10 The binomial Random Graph
We are interested in sliding
from to
. At
what point do certain structures emerge?
Questions:
-
When do we expect to see a triangle?
-
When do we expect
to be connected?
-
When does
have a component that spans
vertices?
-
When does
have a Hamiltonian cycle?
Fix some . When
do we expect ,
?
Definition 10.1 (m).
For
a graph, define
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Theorem 10.2 (Bollobás, 80s).
Assuming that:
Then |
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Proof.
We use a second moment argument. Let
where .
Then
Since , we will
compute .
𝟙
Now note
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when .
So
10.1 The Giant Component
Proposition 10.3.
Let .
If , then
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and
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Remark.
This “threshold” for
looks different from what we saw from the threshold of containing a given
– it has a “sharp”
jump from
to .
Definition 10.4 (c1).
For a graph ,
let
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Theorem 10.5 (Erdös-Renyi).
Let .
If then
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Definition 10.6 (DFS-sequence).
Let be
a graph. I define the DFS-sequence from
as follows. We process the graph using a Depth first search. We also imagine
have an order on them. As we saw before, we maintain a partition
where
is a path.
We start with ,
,
. We
then repeat the following:
-
(1)
If
is empty, move a vertex from
to .
-
(2)
If
is not empty, let
be the last vertex of ,
and now query the edges from
to
in order until I get a “yes”.
Now move the neighbour of
to the end of
and append to our DFS sequence the outcomes of the queries.
If
has no neighbour in ,
then move
to
and append all the “no” outcomes to the DFS-sequence.
Note.
-
We don’t query an edge more than once.
-
We don’t query all edges in .
But if
is not queried, then
are in the same component.
Lemma 10.7.
Let be a
graph on vertices, with
a component of size . Let
be the DFS-sequence.
Then there exists
such that
Proof.
Let be a
component of size ,
. Say we first
encounter at
time . From
time up until
, we only query
edges incident to .
So we make at most
exposures. And during this time we must have seen
’s
since
is a component. □
Lemma 10.8.
For ,
let .
Then if ,
Proof.
Let
be the corresponding DFS-sequence. Since we never query an edge more than once in the definition of
the DFS-sequence, we have that
is a sequence of iid random variables, with each being .
So let
for .
We will use Chernoof’s inequality:
Let
defined by
Then for ,
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Note so
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Note that if ,
then the above is .
Then union bound over all
choices for
to see no such
satisfies .
□
Theorem 10.9 (Ajtai, Komlós, Szemeredi).
For
, if
where
, then
where
.
Lemma 10.10.
Let
be an vertex graph
with DFS-sequence
and
where .
Then
where .
Proof.
If at time we
have , then since
, assuming for contradiction
, for all steps. Then there
must have been some step
in the algorithm where
and .
Then
contradiction.
So we may assume . So
we can assume at time
that the algorithm is still running. We have
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since, each time we see
we move a vertex from
into ,
and it never returns. Now
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So combining we have
Also note
so
contradiction. □
Reminder of Chernoff bound:
Lemma 10.11.
Let .
Then, for ,
we have
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Proof of Theorem 10.9.
Let ,
where ,
for . Let
DFS-sequence. This is a sequence of iid Bernoulli random variables with probability
. So set
and
note
since . So apply Lemma 10.10
to see that with probability ,
, where
.
□
Theorem 10.12.
Let .
Let be some
function with .
Let .
Then
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