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 8 clarification after Victor's comment; fix variable names edited Aug 6 at 3:41 François G. Dorais♦ 38k55 gold badges123123 silver badges205205 bronze badges Very recently, Hao Huang proved the Sensitivity Conjecture, which had been open for 30 years or so. Huang's proof is surprisingly short and easy. Here is Huang's preprint, a discussion on Scott Aaronson's blog, and a one-page streamlined proof by Don Knuth. Huang's proof relies on the existence of a $$2^n \times 2^n$$ Hermitian unitary matrix $$U_n$$ ($$= A_n/\sqrt n$$, where $$A_n$$ is from Huang's paper) such that, for some enumeration $$v_1,v_2,\ldots,v_{2^n}$$ of the vertices of the hypercube graph $$Q_n$$, we have that the $$(i,j)$$th entry of $$U_n$$ is $$0$$ unless $$(v_i,v_j)$$ is an edge of $$Q_n$$. (Huang's matrix is actually symmetric but the more general Hermitian case also handles the skew-symmetric Klee-Minty matrix described by Knuth in a footnote.) We can view this Hermitian unitary matrix as a (complex-valued) weighting on the edges of the graph. In the general case, for a graph $$G = (V,E)$$, lets say that a weighting $$w:E \to \mathbb{C}$$ is Hermitian unitary if the matrix $$U$$ whose entries are $$u_{ij} = \begin{cases} w(u_i,u_j) & \text{if (u_i,u_j) \in E,} \\ 0 & \text{otherwise,} \end{cases}$$$$u_{ij} = \begin{cases} w(v_i,v_j) & \text{if (v_i,v_j) \in E,} \\ 0 & \text{otherwise,} \end{cases}$$ is Hermitian unitary for any (and hence all) enumerations $$v_1,v_2,\ldots,v_n$$ of $$V$$. In other words: Hermitian: $$w(u,v) = \overline{w(v,u)}$$ for all $$(u,v) \in E$$ Unitary: for all $$u,v \in V$$, $$\sum_{\substack{t \in V \\ (u,t),(v,t) \in E}} w(u,t)\overline{w(v,t)} = \sum_{\substack{t \in V \\ (u,t),(t,v) \in E}} w(u,t)w(t,v) = \begin{cases} 1 & \text{if u = v,} \\ 0 & \text{otherwise.} \end{cases}$$ Theorem. Suppose $$G = (V,E)$$ has a Hermitian unitary weighting $$w$$. Then for every set $$H \subseteq V$$ with $$|H| > n/2$$, there is a $$v \in H$$ which is connected to at least $$1/\Vert U\Vert_{\infty}$$ vertices inside $$H$$, where $$\|U\|_\infty$$ is the maximum absolute value $$|w(u,v)|$$ ranging over edges $$uv$$ of $$G$$. (In the case of Huang's proof, $$U = \frac{1}{\sqrt{n}}A_n$$ has all entries in $$\{0,\pm1/\sqrt{n}\}$$ and the result follows immediately. In the Klee-Minty case the entries of $$U = \frac{i}{\sqrt n}\widehat{A}_n$$ are all in $$\{0,\pm i/\sqrt{n}\},$$ using notation from Knuth.) It is surprising that this was not noticed for 30 years until Huang put the pieces together. (The Klee-Minty cube has been around since 1973!) Why this was never noticed is interesting but, even after this fact, there are still some follow-up questions: Which graphs have Hermitian unitary weightings as above? Is this rare, common, neither? Is there an algorithm to build such edge weights, given that one exists? How hard is it to compute the minimum value of $$\Vert U \Vert_\infty,$$ given that a Hermitian unitary weighting exists? And probably many more related questions... Since there is strong evidence that connections between existing literature are missing, this is an opportune time to bridge these gaps. Since the theorem above is somewhat more general than what Huang proved, here is a quick proof: Since $$U$$ is Hermitian, it is unitarily diagonalizable with real eigenvalues. Since $$U$$ is unitary, the only possible eigenvalues are $$\pm1.$$ One of the two eigenspaces $$E_{\pm} = \{ x : Ux = \pm x \}$$ has dimension at least $$n/2.$$ Replacing $$U$$ by $$-U$$ if necessary, we may suppose $$\dim E_{+} \geq n/2.$$ Since $$|V - H| < n/2,$$ we can find an eigenvector $$x \in E_{+}$$ such that $$x_i = 0$$ when $$v_i \notin H.