Zhegalkin polynomial

Using the method of indeterminate coefficients, a linear system consisting of all the tuples of the function and their values is generated. Solving the linear system gives the coefficients of the Zhegalkin polynomial.

Given the Boolean function ${\displaystyle f(A,B,C)={\bar {A}}{\bar {B}}{\bar {C}}+{\bar {A}}B{\bar {C}}+A{\bar {B}}{\bar {C}}+ABC}$, express it as a Zhegalkin polynomial. This function can be expressed as a column vector

${\displaystyle {\vec {f}}={\begin{pmatrix}1\\0\\1\\0\\1\\0\\0\\1\end{pmatrix}}}$.

This vector should be the output of left-multiplying a vector of undetermined coefficients

${\displaystyle {\vec {c}}={\begin{pmatrix}c_{0}\\c_{1}\\c_{2}\\c_{3}\\c_{4}\\c_{5}\\c_{6}\\c_{7}\end{pmatrix}}}$

by an 8x8 logical matrix which represents the possible values that all the possible conjunctions of A, B, C can take. These possible values are given in the following truth table:

${\displaystyle {\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|}A&B&C&\;\;\;\;\;\;\;\;&1&C&B&BC&A&AC&AB&ABC\\\hline 0&0&0&&1&0&0&0&0&0&0&0\\0&0&1&&1&1&0&0&0&0&0&0\\0&1&0&&1&0&1&0&0&0&0&0\\0&1&1&&1&1&1&1&0&0&0&0\\1&0&0&&1&0&0&0&1&0&0&0\\1&0&1&&1&1&0&0&1&1&0&0\\1&1&0&&1&0&1&0&1&0&1&0\\1&1&1&&1&1&1&1&1&1&1&1\end{array}}}$.

The information in the above truth table can be encoded in the following logical matrix:

${\displaystyle S_{3}={\begin{pmatrix}1&0&0&0&0&0&0&0\\1&1&0&0&0&0&0&0\\1&0&1&0&0&0&0&0\\1&1&1&1&0&0&0&0\\1&0&0&0&1&0&0&0\\1&1&0&0&1&1&0&0\\1&0&1&0&1&0&1&0\\1&1&1&1&1&1&1&1\end{pmatrix}}}$

where the 'S' here stands for "Sierpiński", as in Sierpiński triangle, and the subscript 3 gives the exponents of its size: ${\displaystyle 2^{3}\times 2^{3}}$.

It can be proven through mathematical induction and block-matrix multiplication that any such "Sierpiński matrix" ${\displaystyle S_{n}}$ is its own inverse.[nb 1]

Then the linear system is

${\displaystyle S_{3}{\vec {c}}={\vec {f}}}$

which can be solved for ${\displaystyle {\vec {c}}}$:

${\displaystyle {\vec {c}}=S_{3}^{-1}{\vec {f}}=S_{3}{\vec {f}}={\begin{pmatrix}1&0&0&0&0&0&0&0\\1&1&0&0&0&0&0&0\\1&0&1&0&0&0&0&0\\1&1&1&1&0&0&0&0\\1&0&0&0&1&0&0&0\\1&1&0&0&1&1&0&0\\1&0&1&0&1&0&1&0\\1&1&1&1&1&1&1&1\end{pmatrix}}{\begin{pmatrix}1\\0\\1\\0\\1\\0\\0\\1\end{pmatrix}}={\begin{pmatrix}1\\1\\1\oplus 1\\1\oplus 1\\1\oplus 1\\1\oplus 1\\1\oplus 1\oplus 1\\1\oplus 1\oplus 1\oplus 1\end{pmatrix}}={\begin{pmatrix}1\\1\\0\\0\\0\\0\\1\\0\end{pmatrix}}}$,

and the Zhegalkin polynomial corresponding to ${\displaystyle {\vec {c}}}$ is ${\displaystyle 1\oplus C\oplus AB}$.

