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I express multilinear functions in the following format. Is there any ready command to convert them to multilinear functions easily?

Input

posTerms = {2, 4, 9, 13, 19};
negTerms = {6, 11, 26};
IntegerString[posTerms, 2]
IntegerString[negTerms, 2]

{"10", "100", "1001", "1101", "10011"}
{"110", "1011", "11010"}

Intended Output

x_2+x_3+x_4*x_1+x_4*x_3*x_1+x_5*x_2*x_1-(x_3*x_2+x_4*x_2*x_1+x_5*x_4*x_2)

Intended Output in Mathematica

Subscript[p, 2]+Subscript[p, 3]+Subscript[p, 4]*Subscript[p, 1]+Subscript[p, 4]*Subscript[p, 3]*Subscript[p, 1]+Subscript[p, 5]*Subscript[p, 2]*Subscript[p, 1]

-Subscript[p, 3]*Subscript[p, 2]-Subscript[p, 4]*Subscript[p, 2]*Subscript[p, 1]-Subscript[p, 5]*Subscript[p, 4]*Subscript[p, 2]
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2 Answers 2

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How about

FromCoefficientRules[#, Subscript[x, #] & /@ Reverse@Range@Length@#[[1, 1]]] &@
    Thread[PadLeft@IntegerDigits[posTerms, 2] -> 1] - 
   FromCoefficientRules[#, Subscript[x, #] & /@ Reverse@Range@Length@#[[1, 1]]] &@
 Thread[PadLeft@IntegerDigits[negTerms, 2] -> 1]

enter image description here

As belisarius wrote in comment it can be reduced to

Subtract @@ (FromCoefficientRules[#, Subscript[x, #] & /@ Reverse@Range@Length@#[[1, 1]]] &@
     Thread[PadLeft@IntegerDigits[#, 2] -> 1] & /@ {posTerms, negTerms})

Also there is another short solution

Subtract @@ (Total[Times @@ MapIndexed[Subscript[x, #2[[1]]]^# &, Reverse[#]] & /@ 
   IntegerDigits[#, 2]] & /@ {posTerms, negTerms})

It is shorter than belisarius's solution by 13 characters :)

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  • 1
    $\begingroup$ @hhh It is because I incorrectly aligned posTerms with negTerms. See my update, does it work now? $\endgroup$
    – ybeltukov
    Nov 6, 2013 at 22:56
  • $\begingroup$ I cannot fully understand this: suppose I specify the terms with {1,3,7} to mean p_1p_3p_7 -- is it easier to get the multilinear term from it? So here terms Subscript[x, #] & /@ {1, 3, 7} and now just multiplication. Where is the multiplication? $\endgroup$
    – hhh
    Nov 10, 2013 at 19:14
  • $\begingroup$ Moved the new q here. $\endgroup$
    – hhh
    Nov 10, 2013 at 19:23
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Also

-Differences[Tr[Times @@@ (Position[Reverse@#, 1] & /@ # /. {n_Integer} -> 
                            Subscript[x, n])] & /@ (IntegerDigits[#, 2] & /@ {posTerms, negTerms})]

Mathematica graphics

Equivalent: 

Subtract@@Tr/@ Map[Times @@ (Position[Reverse@#, 1] /. {n_Integer} -> 
                           Subscript[x, n]) &, (IntegerDigits[#, 2] & /@ {posTerms, negTerms}), {2}]
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  • $\begingroup$ I like shorter +1. $\endgroup$
    – hhh
    Nov 7, 2013 at 0:21
  • 1
    $\begingroup$ @hhh So here you have a shorter version of ybeltukov's answer too Subtract @@ (FromCoefficientRules[#, Subscript[x, #] & /@ Reverse@Range@Length@#[[1, 1]]] &@ Thread[PadLeft@IntegerDigits[#, 2] -> 1] & /@ {posTerms, negTerms}) $\endgroup$
    – Dr. belisarius
    Nov 7, 2013 at 0:35

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