Embed Size (px)
Citation preview
Shanghai Jiao Tong University
Chihao Zhang
Advanced Algorithms (I)
March 2nd, 2020
Information
Every Monday, 10:00 am - 11:40 am
Zoom @ 123363659
Instructor: 张驰豪 ([email protected])
TA: 杨凤麟 ([email protected])
Office Hour: via Canvas or WeChat Group
References
Probability and Computing Randomized Algorithms The Probabilistic Method
M. Mitzenmacher & E. Upfal R. Motwani & P. Raghavan N. Alon & J. Spencer
Polynomial Identity Testing
Polynomial Identity TestingGiven two polynomials
and F(x) =d
∏i=1
(x − ai) G(x) =d
∑i=0
bixi
Polynomial Identity TestingGiven two polynomials
and F(x) =d
∏i=1
(x − ai) G(x) =d
∑i=0
bixi
Is F(x) ≡ G(x)?
Polynomial Identity TestingGiven two polynomials
and F(x) =d
∏i=1
(x − ai) G(x) =d
∑i=0
bixi
Is F(x) ≡ G(x)?
Example. F(x) = (x − 1)(x − 2)(x + 3); G(x) = x3 − 7x + 6
One can expand and compare the coefficients…F(x)
One can expand and compare the coefficients…F(x)
It takes arithmetic operations.O(d2)
One can expand and compare the coefficients…F(x)
It takes arithmetic operations.O(d2)
Can be improved to using FFT.O(d log d)
One can expand and compare the coefficients…F(x)
It takes arithmetic operations.O(d2)
Can be improved to using FFT.O(d log d)
If random coins are allowed…
One can expand and compare the coefficients…F(x)
It takes arithmetic operations.O(d2)
Can be improved to using FFT.O(d log d)
If random coins are allowed…
• The problem can be solved much faster
One can expand and compare the coefficients…F(x)
It takes arithmetic operations.O(d2)
Can be improved to using FFT.O(d log d)
If random coins are allowed…
• The problem can be solved much faster
• at the cost of making error.
Choosing a uniform number s ∈ {1,2,…,100d}
Choosing a uniform number s ∈ {1,2,…,100d}
Test whether F(s) = G(s)
Choosing a uniform number s ∈ {1,2,…,100d}
Test whether F(s) = G(s)
• If , then it always holds that F(x) ≡ G(x) F(s) = G(s)
Choosing a uniform number s ∈ {1,2,…,100d}
Test whether F(s) = G(s)
• If , then it always holds that F(x) ≡ G(x) F(s) = G(s)
• If , how likely is that ?F(x) ≢ G(x) F(s) ≠ G(s)
Choosing a uniform number s ∈ {1,2,…,100d}
Test whether F(s) = G(s)
• If , then it always holds that F(x) ≡ G(x) F(s) = G(s)
• If , how likely is that ?F(x) ≢ G(x) F(s) ≠ G(s)
Theorem. (Fundamental Theorem of Algebra)
A polynomial of degree has at most roots in d d ℂ
Our algorithm outputs wrong answer only when
Our algorithm outputs wrong answer only when
• ; andF(x) ≢ G(x)
Our algorithm outputs wrong answer only when
• ; andF(x) ≢ G(x)
• is a root of s F(x) − G(x)
Our algorithm outputs wrong answer only when
• ; andF(x) ≢ G(x)
• is a root of s F(x) − G(x)
This happens with probability at most .1
100
Our algorithm outputs wrong answer only when
• ; andF(x) ≢ G(x)
• is a root of s F(x) − G(x)
This happens with probability at most .1
100
It only costs operations to compute and .O(d) F(s) G(s)
Our algorithm outputs wrong answer only when
• ; andF(x) ≢ G(x)
• is a root of s F(x) − G(x)
This happens with probability at most .1
100
It only costs operations to compute and .O(d) F(s) G(s)
One can repeat the algorithm times: t
• error reduces to ;100−t
• cost increases to .O(td)
Multi-variable Polynomials
Multi-variable Polynomials
The idea applies to a more general setting.
Multi-variable Polynomials
The idea applies to a more general setting.
