Wrestling with substitution rules











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2
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Summary



I am struggling with substitution rules.



Example



Here are several cases which are problematic:



Clear[a,α];

{a + 2 b + 1, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2} /. {a + b -> α}


Actual result:



{1 + a + 2 b, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2}


Desired result:



{1 + α +  b, -α, 2 α, α^2}


Question



Currently, the rule is permuted for every case, e.g.



2 a + 2 b -> 2α


Can the alpha substitution rule be generalized That, is there a single rule to handle all cases?






replacement rule







share|improve this question




















  • 2




    You mean: "One to rule them all"? ;)
    – Henrik Schumacher
    Nov 4 at 18:38








  • 1




    Use PolynomialReduce to obtain algebraic "substitutions". In[208]:= PolynomialReduce[{a + 2 b + 1, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2}, a + b - alpha, {a, b}][[All, 2]] Out[208]= {1 + alpha + b, -alpha, 2 alpha, alpha^2}
    – Daniel Lichtblau
    Nov 4 at 19:14






  • 1




    Generally, to apply a relationship broadly, write the corresponding rule such that the LHS of the rule is as simple as possible, e.g., solution posted by @HenrikSchumacher. Since rules are applied to the structure of the internal (FullForm) representation, this will result in the highest number of matches with the LHS.
    – Bob Hanlon
    Nov 4 at 19:38















up vote
2
down vote

favorite












Summary



I am struggling with substitution rules.



Example



Here are several cases which are problematic:



Clear[a,α];

{a + 2 b + 1, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2} /. {a + b -> α}


Actual result:



{1 + a + 2 b, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2}


Desired result:



{1 + α +  b, -α, 2 α, α^2}


Question



Currently, the rule is permuted for every case, e.g.



2 a + 2 b -> 2α


Can the alpha substitution rule be generalized That, is there a single rule to handle all cases?






replacement rule







share|improve this question




















  • 2




    You mean: "One to rule them all"? ;)
    – Henrik Schumacher
    Nov 4 at 18:38








  • 1




    Use PolynomialReduce to obtain algebraic "substitutions". In[208]:= PolynomialReduce[{a + 2 b + 1, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2}, a + b - alpha, {a, b}][[All, 2]] Out[208]= {1 + alpha + b, -alpha, 2 alpha, alpha^2}
    – Daniel Lichtblau
    Nov 4 at 19:14






  • 1




    Generally, to apply a relationship broadly, write the corresponding rule such that the LHS of the rule is as simple as possible, e.g., solution posted by @HenrikSchumacher. Since rules are applied to the structure of the internal (FullForm) representation, this will result in the highest number of matches with the LHS.
    – Bob Hanlon
    Nov 4 at 19:38













up vote
2
down vote

favorite









up vote
2
down vote

favorite











Summary



I am struggling with substitution rules.



Example



Here are several cases which are problematic:



Clear[a,α];

{a + 2 b + 1, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2} /. {a + b -> α}


Actual result:



{1 + a + 2 b, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2}


Desired result:



{1 + α +  b, -α, 2 α, α^2}


Question



Currently, the rule is permuted for every case, e.g.



2 a + 2 b -> 2α


Can the alpha substitution rule be generalized That, is there a single rule to handle all cases?






replacement rule







share|improve this question















Summary



I am struggling with substitution rules.



Example



Here are several cases which are problematic:



Clear[a,α];

{a + 2 b + 1, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2} /. {a + b -> α}


Actual result:



{1 + a + 2 b, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2}


Desired result:



{1 + α +  b, -α, 2 α, α^2}


Question



Currently, the rule is permuted for every case, e.g.



2 a + 2 b -> 2α


Can the alpha substitution rule be generalized That, is there a single rule to handle all cases?






replacement rule




replacement rule






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 4 at 18:35









Henrik Schumacher

44.1k263129




44.1k263129










asked Nov 4 at 18:34









dantopa

47428




47428








  • 2




    You mean: "One to rule them all"? ;)
    – Henrik Schumacher
    Nov 4 at 18:38








  • 1




    Use PolynomialReduce to obtain algebraic "substitutions". In[208]:= PolynomialReduce[{a + 2 b + 1, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2}, a + b - alpha, {a, b}][[All, 2]] Out[208]= {1 + alpha + b, -alpha, 2 alpha, alpha^2}
    – Daniel Lichtblau
    Nov 4 at 19:14






  • 1




    Generally, to apply a relationship broadly, write the corresponding rule such that the LHS of the rule is as simple as possible, e.g., solution posted by @HenrikSchumacher. Since rules are applied to the structure of the internal (FullForm) representation, this will result in the highest number of matches with the LHS.
    – Bob Hanlon
    Nov 4 at 19:38














