浙江大学 算法与程序设计 贪心算法

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.
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Analysis – exact solution cont.
Let A {A1,, Ak } and OPT {B1,, Bm} be sorted. By definition of OPT we have k ≤ m Fact: for every i ≤ k, Ai finishes not later than Bi. Pf. by induction.
nt 1 nt 1 nt nt (1 ) k k
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Upper bound con.
Then , since 1 x e x for all x, with equality if and only if x=0.
1 t nt n0 (1 ) k
Thus
1 t nt n0 (1 ) n0 (e 1/ k )t ne t / k k
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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Matroids
When 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
OPT #4
Algorithm #3
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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!
Greedy algorithm
叶德仕 yedeshi@
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Greedy algorithm’s paradigm
Algorithm is greedy if
it builds up a solution in small steps it chooses a decision at each step myopically to optimize some underlying criterion
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!
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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!
Edge: towns no far than 30 miles
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Set cover
Input. A set of elements B, sets S ,S B Output. A selection of the Si whose union is B. Cost. Number of sets picked.
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Interval scheduling
Input: 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.
1 m
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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.
i-1 i
Ai-1
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
For i = 1 by definition of a step in the algorithm. Suppose that Ai-1 finishes not later than Bi-1.
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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 B contradiction. B
B , and A B, then A .
We say that ℓ is hereditary if it satisfies this property. Note that empty set is necessarily a member of ℓ.
, then there is some element x B A such that A {x} . We say that M satisfies the exchage property. 19
The solution is exact:
Let A be the set of intervals obtained by the algorithm, and OPT be the largest set of pairwise nonoverlapping intervals. We show that A must be as large as OPT
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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
OPT #4 Algorithm #1
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Rule 2
Select the interval which is shortest (but not overlapping the already chosen intervals) Underestimated solution!
OPT #2 Algorithm #1
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
If
A , B , and | A | | B |
பைடு நூலகம்
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.
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
Bk-1 Bk Bk+1
Ak-1
Ak
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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