Wrestling with substitution rules

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1
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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























  • You mean: "One to rule them all"? ;)
    – Henrik Schumacher
    1 hour ago











  • 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
    46 mins ago






  • 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
    23 mins ago














up vote
1
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























  • You mean: "One to rule them all"? ;)
    – Henrik Schumacher
    1 hour ago











  • 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
    46 mins ago






  • 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
    23 mins ago












up vote
1
down vote

favorite









up vote
1
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 1 hour ago









Henrik Schumacher

43.9k263129




43.9k263129










asked 1 hour ago









dantopa

46928




46928











  • You mean: "One to rule them all"? ;)
    – Henrik Schumacher
    1 hour ago











  • 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
    46 mins ago






  • 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
    23 mins ago
















  • You mean: "One to rule them all"? ;)
    – Henrik Schumacher
    1 hour ago











  • 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
    46 mins ago






  • 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
    23 mins ago















You mean: "One to rule them all"? ;)
– Henrik Schumacher
1 hour ago





You mean: "One to rule them all"? ;)
– Henrik Schumacher
1 hour ago













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
46 mins ago




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
46 mins ago




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
23 mins ago




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
23 mins ago










2 Answers
2






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up vote
3
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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
    2
    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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      active

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      2 Answers
      2






      active

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      active

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      active

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      up vote
      3
      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
        3
        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
          3
          down vote










          up vote
          3
          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 1 hour ago









          Henrik Schumacher

          43.9k263129




          43.9k263129




















              up vote
              2
              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
                2
                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
                  2
                  down vote










                  up vote
                  2
                  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 43 mins ago









                  kglr

                  169k8192396




                  169k8192396



























                       

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