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

          5,3631513




          5,3631513












          • $begingroup$
            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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          張江高科駅

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          張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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