How many Euler diagrams with $n$ sets exist?

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Does anyone have any thoughts on this? I have been struggling with it and I'm not sure if it's a hard problem, or easy and I'm just not getting it?



For $n=2$ sets (say $A$ and $B$), it's obviously 4:



  1. The one where $A cap B = emptyset$

  2. The one in which $A cap B neq emptyset$, but neither set is a subset of the other

  3. The one where $A subset B$

  4. The one where $B subset A$

(I'm assuming $A neq B$ in all cases.)



For $n ge2 $, though it seems rather difficult. I can't think of a systematic way to count them. Originally I thought I could look at all distinct pairs and allow each pair to take on one of three values (representing disjoint, intersecting, and subset). But I've already found a case where two different diagrams can have the same representation this way.



Any ideas? Thanks.










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  • 1




    One possible idea: looking at the second figure at en.wikipedia.org/wiki/Euler_diagram , maybe try to count the number of ways to shade in regions of the $n$-fold Venn diagram?
    – Greg Martin
    Jul 11 '14 at 1:47










  • Exactly which configurations are disallowed? For $n$ sets there are $2^n$ ways in which an element can be contained/not contained in each of the sets. Ignoring the elements in none of the sets we can assign the $2^n-1$ regions to be empty/not empty in $2^2^n-1$ ways, some of which will be disallowed. It will probably be easier to count the disallowed configurations.
    – deinst
    Jul 11 '14 at 13:04










  • Sounds good. Thanks, both of you. Surprisingly, I have a textbook problem in which it asks to sketch all of the "functionally different" configurations for $n=3$. (So for example, the set where $A cap B neq emptyset$ and $C$ is disjoint from the other two would be functionally equivalent to the set where $B cap C neq emptyset$ and $A$ is disjoint). But there will still be too many to sketch I believe and some are difficult to draw. It's just a strange question.
    – user2556975
    Jul 11 '14 at 14:19














up vote
2
down vote

favorite












Does anyone have any thoughts on this? I have been struggling with it and I'm not sure if it's a hard problem, or easy and I'm just not getting it?



For $n=2$ sets (say $A$ and $B$), it's obviously 4:



  1. The one where $A cap B = emptyset$

  2. The one in which $A cap B neq emptyset$, but neither set is a subset of the other

  3. The one where $A subset B$

  4. The one where $B subset A$

(I'm assuming $A neq B$ in all cases.)



For $n ge2 $, though it seems rather difficult. I can't think of a systematic way to count them. Originally I thought I could look at all distinct pairs and allow each pair to take on one of three values (representing disjoint, intersecting, and subset). But I've already found a case where two different diagrams can have the same representation this way.



Any ideas? Thanks.










share|cite|improve this question

















  • 1




    One possible idea: looking at the second figure at en.wikipedia.org/wiki/Euler_diagram , maybe try to count the number of ways to shade in regions of the $n$-fold Venn diagram?
    – Greg Martin
    Jul 11 '14 at 1:47










  • Exactly which configurations are disallowed? For $n$ sets there are $2^n$ ways in which an element can be contained/not contained in each of the sets. Ignoring the elements in none of the sets we can assign the $2^n-1$ regions to be empty/not empty in $2^2^n-1$ ways, some of which will be disallowed. It will probably be easier to count the disallowed configurations.
    – deinst
    Jul 11 '14 at 13:04










  • Sounds good. Thanks, both of you. Surprisingly, I have a textbook problem in which it asks to sketch all of the "functionally different" configurations for $n=3$. (So for example, the set where $A cap B neq emptyset$ and $C$ is disjoint from the other two would be functionally equivalent to the set where $B cap C neq emptyset$ and $A$ is disjoint). But there will still be too many to sketch I believe and some are difficult to draw. It's just a strange question.
    – user2556975
    Jul 11 '14 at 14:19












up vote
2
down vote

favorite









up vote
2
down vote

favorite











Does anyone have any thoughts on this? I have been struggling with it and I'm not sure if it's a hard problem, or easy and I'm just not getting it?



For $n=2$ sets (say $A$ and $B$), it's obviously 4:



  1. The one where $A cap B = emptyset$

  2. The one in which $A cap B neq emptyset$, but neither set is a subset of the other

  3. The one where $A subset B$

  4. The one where $B subset A$

(I'm assuming $A neq B$ in all cases.)



