What is Graph Isomorphism and Graph Invariant? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Isomorphism of Complete GraphsGraph isomorphism and existence of nontrivial automorphismsWhat's the difference between the automorphism and isomorphism of graph?Vertex degree and graph isomorphismisomorphism of graph definitionNon-isomorphic graphs with four total vertices, arranged by sizeIsomorphism with Directional GraphsUnderstanding graph automorphism as opposed to isomorphism.Graph Theory | Euler's formula and Graph IsomorphismGraph isomorphism representative for small graphs

Did the UK government pay "millions and millions of dollars" to try to snag Julian Assange?

Mortgage adviser recommends a longer term than necessary combined with overpayments

Circular reasoning in L'Hopital's rule

How to support a colleague who finds meetings extremely tiring?

Python - Fishing Simulator

One-dimensional Japanese puzzle

Match Roman Numerals

The following signatures were invalid: EXPKEYSIG 1397BC53640DB551

How to politely respond to generic emails requesting a PhD/job in my lab? Without wasting too much time

60's-70's movie: home appliances revolting against the owners

Why not take a picture of a closer black hole?

"... to apply for a visa" or "... and applied for a visa"?

how can a perfect fourth interval be considered either consonant or dissonant?

What do I do when my TA workload is more than expected?

Why did Peik Lin say, "I'm not an animal"?

should truth entail possible truth

Can the DM override racial traits?

Working through the single responsibility principle (SRP) in Python when calls are expensive

Do working physicists consider Newtonian mechanics to be "falsified"?

How to determine omitted units in a publication

Why doesn't a hydraulic lever violate conservation of energy?

Would an alien lifeform be able to achieve space travel if lacking in vision?

Word to describe a time interval

Are there continuous functions who are the same in an interval but differ in at least one other point?



What is Graph Isomorphism and Graph Invariant?



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Isomorphism of Complete GraphsGraph isomorphism and existence of nontrivial automorphismsWhat's the difference between the automorphism and isomorphism of graph?Vertex degree and graph isomorphismisomorphism of graph definitionNon-isomorphic graphs with four total vertices, arranged by sizeIsomorphism with Directional GraphsUnderstanding graph automorphism as opposed to isomorphism.Graph Theory | Euler's formula and Graph IsomorphismGraph isomorphism representative for small graphs










2












$begingroup$


While I was reading Reinhard Diestel text on graphs, I came across this paragraph.




Let G = (V, E) and G' = (V' , E' ) be two graphs.
We call G and G' isomorphic,and write G $simeq $ G' , if there exists a bijection φ: V → V with xy ∈ E $ leftrightarrow $ φ(x)φ(y) ∈ E for all x, y ∈ V . Such a map φ is called
an isomorphism; if G = G' , it is called an automorphism.



We do not normally distinguish between isomorphic graphs. Thus, we usually write
G = G' rather than G $ simeq $ G , speak of the complete graph on 17 vertices,
and so on.



A map taking graphs as arguments is called a graph invariant
if it assigns equal values to isomorphic graphs. The number of vertices
and the number of edges of a graph are two simple graph invariants; the
greatest number of pairwise adjacent vertices is another.




What does this paragraph mean and what are graph invariants, in simple explanations.










share|cite|improve this question









$endgroup$











  • $begingroup$
    A graph invariant is a map on graphs that acts the same on two isomorphic graphs. They define it there. What about it confuses you?
    $endgroup$
    – davidlowryduda
    Jun 22 '15 at 15:20











  • $begingroup$
    The mathematical defination of isomorphism, given above.
    $endgroup$
    – Ankit Panda
    Jun 22 '15 at 15:21










  • $begingroup$
    examples of graph invariants are things like max-degree (an isomorphism preserves max degree), chromatic number, number of vertices, number of edges, length of longest path, diameter, edge density, number of connected components, and so forth.
    $endgroup$
    – TravisJ
    Jun 22 '15 at 20:00















2












$begingroup$


While I was reading Reinhard Diestel text on graphs, I came across this paragraph.




