Manipulating a general length function












5












$begingroup$


As a minimal example, define a function like s[1,2,3]; this is a function of three arguments in this case, but I'd like to apply this general idea to functions like s[3,4,5,6,7] which don't necessarily start at 1.



From an identity from theory, I need to re-express this as as sum of 2-argument functions that cycle over all of the possible combinations of the original function, but don't duplicate. Explicitly, this would be the same as writing



s[1,2,3]=f[1,2]+f[1,3]+f[2,3],



or in the longer more general case (and to illustrate my point)



s[3,4,5,6,7] = f[3,4]+f[3,5]+f[3,6]+f[3,7]+f[4,5]+f[4,6]+f[4,7]+f[5,6]+f[5,7]+f[6,7]


I was previously using the definition with simple replacement rules for each of the these chains, with general cases of s[i_,j_,k_] and such similar rules. However, this isn't optimal in larger problems, where I have many thousands of these results, and leads to remarkably long evaluation times.



To see my functional but slow solution, I have attached a partial screenshot here, but naturally, I think there must be a better method. I was considering using Partitions, as in this SE post or BlankSpaces, but I struggle to implement more complex uses of these. Any guidance or other ideas would be gratefully appreciated. Also, a title suggestion would be great as well!






list-manipulation performance-tuning pattern-matching







share|improve this question









$endgroup$












  • $begingroup$
    Look into Subsets[listOfArguments, {2}]
    $endgroup$
    – MarcoB
    8 hours ago
















5












$begingroup$


As a minimal example, define a function like s[1,2,3]; this is a function of three arguments in this case, but I'd like to apply this general idea to functions like s[3,4,5,6,7] which don't necessarily start at 1.



From an identity from theory, I need to re-express this as as sum of 2-argument functions that cycle over all of the possible combinations of the original function, but don't duplicate. Explicitly, this would be the same as writing



s[1,2,3]=f[1,2]+f[1,3]+f[2,3],



or in the longer more general case (and to illustrate my point)



s[3,4,5,6,7] = f[3,4]+f[3,5]+f[3,6]+f[3,7]+f[4,5]+f[4,6]+f[4,7]+f[5,6]+f[5,7]+f[6,7]


I was previously using the definition with simple replacement rules for each of the these chains, with general cases of s[i_,j_,k_] and such similar rules. However, this isn't optimal in larger problems, where I have many thousands of these results, and leads to remarkably long evaluation times.



To see my functional but slow solution, I have attached a partial screenshot here, but naturally, I think there must be a better method. I was considering using Partitions, as in this SE post or BlankSpaces, but I struggle to implement more complex uses of these. Any guidance or other ideas would be gratefully appreciated. Also, a title suggestion would be great as well!






list-manipulation performance-tuning pattern-matching







share|improve this question









$endgroup$












  • $begingroup$
    Look into Subsets[listOfArguments, {2}]
    $endgroup$
    – MarcoB
    8 hours ago














5












5








5





$begingroup$


As a minimal example, define a function like s[1,2,3]; this is a function of three arguments in this case, but I'd like to apply this general idea to functions like s[3,4,5,6,7] which don't necessarily start at 1.



From an identity from theory, I need to re-express this as as sum of 2-argument functions that cycle over all of the possible combinations of the original function, but don't duplicate. Explicitly, this would be the same as writing



s[1,2,3]=f[1,2]+f[1,3]+f[2,3],



or in the longer more general case (and to illustrate my point)



s[3,4,5,6,7] = f[3,4]+f[3,5]+f[3,6]+f[3,7]+f[4,5]+f[4,6]+f[4,7]+f[5,6]+f[5,7]+f[6,7]


I was previously using the definition with simple replacement rules for each of the these chains, with general cases of s[i_,j_,k_] and such similar rules. However, this isn't optimal in larger problems, where I have many thousands of these results, and leads to remarkably long evaluation times.



To see my functional but slow solution, I have attached a partial screenshot here, but naturally, I think there must be a better method. I was considering using Partitions, as in this SE post or BlankSpaces, but I struggle to implement more complex uses of these. Any guidance or other ideas would be gratefully appreciated. Also, a title suggestion would be great as well!






list-manipulation performance-tuning pattern-matching







share|improve this question









$endgroup$




As a minimal example, define a function like s[1,2,3]; this is a function of three arguments in this case, but I'd like to apply this general idea to functions like s[3,4,5,6,7] which don't necessarily start at 1.



