Horner's method

Evaluate ${\displaystyle f(x)=2x^{3}-6x^{2}+2x-1}$ for ${\displaystyle x=3.}$

We use synthetic division as follows:

 x0│   x3    x2    x1    x0
 3 │   2    −6     2    −1
   │         6     0     6
   └────────────────────────
       2     0     2     5

The entries in the third row are the sum of those in the first two. Each entry in the second row is the product of the x-value (3 in this example) with the third-row entry immediately to the left. The entries in the first row are the coefficients of the polynomial to be evaluated. Then the remainder of ${\displaystyle f(x)}$ on division by ${\displaystyle x-3}$ is 5.

But by the polynomial remainder theorem, we know that the remainder is ${\displaystyle f(3)}$. Thus ${\displaystyle f(3)=5}$

In this example, if ${\displaystyle a_{3}=2,a_{2}=-6,a_{1}=2,a_{0}=-1}$ we can see that ${\displaystyle b_{3}=2,b_{2}=0,b_{1}=2,b_{0}=5}$, the entries in the third row. So, synthetic division is based on Horner's method.

As a consequence of the polynomial remainder theorem, the entries in the third row are the coefficients of the second-degree polynomial, the quotient of ${\displaystyle f(x)}$ on division by ${\displaystyle x-3}$. The remainder is 5. This makes Horner's method useful for polynomial long division.

Divide ${\displaystyle x^{3}-6x^{2}+11x-6}$ by ${\displaystyle x-2}$:

 2 │   1    −6    11    −6
   │         2    −8     6
   └────────────────────────
       1    −4     3     0

The quotient is ${\displaystyle x^{2}-4x+3}$.

Let ${\displaystyle f_{1}(x)=4x^{4}-6x^{3}+3x-5}$ and ${\displaystyle f_{2}(x)=2x-1}$. Divide ${\displaystyle f_{1}(x)}$ by ${\displaystyle f_{2}\,(x)}$ using Horner's method.

  0.5 │ 4  -6   0   3  -5
      │     2  -2  -1   1
└───────────────────────
        2  -2  -1   1  -2

The third row is the sum of the first two rows, divided by 2. Each entry in the second row is the product of 1 with the third-row entry to the left. The answer is

${\displaystyle {\frac {f_{1}(x)}{f_{2}(x)}}=2x^{3}-2x^{2}-x+1-{\frac {4}{2x-1}}.}$

Evaluation using the monomial form of a degree-n polynomial requires at most n additions and (n² + n)/2 multiplications, if powers are calculated by repeated multiplication and each monomial is evaluated individually. (This can be reduced to n additions and 2n − 1 multiplications by evaluating the powers of x iteratively.) If numerical data are represented in terms of digits (or bits), then the naive algorithm also entails storing approximately 2n times the number of bits of x (the evaluated polynomial has approximate magnitude xⁿ, and one must also store xⁿ itself). By contrast, Horner's method requires only n additions and n multiplications, and its storage requirements are only n times the number of bits of x. Alternatively, Horner's method can be computed with n fused multiply–adds. Horner's method can also be extended to evaluate the first k derivatives of the polynomial with kn additions and multiplications.[3]

Horner's method is optimal, in the sense that any algorithm to evaluate an arbitrary polynomial must use at least as many operations. Alexander Ostrowski proved in 1954 that the number of additions required is minimal.[4] Victor Pan proved in 1966 that the number of multiplications is minimal.[5] However, when x is a matrix, Horner's method is not optimal.

This assumes that the polynomial is evaluated in monomial form and no preconditioning of the representation is allowed, which makes sense if the polynomial is evaluated only once. However, if preconditioning is allowed and the polynomial is to be evaluated many times, then faster algorithms are possible. They involve a transformation of the representation of the polynomial. In general, a degree-n polynomial can be evaluated using only ⌊n/2⌋+2 multiplications and n additions.[6]

A disadvantage of Horner's rule is that all of the operations are sequentially dependent, so it is not possible to take advantage of instruction level parallelism on modern computers. In most applications where the efficiency of polynomial evaluation matters, many low-order polynomials are evaluated simultaneously (for each pixel or polygon in computer graphics, or for each grid square in a numerical simulation), so it is not necessary to find parallelism within a single polynomial evaluation.

