Planar Graphs and Euler's Formula - TCE Mathematics Applications (Tasmania)
“Recognise planar graphs and apply Euler's formula relating vertices, edges and faces.”
You must decide whether a graph is planar (can be redrawn with no edges crossing), identify its faces, and apply Euler's formula to connect the number of vertices, edges, and faces.
What this dot point is asking
A graph drawing often has edges that cross, but the crossings may be an accident of how it was drawn. The real question is whether the same connections can be redrawn flat with no crossings at all.
A graph is planar if it can be drawn in the plane so that no two edges cross. The drawing with no crossings divides the plane into regions called faces, and one of these faces is always the unbounded outer region.
Counting faces
Faces are the enclosed regions of a planar drawing, plus the single infinite region outside the graph. Beginners forget the outer face, so always add one for the outside when you count.
For any connected planar graph,
where is the number of vertices, the number of edges, and the number of faces (including the outer face). Rearranged, .
A connected planar graph has 5 vertices and 6 edges. How many faces does it have, and verify the count.
Apply the formula
.
So the drawing has 3 faces: 2 enclosed regions plus the outer region. Substituting back: , which confirms the relationship holds.
Why planarity matters
Planar graphs model situations where crossings are physically awkward or forbidden, such as circuit boards where wires must not touch, road networks without overpasses, or seating plans. Knowing a graph is planar tells you a clean, crossing-free layout exists.
A graph of 4 vertices is drawn as a square with both diagonals, so the diagonals appear to cross in the middle.
Test for planarity
Although drawn with a crossing, this graph can be redrawn by placing one vertex inside a triangle formed by the other three, connecting it to all three. No edges then cross, so the graph is planar.
With and , Euler gives faces in the crossing-free drawing.
Vertices and edges are easy to count from the connection list, even from a messy drawing. Rather than struggling to count faces directly, use and let the formula give the face count.
Using Euler's formula to find any missing count
Because the formula links three quantities, knowing any two gives the third. Rearrange to suit the unknown: for faces, for edges, or for vertices. TASC questions exploit this by giving, for instance, the number of faces and edges and asking for the vertices, as in the worked exam answer above. The key is to substitute carefully and watch the signs, since the and terms are where arithmetic slips occur.
A connected planar graph has 7 vertices and 5 faces. How many edges does it have?
Rearrange and substitute
.
The graph has 10 edges. Check with the original form: , which holds, so the count is consistent.
Testing for non-planarity
Some graphs simply cannot be drawn without crossings no matter how they are rearranged. The two classic non-planar graphs are (five vertices each joined to every other) and (the "three houses, three utilities" graph). When a TASC question asks you to show a graph is non-planar, demonstrating that it contains one of these structures, or that no crossing-free redrawing is possible, is the expected argument. For such graphs Euler's formula in the simple form does not apply, because the formula assumes a crossing-free planar drawing exists.
Connected requirement
Euler's formula in the form applies to connected planar graphs. If the graph splits into separate pieces, the formula must be adjusted (each extra component subtracts one from the total), so first confirm the graph is in one connected piece before applying it.
The most frequent error is counting only the enclosed regions and ignoring the infinite region outside the graph. That outer face is always one of the faces, so a drawing with 2 enclosed regions has , not 2. Add one for the outside, then check against Euler's formula.
A complete answer states whether the graph is planar (and redraws it without crossings if needed), counts vertices and edges, includes the outer face when counting faces, and uses to confirm or to find a missing count.
Exam-style questions
Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.
Practice5 marksA planar connected network (graph) has 5 faces and 8 edges. a) Use Euler's formula to determine the number of vertices in this network. b) Draw the planar connected network in the space below and label the vertices. c) Describe a Hamiltonian circuit that can be travelled on the network in part b).Show worked answer →
a) (2 marks) Euler's formula for a connected planar graph is v - e + f = 2. Substitute e = 8 and f = 5: v - 8 + 5 = 2, so v - 3 = 2, giving v = 5 vertices.
b) (2 marks) Draw any connected planar graph with 5 vertices and 8 edges so that no edges cross. Counting the regions it splits the plane into (including the outer region) must give 5 faces. Label the 5 vertices.
c) (1 mark) A Hamiltonian circuit visits every vertex exactly once and returns to the start. Trace such a closed route through all 5 vertices on your graph and write it as a sequence of vertex labels.
Practice4 marksA network has 12 vertices and 20 edges. a) Use Euler's formula to calculate the number of faces. c) Draw an example of a network with 5 vertices and 7 edges for which Euler's formula does not apply, and state why it does not apply.Show worked answer →
a) (1 mark) Euler's formula rearranges to . With and : faces.
c) (3 marks) Draw a network with 5 vertices and 7 edges that is either not connected or not planar (cannot be drawn without edges crossing). Euler's formula holds only for connected planar graphs, so if the graph is disconnected (or non-planar) the relationship breaks down. State which condition your example violates.