Deciding One to One property of Boolean maps: Condition and algorithm in terms of implicants
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arXiv
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| Format: | Preprint |
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2023
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| _version_ | 1866908525729415168 |
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| author | Sule, Virendra |
| author_facet | Sule, Virendra |
| contents | This paper addresses the computational problem of deciding invertibility (or one to one-ness) of a Boolean map $F$ in $n$-Boolean variables. This problem is a special case of deciding invertibilty of a map $F:\mathbb{F}_{q}^n\rightarrow\mathbb{F}_{q}^n$ over the finite field $\mathbb{F}_q$ for $q=2$. Algebraic condition for invertibility of $F$ is well known to be equivalent to invertibility of the Koopman operator of $F$ as shown in \cite{RamSule}. In this paper a condition for invertibility is derived in the special case of Boolean maps $F:B_0^n\rightarrow B_0^n$ where $B_0$ is the two element Boolean algebra in terms of \emph{implicants} of Boolean equations defined by the map. This condition is then extended to the case of general maps in $n$ variables and $m\geq n$ equations. Hence this condition answers the special case of invertibility of maps $F$ defined over the binary field $\mathbb{F}_2$ alternatively, in terms of implicants instead of the Koopman operator. The problem of deciding invertibility of a map $F$ (or that of finding its Garden of Eden (GOE)) over finite fields is distinct from the satisfiability problem (SAT) or the problem of deciding consistency of polynomial equations over finite fields. Hence the well known algorithms for deciding SAT or of solvability using Grobner basis for checking membership in an ideal generated by polynomials is not known to answer the question of invertibility of a map. Similarly it appears that algorithms for satisfiability or polynomial solvability are not useful for computation of GOE of $F$ even for maps over the binary field $\mathbb{F}_2$. |
| format | Preprint |
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arxiv_https___arxiv_org_abs_2307_07788 |
| institution | arXiv |
| publishDate | 2023 |
| record_format | arxiv |
| spellingShingle | Deciding One to One property of Boolean maps: Condition and algorithm in terms of implicants Sule, Virendra Symbolic Computation Computational Complexity I.1.2; G.2.0 This paper addresses the computational problem of deciding invertibility (or one to one-ness) of a Boolean map $F$ in $n$-Boolean variables. This problem is a special case of deciding invertibilty of a map $F:\mathbb{F}_{q}^n\rightarrow\mathbb{F}_{q}^n$ over the finite field $\mathbb{F}_q$ for $q=2$. Algebraic condition for invertibility of $F$ is well known to be equivalent to invertibility of the Koopman operator of $F$ as shown in \cite{RamSule}. In this paper a condition for invertibility is derived in the special case of Boolean maps $F:B_0^n\rightarrow B_0^n$ where $B_0$ is the two element Boolean algebra in terms of \emph{implicants} of Boolean equations defined by the map. This condition is then extended to the case of general maps in $n$ variables and $m\geq n$ equations. Hence this condition answers the special case of invertibility of maps $F$ defined over the binary field $\mathbb{F}_2$ alternatively, in terms of implicants instead of the Koopman operator. The problem of deciding invertibility of a map $F$ (or that of finding its Garden of Eden (GOE)) over finite fields is distinct from the satisfiability problem (SAT) or the problem of deciding consistency of polynomial equations over finite fields. Hence the well known algorithms for deciding SAT or of solvability using Grobner basis for checking membership in an ideal generated by polynomials is not known to answer the question of invertibility of a map. Similarly it appears that algorithms for satisfiability or polynomial solvability are not useful for computation of GOE of $F$ even for maps over the binary field $\mathbb{F}_2$. |
| title | Deciding One to One property of Boolean maps: Condition and algorithm in terms of implicants |
| topic | Symbolic Computation Computational Complexity I.1.2; G.2.0 |
| url | https://arxiv.org/abs/2307.07788 |