 3 1s and 3 0s =
000111x, 111000x, 00001111, 1111000,
1000111, 0111000: 8 possibilities
• 47+47-8=86