Let for some field ,F, G ∈ 𝔽(x1, …, xn) 𝔽
Multi-variable Polynomials
The idea applies to a more general setting.
Let for some field ,F, G ∈ 𝔽(x1, …, xn) 𝔽
Is F(x1, …, xn) ≡ G(x1, …, xn)?
Theorem. (Schwartz-Zippel Theorem)
Let be a non-zero multivariate polynomial of degree at most . For any set , it
holds that
Q ∈ 𝔽[x1, …, xn]d U ⊆ 𝔽
Prr1,…,rn∈RU
[Q(r1, …, rn) = 0] ≤d
|U |
Proof of Schwartz-Zippel
Proof of Schwartz-Zippel
Induction on , case is Fundamental Theorem of Algebra.
n n = 1
Proof of Schwartz-Zippel
Induction on , case is Fundamental Theorem of Algebra.
n n = 1
Assuming it holds for smaller …n
Proof of Schwartz-Zippel
Induction on , case is Fundamental Theorem of Algebra.
n n = 1
Assuming it holds for smaller …n
Q(x1, …, xn) =k
∑i=0
xi1 ⋅ Qi(x2, …, xn)
Proof of Schwartz-Zippel
Induction on , case is Fundamental Theorem of Algebra.
n n = 1
Assuming it holds for smaller …n
Q(x1, …, xn) =k
∑i=0
xi1 ⋅ Qi(x2, …, xn)
Pr [Q = 0] ≤ Pr [Qk = 0] + Pr [Q = 0 |Qk ≠ 0] ≤d − k|U |
+k
|U |
Our randomized algorithm generalizes to the multi-variable setting by Schwartz-Zippel theorem.
Our randomized algorithm generalizes to the multi-variable setting by Schwartz-Zippel theorem.
If the polynomials are given in product form, one scan of and is sufficient to evaluate them.F G
Our randomized algorithm generalizes to the multi-variable setting by Schwartz-Zippel theorem.
If the polynomials are given in product form, one scan of and is sufficient to evaluate them.F G
Linear time algorithm with at most error!1 %
Our randomized algorithm generalizes to the multi-variable setting by Schwartz-Zippel theorem.
If the polynomials are given in product form, one scan of and is sufficient to evaluate them.F G
Linear time algorithm with at most error!1 %
It is a wide open problem in the complexity theory that whether this can be done in deterministic polynomial time.
Some Complexity Theory
Some Complexity Theory
Problems solvable in deterministic polynomial-time: P
Some Complexity Theory
Problems solvable in deterministic polynomial-time: P
Problems solvable in randomized polynomial-time: BPP
Some Complexity Theory
Problems solvable in deterministic polynomial-time: P
Problems solvable in randomized polynomial-time: BPP
Is ?BPP = P
Min-Cut in a Graph
Min-Cut in a GraphA cut in a graph is a set of edges whose removal disconnects .
G = (V, E) C ⊆ EG
Min-Cut in a GraphA cut in a graph is a set of edges whose removal disconnects .
G = (V, E) C ⊆ EG
How to find the minimum cut?
Min-Cut in a GraphA cut in a graph is a set of edges whose removal disconnects .
G = (V, E) C ⊆ EG
How to find the minimum cut?
It can be solved using max-flow techniques
Min-Cut in a GraphA cut in a graph is a set of edges whose removal disconnects .
G = (V, E) C ⊆ EG
How to find the minimum cut?
It can be solved using max-flow techniques
With the fastest max-flow algorithm, it takes time.
O(n × mn)
Karger’s Min-Cut Algorithm
Karger’s Min-Cut Algorithm
David Karger
Karger’s Min-Cut Algorithm
David Karger
Using random bits, Karger found a much simpler algorithm
Karger’s Min-Cut Algorithm
David Karger
Using random bits, Karger found a much simpler algorithm
The only operation required is edge contraction
Edge Contraction
Edge Contraction
1
2
3
Edge Contraction
1
2
3
1
2
3
Edge Contraction
1
2
3
1
2
3
Edge Contraction
1
2
3
1
2
3 1/2 3
Edge Contraction
1
2
3
1
2
3 1/2 3
Edge Contraction
1
2
3
1
2
3 1/2 3
• no self-loop
Edge Contraction
1
2
3
1
2
3 1/2 3
• parallel edges may exist
• no self-loop
The Algorithm
The Algorithm
Karger’s Min-cut Algorithm
1. Randomly choose an edge and contract it until only two vertices remains.
2. Output remaining edges.
Analysis
AnalysisThe algorithm contracts pair of vertices in total.n − 2
AnalysisThe algorithm contracts pair of vertices in total.n − 2
Fix an minimum cut , we bound the probability that it survives.
