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algorithms – Counterexample to greedy solution for set cover problem

I am looking for the answer to the exercise 1-6 in “The Algorithm Design Manual” book. it is stated as follows:

1-6. The set cover problem is as follows: given a set of subsets
S1,…,Sm of the universal set U = {1,…,n}, find the smallest subset
of enter image description here subsets such that enter
image description here. For example, there are the following
subsets, S1 = {1,3,5}, S2 = {2,4}, S3 = {1,4}, and S4 = {2,5} The set
cover would then be S1 and S2.

Find a counterexample for the following algorithm: Select the largest
subset for the cover, and then delete all its elements from the
universal set. Repeat by adding the subset containing the largest
number of uncovered elements until all are covered.

My answer is as follows:

S = {{1, 2}, {3, 4}, {2, 5}, {5}}, U={1,2,3,4,5} algorithm will use
following sets: {{1, 2}, {3, 4}, {2, 5}}, while {{1, 2}, {3, 4}, {5}}
would be more minimal

do you think its correct? I assume here that algorithm minimizes number of uncovered elements in each iteration and not the entire size of set to choose – thats why it might choose {2,5} instead of {5}.

I have found this answer on internet:

One counter-example consists of a series of subsets that increase in
size exponentially, plus 2 additional subsets that each cover half of
the elements. Example:
S1 = {1,2}

S2 = {3,4,5,6}

S3 = {7,8,9,10,11,12,13,14,15,16}

S4 = {1,2,3,4,5,6,7,8}

S5 = {9,10,11,12,13,14,15,16}
The greedy algorithm will choose.
S3,S2,S1, while the optimal solution is simply S4,S5

To my understandment, algorithm would choose S3,S4 and not S3,S2,S1. S4 has more uncovered elements than S2.

graphs – A question about the work per recursive call in FPT vertex cover of size k algorithm

You do not need to create any new graph.

You can simply maintain an additional array $VT$ of size $|V| = n$ such that $VT(i) = 1$ if the vertex $i$ is deleted from the graph. Otherwise, $VT(i) = 0$.

While accessing an edge $(i,j)$, first the algorithm can check if $VT(i)$ and $VT(j)$ are both $0$ or not. If both are $0$ then the algorithm access that edge. Otherwise, it does not.

In addition to it, maintain the count of edges a vertex $i$ is currently connected to. Suppose the count is stored in an array $C$ of size $n$. In the beginning of the algorithm $C(i)$ is initialized to the $degree(i)$.


Accessing an Undeleted Edge: Go to the adjacency list of that vertex which has $VT(i) = 0$ and $C(i) neq 0$. Since we know that at least one edge in the list is not deleted, the algorithm will find an undelete edge in $O(n)$ time. Updating the count array after deleting a vertex takes $O(n)$ time. Overall operation is $O(n)$ time.

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algorithms – Approximation of Set Cover

I wonder why do we say $log n$ is the best possible approximation factor for Set Cover Algorithm? We already know there exists a 2-approximation algorithm for vertex cover, which is obviously better than $log n$. So we can use Vertex Cover algorithm to solve Set Cover problem and get a better approximation ratio. Can anyone explain this to me?

Thanks in advance

Is there any polytime reduction from feedback vertex set to vertex cover?

I know that feedback vertex set (FVS) problem is $mathrm{NP}$-complete since there is a simple and nice polytime reduction from vertex cover (VC) problem to FVS.

Specifically, given a undirected graph $G$, we construct $G’$ as follow:

  1. For every $uv in E(G)$, we construct a new vertex $e_{uv}$, and let $V_{0} = { e_{uv} : uv in E(G) }$.
  2. Let $V(G’) = V(G) cup V_{0}$
  3. An edge is in $E(G’)$ if it is in $E(G)$ or one of its endpoint is the vertex $e_{uv} in V_{0}$ and another endpoint of it is the endpoint of $e_{uv}$. Namely, $E(G’) = E(G) cup { ue_{uv}, ve_{uv} : e_{uv} in V_{0} }$.

It is easy to see that $G$ has a vertex cover of size at most $k$ iff $G’$ has a feedback vertex set of size at most $k$.

Of course, there is a polytime reduction from FVS to VC. I want to know whether there is a reduction with a simple construction. And I hope that the size of the vertex cover and the size of feedback vertex set are equalent, if possible.

dnd 5e – Can archers bypass partial cover by arcing their shot?

There are two basic ways an archer fires at a target. In close quarters engagements, archers (and anyone using a projectile weapon) would likely use “direct fire”, ie. fire at an angle nearly parallel to the ground. At longer distances and especially when targets are hiding behind terrain and walls, archers instead use “indirect fire”, ie. firing at angle greater than 45 degrees, in order to lob arrows over and behind cover.

Imagine an archer firing at a range of 100 feet on a creature using 5-foot tall wall for cover.

The rules for cover on a grid state the following:

Choose a corner of the attacker’s space or the point of origin of an area of effect. Then trace imaginary lines from that corner to every corner of any one square the target occupies. If one or two of those lines are blocked by an obstacle (including another creature), the target has half cover. If three or four of those lines are blocked but the attack can still reach the target (such as when the target is behind an arrow slit), the target has three-quarters cover.

Using these rules, it’s easy to see how the wall could provide half or three-quarters cover against direct fire. The trajectory of the arrow will always intersect with the wall, and if enough lines from the archer intersect with the wall, then partial cover is granted. This is consistent with a physical understanding of the scenario, because the arrows will follow a nearly straight line from the archer to their target.

But what if the archer chooses to fire indirectly at their target? In the physical world a wall would provide no cover against an attack that falls from above. Drawing lines from the archer, however, results in the same result as direct fire, granting partial cover in a way which is inconsistent with reality.

Are there any rules that would allow the archer to use indirect fire to bypass partial cover?

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