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$\begin{align*} &S = \left \{ G_i \;\; | \; G_i \in \text{ lebeled trees with 4 vertices} \right \} \\ &\text{Relation } \quad R = \left \{ {\color{red}{\left ( G_i,G_j \right )}} \; | G_i,G_j \in S \;\; \text{and} \;\; G_i,G_j \;\; \text{are} \;\; \text{isomorphic to each other} \right \} \end{align*}$

No of equivalent classes of $R$ ?
asked in Combinatory by Veteran (56.9k points) 36 194 500
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Considering that the equivalence class of an element a is the subset of equivalence relation S of all elements related to a. then i think the answer should be 1 + 6 + 15 + 20 + 15 + 6 + 1 = 64 ?
@debashish, is it 14?
oh sorry! i thought binary trees.
And i thought it to be a graph. @anusha what's ur method for binary trees ?
now am getting 3 as answer :P .
@gatesjt.. yeah i will let u know if this answer is correct :D

I think it is $2$. though not fully sure.

There are $16$ ($\large2^{4-2}$) labeled trees with $4$ vertices.

 

  • The first $12$ are in one class of isomorphism
  • And the last $4$ are other class of isomorphism
oh yes correct! :) i did a mistake
How come you are comparing trees? For isomorphism, the labels must also be same, right?
Sir is isomorphism is checked only in terms of their structure? Shouldn't it be checked for adjacent colors also?
@debashish,i thought the answer is only 1 as we can get tree with nonly 3 edges
so,you are saying that first 12 are in one class of isomorphism,

whya bove 12 not isomorphic to bottom 4??

is it because there degrees are different..right??
Yes first see no. Of vertices and edges, then check for degree sequence then check for same cycle length. 2 equivalence classes. Btw good question.

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