Using this method, the canonical disjunctive normal form (a fully expanded disjunctive normal form) is computed first. Then the negations in this expression are replaced by an equivalent expression using the mod 2 sum of the variable and 1. The disjunction signs are changed to addition mod 2, the brackets are opened, and the resulting Boolean expression is simplified. This simplification results in the Zhegalkin polynomial.

Computing the Zhegalkin polynomial for an example function P by the table method

Let ${\displaystyle c_{0},...,c_{2^{n}-1}}$ be the outputs of a truth table for the function P of n variables, such that the index of the ${\displaystyle c_{i}}$'s corresponds to the binary indexing of the minterms.[nb 2] Define a function ζ recursively by:

${\displaystyle \zeta (c_{i}):=c_{i}}$

${\displaystyle \zeta (c_{0},...,c_{k}):=\zeta (c_{0},...,c_{k-1})\oplus \zeta (c_{1},...,c_{k})}$.

Note that

${\displaystyle \zeta (c_{0},...,c_{m})=\bigoplus _{k=0}^{m}{m \choose k}_{2}c_{k}}$

where ${\displaystyle {m \choose k}_{2}}$ is the binomial coefficient reduced modulo 2.

Then

${\displaystyle g_{i}=\zeta (c_{0},...,c_{i})}$

is the i ^th coefficient of a Zhegalkin polynomial whose literals in the i ^th monomial are the same as the literals in the i ^th minterm, except that the negative literals are removed (or replaced by 1).

The ζ-transformation is its own inverse, so the same kind of table can be used to compute the coefficients ${\displaystyle c_{0},...,c_{2^{n}-1}}$ given the coefficients ${\displaystyle g_{0},...,g_{2^{n}-1}}$. Just let

${\displaystyle c_{i}=\zeta (g_{0},...,g_{i})}$.

In terms of the table in the figure, copy the outputs of the truth table (in the column labeled P) into the leftmost column of the triangular table. Then successively compute columns from left to right by applying XOR to each pair of vertically adjacent cells in order to fill the cell immediately to the right of the top cell of each pair. When the entire triangular table is filled in then the top row reads out the coefficients of a linear combination which, when simplified (removing the zeroes), yields the Zhegalkin polynomial.

To go from a Zhegalkin polynomial to a truth-table, it is possible to fill out the top row of the triangular table with the coefficients of the Zhegalkin polynomial (putting in zeroes for any combinations of positive literals not in the polynomial). Then successively compute rows from top to bottom by applying XOR to each pair of horizontally adjacent cells in order to fill the cell immediately to the bottom of the leftmost cell of each pair. When the entire triangular table is filled then the leftmost column of it can be copied to column P of the truth table.

As an aside, this method of calculation corresponds to the method of operation of the elementary cellular automaton called Rule 102. For example, start such a cellular automaton with eight cells set up with the outputs of the truth table (or the coefficients of the canonical disjunctive normal form) of the Boolean expression: 10101001. Then run the cellular automaton for seven more generations while keeping a record of the state of the leftmost cell. The history of this cell then turns out to be: 11000010, which shows the coefficients of the corresponding Zhegalkin polynomial.[3][4]

Using the Pascal method to compute the Zhegalkin polynomial for the Boolean function ${\displaystyle {\bar {a}}{\bar {b}}{\bar {c}}+{\bar {a}}b{\bar {c}}+{\bar {a}}bc+ab{\bar {c}}}$. The line in Russian at the bottom says:
${\displaystyle \oplus }$ – bitwise operation "Exclusive OR"

The most economical in terms of the amount of computation and expedient for constructing the Zhegalkin polynomial manually is the Pascal method.

We build a table consisting of ${\displaystyle 2^{N}}$ columns and ${\displaystyle N+1}$ rows, where N is the number of variables in the function. In the top row of the table we place the vector of function values, that is, the last column of the truth table.

Each row of the resulting table is divided into blocks (black lines in the figure). In the first line, the block occupies one cell, in the second line — two, in the third — four, in the fourth — eight, and so on. Each block in a certain line, which we will call "lower block", always corresponds to exactly two blocks in the previous line. We will call them "left upper block" and "right upper block".