  • 2




    You mean: "One to rule them all"? ;)
    – Henrik Schumacher
    Nov 4 at 18:38








  • 1




    Use PolynomialReduce to obtain algebraic "substitutions". In[208]:= PolynomialReduce[{a + 2 b + 1, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2}, a + b - alpha, {a, b}][[All, 2]] Out[208]= {1 + alpha + b, -alpha, 2 alpha, alpha^2}
    – Daniel Lichtblau
    Nov 4 at 19:14






  • 1




    Generally, to apply a relationship broadly, write the corresponding rule such that the LHS of the rule is as simple as possible, e.g., solution posted by @HenrikSchumacher. Since rules are applied to the structure of the internal (FullForm) representation, this will result in the highest number of matches with the LHS.
    – Bob Hanlon
    Nov 4 at 19:38








2




2




You mean: "One to rule them all"? ;)
– Henrik Schumacher
Nov 4 at 18:38






You mean: "One to rule them all"? ;)
– Henrik Schumacher
Nov 4 at 18:38






1




1




Use PolynomialReduce to obtain algebraic "substitutions". In[208]:= PolynomialReduce[{a + 2 b + 1, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2}, a + b - alpha, {a, b}][[All, 2]] Out[208]= {1 + alpha + b, -alpha, 2 alpha, alpha^2}
– Daniel Lichtblau
Nov 4 at 19:14




Use PolynomialReduce to obtain algebraic "substitutions". In[208]:= PolynomialReduce[{a + 2 b + 1, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2}, a + b - alpha, {a, b}][[All, 2]] Out[208]= {1 + alpha + b, -alpha, 2 alpha, alpha^2}
– Daniel Lichtblau
Nov 4 at 19:14




1




1




Generally, to apply a relationship broadly, write the corresponding rule such that the LHS of the rule is as simple as possible, e.g., solution posted by @HenrikSchumacher. Since rules are applied to the structure of the internal (FullForm) representation, this will result in the highest number of matches with the LHS.
– Bob Hanlon
Nov 4 at 19:38




Generally, to apply a relationship broadly, write the corresponding rule such that the LHS of the rule is as simple as possible, e.g., solution posted by @HenrikSchumacher. Since rules are applied to the structure of the internal (FullForm) representation, this will result in the highest number of matches with the LHS.
– Bob Hanlon
Nov 4 at 19:38










2 Answers
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6
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{a + 2 b + 1, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2} /. {a -> α - b} // Simplify



{1 + b + α, -α, 2 α, α^2}







share|improve this answer




























    up vote
    5
    down vote













    Also



    Simplify[{a + 2 b + 1, -a - b, 2 a + 2 b,  a^2 + 2 a b + b^2}, {a + b == α}]



    {1 + b + α, -α, 2 α, α^2}







    share|improve this answer





















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      2 Answers
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      2 Answers
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      up vote
      6
      down vote













      {a + 2 b + 1, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2} /. {a -> α - b} // Simplify



      {1 + b + α, -α, 2 α, α^2}







      share|improve this answer

























        up vote
        6
        down vote













        {a + 2 b + 1, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2} /. {a -> α - b} // Simplify



        {1 + b + α, -α, 2 α, α^2}







        share|improve this answer























          up vote
          6
          down vote










          up vote
          6
          down vote









          {a + 2 b + 1, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2} /. {a -> α - b} // Simplify



          {1 + b + α, -α, 2 α, α^2}







          share|improve this answer












          {a + 2 b + 1, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2} /. {a -> α - b} // Simplify



          {1 + b + α, -α, 2 α, α^2}








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 4 at 18:37









          Henrik Schumacher

          44.1k263129




          44.1k263129






















              up vote
              5
              down vote













              Also



              Simplify[{a + 2 b + 1, -a - b, 2 a + 2 b,  a^2 + 2 a b + b^2}, {a + b == α}]



              {1 + b + α, -α, 2 α, α^2}







              share|improve this answer

























                up vote
                5
                down vote













                Also



                Simplify[{a + 2 b + 1, -a - b, 2 a + 2 b,  a^2 + 2 a b + b^2}, {a + b == α}]



                {1 + b + α, -α, 2 α, α^2}







                share|improve this answer























                  up vote
                  5
                  down vote










                  up vote
                  5
                  down vote









                  Also



                  Simplify[{a + 2 b + 1, -a - b, 2 a + 2 b,  a^2 + 2 a b + b^2}, {a + b == α}]



                  {1 + b + α, -α, 2 α, α^2}







                  share|improve this answer












                  Also



                  Simplify[{a + 2 b + 1, -a - b, 2 a + 2 b,  a^2 + 2 a b + b^2}, {a + b == α}]



                  {1 + b + α, -α, 2 α, α^2}








                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 4 at 19:18









                  kglr

                  170k8193396




                  170k8193396






























                       

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