1 greedy algorithm 叶德仕 [email protected]. 2 greedy algorithm’s paradigm algorithm is greedy...
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Greedy algorithm’s paradigm
Algorithm is greedy ifit builds up a solution in small stepsit chooses a decision at each step myopically to optimize some underlying criterion
Analyzing optimal greedy algorithms by showing that:
in every step it is not worse than any other algorithm, or every algorithm can be gradually transformed to thegreedy one without hurting its quality
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Interval schedulingInput: set of intervals on the line, represented by pairs of points (ends of intervals). In another word, the ith interval, starts at time si and finish at
fi.
Output: finding the largest set of intervals such that none two of them overlap. Or the maximum number of intervals that without overlap.
Greedy algorithm:Select intervals one after another using some rule
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Rule 1
Select the interval which starts earliest (but not overlapping the already chosen intervals)
Underestimated solution!
Algorithm #1
OPT #4
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Rule 2
Select the interval which is shortest (but not overlapping the already chosen intervals)
Underestimated solution!
Algorithm #1
OPT #2
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Rule 3
Select the interval with the fewest conflicts with other remaining intervals (but still not overlapping the already chosen intervals)
Underestimated solution!
Algorithm #3
OPT #4
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Rule 4
Select the interval which ends first (but still not overlapping the already chosen intervals)
Quite a nature idea: we ensure that our resource become free as soon as possible while still satisfying one request
Hurray! Exact solution!
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f1 smallest
Algorithm #3
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Analysis - exact solution
Algorithm gives non-overlapping intervals:obvious, since we always choose an interval which does not overlap the previously chosen intervals
The solution is exact:Let A be the set of intervals obtained by the algorithm,
and OPT be the largest set of pairwise non-overlapping intervals.
We show that A must be as large as OPT
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Analysis – exact solution cont.
Let and be sorted. By definition of OPT we have k ≤ m
Fact: for every i ≤ k, Ai finishes not later than Bi.
Pf. by induction.For i = 1 by definition of a step in the algorithm.
Suppose that Ai-1 finishes not later than Bi-1.
1{ , , }kA A A 1{ , , }mOPT B B
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Analysis con.
From the definition of a step in the algorithm we get that Ai is the first interval that finishes after Ai-1 and does not verlap it.
If Bi finished before Ai then it would overlap some of the previous A1 ,…, Ai-1 and
consequently - by the inductive assumption - it would overlap Bi-1, which would be a contradiction.
Bi-1
Ai-1
Bi
Ai
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Analysis con.
Theorem: A is the exact solution.
Proof: we show that k = m.
Suppose to the contrary that k < m. We have that Ak finishes not later than Bk
Hence we could add Bk+1 to A and obtain bigger solution by the algorithm-contradiction Bk
Ak
Bk-1
Ak-1
Bk+1
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Time complexity
Sorting intervals according to the right-most ends
For every consecutive interval:
If the left-most end is after the right-most end of the last selected interval then we select this interval
Otherwise we skip it and go to the next interval
Time complexity: O(n log n + n) = O(n log n)
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Planning of schools
A collection of towns. We want to plan schools in towns.
Each school should be in a town
No one should have to travel more than 30 miles to reach one of them.
Edge: towns no far than 30
miles
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Set cover
Input. A set of elements B, sets
Output. A selection of the Si whose union is B.
Cost. Number of sets picked.
1, mS S B
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Greedy
Greedy: first choose a set that covers the largest number of elements.
example: place a school at town a, since this covers the largest number of other towns.
Greedy #4
OPT #3
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Upper bound
Theorem. Suppose B contains n elements that the optimal cover consist of k sets. Then the greedy algorithm will use at most k ln n sets.
Pf. Let nt be the number of elements still not covered after t iterations of the greedy algorithm (n0=n). Since these remaining elements are covered by the optimal k sets, there must be some set with at least nt /k of them. Therefore, the greedy algorithm will ensure that
1
1(1 )t
t t t
nn n n
k k
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Upper bound con.Then , since for all x,
with equality if and only if x=0.
Thus
At t=k ln n, therefore, nt is strictly less than ne-ln n =1, which means no elements remains to be covered.Consequently, the approximation ratio is at most ln n
0
1(1 )t
tn nk
1 xx e
1/ /0 0
1(1 ) ( )t k t t k
tn n n e nek
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MatroidsWhen will the greedy algorithm yields optimal solutions?Matroids [Hassler Whitney]: A matroid is an ordered pair M=(S, ℓ) satisfying the following conditions.
S is a finite nonempty set
ℓ is a nonempty family of subsets of S, called the independent subsets of S, such that if
We say that ℓ is hereditary if it satisfies this property. Note that empty set is necessarily a member of ℓ.
If , then there is some element such that . We say that M satisfies the
exchage property.
