Probability theory and average-case complexity
Review of probability theory
Review of probability theory:
outcome
• Examples:
– Rolling a die and getting 1
– Rolling a die and getting 6
– Flipping three coins and getting H, H, T
– Drawing two cards and getting 7 of hearts, 9 of clubs
• NOTexamples:
– Rolling a 6-sided die and getting an even number
(this is more than one outcome—3 to be exact!) – Drawing a card and getting an ace (4 outcomes!)
Review of probability theory:
event
• Defn: one or more possible outcomes
• Examples:
– Rolling a die and getting 1
– Rolling a die and getting an even number
– Flipping three coins and getting at least 2 “heads” – Drawing five cards and getting one of each suit
Review of probability theory:
• Defn: A set of events
• Examples:
Flipping 3 coins
Drawing a card
sample space
Rolling 2 dice
Review of probability theory:
probability distribution
• Idea:takeasamplespaceSandaddprobabilities for each event
• Defn: mapping from events of S to real numbers –Pr ≥0foranyeventAinS
–∑∈Pr =1
• Example:
0.75*0.75
0.753
Flipping 3 biased coins 75% heads, 25% tails
0.75*0.25 0.25*0.75
0.75*0.252
0.75
0.752*0.25 0.752*0.25
0.25 0.25*0.25
0.752*0.25
0.75*0.252 0.75*0.252
0.253
Review of probability theory:
probability of an event A
• Defn: Pr = ∑∈ Pr • Example:
– Pr(roll a die and get even number) = Pr(roll a 2) + Pr(roll a 4) + Pr(roll a 6)
Review of probability theory:
random variable
• Idea:turneventsintonumbers
• Let S be a sample space
• Defn: mapping from events to real numbers
• Example:
X = the number on a die after a roll
event “rolling a 1” -> 1 event “rolling a 2” -> 2 …
event “rolling a 6” -> 6
Technically, X is a function, so we can write: X(rolling a 1) = 1 X(rolling a 2) = 2
…
X(rolling a 6) = 6
Review of probability theory:
expected value of a random variable • Idea:“average”valueoftherandomvariable
• Remember:randomvariableXisamappingfrom events in a sample space S to numbers
• Defn:ExpectedvalueofX=E X = Pr =
∈
• Short form: E X = ∑ Pr =
Here, x = X(event)
• Example: X = number on die after rolling
= Pr= =11+21+⋯+61=7 66 62
Expected running time of an algorithm
Expected running time of an algorithm
• Let A be an algorithm.
• Let Sn be the sample space of all inputs of size n.
• To talk about expected (/average) running time, we must specify how we measure running time.
– We want to turn each input into a number (runtime). – Random variables do that…
• We must also specify how likely each input is.
– We do this by specifying a probability distribution over Sn.
Expected running time of an algorithm • Recall: algorithm A, sample space Sn
• We define a random variable
tn(I) = number of steps taken by A on input I
• We then obtain:
=Pr ∈
• In this equation, I is an input in Sn, and
Pr(I) is the probability of input I according to the probability distribution we defined over Sn
• is the average running time of A, given Sn
Example time!
Example time: searching an array
• Let L be an array containing 8 distinct keys Search(k, L[1..8]):
for i = 1..8
if L[i].key == k then return true
return false
• What should our sample space S9 of inputs be? • Hard to reason about all possible inputs.
– (In fact, there are uncountably infinitely many!)
• Cangroupinputsbyhowmanystepstheytake!
Grouping inputs by how long they take
Search(k, L[1..8]):
for i = 1..8
if L[i].key == k then return true
return false
• What causes us to return in loop iteration 1? • How about iteration 2? 3? … 8? After loop? • S9 = { L[1]=k, L[2]=k, …, L[8]=k, k not in L }
• NowweneedarandomvariableforS9!
Using a random variable to capture running time
Search(k, L[1..8]):
For simplicity: assume each iteration takes 2 steps. Do we have enough information to
for i = 1..8
if L[i].key == k then return true
compute an answer? return false 1 step
• S9 = { L[1]=k, L[2]=k, …, L[8]=k, k not in L }
• Let T(e) = running time for event e in S9
• T(L[1]=k) = 2, T(L[2]=k) = 4, …, T(L[i]=k) = 2i • T(knotinL)=2*8+1=17 •Wethenobtain: =∑∈()
What about a probability distribution?
• We have a sample space and a random variable. • Now,weneedaprobabilitydistribution.
• Thisisgiventousintheproblemstatement.
–Foreachi,Pr = = –Pr =
• If you don’t get a probability distribution from the problem statement, you have to figure out how likely each input is, and come up with your own.
Computing the average running time
•Wenowknow: =∑∈()
• T(e) = running time for event e in S9
• T(L[i]=k)=2i T(knotinL)=17
• S9 = { L[1]=k, L[2]=k, …, L[8]=k, k not in L} • Probability distribution:
– For each i, Pr = = –Pr =
•Therefore: =1 =1 = +
⋯+8 =8 = + ()
The final answer
• Recall: T(L[i]=k) = 2i T(k not in L) = 17
– For each i, Pr = = –Pr =
• = 1 = 1 = + ⋯ +
8 = 8 = +
=++
⋯ + + = 13
• Thus,theaveragerunningtimeis13.
Slightly harder problem: L[1..n]
Search(k, L[1..n]):
for i = 1..n
if L[i].key == k then return true
return false
• Problem: what is the average running time of Search, given the following probabilities?
– Pr = =
–Pr =
Computing E[T]: part 1
Search(k, L[1..n]):
for i = 1..n
if L[i].key == k then return true
return false
• What is our sample space?
– Sn+1 = { L[1]=k, L[2]=k, …, L[n]=k, k not in L }
• What is our random variable?
– Let T(e) = running time for event e in Sn+1
• What is the running time of each event? – T(L[i]=k) = 2i, T(k not in L)=2n+1
Computing E[T]: part 2
• Whatweknow:
– Sn+1 = { L[1]=k, L[2]=k, …, L[n]=k, k not in L } – T(L[i]=k) = 2i, T(k not in L)=2n+1
–Pr= andPr==
• Nowwecancompute =∑∈.
–=
T L 1 = k Pr L 1 = k +
T L 2 = k Pr L 2 = k + ⋯ + T L n = k Pr L n = k +
T Pr()
– =2 +4 +⋯+2n + 2n+1
Computing E[T]: part 3 – =2 +4 +⋯+2n + 2n+1
– = 1+2+⋯+ + 2+1
– =()+=+=+1
– Thus, Search(k, L[1..n]) has expected (or average) running time 3n/2+1 for the given probabilities.