Multiply Two Integer Polynomials The 2019 Stack Overflow Developer Survey Results Are InTips for golfing in PythonTips for golfing in <all languages>Discrete Convolution or Polynomial MultiplicationPretty-printing polynomialsPrime polynomialsSymbolic Differentiation of PolynomialsSymbolic Integration of PolynomialsIrreducible polynomials over GF(5)PolynomialceptionSelf Referential PolynomialsAdd up two algebraic numbersМногочлены Чебышёва (Chebyshev Polynomials)Decompose Polynomials
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Multiply Two Integer Polynomials
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Multiply Two Integer Polynomials
The 2019 Stack Overflow Developer Survey Results Are InTips for golfing in PythonTips for golfing in <all languages>Discrete Convolution or Polynomial MultiplicationPretty-printing polynomialsPrime polynomialsSymbolic Differentiation of PolynomialsSymbolic Integration of PolynomialsIrreducible polynomials over GF(5)PolynomialceptionSelf Referential PolynomialsAdd up two algebraic numbersМногочлены Чебышёва (Chebyshev Polynomials)Decompose Polynomials
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Your task is to take two single-variable integer polynomial expressions and multiply them into their unsimplified first-term-major left-to-right expansion (A.K.A. FOIL in the case of binomials). Do not combine like terms or reorder the result. To be more explicit about the expansion, multiply the first term in the first expression by each term in the second, in order, and continue in the first expression until all terms have been multiplied by all other terms. Expressions will be given in a simplified LaTeX variant.
Each expression will be a sequence of terms separated by +
(with exactly one space on each side) Each term will conform to the following regular expression: (PCRE notation)
-?d+x^d+
In plain English, the term is an optional leading -
followed by one or more digits followed by x
and a nonnegative integer power (with ^
)
An example of a full expression:
6x^3 + 1337x^2 + -4x^1 + 2x^0
When plugged into LaTeX, you get $6x^3 + 1337x^2 + -4x^1 + 2x^0$
The output should also conform to this format.
Since brackets do not surround exponents in this format, LaTeX will actually render multi-digit exponents incorrectly. (e.g. 4x^3 + -2x^14 + 54x^28 + -4x^5
renders as $4x^3 + -2x^14 + 54x^28 + -4x^5$) You do not need to account for this and you should not include the brackets in your output.
Example Test Cases
5x^4
3x^23
15x^27
6x^2 + 7x^1 + -2x^0
1x^2 + -2x^3
6x^4 + -12x^5 + 7x^3 + -14x^4 + -2x^2 + 4x^3
3x^1 + 5x^2 + 2x^4 + 3x^0
3x^0
9x^1 + 15x^2 + 6x^4 + 9x^0
4x^3 + -2x^14 + 54x^28 + -4x^5
-0x^7
0x^10 + 0x^21 + 0x^35 + 0x^12
4x^3 + -2x^4 + 0x^255 + -4x^5
-3x^4 + 2x^2
-12x^7 + 8x^5 + 6x^8 + -4x^6 + 0x^259 + 0x^257 + 12x^9 + -8x^7
Rules and Assumptions
- You may assume that all inputs conform to this exact format. Behavior for any other format is undefined for the purposes of this challenge.
- It should be noted that any method of taking in the two polynomials is valid, provided that both are read in as strings conforming to the above format.
- The order of the polynomials matters due to the expected order of the product expansion.
- You must support input coefficients between $-128$ and $127$ and input exponents up to $255$.
- Output coefficents between $-16,256$ and $16,384$ and exponents up to $510$ must therefore be supported.
- You may assume each input polynomial contains no more than 16 terms
- Therefore you must (at minimum) support up to 256 terms in the output
- Terms with zero coefficients should be left as is, with exponents being properly combined
- Negative zero is allowed in the input, but is indistinguishable from positive zero semantically. Always output positive zero. Do not omit zero terms.
Happy Golfing! Good luck!
code-golf math parsing
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add a comment |
$begingroup$
Your task is to take two single-variable integer polynomial expressions and multiply them into their unsimplified first-term-major left-to-right expansion (A.K.A. FOIL in the case of binomials). Do not combine like terms or reorder the result. To be more explicit about the expansion, multiply the first term in the first expression by each term in the second, in order, and continue in the first expression until all terms have been multiplied by all other terms. Expressions will be given in a simplified LaTeX variant.
Each expression will be a sequence of terms separated by +
(with exactly one space on each side) Each term will conform to the following regular expression: (PCRE notation)
-?d+x^d+
In plain English, the term is an optional leading -
followed by one or more digits followed by x
and a nonnegative integer power (with ^
)
An example of a full expression:
6x^3 + 1337x^2 + -4x^1 + 2x^0
When plugged into LaTeX, you get $6x^3 + 1337x^2 + -4x^1 + 2x^0$
The output should also conform to this format.
Since brackets do not surround exponents in this format, LaTeX will actually render multi-digit exponents incorrectly. (e.g. 4x^3 + -2x^14 + 54x^28 + -4x^5
renders as $4x^3 + -2x^14 + 54x^28 + -4x^5$) You do not need to account for this and you should not include the brackets in your output.
Example Test Cases
5x^4
3x^23
15x^27
6x^2 + 7x^1 + -2x^0
1x^2 + -2x^3
6x^4 + -12x^5 + 7x^3 + -14x^4 + -2x^2 + 4x^3
3x^1 + 5x^2 + 2x^4 + 3x^0
3x^0
9x^1 + 15x^2 + 6x^4 + 9x^0
4x^3 + -2x^14 + 54x^28 + -4x^5
-0x^7
0x^10 + 0x^21 + 0x^35 + 0x^12
4x^3 + -2x^4 + 0x^255 + -4x^5
-3x^4 + 2x^2
-12x^7 + 8x^5 + 6x^8 + -4x^6 + 0x^259 + 0x^257 + 12x^9 + -8x^7
Rules and Assumptions
- You may assume that all inputs conform to this exact format. Behavior for any other format is undefined for the purposes of this challenge.
- It should be noted that any method of taking in the two polynomials is valid, provided that both are read in as strings conforming to the above format.
- The order of the polynomials matters due to the expected order of the product expansion.
- You must support input coefficients between $-128$ and $127$ and input exponents up to $255$.
- Output coefficents between $-16,256$ and $16,384$ and exponents up to $510$ must therefore be supported.
- You may assume each input polynomial contains no more than 16 terms
- Therefore you must (at minimum) support up to 256 terms in the output
- Terms with zero coefficients should be left as is, with exponents being properly combined
- Negative zero is allowed in the input, but is indistinguishable from positive zero semantically. Always output positive zero. Do not omit zero terms.
Happy Golfing! Good luck!
code-golf math parsing
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related
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– H.PWiz
7 hours ago
1
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@LuisfelipeDejesusMunoz I imagine not. Parsing is an integral part of the challenge and the OP says -- "It should be noted that any method of taking in the two polynomials is valid, provided that both are read in as strings conforming to the above format." (emphasis added)
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– Giuseppe
6 hours ago
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Your regex is wrong:^
should be^
.
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– Erik the Outgolfer
6 hours ago
add a comment |
$begingroup$
Your task is to take two single-variable integer polynomial expressions and multiply them into their unsimplified first-term-major left-to-right expansion (A.K.A. FOIL in the case of binomials). Do not combine like terms or reorder the result. To be more explicit about the expansion, multiply the first term in the first expression by each term in the second, in order, and continue in the first expression until all terms have been multiplied by all other terms. Expressions will be given in a simplified LaTeX variant.
