Min cut of at most k edges

I have been studying for my algorithms exam and whilst doing previous exams found this question for which I am not sure how to handle. Given a Graph with integer capacities, find an efficient algorithm to determine whether the graph has a min cut of at most k edges (in the question k = 100) or not.

My thinking was computing a min cut, then for every edge on that cut, increase it by 1 and compute the min cut again – if the max flow has risen, that means that that edge exists in every single min cut. And If I found there to be over 100 of these edges, then there is no min cut of at most k edges.

However, I don’t think that if I found there to be less 100 of these edges, there’s necessarily a min cut of at most 100 edges.

Anyone care to give a helping hand? Thank you!

how to maximise the no of edges selected in the graph in form of cycles of unbounded length?

Recently i started looking the perfect cycle cover algorithm related to kidney exchange problem where it is considered as NP-Complete for cycles restricted to a length>2. However if cycles are not restricted by removing the length constraint it is considered as solvable in polynomial time ,But if perfect cycle cover doesn’t exist and just we need to maximise the no of edges selected in form of cycles(So as to maximise the no of kidney exchanges), also no restriction on cycle’s length, then how do we do it?

I am thinking of finding all cycles in the graph and then one by one remove the cycle’s edges from the graph and then find the cycles in the residual graph and so on till no more cycles exist and then sum up all the cycle’s lengths continuing this for all cycles and find the solution with maximum edges selected. But since this is not a good solution i want to know how to solve this maximising problem

Are edges in a minimum spanning tree not heavier than respective edges in another spanning tree?

Let $ G$ be an undirected connected weighted graph, and let $ T$ be a minimum spanning tree of $ G$ with edge weights: $ w_1 \le w_2 \le … \le w_{n-1}$ .

Now let $ T’$ be some other spanning tree of $ G$ (doesn’t have to be minimum) with edge weights: $ w’_1 \le w’_2 \le … \le w’_{n-1}$ .

I need to prove\disprove the following claim: for every $ i$ : $ w_i \le w’_i$ .

I’ve tried to find a counterexample, but I wasn’t successful. So I’m quite sure the claim is correct. However, I had trouble to prove it formally.

I assume the contrary, which means there is an $ i$ that satisfies $ w’_i < w_i$ and I take the first one that does. I tried adding it to the tree $ T$ or trying to find a minimum spanning tree with this edge using Kruskal, but no luck.

Adding edges to a DAG to make it strongly connected with minimum cost

I have a weighted DAG and a function computing the weight of edges that is not connected in the DAG. The weight of u to v equals to the weight of v to u.

I want to connect edges to make the DAG strongly connected with minimum total weights of added edges.

I know that the minimum number of added edges equals to $ \text{max}(|source|, |sink|)$ but which vertex connects to which vertex that the total weights is minimum?

Assign weights to the edges in a DAG so that, for all S and T, all paths from S to T have equal weight

I have a DAG, and on each edge, I have a minimum and maximum weight. I would like to assign (or determine it’s impossible to assign) exact weights to each edge so that

  1. Each edge’s weight is between its min and max
  2. For any two nodes $ S$ and $ T$ in the DAG, all paths from $ S$ to $ T$ have equal weight (where the weight of a path is the sum of all the weights of the edges in that path)
  3. [Bonus points] The total weight of all edges in the graph is minimized

I know I can set this up as a linear program, but that feels like overkill. There has to be a simpler way to do this. I’ve tried to think of a way to reduce it to a min-cost flow problem, but I’ve had no success so far. I think that might not be the right direction. Any ideas?

Help making the mouse cross the edges screen

Hello guys im playing fps games on Android with otg and mouse, and im triying to Cross the screen with the pointer of the mouse but is imposible… Any idea? I tried octopus APK but there is something strange… It can be crossed on fps Mode and it works perfect if u dont Cross the screen, If you do It the movement of the x axis is much faster till is total unplayable. Thanks in advance!

Help making the mouse cross the edges screen

Hello guys im playing fps games on Android with otg and mouse, and im triying to Cross the screen with the pointer of the mouse but is imposible… Any idea? I tried octopus APK but there is something strange… It can be crossed on fps Mode and it works perfect if u dont Cross the screen, If you do It the movement of the x axis is much faster till is total unplayable. Thanks in advance!

