Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Max-flow min-cut theorem
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
==Example== [[File:Max_flow.svg|thumb|right|A maximal flow in a network. Each edge is labeled with ''f/c'', where ''f'' is the flow over the edge and ''c'' is the edge's capacity. The flow value is 5. There are several minimal ''s''-''t'' cuts with capacity 5; one is ''S''={''s'',''p''} and ''T''={''o'', ''q'', ''r'', ''t''}.]] The figure on the right shows a flow in a network. The numerical annotation on each arrow, in the form ''f''/''c'', indicates the flow (''f'') and the capacity (''c'') of the arrow. The flows emanating from the source total five (2+3=5), as do the flows into the sink (2+3=5), establishing that the flow's value is 5. One ''s''-''t'' cut with value 5 is given by ''S''={''s'',''p''} and ''T''={''o'', ''q'', ''r'', ''t''}. The capacities of the edges that cross this cut are 3 and 2, giving a cut capacity of 3+2=5. (The arrow from ''o'' to ''p'' is not considered, as it points from ''T'' back to ''S''.) The value of the flow is equal to the capacity of the cut, showing that the flow is a maximal flow and the cut is a minimal cut. Note that the flow through each of the two arrows that connect ''S'' to ''T'' is at full capacity; this is always the case: a minimal cut represents a 'bottleneck' of the system.
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)