CS代考程序代写 algorithm computational biology distributed system Lecture 11: NP-completeness + Coping with hardness

Lecture 11: NP-completeness + Coping with hardness
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Definition of the class NP
Class NP: Decision problems for which YES-instances have a poly-time certifier.
Certification algorithm intuition.
– Certifier views things from “managerial” viewpoint.
– Certifier does not solve the problem by its own;
rather, it checks if a proposed proof t is a valid solution.
Definition: Algorithm C(s,t) is a certifier for problem X if for every input instance s and proposed proof t, C(s, t) = `yes’ if and only if t is a valid solution to s.
“certificate” or “witness”
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Class NP-hard
Class NP-complete: A problem in NP such that every problem in NP polynomially reduces to it.
Class NP-hard:
A decision problem such that every problem in NP polynomially reduces to it.
not necessarily in NP
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Establishing NP-Completeness
Remark: Once we establish first “natural” NP-complete problem, others fall like dominoes.
Recipe to establish NP-completeness of problem Y. – Step1. ShowthatYisinNP.
– Step 2. Choose an NP-complete problem X. – Step 3. Prove that X £ p Y.
Justification: If X is an NP-complete problem, and Y is a problem in NP with the property that X £ P Y then Y is NP-complete.
Proof: LetWbeanyprobleminNP. ThenW £P X £P Y. – Bytransitivity,W£P Y.
– Hence Y is NP-complete. ▪ The University of Sydney
by definition of NP-complete
by assumption
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Reduction Template for NP-Completeness Proofs
Algorithm for X
Special Instance of Y
f(I)
Yes for f(I) No for f(I)
Transforms instance of X to instance of Y
Algorithm for Y
Instance of X
I
Yes for I No for I
Step 2
Step 1
Step 1: Construct a polynomial-time algorithm that transforms instance I of X to a special instance f(I) of Y
Step 2: Prove correctness, which boils down to showing
Answer for I is Yes iff answer for f(I) is Yes
(⇒) Transform a YES-certificate for I into a YES-certificate for f(I)
(⇐) Transform a YES-certificate for f(I) into a YES-certificate for I The University of Sydney
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Taxonomy of NP-Complete Problems
Six Basic genres
§ Constraint satisfaction problems: SAT, 3-SAT.
§ Packing problems: SET-PACKING, INDEPENDENT SET. § Covering problems: SET-COVER, VERTEX-COVER.
§ Sequencing problems: HAMILTONIAN-CYCLE, TSP.
3-SAT £P DIR HAMILTONIAN CYCLE £P HAMILTONIAN CYCLE £P TSP § Partitioning problems: 3D-MATCHING, 3-COLOR.
§ Numerical problems: SUBSET-SUM, KNAPSACK. The University of Sydney
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Vertex Cover Reduces to Subset Sum
Proof: Given an instance (G, k) of VERTEX-COVER, construct an instance (S, t) of SUBSET-SUM that has a subset of S summing to t iff G has a vertex cover of size at least k.
Number edges from 0 to |E|-1. Set S of integers:
– For the i-th edge, add an integer bi
– For each vertex v, add an integer av
bi = 4i
av = 4|E| + X 4i
i:i-th edge is incident to v
|EX|1 t=k·4|E| +
2·4i
Subset-Sum can be solved in O(nT) using DP WhereT=totalsumofintegersinS
Does this mean P = NP? Page 7
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3-SAT Reduces to 3-COLOR
Theorem: 3-SAT £P 3-COLOR.
Proof: Given an arbitrary instance F of 3-SAT, we construct an instance G of 3-
COLOR that has a 3-coloring iff F is satisfiable. Construction of G:
1. Truth gadget. Introduce 3 special vertices called T, F and O connected to each other. Call their colors T, F and O.
2. Variable gadget. Introduce 2 literal vertices per variable and connect them to each other and O. Literals with same color as T is assigned true, those with same color as F is assigned false. Thus, 3-coloring = consistent truth assignment (every variable is either true or false).
O
TF
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x1 x1
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3-SAT Reduces to 3-COLOR
Construction of G:
3.
– –
Clause gadget.
For each clause, add 5 new vertices, connecting them and the corresponding variable vertices with T in the following manner.
Suppose both left and middle variable vertices colored with F.
T
(x1 Úx2 Úx3)Ù(x1 Úx2 Úx3)Ù(x1 Úx2 Úx4)Ù(x1 Úx3 Úx4)
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8.5 Sequencing Problems
Six Basic genres
§ Packing problems: SET-PACKING, INDEPENDENT SET. § Covering problems: SET-COVER, VERTEX-COVER.
§ Constraint satisfaction problems: SAT, 3-SAT.
§ Sequencing problems: HAMILTONIAN-CYCLE, TSP.
3-SAT £P DIR HAMILTONIAN CYCLE £P HAMILTONIAN CYCLE £P TSP § Partitioning problems: 3D-MATCHING, 3-COLOR.
§ Numerical problems: SUBSET-SUM, KNAPSACK. The University of Sydney
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Hamiltonian Cycle
HAM-CYCLE: given an undirected graph G = (V, E), does there exist a simple cycle G that contains every node in V.
The University of Sydney Vertices and faces of a dodecahedron: Yes Page 11

