Solved Euler circuit and path worksheet: Part 1: For each of ... - Free Printable

Let’s solve this Euler Circuit and Path Worksheet step by step.

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## 🔍 What are Euler Circuits and Euler Paths?

Before solving, recall:

- Euler Circuit: A path that starts and ends at the *same vertex*, traversing *every edge exactly once*.
- Euler Path: A path that starts and ends at *different vertices*, traversing *every edge exactly once*.
- Not Traceable: Cannot traverse every edge without repeating or lifting pen.

📌 Key Theorem (Euler’s Theorem):

> A connected graph has:
> - An Euler circuit ⇔ all vertices have even degree.
> - An Euler path (but not circuit) ⇔ exactly two vertices have odd degree.
> - Neither ⇔ more than two vertices have odd degree.

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## ✔ Let’s analyze each graph:

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A. (Already Solved)

✔ Given: Euler path
→ Has 2 odd-degree vertices (start and end), so it’s an Euler path.

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B. Triangle inside a circle — all vertices connected

Vertices: 3 outer triangle vertices + 3 on circle? Wait — actually, looking at the diagram, it’s likely a triangle inscribed in a circle with chords connecting all points? But simpler: it’s probably 3 vertices, each connected to the other two → complete graph K₃.

Each vertex has degree 2 (connected to 2 others).

✔ All degrees even → Euler circuit

✔️ Answer: a. Euler circuit

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C. Square with diagonals — like a diamond with X inside

Vertices: 4 corners.

Each corner connects to 2 adjacent corners + 1 diagonal = degree 3.

So all 4 vertices have odd degree (3) → more than 2 odd-degree vertices.

✘ Not traceable.

✔️ Answer: c. Not traceable

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D. Circle with a diameter line

Vertices: 2 endpoints of diameter + maybe the circle is just an edge? Actually, in graph terms, this is usually interpreted as two vertices connected by two edges (the semicircles) and one straight edge? Or perhaps it’s a multigraph.

But standard interpretation: two vertices, each connected by 3 edges (top arc, bottom arc, straight line). So each vertex has degree 3 → odd.

Exactly two vertices with odd degree → Euler path.

But wait — can you trace it? Start at one end, go along top arc, back along diameter, then bottom arc? That would repeat the diameter.

Actually — let’s think differently.

If it’s drawn as a single circle with a diameter, the “vertices” are the two endpoints of the diameter. The circle itself is made of two arcs. So total edges: 3 (top arc, bottom arc, diameter).

Each vertex has degree 3 → odd → Euler path exists.

You can start at one vertex, go along top arc to other vertex, then diameter back, then bottom arc — but that uses diameter twice? No — wait, you must use each edge once.

Actually, possible path:

Start at vertex A → go along top arc to B → go along diameter to A → go along bottom arc to B. Ends at B. Used all 3 edges once.

✔ Euler path (starts at A, ends at B).

✔️ Answer: b. Euler path

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E. Two triangles sharing a vertex? Or linked triangles?

Looks like two triangles connected at a vertex — like a bowtie or hourglass.

Vertices: 5 total? Let’s count degrees.

Assume it's two triangles sharing one common vertex.

Common vertex: connected to 4 others → degree 4 (even)

Other 4 vertices: each connected to 2 others → degree 2 (even)

✔ All degrees even → Euler circuit

✔️ Answer: a. Euler circuit

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F. Three lines from one center point — like a “Y”

Vertices: 1 center + 3 endpoints.

Center vertex: degree 3 (odd)

Each endpoint: degree 1 (odd)

→ Total of 4 vertices with odd degree → violates Euler condition.

✘ Not traceable.

✔️ Answer: c. Not traceable

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G. Circle with a line sticking out — like a lollipop

Vertices: center of circle? Or just the junction point?

Typically: 1 central vertex where circle and stick meet.

The circle: if we consider it as a loop, then the central vertex has degree 2 (from circle) + 1 (from stick) = degree 3.

The endpoint of the stick: degree 1.

So: 2 vertices with odd degree → Euler path.

You can start at stick endpoint → go to center → traverse circle → end back at center? But you’d be stuck.

Wait — better: start at center → go around circle → back to center → go out stick → end at tip.

That uses: circle (one edge? or multiple?) — problem is, if the circle is one edge (loop), then degree of center is 2 (for loop) + 1 (stick) = 3.

But a loop contributes 2 to degree.

So: center vertex: loop gives +2, stick gives +1 → total degree 3.

Stick endpoint: degree 1.

So 2 odd-degree vertices → Euler path.

Can you trace? Start at stick tip → to center → go around circle → back to center → done? But you ended at center, started at tip — that’s fine for Euler path.

Yes.

✔️ Answer: b. Euler path

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H. Three circles connected in a chain — like O-O-O

This is tricky. If each circle is a separate loop, and they’re connected by single edges?

Actually, likely: three circles sharing connection points — but standard interpretation is three disjoint cycles? Then not connected → not traceable.

But if they’re connected end-to-end — like each circle shares a vertex with next.

Assume: 3 circles in a row, sharing vertices.

Vertices: Let’s say left circle has 2 vertices (shared with middle), middle has 2 shared, right has 2.

Actually, simpler: if it’s 3 separate loops with no connection — disconnected graph → cannot trace entire graph.

Even if connected, if each “circle” is a loop, then internal vertices have even degree, but connection points might be odd.

But typically, in such diagrams, if circles are separate, it’s not connected.

So — cannot trace all edges without lifting pen.

✘ Not traceable.

✔️ Answer: c. Not traceable

*(Note: If the circles are connected at points, it might be traceable, but standard worksheet assumes disconnected if not explicitly connected.)*

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I. Square with both diagonals — same as C?

Wait — C was square with diagonals, I is also square with diagonals? Maybe typo? Or different.

Actually, looking again — C had 4 outer vertices with diagonals. I is same? Probably identical.

But let’s check: 4 vertices, each connected to 2 neighbors + 1 diagonal → degree 3 → all odd → 4 odd-degree vertices → not traceable.

Same as C.

✔️ Answer: c. Not traceable

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## ✍️ J. Explain how you can predict if a network is specifically an Euler Circuit.

> Answer:
>
> A connected graph has an Euler circuit if and only if every vertex has even degree. This is known as Euler’s Theorem.
>
> To predict whether a network is an Euler circuit:
> 1. Ensure the graph is connected (you can reach any vertex from any other).
> 2. Count the degree (number of edges incident) of each vertex.
> 3. If ALL vertices have even degree, then the graph contains an Euler circuit.
> 4. If any vertex has odd degree, it does not have an Euler circuit (though it may have an Euler path if exactly two vertices have odd degree).
>
> Example: In graph B (triangle), each vertex has degree 2 (even) → Euler circuit. In graph D, two vertices have degree 3 (odd) → Euler path, not circuit.

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## ✔ Final Answers:

| Graph | Answer |
|-------|------------------|
| A | b. Euler path |
| B | a. Euler circuit |
| C | c. Not traceable |
| D | b. Euler path |
| E | a. Euler circuit |
| F | c. Not traceable |
| G | b. Euler path |
| H | c. Not traceable |
| I | c. Not traceable |

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## 🧠 Bonus Tip:

When tracing, always start at a vertex with odd degree if you’re looking for an Euler path. For Euler circuit, start anywhere — you’ll return to start.

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Let me know if you want me to sketch the actual paths!