$$ Scaling $$x$$ if necessary, we may assume that $$\|x\|_\infty = 1$$ and that there is a $$j$$ with $$x_j = 1.$$ Then $$1 = |x_j| = \Big|\sum_{k=1}^n w(v_j,v_k)x_k\Big| \leq \sum_{k=1}^n |w(v_j,v_k)||x_k| \leq \Vert U\Vert_\infty |\{ k \mid w(v_j,v_k) \neq 0, x_k \neq 0\}|.$$ Now $$v_j \in H$$ since $$x_j \neq 0$$ and since $$\{ k \mid w(v_j,v_k) \neq 0, x_k \neq 0\} \subseteq \{ k \mid (v_j,v_k) \in E, v_k \in H \}$$ we see that $$v_j$$ is connected to at least $$1/\Vert U \Vert_\infty$$ vertices in $$H.$$ Very recently, Hao Huang proved the Sensitivity Conjecture, which had been open for 30 years or so. Huang's proof is surprisingly short and easy. Here is Huang's preprint, a discussion on Scott Aaronson's blog, and a one-page streamlined proof by Don Knuth. Huang's proof relies on the existence of a $$2^n \times 2^n$$ Hermitian unitary matrix $$U_n$$ ($$= A_n/\sqrt n$$, where $$A_n$$ is from Huang's paper) such that, for some enumeration $$v_1,v_2,\ldots,v_{2^n}$$ of the vertices of the hypercube graph $$Q_n$$, we have that the $$(i,j)$$th entry of $$U_n$$ is $$0$$ unless $$(v_i,v_j)$$ is an edge of $$Q_n$$. (Huang's matrix is actually symmetric but the more general Hermitian case also handles the skew-symmetric Klee-Minty matrix described by Knuth in a footnote.) We can view this Hermitian unitary matrix as a (complex-valued) weighting on the edges of the graph. In the general case, for a graph $$G = (V,E)$$, lets say that a weighting $$w:E \to \mathbb{C}$$ is Hermitian unitary if the matrix $$U$$ whose entries are $$u_{ij} = \begin{cases} w(u_i,u_j) & \text{if (u_i,u_j) \in E,} \\ 0 & \text{otherwise,} \end{cases}$$ is Hermitian unitary for any (and hence all) enumerations $$v_1,v_2,\ldots,v_n$$ of $$V$$. Theorem. Suppose $$G = (V,E)$$ has a Hermitian unitary weighting $$w$$. Then for every set $$H \subseteq V$$ with $$|H| > n/2$$, there is a $$v \in H$$ which is connected to at least $$1/\Vert U\Vert_{\infty}$$ vertices inside $$H$$, where $$\|U\|_\infty$$ is the maximum absolute value $$|w(u,v)|$$ ranging over edges $$uv$$ of $$G$$. (In the case of Huang's proof, $$U = \frac{1}{\sqrt{n}}A_n$$ has all entries in $$\{0,\pm1/\sqrt{n}\}$$ and the result follows immediately. In the Klee-Minty case the entries of $$U = \frac{i}{\sqrt n}\widehat{A}_n$$ are all in $$\{0,\pm i/\sqrt{n}\},$$ using notation from Knuth.) It is surprising that this was not noticed for 30 years until Huang put the pieces together. (The Klee-Minty cube has been around since 1973!) Why this was never noticed is interesting but, even after this fact, there are still some follow-up questions: Which graphs have Hermitian unitary weightings as above? Is this rare, common, neither? Is there an algorithm to build such edge weights, given that one exists? How hard is it to compute the minimum value of $$\Vert U \Vert_\infty,$$ given that a Hermitian unitary weighting exists? And probably many more related questions... Since there is strong evidence that connections between existing literature are missing, this is an opportune time to bridge these gaps. Since the theorem above is somewhat more general than what Huang proved, here is a quick proof: Since $$U$$ is Hermitian, it is unitarily diagonalizable with real eigenvalues. Since $$U$$ is unitary, the only possible eigenvalues are $$\pm1.$$ One of the two eigenspaces $$E_{\pm} = \{ x : Ux = \pm x \}$$ has dimension at least $$n/2.$$ Replacing $$U$$ by $$-U$$ if necessary, we may suppose $$\dim E_{+} \geq n/2.$$ Since $$|V - H| < n/2,$$ we can find an eigenvector $$x \in E_{+}$$ such that $$x_i = 0$$ when $$v_i \notin H.$$ Scaling $$x$$ if necessary, we may assume that $$\|x\|_\infty = 1$$ and that there is a $$j$$ with $$x_j = 1.