C
AnalysisThe algorithm contracts pair of vertices in total.n − 2
Fix an minimum cut , we bound the probability that it survives.
C
Assume the removal of separates and .
C S ⊆ VS̄ = V∖S
AnalysisThe algorithm contracts pair of vertices in total.n − 2
Fix an minimum cut , we bound the probability that it survives.
C
Assume the removal of separates and .
C S ⊆ VS̄ = V∖S
All contractions happen within or .S S̄
For , let be the event that “ -th contraction avoids ”
i = 1,…, n − 2 Ai iC
For , let be the event that “ -th contraction avoids ”
i = 1,…, n − 2 Ai iC
We need to bound
For , let be the event that “ -th contraction avoids ”
i = 1,…, n − 2 Ai iC
We need to bound
Pr [n−2
⋂i=1
Ai] =n−2
∏i=1
Pr Ai
i−1
⋂j=1
Aj
For , let be the event that “ -th contraction avoids ”
i = 1,…, n − 2 Ai iC
We need to bound
Pr [n−2
⋂i=1
Ai] =n−2
∏i=1
Pr Ai
i−1
⋂j=1
Aj
We assume |C | = k
In -th contraction,i
In -th contraction,i
• the graph contains vertices;
• each vertex is of degree at least .
n − i + 1
k
In -th contraction,i
• the graph contains vertices;
• each vertex is of degree at least .
n − i + 1
k
Therefore, conditional on that still survives,C
In -th contraction,i
• the graph contains vertices;
• each vertex is of degree at least .
n − i + 1
k
Therefore, conditional on that still survives,C
Pr Ai
i−1
⋂j=1
Aj ≥ 1 −2k
k(n − i + 1)=
n − i − 1n − i + 1
Therefore
Therefore
Pr [n−2
⋂i=1
Ai] =n−2
∏i=1
Pr Ai
i−1
⋂j=1
Aj
≥n−2
∏i=1
n − i − 1n − i + 1
=n − 2
n⋅
n − 3n − 1
⋯13
=2
n(n − 1)
So if we repeat the algorithm times, the minimum cut survives with probability at least
50n2
So if we repeat the algorithm times, the minimum cut survives with probability at least
50n2
1 − (1 −2
n(n − 1) )50n2
≥ 1 − e−100
So if we repeat the algorithm times, the minimum cut survives with probability at least
50n2
1 − (1 −2
n(n − 1) )50n2
≥ 1 − e−100
How about the time cost?
So if we repeat the algorithm times, the minimum cut survives with probability at least
50n2
1 − (1 −2
n(n − 1) )50n2
≥ 1 − e−100
How about the time cost?
If we store the graph in a adjacency matrix, one needs to contract an edge…O(n)
Karger-Stein’s Trick
Karger-Stein’s Trick
Recall Pr [n−2
⋂i=1
Ai] =n − 2
n⋅
n − 3n − 1
⋯13
Karger-Stein’s Trick
Recall Pr [n−2
⋂i=1
Ai] =n − 2
n⋅
n − 3n − 1
⋯13
The more we contracts, the easier gets hitC
Karger-Stein’s Trick
Recall Pr [n−2
⋂i=1
Ai] =n − 2
n⋅
n − 3n − 1
⋯13
The more we contracts, the easier gets hitC
Idea: Make a copy before it becomes too bad!
Karger-Stein’s Trick
Recall Pr [n−2
⋂i=1
Ai] =n − 2
n⋅
n − 3n − 1
⋯13
The more we contracts, the easier gets hitC
Idea: Make a copy before it becomes too bad!
The success probability can be improved to Ω ( 1log n )
