# Subsign

## Sub`sign´

v. t. | 1. | To sign beneath; to subscribe. [imp. & p. p. Subsigned |

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v. t. | 1. | To sign beneath; to subscribe. [imp. & p. p. Subsigned |

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Each memory signature is in the format of "MalwareName=subsig1, subsig2, ..., subsigN".

Now from Theorem 3, there exist n complete subsigned graphs such that (i), (ii), and (iii) hold.

Then by Theorem 3, it can be written as the union of n complete subsigned graphs [S.sub.1], [S.sub.2], ..., [S.sub.k] of marked vertices such that for each i, j = 1, ..., n, (i), (ii), and (iii) hold.

A 2-path product signed graph S[??]S of a given signed graph S is all negative if and only if S is either a cycle of length 4m or a signed path and S does not contain a subsigned path [mathematical expression not reproducible].

To produce all negative 2-path product signed graph S[??]S, S can not have subsigned path [mathematical expression not reproducible], on any subsigned path since then uv will be a positive edge in S[??]S.

A signed graph S is sign-compatible if and only if S does not contain a subsigned graph isomorphic to either of the two signed graphs in Figure 3, [S.sub.1] formed by taking the path [P.sub.4] : x, u, v, y with both the edges xu and vy negative and the edge uv positive, and [S.sub.2] formed by taking [S.sub.1] and identifying the vertices x and y.

(ii) S does not consist of the canonically marked subsigned path [mathematical expression not reproducible].

Let if possible S consist of the canonically marked subsigned path [mathematical expression not reproducible].

Then S[??]S must consist of subsigned graph isomorphic to Figure 3, which is not possible as then either (i) or (ii) does not hold true.

(ii) S does not contain a subsigned path [mathematical expression not reproducible], of vertices u, w, v where [[mu].sub.1] [member of] {+, -};

(iii) if there exist a subsigned path [mathematical expression not reproducible] of vertices u, w, v in S; then either [d.sup.-](u) = 0 or [d.sup.-](v) = 0, where [[mu].sub.1] [member of] {+, -};

Let us suppose S contains a subsigned graph [mathematical expression not reproducible]; then clearly uv is a positive edge in S[??]S such that [d.sup.-](u) = 0 and [d.sup.-](v) = 0, which is a contradiction to the fact that S[??]S is C-sign-compatible.

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