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This matrix can be block-partitioned into \(q \times q\) blocks, with$$\begin{aligned} (M_q)_{ij} = \left\{ \begin{array}{ll} q I_q - \mathbf{1}_{q \times q} , &{} \text{ if } j-i \equiv 0 ~\mathrm{mod}~q , \\ 0, &{} \text{ otherwise } . \end{array}\right. \end{aligned}$$ Now consider the subcase that \(l \equiv 0 ~\mathrm{mod}~q\). Then \(B^\top B\) consists of l/q repetitions of \(M_q\) on each block row and block column of size \(q^2 \times q^2\). Therefore each such row and column block gives \((q-1)q\) eigenvalues of magnitude l and rest of the eigenvalues will be zero. Next, if \(l ~\mathrm{mod}~q =: k \ne 0\), then \(B^\top B\) consists of \((l-k)/q\) repetitions of \(M_q\) on each block row and column of size \(q^2 \times q^2\). In addition it contains first k column blocks of \(M_q\) concatenated l times column-wise as the last block of columns. It contains first k row blocks of \(M_q\) concatenated \(l-k\) times row-wise as the last block of rows. The extra k rows and columns of \(B^\top B\) give \((q-1)k\) eigenvalues of magnitude \(l+(q-k)\). Another set of row and column blocks gives \((q-1)(q-k)\) eigenvalues of magnitude \(l-k\). The remaining eigenvalues are 0. \(\square \) Remark It has been pointed out to the authors by the referee that the construction in Theorem 3 can be viewed as an explicit realization of the results of [20] (in particular [20, Corollary 2.2]). For any positive integers m and n, the complete bipartite graph \(K_{m,n}\) is immediately seen to be a Ramanujan bigraph as its eigenvalues are \(\{\pm \sqrt{mn},\,0\}\). The Ramanujan bigraphs constructed in Items (1) and (2) of Theorem 3 are explicit constructions of q-coverings of \(K_{l,q}\) and \(K_{q,l}\), respectively. We also note that when \(l = q\), Theorem 3 leads to the following corollary. Corollary 1 For every prime number q, the matrix B(q, q) defined in (6) is the biadjacency matrix of a \((q^2,q^2)\)-bipartite Ramanujan graph with degree q. In this case \(\lambda _1 = q\), and \(\lambda _2 = \sqrt{q}\) with a multiplicity of \(q(q-1)\). The remaining \(q-1\) eigenvalues of B(q, q) are zero. This leads to a new class of \((q^2,q^2)\)-bipartite. 06Q Q Q Q Q Q Q Q Q

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Configuring Q-in-Q Tunneling and VLAN Q-in-Q Tunneling and VLAN

Each clause is a disjunction of variables and their negations. Each variable or its negation appears in each clause exactly once. 11 Examples: Conjunctive Normal Formp  q  p  q. p  q  (p  q)  (p  q). p | q  p  q. p  q  (p  q)  (p  q)  (p  q). 12 Output Tables An output table shows the output of the circuit for every possible combination of inputs. Inputs Output 1 13 Designing a Circuit Write an output table for the circuit.Write the expression in disjunctive normal form. Simplify the expression as much as possible. Write the circuit using AND-, OR-, and NOT-gates. 14 Example: Designing a CircuitDesign a circuit for (p  q). Inputs Output p q 1 15 Example: Designing a Circuit(p  q) is equivalent to p  q. Draw the circuit using an AND-gate and a NOT-gate. 16 Example: Designing a CircuitDesign a circuit for (p  q)  (q  r). Inputs Output p q r 1 17 Example: Designing a Circuit(p  q)  (q  r) is equivalent to (p  q  r)  (p  q  r)  (p  q  r). Does this simplify? In any case, we can draw a circuit, although it may not be optimal. 18 Puzzle Three men apply for a job.They are equally well qualified, so the employer needs a way to choose one. He tells them “On