The construction starts from the second line. The contents of the left upper blocks are transferred without change into the corresponding cells of the lower block (green arrows in the figure). Then, the operation "addition modulo two" is performed bitwise over the right upper and left upper blocks and the result is transferred to the corresponding cells of the right side of the lower block (red arrows in the figure). This operation is performed with all lines from top to bottom and with all blocks in each line. After the construction is completed, the bottom line contains a string of numbers, which are the coefficients of the Zhegalkin polynomial, written in the same sequence as in the triangle method described above.

Graphic representation of the coefficients of the Zhegalkin polynomial for functions with different numbers of variables.

According to the truth table, it is easy to calculate the individual coefficients of the Zhegalkin polynomial. To do this, sum up modulo 2 the values of the function in those rows of the truth table where variables that are not in the conjunction (that corresponds to the coefficient being calculated) take zero values.

Suppose, for example, that we need to find the coefficient of the xz conjunction for the function of three variables ${\displaystyle f(x,y,z)}$. There is no variable y in this conjunction. Find the input sets in which the variable y takes a zero value. These are the sets 0, 1, 4, 5 (000, 001, 100, 101). Then the coefficient at conjunction xz is

${\displaystyle a_{5}=f_{0}\oplus f_{1}\oplus f_{4}\oplus f_{5}=f(0,0,0)\oplus f(0,0,1)\oplus f(1,0,0)\oplus f(1,0,1)}$

Since there are no variables with the constant term,

${\displaystyle a_{0}=f_{0}}$.

For a term which includes all variables, the sum includes all values of the function:

${\displaystyle a_{N-1}=f_{0}\oplus f_{1}\oplus f_{2}\oplus ...\oplus f_{N-2}\oplus f_{N-1}}$

Let us graphically represent the coefficients of the Zhegalkin polynomial as sums modulo 2 of values of functions at certain points. To do this, we construct a square table, where each column represents the value of the function at one of the points, and the row is the coefficient of the Zhegalkin polynomial. The point at the intersection of some column and row means that the value of the function at this point is included in the sum for the given coefficient of the polynomial (see figure). We call this table ${\displaystyle T_{N}}$, where N is the number of variables of the function.

There is a pattern that allows you to get a table for a function of N variables, having a table for a function of ${\displaystyle N-1}$ variables. The new table ${\displaystyle T_{N}+1}$ is arranged as a 2 × 2 matrix of ${\displaystyle T_{N}}$ tables, and the right upper block of the matrix is cleared.

Consider the columns of a table ${\displaystyle T_{N}}$ as corresponding to elements of a Boolean lattice of size ${\displaystyle 2^{N}}$. For each column ${\displaystyle f_{M}}$ express number M as a binary number ${\displaystyle M_{2}}$, then ${\displaystyle f_{M}\leq f_{K}}$ if and only if ${\displaystyle M_{2}\vee K_{2}=K_{2}}$, where ${\displaystyle \vee }$ denotes bitwise OR.

If the rows of table ${\displaystyle T_{N}}$ are numbered, from top to bottom, with the numbers from 0 to ${\displaystyle 2^{N}-1}$, then the tabular content of row number R is the ideal generated by element ${\displaystyle f_{R}}$ of the lattice.

Note incidentally that the overall pattern of a table ${\displaystyle T_{N}}$ is that of a logical matrix Sierpiński triangle. Also, the pattern corresponds to an elementary cellular automaton called Rule 60, starting with the leftmost cell set to 1 and all other cells cleared.

Converting a Karnaugh map to a Zhegalkin polynomial.