Each expression will be a sequence of terms separated by +
(with exactly one space on each side) Each term will conform to the following regular expression: (PCRE notation)
-?d+x^d+
In plain English, the term is an optional leading -
followed by one or more digits followed by x
and a nonnegative integer power (with ^
)
An example of a full expression:
6x^3 + 1337x^2 + -4x^1 + 2x^0
When plugged into LaTeX, you get $6x^3 + 1337x^2 + -4x^1 + 2x^0$
The output should also conform to this format.
Since brackets do not surround exponents in this format, LaTeX will actually render multi-digit exponents incorrectly. (e.g. 4x^3 + -2x^14 + 54x^28 + -4x^5
renders as $4x^3 + -2x^14 + 54x^28 + -4x^5$) You do not need to account for this and you should not include the brackets in your output.
Example Test Cases
5x^4
3x^23
15x^27
6x^2 + 7x^1 + -2x^0
1x^2 + -2x^3
6x^4 + -12x^5 + 7x^3 + -14x^4 + -2x^2 + 4x^3
3x^1 + 5x^2 + 2x^4 + 3x^0
3x^0
9x^1 + 15x^2 + 6x^4 + 9x^0
4x^3 + -2x^14 + 54x^28 + -4x^5
-0x^7
0x^10 + 0x^21 + 0x^35 + 0x^12
4x^3 + -2x^4 + 0x^255 + -4x^5
-3x^4 + 2x^2
-12x^7 + 8x^5 + 6x^8 + -4x^6 + 0x^259 + 0x^257 + 12x^9 + -8x^7
Rules and Assumptions
- You may assume that all inputs conform to this exact format. Behavior for any other format is undefined for the purposes of this challenge.
- It should be noted that any method of taking in the two polynomials is valid, provided that both are read in as strings conforming to the above format.
- The order of the polynomials matters due to the expected order of the product expansion.
- You must support input coefficients between $-128$ and $127$ and input exponents up to $255$.
- Output coefficents between $-16,256$ and $16,384$ and exponents up to $510$ must therefore be supported.
- You may assume each input polynomial contains no more than 16 terms
- Therefore you must (at minimum) support up to 256 terms in the output
- Terms with zero coefficients should be left as is, with exponents being properly combined
- Negative zero is allowed in the input, but is indistinguishable from positive zero semantically. Always output positive zero. Do not omit zero terms.
Happy Golfing! Good luck!
code-golf math parsing
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Your task is to take two single-variable integer polynomial expressions and multiply them into their unsimplified first-term-major left-to-right expansion (A.K.A. FOIL in the case of binomials). Do not combine like terms or reorder the result. To be more explicit about the expansion, multiply the first term in the first expression by each term in the second, in order, and continue in the first expression until all terms have been multiplied by all other terms. Expressions will be given in a simplified LaTeX variant.
Each expression will be a sequence of terms separated by +
(with exactly one space on each side) Each term will conform to the following regular expression: (PCRE notation)
-?d+x^d+
In plain English, the term is an optional leading -
followed by one or more digits followed by x
and a nonnegative integer power (with ^
)
An example of a full expression:
6x^3 + 1337x^2 + -4x^1 + 2x^0
When plugged into LaTeX, you get $6x^3 + 1337x^2 + -4x^1 + 2x^0$
The output should also conform to this format.
Since brackets do not surround exponents in this format, LaTeX will actually render multi-digit exponents incorrectly. (e.g. 4x^3 + -2x^14 + 54x^28 + -4x^5
renders as $4x^3 + -2x^14 + 54x^28 + -4x^5$) You do not need to account for this and you should not include the brackets in your output.
Example Test Cases
5x^4
3x^23
15x^27
6x^2 + 7x^1 + -2x^0
1x^2 + -2x^3
6x^4 + -12x^5 + 7x^3 + -14x^4 + -2x^2 + 4x^3
3x^1 + 5x^2 + 2x^4 + 3x^0
3x^0
9x^1 + 15x^2 + 6x^4 + 9x^0
4x^3 + -2x^14 + 54x^28 + -4x^5
-0x^7
0x^10 + 0x^21 + 0x^35 + 0x^12
4x^3 + -2x^4 + 0x^255 + -4x^5
-3x^4 + 2x^2
-12x^7 + 8x^5 + 6x^8 + -4x^6 + 0x^259 + 0x^257 + 12x^9 + -8x^7
Rules and Assumptions
- You may assume that all inputs conform to this exact format. Behavior for any other format is undefined for the purposes of this challenge.
- It should be noted that any method of taking in the two polynomials is valid, provided that both are read in as strings conforming to the above format.
- The order of the polynomials matters due to the expected order of the product expansion.
- You must support input coefficients between $-128$ and $127$ and input exponents up to $255$.
- Output coefficents between $-16,256$ and $16,384$ and exponents up to $510$ must therefore be supported.
- You may assume each input polynomial contains no more than 16 terms
- Therefore you must (at minimum) support up to 256 terms in the output
- Terms with zero coefficients should be left as is, with exponents being properly combined
- Negative zero is allowed in the input, but is indistinguishable from positive zero semantically. Always output positive zero. Do not omit zero terms.
Happy Golfing! Good luck!
code-golf math parsing
code-golf math parsing
asked 7 hours ago
BeefsterBeefster
2,6271244
2,6271244
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related
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– H.PWiz
7 hours ago
1
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@LuisfelipeDejesusMunoz I imagine not. Parsing is an integral part of the challenge and the OP says -- "It should be noted that any method of taking in the two polynomials is valid, provided that both are read in as strings conforming to the above format." (emphasis added)
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– Giuseppe
6 hours ago
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Your regex is wrong:^
should be^
.
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– Erik the Outgolfer
6 hours ago
add a comment |
$begingroup$
related
$endgroup$
– H.PWiz
7 hours ago
1
$begingroup$
@LuisfelipeDejesusMunoz I imagine not. Parsing is an integral part of the challenge and the OP says -- "It should be noted that any method of taking in the two polynomials is valid, provided that both are read in as strings conforming to the above format." (emphasis added)
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– Giuseppe
6 hours ago
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Your regex is wrong:^
should be^
.
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– Erik the Outgolfer
6 hours ago
$begingroup$
related
$endgroup$
– H.PWiz
7 hours ago
$begingroup$
related
$endgroup$
– H.PWiz
7 hours ago
1
1
$begingroup$
@LuisfelipeDejesusMunoz I imagine not. Parsing is an integral part of the challenge and the OP says -- "It should be noted that any method of taking in the two polynomials is valid, provided that both are read in as strings conforming to the above format." (emphasis added)
$endgroup$
– Giuseppe
6 hours ago
$begingroup$
@LuisfelipeDejesusMunoz I imagine not. Parsing is an integral part of the challenge and the OP says -- "It should be noted that any method of taking in the two polynomials is valid, provided that both are read in as strings conforming to the above format." (emphasis added)
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– Giuseppe
6 hours ago
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Your regex is wrong:
^
should be ^
.$endgroup$
– Erik the Outgolfer
6 hours ago
$begingroup$
Your regex is wrong:
^
should be ^
.$endgroup$
– Erik the Outgolfer
6 hours ago
add a comment |
12 Answers
12
active
oldest
votes
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R, 159 153 bytes
function(P,Q,a=h(P),b=h(Q))paste0(b[1,]%o%a[1,],"x^",outer(b[2,],a[2,],"+"),collapse=" + ")
h=function(s,`/`=strsplit)sapply(el(s/" \+ ")/"x\^",strtoi)
Try it online!
I really wanted to use outer
, so there's almost surely a more efficient approach.
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add a comment |
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Pyth - 39 bytes
LmsMcdK"x^"%2cb)j" + "m++*FhdKsedCM*FyM
Try it online.
