Given the roots of a polynomial separated by spaces as input, output the expanded form of the polynomial.
For example, the input
1 2
represents this equation:
(x-1)(x-2)
And should output:
x^2-3x+2
The exact format of output is not important, it can be:
1x^2+-3x^1+2x^0
or:
0 0 0
1x^3+0x^2+0x^1+0
or:
3 14 15 92
1x^4+-124x^3+3241x^2+-27954x^1+57960
eval and likes are disallowed.
ljU"1@_hX+]tn:Pqv'+%gx^%g'wYD
Input is an array with the roots.
EDITS:
X+ function has been removed, as Y+ includes its functionality. So in the above code replace X+ by Y+.YD function, w in the above code should be removed.The following link includes those changes.
This applies repeated convolution with terms of the form [1, -r] where r is a root.
l % push number 1
jU % take input string. Convert to number array
" % for each root r
1 % push number 1
@_ % push -r
h % concatenate horizontally
X+ % convolve. This gradually builds array of coefficients
] % end for each
tn:Pq % produce array [n-1,n-2,...,0], where n is the number of roots
v % concatenate vertically with array of coefficients
'+%gx^%g' % format string, sprintf-style
w % swap
YD % sprintf. Implicitly display
Æṛ‘Ė’Uj€“x^”j”+
This uses Æṛ to construct the coefficients of a monic polynomial with given roots. Try it online!
Æṛ‘Ė’Uj€“x^”j”+ Main link. Argument: A (list of roots)
Æṛ Yield the coefficients of a monic polynomial with roots A.
‘ Increment each root by 1.
Ė Enumerate the roots, yielding
[[1, coeff. of x^0 + 1], ... [n + 1, coeff. of x^n + 1]].
’ Decrement all involved integers, yielding
[[0, coeff. of x^0], ... [n, coeff. of x^n]].
U Upend to yield [[coeff. of x^0, 0], ... [coeff. of x^n, n]].
j€“x^” Join each pair, separating by 'x^'.
j”+ Join the pairs, separating by '+'.
1WW;ð0;_×µ/‘Ė’Uj€“x^”j”+
This uses no polynomial-related built-ins. Try it online!
1WW;ð0;_×µ/‘Ė’Uj€“x^”j”+ Main link. Argument: A (list of roots)
1WW Yield [[1]].
; Concatenate with A.
ð µ/ Reduce [[1]] + A by the following, dyadic chain:
0; Prepend a zero to the left argument (initially [1]).
This multiplies the left argument by "x".
× Take the product of both, unaltered arguments.
This multiplies the left argument by "r", where r is
the root specified in the right argument.
_ Subtract.
This computes the left argument multiplied by "x-r".
‘Ė’Uj€“x^”j”+ As before.
Anonymous function, input is an array of the roots.
Prints from lowest power first, so calling f[[1,2]] (assuming you assigned the function to f) returns the string "2x^0+-3x^1+1x^2".
->x{j=-1
x.map{|r|[-r,1]}.reduce{|a,b|q=a.map{|c|b=[0]+b
b.map{|e|e*c}[1..-1]}
q.pop.zip(*q).map{|e|(e-[p]).reduce(:+)}}.map{|e|"#{e}x^#{j+=1}"}.join('+')}
import functools
import operator
print([('+'.join(["x^"+str(len(R))]+[str(q)+"x^"+str(r)if r>0else"{0:g}".format(q)for r,q in enumerate([sum(functools.reduce(operator.mul,(-int(R[n])for n,m in enumerate(j)if int(m)==1),1)for j in[(len(R)-len(bin(i)[2:]))*'0'+bin(i)[2:]for i in range(1,2**len(R))]if sum(1-int(k) for k in j)==p)for p in range(len(R))]) ][::-1]))for R in[input().split()]][0])
Check This link, Will help understand the reason behind those two imports.
from operator import*, from functools import* save a few bytes
\$\endgroup\$
import functools,operator
\$\endgroup\$
Commented
Mar 26, 2016 at 22:11
l%r=zipWith(-)(0:l)$map(r*)l++[0]
f e='0':do(c,i)<-zip(foldl(%)[1]e)[0..];'+':show c++"x^"++show i
prints the lower powers first, with an additional 0+ at the start. for example:
>f [1,1]
"0+1x^0+-2x^1+1x^2"
The function computes the coefficients by progressively adding more roots, like convolutions, but without the builtin.
Then we use the list monad to implicitly concat all of the different monomials.
lambda N:prod(x-t for t in N).expand()
This defines an unnamed lambda that takes an iterable of roots as input and computes the product (x-x_n) for x_n in roots, then expands it.
Expand@Product[x-k,{k,#}]&
Mathematica has powerful polynomial builtins.
f = Expand@Product[x-k,{k,#}]&
f@{3, 14, 15, 92}
x^4 - 124 x^3 + 3241 x^2 - 27954 x + 57960
a=>a.map(x=>r.map((n,i)=>(r[i]-=x*a,a=n),++j,r.push(a=0)),r=[j=1])&&r.map(n=>n+`x^`+--j).join`+`
numpy.poly? \$\endgroup\$