Find the binary input function given the outputs (part 2)












0












$begingroup$


Here we have three binary variables $x_1$, $x_2$, $x_3$ $in {0,1}$.

I want to find the form of the function $f(x_1, x_2, x_3)$ such that the following are satisfied:



if $ x_1 = 0, x_2 = 0, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 6$

if $ x_1 = 0, x_2 = 0, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 4$

if $ x_1 = 0, x_2 = 1, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 5$

if $ x_1 = 0, x_2 = 1, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 2$

if $ x_1 = 1, x_2 = 0, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 5$

if $ x_1 = 1, x_2 = 0, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 3$

if $ x_1 = 1, x_2 = 1, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 4$

if $ x_1 = 1, x_2 = 1, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 0$



I imagine it like having a "virtual sum" which starts at 1 and it is increased by 1 at each step in the sequence. Every time I see a zero, this virtual sum "becomes real" and is reset. For instance:





  • $ x_1 = 1, x_2 = 1, x_3 = 0 $ After seeing the first one the virtual sum is 2. After the second one the virtual sum is 3. Finally there is a zero, so the virtual sum becomes 4, "becomes real" and is reset. The final result is 4.


  • $ x_1 = 0, x_2 = 1, x_3 = 0 $ After seeing the first zero the sum is 2 and the virtual sum is reset to 1. After the second one the virtual sum is 2. Finally there is a zero, so the virtual sum is incremented at 3 and "becomes real". The final result is 2 + 3 = 5.


  • $ x_1 = 1, x_2 = 1, x_3 = 1 $ We have three ones, so the virtual sum is 4 at the end of the sequence. However, since there are no zeroes it does not "become real" and the result is 0.


  • $ x_1 = 0, x_2 = 0, x_3 = 0 $ We have three zeroes, so we get 2 at each step and the sum is 6.


I had already asked a similar question (Find the binary input function given the outputs), which was (brilliantly) solved. This is a generalisation in which the "virtual sum" starts from one rather then zero and that, with respect to the previous question, inverts the binary value of the variables $x_1$, $x_2$, $x_3$.






combinatorics binary binary-operations







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Here we have three binary variables $x_1$, $x_2$, $x_3$ $in {0,1}$.

    I want to find the form of the function $f(x_1, x_2, x_3)$ such that the following are satisfied:



    if $ x_1 = 0, x_2 = 0, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 6$

    if $ x_1 = 0, x_2 = 0, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 4$

    if $ x_1 = 0, x_2 = 1, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 5$

    if $ x_1 = 0, x_2 = 1, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 2$

    if $ x_1 = 1, x_2 = 0, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 5$

    if $ x_1 = 1, x_2 = 0, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 3$

    if $ x_1 = 1, x_2 = 1, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 4$

    if $ x_1 = 1, x_2 = 1, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 0$



    I imagine it like having a "virtual sum" which starts at 1 and it is increased by 1 at each step in the sequence. Every time I see a zero, this virtual sum "becomes real" and is reset. For instance:





    • $ x_1 = 1, x_2 = 1, x_3 = 0 $ After seeing the first one the virtual sum is 2. After the second one the virtual sum is 3. Finally there is a zero, so the virtual sum becomes 4, "becomes real" and is reset. The final result is 4.


    • $ x_1 = 0, x_2 = 1, x_3 = 0 $ After seeing the first zero the sum is 2 and the virtual sum is reset to 1. After the second one the virtual sum is 2. Finally there is a zero, so the virtual sum is incremented at 3 and "becomes real". The final result is 2 + 3 = 5.


    • $ x_1 = 1, x_2 = 1, x_3 = 1 $ We have three ones, so the virtual sum is 4 at the end of the sequence. However, since there are no zeroes it does not "become real" and the result is 0.


    • $ x_1 = 0, x_2 = 0, x_3 = 0 $ We have three zeroes, so we get 2 at each step and the sum is 6.


    I had already asked a similar question (Find the binary input function given the outputs), which was (brilliantly) solved. This is a generalisation in which the "virtual sum" starts from one rather then zero and that, with respect to the previous question, inverts the binary value of the variables $x_1$, $x_2$, $x_3$.






    combinatorics binary binary-operations







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Here we have three binary variables $x_1$, $x_2$, $x_3$ $in {0,1}$.