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add a comment |
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Haskell, 124 bytes
import Data.Lists
s=splitOn
z=map(map read.s"x^").s"+"
a#b=intercalate" + "[shows(u*p)"x^"++show(v+q)|[u,v]<-z a,[p,q]<-z b]
Note: TIO lacks Data.Lists
, so I import Data.Lists.Split
and Data.List
: Try it online!
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add a comment |
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Ruby, 102 bytes
->a,ba.scan(w=/(-?d+)x.(d+)/).product(b.scan w).mapx,y*?+
Try it online!
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add a comment |
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JavaScript, 112 bytes
I found three alternatives with the same length. Call with currying syntax.
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(a=>P(B).map(b=>a[0]*b[0]+'x^'+(a[1]- -b[1]))).join` + `
f=
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(a=>P(B).map(b=>a[0]*b[0]+'x^'+(a[1]- -b[1]))).join` + `
console.log( f('5x^4')('3x^23') )
console.log( f('6x^2 + 7x^1 + -2x^0')('1x^2 + -2x^3') )
console.log( f('3x^1 + 5x^2 + 2x^4 + 3x^0')('3x^0') )
console.log( f('4x^3 + -2x^14 + 54x^28 + -4x^5')('-0x^7') )
console.log( f('4x^3 + -2x^4 + 0x^255 + -4x^5')('-3x^4 + 2x^2') )
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(([c,e])=>P(B).map(([C,E])=>c*C+'x^'+(e- -E))).join` + `
f=
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(([c,e])=>P(B).map(([C,E])=>c*C+'x^'+(e- -E))).join` + `
console.log( f('5x^4')('3x^23') )
console.log( f('6x^2 + 7x^1 + -2x^0')('1x^2 + -2x^3') )
console.log( f('3x^1 + 5x^2 + 2x^4 + 3x^0')('3x^0') )
console.log( f('4x^3 + -2x^14 + 54x^28 + -4x^5')('-0x^7') )
console.log( f('4x^3 + -2x^4 + 0x^255 + -4x^5')('-3x^4 + 2x^2') )
A=>B=>(P=x=>[...x.matchAll(/(S+)x.(S+)/g)])(A).flatMap(a=>P(B).map(b=>a[1]*b[1]+'x^'+(a[2]- -b[2]))).join` + `
f=
A=>B=>(P=x=>[...x.matchAll(/(S+)x.(S+)/g)])(A).flatMap(a=>P(B).map(b=>a[1]*b[1]+'x^'+(a[2]- -b[2]))).join` + `
console.log( f('5x^4')('3x^23') )
console.log( f('6x^2 + 7x^1 + -2x^0')('1x^2 + -2x^3') )
console.log( f('3x^1 + 5x^2 + 2x^4 + 3x^0')('3x^0') )
console.log( f('4x^3 + -2x^14 + 54x^28 + -4x^5')('-0x^7') )
console.log( f('4x^3 + -2x^4 + 0x^255 + -4x^5')('-3x^4 + 2x^2') )
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split' + ' => split'+'
to save 2 bytes
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– Luis felipe De jesus Munoz
4 hours ago
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@Arnauld Seems fine without them
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– Embodiment of Ignorance
2 hours ago
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@EmbodimentofIgnorance My bad, I misread Luis' comment. I thought it was about thejoin
.
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– Arnauld
2 hours ago
add a comment |
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SNOBOL4 (CSNOBOL4), 192 176 bytes
P =INPUT
Q =INPUT
D =SPAN(1234567890)
P P D . K ARB D . W REM . P :F(O)
B =Q
B B D . C ARB D . E REM . B :F(P)
O =O ' + ' K * C 'x^' W + E :(B)
O O POS(3) REM . OUTPUT
END
Try it online!
Explanation to come when I'm off my phone.
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add a comment |
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JavaScript (Babel Node), 118 bytes
Takes input as (a)(b)
.
a=>b=>(g=s=>[...s.matchAll(/(-?d+)x.(d+)/g)])(a).flatMap(([_,x,p])=>g(b).map(([_,X,P])=>x*X+'x^'+-(-p-P))).join` + `
Try it online!
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add a comment |
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Haskell, 133 bytes
f""=[]
f t|[(a,_:_:u)]<-reads t,[(i,v)]<-reads u=(a,i):f(drop 3v)
p!q=drop 3$do(a,i)<-f p;(b,j)<-f q;" + "++shows(a*b)"x^"++show(i+j)
Try it online!
f
parses a polynomial from a string, !
multiplies two of them and formats the result.
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add a comment |
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Python 2, 193 bytes
import re
f=re.finditer
lambda a,b:' + '.join(' + '.join(`int(m.group(1))*int(n.group(1))`+'x^'+`int(m.group(2))+int(n.group(2))`for n in f('(-?d+)x^(d+)',b))for m in f('(-?d+)x^(d+)',a))
Try it online!
Side note: First time doing a code golf challenge, so sorry if the attempt sucks haha
New contributor
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1
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Welcome to PPCG! I'm not much of a python programmer, but if your submission is longer than a SNOBOL one, there's probably room for improvement, heheh. Perhaps you can find help at Tips for Golfing in Python or Tips for Golfing in <all languages>! Hope you enjoy the time you spend here :-)
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– Giuseppe
4 hours ago
add a comment |
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Retina, 110 bytes
SS+(?=.*n(.+))
$1#$&
|" + "L$v` (-?)(d+)x.(d+).*?#(-?)(d+)x.(d+)
$1$4$.($2*$5*)x^$.($3*_$6*
--|-(0)
$1
Try it online! Explanation:
SS+(?=.*n(.+))
$1#$&
Prefix each term in the first input with a #
, a copy of the second input, and a space. This means that all of the terms in copies of the second input are preceded by a space and none of the terms from the first input are.
|" + "L$v` (-?)(d+)x.(d+).*?#(-?)(d+)x.(d+)
$1$4$.($2*$5*)x^$.($3*_$6*
Match all of the copies of terms in the second input and their corresponding term from the first input. Concatenate any -
signs, multiply the coefficients, and add the indices. Finally join all of the resulting substitutions with the string +
.
--|-(0)
$1
Delete any pairs of -
s and convert -0
to 0
.
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add a comment |
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Jelly, 28 bytes
ṣ”+ṣ”xV$€)p/ZPSƭ€j⁾x^Ʋ€j“ +
Try it online!
Full program. Takes the two polynomials as a list of two strings.
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add a comment |
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C# (Visual C# Interactive Compiler), 192 190 188 bytes
n=>m=>string.Join(g="+",from a in n.Split(g)from b in m.Split(g)select f(a.Split(p="x^")[0])*f(b.Split(p)[0])+p+(f(a.Split(p)[1])+f(b.Split(p)[1])));Func<string,int>f=int.Parse;string p,g;
Query syntax seems to be a byte shorter than method syntax.
At least I beat SNOBOL.
Try it online!
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Sorry, not beating SNOBOL any more!
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– Giuseppe
1 hour ago
add a comment |
Your Answer
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12 Answers
12
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votes
12 Answers
12
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$begingroup$
R, 159 153 bytes
function(P,Q,a=h(P),b=h(Q))paste0(b[1,]%o%a[1,],"x^",outer(b[2,],a[2,],"+"),collapse=" + ")
h=function(s,`/`=strsplit)sapply(el(s/" \+ ")/"x\^",strtoi)
Try it online!
I really wanted to use outer
, so there's almost surely a more efficient approach.
