Sentence examples for equal to the unique from inspiring English sources

If A ∩ B ≠ ∅, then z = y ∈ A ∩ B is the unique fixed point of S, T, T ∘ S : A ∪ B → A ∪ B which is in A ∩ B. Proof If D = 0, i.e. A and B intersect, then this result reduces to Theorem 2.3, with the best proximity points being coincident and equal to the unique fixed point.

The sequences T 2 n x n ∈ N 0 and T 2 n + 1 x n ∈ N 0 converge to z and y for all x ∈ A, respectively, to y and z for all x ∈ B. If A∩ B ≠ ∅ then z = y ∈ A∩B is the unique fixed point of T A ∪ B → A ∪B. Proof: If D = 0, i.e., A and B intersect then this result reduces to Theorem 2.2 with the best proximity points being coincident and equal to the unique fixed point.

Namely, ϕ(q~, a) is equal to the unique state q such that π(q~, a, q)>0 and ρ(q~, a) is π(q~, a, q).

Here is equal to the number of unique MAC keys that expects to use for authenticating its messages.

where is equal to the number of unique MAC keys that the CA expects to use for authenticating the certificates and messages it generates during its operational lifetime.

As a result, the maximum number of resolvable sources is equal to the number of unique positive lags in the combined co-array.

As stated in the first example, the maximum number of resolvable sources using single-frequency sparse reconstruction is equal to the number of unique positive lags in the difference co-array, which is eight in this case.

Note that in the case of simple flooding (where every node retransmits every received message once), this value would be equal to the amount of unique DENM messages (6×350=2100).

This is to be expected from theory; recall from the design of the experiment that in the first generation n g random solutions are generated whereas the repeated random draws based method performs n h =(1−r f )×n p ×n g +r f ×n p solutions which is equal to the number of unique solutions created by the genetic algorithm across all generations.

In this graph, nodes are clone contigs which are initialized to be sequences of one clone each: Oi = 〈Ci〉; weighted edges connect the nodes with weight Wij equal to the count of unique k-mers shared between the two clones Ci and Cj.

To this end, the coverage evenness score — equal to the number of unique reference positions matched by the first bases of reads extending past the breakpoint and any overlap or read-only junction bases — is recalculated based on the re-aligned reads that match best to each junction.