If, however, one is evaluating a single polynomial of very high order, it may be useful to break it up as follows:

${\displaystyle {\begin{aligned}p(x)&=\sum _{i=0}^{n}a_{i}x^{i}\\&=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\cdots +a_{n}x^{n}\\&=\left(a_{0}+a_{2}x^{2}+a_{4}x^{4}+\cdots \right)+\left(a_{1}x+a_{3}x^{3}+a_{5}x^{5}+\cdots \right)\\&=\left(a_{0}+a_{2}x^{2}+a_{4}x^{4}+\cdots \right)+x\left(a_{1}+a_{3}x^{2}+a_{5}x^{4}+\cdots \right)\\&=\sum _{i=0}^{\lfloor n/2\rfloor }a_{2i}x^{2i}+x\sum _{i=0}^{\lfloor n/2\rfloor }a_{2i+1}x^{2i}\\&=p_{0}(x^{2})+xp_{1}(x^{2}).\\\end{aligned}}}$

More generally, the summation can be broken into k parts:

${\displaystyle {\begin{aligned}p(x)&=\sum _{i=0}^{n}a_{i}x^{i}\\&=\sum _{j=0}^{k-1}x^{j}\sum _{i=0}^{\lfloor n/k\rfloor }a_{ki+j}x^{ki}\\&=\sum _{j=0}^{k-1}x^{j}p_{j}(x^{k})\\\end{aligned}}}$

where the inner summations may be evaluated using separate parallel instances of Horner's method. This requires slightly more operations than the basic Horner's method, but allows k-way SIMD execution of most of them. Modern compilers generally evaluate polynomials this way when advantageous, although for floating-point calculations this require enabling (unsafe) reassociative math.

Horner's method is a fast, code-efficient method for multiplication and division of binary numbers on a microcontroller with no hardware multiplier. One of the binary numbers to be multiplied is represented as a trivial polynomial, where (using the above notation) ${\displaystyle a_{i}=1}$, and ${\displaystyle x=2}$. Then, x (or x to some power) is repeatedly factored out. In this binary numeral system (base 2), ${\displaystyle x=2}$, so powers of 2 are repeatedly factored out.

For example, to find the product of two numbers (0.15625) and m:

${\displaystyle {\begin{aligned}(0.15625)m&=(0.00101_{b})m=(2^{-3}+2^{-5})m=(2^{-3})m+(2^{-5})m\\&=2^{-3}(m+(2^{-2})m)=2^{-3}(m+2^{-2}(m)).\end{aligned}}}$

To find the product of two binary numbers d and m:

1. A register holding the intermediate result is initialized to d.

2. Begin with the least significant (rightmost) non-zero bit in m.

2b. Count (to the left) the number of bit positions to the next most significant non-zero bit. If there are no more-significant bits, then take the value of the current bit position.

2c. Using that value, perform a left-shift operation by that number of bits on the register holding the intermediate result

3. If all the non-zero bits were counted, then the intermediate result register now holds the final result. Otherwise, add d to the intermediate result, and continue in step 2 with the next most significant bit in m.