$endgroup$
add a comment |
$begingroup$
R, 159 153 bytes
function(P,Q,a=h(P),b=h(Q))paste0(b[1,]%o%a[1,],"x^",outer(b[2,],a[2,],"+"),collapse=" + ")
h=function(s,`/`=strsplit)sapply(el(s/" \+ ")/"x\^",strtoi)
Try it online!
I really wanted to use outer
, so there's almost surely a more efficient approach.
$endgroup$
add a comment |
$begingroup$
R, 159 153 bytes
function(P,Q,a=h(P),b=h(Q))paste0(b[1,]%o%a[1,],"x^",outer(b[2,],a[2,],"+"),collapse=" + ")
h=function(s,`/`=strsplit)sapply(el(s/" \+ ")/"x\^",strtoi)
Try it online!
I really wanted to use outer
, so there's almost surely a more efficient approach.
$endgroup$
R, 159 153 bytes
function(P,Q,a=h(P),b=h(Q))paste0(b[1,]%o%a[1,],"x^",outer(b[2,],a[2,],"+"),collapse=" + ")
h=function(s,`/`=strsplit)sapply(el(s/" \+ ")/"x\^",strtoi)
Try it online!
I really wanted to use outer
, so there's almost surely a more efficient approach.
edited 6 hours ago
answered 7 hours ago
GiuseppeGiuseppe
17.7k31153
17.7k31153
add a comment |
add a comment |
$begingroup$
Pyth - 39 bytes
LmsMcdK"x^"%2cb)j" + "m++*FhdKsedCM*FyM
Try it online.
$endgroup$
add a comment |
$begingroup$
Pyth - 39 bytes
LmsMcdK"x^"%2cb)j" + "m++*FhdKsedCM*FyM
Try it online.
$endgroup$
add a comment |
$begingroup$
Pyth - 39 bytes
LmsMcdK"x^"%2cb)j" + "m++*FhdKsedCM*FyM
Try it online.
$endgroup$
Pyth - 39 bytes
LmsMcdK"x^"%2cb)j" + "m++*FhdKsedCM*FyM
Try it online.
answered 7 hours ago
MaltysenMaltysen
21.3k445116
21.3k445116
add a comment |
add a comment |
$begingroup$
Haskell, 124 bytes
import Data.Lists
s=splitOn
z=map(map read.s"x^").s"+"
a#b=intercalate" + "[shows(u*p)"x^"++show(v+q)|[u,v]<-z a,[p,q]<-z b]
Note: TIO lacks Data.Lists
, so I import Data.Lists.Split
and Data.List
: Try it online!
$endgroup$
add a comment |
$begingroup$
Haskell, 124 bytes
import Data.Lists
s=splitOn
z=map(map read.s"x^").s"+"
a#b=intercalate" + "[shows(u*p)"x^"++show(v+q)|[u,v]<-z a,[p,q]<-z b]
Note: TIO lacks Data.Lists
, so I import Data.Lists.Split
and Data.List
: Try it online!
$endgroup$
add a comment |
$begingroup$
Haskell, 124 bytes
import Data.Lists
s=splitOn
z=map(map read.s"x^").s"+"
a#b=intercalate" + "[shows(u*p)"x^"++show(v+q)|[u,v]<-z a,[p,q]<-z b]
Note: TIO lacks Data.Lists
, so I import Data.Lists.Split
and Data.List
: Try it online!
$endgroup$
Haskell, 124 bytes
import Data.Lists
s=splitOn
z=map(map read.s"x^").s"+"
a#b=intercalate" + "[shows(u*p)"x^"++show(v+q)|[u,v]<-z a,[p,q]<-z b]
Note: TIO lacks Data.Lists
, so I import Data.Lists.Split
and Data.List
: Try it online!
answered 7 hours ago
niminimi
32.7k32489
32.7k32489
add a comment |
add a comment |
$begingroup$
Ruby, 102 bytes
->a,ba.scan(w=/(-?d+)x.(d+)/).product(b.scan w).mapx,y*?+
Try it online!
$endgroup$
add a comment |
$begingroup$
Ruby, 102 bytes
->a,ba.scan(w=/(-?d+)x.(d+)/).product(b.scan w).mapx,y*?+
Try it online!
$endgroup$
add a comment |
$begingroup$
Ruby, 102 bytes
->a,ba.scan(w=/(-?d+)x.(d+)/).product(b.scan w).mapx,y*?+
Try it online!
$endgroup$
Ruby, 102 bytes
->a,ba.scan(w=/(-?d+)x.(d+)/).product(b.scan w).mapx,y*?+
Try it online!
edited 6 hours ago
answered 7 hours ago
G BG B
8,2661429
8,2661429
add a comment |
add a comment |
$begingroup$
JavaScript, 112 bytes
I found three alternatives with the same length. Call with currying syntax.
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(a=>P(B).map(b=>a[0]*b[0]+'x^'+(a[1]- -b[1]))).join` + `
f=
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(a=>P(B).map(b=>a[0]*b[0]+'x^'+(a[1]- -b[1]))).join` + `
console.log( f('5x^4')('3x^23') )
console.log( f('6x^2 + 7x^1 + -2x^0')('1x^2 + -2x^3') )
console.log( f('3x^1 + 5x^2 + 2x^4 + 3x^0')('3x^0') )
console.log( f('4x^3 + -2x^14 + 54x^28 + -4x^5')('-0x^7') )
console.log( f('4x^3 + -2x^4 + 0x^255 + -4x^5')('-3x^4 + 2x^2') )
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(([c,e])=>P(B).map(([C,E])=>c*C+'x^'+(e- -E))).join` + `
f=
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(([c,e])=>P(B).map(([C,E])=>c*C+'x^'+(e- -E))).join` + `
console.log( f('5x^4')('3x^23') )
console.log( f('6x^2 + 7x^1 + -2x^0')('1x^2 + -2x^3') )
console.log( f('3x^1 + 5x^2 + 2x^4 + 3x^0')('3x^0') )
console.log( f('4x^3 + -2x^14 + 54x^28 + -4x^5')('-0x^7') )
console.log( f('4x^3 + -2x^4 + 0x^255 + -4x^5')('-3x^4 + 2x^2') )
A=>B=>(P=x=>[...x.matchAll(/(S+)x.(S+)/g)])(A).flatMap(a=>P(B).map(b=>a[1]*b[1]+'x^'+(a[2]- -b[2]))).join` + `
f=
A=>B=>(P=x=>[...x.matchAll(/(S+)x.(S+)/g)])(A).flatMap(a=>P(B).map(b=>a[1]*b[1]+'x^'+(a[2]- -b[2]))).join` + `
console.log( f('5x^4')('3x^23') )
console.log( f('6x^2 + 7x^1 + -2x^0')('1x^2 + -2x^3') )
console.log( f('3x^1 + 5x^2 + 2x^4 + 3x^0')('3x^0') )
console.log( f('4x^3 + -2x^14 + 54x^28 + -4x^5')('-0x^7') )
console.log( f('4x^3 + -2x^4 + 0x^255 + -4x^5')('-3x^4 + 2x^2') )
$endgroup$
$begingroup$
split' + ' => split'+'
to save 2 bytes
$endgroup$
– Luis felipe De jesus Munoz
4 hours ago
$begingroup$
@Arnauld Seems fine without them
$endgroup$
– Embodiment of Ignorance
2 hours ago
$begingroup$
@EmbodimentofIgnorance My bad, I misread Luis' comment. I thought it was about thejoin
.
$endgroup$
– Arnauld
2 hours ago
add a comment |
$begingroup$
JavaScript, 112 bytes
I found three alternatives with the same length. Call with currying syntax.
