浙江大学 算法与程序设计 贪心算法
相关主题
- 1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。
2
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.
Time complexity: O(n log n + n) = O(n log n)
13
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
14
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.
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
18
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
7
f1 smallest
Algorithm #3
8
Analysis - exact solution
Algorithm gives non-overlapping intervals:
obvious, since we always choose an interval which does not overlap the previously chosen intervals
Greedy #4 OPT #3
16
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
12
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
OPT #4 Algorithm #1
4
Rule 2
Select the interval which is shortest (but not overlapping the already chosen intervals) Underestimated solution!
OPT #2 Algorithm #1
For i = 1 by definition of a step in the algorithm. Suppose that Ai-1 finishes not later than Bi-1.
10
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
Greedy algorithm:
Select intervals one after another using some rule
3
Rule 1
SelectFra Baidu bibliotekthe interval which starts earliest (but not overlapping the already chosen intervals) Underestimated solution!
9
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.
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
OPT #4
Algorithm #3
6
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!
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
i-1 i
Ai-1
Ai
11
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
5
Rule 3
Select the interval with the fewest conflicts with other remaining intervals (but still not overlapping the already chosen intervals) Underestimated solution!
1 m
15
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.
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
Greedy algorithm
叶德仕 yedeshi@gmail.com
1
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
If
A , B , and | A | | B |
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.
nt 1 nt 1 nt nt (1 ) k k
17
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
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.
Time complexity: O(n log n + n) = O(n log n)
13
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
14
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.
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
18
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
7
f1 smallest
Algorithm #3
8
Analysis - exact solution
Algorithm gives non-overlapping intervals:
obvious, since we always choose an interval which does not overlap the previously chosen intervals
Greedy #4 OPT #3
16
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
12
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
OPT #4 Algorithm #1
4
Rule 2
Select the interval which is shortest (but not overlapping the already chosen intervals) Underestimated solution!
OPT #2 Algorithm #1
For i = 1 by definition of a step in the algorithm. Suppose that Ai-1 finishes not later than Bi-1.
10
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
Greedy algorithm:
Select intervals one after another using some rule
3
Rule 1
SelectFra Baidu bibliotekthe interval which starts earliest (but not overlapping the already chosen intervals) Underestimated solution!
9
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.
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
OPT #4
Algorithm #3
6
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!
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
i-1 i
Ai-1
Ai
11
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
5
Rule 3
Select the interval with the fewest conflicts with other remaining intervals (but still not overlapping the already chosen intervals) Underestimated solution!
1 m
15
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.
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
Greedy algorithm
叶德仕 yedeshi@gmail.com
1
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
If
A , B , and | A | | B |
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.
nt 1 nt 1 nt nt (1 ) k k
17
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