$$ Then $$1 = |x_j| = \Big|\sum_{k=1}^n w(v_j,v_k)x_k\Big| \leq \sum_{k=1}^n |w(v_j,v_k)||x_k| \leq \Vert U\Vert_\infty |\{ k \mid w(v_j,v_k) \neq 0, x_k \neq 0\}|.$$ Now $$v_j \in H$$ since $$x_j \neq 0$$ and since $$\{ k \mid w(v_j,v_k) \neq 0, x_k \neq 0\} \subseteq \{ k \mid (v_j,v_k) \in E, v_k \in H \}$$ we see that $$v_j$$ is connected to at least $$1/\Vert U \Vert_\infty$$ vertices in $$H.$$ Very recently, Hao Huang proved the Sensitivity Conjecture, which had been open for 30 years or so. Huang's proof is surprisingly short and easy. Here is Huang's preprint, a discussion on Scott Aaronson's blog, and a one-page streamlined proof by Don Knuth. Huang's proof relies on the existence of a $$2^n \times 2^n$$ Hermitian unitary matrix $$U_n$$ ($$= A_n/\sqrt n$$, where $$A_n$$ is from Huang's paper) such that, for some enumeration $$v_1,v_2,\ldots,v_{2^n}$$ of the vertices of the hypercube graph $$Q_n$$, we have that the $$(i,j)$$th entry of $$U_n$$ is $$0$$ unless $$(v_i,v_j)$$ is an edge of $$Q_n$$. (Huang's matrix is actually symmetric but the more general Hermitian case also handles the skew-symmetric Klee-Minty matrix described by Knuth in a footnote.) We can view this Hermitian unitary matrix as a (complex-valued) weighting on the edges of the graph. In the general case, for a graph $$G = (V,E)$$, lets say that a weighting $$w:E \to \mathbb{C}$$ is Hermitian unitary if the matrix $$U$$ whose entries are $$u_{ij} = \begin{cases} w(v_i,v_j) & \text{if (v_i,v_j) \in E,} \\ 0 & \text{otherwise,} \end{cases}$$ is Hermitian unitary for any (and hence all) enumerations $$v_1,v_2,\ldots,v_n$$ of $$V$$. In other words: Hermitian: $$w(u,v) = \overline{w(v,u)}$$ for all $$(u,v) \in E$$ Unitary: for all $$u,v \in V$$, $$\sum_{\substack{t \in V \\ (u,t),(v,t) \in E}} w(u,t)\overline{w(v,t)} = \sum_{\substack{t \in V \\ (u,t),(t,v) \in E}} w(u,t)w(t,v) = \begin{cases} 1 & \text{if u = v,} \\ 0 & \text{otherwise.} \end{cases}$$ Theorem. Suppose $$G = (V,E)$$ has a Hermitian unitary weighting $$w$$. Then for every set $$H \subseteq V$$ with $$|H| > n/2$$, there is a $$v \in H$$ which is connected to at least $$1/\Vert U\Vert_{\infty}$$ vertices inside $$H$$, where $$\|U\|_\infty$$ is the maximum absolute value $$|w(u,v)|$$ ranging over edges $$uv$$ of $$G$$. (In the case of Huang's proof, $$U = \frac{1}{\sqrt{n}}A_n$$ has all entries in $$\{0,\pm1/\sqrt{n}\}$$ and the result follows immediately. In the Klee-Minty case the entries of $$U = \frac{i}{\sqrt n}\widehat{A}_n$$ are all in $$\{0,\pm i/\sqrt{n}\},$$ using notation from Knuth.) It is surprising that this was not noticed for 30 years until Huang put the pieces together. (The Klee-Minty cube has been around since 1973!) Why this was never noticed is interesting but, even after this fact, there are still some follow-up questions: Which graphs have Hermitian unitary weightings as above? Is this rare, common, neither? Is there an algorithm to build such edge weights, given that one exists? How hard is it to compute the minimum value of $$\Vert U \Vert_\infty,$$ given that a Hermitian unitary weighting exists? And probably many more related questions... Since there is strong evidence that connections between existing literature are missing, this is an opportune time to bridge these gaps. Since the theorem above is somewhat more general than what Huang proved, here is a quick proof: Since $$U$$ is Hermitian, it is unitarily diagonalizable with real eigenvalues. Since $$U$$ is unitary, the only possible eigenvalues are $$\pm1.$$ One of the two eigenspaces $$E_{\pm} = \{ x : Ux = \pm x \}$$ has dimension at least $$n/2.$$ Replacing $$U$$ by $$-U$$ if necessary, we may suppose $$\dim E_{+} \geq n/2.$$ Since $$|V - H| < n/2,$$ we can find an eigenvector $$x \in E_{+}$$ such that $$x_i = 0$$ when $$v_i \notin H.