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(a=>P(B).map(b=>a[0]*b[0]+'x^'+(a[1]- -b[1]))).join` + `
f=
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(a=>P(B).map(b=>a[0]*b[0]+'x^'+(a[1]- -b[1]))).join` + `
console.log( f('5x^4')('3x^23') )
console.log( f('6x^2 + 7x^1 + -2x^0')('1x^2 + -2x^3') )
console.log( f('3x^1 + 5x^2 + 2x^4 + 3x^0')('3x^0') )
console.log( f('4x^3 + -2x^14 + 54x^28 + -4x^5')('-0x^7') )
console.log( f('4x^3 + -2x^4 + 0x^255 + -4x^5')('-3x^4 + 2x^2') )
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(([c,e])=>P(B).map(([C,E])=>c*C+'x^'+(e- -E))).join` + `
f=
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(([c,e])=>P(B).map(([C,E])=>c*C+'x^'+(e- -E))).join` + `
console.log( f('5x^4')('3x^23') )
console.log( f('6x^2 + 7x^1 + -2x^0')('1x^2 + -2x^3') )
console.log( f('3x^1 + 5x^2 + 2x^4 + 3x^0')('3x^0') )
console.log( f('4x^3 + -2x^14 + 54x^28 + -4x^5')('-0x^7') )
console.log( f('4x^3 + -2x^4 + 0x^255 + -4x^5')('-3x^4 + 2x^2') )
A=>B=>(P=x=>[...x.matchAll(/(S+)x.(S+)/g)])(A).flatMap(a=>P(B).map(b=>a[1]*b[1]+'x^'+(a[2]- -b[2]))).join` + `
f=
A=>B=>(P=x=>[...x.matchAll(/(S+)x.(S+)/g)])(A).flatMap(a=>P(B).map(b=>a[1]*b[1]+'x^'+(a[2]- -b[2]))).join` + `
console.log( f('5x^4')('3x^23') )
console.log( f('6x^2 + 7x^1 + -2x^0')('1x^2 + -2x^3') )
console.log( f('3x^1 + 5x^2 + 2x^4 + 3x^0')('3x^0') )
console.log( f('4x^3 + -2x^14 + 54x^28 + -4x^5')('-0x^7') )
console.log( f('4x^3 + -2x^4 + 0x^255 + -4x^5')('-3x^4 + 2x^2') )
$endgroup$
$begingroup$
split' + ' => split'+'
to save 2 bytes
$endgroup$
– Luis felipe De jesus Munoz
4 hours ago
$begingroup$
@Arnauld Seems fine without them
$endgroup$
– Embodiment of Ignorance
2 hours ago
$begingroup$
@EmbodimentofIgnorance My bad, I misread Luis' comment. I thought it was about thejoin
.
$endgroup$
– Arnauld
2 hours ago
add a comment |
$begingroup$
JavaScript, 112 bytes
I found three alternatives with the same length. Call with currying syntax.
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(a=>P(B).map(b=>a[0]*b[0]+'x^'+(a[1]- -b[1]))).join` + `
f=
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(a=>P(B).map(b=>a[0]*b[0]+'x^'+(a[1]- -b[1]))).join` + `
console.log( f('5x^4')('3x^23') )
console.log( f('6x^2 + 7x^1 + -2x^0')('1x^2 + -2x^3') )
console.log( f('3x^1 + 5x^2 + 2x^4 + 3x^0')('3x^0') )
console.log( f('4x^3 + -2x^14 + 54x^28 + -4x^5')('-0x^7') )
console.log( f('4x^3 + -2x^4 + 0x^255 + -4x^5')('-3x^4 + 2x^2') )
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(([c,e])=>P(B).map(([C,E])=>c*C+'x^'+(e- -E))).join` + `
f=
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(([c,e])=>P(B).map(([C,E])=>c*C+'x^'+(e- -E))).join` + `
console.log( f('5x^4')('3x^23') )
console.log( f('6x^2 + 7x^1 + -2x^0')('1x^2 + -2x^3') )
console.log( f('3x^1 + 5x^2 + 2x^4 + 3x^0')('3x^0') )
console.log( f('4x^3 + -2x^14 + 54x^28 + -4x^5')('-0x^7') )
console.log( f('4x^3 + -2x^4 + 0x^255 + -4x^5')('-3x^4 + 2x^2') )
A=>B=>(P=x=>[...x.matchAll(/(S+)x.(S+)/g)])(A).flatMap(a=>P(B).map(b=>a[1]*b[1]+'x^'+(a[2]- -b[2]))).join` + `
f=
A=>B=>(P=x=>[...x.matchAll(/(S+)x.(S+)/g)])(A).flatMap(a=>P(B).map(b=>a[1]*b[1]+'x^'+(a[2]- -b[2]))).join` + `
console.log( f('5x^4')('3x^23') )
console.log( f('6x^2 + 7x^1 + -2x^0')('1x^2 + -2x^3') )
console.log( f('3x^1 + 5x^2 + 2x^4 + 3x^0')('3x^0') )
console.log( f('4x^3 + -2x^14 + 54x^28 + -4x^5')('-0x^7') )
console.log( f('4x^3 + -2x^4 + 0x^255 + -4x^5')('-3x^4 + 2x^2') )
$endgroup$
JavaScript, 112 bytes
I found three alternatives with the same length. Call with currying syntax.
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(a=>P(B).map(b=>a[0]*b[0]+'x^'+(a[1]- -b[1]))).join` + `
f=
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(a=>P(B).map(b=>a[0]*b[0]+'x^'+(a[1]- -b[1]))).join` + `
console.log( f('5x^4')('3x^23') )
console.log( f('6x^2 + 7x^1 + -2x^0')('1x^2 + -2x^3') )
console.log( f('3x^1 + 5x^2 + 2x^4 + 3x^0')('3x^0') )
console.log( f('4x^3 + -2x^14 + 54x^28 + -4x^5')('-0x^7') )
console.log( f('4x^3 + -2x^4 + 0x^255 + -4x^5')('-3x^4 + 2x^2') )
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(([c,e])=>P(B).map(([C,E])=>c*C+'x^'+(e- -E))).join` + `
f=
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(([c,e])=>P(B).map(([C,E])=>c*C+'x^'+(e- -E))).join` + `
console.log( f('5x^4')('3x^23') )
console.log( f('6x^2 + 7x^1 + -2x^0')('1x^2 + -2x^3') )
console.log( f('3x^1 + 5x^2 + 2x^4 + 3x^0')('3x^0') )
console.log( f('4x^3 + -2x^14 + 54x^28 + -4x^5')('-0x^7') )
console.log( f('4x^3 + -2x^4 + 0x^255 + -4x^5')('-3x^4 + 2x^2') )
A=>B=>(P=x=>[...x.matchAll(/(S+)x.(S+)/g)])(A).flatMap(a=>P(B).map(b=>a[1]*b[1]+'x^'+(a[2]- -b[2]))).join` + `
f=
A=>B=>(P=x=>[...x.matchAll(/(S+)x.(S+)/g)])(A).flatMap(a=>P(B).map(b=>a[1]*b[1]+'x^'+(a[2]- -b[2]))).join` + `
console.log( f('5x^4')('3x^23') )
console.log( f('6x^2 + 7x^1 + -2x^0')('1x^2 + -2x^3') )
console.log( f('3x^1 + 5x^2 + 2x^4 + 3x^0')('3x^0') )
console.log( f('4x^3 + -2x^14 + 54x^28 + -4x^5')('-0x^7') )
console.log( f('4x^3 + -2x^4 + 0x^255 + -4x^5')('-3x^4 + 2x^2') )
f=
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(a=>P(B).map(b=>a[0]*b[0]+'x^'+(a[1]- -b[1]))).join` + `