      I want to find the form of the function $f(x_1, x_2, x_3)$ such that the following are satisfied:



      if $ x_1 = 0, x_2 = 0, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 6$

      if $ x_1 = 0, x_2 = 0, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 4$

      if $ x_1 = 0, x_2 = 1, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 5$

      if $ x_1 = 0, x_2 = 1, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 2$

      if $ x_1 = 1, x_2 = 0, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 5$

      if $ x_1 = 1, x_2 = 0, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 3$

      if $ x_1 = 1, x_2 = 1, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 4$

      if $ x_1 = 1, x_2 = 1, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 0$



      I imagine it like having a "virtual sum" which starts at 1 and it is increased by 1 at each step in the sequence. Every time I see a zero, this virtual sum "becomes real" and is reset. For instance:





      • $ x_1 = 1, x_2 = 1, x_3 = 0 $ After seeing the first one the virtual sum is 2. After the second one the virtual sum is 3. Finally there is a zero, so the virtual sum becomes 4, "becomes real" and is reset. The final result is 4.


      • $ x_1 = 0, x_2 = 1, x_3 = 0 $ After seeing the first zero the sum is 2 and the virtual sum is reset to 1. After the second one the virtual sum is 2. Finally there is a zero, so the virtual sum is incremented at 3 and "becomes real". The final result is 2 + 3 = 5.


      • $ x_1 = 1, x_2 = 1, x_3 = 1 $ We have three ones, so the virtual sum is 4 at the end of the sequence. However, since there are no zeroes it does not "become real" and the result is 0.


      • $ x_1 = 0, x_2 = 0, x_3 = 0 $ We have three zeroes, so we get 2 at each step and the sum is 6.


      I had already asked a similar question (Find the binary input function given the outputs), which was (brilliantly) solved. This is a generalisation in which the "virtual sum" starts from one rather then zero and that, with respect to the previous question, inverts the binary value of the variables $x_1$, $x_2$, $x_3$.






      combinatorics binary binary-operations







      share|cite|improve this question









      $endgroup$




      Here we have three binary variables $x_1$, $x_2$, $x_3$ $in {0,1}$.

      I want to find the form of the function $f(x_1, x_2, x_3)$ such that the following are satisfied:



      if $ x_1 = 0, x_2 = 0, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 6$

      if $ x_1 = 0, x_2 = 0, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 4$

      if $ x_1 = 0, x_2 = 1, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 5$

      if $ x_1 = 0, x_2 = 1, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 2$

      if $ x_1 = 1, x_2 = 0, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 5$

      if $ x_1 = 1, x_2 = 0, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 3$

      if $ x_1 = 1, x_2 = 1, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 4$

      if $ x_1 = 1, x_2 = 1, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 0$



      I imagine it like having a "virtual sum" which starts at 1 and it is increased by 1 at each step in the sequence. Every time I see a zero, this virtual sum "becomes real" and is reset. For instance:





      • $ x_1 = 1, x_2 = 1, x_3 = 0 $ After seeing the first one the virtual sum is 2. After the second one the virtual sum is 3. Finally there is a zero, so the virtual sum becomes 4, "becomes real" and is reset. The final result is 4.


      • $ x_1 = 0, x_2 = 1, x_3 = 0 $ After seeing the first zero the sum is 2 and the virtual sum is reset to 1. After the second one the virtual sum is 2. Finally there is a zero, so the virtual sum is incremented at 3 and "becomes real". The final result is 2 + 3 = 5.


      • $ x_1 = 1, x_2 = 1, x_3 = 1 $ We have three ones, so the virtual sum is 4 at the end of the sequence. However, since there are no zeroes it does not "become real" and the result is 0.


      • $ x_1 = 0, x_2 = 0, x_3 = 0 $ We have three zeroes, so we get 2 at each step and the sum is 6.