Hamiltonian Cycle
HAM-CYCLE: given an undirected graph G = (V, E), does there exist a simple cycle G that contains every node in V.
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1 1′
2 2′
3 3′
4 4′
5
Bipartite graph with odd number of nodes: No (Why?)
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Eulerian Cycle
EULER-CYCLE: given an undirected graph G = (V, E), does there exist a cycle G that contains every edge in G.
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1 1′ 2 2′
3 3′ 4 4′
5
Graph with odd-degree nodes: No
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Eulerian Cycle
EULER-CYCLE: given an undirected graph G = (V, E), does there exist a cycle G that contains every edge in G exactly once.
EULER-CYCLE is in P
Theorem G has an Eulerian cycle iff every vertex has even degree. But HAM-CYCLE and DIR-HAM-CYCLE are NP-complete!
What are the certificates?
Will show 3-SAT £ P DIR-HAM-CYCLE £ P HAM-CYCLE
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Page 14

Directed Hamiltonian Cycle
DIR-HAM-CYCLE: Given a directed graph G = (V, E), does there exist a simple directed cycle G that contains every node in V?
Theorem: DIR-HAM-CYCLE £ P HAM-CYCLE.
Proof idea: Given a directed graph G = (V, E), construct an undirected graph
G’ with 3n vertices.
For each vertex v, add vertices vin, v, vout. For directed edge (u,v) , add edge (uout, vin)
a
bv c
G
aout d
din
ein
e
bout cout
vin v vout
G’
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Directed Hamiltonian Cycle
Claim: G has a Hamiltonian cycle iff G’ does.
Proof:
Þ –
– Then G’ has an undirected Hamiltonian cycle (same order).
Suppose G has a directed Hamiltonian cycle G.
Ü –
– G’ must visit nodes in G’ using one of two orders:
Suppose G’ has an undirected Hamiltonian cycle G’.
…, B, G, R, B, G, R, B, G, R, B, …
…, B, R, G, B, R, G, B, R, G, B, …
– Blue nodes in G’ make up directed Hamiltonian cycle G in G, or reverse
of one. ▪
aout
bout vin cout
din
ein
d dout
v vout G’
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3-SAT Reduces to Directed Hamiltonian Cycle
Theorem: 3-SAT £ P DIR-HAM-CYCLE.
Proof: Given an instance F of 3-SAT, we construct an instance of DIR-HAM-
CYCLE that has a Hamiltonian cycle iff F is satisfiable.
Construction. First, create graph that has 2n Hamiltonian cycles which
correspond in a natural way to 2n possible truth assignments.
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3-SAT Reduces to Directed Hamiltonian Cycle
Construction: Given a 3-SAT instance F with n variables xi and k clauses.
– Construct G to have 2n Hamiltonian cycles.
– Variable gadget: Path i corresponds to variable xi and has 3k + 3 vertices (2 endpoints, 2 vertices per clause and a filler vertex between clauses and between first/last clause and endpoint.
– Intuition: Traverse path i from left to right Û set variable xi = 1.
– Can’t skip rows!
s
x2
x3
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3k + 3
t
x1