In general, for a binary number with bit values (${\displaystyle d_{3}d_{2}d_{1}d_{0}}$) the product is

${\displaystyle (d_{3}2^{3}+d_{2}2^{2}+d_{1}2^{1}+d_{0}2^{0})m=d_{3}2^{3}m+d_{2}2^{2}m+d_{1}2^{1}m+d_{0}2^{0}m.}$

At this stage in the algorithm, it is required that terms with zero-valued coefficients are dropped, so that only binary coefficients equal to one are counted, thus the problem of multiplication or division by zero is not an issue, despite this implication in the factored equation:

${\displaystyle =d_{0}\left(m+2{\frac {d_{1}}{d_{0}}}\left(m+2{\frac {d_{2}}{d_{1}}}\left(m+2{\frac {d_{3}}{d_{2}}}(m)\right)\right)\right).}$

The denominators all equal one (or the term is absent), so this reduces to

${\displaystyle =d_{0}(m+2{d_{1}}(m+2{d_{2}}(m+2{d_{3}}(m)))),}$

or equivalently (as consistent with the "method" described above)

${\displaystyle =d_{3}(m+2^{-1}{d_{2}}(m+2^{-1}{d_{1}}(m+{d_{0}}(m)))).}$

In binary (base-2) math, multiplication by a power of 2 is merely a register shift operation. Thus, multiplying by 2 is calculated in base-2 by an arithmetic shift. The factor (2⁻¹) is a right arithmetic shift, a (0) results in no operation (since 2⁰ = 1 is the multiplicative identity element), and a (2¹) results in a left arithmetic shift. The multiplication product can now be quickly calculated using only arithmetic shift operations, addition and subtraction.

The method is particularly fast on processors supporting a single-instruction shift-and-addition-accumulate. Compared to a C floating-point library, Horner's method sacrifices some accuracy, however it is nominally 13 times faster (16 times faster when the "canonical signed digit" (CSD) form is used) and uses only 20% of the code space.[7]

Horner's method can be used to convert between different positional numeral systems – in which case x is the base of the number system, and the a_i coefficients are the digits of the base-x representation of a given number – and can also be used if x is a matrix, in which case the gain in computational efficiency is even greater. In fact, when x is a matrix, further acceleration is possible which exploits the structure of matrix multiplication, and only ${\displaystyle {\sqrt {n}}}$ instead of n multiplies are needed (at the expense of requiring more storage) using the 1973 method of Paterson and Stockmeyer.[8]

Polynomial root finding using Horner's method

Consider the polynomial

${\displaystyle p_{6}(x)=(x+8)(x+5)(x+3)(x-2)(x-3)(x-7)}$

which can be expanded to

${\displaystyle p_{6}(x)=x^{6}+4x^{5}-72x^{4}-214x^{3}+1127x^{2}+1602x-5040.}$

From the above we know that the largest root of this polynomial is 7 so we are able to make an initial guess of 8. Using Newton's method the first zero of 7 is found as shown in black in the figure to the right. Next ${\displaystyle p(x)}$ is divided by ${\displaystyle (x-7)}$ to obtain

${\displaystyle p_{5}(x)=x^{5}+11x^{4}+5x^{3}-179x^{2}-126x+720}$

which is drawn in red in the figure to the right. Newton's method is used to find the largest zero of this polynomial with an initial guess of 7. The largest zero of this polynomial which corresponds to the second largest zero of the original polynomial is found at 3 and is circled in red. The degree 5 polynomial is now divided by ${\displaystyle (x-3)}$ to obtain

${\displaystyle p_{4}(x)=x^{4}+14x^{3}+47x^{2}-38x-240}$

which is shown in yellow. The zero for this polynomial is found at 2 again using Newton's method and is circled in yellow. Horner's method is now used to obtain

${\displaystyle p_{3}(x)=x^{3}+16x^{2}+79x+120}$

which is shown in green and found to have a zero at −3. This polynomial is further reduced to

${\displaystyle p_{2}(x)=x^{2}+13x+40}$

which is shown in blue and yields a zero of −5. The final root of the original polynomial may be found by either using the final zero as an initial guess for Newton's method, or by reducing ${\displaystyle p_{2}(x)}$ and solving the linear equation. As can be seen, the expected roots of −8, −5, −3, 2, 3, and 7 were found.