

 This page uses content from Wikipedia. The original article was at Polynomial long division. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)

In algebra, polynomial long division is an algorithm for dividing a polynomial by another polynomial of the same or lower degree.
Let us suppose a polynomial is represented by a vector, 



x


{\displaystyle x}

 (i.e., an ordered collection of coefficients) so that the 



i


{\displaystyle i}

th element keeps the coefficient of 




x

i




{\displaystyle x^{i}}

, and the multiplication by a monomial is a shift of the vector's elements "towards right" (injecting ones from left) followed by a multiplication of each element by the coefficient of the monomial.

Then a pseudocode for the polynomial long division using the conventions described above could be:

degree(P):
  return the index of the last non-zero element of P;
         if all elements are 0, return -∞

polynomial_long_division(N, D) returns (q, r):
  // N, D, q, r are vectors
  if degree(D) < 0 then error
  q ← 0
  while degree(N) ≥ degree(D)
    d ← D shifted right by (degree(N) - degree(D))
    q(degree(N) - degree(D)) ← N(degree(N)) / d(degree(d))
    // by construction, degree(d) = degree(N) of course
    d ← d * q(degree(N) - degree(D))
    N ← N - d
  endwhile
  r ← N
  return (q, r)

Note: vector * scalar multiplies each element of the vector by the scalar; vectorA - vectorB subtracts each element of the vectorB from the element of the vectorA with "the same index". The vectors in the pseudocode are zero-based.

 Error handling (for allocations or for wrong inputs) is not mandatory.
 Conventions can be different; in particular, note that if the first coefficient in the vector is the highest power of x for the polynomial represented by the vector, then the algorithm becomes simpler.

Example for clarification


This example is from Wikipedia, but changed to show how the given pseudocode works.

      0    1    2    3
   ----------------------
N:  -42    0  -12    1        degree = 3
D:   -3    1    0    0        degree = 1

   d(N) - d(D) = 2, so let's shift D towards right by 2:

N:  -42    0  -12    1
d:    0    0   -3    1

   N(3)/d(3) = 1, so d is unchanged. Now remember that "shifting by 2"
   is like multiplying by x2, and the final multiplication
   (here by 1) is the coefficient of this monomial. Let's store this
   into q:
                               0     1     2
                              ---------------
                          q:   0     0     1

   now compute N - d, and let it be the "new" N, and let's loop

N:  -42    0   -9    0        degree = 2
D:   -3    1    0    0        degree = 1

   d(N) - d(D) = 1, right shift D by 1 and let it be d

N:  -42    0   -9    0
d:    0   -3    1    0        * -9/1 = -9

                          q:   0    -9     1

d:    0   27   -9    0        

   N ← N - d

N:  -42  -27    0    0        degree = 1
D:   -3    1    0    0        degree = 1

   looping again... d(N)-d(D)=0, so no shift is needed; we
   multiply D by -27 (= -27/1) storing the result in d, then

                          q:  -27   -9     1

   and

N:  -42  -27    0    0        -
d:   81  -27    0    0        =
N: -123    0    0    0        (last N)

    d(N) < d(D), so now r ← N, and the result is:

       0   1  2
   -------------
q:   -27  -9  1   →  x2 - 9x - 27
r:  -123   0  0   →          -123



Related task

   Polynomial derivative


let rec shift n l = if n <= 0 then l else shift (pred n) (l @ [0.0])
let rec pad n l = if n <= 0 then l else pad (pred n) (0.0 :: l)
let rec norm = function | 0.0 :: tl -> norm tl | x -> x
let deg l = List.length (norm l) - 1
 
let zip op p q =
  let d = (List.length p) - (List.length q) in
  List.map2 op (pad (-d) p) (pad d q)