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How to best multiply by 3 a natural number given in binary, using combinatorial logic? How to divide by 3? wasn't a good fit with StackExchange, starting with essentially asking two questions, if obviously related ones.
I split off How to divide a natural number by 3 using combinatorial logic, dividend and quotient represent in binary? which got closed with a prompt to Add details and clarify the problem you’re solving., staying that way subsequent to at least one re-evaluation.

I could easily see how the question is broad, as in could be answered by an entire book, or has many valid answers (one of the old "close choices").

With no comment cluing me in:
Following a sketch of the common division by repeated subtraction, is there anything unclear about

With both dividend and quotient in binary representation, does division by the constant 3 allow a combinatorial logic circuit significantly simpler or faster?

such that it cannot be answered? What detail/type of detail would make a difference between can be and should not be answered as-is?

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  • \$\begingroup\$ (entire book, or has many valid answers not considering it that broad - diverse valid answers and chapter in a book or article in a periodical - or (Q&)A on SE?) \$\endgroup\$
    – greybeard
    Mar 2 at 11:49
  • \$\begingroup\$ (If I was positive the question should not be closed - or one of my edits resolved original close reason(s) (which?) -, I would have flagged it for re-evaluation.) \$\endgroup\$
    – greybeard
    Mar 2 at 11:55
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    \$\begingroup\$ For what it's worth, I have a difficult time reading your posts because of some unusual choices in wording. For example, the main question you quoted really needs the words "to be" inserted between circuit and significantly: "...does X allow a combinatorial logic circuit to be significantly simpler or faster?" I had to read it a couple of times to figure out what was meant. Stuff like that causes me to often just skip or down-vote posts. \$\endgroup\$
    – JYelton
    Mar 4 at 16:00
  • \$\begingroup\$ (@JYelton I am thinking that is there, as it happens. Both seem to work-?) \$\endgroup\$
    – greybeard
    Mar 4 at 17:44
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    \$\begingroup\$ I think even your title is confusing. Consider this: How to multiply by 3 a natural number given in binary, using combinatorial logic?... does it actually mean this... How to multiply a natural binary number by 3 using combinatorial logic <-- it took me a couple of minutes to figure that out and I'm still not sure I caught the essence. So, like @JYelton I tend to bypass questions that don't appear to make a lot of sense. I mean, you might be dyslexic or something related and I apologize if my words might seem picky but, you have to think about things from the reader's point of view. \$\endgroup\$
    – Andy aka
    Mar 9 at 15:35
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    \$\begingroup\$ In your 2nd question, you says this: For general divisors, both restoring(non-performing/subtracting(/assigning)) and non-restoring division can be implemented in gates... and that really hurt my head so, did you just mean this: general division can be implemented using gates...? I mean you have an inclusive set of opposite conditions that appear to bring nothing to the party in your version and that's the bit that made it hard for me. \$\endgroup\$
    – Andy aka
    Mar 9 at 15:43
  • \$\begingroup\$ did you just mean this: general division can be implemented using gates No. I used general divisors where I didn't think non-special, unrestricted 2nd operands of division better. \$\endgroup\$
    – greybeard
    Mar 9 at 15:59
  • \$\begingroup\$ @Andyaka [the question has] an inclusive set of opposite conditions You've lost me there. Please single out one pair of opposite (contradicting?) conditions. \$\endgroup\$
    – greybeard
    Mar 9 at 16:08
  • \$\begingroup\$ both restoring(non-performing/subtracting(/assigning)) and non-restoring <-- seems to say the solution must be valid for \$X\$ (non-restoring) and \$\overline{X}\$ (restoring) i.e. it is an unnecessary requirement (logically) @greybeard <-- of course I may have misinterpreted entirely what you meant. \$\endgroup\$
    – Andy aka
    Mar 9 at 17:41
  • \$\begingroup\$ @Andyaka Both variant A and variant B can be implemented… - I see no solution must be valid for X and not X. (And that's still in the "context part" preceding the question.) \$\endgroup\$
    – greybeard
    Mar 9 at 17:45
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    \$\begingroup\$ That's the problem. It's how others read your words. It seems to say you want Both restoring and non-restoring and that means logical 1 \$\endgroup\$
    – Andy aka
    Mar 9 at 17:46

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I think the topic of the question is on topic on this site: it is about - "a specific electronics design problem" https://electronics.stackexchange.com/help/on-topic

It has a specific question and clearly states the problem so that is good. It was closed for not having enough documentation before edits, with no diagram it would be hard for a user to understand. The edits make it clearer and I think it's acceptable.

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