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I'm learning fuzzy logic and more or less understand the basic concept, but i'm having a hard time understanding how to apply it to a method. I tried browsing online for explanation on how to use it, but only found some implementation and test case using the basic form of 4 rules and 3 variables, and 2 rules per variable. Anyway this is an example case, i will use Tsukamoto method.

In this case actually i have 6 rules and 3 variables with 3 rules ver variable, but i will only explain 1 of the variables because i think the rest will have the same solution. I have 3 variables one of them is "size", the range is for small it's 0-2 and for large it's 7-10. The current condition is size = 6.5. The rules is as follow(simplified to only use this variable):

  • [R1] size = small
  • [R2] size = medium
  • [R3] size = large

What i want to know is:

  • how do i define the formula for medium(the middle rule if the case is different)?
  • what if the rule is more than 3 (i.e. small, medium, large, ex-large)?

What i understands if the rule is only 2 i can use this formula

  • small[x]=(max-x)/(max-min)
  • large[x]=(x-min)/(max-min)

My current approach to this problem is as follow:

small[x]=1; x<=2

medium[x]=(max-x)/(max-min); 2 < x < 7

large[x]=0; x>=7

Is this correct? Also can you refer me to some source to study this? As i mentioned before i can only find some implementation and basic explanation, it's either there is no online source for this or i don't know what to search for. Sorry if it's hard to understand i can edit and post the whole problem if you want, thanks in advance.

Extra question: what is the name of algorithm which can be used to solve the crossing bridge puzzle(the one with timer, max person, and stuff)?i forgot the name.

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    $\begingroup$ Is the question about understanding the Tsukamoto method? The method applies to n multivariate rules, but first, it is not clear how you applied the method here. $\endgroup$ Jun 4, 2018 at 14:29
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    $\begingroup$ No it's the membership graph and function that I don't understand. I am confused at what function/formula should i use and what values should i add to the function if the value of a variable is a range in above example large = 7-10. I know a bit to proceed after that, only the basic though but I want to understand this one first. $\endgroup$
    – Dizz
    Jun 4, 2018 at 15:27

2 Answers 2

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After i did a bit more study on this, i found this solution.

Membership Graph

μsmall[x]

  • 1; -> x≤0

  • (20-x)/(20-0); -> 0

  • 0; -> x≥20

μmedium[x]

  • 0; -> x≤20 or x≥70

  • (x-20)/(45-20); -> 20

  • (70-x)/(70-45); -> 45

μlarge[x]

  • 0; -> x≤70
  • (x-70)/(100-70); -> 70
  • 1; -> x≥100

For the final result i need to add 3 more rules for the function to work. So it become 9 rules and 3 variables, with 3 rules for each input variables.

To answer both of my question:

  1. Add a triangle graph and use the formula.

  2. Add a triangle graph for each rules.

What i learn from this is the total rules needed to determine the output is the number of input variables rules multiplied(i.e. v1 = 3 rules, v2 3 rules, total rules = 3x3=9 rules). And i'm not really sure about this one but you can combine multiple kind of graphs.

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  • $\begingroup$ I know this is a bad practice because it's basically asking clarification for my answer. But because i'm still new at this and not really sure about my answer, i will mark my answer as accepted after someone can verify it. $\endgroup$
    – Dizz
    Jun 6, 2018 at 7:36
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I think maybe you are overthinking it, it's been a while since I have done fuzzy logic but the answer you have given seems to be correct from what I can tell. At university when I have done this in the past you only usually give the answer you have given in that format exactly then just do a brief explanation. I can't add this as a comment due to a low score here as I am new, but hopefully, my personal experience helps somewhat here, if not I can provide more sources of information to help clarify, but honestly I do find the trick with this is to not overthink it.

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