For $n ge2 $, though it seems rather difficult. I can't think of a systematic way to count them. Originally I thought I could look at all distinct pairs and allow each pair to take on one of three values (representing disjoint, intersecting, and subset). But I've already found a case where two different diagrams can have the same representation this way.



Any ideas? Thanks.










share|cite|improve this question













Does anyone have any thoughts on this? I have been struggling with it and I'm not sure if it's a hard problem, or easy and I'm just not getting it?



For $n=2$ sets (say $A$ and $B$), it's obviously 4:



  1. The one where $A cap B = emptyset$

  2. The one in which $A cap B neq emptyset$, but neither set is a subset of the other

  3. The one where $A subset B$

  4. The one where $B subset A$

(I'm assuming $A neq B$ in all cases.)



For $n ge2 $, though it seems rather difficult. I can't think of a systematic way to count them. Originally I thought I could look at all distinct pairs and allow each pair to take on one of three values (representing disjoint, intersecting, and subset). But I've already found a case where two different diagrams can have the same representation this way.



Any ideas? Thanks.







combinatorics elementary-set-theory






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asked Jul 11 '14 at 1:06









user2556975

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111







  • 1




    One possible idea: looking at the second figure at en.wikipedia.org/wiki/Euler_diagram , maybe try to count the number of ways to shade in regions of the $n$-fold Venn diagram?
    – Greg Martin
    Jul 11 '14 at 1:47










  • Exactly which configurations are disallowed? For $n$ sets there are $2^n$ ways in which an element can be contained/not contained in each of the sets. Ignoring the elements in none of the sets we can assign the $2^n-1$ regions to be empty/not empty in $2^2^n-1$ ways, some of which will be disallowed. It will probably be easier to count the disallowed configurations.
    – deinst
    Jul 11 '14 at 13:04










  • Sounds good. Thanks, both of you. Surprisingly, I have a textbook problem in which it asks to sketch all of the "functionally different" configurations for $n=3$. (So for example, the set where $A cap B neq emptyset$ and $C$ is disjoint from the other two would be functionally equivalent to the set where $B cap C neq emptyset$ and $A$ is disjoint). But there will still be too many to sketch I believe and some are difficult to draw. It's just a strange question.
    – user2556975
    Jul 11 '14 at 14:19












  • 1




    One possible idea: looking at the second figure at en.wikipedia.org/wiki/Euler_diagram , maybe try to count the number of ways to shade in regions of the $n$-fold Venn diagram?
    – Greg Martin
    Jul 11 '14 at 1:47










  • Exactly which configurations are disallowed? For $n$ sets there are $2^n$ ways in which an element can be contained/not contained in each of the sets. Ignoring the elements in none of the sets we can assign the $2^n-1$ regions to be empty/not empty in $2^2^n-1$ ways, some of which will be disallowed. It will probably be easier to count the disallowed configurations.
    – deinst
    Jul 11 '14 at 13:04










  • Sounds good. Thanks, both of you. Surprisingly, I have a textbook problem in which it asks to sketch all of the "functionally different" configurations for $n=3$. (So for example, the set where $A cap B neq emptyset$ and $C$ is disjoint from the other two would be functionally equivalent to the set where $B cap C neq emptyset$ and $A$ is disjoint). But there will still be too many to sketch I believe and some are difficult to draw. It's just a strange question.
    – user2556975
    Jul 11 '14 at 14:19







1




1




One possible idea: looking at the second figure at en.wikipedia.org/wiki/Euler_diagram , maybe try to count the number of ways to shade in regions of the $n$-fold Venn diagram?
– Greg Martin
Jul 11 '14 at 1:47




One possible idea: looking at the second figure at en.wikipedia.org/wiki/Euler_diagram , maybe try to count the number of ways to shade in regions of the $n$-fold Venn diagram?
– Greg Martin
Jul 11 '14 at 1:47












Exactly which configurations are disallowed? For $n$ sets there are $2^n$ ways in which an element can be contained/not contained in each of the sets. Ignoring the elements in none of the sets we can assign the $2^n-1$ regions to be empty/not empty in $2^2^n-1$ ways, some of which will be disallowed. It will probably be easier to count the disallowed configurations.
– deinst
Jul 11 '14 at 13:04




