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I have the following homework (AI-related): Which boolean function does the following TLU (threshold logic unit) implement? The TLU (threshold logic unit) has five inputs. Its weight vector is (1.1, 3.1, -1, -2, 0.5), and threshold is 1.

So in other words:

  • There are 5 "inputs" that can be either 1 or 0.
  • Inputs are multiplied by corresponding "weights" (i.e. input number 1 is multiplied by 1.1 etc.)
  • after multiplications, results are summed up and that sum needs to be more than 1.

So for example, one function to implement this would be when inputs 1 and 2 are true and inputs 3, 4 and 5 are 0.

I know how to create a truth table and look up the entire set of boolean functions but is there another, more "intelligent" way of finding the function than "brute force" manner I'm using?

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Yes, this one is easy enough to analyze without a 'brute force' truth table.

Let's just call the inputs $1$, $2$, $3$, $4$, $5$.

First, it is easy to see that the value of $5$ will never make a difference as to whether the threshold gets passed or not, so $5$ is a "don't care": we can completely ignore it in our analysis and hence it need not show up in our boolean expression.

Second, focusing just on $1$, $2$, $3$, $4$, you can quickly observe that the threshold will only be passed if at least one of $1$ or $2$ are 'on'.

Now, if $1$ is on but $2$ is not, then the threshold will be passed as long as neither $3$ nor $4$ are on. This gives us the term $1 \land \neg 2 \land \neg 3 \land \neg 4$. But actually, it's ok if $2$ would be on as well. So, this term can be simplified to just $1 \land \neg 3 \land \neg 4$

If $2$ is on but $1$ is not, then the threshold is passed as long as not both $3$ and $4$ are on. This gives us $\neg 1 \land 2 \land \neg (3 \land 4)$. And again, since it's actually ok for $1$ to be on, we get $2 \land \neg (3 \land 4)$

Finally, if both $1$ and $2$ are on, then the threshold will be passed no matter what. So: $1 \land 2$

Since these are the only 3 situations in which the threshold will be passed, we thus get:

$$(1 \land \neg 3 \land \neg 4) \lor (2 \land \neg (3 \land 4)) \lor (1 \land 2)$$

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  • $\begingroup$ You're genius! Thanks. $\endgroup$ – Marso Feb 23 '18 at 4:52
  • $\begingroup$ @Marso You're welcome! :) $\endgroup$ – Bram28 Feb 23 '18 at 17:09

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