Q Q - What does Q Q stand for? The Free Dictionary

HOLD THE LINEAs recorded by Toto(From the 1978 Album TOTO)Words and Music by David PaichTranscribed by SlowhandGtr I (E A D G B E) - 'Steve Lukather - Electric Guitar w/ distortion'Gtr II (E A D G B E) - 'Steve Lukather - Electric Guitar w/ distortion'Gtr III (E A D G B E) - 'Steve Lukather - Electric Guitar w/ light distortion' Intro Moderate Rock Q=104 F#5 C#D5 E5 F#5 C#D5 4/4 Gtr I |-3-| |-3-| Wx4 Q Q Q Q Q Q Q H Q Q Q Q Q|------------|---------------|-------------|---------------||------------|---------------|-------------|---------------||------------|---------------|-------------|---------------||------------|-11\-----6-7-7-|-9\----------|-11\-----6-7-7-||------------|--9\-----4-5-5-|-7\----------|--9\-----4-5-5-||------------|---------------|-------------|---------------| 🎸Accuracy Rating: - Votes: - 1st - 3rd Verses $ E5 N.C. PM--------------------------| |-3-| |-3-| |-3-| |-3-| |-3-| Q Q a Q Q Q Q E E E Q E E E Q E E E Q Q Q|----------------||---------------------|----------------||----------------||o--------------------|----------------||----------------||---------------------|----------------||-9\-------------||---------------------|----------------||-7\-----6s7-----||o------------------7-|-9------7-9-----||------------9-8-||--7\----7-8-9----9---|------9-----9-8-| PM---------------------------------------------------| |-3-| |-3-| |-3-| |-3-| |-3-| |-3-| Q E E E Q E E E Q E E E Q Q Q Q E E E Q E E E|-------------------|----------------|--------------------||-------------------|----------------|--------------------||-------------------|----------------|--------------------||-------------------|----------------|--------------------||-----------------7-|-9------7-9-----|--------------------||-7----7-8-9----9---|------9-----9-8-|-7----7-7-9----9-9--| PM-----------------------------------------------------| |--3--| |-3-| |-3-| |-3-| |--3--| |-3-| Q E E E Q Q Q Q E E E Q E E E Q E E E Q Q Q|-------------------|-------------------|-------------------||-------------------|-------------------|-------------------||-------------------|-------------------|-------------------||-------------------|-------------------|-------------------||-------------7-----|-------------------|-------------7-----||-10----10-10---9-8-|-7----7-7-9----9-9-|-10----10-10---9-8-| Chorus D5 E5 F#5 C#D5 E5 Gtrs I, II PM--------------| |-3-| |-3-| |-3-| |-3-| |-3-| |-3--| |-3-| Q E E E Q E E E E E E E E E E E E E E E Q Q Q Q Q Q Q H|-------------------|--------------------------||---------------|-------------||-------------------|--------------------------||---------------|-------------||-------------------|--------------------------||---------------|-------------||-------------------|-------7-7-7-9-9-9-9------||-11\-----6-7-7-|-9\----------||-------------------|-5-5-5-5-5-5-7-7-7-7----7-||--9\-----4-5-5-|-7\----------||-7----7-7-9----9-9-|---------------------/9---||---------------|-------------| To Coda F#5 C#D5 E5 F#5 C#D5 E5 |-3-| |-3-| |-3-| Q Q Q Q Q Q Q Q E E E Q Q Q Q Q Q Q H|---------------|-----------------|---------------|-------------||---------------|-----------------|---------------|-------------||---------------|-----------------|---------------|-------------||-11\-----6-7-7-|-9\--------------|-11\-----6-7-7-|-9\----------||--9\-----4-5-5-|-7\------------7-|--9\-----4-5-5-|-7\----------||---------------|-------------9---|---------------|-------------| 1. F#5 C#D5 E5 |-3-| |-3-| Q Q Q Q Q Q Q a Q Q Q|----------------|-----------------|||----------------|----------------o|||----------------|-----------------|||-11\-----6-7-7--|-9\--------------|||--9\-----4-5-5--|-7\-----6s7-----o|||----------------|------------9-8--|||| W W|----------------|-----------------|||----------------|----------------o|||----------------|-----------------|||----------------|-----------------|||----------------|----------------o|||----------------|-----------------|| Guitar Solo 2. 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\(B^\top B\) and determining its eigenvalues. Throughout we make use of the fact that \(P^\top = P^{-1}\). We begin with the case \(l \le q\). Use block-partition notation to divide \(B B^\top \) into l blocks of size \(q \times q\). Then$$\begin{aligned} (B B^\top )_{ij}= & {} \sum _{s=1}^q P^{(i-1)(s-1)} (P^\top )^{(s-1)(j-1)} \\= & {} \sum _{s=1}^q P^{(i-j)(s-1)} = \sum _{s=0}^{q-1} P^{(i-j)s} . \end{aligned}$$It readily follows that \( (B B^\top )_{ii} = q I_q , i = 1 , \ldots , q . \) Now observe that, for any nonzero integer k, the set of numbers \(ks ~\mathrm{mod}~q\) as s varies over \(\{ 0 , \ldots , q-1 \}\) equals \(\{ 0 , \ldots , q-1 \}\). (This is where we use the fact that q is a prime number.) Therefore, whenever \(i \ne j\), we have that \( (B B^\top )_{ij} = \sum _{s=0}^{q-1} P^s = \mathbf{1}_{q \times q} , \) where \(\mathbf{1}_{q \times q}\) denotes the \(q \times q\) matrix whose entries are all equal to one. We observe that ql is an eigenvalue of \(B B^\top \), with normalized eigenvector \((1/\sqrt{ql}) \mathbf{1}_{ql}\). Therefore if we define \(M_l = B B^\top - \mathbf{1}_{ql \times ql}\) and partition it commensurately with B, we see that the off-diagonal blocks of \(M_l\) are all equal to zero, while the diagonal blocks are all identical and equal to \(q I_q - \mathbf{1}_{q \times q}\). This is the Laplacian matrix of a fully connected graph with q vertices, and thus has \(q-1\) eigenvalues of q and one eigenvalue of 0. Therefore \(M_l = B B^\top - \mathbf{1}_{ql \times ql}\) has \(l(q-1)\) eigenvalues of q and l eigenvalues of 0. Moreover, \(\mathbf{1}_{ql}\) is an eigenvector of M corresponding to the eigenvalue zero. Therefore \(B B^\top = M_l + \mathbf{1}_{ql} \mathbf{1}_{ql}^\top \) has a single eigenvalue of ql, \(l(q-1)\) eigenvalues of q, and \(l-1\) eigenvalues of 0. This is equivalent to the claim about singular values of \(B^\top \). Now we study the case where \(l \ge q\). Let \(M_q \in \{0,1\}^{q^2 \times q^2}\) denote the matrix in the previous case with \(l = q\).