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CSC 373H1 / L0101   Lecture Notes -- Proof of Prim's Alg.       Winter 2012
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A proof of Prim's algorithm showing that each step "is promising".
 
Theorem: Given a connected undirected graph G = (V, E), with
edge weights w(e), Prim's algorithm constructs a MST of G.

Proof:  We use the loop invariant:

  L(k): Let T_k = (S_k, F_k) be the subgraph that Prim has constructed 
  with k = |S_k| (after the execution of the loop (k-1) times).  Then 
  there exists an MST T=(V,F) of G=(V, E) s.t. T_k is a subgraph of T.
 
That is, T_k "is promising".

We prove L(k) is true for k = 1 to |V| using induction.
 
For the base case, k=1,  T_1 = ({v}, {}).  Given the
existence of an MST T for G, T_1 must be a subgraph of T.  
We leave the existence of an MST for G to the reader.
 
Let 1< k <= |V|.  

Assume L(k-1) is true.  We need to prove L(k) is true.

Since L(k-1) is assumed true, let T_(k-1) = (S_(k-1), F_(k-1)) be
the subgraph generated during the execution of the algorithm 
with |S_(k-1)| = k-1.  And let T be an MST of G which contains T_(k-1).  
(Such a T exists according to L(k-1).)

Let e = (u, v), with u in S_(k-1) and v in V \ S_{k-1), be 
the k-th edge added by Prim.
 
If edge e is in the MST T = (V, F), then L(k) follows.
So we are left with considering the case e is not in F.

Suppose e = (u,v) is not in F.  Since T is a spanning tree, 
there exists a simple path P = (u, a1, a2, ... an, v) in T,
from u to v.  Let f = (a,b) = (a(i), a(i+1)) be the first edge
on this path that has a(i) in S_(k-1) and a(i+1) in V\S_(k-1).
Note u in S_(k-1) and v is in the complement, so such an edge
must exist.  Also, all the edges in P are in T and e
is not in T, so f does not equal e.
 
Claim: The graph Tp = (V, (F\{f})U{e})) is a tree.
Pf Omitted: Left to reader to show: 1) Tp has the right number
of edges, namely |V|-1, and 2) Tp is connected.
Together these imply Tp is a tree (Property 2 in the MST slides.)
Hint: To show Tp is connected, use the simple cycle
C = (u, a1, a2, ... an, v, u) in the graph A=(V, F U {e}),
where P = (u, a1, ... an, v) is as constructed above.

Finally consider the weight of this tree Tp, 
(*)  w(Tp) = w(T) - w(f) + w(e)

There are three cases:

a) w(f) < w(e):  Impossible, since Prim's alg would not have chosen 
e in the presence of the edge , for which f = (a, b) with w(f) < w(e), 
a is in S_(k-1), and b in V\S_(k-1).

b) w(f) = w(e):  Then (*) implies w(Tp) = w(T) and, since we
know T is a MST, then Tp must be an MST too. Therefore L(k) holds
with this modified MST Tp. 

c) w(f) > w(e): Impossible, since in this case (*) implies 
w(Tp) < w(T), but this contradicts T being a MST.

Therefore L(k) must be true.
 
It follows by induction that L(k) is true for k = 1 to |V|.

Finally, L(|V|) implies that the computed subgraph is an MST for G.
 
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Note: At each stage of Prim's algorithm T_k is a subtree.  We
carefully didn't make use of this above (and we didn't prove it).
Because of this, the same loop invariant (with "Prim" replaced by Kruskal)
applies to Kruskal's algorithm.  The proof of Kruskal's algorithm then
requires only a small modification.