      I had already asked a similar question (Find the binary input function given the outputs), which was (brilliantly) solved. This is a generalisation in which the "virtual sum" starts from one rather then zero and that, with respect to the previous question, inverts the binary value of the variables $x_1$, $x_2$, $x_3$.






      combinatorics binary binary-operations




      combinatorics binary binary-operations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 16 at 15:58









      aprosperoaprospero

      103




      103






















          1 Answer
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          0












          $begingroup$

          How about (completed) $$f(x_1,x_2,x_3) = 6cdot (1-x_1)(1-x_2)(1-x_3) + 4cdot (1-x_1)(1-x_2)x_3 \+ 5cdot (1-x_1)x_2(1-x_3) + 2cdot (1-x_1)x_2x_3
          +5cdot x_1(1-x_2)(1-x_3) + 3cdot x_1(1-x_2)x_3 + 4cdot x_1x_2(1-x_3) + 0cdot x_1x_2x_3.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Is the complete formula $f(x_1,x_2,x_3) = 6(1-x_1)(1-x_2)(1-x_3) + 4(1-x_1)(1-x_2)x_3 + 2(1-x_1)x_2x_3$? Because it doesn't work for many inputs. Try (0,1,0) for example.
            $endgroup$
            – aprospero
            Jan 16 at 16:26










          • $begingroup$
            No I guess you got the idea. You need to complete it.
            $endgroup$
            – Wuestenfux
            Jan 16 at 16:30










          • $begingroup$
            Oooh! I got the point. So it is enough to hardcode the output sequence I want and then simplify the expression. Wow, did not come into my mind! Also I can easily generalise at 4, 5 variables and search for a pattern. Thank you @Wuestenfux, really helpful!
            $endgroup$
            – aprospero
            Jan 16 at 17:06











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          1 Answer
          1






          active

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          active

          oldest

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          active

          oldest

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          0












          $begingroup$

          How about (completed) $$f(x_1,x_2,x_3) = 6cdot (1-x_1)(1-x_2)(1-x_3) + 4cdot (1-x_1)(1-x_2)x_3 \+ 5cdot (1-x_1)x_2(1-x_3) + 2cdot (1-x_1)x_2x_3
          +5cdot x_1(1-x_2)(1-x_3) + 3cdot x_1(1-x_2)x_3 + 4cdot x_1x_2(1-x_3) + 0cdot x_1x_2x_3.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Is the complete formula $f(x_1,x_2,x_3) = 6(1-x_1)(1-x_2)(1-x_3) + 4(1-x_1)(1-x_2)x_3 + 2(1-x_1)x_2x_3$? Because it doesn't work for many inputs. Try (0,1,0) for example.
            $endgroup$
            – aprospero
            Jan 16 at 16:26










          • $begingroup$
            No I guess you got the idea. You need to complete it.
            $endgroup$
            – Wuestenfux
            Jan 16 at 16:30










          • $begingroup$
            Oooh! I got the point. So it is enough to hardcode the output sequence I want and then simplify the expression. Wow, did not come into my mind! Also I can easily generalise at 4, 5 variables and search for a pattern. Thank you @Wuestenfux, really helpful!
            $endgroup$
            – aprospero
            Jan 16 at 17:06
















          0












          $begingroup$

          How about (completed) $$f(x_1,x_2,x_3) = 6cdot (1-x_1)(1-x_2)(1-x_3) + 4cdot (1-x_1)(1-x_2)x_3 \+ 5cdot (1-x_1)x_2(1-x_3) + 2cdot (1-x_1)x_2x_3
          +5cdot x_1(1-x_2)(1-x_3) + 3cdot x_1(1-x_2)x_3 + 4cdot x_1x_2(1-x_3) + 0cdot x_1x_2x_3.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Is the complete formula $f(x_1,x_2,x_3) = 6(1-x_1)(1-x_2)(1-x_3) + 4(1-x_1)(1-x_2)x_3 + 2(1-x_1)x_2x_3$? Because it doesn't work for many inputs. Try (0,1,0) for example.
            $endgroup$
            – aprospero
            Jan 16 at 16:26










          • $begingroup$
            No I guess you got the idea. You need to complete it.
            $endgroup$
            – Wuestenfux
            Jan 16 at 16:30










          • $begingroup$
            Oooh! I got the point. So it is enough to hardcode the output sequence I want and then simplify the expression. Wow, did not come into my mind! Also I can easily generalise at 4, 5 variables and search for a pattern. Thank you @Wuestenfux, really helpful!
            $endgroup$
            – aprospero
            Jan 16 at 17:06