Let G = (V, E) and G' = (V' , E' ) be two graphs.
We call G and G' isomorphic,and write G $simeq $ G' , if there exists a bijection φ: V → V with xy ∈ E $ leftrightarrow $ φ(x)φ(y) ∈ E for all x, y ∈ V . Such a map φ is called
an isomorphism; if G = G' , it is called an automorphism.



We do not normally distinguish between isomorphic graphs. Thus, we usually write
G = G' rather than G $ simeq $ G , speak of the complete graph on 17 vertices,
and so on.



A map taking graphs as arguments is called a graph invariant
if it assigns equal values to isomorphic graphs. The number of vertices
and the number of edges of a graph are two simple graph invariants; the
greatest number of pairwise adjacent vertices is another.




What does this paragraph mean and what are graph invariants, in simple explanations.










share|cite|improve this question









$endgroup$











  • $begingroup$
    A graph invariant is a map on graphs that acts the same on two isomorphic graphs. They define it there. What about it confuses you?
    $endgroup$
    – davidlowryduda
    Jun 22 '15 at 15:20











  • $begingroup$
    The mathematical defination of isomorphism, given above.
    $endgroup$
    – Ankit Panda
    Jun 22 '15 at 15:21










  • $begingroup$
    examples of graph invariants are things like max-degree (an isomorphism preserves max degree), chromatic number, number of vertices, number of edges, length of longest path, diameter, edge density, number of connected components, and so forth.
    $endgroup$
    – TravisJ
    Jun 22 '15 at 20:00













2












2








2





$begingroup$


While I was reading Reinhard Diestel text on graphs, I came across this paragraph.




Let G = (V, E) and G' = (V' , E' ) be two graphs.
We call G and G' isomorphic,and write G $simeq $ G' , if there exists a bijection φ: V → V with xy ∈ E $ leftrightarrow $ φ(x)φ(y) ∈ E for all x, y ∈ V . Such a map φ is called
an isomorphism; if G = G' , it is called an automorphism.



We do not normally distinguish between isomorphic graphs. Thus, we usually write
G = G' rather than G $ simeq $ G , speak of the complete graph on 17 vertices,
and so on.



A map taking graphs as arguments is called a graph invariant
if it assigns equal values to isomorphic graphs. The number of vertices
and the number of edges of a graph are two simple graph invariants; the
greatest number of pairwise adjacent vertices is another.




What does this paragraph mean and what are graph invariants, in simple explanations.










share|cite|improve this question









$endgroup$




While I was reading Reinhard Diestel text on graphs, I came across this paragraph.




Let G = (V, E) and G' = (V' , E' ) be two graphs.
We call G and G' isomorphic,and write G $simeq $ G' , if there exists a bijection φ: V → V with xy ∈ E $ leftrightarrow $ φ(x)φ(y) ∈ E for all x, y ∈ V . Such a map φ is called
an isomorphism; if G = G' , it is called an automorphism.



We do not normally distinguish between isomorphic graphs. Thus, we usually write
G = G' rather than G $ simeq $ G , speak of the complete graph on 17 vertices,
and so on.



A map taking graphs as arguments is called a graph invariant
if it assigns equal values to isomorphic graphs. The number of vertices
and the number of edges of a graph are two simple graph invariants; the
greatest number of pairwise adjacent vertices is another.




