6.889: Algorithms for Planar Graphs and Beyond
Problem Set 1  Solutions
1.
Solution:
Since every face has size at least three, and each edge is in exactly two faces,
3
f
≤
2
m
(here,
f
is the number of faces).
Substituting into Euler’s formula we get 2 =
n

m
+
f
≤
n

m/
3. Or, equivalently,
m
≤
3
n

6.
2.
Solution:
The crucial observation is that in
any
planar graph with no self loops and no
parallel edges, there always exists a node whose degree is at most 5. To see this, note that
by the sparsity of planar graphs,
∑
v
∈
V
degree(
v
) = 2
m
≤
6
n

12.
The algorithm is:
1:
T
← ∅
2:
while
G
is not empty
do
3:
choose an arbitrary node
v
s.t.
deg
(
v
)
≤
5
4:
let
e
be the minimal weight edge incident to
v
5:
T
←
T
∪ {
e
}
6:
contract
e
, eliminating any parallel edges that occur by only keeping the one with
minimal weight.
7:
end while
First, we may assume that
G
contains no self loops and parallel edges since we can detect and
delete them in linear time (for parallel edges delete all but the lightest edge). The correctness
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 Fall '11
 ErikDemaine
 Algorithms, Graph Theory, 2m, constant time, g∗, parallel edges

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