, , .B and A B then A
| |, , | | BA B and A x B A { }A x
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Max independent
Theorem. All maximal independent subsets in a matroid have the same size.
Pf. Suppose to the contrary that A is a maximal independent subset of M and there exists another larger maximal independent subset B of M. Then, the exchange property implies that A is extendible to a larger independent set A {x}∪ for some x B - A∈ , contradicting the assumption that A is maximal.
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Weighted Matroid
We say that a matroid M = (S,ℓ) is weighted if there is an associated weight function w that assigns a strictly positive weight w(x) to each element x ∈ S. The weight function w extends to subsets of S by summation:
for any A ⊆ S.
( ) ( )x A
w A w x
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Greedy algorithms on a weighted matroid
Many problems for which a greedy approach provides optimal solutions can be formulated in terms of finding a maximum-weight independent subset in a weighted matroid. That is, we are given a weighted matroid M = (S,ℓ), and we wish to find an independent set A ℓ∈ such that w(A) is maximized. We call such a subset that is independent and has maximum possible weight an optimal subset of the matroid. Because the weight w(x) of any element x ∈ S is positive, an optimal subset is always a maximal independent subset-it always helps to make A as large as possible.
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Greedy algorithm
GREEDY(M, w)
1. A ← Ø
2. sort S[M] into monotonically decreasing order by weight w
3. for each x ∈ S[M], taken in monotonically decreasing order by weight w(x)
4. do if A {∪ x} ℓ [∈ M]
5. then A ← A {∪ x}
6. return A
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LemmaLemma 1. Suppose that M = (S,ℓ) is a weighted matroid with weight function w and that S is sorted into monotonically decreasing order by weight. Let x be the first element of S such that {x} is independent, if any such x exists. If x exists, then there exists an optimal subset A of S that contains x.Pf.If no such x exists, then the only independent subset is the empty set and we're done. Otherwise, let B be any nonempty optimal subset. Assume that x ∉ B; otherwise, we let A = B and we're done.No element of B has weight greater than w(x). To see this, observe that y ∈ B implies that {y} is independent, since B ℓ∈ and ℓ is hereditary. Our choice of x therefore ensures that w(x) ≥ w(y) for any y ∈ B.
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Lemma
Construct the set A as follows. Begin with A = {x}. By the choice of x, A is independent. Using the exchange property, repeatedly find a new element of B that can be added to A until |A| = |B| while preserving the independence of A. Then, A = B - {y}
{∪ x} for some y ∈ B, and so w(A)=w(B) - w(y) + w(x) ≥ w(B).
Because B is optimal, A must also be optimal, and because x ∈ A, the lemma is proven.
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Lemma
Lemma 2. Let M = (S,ℓ) be any matroid. If x is an element of S that is an extension of some independent subset A of S, then x is also an extension of Ø.
Pf. Since x is an extension of A, we have that A ∪{x} is independent. Since ℓ is hereditary, {x} must be independent. Thus, x is an extension of Ø.
It is shown that if an element is not an option initially, then it cannot be an option later.
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Corollary
Corollary Let M = (S,ℓ) be any matroid. If x is an element of S such that x is not an extension of Ø, then x is not an extension of any independent subset A of S.Any element that cannot be used immediately can never be used. Therefore, GREEDY cannot make an error by passing over any initial elements in S that are not an extension of Ø, since they can never be used.
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Lemma 3. Let x be the first element of S chosen by GREEDY for the weighted matroid M = (S,ℓ). The remaining problem of finding a maximum-weight independent subset containing x reduces to finding a maximum-weight independent subset of the weighted matroid M′ = (S′,ℓ), where
S′ ={y ∈ S : {x, y} ℓ},∈ ℓ′ ={B ⊆ S - {x} : B {∪ x} ℓ},∈
and the weight function for M′ is the weight function for M, restricted to S′. (We call M′ the contraction of M by the element x.)
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Proof
If A is any maximum-weight independent subset of M containing x, then A′ = A - {x} is an independent subset of M′. Conversely, any independent subset A′ of M′ yields an independent subset A = A′ {∪ x} of M. Since we have in both cases that w(A) = w(A′) + w(x), a maximum-weight solution in M containing x yields a maximum-weight solution in M′, and vice versa.
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Theorem
Theorem. If M = (S,ℓ) is a weighted matroid with weight function w, then GREEDY(M, w) returns an optimal subset.
Pf. By Corollary, any elements that are passed over initially because they are not extensions of Ø can be forgotten about, since they can never be useful.
Once the first element x is selected, Lemma 1 implies that GREEDY does not err by adding x to A, since there exists an optimal subset containing x.
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Theorem con.
Finally, Lemma 3 implies that the remaining problem is one of finding an optimal subset in the matroid M′ that is the contraction of M by x. After the procedure GREEDY sets A to {x}, all of its remaining steps can be interpreted as acting in the matroid M′ = (S′,ℓ′), because B is independent in M′ if and only if B {∪ x} is independent in M, for all sets B
ℓ′.∈ Thus, the subsequent operation of GREEDY will find a maximum-weight independent subset for M′, and the overall operation of GREEDY will find a maximum-weight independent subset for M.