What does this paragraph mean and what are graph invariants, in simple explanations.







graph-theory graph-isomorphism






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jun 22 '15 at 15:17









Ankit PandaAnkit Panda

88110




88110











  • $begingroup$
    A graph invariant is a map on graphs that acts the same on two isomorphic graphs. They define it there. What about it confuses you?
    $endgroup$
    – davidlowryduda
    Jun 22 '15 at 15:20











  • $begingroup$
    The mathematical defination of isomorphism, given above.
    $endgroup$
    – Ankit Panda
    Jun 22 '15 at 15:21










  • $begingroup$
    examples of graph invariants are things like max-degree (an isomorphism preserves max degree), chromatic number, number of vertices, number of edges, length of longest path, diameter, edge density, number of connected components, and so forth.
    $endgroup$
    – TravisJ
    Jun 22 '15 at 20:00
















  • $begingroup$
    A graph invariant is a map on graphs that acts the same on two isomorphic graphs. They define it there. What about it confuses you?
    $endgroup$
    – davidlowryduda
    Jun 22 '15 at 15:20











  • $begingroup$
    The mathematical defination of isomorphism, given above.
    $endgroup$
    – Ankit Panda
    Jun 22 '15 at 15:21










  • $begingroup$
    examples of graph invariants are things like max-degree (an isomorphism preserves max degree), chromatic number, number of vertices, number of edges, length of longest path, diameter, edge density, number of connected components, and so forth.
    $endgroup$
    – TravisJ
    Jun 22 '15 at 20:00















$begingroup$
A graph invariant is a map on graphs that acts the same on two isomorphic graphs. They define it there. What about it confuses you?
$endgroup$
– davidlowryduda
Jun 22 '15 at 15:20





$begingroup$
A graph invariant is a map on graphs that acts the same on two isomorphic graphs. They define it there. What about it confuses you?
$endgroup$
– davidlowryduda
Jun 22 '15 at 15:20













$begingroup$
The mathematical defination of isomorphism, given above.
$endgroup$
– Ankit Panda
Jun 22 '15 at 15:21




$begingroup$
The mathematical defination of isomorphism, given above.
$endgroup$
– Ankit Panda
Jun 22 '15 at 15:21












$begingroup$
examples of graph invariants are things like max-degree (an isomorphism preserves max degree), chromatic number, number of vertices, number of edges, length of longest path, diameter, edge density, number of connected components, and so forth.
$endgroup$
– TravisJ
Jun 22 '15 at 20:00




$begingroup$
examples of graph invariants are things like max-degree (an isomorphism preserves max degree), chromatic number, number of vertices, number of edges, length of longest path, diameter, edge density, number of connected components, and so forth.
$endgroup$
– TravisJ
Jun 22 '15 at 20:00










1 Answer
1






active

oldest

votes


















3












$begingroup$

In simple terms, two graphs are isomorphic to each other so long as there is a bijection between the two vertex sets and such that the bijection preserves edges. That is, two graphs $G$ and $G'$ are isomorphic if there exists a bijection, $phi: V(G) to V(G')$, and if that bijection also preserves edges. Explicitly, this means that if $uv$ is an edge in $E(G)$, then $phi (u) phi (v)$ is an edge in $E(G')$.



One can visualize a graph isomorphism in two ways.
1. Start with a graph and move around vertices in what ever way you want while keeping all the edges in tact. The result, a graph that looks different, but isn't really.
2. A re-labeling of the vertices of $G$. This produces a graph that looks the same, but the vertices are called something else.



Automorphisms are a little bit trickier. The way I look at an automorphism is a moving around of vertices (while keeping edges in tact, as in (1) above) with the caveat that the graph must look exactly the same as before. For example, take $C_4$ with vertex set $V(G) = u_1,u_2,u_3,u_4$ and edge set $E(G) = u_1u_2,u_2u_3,u_3u_4,u_4u_1$ and consider the following bijection.

