Graphs Homeographic plane

[late sleet]

please 5(d,e,f)

[late sleet]

PLease

[ionic sigil]

(e) Planar – can redraw with no edge crossings.
(d) Non-planar – contains a K3,3 subgraph
(f) Non-planar – contains a K5 subgraph

[late sleet]

how you find K3,3 and K5 subgraphs in d and f

Planar like this?

[late sleet]

please RYM

[ionic sigil]

there are two sets of three vertices and three disjoint paths joining every vertex in the first set to every vertex in the second set.

understood?

(d) i mean

[late sleet]

it is desciption K3,3?

you can't draw? To show (e)

[ionic sigil]

(d) cannot be drawn without an edge crossing

[late sleet]

but it don't have K3,3 or K5?

[ionic sigil]

u r missing some lines

[late sleet]

yes, i understand it now

[ionic sigil]

by Kuratowski’s theorem

Do u know it?

[late sleet]

Sorry, I was busy