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fiksx

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- TL;DR Summary
- classify isomoprhism of graph

N and k are positive integers satisfying $$ 1<=k < n$$

An undirected graph $$G_{n,k}= (V_{n,k} ,E_{n,k})$$ is defined as follows.

$$V_{n,k}={1,2,3,...n}$$

$$E_{n,k}={\{\{u,v\}|u-v \equiv k \, (mod \, n) \, or \, u-v \equiv -k \, mod \, n}$$

However, $$x \equiv y \, (mod \, n) $$ indicates that x and y are congruent modulo n. For example, $$12 \equiv 5 \equiv -2 (mod 7)$$

i need to Classify graph G by all k that is isomorphism to each other for $$G_{16,k} (1<=k<=15) $$ Example {1,2},{3},{4,5,6}my attempt:

$$G_{16,k} (1<=k<=15) $$

Classify graph G by all k that is isomorphism

Example {1,2},{3},{4,5,6}

I see that by definition for example $$G_{5,1}$$

we can see that

$$V={1,2,3,4,5}$$

Since $$ 6 \equiv 1 \equiv -4 \, mod \, 5$$ we know that$$E=\{(1,5),(5,4),(4,3),(3,2),(2,1) \} $$

But how can i classify k that isomorphism to each other for $$G_{16,k}$$

I thought that $$\{2,4,6,8,10,12,14,16 \} $$ and $$\{1,3,5,7,9,11,13,15 \} $$ will be isomorphism to each other, am i right?

i.e $$G_{2,4} \& G_{4,6}$$ will have same vertex and degree(?)

An undirected graph $$G_{n,k}= (V_{n,k} ,E_{n,k})$$ is defined as follows.

$$V_{n,k}={1,2,3,...n}$$

$$E_{n,k}={\{\{u,v\}|u-v \equiv k \, (mod \, n) \, or \, u-v \equiv -k \, mod \, n}$$

However, $$x \equiv y \, (mod \, n) $$ indicates that x and y are congruent modulo n. For example, $$12 \equiv 5 \equiv -2 (mod 7)$$

i need to Classify graph G by all k that is isomorphism to each other for $$G_{16,k} (1<=k<=15) $$ Example {1,2},{3},{4,5,6}my attempt:

$$G_{16,k} (1<=k<=15) $$

Classify graph G by all k that is isomorphism

Example {1,2},{3},{4,5,6}

I see that by definition for example $$G_{5,1}$$

we can see that

$$V={1,2,3,4,5}$$

Since $$ 6 \equiv 1 \equiv -4 \, mod \, 5$$ we know that$$E=\{(1,5),(5,4),(4,3),(3,2),(2,1) \} $$

But how can i classify k that isomorphism to each other for $$G_{16,k}$$

I thought that $$\{2,4,6,8,10,12,14,16 \} $$ and $$\{1,3,5,7,9,11,13,15 \} $$ will be isomorphism to each other, am i right?

i.e $$G_{2,4} \& G_{4,6}$$ will have same vertex and degree(?)