Exactly which configurations are disallowed? For $n$ sets there are $2^n$ ways in which an element can be contained/not contained in each of the sets. Ignoring the elements in none of the sets we can assign the $2^n-1$ regions to be empty/not empty in $2^2^n-1$ ways, some of which will be disallowed. It will probably be easier to count the disallowed configurations.
– deinst
Jul 11 '14 at 13:04












Sounds good. Thanks, both of you. Surprisingly, I have a textbook problem in which it asks to sketch all of the "functionally different" configurations for $n=3$. (So for example, the set where $A cap B neq emptyset$ and $C$ is disjoint from the other two would be functionally equivalent to the set where $B cap C neq emptyset$ and $A$ is disjoint). But there will still be too many to sketch I believe and some are difficult to draw. It's just a strange question.
– user2556975
Jul 11 '14 at 14:19




Sounds good. Thanks, both of you. Surprisingly, I have a textbook problem in which it asks to sketch all of the "functionally different" configurations for $n=3$. (So for example, the set where $A cap B neq emptyset$ and $C$ is disjoint from the other two would be functionally equivalent to the set where $B cap C neq emptyset$ and $A$ is disjoint). But there will still be too many to sketch I believe and some are difficult to draw. It's just a strange question.
– user2556975
Jul 11 '14 at 14:19










2 Answers
2






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up vote
1
down vote













In the past month and an half I have spent more hours that I care to admit on this problem.



The original answer is still valid: 2^(2^n - 1) - 1 is an upper boundary on the number of possible Euler diagram. This is sequence A077585 in the OEIS.



A better upper limit is the the number of "covers" of an N set, sequence A003464. This eliminate the cases that reduce to a lower number of "characteristics". For example, the case with two separate regions B and C and no A will be eliminated from the count n=3;



An ever better upper limit is calculated with an iterative "procedure" that tracks how many regions are added when a new characteristic is added. I will upload this sequence to the OEIS, but it is related to A007018.



A lower limit is given by this same procedure, but removing repetitions.
This catches most of the "equivalent" arrangements, but has a number of false positives, as the example below



false positive



A different lower limit is given by the number of possible arrangements of circles in a projective plane, sequence A250001 on OEIS.
This however misses the possibilities of non-well formed Euler diagrams (e.g. with partially overlapping boundaries) like this



overlapping



Moreover the sequence is not defined after n=4, while the iterative-no repetition could be easily extended to n=6.



In summary, this is the table that i came up with



enter image description here






share|cite|improve this answer
















  • 1




    "In the past month and an half I have spent more hours that I care to admit on this problem.", welcome to the world of mathematics.
    – Asaf Karagila♦
    Feb 27 '15 at 5:58

















up vote
1
down vote













I stumbled on the same problem while trying to have an automatic Euler diagram generator and I think the answer of deinst is correct.



Given $n$ “characteristics” there are
$$V(n) = 2^n-1$$ different subsets (i.e. the Venn diagram sets, excluding the "rest of universe").



$$V(2)=3$$ $A setminus B \ B setminus A,\ Acap B$



$$V(3)=7$$ $A setminus (Bcup C)\ B setminus (Acup C)\ Csetminus (Acup B)\ (Acap B) setminus C\ (Acap C) setminus B\ (Bcap C) setminus A\ Acap B cap C$



Then we can assign each of these regions to exist / not exist in our Euler diagram. The number of combinations excluding the empty Universe is
$$E(V) = 2^V - 1 \E(n) = 2^(2^n -1)-1$$



I use a short notation, in which a letter indicates that the zone has only that characteristic. So for 3 characteristics $A triangleq A setminus (B cup C) qquad AB triangleq (Acap B)setminus C$




For n = 2, E= 7
1) A, B=∅
2) B, A=∅
3) AB, A = B ≠ ∅
4) A+B, A∩B=∅
5) A+AB, B⊂A
6) B+AB, A⊂B
7) A+B+AB A∩B≠∅, but neither set is a subset of the other

Of these, the first 3 reduce to cases with only "1" characteristic
and the last 4 are the cases you listed in the question.


Note that for three characteristics the number is 127, so sketching all of them would take quite some time.



Much more difficult is to define when two diagrams are “functionally equivalent” as cases 5 and 6 above.






