1.Suppose ALG is an α-approximation algorithm for an optimization problem Π whose approximation ratio is tight. Then for every ?>0 there is no (α??)-approximation algorithm for Π unless P = NP.
F, 按照我的理解，可能會有近似比更小的算法。

2.As we know there is a 2-approximation algorithm for the Vertex Cover problem. Then we must be able to obtain a 2-approximation algorithm for the Clique problem, since the Clique problem can be polynomially reduced to the Vertex Cover problem.
T， reduce to 就算<=，如果CP<=VC, VC有近似比為2的算法，那么CP也有

3.For the bin-packing problem: let S=∑Si. Which of the following statements is FALSE?

A.The number of bins used by the next-fit heuristic is never more than ?2S?

B.The number of bins used by the first-fit heuristic is never more than ?2S?

C.The next-fit heuristic leaves at most one bin less than half full

D.The first-fit heuristic leaves at most one bin less than half full

NF近似比是2，其他的近似比都比2小。Next fit可能有多個半空的bit，因為如果永遠往前放，不會回頭放之前的，所以是C，而FF會檢查之前所有位，因此如果有兩個半空的，它們會放在一起。

4.To approximate a maximum spanning tree T of an undirected graph G=(V,E) with distinct edge weights w(u,v) on each edge (u,v)∈E, let’s denote the set of maximum-weight edges incident on each vertex by S. Also let w(E′)=∑?(u,v)∈E?′??， w(u,v) for any edge set E′. Which of the following statements is TRUE?

A.S=T for any graph G

B.S≠T for any graph G

C.w(T)≥w(S)/2 for any graph G

D.None of the above

C, 題目的意思是，如果把每個點最大權值的邊加入一個集合，那么這個集合的權值和最大生成樹權值之比是多少。注意，點的最大權值邊集合意味著集合里相同的邊最多出現(xiàn)一次。

很容易證明，S里面不存在環(huán)，因此T一定包含S.所以w(T)>=w(S)

假如存在環(huán)，設邊為e1,e2,e3,…ej, 點為p1, …pj

由于e1在S中，因此w(e1)>w(ej)，由于e2在S中，因此w(e2)>w(e1),…，最后得到的是w(ej)>w(e1)，矛盾。

5.Assume that you are a real world Chinese postman, which have learned an awesome course “Advanced Data Structures and Algorithm Analysis” (ADS). Given a 2-dimensional map indicating N positions pi(xi?? ,y?i) of your post office and all the addresses you must visit, you’d like to find a shortest path starting and finishing both at your post office, and visit all the addresses at least once in the circuit. Fortunately, you have a magic item “Bamboo copter & Hopter” from “Doraemon”, which makes sure that you can fly between two positions using the directed distance (or displacement).

Bamboo.jpg (“Bamboo copter & Hopter”, japan12.com/bamboo-copter-hopter)

However, reviewing the knowledge in the ADS course, it is an NPC problem! Wasting too much time in finding the shortest path is unwise, so you decide to design a 2?approximation algorithm as follows, to achieve an acceptable solution.

Compute a minimum spanning tree T connecting all the addresses.
Regard the post office as the root of T.
Start at the post office.
Visit the addresses in order of a _____ of T.
Finish at the post office.
There are several methods of traversal which can be filled in the blank of the above algorithm. Assume that P≠NP, how many methods of traversal listed below can fulfill the requirement?

Level-Order Traversal
Pre-Order Traversal
Post-Order Traversal

A.0

B.1

C.2

D.3

C，不知道題目再說什么

6.An approximation scheme that runs in $O(n^2/?)$ for any fixed ?>0 is a fully polynomial-time approximation scheme.

T, 如果$1/?$也是多項式級別的，那么就算是full

7.An approximation scheme that runs in $O(n^2{3}^{?})$ for any fixed ?>0 is a polynomial-time approximation scheme.

T, 只要是n是多項式級別的就可以。

8.In the bin packing problem, we are asked to pack a list of items L to the minimum number of bins of capacity 1. For the instance L, let FF(L) denote the number of bins used by the algorithm First Fit. The instance L′ is derived from L by deleting one item from L. Then FF(L?′) is at most of FF(L).

F, 如果是NF則是F, yds說的。

9.For the 0-1 version of the Knapsack problem, if we are greedy on taking the maximum profit or profit density, then the resulting profit must be bounded below by the optimal solution minus the maximum profit.

Popt<Pfrac<Pgre+pmax，最優(yōu)解一定小于物體可分情況下的解。而物體可分情況下的解，可以看成greedy的解+一部分不完整的物體。不完整的物體權值一定小于最大權值。

(感覺證明有問題)

10.An (1+?)-approximation scheme of time complexity $(n+1/?{)}^3$ is a PTAS but not an FPTAS.

F

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