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unable to get the below logic

-1

$begingroup$

I stuck at a problem Flawed flow

I am not getting how F[u]=0 guarantees acyclic graph in below approach
I found a comment in editorial But it is not very clear

comment -> if there's no vertex v that f[v]=0 then we have an cycle in graph, because every vertex in graph have at least one incoming edge. (Consider longest simple path in remain graph, we know that head vertex in path has an incoming edge and the head of that edge is also in path).

Please someone explain what is the correct reason for above mentioned statement ?

The key element to solving the task is the following observation: if we know all the incoming edges of a vertex, all the remaining edges must be outgoing. The source has no incoming edges, so we already know that all its edges are outgoing. For all other vertices except the sink the amount of incoming and outcoming flow is the same, and is equal to half of the sum of the flow along its incident edges. The algorithm then is to repeatedly direct all the flow from the vertices for which all the incoming edges are known. This can be done with a single BFS:

for all v from 2 to n-1
 f[v] := sum(flow(v,u))/2;
put source in queue
while queue is not empty
 v := pop(queue)
 for all edges (v, u)
 if (v, u) is not directed yet
 direct v -> u
 f[u] = f[u] - flow(v,u)
 if u not sink and f[u] = 0
 push(queue, u)

As the flow contains no cycles, we can sort the vertices topologically. Then

we can be sure that, until all edge directions are known, we can put at least the first vertex with unknown edges in the queue, as all of its incoming edges will be from vertices with lower indices, but we took the first vertex with unknown edges.

Time: O(n + m)

Memory: O(n + m)

Implementation: c++

graph-theory network-flow

share|cite|improve this question

asked Mar 31 at 7:19

aka1234aka1234

62

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hello, this is the mathematics site so please ask your question on a programming site, not here.
  $endgroup$
  – Stallmp
  Mar 31 at 7:30
  

  
  
  
  

add a comment | 

-1

$begingroup$

I stuck at a problem Flawed flow

I am not getting how F[u]=0 guarantees acyclic graph in below approach
I found a comment in editorial But it is not very clear

comment -> if there's no vertex v that f[v]=0 then we have an cycle in graph, because every vertex in graph have at least one incoming edge. (Consider longest simple path in remain graph, we know that head vertex in path has an incoming edge and the head of that edge is also in path).

Please someone explain what is the correct reason for above mentioned statement ?

The key element to solving the task is the following observation: if we know all the incoming edges of a vertex, all the remaining edges must be outgoing. The source has no incoming edges, so we already know that all its edges are outgoing. For all other vertices except the sink the amount of incoming and outcoming flow is the same, and is equal to half of the sum of the flow along its incident edges. The algorithm then is to repeatedly direct all the flow from the vertices for which all the incoming edges are known. This can be done with a single BFS:

for all v from 2 to n-1
 f[v] := sum(flow(v,u))/2;
put source in queue
while queue is not empty
 v := pop(queue)
 for all edges (v, u)
 if (v, u) is not directed yet
 direct v -> u
 f[u] = f[u] - flow(v,u)
 if u not sink and f[u] = 0
 push(queue, u)

As the flow contains no cycles, we can sort the vertices topologically. Then

we can be sure that, until all edge directions are known, we can put at least the first vertex with unknown edges in the queue, as all of its incoming edges will be from vertices with lower indices, but we took the first vertex with unknown edges.

Time: O(n + m)

Memory: O(n + m)

Implementation: c++

graph-theory network-flow

share|cite|improve this question

asked Mar 31 at 7:19

aka1234aka1234

62

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hello, this is the mathematics site so please ask your question on a programming site, not here.
  $endgroup$
  – Stallmp
  Mar 31 at 7:30
  

  
  
  
  

add a comment | 

$begingroup$

I stuck at a problem Flawed flow

I am not getting how F[u]=0 guarantees acyclic graph in below approach
I found a comment in editorial But it is not very clear

comment -> if there's no vertex v that f[v]=0 then we have an cycle in graph, because every vertex in graph have at least one incoming edge. (Consider longest simple path in remain graph, we know that head vertex in path has an incoming edge and the head of that edge is also in path).

Please someone explain what is the correct reason for above mentioned statement ?

The key element to solving the task is the following observation: if we know all the incoming edges of a vertex, all the remaining edges must be outgoing. The source has no incoming edges, so we already know that all its edges are outgoing. For all other vertices except the sink the amount of incoming and outcoming flow is the same, and is equal to half of the sum of the flow along its incident edges. The algorithm then is to repeatedly direct all the flow from the vertices for which all the incoming edges are known. This can be done with a single BFS:

for all v from 2 to n-1
 f[v] := sum(flow(v,u))/2;
put source in queue
while queue is not empty
 v := pop(queue)
 for all edges (v, u)
 if (v, u) is not directed yet
 direct v -> u
 f[u] = f[u] - flow(v,u)
 if u not sink and f[u] = 0
 push(queue, u)

As the flow contains no cycles, we can sort the vertices topologically. Then

we can be sure that, until all edge directions are known, we can put at least the first vertex with unknown edges in the queue, as all of its incoming edges will be from vertices with lower indices, but we took the first vertex with unknown edges.

Time: O(n + m)

Memory: O(n + m)

Implementation: c++

graph-theory network-flow

share|cite|improve this question

asked Mar 31 at 7:19

aka1234aka1234

62

$endgroup$

for all v from 2 to n-1
 f[v] := sum(flow(v,u))/2;
put source in queue
while queue is not empty
 v := pop(queue)
 for all edges (v, u)
 if (v, u) is not directed yet
 direct v -> u
 f[u] = f[u] - flow(v,u)
 if u not sink and f[u] = 0
 push(queue, u)

share|cite|improve this question

asked Mar 31 at 7:19

aka1234aka1234

62

share|cite|improve this question

asked Mar 31 at 7:19

aka1234aka1234

62

asked Mar 31 at 7:19

aka1234aka1234

62

asked Mar 31 at 7:19

aka1234aka1234

62