Unfortunately, version vectors are not safe in the presenceof Byzantine nodes, as shown in Figure 1. This is because aByzantine node may generate several distinct updates withthe same sequence number, and send them to different nodes(this failure mode is known as equivocation). Subsequently,when correct nodes p and q exchange version vectors, theymay believe that they have delivered the same set of updatesbecause their version vectors are identical, even though theyhave in fact delivered different updates.

Comparing version vectors betweenpayloads is an inclusion check without the need to perform aDAG-walking

Sweden Poised to Miss the Long-Term Climate Target It Pioneered

A very creative concept that i`ve never seen before in a video game.

A cognitive signature™ encodes the exact structure of a graph.●It is a lossless encoding, similar to a Gödel numbering. *●For unlabeled graphs, integers are sufficient for a cognitive signature.●For example, 0 maps to and from an empty graph with no nodes or arcs.●1, 2, 3, 4, 5, and 6 can be mapped to and from the following graphs:●To encode the structure of conceptual graphs in Cognitive Memory, the cognitive signatures are based on generalized combinatorial maps. **By contrast, a word vector encodes labels, but not structure.●A word vector is a “bag of labels” that ignores the graph connections.●Word vectors are often used for measuring the similarity of documents.●But they discard the structural information necessary for reasoning, question answering, and language understanding.

What does a pair of orthonormal vectors in 2-D Euclidean space look like? Let u = (x1, y1) and v = (x2, y2). Consider the restrictions on x1, x2, y1, y2 required to make u and v form an orthonormal pair. From the orthogonality restriction, u • v = 0. From the unit length restriction on u, ||u|| = 1. From the unit length restriction on v, ||v|| = 1. Expanding these terms gives 3 equations: x 1 x 2 + y 1 y 2 = 0 {\displaystyle x_{1}x_{2}+y_{1}y_{2}=0\quad } x 1 2 + y 1 2 = 1 {\displaystyle {\sqrt {{x_{1}}^{2}+{y_{1}}^{2}}}=1} x 2 2 + y 2 2 = 1 {\displaystyle {\sqrt {{x_{2}}^{2}+{y_{2}}^{2}}}=1} Converting from Cartesian to polar coordinates, and considering Equation ( 2 ) {\displaystyle (2)} and Equation ( 3 ) {\displaystyle (3)} immediately gives the result r1 = r2 = 1. In other words, requiring the vectors be of unit length restricts the vectors to lie on the unit circle. After substitution, Equation ( 1 ) {\displaystyle (1)} becomes cos ⁡ θ 1 cos ⁡ θ 2 + sin ⁡ θ 1 sin ⁡ θ 2 = 0 {\displaystyle \cos \theta _{1}\cos \theta _{2}+\sin \theta _{1}\sin \theta _{2}=0} . Rearranging gives tan ⁡ θ 1 = − cot ⁡ θ 2 {\displaystyle \tan \theta _{1}=-\cot \theta _{2}} . Using a trigonometric identity to convert the cotangent term gives tan ⁡ ( θ 1 ) = tan ⁡ ( θ 2 + π 2 ) {\displaystyle \tan(\theta _{1})=\tan \left(\theta _{2}+{\tfrac {\pi }{2}}\right)} ⇒ θ 1 = θ 2 + π 2 {\displaystyle \Rightarrow \theta _{1}=\theta _{2}+{\tfrac {\pi }{2}}} It is clear that in the plane, orthonormal vectors are simply radii of the unit circle whose difference in angles equals 90°.