A Fullerene is a 3-regular planar graph with faces of degree 5 and 6. Let P be the number of
pentagons (degree 5 faces) and H be the number of hexagons (degree 6 faces). Given that a
Fullerene has v vertices, determine P and H as functions of v.
Solution
Here there are P faces of degree five( pentagons ) and H faces of degree six( Hexagons)
Then there are F = P+H faces
we know from eulers formula that F= 2 + E - V
By 3 Regularity we know 2E = 3V
so P+H = F = 2 + 3V/2 -V
= 2 + V/2
Further by handshaking for planar graphs we know
2E = 5P + 6H
so , 3V= 2 E = 5P + 6H
= 5(P+H) + H
= 5(2+ v/2) + H
This tells us that
v = 2H + 20,
so H = V/2 - 10
as P + H = 2+V/2
we conclude
P = 2+ V/2 - H
= 2+ V/2 - (V/2 - 10)
= 12.
A Fullerene is a 3-regular planar graph with faces of degree 5 and 6.pdf
1. A Fullerene is a 3-regular planar graph with faces of degree 5 and 6. Let P be the number of
pentagons (degree 5 faces) and H be the number of hexagons (degree 6 faces). Given that a
Fullerene has v vertices, determine P and H as functions of v.
Solution
Here there are P faces of degree five( pentagons ) and H faces of degree six( Hexagons)
Then there are F = P+H faces
we know from eulers formula that F= 2 + E - V
By 3 Regularity we know 2E = 3V
so P+H = F = 2 + 3V/2 -V
= 2 + V/2
Further by handshaking for planar graphs we know
2E = 5P + 6H
so , 3V= 2 E = 5P + 6H
= 5(P+H) + H
= 5(2+ v/2) + H
This tells us that
v = 2H + 20,
so H = V/2 - 10
as P + H = 2+V/2
we conclude
P = 2+ V/2 - H
= 2+ V/2 - (V/2 - 10)
= 12