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    2 Answers
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    2 Answers
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    active

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    up vote
    1
    down vote













    In the past month and an half I have spent more hours that I care to admit on this problem.



    The original answer is still valid: 2^(2^n - 1) - 1 is an upper boundary on the number of possible Euler diagram. This is sequence A077585 in the OEIS.



    A better upper limit is the the number of "covers" of an N set, sequence A003464. This eliminate the cases that reduce to a lower number of "characteristics". For example, the case with two separate regions B and C and no A will be eliminated from the count n=3;



    An ever better upper limit is calculated with an iterative "procedure" that tracks how many regions are added when a new characteristic is added. I will upload this sequence to the OEIS, but it is related to A007018.



    A lower limit is given by this same procedure, but removing repetitions.
    This catches most of the "equivalent" arrangements, but has a number of false positives, as the example below



    false positive



    A different lower limit is given by the number of possible arrangements of circles in a projective plane, sequence A250001 on OEIS.
    This however misses the possibilities of non-well formed Euler diagrams (e.g. with partially overlapping boundaries) like this



    overlapping



    Moreover the sequence is not defined after n=4, while the iterative-no repetition could be easily extended to n=6.



    In summary, this is the table that i came up with



    enter image description here






    share|cite|improve this answer
















    • 1




      "In the past month and an half I have spent more hours that I care to admit on this problem.", welcome to the world of mathematics.
      – Asaf Karagila♦
      Feb 27 '15 at 5:58














    up vote
    1
    down vote













    In the past month and an half I have spent more hours that I care to admit on this problem.



    The original answer is still valid: 2^(2^n - 1) - 1 is an upper boundary on the number of possible Euler diagram. This is sequence A077585 in the OEIS.



    A better upper limit is the the number of "covers" of an N set, sequence A003464. This eliminate the cases that reduce to a lower number of "characteristics". For example, the case with two separate regions B and C and no A will be eliminated from the count n=3;



    An ever better upper limit is calculated with an iterative "procedure" that tracks how many regions are added when a new characteristic is added. I will upload this sequence to the OEIS, but it is related to A007018.



    A lower limit is given by this same procedure, but removing repetitions.
    This catches most of the "equivalent" arrangements, but has a number of false positives, as the example below



    false positive



    A different lower limit is given by the number of possible arrangements of circles in a projective plane, sequence A250001 on OEIS.
    This however misses the possibilities of non-well formed Euler diagrams (e.g. with partially overlapping boundaries) like this



    overlapping



    Moreover the sequence is not defined after n=4, while the iterative-no repetition could be easily extended to n=6.



    In summary, this is the table that i came up with



    enter image description here






    share|cite|improve this answer
















    • 1




      "In the past month and an half I have spent more hours that I care to admit on this problem.", welcome to the world of mathematics.
      – Asaf Karagila♦
      Feb 27 '15 at 5:58












    up vote
    1
    down vote










    up vote
    1
    down vote









    In the past month and an half I have spent more hours that I care to admit on this problem.



    The original answer is still valid: 2^(2^n - 1) - 1 is an upper boundary on the number of possible Euler diagram. This is sequence A077585 in the OEIS.



    A better upper limit is the the number of "covers" of an N set, sequence A003464. This eliminate the cases that reduce to a lower number of "characteristics". For example, the case with two separate regions B and C and no A will be eliminated from the count n=3;



    An ever better upper limit is calculated with an iterative "procedure" that tracks how many regions are added when a new characteristic is added. I will upload this sequence to the OEIS, but it is related to A007018.



    A lower limit is given by this same procedure, but removing repetitions.
    This catches most of the "equivalent" arrangements, but has a number of false positives, as the example below



    false positive



    A different lower limit is given by the number of possible arrangements of circles in a projective plane, sequence A250001 on OEIS.
    This however misses the possibilities of non-well formed Euler diagrams (e.g. with partially overlapping boundaries) like this



    overlapping



    Moreover the sequence is not defined after n=4, while the iterative-no repetition could be easily extended to n=6.



    In summary, this is the table that i came up with



    enter image description here






    share|cite|improve this answer












    In the past month and an half I have spent more hours that I care to admit on this problem.