console.log( f('5x^4')('3x^23') )
console.log( f('6x^2 + 7x^1 + -2x^0')('1x^2 + -2x^3') )
console.log( f('3x^1 + 5x^2 + 2x^4 + 3x^0')('3x^0') )
console.log( f('4x^3 + -2x^14 + 54x^28 + -4x^5')('-0x^7') )
console.log( f('4x^3 + -2x^4 + 0x^255 + -4x^5')('-3x^4 + 2x^2') )
f=
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(a=>P(B).map(b=>a[0]*b[0]+'x^'+(a[1]- -b[1]))).join` + `
console.log( f('5x^4')('3x^23') )
console.log( f('6x^2 + 7x^1 + -2x^0')('1x^2 + -2x^3') )
console.log( f('3x^1 + 5x^2 + 2x^4 + 3x^0')('3x^0') )
console.log( f('4x^3 + -2x^14 + 54x^28 + -4x^5')('-0x^7') )
console.log( f('4x^3 + -2x^4 + 0x^255 + -4x^5')('-3x^4 + 2x^2') )
f=
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(([c,e])=>P(B).map(([C,E])=>c*C+'x^'+(e- -E))).join` + `
console.log( f('5x^4')('3x^23') )
console.log( f('6x^2 + 7x^1 + -2x^0')('1x^2 + -2x^3') )
console.log( f('3x^1 + 5x^2 + 2x^4 + 3x^0')('3x^0') )
console.log( f('4x^3 + -2x^14 + 54x^28 + -4x^5')('-0x^7') )
console.log( f('4x^3 + -2x^4 + 0x^255 + -4x^5')('-3x^4 + 2x^2') )
f=
A=>B=>(P=x=>x.split` + `.map(x=>x.split`x^`))(A).flatMap(([c,e])=>P(B).map(([C,E])=>c*C+'x^'+(e- -E))).join` + `
console.log( f('5x^4')('3x^23') )
console.log( f('6x^2 + 7x^1 + -2x^0')('1x^2 + -2x^3') )
console.log( f('3x^1 + 5x^2 + 2x^4 + 3x^0')('3x^0') )
console.log( f('4x^3 + -2x^14 + 54x^28 + -4x^5')('-0x^7') )
console.log( f('4x^3 + -2x^4 + 0x^255 + -4x^5')('-3x^4 + 2x^2') )
f=
A=>B=>(P=x=>[...x.matchAll(/(S+)x.(S+)/g)])(A).flatMap(a=>P(B).map(b=>a[1]*b[1]+'x^'+(a[2]- -b[2]))).join` + `
console.log( f('5x^4')('3x^23') )
console.log( f('6x^2 + 7x^1 + -2x^0')('1x^2 + -2x^3') )
console.log( f('3x^1 + 5x^2 + 2x^4 + 3x^0')('3x^0') )
console.log( f('4x^3 + -2x^14 + 54x^28 + -4x^5')('-0x^7') )
console.log( f('4x^3 + -2x^4 + 0x^255 + -4x^5')('-3x^4 + 2x^2') )
f=
A=>B=>(P=x=>[...x.matchAll(/(S+)x.(S+)/g)])(A).flatMap(a=>P(B).map(b=>a[1]*b[1]+'x^'+(a[2]- -b[2]))).join` + `
console.log( f('5x^4')('3x^23') )
console.log( f('6x^2 + 7x^1 + -2x^0')('1x^2 + -2x^3') )
console.log( f('3x^1 + 5x^2 + 2x^4 + 3x^0')('3x^0') )
console.log( f('4x^3 + -2x^14 + 54x^28 + -4x^5')('-0x^7') )
console.log( f('4x^3 + -2x^4 + 0x^255 + -4x^5')('-3x^4 + 2x^2') )
edited 5 hours ago
answered 5 hours ago
darrylyeodarrylyeo
5,2641034
5,2641034
$begingroup$
split' + ' => split'+'
to save 2 bytes
$endgroup$
– Luis felipe De jesus Munoz
4 hours ago
$begingroup$
@Arnauld Seems fine without them
$endgroup$
– Embodiment of Ignorance
2 hours ago
$begingroup$
@EmbodimentofIgnorance My bad, I misread Luis' comment. I thought it was about thejoin
.
$endgroup$
– Arnauld
2 hours ago
add a comment |
$begingroup$
split' + ' => split'+'
to save 2 bytes
$endgroup$
– Luis felipe De jesus Munoz
4 hours ago
$begingroup$
@Arnauld Seems fine without them
$endgroup$
– Embodiment of Ignorance
2 hours ago
$begingroup$
@EmbodimentofIgnorance My bad, I misread Luis' comment. I thought it was about thejoin
.
$endgroup$
– Arnauld
2 hours ago
$begingroup$
split' + ' => split'+'
to save 2 bytes$endgroup$
– Luis felipe De jesus Munoz
4 hours ago
$begingroup$
split' + ' => split'+'
to save 2 bytes$endgroup$
– Luis felipe De jesus Munoz
4 hours ago
$begingroup$
@Arnauld Seems fine without them
$endgroup$
– Embodiment of Ignorance
2 hours ago
$begingroup$
@Arnauld Seems fine without them
$endgroup$
– Embodiment of Ignorance
2 hours ago
$begingroup$
@EmbodimentofIgnorance My bad, I misread Luis' comment. I thought it was about the
join
.$endgroup$
– Arnauld
2 hours ago
$begingroup$
@EmbodimentofIgnorance My bad, I misread Luis' comment. I thought it was about the
join
.$endgroup$
– Arnauld
2 hours ago
add a comment |
$begingroup$
SNOBOL4 (CSNOBOL4), 192 176 bytes
P =INPUT
Q =INPUT
D =SPAN(1234567890)
P P D . K ARB D . W REM . P :F(O)
B =Q
B B D . C ARB D . E REM . B :F(P)
O =O ' + ' K * C 'x^' W + E :(B)
O O POS(3) REM . OUTPUT
END
Try it online!
Explanation to come when I'm off my phone.
$endgroup$
add a comment |
$begingroup$
SNOBOL4 (CSNOBOL4), 192 176 bytes
P =INPUT
Q =INPUT
D =SPAN(1234567890)
P P D . K ARB D . W REM . P :F(O)
B =Q
B B D . C ARB D . E REM . B :F(P)
O =O ' + ' K * C 'x^' W + E :(B)
O O POS(3) REM . OUTPUT
END
Try it online!
Explanation to come when I'm off my phone.
$endgroup$
add a comment |
$begingroup$
SNOBOL4 (CSNOBOL4), 192 176 bytes
P =INPUT
Q =INPUT
D =SPAN(1234567890)
P P D . K ARB D . W REM . P :F(O)
B =Q
B B D . C ARB D . E REM . B :F(P)
O =O ' + ' K * C 'x^' W + E :(B)
O O POS(3) REM . OUTPUT
END
Try it online!
Explanation to come when I'm off my phone.
$endgroup$
SNOBOL4 (CSNOBOL4), 192 176 bytes
P =INPUT
Q =INPUT
D =SPAN(1234567890)
P P D . K ARB D . W REM . P :F(O)
B =Q
B B D . C ARB D . E REM . B :F(P)
O =O ' + ' K * C 'x^' W + E :(B)
O O POS(3) REM . OUTPUT
END
Try it online!
Explanation to come when I'm off my phone.
edited 1 hour ago
answered 6 hours ago
GiuseppeGiuseppe
17.7k31153
17.7k31153
add a comment |
add a comment |
$begingroup$
JavaScript (Babel Node), 118 bytes
Takes input as (a)(b)
.
a=>b=>(g=s=>[...s.matchAll(/(-?d+)x.(d+)/g)])(a).flatMap(([_,x,p])=>g(b).map(([_,X,P])=>x*X+'x^'+-(-p-P))).join` + `
Try it online!