From an identity from theory, I need to re-express this as as sum of 2-argument functions that cycle over all of the possible combinations of the original function, but don't duplicate. Explicitly, this would be the same as writing



s[1,2,3]=f[1,2]+f[1,3]+f[2,3],



or in the longer more general case (and to illustrate my point)



s[3,4,5,6,7] = f[3,4]+f[3,5]+f[3,6]+f[3,7]+f[4,5]+f[4,6]+f[4,7]+f[5,6]+f[5,7]+f[6,7]


I was previously using the definition with simple replacement rules for each of the these chains, with general cases of s[i_,j_,k_] and such similar rules. However, this isn't optimal in larger problems, where I have many thousands of these results, and leads to remarkably long evaluation times.



To see my functional but slow solution, I have attached a partial screenshot here, but naturally, I think there must be a better method. I was considering using Partitions, as in this SE post or BlankSpaces, but I struggle to implement more complex uses of these. Any guidance or other ideas would be gratefully appreciated. Also, a title suggestion would be great as well!






list-manipulation performance-tuning pattern-matching




list-manipulation performance-tuning pattern-matching






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 8 hours ago









BradBrad

768




768












  • $begingroup$
    Look into Subsets[listOfArguments, {2}]
    $endgroup$
    – MarcoB
    8 hours ago


















  • $begingroup$
    Look into Subsets[listOfArguments, {2}]
    $endgroup$
    – MarcoB
    8 hours ago
















$begingroup$
Look into Subsets[listOfArguments, {2}]
$endgroup$
– MarcoB
8 hours ago




$begingroup$
Look into Subsets[listOfArguments, {2}]
$endgroup$
– MarcoB
8 hours ago










2 Answers
2






active

oldest

votes


















7












$begingroup$

Something like the following?



ClearAll[s]

s[seq__] := Total[f@@@Subsets[List[seq], {2}]]





share|improve this answer









$endgroup$













  • $begingroup$
    Hi @MarcoB - Subsets was indeed the function I was looking for. I was thinking perhaps of more of a general replacement rule at the end, where in my final stage I remove all of the (previously defined) functions s, in exchange for these functions p. I will try and play around and see if I can figure out how to create a replacement rule based on this method. Thank you for your help:
    $endgroup$
    – Brad
    7 hours ago












  • $begingroup$
    After changing my definition of recursion to work with a symbolic sX, which actually makes evaluations much much faster overall, I can utilise this to get exactly what I need. Thank you all for your help, this is the fastest solution I found.
    $endgroup$
    – Brad
    6 hours ago



















7












$begingroup$

ClearAll[s]

s[f_] := Total @ Subsets[f @ ##, {2}] &;



s[f][a, b, c]



f[a, b] + f[a, c] + f[b, c]




s[f][3, 4, 5, 6, 7]



f[3, 4] + f[3, 5] + f[3, 6] + f[3, 7] + f[4, 5] + f[4, 6] + f[4, 7] +
f[5, 6] + f[5, 7] + f[6, 7]







share|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7












    $begingroup$

    Something like the following?



    ClearAll[s]
    
s[seq__] := Total[f@@@Subsets[List[seq], {2}]]





    share|improve this answer









    $endgroup$













    • $begingroup$
      Hi @MarcoB - Subsets was indeed the function I was looking for. I was thinking perhaps of more of a general replacement rule at the end, where in my final stage I remove all of the (previously defined) functions s, in exchange for these functions p. I will try and play around and see if I can figure out how to create a replacement rule based on this method. Thank you for your help:
      $endgroup$
      – Brad
      7 hours ago












    • $begingroup$
      After changing my definition of recursion to work with a symbolic sX, which actually makes evaluations much much faster overall, I can utilise this to get exactly what I need. Thank you all for your help, this is the fastest solution I found.
      $endgroup$
      – Brad
      6 hours ago
















    7












    $begingroup$

    Something like the following?



    ClearAll[s]
    
s[seq__] := Total[f@@@Subsets[List[seq], {2}]]





    share|improve this answer









    $endgroup$













    • $begingroup$
      Hi @MarcoB - Subsets was indeed the function I was looking for. I was thinking perhaps of more of a general replacement rule at the end, where in my final stage I remove all of the (previously defined) functions s, in exchange for these functions p. I will try and play around and see if I can figure out how to create a replacement rule based on this method. Thank you for your help:
      $endgroup$
      – Brad
      7 hours ago












    • $begingroup$
      After changing my definition of recursion to work with a symbolic sX, which actually makes evaluations much much faster overall, I can utilise this to get exactly what I need. Thank you all for your help, this is the fastest solution I found.
      $endgroup$
      – Brad
      6 hours ago














    7












    7








    7





    $begingroup$

    Something like the following?



    ClearAll[s]
    
s[seq__] := Total[f@@@Subsets[List[seq], {2}]]





    share|improve this answer









    $endgroup$



    Something like the following?