1. $phi: V(G) to V(G')$ where $phi$ is the permuation $(u_1u_2u_3u_4)$ that sends $u_1 to u_2$ $u_2 to u_3$ $u_3 to u_4$, and $u_4 to u_1$. (If one is unfamiliar with cycle notation). This is an isomorphism since every edge is preserved, and indeed it is also an automorphism since the resulting graph looks exactly the same as the regular graph. (the permuation above is really just a rotation by 90 degrees if you lay out the original as a "square", as is tradition.






share|cite|improve this answer









$endgroup$













    Your Answer








    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "69"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader:
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    ,
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );













    draft saved

    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1335035%2fwhat-is-graph-isomorphism-and-graph-invariant%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    In simple terms, two graphs are isomorphic to each other so long as there is a bijection between the two vertex sets and such that the bijection preserves edges. That is, two graphs $G$ and $G'$ are isomorphic if there exists a bijection, $phi: V(G) to V(G')$, and if that bijection also preserves edges. Explicitly, this means that if $uv$ is an edge in $E(G)$, then $phi (u) phi (v)$ is an edge in $E(G')$.



    One can visualize a graph isomorphism in two ways.
    1. Start with a graph and move around vertices in what ever way you want while keeping all the edges in tact. The result, a graph that looks different, but isn't really.
    2. A re-labeling of the vertices of $G$. This produces a graph that looks the same, but the vertices are called something else.



    Automorphisms are a little bit trickier. The way I look at an automorphism is a moving around of vertices (while keeping edges in tact, as in (1) above) with the caveat that the graph must look exactly the same as before. For example, take $C_4$ with vertex set $V(G) = u_1,u_2,u_3,u_4$ and edge set $E(G) = u_1u_2,u_2u_3,u_3u_4,u_4u_1$ and consider the following bijection.

    1. $phi: V(G) to V(G')$ where $phi$ is the permuation $(u_1u_2u_3u_4)$ that sends $u_1 to u_2$ $u_2 to u_3$ $u_3 to u_4$, and $u_4 to u_1$. (If one is unfamiliar with cycle notation). This is an isomorphism since every edge is preserved, and indeed it is also an automorphism since the resulting graph looks exactly the same as the regular graph. (the permuation above is really just a rotation by 90 degrees if you lay out the original as a "square", as is tradition.






    share|cite|improve this answer









    $endgroup$

















      3












      $begingroup$

      In simple terms, two graphs are isomorphic to each other so long as there is a bijection between the two vertex sets and such that the bijection preserves edges. That is, two graphs $G$ and $G'$ are isomorphic if there exists a bijection, $phi: V(G) to V(G')$, and if that bijection also preserves edges. Explicitly, this means that if $uv$ is an edge in $E(G)$, then $phi (u) phi (v)$ is an edge in $E(G')$.



      One can visualize a graph isomorphism in two ways.
      1. Start with a graph and move around vertices in what ever way you want while keeping all the edges in tact. The result, a graph that looks different, but isn't really.
      2. A re-labeling of the vertices of $G$. This produces a graph that looks the same, but the vertices are called something else.



      Automorphisms are a little bit trickier. The way I look at an automorphism is a moving around of vertices (while keeping edges in tact, as in (1) above) with the caveat that the graph must look exactly the same as before. For example, take $C_4$ with vertex set $V(G) = u_1,u_2,u_3,u_4$ and edge set $E(G) = u_1u_2,u_2u_3,u_3u_4,u_4u_1$ and consider the following bijection.

      1. $phi: V(G) to V(G')$ where $phi$ is the permuation $(u_1u_2u_3u_4)$ that sends $u_1 to u_2$ $u_2 to u_3$ $u_3 to u_4$, and $u_4 to u_1$. (If one is unfamiliar with cycle notation). This is an isomorphism since every edge is preserved, and indeed it is also an automorphism since the resulting graph looks exactly the same as the regular graph. (the permuation above is really just a rotation by 90 degrees if you lay out the original as a "square", as is tradition.






      share|cite|improve this answer









      $endgroup$















        3












        3








        3





        $begingroup$

        In simple terms, two graphs are isomorphic to each other so long as there is a bijection between the two vertex sets and such that the bijection preserves edges. That is, two graphs $G$ and $G'$ are isomorphic if there exists a bijection, $phi: V(G) to V(G')$, and if that bijection also preserves edges. Explicitly, this means that if $uv$ is an edge in $E(G)$, then $phi (u) phi (v)$ is an edge in $E(G')$.