$endgroup$
add a comment |
$begingroup$
JavaScript (Babel Node), 118 bytes
Takes input as (a)(b)
.
a=>b=>(g=s=>[...s.matchAll(/(-?d+)x.(d+)/g)])(a).flatMap(([_,x,p])=>g(b).map(([_,X,P])=>x*X+'x^'+-(-p-P))).join` + `
Try it online!
$endgroup$
add a comment |
$begingroup$
JavaScript (Babel Node), 118 bytes
Takes input as (a)(b)
.
a=>b=>(g=s=>[...s.matchAll(/(-?d+)x.(d+)/g)])(a).flatMap(([_,x,p])=>g(b).map(([_,X,P])=>x*X+'x^'+-(-p-P))).join` + `
Try it online!
$endgroup$
JavaScript (Babel Node), 118 bytes
Takes input as (a)(b)
.
a=>b=>(g=s=>[...s.matchAll(/(-?d+)x.(d+)/g)])(a).flatMap(([_,x,p])=>g(b).map(([_,X,P])=>x*X+'x^'+-(-p-P))).join` + `
Try it online!
answered 6 hours ago
ArnauldArnauld
80.7k797334
80.7k797334
add a comment |
add a comment |
$begingroup$
Haskell, 133 bytes
f""=[]
f t|[(a,_:_:u)]<-reads t,[(i,v)]<-reads u=(a,i):f(drop 3v)
p!q=drop 3$do(a,i)<-f p;(b,j)<-f q;" + "++shows(a*b)"x^"++show(i+j)
Try it online!
f
parses a polynomial from a string, !
multiplies two of them and formats the result.
$endgroup$
add a comment |
$begingroup$
Haskell, 133 bytes
f""=[]
f t|[(a,_:_:u)]<-reads t,[(i,v)]<-reads u=(a,i):f(drop 3v)
p!q=drop 3$do(a,i)<-f p;(b,j)<-f q;" + "++shows(a*b)"x^"++show(i+j)
Try it online!
f
parses a polynomial from a string, !
multiplies two of them and formats the result.
$endgroup$
add a comment |
$begingroup$
Haskell, 133 bytes
f""=[]
f t|[(a,_:_:u)]<-reads t,[(i,v)]<-reads u=(a,i):f(drop 3v)
p!q=drop 3$do(a,i)<-f p;(b,j)<-f q;" + "++shows(a*b)"x^"++show(i+j)
Try it online!
f
parses a polynomial from a string, !
multiplies two of them and formats the result.
$endgroup$
Haskell, 133 bytes
f""=[]
f t|[(a,_:_:u)]<-reads t,[(i,v)]<-reads u=(a,i):f(drop 3v)
p!q=drop 3$do(a,i)<-f p;(b,j)<-f q;" + "++shows(a*b)"x^"++show(i+j)
Try it online!
f
parses a polynomial from a string, !
multiplies two of them and formats the result.
answered 5 hours ago
LynnLynn
50.7k898233
50.7k898233
add a comment |
add a comment |
$begingroup$
Python 2, 193 bytes
import re
f=re.finditer
lambda a,b:' + '.join(' + '.join(`int(m.group(1))*int(n.group(1))`+'x^'+`int(m.group(2))+int(n.group(2))`for n in f('(-?d+)x^(d+)',b))for m in f('(-?d+)x^(d+)',a))
Try it online!
Side note: First time doing a code golf challenge, so sorry if the attempt sucks haha
New contributor
$endgroup$
1
$begingroup$
Welcome to PPCG! I'm not much of a python programmer, but if your submission is longer than a SNOBOL one, there's probably room for improvement, heheh. Perhaps you can find help at Tips for Golfing in Python or Tips for Golfing in <all languages>! Hope you enjoy the time you spend here :-)
$endgroup$
– Giuseppe
4 hours ago
add a comment |
$begingroup$
Python 2, 193 bytes
import re
f=re.finditer
lambda a,b:' + '.join(' + '.join(`int(m.group(1))*int(n.group(1))`+'x^'+`int(m.group(2))+int(n.group(2))`for n in f('(-?d+)x^(d+)',b))for m in f('(-?d+)x^(d+)',a))
Try it online!
Side note: First time doing a code golf challenge, so sorry if the attempt sucks haha
New contributor
$endgroup$
1
$begingroup$
Welcome to PPCG! I'm not much of a python programmer, but if your submission is longer than a SNOBOL one, there's probably room for improvement, heheh. Perhaps you can find help at Tips for Golfing in Python or Tips for Golfing in <all languages>! Hope you enjoy the time you spend here :-)
$endgroup$
– Giuseppe
4 hours ago
add a comment |
$begingroup$
Python 2, 193 bytes
import re
f=re.finditer
lambda a,b:' + '.join(' + '.join(`int(m.group(1))*int(n.group(1))`+'x^'+`int(m.group(2))+int(n.group(2))`for n in f('(-?d+)x^(d+)',b))for m in f('(-?d+)x^(d+)',a))
Try it online!
Side note: First time doing a code golf challenge, so sorry if the attempt sucks haha
New contributor
$endgroup$
Python 2, 193 bytes
import re
f=re.finditer
lambda a,b:' + '.join(' + '.join(`int(m.group(1))*int(n.group(1))`+'x^'+`int(m.group(2))+int(n.group(2))`for n in f('(-?d+)x^(d+)',b))for m in f('(-?d+)x^(d+)',a))
Try it online!
Side note: First time doing a code golf challenge, so sorry if the attempt sucks haha
New contributor
New contributor
answered 4 hours ago
GotCubesGotCubes
1
1
New contributor
New contributor
1
$begingroup$
Welcome to PPCG! I'm not much of a python programmer, but if your submission is longer than a SNOBOL one, there's probably room for improvement, heheh. Perhaps you can find help at Tips for Golfing in Python or Tips for Golfing in <all languages>! Hope you enjoy the time you spend here :-)
$endgroup$
– Giuseppe
4 hours ago
add a comment |
1
$begingroup$
Welcome to PPCG! I'm not much of a python programmer, but if your submission is longer than a SNOBOL one, there's probably room for improvement, heheh. Perhaps you can find help at Tips for Golfing in Python or Tips for Golfing in <all languages>! Hope you enjoy the time you spend here :-)
$endgroup$
– Giuseppe
4 hours ago
1
1
$begingroup$
Welcome to PPCG! I'm not much of a python programmer, but if your submission is longer than a SNOBOL one, there's probably room for improvement, heheh. Perhaps you can find help at Tips for Golfing in Python or Tips for Golfing in <all languages>! Hope you enjoy the time you spend here :-)
$endgroup$
– Giuseppe
4 hours ago
$begingroup$
Welcome to PPCG! I'm not much of a python programmer, but if your submission is longer than a SNOBOL one, there's probably room for improvement, heheh. Perhaps you can find help at Tips for Golfing in Python or Tips for Golfing in <all languages>! Hope you enjoy the time you spend here :-)
$endgroup$
– Giuseppe
4 hours ago
add a comment |
$begingroup$
Retina, 110 bytes
SS+(?=.*n(.+))
$1#$&
|" + "L$v` (-?)(d+)x.(d+).*?#(-?)(d+)x.(d+)
$1$4$.($2*$5*)x^$.($3*_$6*
--|-(0)
$1
Try it online! Explanation:
SS+(?=.*n(.+))
$1#$&
Prefix each term in the first input with a #
, a copy of the second input, and a space. This means that all of the terms in copies of the second input are preceded by a space and none of the terms from the first input are.
|" + "L$v` (-?)(d+)x.(d+).*?#(-?)(d+)x.(d+)
$1$4$.($2*$5*)x^$.($3*_$6*
Match all of the copies of terms in the second input and their corresponding term from the first input. Concatenate any -
signs, multiply the coefficients, and add the indices. Finally join all of the resulting substitutions with the string +
.