The figure shows a function of three variables, P(A, B, C) represented as a Karnaugh map, which the reader may consider as an example of how to convert such maps into Zhegalkin polynomials; the general procedure is given in the following steps:

• We consider all the cells of the Karnaugh map in ascending order of the number of units in their codes. For the function of three variables, the sequence of cells will be 000–100–010–001–110–101–011–111. Each cell of the Karnaugh map is associated with a member of the Zhegalkin polynomial depending on the positions of the code in which there are ones. For example, cell 111 corresponds to the member ABC, cell 101 corresponds to the member AC, cell 010 corresponds to the member B, and cell 000 corresponds to member 1.
• If the cell in question is 0, go to the next cell.
• If the cell in question is 1, add the corresponding term to the Zhegalkin polynomial, then invert all cells in the Karnaugh map where this term is 1 (or belonging to the ideal generated by this term, in a Boolean lattice of monomials) and go to the next cell. For example, if, when examining cell 110, a one appears in it, then the term AB is added to the Zhegalkin polynomial and all cells of the Karnaugh map are inverted, for which A = 1 and B = 1. If unit is in cell 000, then a term 1 is added to the Zhegalkin polynomial and the entire Karnaugh map is inverted.
• The transformation process can be considered complete when, after the next inversion, all cells of the Karnaugh map become zero, or a don't care condition.

As an example, consider the three-variable case. The following table shows the divisibility relation:

Then

${\displaystyle g(000)=f(000)}$

${\displaystyle g(001)=f(000)\oplus f(001)}$

${\displaystyle g(010)=f(000)\oplus f(010)}$

${\displaystyle g(011)=f(000)\oplus f(001)\oplus f(010)\oplus f(011)}$

${\displaystyle g(100)=f(000)\oplus f(100)}$

${\displaystyle g(101)=f(000)\oplus f(001)\oplus f(100)\oplus (101)}$

${\displaystyle g(110)=f(000)\oplus f(010)\oplus f(100)\oplus f(110)}$

${\displaystyle g(111)=f(000)\oplus f(001)\oplus f(010)\oplus f(011)\oplus f(100)\oplus f(101)\oplus f(110)\oplus f(111)}$

The above system of equations can be solved for f, and the result can be summarized as being obtainable by exchanging g and f throughout the above system.

The table below shows the binary numbers along with their associated Zhegalkin monomials and Boolean minterms:

Boolean minterm	ABC	Zhegalkin monomial
${\displaystyle {\bar {A}}{\bar {B}}{\bar {C}}}$	000	1
${\displaystyle {\bar {A}}{\bar {B}}C}$	001	C
${\displaystyle {\bar {A}}B{\bar {C}}}$	010	B
${\displaystyle {\bar {A}}BC}$	011	BC
${\displaystyle A{\bar {B}}{\bar {C}}}$	100	A
${\displaystyle A{\bar {B}}C}$	101	AC
${\displaystyle AB{\bar {C}}}$	110	AB
${\displaystyle ABC}$	111	ABC

The Zhegalkin monomials are naturally ordered by divisibility, whereas the Boolean minterms do not so naturally order themselves; each one represents an exclusive eighth of the three-variable Venn diagram. The ordering of the monomials transfers to the bit strings as follows: given ${\displaystyle a_{1}a_{2}a_{3}}$ and ${\displaystyle b_{1}b_{2}b_{3}}$, a pair of bit triplets, then ${\displaystyle a_{1}a_{2}a_{3}\leq b_{1}b_{2}b_{3}\leftrightarrow a_{1}\leq b_{1}\wedge a_{2}\leq b_{2}\wedge a_{3}\leq b_{3}}$.

The correspondence between a three-variable Boolean sum-of-minterms and a Zhegalkin polynomial is then:

${\displaystyle f(000){\bar {A}}{\bar {B}}{\bar {C}}\vee f(001){\bar {A}}{\bar {B}}C\vee f(010){\bar {A}}B{\bar {C}}\vee f(011){\bar {A}}BC\vee f(100)A{\bar {B}}{\bar {C}}\vee f(101)A{\bar {B}}C\vee f(110)AB{\bar {C}}\vee f(111)ABC}$

${\displaystyle \qquad \qquad \equiv g(000)\oplus g(001)C\oplus g(010)B\oplus g(011)BC\oplus g(100)A\oplus g(101)AC\oplus g(110)AB\oplus g(111)ABC}$.