$$ Scaling $$x$$ if necessary, we may assume that $$\|x\|_\infty = 1$$ and that there is a $$j$$ with $$x_j = 1.$$ Then $$1 = |x_j| = \Big|\sum_{k=1}^n w(v_j,v_k)x_k\Big| \leq \sum_{k=1}^n |w(v_j,v_k)||x_k| \leq \Vert U\Vert_\infty |\{ k \mid w(v_j,v_k) \neq 0, x_k \neq 0\}|.$$ Now $$v_j \in H$$ since $$x_j \neq 0$$ and since $$\{ k \mid w(v_j,v_k) \neq 0, x_k \neq 0\} \subseteq \{ k \mid (v_j,v_k) \in E, v_k \in H \}$$ we see that $$v_j$$ is connected to at least $$1/\Vert U \Vert_\infty$$ vertices in $$H.$$ 7 clarification after Victor's comment edited Aug 6 at 3:28 François G. Dorais♦ 38k55 gold badges123123 silver badges205205 bronze badges Very recently, Hao Huang proved the Sensitivity Conjecture, which had been open for 30 years or so. Huang's proof is surprisingly short and easy. Here is Huang's preprint, a discussion on Scott Aaronson's blog, and a one-page streamlined proof by Don Knuth. Huang's proof relies on the existence of an edge weighing $$w$$ on the hypercube graph $$Q_n$$. Where if one forms thea $$2^n\times2^n$$$$2^n \times 2^n$$ Hermitian unitary matrix $$U_n$$ ($$= A_n/\sqrt n$$, where $$A_n$$ is from Huang's paper) where the $$(i,j)$$-th entry of $$U_n$$ is $$w(v_i,v_j)$$such that, for some enumeration $$v_1,\ldots,v_{2^n}$$$$v_1,v_2,\ldots,v_{2^n}$$ of the vertices of the hypercube graph $$Q_n$$, thenwe have that the $$(i,j)$$th entry of $$U_n$$ is both unitary and Hermitian$$0$$ unless $$(v_i,v_j)$$ is an edge of $$Q_n$$. (Huang's matrix is actually symmetric but the more general Hermitian case also covershandles the skew-symmetric Klee-Minty matrix described by Knuth in a footnote.) It's easy to see that Huang's result holds for any graph $$G=(V,E)$$ with a weighting $$w$$We can view this Hermitian unitary matrix as described above: Theorem. Suppose there is a weighting $$w$$ of(complex-valued) weighting on the edges of the graph. In the general case, for a graph $$G = (V,E)$$ such, lets say that the matrixa weighting $$U$$$$w:E \to \mathbb{C}$$ is both Hermitian and unitary, where if the $$(i,j)$$-th entry ofmatrix $$U$$ iswhose entries are $$w(v_i,v_j)$$$$u_{ij} = \begin{cases} w(u_i,u_j) & \text{if (u_i,u_j) \in E,} \\ 0 & \text{otherwise,} \end{cases}$$ is Hermitian unitary for some enumerationany (and hence all) enumerations $$v_1,v_2,\ldots,v_n$$ of $$V$$. Theorem. Suppose $$G = (V,E)$$ has a Hermitian unitary weighting $$w$$. Then for every set $$H \subseteq V$$ with $$|H| > n/2$$, there is a $$v \in H$$ which is connected to at least $$1/\Vert U\Vert_{\infty}$$ vertices inside $$H$$, where $$\|U\|_\infty$$ is the maximum absolute value $$|w(u,v)|$$ ranging over edges $$uv$$ of $$G$$. (In the case of Huang's proof, $$U = \frac{1}{\sqrt{n}}A_n$$ has all entries in $$\{0,\pm1/\sqrt{n}\}$$ and the result follows immediately. In the Klee-Minty case the entries of $$U = \frac{i}{\sqrt n}\widehat{A}_n$$ are all in $$\{0,\pm i/\sqrt{n}\},$$ using notation from Knuth.) It is surprising that this was not noticed for 30 years until Huang put the pieces together. (The Klee-Minty cube has been around since 1973!) Why this was never noticed is interesting but, even after this fact, there are still some follow-up questions: Which graphs have Hermitian unitary edge weightsweightings as above? Is this rare, common, neither? Is there an algorithm to build such edge weights, given that one exists? How hard is it to compute the minimum value of $$\Vert U \Vert_\infty,$$ given that such a Hermitian unitary weighting exists? And probably many more related questions... Since there is strong evidence that connections between existing literature are missing, this is an opportune time to fixbridge these lacunasgaps. Since the theorem above is somewhat more general than what Huang proved, here is a quick proof: Since $$U$$ is Hermitian, it is unitarily diagonalizable with real eigenvalues. Since $$U$$ is unitary, the only possible eigenvalues are $$\pm1.