QUESTION IMAGE
Question
the following graph is planar. (trust me.) if it is drawn with no intersecting edges, into how many regions (\faces\) does it cut up the plane?
2
3
faces =
1
4
8
5
7
6
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Step1: Count Vertices (V)
Vertices: 1,2,3,4,5,6,7,8 → \( V = 8 \)
Step2: Count Edges (E)
Count each edge:
- 1-8, 2-8, 2-3, 2-4, 2-7, 3-4, 3-5, 3-7, 4-5, 4-6, 4-7, 5-6, 5-7, 6-7, 8-5, 8-6, 8-7 (Wait, better way: list all connections. Let's recount carefully:
From 1: 1-8 (1)
From 2: 2-8, 2-3, 2-4, 2-7 (4) → total 1+4=5
From 3: 3-4, 3-5, 3-7 (3) → total 5+3=8
From 4: 4-5, 4-6, 4-7 (3) → total 8+3=11
From 5: 5-6, 5-7 (2) → total 11+2=13
From 6: 6-7 (1) → total 13+1=14
From 8: 8-5, 8-6, 8-7 (3) → Wait, earlier missed? Wait no, 2-8,1-8, then 8-5,8-6,8-7. Wait, let's list all edges:
1-8, 2-8, 2-3, 2-4, 2-7, 3-4, 3-5, 3-7, 4-5, 4-6, 4-7, 5-6, 5-7, 6-7, 8-5, 8-6, 8-7. Wait, that's 17? Wait no, maybe better to use hand count:
Looking at the graph:
- 1 connected to 8 (1)
- 2 connected to 8,3,4,7 (4) → 1+4=5
- 3 connected to 2,4,5,7 (4? Wait 3-2, 3-4, 3-5, 3-7 → 4) → 5+4=9
- 4 connected to 2,3,5,6,7 (5? 4-2,4-3,4-5,4-6,4-7 → 5) → 9+5=14
- 5 connected to 3,4,6,7,8 (5? 5-3,5-4,5-6,5-7,5-8 → 5) → 14+5=19
- 6 connected to 4,5,7,8 (4? 6-4,6-5,6-7,6-8 → 4) → 19+4=23? No, this is wrong. Let's use Euler's formula: \( F = E - V + 2 \), but we need correct E.
Alternative: Let's count edges properly:
Nodes: 8. Let's list adjacency:
1: [8] → 1 edge
2: [8,3,4,7] → 4 edges (total 1+4=5)
3: [2,4,5,7] → 4 edges (5+4=9)
4: [2,3,5,6,7] → 5 edges (9+5=14)
5: [3,4,6,7,8] → 5 edges (14+5=19)
6: [4,5,7,8] → 4 edges (19+4=23)
7: [2,3,4,5,6,8] → 6 edges (23+6=29? No, this is overcounting because each edge is shared. Wait, no: each edge is between two nodes, so we should count each edge once. Let's use a better approach:
List all unique edges:
1-8,
2-8, 2-3, 2-4, 2-7,
3-4, 3-5, 3-7,
4-5, 4-6, 4-7,
5-6, 5-7, 5-8,
6-7, 6-8,
7-8 (Wait, 7-8? Wait node 7 and 8: is there an edge? In the graph, 8 is connected to 1,2,5,6,7? Let's see the graph: 8 is connected to 1 (vertical), 2 (line), 5 (line), 6 (line), 7 (line). So 8-1, 8-2, 8-5, 8-6, 8-7 → 5 edges from 8.
2 is connected to 8,3,4,7 → 4 edges.
3 is connected to 2,4,5,7 → 4 edges.
4 is connected to 2,3,5,6,7 → 5 edges.
5 is connected to 3,4,6,7,8 → 5 edges.
6 is connected to 4,5,7,8 → 4 edges.
7 is connected to 2,3,4,5,6,8 → 6 edges.
1 is connected to 8 → 1 edge.
Now, sum all edges (each edge counted twice, once from each end), so total edges \( E = \frac{1 + 4 + 4 + 5 + 5 + 4 + 6}{2} \)? Wait no: 1 (from 1) + 4 (2) +4 (3)+5(4)+5(5)+4(6)+6(7)+5(8) → Wait 8 nodes: 1,2,3,4,5,6,7,8. Their degrees:
1:1, 2:4, 3:4, 4:5, 5:5, 6:4, 7:6, 8:5.
Sum of degrees: 1+4+4+5+5+4+6+5 = 34. So \( E = \frac{34}{2} = 17 \). Yes, because each edge contributes to two degrees. So \( E = 17 \).
Step3: Apply Euler's Formula for Planar Graphs
Euler's formula: \( V - E + F = 2 \) (for connected planar graphs; check if connected: all nodes are connected through 8,2, etc. So connected.)
We have \( V = 8 \), \( E = 17 \).
Solve for \( F \):
\( F = E - V + 2 \)
\( F = 17 - 8 + 2 = 11 \)? Wait no, wait Euler's formula is \( V - E + F = 2 \) → \( F = E - V + 2 \). Wait 17-8=9, 9+2=11? But wait, maybe I miscounted E. Let's recount degrees:
Node 1: degree 1 (correct, only 1-8)
Node 2: connected to 8,3,4,7 → degree 4 (correct)
Node 3: connected to 2,4,5,7 → degree 4 (correct)
Node 4: connected to 2,3,5,6,7 → degree 5 (correct)
Node 5: connected to 3,4,6,7,8 → degree 5 (correct)
Node 6: connected to 4,5,7,8 → degree 4 (correct)
Node 7: connected to 2,3,4,5,6,8 → degree 6 (correct)
Node 8: connected to 1,2,5,6,7 → degree 5 (correct)
Sum: 1+4=5, +4=9, +5=14, +5=19, +4=23, +6=29, +5=34. So \( E = 34/2 = 17 \). Correct.
Then \( F = 17 - 8 + 2 = 11 \)? Wait but…
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