    The original answer is still valid: 2^(2^n - 1) - 1 is an upper boundary on the number of possible Euler diagram. This is sequence A077585 in the OEIS.



    A better upper limit is the the number of "covers" of an N set, sequence A003464. This eliminate the cases that reduce to a lower number of "characteristics". For example, the case with two separate regions B and C and no A will be eliminated from the count n=3;



    An ever better upper limit is calculated with an iterative "procedure" that tracks how many regions are added when a new characteristic is added. I will upload this sequence to the OEIS, but it is related to A007018.



    A lower limit is given by this same procedure, but removing repetitions.
    This catches most of the "equivalent" arrangements, but has a number of false positives, as the example below



    false positive



    A different lower limit is given by the number of possible arrangements of circles in a projective plane, sequence A250001 on OEIS.
    This however misses the possibilities of non-well formed Euler diagrams (e.g. with partially overlapping boundaries) like this



    overlapping



    Moreover the sequence is not defined after n=4, while the iterative-no repetition could be easily extended to n=6.



    In summary, this is the table that i came up with



    enter image description here







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Feb 27 '15 at 5:01









    Maurizio De Leo

    1237




    1237







    • 1




      "In the past month and an half I have spent more hours that I care to admit on this problem.", welcome to the world of mathematics.
      – Asaf Karagila♦
      Feb 27 '15 at 5:58












    • 1




      "In the past month and an half I have spent more hours that I care to admit on this problem.", welcome to the world of mathematics.
      – Asaf Karagila♦
      Feb 27 '15 at 5:58







    1




    1




    "In the past month and an half I have spent more hours that I care to admit on this problem.", welcome to the world of mathematics.
    – Asaf Karagila♦
    Feb 27 '15 at 5:58




    "In the past month and an half I have spent more hours that I care to admit on this problem.", welcome to the world of mathematics.
    – Asaf Karagila♦
    Feb 27 '15 at 5:58










    up vote
    1
    down vote













    I stumbled on the same problem while trying to have an automatic Euler diagram generator and I think the answer of deinst is correct.



    Given $n$ “characteristics” there are
    $$V(n) = 2^n-1$$ different subsets (i.e. the Venn diagram sets, excluding the "rest of universe").



    $$V(2)=3$$ $A setminus B \ B setminus A,\ Acap B$



    $$V(3)=7$$ $A setminus (Bcup C)\ B setminus (Acup C)\ Csetminus (Acup B)\ (Acap B) setminus C\ (Acap C) setminus B\ (Bcap C) setminus A\ Acap B cap C$



    Then we can assign each of these regions to exist / not exist in our Euler diagram. The number of combinations excluding the empty Universe is
    $$E(V) = 2^V - 1 \E(n) = 2^(2^n -1)-1$$



    I use a short notation, in which a letter indicates that the zone has only that characteristic. So for 3 characteristics $A triangleq A setminus (B cup C) qquad AB triangleq (Acap B)setminus C$




    For n = 2, E= 7
    1) A, B=∅
    2) B, A=∅
    3) AB, A = B ≠ ∅
    4) A+B, A∩B=∅
    5) A+AB, B⊂A
    6) B+AB, A⊂B
    7) A+B+AB A∩B≠∅, but neither set is a subset of the other

    Of these, the first 3 reduce to cases with only "1" characteristic
    and the last 4 are the cases you listed in the question.


    Note that for three characteristics the number is 127, so sketching all of them would take quite some time.



    Much more difficult is to define when two diagrams are “functionally equivalent” as cases 5 and 6 above.






    share|cite|improve this answer


























      up vote
      1
      down vote













      I stumbled on the same problem while trying to have an automatic Euler diagram generator and I think the answer of deinst is correct.