    ClearAll[s]
    
s[seq__] := Total[f@@@Subsets[List[seq], {2}]]






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 8 hours ago









    MarcoBMarcoB

    36.6k556112




    36.6k556112












    • $begingroup$
      Hi @MarcoB - Subsets was indeed the function I was looking for. I was thinking perhaps of more of a general replacement rule at the end, where in my final stage I remove all of the (previously defined) functions s, in exchange for these functions p. I will try and play around and see if I can figure out how to create a replacement rule based on this method. Thank you for your help:
      $endgroup$
      – Brad
      7 hours ago












    • $begingroup$
      After changing my definition of recursion to work with a symbolic sX, which actually makes evaluations much much faster overall, I can utilise this to get exactly what I need. Thank you all for your help, this is the fastest solution I found.
      $endgroup$
      – Brad
      6 hours ago


















    • $begingroup$
      Hi @MarcoB - Subsets was indeed the function I was looking for. I was thinking perhaps of more of a general replacement rule at the end, where in my final stage I remove all of the (previously defined) functions s, in exchange for these functions p. I will try and play around and see if I can figure out how to create a replacement rule based on this method. Thank you for your help:
      $endgroup$
      – Brad
      7 hours ago












    • $begingroup$
      After changing my definition of recursion to work with a symbolic sX, which actually makes evaluations much much faster overall, I can utilise this to get exactly what I need. Thank you all for your help, this is the fastest solution I found.
      $endgroup$
      – Brad
      6 hours ago
















    $begingroup$
    Hi @MarcoB - Subsets was indeed the function I was looking for. I was thinking perhaps of more of a general replacement rule at the end, where in my final stage I remove all of the (previously defined) functions s, in exchange for these functions p. I will try and play around and see if I can figure out how to create a replacement rule based on this method. Thank you for your help:
    $endgroup$
    – Brad
    7 hours ago






    $begingroup$
    Hi @MarcoB - Subsets was indeed the function I was looking for. I was thinking perhaps of more of a general replacement rule at the end, where in my final stage I remove all of the (previously defined) functions s, in exchange for these functions p. I will try and play around and see if I can figure out how to create a replacement rule based on this method. Thank you for your help:
    $endgroup$
    – Brad
    7 hours ago














    $begingroup$
    After changing my definition of recursion to work with a symbolic sX, which actually makes evaluations much much faster overall, I can utilise this to get exactly what I need. Thank you all for your help, this is the fastest solution I found.
    $endgroup$
    – Brad
    6 hours ago




    $begingroup$
    After changing my definition of recursion to work with a symbolic sX, which actually makes evaluations much much faster overall, I can utilise this to get exactly what I need. Thank you all for your help, this is the fastest solution I found.
    $endgroup$
    – Brad
    6 hours ago











    7












    $begingroup$

    ClearAll[s]
    
s[f_] := Total @ Subsets[f @ ##, {2}] &;
    

    
s[f][a, b, c]



    f[a, b] + f[a, c] + f[b, c]




    s[f][3, 4, 5, 6, 7]



    f[3, 4] + f[3, 5] + f[3, 6] + f[3, 7] + f[4, 5] + f[4, 6] + f[4, 7] +
    f[5, 6] + f[5, 7] + f[6, 7]







    share|improve this answer











    $endgroup$


















      7












      $begingroup$

      ClearAll[s]
      
s[f_] := Total @ Subsets[f @ ##, {2}] &;
      

      
s[f][a, b, c]



      f[a, b] + f[a, c] + f[b, c]




      s[f][3, 4, 5, 6, 7]



      f[3, 4] + f[3, 5] + f[3, 6] + f[3, 7] + f[4, 5] + f[4, 6] + f[4, 7] +
      f[5, 6] + f[5, 7] + f[6, 7]







      share|improve this answer











      $endgroup$
















        7












        7








        7





        $begingroup$

        ClearAll[s]
        
s[f_] := Total @ Subsets[f @ ##, {2}] &;
        

        
s[f][a, b, c]



        f[a, b] + f[a, c] + f[b, c]




        s[f][3, 4, 5, 6, 7]



        f[3, 4] + f[3, 5] + f[3, 6] + f[3, 7] + f[4, 5] + f[4, 6] + f[4, 7] +
        f[5, 6] + f[5, 7] + f[6, 7]







        share|improve this answer











        $endgroup$



        ClearAll[s]
        
s[f_] := Total @ Subsets[f @ ##, {2}] &;
        

        
s[f][a, b, c]



        f[a, b] + f[a, c] + f[b, c]




        s[f][3, 4, 5, 6, 7]



        f[3, 4] + f[3, 5] + f[3, 6] + f[3, 7] + f[4, 5] + f[4, 6] + f[4, 7] +
        f[5, 6] + f[5, 7] + f[6, 7]








        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 7 hours ago

























        answered 7 hours ago









        kglrkglr

        186k10203422




        186k10203422






























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