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g
acceleration of gravity
Предыдущее
Следующее
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1. is that for any connected weighted graph
G
,
2. Yes, see you
G
!
3. there exists a minimum spanning tree of
G
such
4. So what I need to proof is that for all
G
,
5. What is
G
?
6. with the same vertices of
G
. So the only difference
7. So this is the minimum spanning tree for the graph
G
8. So what I need to proof is that for all
G
,
9. For all
G
, I still need to prove there
10. So we know a subgraph of
G
that touches all the different edges
11. So we say for all
G
and for all sets
12. We said that a connected graph,
G
-- that's
13. because we showed that for all graphs
G
,
14. There does not exist a connected graph
G
that has no ST.
15. that covers all the vertices of
G
?
16. So this is the minimum spanning tree for the graph
G
17. Jade and Althea kissing on a tree. K-I-S-S-I-N-
G
!
18. And we assume that it has the same vertices of
G
of course.
19. connected, and with the same vertices as
G
.
20. It's defined as the spanning tree of
G
such
21. What
G
?
22. Do we still have a connected subgraph of
G
23. So all vertices in
G
are still connected
24. She's really walking in S&
G
25. It's defined as the spanning tree of
G
such
26. What is
G
?
27. So I have to prove that for all
G
,
28. Do we still have a connected subgraph of
G
29. there exists a minimum spanning tree of
G
.
30. So we already know that T is a connected subgraph of
G
31. let T be a connected subgraph of
G
, but with a property
32. So let this be a connected subgraph of
G
.
33. Well, of course it cover all the vertices of
G
34. a connected graph
G
.
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