Extract the Coefficients of a Linear Equation

Question

Given a linear equation mx+n, return n and m. However, the equation may be in any of the following formats:

5x+2 -> 5 2
5x-2 -> 5 -2
5x   -> 5 0
-2   -> 0 -2
x    -> 1 0

Here are the formats, where each ? stands for a positive integer written without leading zeroes

?x+?
?x-?
-?x+?
-?x-?
x+?
x-?
?
-?
?x
-?x
x

All of these cases must be handled.

Specifications:

• You can assume that the equation is in one of the above formats, i.e. it matches the regex ^(-?[123456789]\d*)?x([+-][123456789]\d*)?|-?[123456789]\d*$.

Test Cases:

-2x+3 -> -2 3
44x   -> 44 0
-123  -> 0 -123
x     -> 1 0
-1x   -> -1 0

Answer 1

Python 2, 55 bytes

j=1j
c=eval(input().replace(*'xj'))
print c.imag,c.real

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Uses Python's built-in code evaluation. The input is formatted like a complex number by replacing x withj, Python's complex unit. Because only literals like 2j are recognized, but not j or -j, the variable j is assigned as 1j to cover those.

Unfortunately, Python doesn't seem to have a built-in to convert a complex number to a pair of reals.

Answer 2

Mathematica, 16 bytes

Inspired by xnor's Python 2 answer:

ReIm[I#/.x->-I]&

Takes input as a literal expression (not a string), and returns a pair of numbers. It works by making x a complex number then taking the real and imaginary parts — the only non-obvious bit is multiplying by i to start with, to get the output in the right order.

We can also use

ReIm[x=-I;I#]&

for 14 bytes (tied with Jelly!), but setting x to equal -i before taking the input, instead of just replacing x afterwards, feels like cheating…

Answer 3

JavaScript (ES6), 53 48 bytes

s=>([a,b]=s.split(/x\+?/),1/b?[a||1,b||0]:[0,s])

Test cases

let f =

s=>([a,b]=s.split(/x\+?/),1/b?[a||1,b||0]:[0,s])

console.log(f("-2x+3")) // -> -2 3
console.log(f("44x"))   // -> 44 0
console.log(f("-123"))  // -> 0 -123
console.log(f("x"))     // -> 1 0
console.log(f("-1x"))   // -> -1 0

Answer 4

sed, 44 42 bytes

s:+::
s:x: :
t
s:^:0 :
:
s:^ :1 :
s: $: 0:

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I/O: one per line.

-2 bytes thanks to @KritixiLithos.

Answer 5

Jelly, 18 14 bytes

”x;ṣ”xVṫ-µ¬ṂW+

Test suite at Try it online!

How?

”x;ṣ”xVṫ-µ¬ṂW+ - Main link: list of characters s  examples: "5x-2"         "x"       "-123"
”x             - literal 'x'
  ;            - concatenate with s                        "x5x-2"        "xx"      "x-123"
   ṣ”x         - split on 'x's                        ["","5","-2]  ["","",""]  ["","-123"]
      V        - evaluate as Jelly code (vectorises)      [0,5,-2]     [0,0,0]     [0,-123]
       ṫ-      - tail from index -1 inclusive               [5,-2]       [0,0]     [0,-123]
         µ     - monadic chain separation (call that z)
          ¬    - not z                                       [0,0]       [1,1]        [1,0]
           Ṃ   - minimum                                         0           1            0
            W  - wrap in a list                                [0]         [1]          [0]
             + - add to z (vectorises)                      [5,-2]       [1,0]     [0,-123]

Answer 6

PHP, 73 Bytes

preg_match("#((.*)x)?\+?(.*)#",$argn,$t);echo$t[1]?$t[2]?:1:0," ",+$t[3];

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PHP, 91 Bytes

works also with -x

preg_match("#((-)?(\d+)?x)?\+?(-?\d+)?#",$argn,$t);echo$t[2],$t[1]?$t[3]?:1:0," ",$t[4]?:0;

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Answer 7

Mathematica, 23 bytes

Coefficient[#,x,{1,0}]&

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copy and paste with ctrl-v this code

Coefficient[#,x,{1,0}]&[-2x+3]

and press shift+enter to run

Answer 8

Python 2, 71 bytes

lambda s:([0,]+[int(x or`1-i`)for i,x in enumerate(s.split('x'))])[-2:]

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Answer 9

Mathematica, 33 bytes

#~CoefficientList~x/.{a_}:>{a,0}&

Pure function taking an expression in the expected format (note: not a string, but a pure expression like -2x+3), and returning an ordered pair of integers with the constant coefficient appearing first (for example, -2x+3 returns {3,-2}).

The builtin CoefficientList (which works for polynomials of any degree) does the heavy lifting; its default behavior is to return constants as length-1 lists, so /.{a_}:>{a,0} overrides that and makes the 0 coefficient of x appear explicitly.

Answer 10

Retina, 38 bytes

x$
x+0
^[^+-]*$
0x+$+
^[^-\d]
1$+
-
+-

Definitely can be improved.

Possibly stretching how we are allowed to format output. It just outputs a complete equation, as in n and m separated by x+

Try it online!

Answer 11

Octave, 45 bytes

@(a)[imag(u=eval(strrep(a,'x','j'))),real(u)]

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Evaluate the expression, with x replaced by complex j. Octave is very flexible when it comes to complex expressions, with j,1j,i and 1i all equivalent (with the advantage that 1j and 1i cannot be overwritten by variables, but that's irrelevant for this challenge). Inline expression is used to build a list of [imag(u) real(u)] with u the complex number.

I initially wanted to try a more idiomatic approach, using the built-in sym2poly. For example (already at 47 bytes to make sure 0*x is handled):

@(a)sym2poly([strrep(a,'x','*x') '+x^2']))(2:3)

This one however failed on the case of x without coefficient. In the end a real/complex setup like xnor turned out to be optimal.

Answer 12

MATL, 12 bytes

120'i'ZtU&Zj

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Port of my Octave answer, but of course adapted for compactness in MATL. Explanation:

120'i'Zt     % Replace character 120 ('x') by 'i' in input (implicit).
             % (Used 120 instead of 'x' to save a separator between 'x' and 'i')
        U    % Convert string to complex number
         &Zj % Split complex number in real and imaginary part. Implicit display.

Answer 13

C (gcc),145 138 112 bytes

f(char*s){int a,b,i=0;sscanf(s,"%dx%d",&a,&b);if(L!=0)i=(int)(L-s);printf("%d %d",s[i]=='x'?a:0,s[i]=='x'?b:a);}

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void f(char*s)
{
    int a,b,i=0;
    sscanf(s,"%dx%d",&a,&b);

    if(L!=0)
      i=(int)(L-s);

    printf("%d %d",s[i]=='x'?a:0,s[i]=='x'?b:a);
}

Can definitely be shortened, but dont see it now!