3-SAT Reduces to Directed Hamiltonian Cycle
– Construction. Given 3-SAT instance F with n variables xi and k clauses. – For clause j: add a node and 6 edges (2 per literal)
C1 =x1 Vx2 Vx3
clause node
clause node
s
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t
C2 =x1 Vx2 Vx3
x1
x2
x3

3-SAT Reduces to Directed Hamiltonian Cycle
– Construction. Given 3-SAT instance F with n variables xi and k clauses.
– For clause j: add a node and 6 edges (2 per literal)
– If literal xi appears in clause j, add edge from 3j-th vertex of path i to clause node and from clause node to (3j+1)-th vertex of path i
– If 𝑥! appears, add edge from (3j+1)-th vertex of path i to clause node !
and from clause node to 3j-th vertex of path i
C1 =x1 Vx2 Vx3
clause node
clause node
s
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t
C2 =x1 Vx2 Vx3
x1
x2
x3

3-SAT Reduces to Directed Hamiltonian Cycle
– Construction. Given 3-SAT instance F with n variables xi and k clauses.
– For clause j: add a node and 6 edges (2 per literal)
– If literal xi appears in clause j, add edge from 3j-th vertex of path i to clause node and from clause node to (3j+1)-th vertex of path i
– If 𝑥! appears, add edge from (3j+1)-th vertex of path i to clause node !
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and from clause node to 3j-th vertex of path i
C1 =x1 Vx2 Vx3
clause node
clause node
C2 =x1 Vx2 Vx3
s
t
x1
x2
x3

3-SAT Reduces to Directed Hamiltonian Cycle
Claim: F is satisfiable iff G has a Hamiltonian cycle.
Proof: Þ
– Suppose 3-SAT instance has satisfying assignment x*. – Then, define Hamiltonian cycle in G as follows:
• if x*i = 1, traverse row i from left to right
• if x*i = 0, traverse row i from right to left
• for each clause Cj , there will be at least one row i in which we are going in “correct” direction to splice node Cj into tour
s
x1
x2
The University of Sydney t Page 22
x3

