use the graph on the right to find a hamilton circuit that begins as a, b,.... (use a comma to separate…

Explanation:

Step1: Recall Hamilton Circuit Definition

A Hamilton circuit is a path in a graph that visits each vertex exactly once and returns to the starting vertex. The graph has vertices ( A, B, C, D, E, F, G ). We start with ( A, B ).

Step2: Choose Next Vertex from B

From ( B ), possible neighbors (excluding ( A ) to avoid repetition early, but we can check) are ( C, D, F, G ). Let's pick ( F ) (arbitrary, but need to cover all). Now we have ( A, B, F ).

Step3: Next from F

From ( F ), neighbors are ( A, B, C, D ). ( A, B ) used, so pick ( C ). Now ( A, B, F, C ).

Step4: Next from C

From ( C ), neighbors are ( E, D, F, B ). ( F, B ) used, pick ( E ). Now ( A, B, F, C, E ).

Step5: Next from E

From ( E ), only neighbor left (not used) is ( D )? Wait, ( E ) is connected to ( A ) and ( C )? Wait, looking at the graph: ( E ) is connected to ( A ) and ( C )? Wait, maybe I misread. Let's re-examine the graph: vertices ( A, B, C, D, E, F, G ). Edges: ( A - B, A - C, A - E, B - C, B - D, B - G, C - D, C - E, D - G, D - F, E - F, F - B, F - A, F - C, F - D )? Wait, maybe better to list connections properly. Alternatively, a possible circuit: ( A, B, G, D, F, E, C, A )? No, wait, start with ( A, B ). Let's try: ( A, B, D, G, B )? No, repetition. Wait, correct approach: start at ( A ), go to ( B ), then from ( B ) to ( G ) (since ( B ) is connected to ( G )), then ( G ) to ( D ), ( D ) to ( F ), ( F ) to ( E ), ( E ) to ( C ), ( C ) to ( A ). Wait, but let's check each vertex: ( A, B, G, D, F, E, C, A ). Wait, but we need to visit each vertex once. Wait, vertices are ( A, B, C, D, E, F, G ) (7 vertices). So the circuit should have 7 vertices plus back to ( A ). Wait, no: Hamilton circuit visits each vertex exactly once and returns, so length 7 (vertices) + 1 (return). Wait, let's list all vertices: ( A, B, C, D, E, F, G ). So the circuit should be ( A, B, [next], [next], ..., [last], A ), with each vertex once.

Let's try again: ( A, B, F, D, G, B )? No, repetition. Wait, correct path: ( A, B, F, D, G, B ) is wrong. Let's use a better method. From ( B ), after ( A ), go to ( F ), then ( F ) to ( D ), ( D ) to ( G ), ( G ) to ( B )? No, ( B ) used. Wait, maybe: ( A, B, C, E, F, D, G, A ). Wait, check each vertex: ( A, B, C, E, F, D, G ) – all 7 vertices, then back to ( A ). Let's verify edges: ( A - B ) (yes), ( B - C ) (yes), ( C - E ) (yes), ( E - F ) (yes), ( F - D ) (yes), ( D - G ) (yes), ( G ) to ( A )? Wait, ( G ) is connected to ( B ) and ( D ), not ( A ). Oops. So ( G ) can't go to ( A ). So adjust: after ( G ), go to ( B )? No, ( B ) used. Wait, ( D ) is connected to ( C ) too. Let's try: ( A, B, F, C, E, A )? No, missing ( D, G ). Wait, I think I made a mistake in vertex connections. Let's assume the graph is a complete-like graph with ( A ) connected to ( B, C, E, F ); ( B ) connected to ( A, C, D, F, G ); ( C ) connected to ( A, B, D, E, F ); ( D ) connected to ( B, C, F, G ); ( E ) connected to ( A, C, F ); ( F ) connected to ( A, B, C, D, E ); ( G ) connected to ( B, D ).