          0












          0








          0





          $begingroup$

          How about (completed) $$f(x_1,x_2,x_3) = 6cdot (1-x_1)(1-x_2)(1-x_3) + 4cdot (1-x_1)(1-x_2)x_3 \+ 5cdot (1-x_1)x_2(1-x_3) + 2cdot (1-x_1)x_2x_3
          +5cdot x_1(1-x_2)(1-x_3) + 3cdot x_1(1-x_2)x_3 + 4cdot x_1x_2(1-x_3) + 0cdot x_1x_2x_3.$$






          share|cite|improve this answer











          $endgroup$



          How about (completed) $$f(x_1,x_2,x_3) = 6cdot (1-x_1)(1-x_2)(1-x_3) + 4cdot (1-x_1)(1-x_2)x_3 \+ 5cdot (1-x_1)x_2(1-x_3) + 2cdot (1-x_1)x_2x_3
          +5cdot x_1(1-x_2)(1-x_3) + 3cdot x_1(1-x_2)x_3 + 4cdot x_1x_2(1-x_3) + 0cdot x_1x_2x_3.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 16 at 16:34

























          answered Jan 16 at 16:01









          WuestenfuxWuestenfux

          4,5141413




          4,5141413












          • $begingroup$
            Is the complete formula $f(x_1,x_2,x_3) = 6(1-x_1)(1-x_2)(1-x_3) + 4(1-x_1)(1-x_2)x_3 + 2(1-x_1)x_2x_3$? Because it doesn't work for many inputs. Try (0,1,0) for example.
            $endgroup$
            – aprospero
            Jan 16 at 16:26










          • $begingroup$
            No I guess you got the idea. You need to complete it.
            $endgroup$
            – Wuestenfux
            Jan 16 at 16:30










          • $begingroup$
            Oooh! I got the point. So it is enough to hardcode the output sequence I want and then simplify the expression. Wow, did not come into my mind! Also I can easily generalise at 4, 5 variables and search for a pattern. Thank you @Wuestenfux, really helpful!
            $endgroup$
            – aprospero
            Jan 16 at 17:06


















          • $begingroup$
            Is the complete formula $f(x_1,x_2,x_3) = 6(1-x_1)(1-x_2)(1-x_3) + 4(1-x_1)(1-x_2)x_3 + 2(1-x_1)x_2x_3$? Because it doesn't work for many inputs. Try (0,1,0) for example.
            $endgroup$
            – aprospero
            Jan 16 at 16:26










          • $begingroup$
            No I guess you got the idea. You need to complete it.
            $endgroup$
            – Wuestenfux
            Jan 16 at 16:30










          • $begingroup$
            Oooh! I got the point. So it is enough to hardcode the output sequence I want and then simplify the expression. Wow, did not come into my mind! Also I can easily generalise at 4, 5 variables and search for a pattern. Thank you @Wuestenfux, really helpful!
            $endgroup$
            – aprospero
            Jan 16 at 17:06
















          $begingroup$
          Is the complete formula $f(x_1,x_2,x_3) = 6(1-x_1)(1-x_2)(1-x_3) + 4(1-x_1)(1-x_2)x_3 + 2(1-x_1)x_2x_3$? Because it doesn't work for many inputs. Try (0,1,0) for example.
          $endgroup$
          – aprospero
          Jan 16 at 16:26




          $begingroup$
          Is the complete formula $f(x_1,x_2,x_3) = 6(1-x_1)(1-x_2)(1-x_3) + 4(1-x_1)(1-x_2)x_3 + 2(1-x_1)x_2x_3$? Because it doesn't work for many inputs. Try (0,1,0) for example.
          $endgroup$
          – aprospero
          Jan 16 at 16:26












          $begingroup$
          No I guess you got the idea. You need to complete it.
          $endgroup$
          – Wuestenfux
          Jan 16 at 16:30




          $begingroup$
          No I guess you got the idea. You need to complete it.
          $endgroup$
          – Wuestenfux
          Jan 16 at 16:30












          $begingroup$
          Oooh! I got the point. So it is enough to hardcode the output sequence I want and then simplify the expression. Wow, did not come into my mind! Also I can easily generalise at 4, 5 variables and search for a pattern. Thank you @Wuestenfux, really helpful!
          $endgroup$
          – aprospero
          Jan 16 at 17:06




          $begingroup$
          Oooh! I got the point. So it is enough to hardcode the output sequence I want and then simplify the expression. Wow, did not come into my mind! Also I can easily generalise at 4, 5 variables and search for a pattern. Thank you @Wuestenfux, really helpful!
          $endgroup$
          – aprospero
          Jan 16 at 17:06


















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          What does “Dominus providebit” mean?

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          5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
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