3-SAT Reduces to Directed Hamiltonian Cycle
Claim: F is satisfiable iff G has a Hamiltonian cycle.
Proof: Ü
– Suppose G has a Hamiltonian cycle G.
– If G enters clause node Cj , it must depart on mate edge (otherwise, some filler vertex on row i is skipped)
• thus, nodes immediately before and after Cj are connected by an edge e in G
• removing Cj from cycle, and replacing it with edge e yields HamiltoniancycleonG{Cj }
– Continuing in this way, we are left with Hamiltonian cycle G’ in G-{C1,C2, …,Ck}.
– Set x*i = 1 iff G’ traverses row i left to right.
– Since G visits each clause node Cj , at least one of the paths is traversed
in “correct” direction, and each clause is satisfied. ▪ The University of Sydney
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Longest Path
SHORTEST-PATH: Given a digraph G = (V, E), does there exists a simple path of length at most k edges?
LONGEST-PATH: Given a digraph G = (V, E), does there exists a simple path of length at least k edges?
Theorem: 3-SAT £ P LONGEST-PATH.
Proof: Redo the proof for DIR-HAM-CYCLE, ignoring back-edge from t to s. A simple path with n edges must start from s and end at t since s only has outoing edges and t only has incoming edges.
Corollary: Deciding who wins in Catan and Ticket to Ride is NP-hard.
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Travelling Salesperson Problem
TSP: Given a set of n cities and a pairwise distance function d(u, v), is there a tour of length £ D?
All 13,509 cities in US with a population of at least 500 Reference: http://www.tsp.gatech.edu
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Travelling Salesperson Problem
TSP: Given a set of n cities and a pairwise distance function d(u, v), is there a tour of length £ D?
Optimal TSP tour
Reference: http://www.tsp.gatech.edu
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Travelling Salesperson Problem
TSP: Given a set of n cities and a pairwise distance function d(u, v), is there a tour of length £ D?
11,849 holes to drill in a programmed logic array Reference: http://www.tsp.gatech.edu
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Travelling Salesperson Problem
TSP: Given a set of n cities and a pairwise distance function d(u, v), is there a tour of length £ D?
HAM-CYCLE: given a graph G = (V, E), does there exists a simple cycle that contains every node in V?
Theorem: HAM-CYCLE £ P TSP. Proof:
– Given instance G = (V, E) of HAM-CYCLE, create n cities with distance
function
– TSP instance has tour of length £ n iff G is Hamiltonian. ▪
d(u,v)=ìí1 if(u,v)ÎE î2 if(u,v)ÏE
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Algorithms and hardness
Algorithmic techniques:
– Greedy
– Divide & Conquer
– Dynamic programming
– Network flow
Hardness:
– NP-hardness: Polynomial-time algorithm is unlikely Today
– How can we cope with hard problems?
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Algorithms and hardness
Lots of problems that we can solve efficiently:
– MST
– Shortest path
– Scheduling
– Max flow –…
But lots of problems that we can’t solve efficiently:
– All NP-complete problems: Polynomial-time algorithm is unlikely vertex cover, independent set, 3-SAT,…
– But what if we need to solve an NP-complete problem? The University of Sydney
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Cope with NP-complete problems?
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Cope with NP-complete problems?
• Heuristics: Local search, simulated annealing, neural networks…
• Randomized algorithms: Not always correct, but a probability is
guaranteed.
• Approximation algorithms: Not optimal solution, but within a guaranteed
error.
• Fixed-parameter algorithms: Algorithms whose complexity depends on
other parameters than n.
• Efficient exact algorithms: Euclidean TSP
• Restricted instance: trees, bipartite graphs…
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Coping With NP-Completeness
Question: What should I do if I need to solve an NP-complete problem?
Answer: Theory says you’re unlikely to find poly-time algorithm.
Must sacrifice one of three desired features. – Solve problem to optimality.
• Approximation algorithms
• Randomized algorithms
– Solve problem in polynomial time.
• Exact exponential time algorithms
– Solve arbitrary instances of the problem.
• Solve restricted classes of instances • Parametrized algorithms
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10.2 Solving NP-Hard Problems on restricted input instances
For example special cases of graphs: – trees,
– bipartite graphs,
– planar graphs,
–…
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Independent Set on Trees
INDEPENDENT-SET: Given a graph G = (V, E) and an integer k, is there a subset of vertices S Í V such that |S| 3 k, and for each edge at most one of its endpoints is in S?
Problem: Given a tree, find a maximum IS. Key observation: If v is a leaf, there exists
a maximum size independent set containing v. Proof: [exchange argument]
u v
– Consider a max cardinality independent set S.
– IfvÎS,we’redone.
– IfuÏSandvÏS,thenSÈ {v}isindependentÞSnotmaximum. – IF u Î S and v Ï S, then S È {v}/{u} is independent. (exchange) ▪
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Independent Set on Trees: Greedy Algorithm
Theorem: The following greedy algorithm finds a maximum cardinality independent set in forests (and hence trees).
u v
Independent-Set-In-A-Forest(F) { S¬Æ
while (F has at least one edge) {
Let e = (u,v) be an edge such that v is a leaf
Add v to S
Delete from F nodes u and v, and all edges
incident to them.
}
Add remaining vertices to S.
return S }
ß
u v
ß
Proof: Correctness follows from the previous key observation. ▪
Remark. Can implement in O(n) time by considering nodes in postorder. The University of Sydney
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Weighted Independent Set on Trees
Weighted independent set on trees. Given a tree and node weights wv > 0, find an independent set S that maximizes SvÎS wv.
r
u vwx
children(u) = { v, w, x }
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Weighted Independent Set on Trees
Weighted independent set on trees. Given a tree and node weights wv > 0, find an independent set S that maximizes SvÎS wv.
Observation: If (u, v) is an edge such that v is a leaf node, then either OPT includes u, or it includes all leaf nodes incident to u.
Dynamic programming solution: Root tree at some node, say r. – OPTin(u) = max weight independent set
rooted at u, containing u.
– OPTout(u) = max weight independent set rooted at u, not containing u.
r
u vwx
children(u) = { v, w, x }
OPTin(u) = wu + å OPTout(v) v Î children(u)
OPTout (u) = å max {OPTin (v), OPTout (v)} v Î children(u)
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Weighted Independent Set on Trees: DP Algorithm
Theorem: The dynamic programming algorithm find a maximum weighted independent set in trees in O(n) time.
u
Weighted-Independent-Set-In-A-Tree(T) {
Root the tree at a node r
foreach (node u of T in postorder) {
if (u is a leaf) { Min [u] = wu
Mout[u] = 0} else {
Min [u] = SvÎchildren(u) Mout[v]+wv
Mout[u] = SvÎchildren(u) max(Mout[v],Min[v])}
}
return max(Min[r], Mout[r]) }
u
Proof: Takes O(n) time since we visit nodes in postorder and examine each edge exactly once. ▪
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10.1 Solving restricted instances
For example in the case when the vertex cover is small.
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10.1 Finding Small Vertex Covers
VERTEX COVER: Given a graph G = (V, E) and an integer k, is there a subset of vertices S Í V such that |S| £ k, and for each edge (u, v) either u Î S, or v Î S, or both.
16 27
38 49
5 10
k=4
S = { 3, 6, 7, 10 }
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Vertex Cover
VERTEX COVER: Given a graph G = (V, E) and an integer k, is there a subset of vertices S Í V such that |S| £ k, and for each edge (u, v) either u Î S, or v Î S, or both.
Vertex Cover (or Independent Set) arises naturally in many applications:
– dynamic detection of race conditions (distributed systems).
– computational biology
– Biocheimstry
– Pattern recognition
– Computer vision –…
What should we do if we need to solve it?
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Finding Small Vertex Covers
Question: What if k is small?
Brute force: O(knk+1).
– Try all nk Î O(nk) subsets of size k.
– Takes O(kn) time to check whether a subset is a vertex cover.
Aim: Limit exponential dependency on k, e.g., to O(2k kn).
Example: n = 1000, k = 10.
– Brute force. knk+1 = 1034 Þ infeasible. – Better. 2k kn = 107 Þ feasible.
Remark. If k is a constant, algorithm is poly-time; if k is a small constant, then it’s also practical.
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Finding Small Vertex Covers
Theorem: Let (u,v) be an edge of G. G has a vertex cover of size £ k iff at least one of G{u} and G{v} has a vertex cover of size £ k-1.
delete u and all incident edges
Proof:
uv
Þ
– Suppose G has a vertex cover S of size £ k.
– S contains either u or v (or both). Assume it contains u.
– S{u} is a vertex cover of G{u}. Ü
– Suppose S is a vertex cover of G{u} of size £ k-1.
– ThenSÈ{u}isavertexcoverofGofsizek. ▪
G{u}
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Page 52