        One can visualize a graph isomorphism in two ways.
        1. Start with a graph and move around vertices in what ever way you want while keeping all the edges in tact. The result, a graph that looks different, but isn't really.
        2. A re-labeling of the vertices of $G$. This produces a graph that looks the same, but the vertices are called something else.



        Automorphisms are a little bit trickier. The way I look at an automorphism is a moving around of vertices (while keeping edges in tact, as in (1) above) with the caveat that the graph must look exactly the same as before. For example, take $C_4$ with vertex set $V(G) = u_1,u_2,u_3,u_4$ and edge set $E(G) = u_1u_2,u_2u_3,u_3u_4,u_4u_1$ and consider the following bijection.

        1. $phi: V(G) to V(G')$ where $phi$ is the permuation $(u_1u_2u_3u_4)$ that sends $u_1 to u_2$ $u_2 to u_3$ $u_3 to u_4$, and $u_4 to u_1$. (If one is unfamiliar with cycle notation). This is an isomorphism since every edge is preserved, and indeed it is also an automorphism since the resulting graph looks exactly the same as the regular graph. (the permuation above is really just a rotation by 90 degrees if you lay out the original as a "square", as is tradition.






        share|cite|improve this answer









        $endgroup$



        In simple terms, two graphs are isomorphic to each other so long as there is a bijection between the two vertex sets and such that the bijection preserves edges. That is, two graphs $G$ and $G'$ are isomorphic if there exists a bijection, $phi: V(G) to V(G')$, and if that bijection also preserves edges. Explicitly, this means that if $uv$ is an edge in $E(G)$, then $phi (u) phi (v)$ is an edge in $E(G')$.



        One can visualize a graph isomorphism in two ways.
        1. Start with a graph and move around vertices in what ever way you want while keeping all the edges in tact. The result, a graph that looks different, but isn't really.
        2. A re-labeling of the vertices of $G$. This produces a graph that looks the same, but the vertices are called something else.



        Automorphisms are a little bit trickier. The way I look at an automorphism is a moving around of vertices (while keeping edges in tact, as in (1) above) with the caveat that the graph must look exactly the same as before. For example, take $C_4$ with vertex set $V(G) = u_1,u_2,u_3,u_4$ and edge set $E(G) = u_1u_2,u_2u_3,u_3u_4,u_4u_1$ and consider the following bijection.

        1. $phi: V(G) to V(G')$ where $phi$ is the permuation $(u_1u_2u_3u_4)$ that sends $u_1 to u_2$ $u_2 to u_3$ $u_3 to u_4$, and $u_4 to u_1$. (If one is unfamiliar with cycle notation). This is an isomorphism since every edge is preserved, and indeed it is also an automorphism since the resulting graph looks exactly the same as the regular graph. (the permuation above is really just a rotation by 90 degrees if you lay out the original as a "square", as is tradition.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jun 23 '15 at 16:33









        Paddling GhostPaddling Ghost

        1,479614




        1,479614



























            draft saved

            draft discarded
















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid


            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.

            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1335035%2fwhat-is-graph-isomorphism-and-graph-invariant%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Bosc Connection Yimello Approaching Angry The produce zaps the market. 구성 기록되다 변경...

            WordPress Information needed

            Hidroelektrana Sadržaj Povijest | Podjela hidroelektrana | Snaga dobivena u hidroelektranama | Dijelovi hidroelektrane | Uloga hidroelektrana u suvremenom svijetu | Prednosti hidroelektrana | Nedostaci hidroelektrana | Države s najvećom proizvodnjom hidro-električne energije | Deset najvećih hidroelektrana u svijetu | Hidroelektrane u Hrvatskoj | Izvori | Poveznice | Vanjske poveznice | Navigacijski izbornikTechnical Report, Version 2Zajedničkom poslužiteljuHidroelektranaHEP Proizvodnja d.o.o. - Hidroelektrane u Hrvatskoj