--|-(0)
$1
Delete any pairs of -
s and convert -0
to 0
.
$endgroup$
add a comment |
$begingroup$
Retina, 110 bytes
SS+(?=.*n(.+))
$1#$&
|" + "L$v` (-?)(d+)x.(d+).*?#(-?)(d+)x.(d+)
$1$4$.($2*$5*)x^$.($3*_$6*
--|-(0)
$1
Try it online! Explanation:
SS+(?=.*n(.+))
$1#$&
Prefix each term in the first input with a #
, a copy of the second input, and a space. This means that all of the terms in copies of the second input are preceded by a space and none of the terms from the first input are.
|" + "L$v` (-?)(d+)x.(d+).*?#(-?)(d+)x.(d+)
$1$4$.($2*$5*)x^$.($3*_$6*
Match all of the copies of terms in the second input and their corresponding term from the first input. Concatenate any -
signs, multiply the coefficients, and add the indices. Finally join all of the resulting substitutions with the string +
.
--|-(0)
$1
Delete any pairs of -
s and convert -0
to 0
.
$endgroup$
add a comment |
$begingroup$
Retina, 110 bytes
SS+(?=.*n(.+))
$1#$&
|" + "L$v` (-?)(d+)x.(d+).*?#(-?)(d+)x.(d+)
$1$4$.($2*$5*)x^$.($3*_$6*
--|-(0)
$1
Try it online! Explanation:
SS+(?=.*n(.+))
$1#$&
Prefix each term in the first input with a #
, a copy of the second input, and a space. This means that all of the terms in copies of the second input are preceded by a space and none of the terms from the first input are.
|" + "L$v` (-?)(d+)x.(d+).*?#(-?)(d+)x.(d+)
$1$4$.($2*$5*)x^$.($3*_$6*
Match all of the copies of terms in the second input and their corresponding term from the first input. Concatenate any -
signs, multiply the coefficients, and add the indices. Finally join all of the resulting substitutions with the string +
.
--|-(0)
$1
Delete any pairs of -
s and convert -0
to 0
.
$endgroup$
Retina, 110 bytes
SS+(?=.*n(.+))
$1#$&
|" + "L$v` (-?)(d+)x.(d+).*?#(-?)(d+)x.(d+)
$1$4$.($2*$5*)x^$.($3*_$6*
--|-(0)
$1
Try it online! Explanation:
SS+(?=.*n(.+))
$1#$&
Prefix each term in the first input with a #
, a copy of the second input, and a space. This means that all of the terms in copies of the second input are preceded by a space and none of the terms from the first input are.
|" + "L$v` (-?)(d+)x.(d+).*?#(-?)(d+)x.(d+)
$1$4$.($2*$5*)x^$.($3*_$6*
Match all of the copies of terms in the second input and their corresponding term from the first input. Concatenate any -
signs, multiply the coefficients, and add the indices. Finally join all of the resulting substitutions with the string +
.
--|-(0)
$1
Delete any pairs of -
s and convert -0
to 0
.
answered 4 hours ago
NeilNeil
82.7k745179
82.7k745179
add a comment |
add a comment |
$begingroup$
Jelly, 28 bytes
ṣ”+ṣ”xV$€)p/ZPSƭ€j⁾x^Ʋ€j“ +
Try it online!
Full program. Takes the two polynomials as a list of two strings.
$endgroup$
add a comment |
$begingroup$
Jelly, 28 bytes
ṣ”+ṣ”xV$€)p/ZPSƭ€j⁾x^Ʋ€j“ +
Try it online!
Full program. Takes the two polynomials as a list of two strings.
$endgroup$
add a comment |
$begingroup$
Jelly, 28 bytes
ṣ”+ṣ”xV$€)p/ZPSƭ€j⁾x^Ʋ€j“ +
Try it online!
Full program. Takes the two polynomials as a list of two strings.
$endgroup$
Jelly, 28 bytes
ṣ”+ṣ”xV$€)p/ZPSƭ€j⁾x^Ʋ€j“ +
Try it online!
Full program. Takes the two polynomials as a list of two strings.
answered 2 hours ago
Erik the OutgolferErik the Outgolfer
33k429106
33k429106
add a comment |
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 192 190 188 bytes
n=>m=>string.Join(g="+",from a in n.Split(g)from b in m.Split(g)select f(a.Split(p="x^")[0])*f(b.Split(p)[0])+p+(f(a.Split(p)[1])+f(b.Split(p)[1])));Func<string,int>f=int.Parse;string p,g;
Query syntax seems to be a byte shorter than method syntax.
At least I beat SNOBOL.
Try it online!
$endgroup$
$begingroup$
Sorry, not beating SNOBOL any more!
$endgroup$
– Giuseppe
1 hour ago
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 192 190 188 bytes
n=>m=>string.Join(g="+",from a in n.Split(g)from b in m.Split(g)select f(a.Split(p="x^")[0])*f(b.Split(p)[0])+p+(f(a.Split(p)[1])+f(b.Split(p)[1])));Func<string,int>f=int.Parse;string p,g;
Query syntax seems to be a byte shorter than method syntax.
At least I beat SNOBOL.
Try it online!
$endgroup$
$begingroup$
Sorry, not beating SNOBOL any more!
$endgroup$
– Giuseppe
1 hour ago
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 192 190 188 bytes
n=>m=>string.Join(g="+",from a in n.Split(g)from b in m.Split(g)select f(a.Split(p="x^")[0])*f(b.Split(p)[0])+p+(f(a.Split(p)[1])+f(b.Split(p)[1])));Func<string,int>f=int.Parse;string p,g;
Query syntax seems to be a byte shorter than method syntax.
At least I beat SNOBOL.
Try it online!
$endgroup$
C# (Visual C# Interactive Compiler), 192 190 188 bytes
n=>m=>string.Join(g="+",from a in n.Split(g)from b in m.Split(g)select f(a.Split(p="x^")[0])*f(b.Split(p)[0])+p+(f(a.Split(p)[1])+f(b.Split(p)[1])));Func<string,int>f=int.Parse;string p,g;
Query syntax seems to be a byte shorter than method syntax.
At least I beat SNOBOL.
Try it online!
edited 2 hours ago
answered 2 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
2,916127
2,916127
$begingroup$
Sorry, not beating SNOBOL any more!
$endgroup$
– Giuseppe
1 hour ago
add a comment |
$begingroup$
Sorry, not beating SNOBOL any more!
$endgroup$
– Giuseppe
1 hour ago
$begingroup$
Sorry, not beating SNOBOL any more!
$endgroup$
– Giuseppe
1 hour ago
$begingroup$
Sorry, not beating SNOBOL any more!
$endgroup$
– Giuseppe
1 hour ago
add a comment |
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
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$begingroup$
related
$endgroup$
– H.PWiz
7 hours ago
1
$begingroup$
@LuisfelipeDejesusMunoz I imagine not. Parsing is an integral part of the challenge and the OP says -- "It should be noted that any method of taking in the two polynomials is valid, provided that both are read in as strings conforming to the above format." (emphasis added)
$endgroup$
– Giuseppe
6 hours ago
$begingroup$
Your regex is wrong:
^
should be^
.$endgroup$
– Erik the Outgolfer
6 hours ago