Finding Small Vertex Covers: Algorithm
Theorem: The following algorithm determines if G has a vertex cover of size £ k in O(2k kn) time.
uv
G,k
G{u} k-1
boolean Vertex-Cover(G,k) {
if (G contains no edges) return true
if (G contains > k(n-1) edges) return false
let (u,v) be any edge of G
a = Vertex-Cover(G{u},k-1)
b = Vertex-Cover(G{v},k-1)
return a or b
}
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Page 54

Finding Small Vertex Covers: Algorithm
Theorem: The following algorithm determines if G has a vertex cover of size £ k in O(2k kn) time.
uv
G,k
boolean Vertex-Cover(G,k) {
if (G contains no edges) return true
if (G contains > k(n-1) edges) return false
let (u,v) be any edge of G
a = Vertex-Cover(G{u},k-1)
b = Vertex-Cover(G{v},k-1)
return a or b
}
G{u} k-1
G{v}
k-1 £k
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Finding Small Vertex Covers: Algorithm
Theorem: The following algorithm determines if G has a vertex cover of size £ k in O(2k kn) time.
boolean Vertex-Cover(G,k) {
if (G contains no edges) return true
if (G contains > k(n-1) edges) return false
let (u,v) be any edge of G
a = Vertex-Cover(G{u},k-1)
b = Vertex-Cover(G{v},k-1)
return a or b
}
uv
G,k
G{u} k-1
G{v}
k-1 £k
Proof:
– Correctness follows from previous two claims.
– There are £ 2k+1 nodes in the recursion tree; each invocation
takes O(kn) time. ▪ The University of Sydney
Page 56

Finding Small Vertex Covers: Algorithm
Theorem: Vertex cover can be solved in O(2k kn) time. This is fine as long as k is (a small) constant.
What if k is not a small constant?
uv
G,k
G{u} k-1
G{v}
k-1 £k
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