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DGALLEG7
Posts: 13
Posted 11:08 May 14, 2020 |

How do you pick the Maximal and Minimal subset from a set?

 

Basically how do you find the AS bar?????

Do we find all possible subsets of participants first????

 

 

JackStrauss
Posts: 236
Posted 12:27 May 14, 2020 |
DGALLEG7 wrote:

How do you pick the Maximal and Minimal subset from a set?

 

Basically how do you find the AS bar?????

Do we find all possible subsets of participants first????

 

 

I'm currently working on this one, it is tricky. I'll try to post an update later.

 

 

DGALLEG7
Posts: 13
Posted 12:32 May 14, 2020 |

thank you

Last edited by DGALLEG7 at 12:33 May 14, 2020.
lren2
Posts: 5
Posted 12:38 May 14, 2020 |

AS bar is the list of sets NOT within all combination of sets. List out all possible combinations of the set and eliminate all the of sets in AS. for example

{p1,p2,p3,p4} all sets are 

AS all = {p1,p2}, {p1, p3}, {p1, p4}, {p2, p4}, {p2, p3}, {p3, p4}, {p1,p2, p3}, {p1, p2, p4}, {p1,p3,p4}, {p2, p3, p4} 

sets in AS is {{p1,p2,p4}, {p1,p3,p4},{p2,p3}}

so you take out {p1, p2, p4} and {p1, p3, p4} and {p2, p3}. You would also take out ANY sets that have a subset, which means you take out {p1, p2, p3} because {p2,p3} is in AS. Then you are left with just:

{p1,p2}, {p1,p3}, {p1,p4}, {p2, p4}, {p3, p4}

Which means AS bar is 5 because there are 5 more sets left. 

 

Im not sure about maximal or minimal. If anyone can answer that, would be appreciated.

JackStrauss
Posts: 236
Posted 12:41 May 14, 2020 |
lren2 wrote:

AS bar is the list of sets NOT within all combination of sets. List out all possible combinations of the set and eliminate all the of sets in AS. for example

{p1,p2,p3,p4} all sets are 

AS all = {p1,p2}, {p1, p3}, {p1, p4}, {p2, p4}, {p2, p3}, {p3, p4}, {p1,p2, p3}, {p1, p2, p4}, {p1,p3,p4}, {p2, p3, p4} 

sets in AS is {{p1,p2,p4}, {p1,p3,p4},{p2,p3}}

so you take out {p1, p2, p4} and {p1, p3, p4} and {p2, p3}. You would also take out ANY sets that have a subset, which means you take out {p1, p2, p3} because {p2,p3} is in AS. Then you are left with just:

{p1,p2}, {p1,p3}, {p1,p4}, {p2, p4}, {p3, p4}

Which means AS bar is 5 because there are 5 more sets left. 

 

Im not sure about maximal or minimal. If anyone can answer that, would be appreciated.

 

Thanks!

DGALLEG7
Posts: 13
Posted 12:43 May 14, 2020 |

Thank you 

vtoriz
Posts: 2
Posted 12:51 May 14, 2020 |

Do we find all possible subsets of participants first???? yes

Basically how do you find the AS bar?????

AS bar are all the combinations that are not in AS and also the combinations that do not contain an AS combination as a subset.

ex. {p1,p2,p4} is a possible combination but it contains the subset {p2,p4} which is part of AS, therefore you eliminate {p1,p2,p4} as a combination of AS bar.

 

 

JackStrauss
Posts: 236
Posted 13:05 May 14, 2020 |
vtoriz wrote:

Do we find all possible subsets of participants first???? yes

Basically how do you find the AS bar?????

AS bar are all the combinations that are not in AS and also the combinations that do not contain an AS combination as a subset.

ex. {p1,p2,p4} is a possible combination but it contains the subset {p2,p4} which is part of AS, therefore you eliminate {p1,p2,p4} as a combination of AS bar.

 

 

Thank you, V.!

DGALLEG7
Posts: 13
Posted 13:43 May 14, 2020 |

thank you