So starting with ( A, B ):

1. ( A, B )
2. From ( B ), choose ( G ) (since ( G ) is only connected to ( B ) and ( D ), better to take ( G ) early). Now ( A, B, G )
3. From ( G ), only ( D ) left (connected to ( G )). Now ( A, B, G, D )
4. From ( D ), choose ( F ) (connected to ( D )). Now ( A, B, G, D, F )
5. From ( F ), choose ( E ) (connected to ( F )). Now ( A, B, G, D, F, E )
6. From ( E ), choose ( C ) (connected to ( E )). Now ( A, B, G, D, F, E, C )
7. From ( C ), only ( A ) left (connected to ( C ) and not used). Now ( A, B, G, D, F, E, C, A )

Wait, but let's check each vertex: ( A, B, G, D, F, E, C ) – all 7 vertices, then back to ( A ). Yes, each vertex is visited once ( ( A, B, G, D, F, E, C ) – 7 vertices, then ( A ) to close the circuit).

Alternatively, another path: ( A, B, F, C, D, G, E, A )? Wait, ( E ) is connected to ( A ) and ( C ), so ( E ) to ( A ) is okay. Let's check: ( A, B, F, C, D, G, E, A ). Each vertex: ( A, B, F, C, D, G, E ) – all 7, then ( A ). Edges: ( A - B ), ( B - F ), ( F - C ), ( C - D ), ( D - G ), ( G - E ) (wait, is ( G ) connected to ( E )? No, looking at the graph, ( E ) is connected to ( A ) and ( C ), ( G ) is connected to ( B ) and ( D ). So that's invalid. So the first path: ( A, B, G, D, F, E, C, A ) is valid? Wait, ( G ) to ( D ) is edge, ( D ) to ( F ) is edge, ( F ) to ( E ) is edge, ( E ) to ( C ) is edge, ( C ) to ( A ) is edge. Yes.

But maybe a simpler one: ( A, B, C, F, D, G, B )? No, repetition. Wait, the key is to visit each vertex once. Let's list all vertices: ( A, B, C, D, E, F, G ) (7 vertices). So the circuit must have 7 edges (visiting each vertex once) and return to ( A ).

Another possible circuit: ( A, B, D, F, E, C, G, A )? Wait, ( G ) to ( A )? No, ( G ) is not connected to ( A ). ( G ) is connected to ( B ) and ( D ). So ( G ) must connect to ( B ) or ( D ), but ( B ) and ( D ) are already used. Wait, no, in the circuit, after ( G ), we need to connect back to ( A ), but ( G ) is only connected to ( B ) and ( D ), which are already in the path. So that's a problem. Wait, maybe my initial graph analysis is wrong. Let's re-express the graph:

• ( A ) connected to ( B, C, E, F )
• ( B ) connected to ( A, C, D, F, G )
• ( C ) connected to ( A, B, D, E, F )
• ( D ) connected to ( B, C, F, G )
• ( E ) connected to ( A, C, F )
• ( F ) connected to ( A, B, C, D, E )
• ( G ) connected to ( B, D )

So ( G ) has degree 2 (connected to ( B ) and ( D )). So in a Hamilton circuit, ( G ) must be between ( B ) and ( D ) (since it has degree 2, it's a bridge between ( B ) and ( D )). So the path through ( G ) must be ( B - G - D ) or ( D - G - B ).

So let's build the circuit with ( B - G - D ):

Start: ( A, B )

Then ( B - G ): ( A, B, G )

Then ( G - D ): ( A, B, G, D )

Then ( D - F ) (since ( D ) is connected to ( F )): ( A, B, G, D, F )

Then ( F - E ) ( ( F ) to ( E )): ( A, B, G, D, F, E )

Then ( E - C ) ( ( E ) to ( C )): ( A, B, G, D, F, E, C )

Then ( C - A ) ( ( C ) to ( A )): ( A, B, G, D, F, E, C, A )

Yes, this works: each vertex ( A, B, G, D, F, E, C ) is visited once, then back to ( A ).

Answer:

A, B, G, D, F, E, C, A