      Given $n$ “characteristics” there are
      $$V(n) = 2^n-1$$ different subsets (i.e. the Venn diagram sets, excluding the "rest of universe").



      $$V(2)=3$$ $A setminus B \ B setminus A,\ Acap B$



      $$V(3)=7$$ $A setminus (Bcup C)\ B setminus (Acup C)\ Csetminus (Acup B)\ (Acap B) setminus C\ (Acap C) setminus B\ (Bcap C) setminus A\ Acap B cap C$



      Then we can assign each of these regions to exist / not exist in our Euler diagram. The number of combinations excluding the empty Universe is
      $$E(V) = 2^V - 1 \E(n) = 2^(2^n -1)-1$$



      I use a short notation, in which a letter indicates that the zone has only that characteristic. So for 3 characteristics $A triangleq A setminus (B cup C) qquad AB triangleq (Acap B)setminus C$




      For n = 2, E= 7
      1) A, B=∅
      2) B, A=∅
      3) AB, A = B ≠ ∅
      4) A+B, A∩B=∅
      5) A+AB, B⊂A
      6) B+AB, A⊂B
      7) A+B+AB A∩B≠∅, but neither set is a subset of the other

      Of these, the first 3 reduce to cases with only "1" characteristic
      and the last 4 are the cases you listed in the question.


      Note that for three characteristics the number is 127, so sketching all of them would take quite some time.



      Much more difficult is to define when two diagrams are “functionally equivalent” as cases 5 and 6 above.






      share|cite|improve this answer
























        up vote
        1
        down vote










        up vote
        1
        down vote









        I stumbled on the same problem while trying to have an automatic Euler diagram generator and I think the answer of deinst is correct.



        Given $n$ “characteristics” there are
        $$V(n) = 2^n-1$$ different subsets (i.e. the Venn diagram sets, excluding the "rest of universe").



        $$V(2)=3$$ $A setminus B \ B setminus A,\ Acap B$



        $$V(3)=7$$ $A setminus (Bcup C)\ B setminus (Acup C)\ Csetminus (Acup B)\ (Acap B) setminus C\ (Acap C) setminus B\ (Bcap C) setminus A\ Acap B cap C$



        Then we can assign each of these regions to exist / not exist in our Euler diagram. The number of combinations excluding the empty Universe is
        $$E(V) = 2^V - 1 \E(n) = 2^(2^n -1)-1$$



        I use a short notation, in which a letter indicates that the zone has only that characteristic. So for 3 characteristics $A triangleq A setminus (B cup C) qquad AB triangleq (Acap B)setminus C$




        For n = 2, E= 7
        1) A, B=∅
        2) B, A=∅
        3) AB, A = B ≠ ∅
        4) A+B, A∩B=∅
        5) A+AB, B⊂A
        6) B+AB, A⊂B
        7) A+B+AB A∩B≠∅, but neither set is a subset of the other

        Of these, the first 3 reduce to cases with only "1" characteristic
        and the last 4 are the cases you listed in the question.


        Note that for three characteristics the number is 127, so sketching all of them would take quite some time.



        Much more difficult is to define when two diagrams are “functionally equivalent” as cases 5 and 6 above.






        share|cite|improve this answer














        I stumbled on the same problem while trying to have an automatic Euler diagram generator and I think the answer of deinst is correct.



        Given $n$ “characteristics” there are
        $$V(n) = 2^n-1$$ different subsets (i.e. the Venn diagram sets, excluding the "rest of universe").



        $$V(2)=3$$ $A setminus B \ B setminus A,\ Acap B$



        $$V(3)=7$$ $A setminus (Bcup C)\ B setminus (Acup C)\ Csetminus (Acup B)\ (Acap B) setminus C\ (Acap C) setminus B\ (Bcap C) setminus A\ Acap B cap C$



        Then we can assign each of these regions to exist / not exist in our Euler diagram. The number of combinations excluding the empty Universe is
        $$E(V) = 2^V - 1 \E(n) = 2^(2^n -1)-1$$



        I use a short notation, in which a letter indicates that the zone has only that characteristic. So for 3 characteristics $A triangleq A setminus (B cup C) qquad AB triangleq (Acap B)setminus C$




        For n = 2, E= 7
        1) A, B=∅
        2) B, A=∅
        3) AB, A = B ≠ ∅
        4) A+B, A∩B=∅
        5) A+AB, B⊂A
        6) B+AB, A⊂B
        7) A+B+AB A∩B≠∅, but neither set is a subset of the other

        Of these, the first 3 reduce to cases with only "1" characteristic
        and the last 4 are the cases you listed in the question.


        Note that for three characteristics the number is 127, so sketching all of them would take quite some time.



        Much more difficult is to define when two diagrams are “functionally equivalent” as cases 5 and 6 above.







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        edited Sep 10 at 10:32

























        answered Jan 12 '15 at 6:18









        Maurizio De Leo

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