Portable monitor past monitor edges

I have an Asus monitor for a 2016 XPS 13 developer edition laptop. I’ve been trying to get the resolution to look better than the defaults. I ran the following xrandr commands:

xrandr --newmode "1920x1080_60.00"  173.00  1920 2048 2248 2576  1080 1083 1088 1120 -hsync +vsync xrandr --addmode DVI-I-1-1 "1920x1080_60.00" 

The screen looks much better but the edges overflow past the monitor edges. Is there a good way to fix this? The monitor I am using is:

https://www.amazon.com/ASUS-MB169B-1920×1080-Portable-Monitor/dp/B00FE690DI/ref=sr_1_3?keywords=portable%2Bmonitor&qid=1560014011&s=gateway&sr=8-3&th=1

Find the number of edges

Let $ \langle V_1, E_1 \rangle, \langle V_2, E_2 \rangle$ and $ \langle V_3, E_3\rangle$ be any three undirected simple graphs with $ m$ , $ n$ and $ p$ number of edges, respectively such that $ E_2$ and $ E_3$ have no edge in common. Then what would be the number of edges in $ [(E_1\cup E_2)\cup\lbrace (a,b): a\in V_1~\text{and}~b\in V_2 \rbrace]\cap [(E_1\cup E_3)\cup\lbrace (c,d): c\in V_1~\text{and}~d\in V_3 \rbrace]$ ?, where the operations $ \cup$ and $ \cap$ are usual set union and join, respectively. Also, given that $ V_2$ and $ V_3$ have no vertex in common. Or, how to simplify and formulate $ [(E_1\cup E_2)\cup\lbrace (a,b): a\in V_1~\text{and}~b\in V_2 \rbrace]\cap [(E_1\cup E_3)\cup\lbrace (c,d): c\in V_1~\text{and}~d\in V_3 \rbrace]$ to put into a more compact form?

Possible ways to have cross and full edges in a mincut maxflow

I am trying to solve the following problem about maxflow mincut it seems like my conclusion is incorrect and I am wonder where.

There is no graph just following question.

An edge e can be (x) always full, (y) sometimes full, (z) never full; it can be (x’) always crossing, (y’) sometimes crossing, (z’) never crossing. So there are nine possible combinations: (xx’) always full and always crossing, (xy’) always full and sometimes crossing, and so on. Or are there? Maybe some possibilities are impossible. Let’s draw a table:

enter image description here

Possible answers —based on my conclusions

  • always full and always crossing. (true based on the image above)
  • always full and sometimes crossing. (true based on the image above and the fact that we may have several crossing edges one of them may be full, all them if all edges in the min cut is 1 and one of the crossing edges is 2)
  • always full and never crossing. (false seems like there always must be a possibility of one of the ways to use edge with less capacity)
  • sometimes full and always crossing. (true if we have all edges in the mincut that are 1 and crossing edge is two)
  • sometimes full and sometimes crossing.( false since if we have many crossing edges that are 1 and one of them is 2 means that we might want to use the edge 2to flow more)
  • sometimes full and never crossing. (true easy to see and edge that sometime full but never crossing in a graph)
  • never full and always crossing. (true if all edges in a mincut is 1 and crossing edge is two)
  • never full and sometimes crossing. (true if we have many 2 crossing edges)
  • never full and never crossing. (true if we have a graph where one edge is 2 in min cut that never crossing)

Have asked this question on /math.stackexchange.com with no luck.

wording used.

The edge is full means in in any way we pump the flow the edge is always going to be full, basically if we have and C—(1)—A——(1)—-B and there is no other way to get to B other then going thru a then A is always full. Somewhat full if we can have C—(1)—A——(1)—-B and if there is another way to get to b by doing C—(1)—E—(1)—-B then edge is only sometime full. never full if we have C—(1)—E—(2)—-B then most we can flow is 1 but b takes 2 then it never full. The crossing part if we it can be the crossing edge in the min-cut from the cut to the sink. basically it and edge that connects the cut with the rest of flow.

Thanks.