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Minimum spanning tree
Input: weighted graph G = (V,E) every edge in E has its positive weight
Output: finding the spanning tree such that the sum of weights is not bigger than the sum of weights of any other spanning tree
Spanning tree: subgraph with no cycle, andconnected (every two nodes in V are connected by a path)
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Properties of minimum spanning trees MST
Spanning trees:n nodesn - 1 edgesat least 2 leaves (leaf - a node with only one neighbor)
MST cycle property:After adding an edge we obtain exactly one cycle and all the edges from MST in this cycle have no bigger weight than the weight of the added edge
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Optimal substructures
MST T:(Other edges of Gare not shown.)
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Optimal substructures
MST T:(Other edges of Gare not shown.)
Remove any edge (u, v) ∈ T.
u
v
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Optimal substructures
MST T:(Other edges of Gare not shown.)
Remove any edge (u, v) ∈ T. Then, T is partitionedinto two subtrees T1 and T2.
T1
T2
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Optimal substructures
MST T:(Other edges of Gare not shown.)
Remove any edge (u, v) ∈ T. Then, T is partitionedinto two subtrees T1 and T2.
T1
T2
Theorem. The subtree T1 is an MST of G1 = (V1, E1),the subgraph of G induced by the vertices of T1:V1 = vertices of T1,E1 = { (x, y) ∈ E : x, y ∈ V1 }.Similarly for T2.
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Proof of optimal substructure
Proof. Cut and paste:
w(T) = w(u, v) + w(T1) + w(T2).
If T1′ were a lower-weight spanning tree than T1 for G1, then
T′ = {(u, v)} ∪ T1′ ∪ T2
would be a lower-weight spanning tree than T for G.
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Do we also have overlapping subproblems?
Yes.
Great, then dynamic programming may work!
Yes, but MST exhibits another powerful property which leads to an even more efficient algorithm.
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Crucial observation about MST
Consider sets of nodes A and V - ALet F be the set of edges between A and V - ALet a be the smallest weight of an edge from F Theorem:Every MST must contain at least one edge of weight afrom set F
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A A
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Proof of the observation
Let e be the edge in F with the smallest weight - for simplicity assume that there is unique such edge. Suppose to the contrary that e is not in some MST. Choose one such MST.Add e to MST - obtain the cycle, where e is (among) smallest weights. Since two ends of e are in different sets A and V - A, there is another edge f in the cycle and in F. Remove f from the tree (with added edge e) - obtain a spanning tree with the smaller weight(since f has bigger weight than e). This is a contradiction with MST.
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Greedy algorithm finding MST
Kruskal’s algorithm:Sort all edges according to the weightsChoose n - 1 edges one after another as follows:
If a new added edge does not create a cycle with previously selected then we keep it in (partial) MST, otherwise we remove it
Remark: we always have a partial forest
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Greedy algorithm finding MST
Prim’s algorithm:Select a node as a root arbitrarilyChoose n - 1 edges one after another as follows:
Look on all edges incident to the currently build (partial) tree and which do not create a cycle in it, and select one which has the smallest weight
Remark: we always have a connected partial tree
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root
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Example of Prim
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Example of Prim
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Example of Prim
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Example of Prim
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Example of Prim
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Example of Prim
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Example of Prim
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Example of Prim
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Example of Prim
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Example of Prim
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Example of Prim
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Example of Prim
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Why the algorithms work?
Follows from the crucial observationKruskal’s algorithm:
Suppose we add edge {v,w}. This edge has the smallest weight among edges between the set of nodes already connected with v (by a path in selected subgraph) and other nodes.
Prim’s algorithm:Always chooses an edge with the smallest weight among edges between the set of already connected nodes and free nodes.
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Time complexity
There are implementations using
Union-find data structure (Kruskal’s algorithm)
Priority queue (Prim’s algorithm)
achieving time complexity
O(m log n)
where n is the number of nodes and m is the
number of edges
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Best of MST
Best to date:
Karger, Klein, and Tarjan [1993].
Randomized algorithm.
O(V + E) expected time.
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Conclusions
Greedy algorithms for finding minimum
spanning tree in a graph, both in time
O(m log n) :
Kruskal’s algorithm
Prim’s algorithm
Remains to design the efficient data structures!
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Conclusions
Greedy algorithms: algorithms constructing solutions step after step using a local ruleExact greedy algorithm for interval selection problem - in time O(n log n) illustrating “greedy stays ahead” ruleGreedy algorithm may not produce optimal solution such as set cover problemMatroids can help to prove when will greedy can lead to optimal solutionMinimum spanning tree could be solved by greedy method in O(m log n)