The system of equations above may be summarized as a logical matrix equation:

${\displaystyle {\begin{pmatrix}g(000)\\g(001)\\g(010)\\g(011)\\g(100)\\g(101)\\g(110)\\g(111)\end{pmatrix}}={\begin{pmatrix}1&&0&&0&&0&&0&&0&&0&&0\\1&&1&&0&&0&&0&&0&&0&&0\\1&&0&&1&&0&&0&&0&&0&&0\\1&&1&&1&&1&&0&&0&&0&&0\\1&&0&&0&&0&&1&&0&&0&&0\\1&&1&&0&&0&&1&&1&&0&&0\\1&&0&&1&&0&&1&&0&&1&&0\\1&&1&&1&&1&&1&&1&&1&&1\end{pmatrix}}{\begin{pmatrix}f(000)\\f(001)\\f(010)\\f(011)\\f(100)\\f(101)\\f(110)\\f(111)\end{pmatrix}}}$

which N. J. Wildberger calls a Boole–Möbius transformation.

Below is shown the “XOR spreadsheet” form of the transformation, going in the direction of g to f:

${\displaystyle f_{000}}$	${\displaystyle f_{001}}$	${\displaystyle f_{010}}$	${\displaystyle f_{011}}$	${\displaystyle f_{100}}$	${\displaystyle f_{101}}$	${\displaystyle f_{110}}$	${\displaystyle f_{111}}$
${\displaystyle g_{000}}$	${\displaystyle g_{000}\oplus g_{001}}$	${\displaystyle g_{000}\oplus g_{010}}$	${\displaystyle g_{000}\oplus g_{010}\oplus g_{001}\oplus g_{011}}$	${\displaystyle g_{000}\oplus g_{100}}$	${\displaystyle g_{000}\oplus g_{100}\oplus g_{001}\oplus g_{101}}$	${\displaystyle g_{000}\oplus g_{100}\oplus g_{010}\oplus g_{110}}$	${\displaystyle g_{000}\oplus g_{100}\oplus g_{010}\oplus g_{110}\oplus g_{001}\oplus g_{101}\oplus g_{011}\oplus g_{111}}$
${\displaystyle g_{001}}$	${\displaystyle g_{001}\oplus g_{010}}$	${\displaystyle g_{001}\oplus g_{011}}$	${\displaystyle g_{001}\oplus g_{011}\oplus g_{010}\oplus g_{100}}$	${\displaystyle g_{001}\oplus g_{101}}$	${\displaystyle g_{001}\oplus g_{101}\oplus g_{010}\oplus g_{110}}$	${\displaystyle g_{001}\oplus g_{101}\oplus g_{011}\oplus g_{111}}$
${\displaystyle g_{010}}$	${\displaystyle g_{010}\oplus g_{011}}$	${\displaystyle g_{010}\oplus g_{100}}$	${\displaystyle g_{010}\oplus g_{100}\oplus g_{011}\oplus g_{101}}$	${\displaystyle g_{010}\oplus g_{110}}$	${\displaystyle g_{010}\oplus g_{110}\oplus g_{011}\oplus g_{111}}$
${\displaystyle g_{011}}$	${\displaystyle g_{011}\oplus g_{100}}$	${\displaystyle g_{011}\oplus g_{101}}$	${\displaystyle g_{011}\oplus g_{101}\oplus g_{100}\oplus g_{110}}$	${\displaystyle g_{011}\oplus g_{111}}$
${\displaystyle g_{100}}$	${\displaystyle g_{100}\oplus g_{101}}$	${\displaystyle g_{100}\oplus g_{110}}$	${\displaystyle g_{100}\oplus g_{110}\oplus g_{101}\oplus g_{111}}$
${\displaystyle g_{101}}$	${\displaystyle g_{101}\oplus g_{110}}$	${\displaystyle g_{101}\oplus g_{111}}$
${\displaystyle g_{110}}$	${\displaystyle g_{110}\oplus g_{111}}$
${\displaystyle g_{111}}$