$$ One of the two eigenspaces $$E_{\pm} = \{ x : Ux = \pm x \}$$ has dimension at least $$n/2.$$ Replacing $$U$$ by $$-U$$ if necessary, we may suppose $$\dim E_{+} \geq n/2.$$ Since $$|V - H| < n/2,$$ we can find an eigenvector $$x \in E_{+}$$ such that $$x_i = 0$$ when $$v_i \notin H.$$ Scaling $$x$$ if necessary, we may assume that $$\|x\|_\infty = 1$$ and that there is a $$j$$ with $$x_j = 1.$$ Then $$1 = |x_j| = \Big|\sum_{k=1}^n w(v_j,v_k)x_k\Big| \leq \sum_{k=1}^n |w(v_j,v_k)||x_k| \leq \Vert U\Vert_\infty |\{ k \mid w(v_j,v_k) \neq 0, x_k \neq 0\}|.$$ Now $$v_j \in H$$ since $$x_j \neq 0$$ and since $$\{ k \mid w(v_j,v_k) \neq 0, x_k \neq 0\} \subseteq \{ k \mid (v_j,v_k) \in E, v_k \in H \}$$ we see that $$v_j$$ is connected to at least $$1/\Vert U \Vert_\infty$$ vertices in $$H.$$ Very recently, Hao Huang proved the Sensitivity Conjecture, which had been open for 30 years or so. Huang's proof is surprisingly short and easy. Here is Huang's preprint, a discussion on Scott Aaronson's blog, and a one-page streamlined proof by Don Knuth. Huang's proof relies on the existence of an edge weighing $$w$$ on the hypercube graph $$Q_n$$. Where if one forms the $$2^n\times2^n$$ matrix $$U_n$$ ($$= A_n/\sqrt n$$ from Huang's paper) where the $$(i,j)$$-th entry of $$U_n$$ is $$w(v_i,v_j)$$ for some enumeration $$v_1,\ldots,v_{2^n}$$ of the vertices of $$Q_n$$, then $$U_n$$ is both unitary and Hermitian. (Huang's matrix is actually symmetric but the more general Hermitian case also covers the skew-symmetric Klee-Minty matrix described by Knuth in a footnote.) It's easy to see that Huang's result holds for any graph $$G=(V,E)$$ with a weighting $$w$$ as described above: Theorem. Suppose there is a weighting $$w$$ of the edges of the graph $$G = (V,E)$$ such that the matrix $$U$$ is both Hermitian and unitary, where the $$(i,j)$$-th entry of $$U$$ is $$w(v_i,v_j)$$ for some enumeration $$v_1,v_2,\ldots,v_n$$ of $$V$$. Then for every set $$H \subseteq V$$ with $$|H| > n/2$$, there is a $$v \in H$$ which is connected to at least $$1/\Vert U\Vert_{\infty}$$ vertices inside $$H$$, where $$\|U\|_\infty$$ is the maximum absolute value $$|w(u,v)|$$ ranging over edges $$uv$$ of $$G$$. (In the case of Huang's proof, $$U = \frac{1}{\sqrt{n}}A_n$$ has all entries in $$\{0,\pm1/\sqrt{n}\}$$ and the result follows immediately. In the Klee-Minty case the entries of $$U = \frac{i}{\sqrt n}\widehat{A}_n$$ are all in $$\{0,\pm i/\sqrt{n}\},$$ using notation from Knuth.) It is surprising that this was not noticed for 30 years until Huang put the pieces together. (The Klee-Minty cube has been around since 1973!) Why this was never noticed is interesting but, even after this fact, there are still some follow-up questions: Which graphs have Hermitian unitary edge weights as above? Is there an algorithm to build such edge weights, given that one exists? How hard is it to compute the minimum value of $$\Vert U \Vert_\infty,$$ given that such a weighting exists? And probably many more related questions... Since there is strong evidence that connections between existing literature are missing, this is an opportune time to fix these lacunas. Since the theorem above is somewhat more general than what Huang proved, here is a quick proof: Since $$U$$ is Hermitian, it is unitarily diagonalizable with real eigenvalues. Since $$U$$ is unitary, the only possible eigenvalues are $$\pm1.$$ One of the two eigenspaces $$E_{\pm} = \{ x : Ux = \pm x \}$$ has dimension at least $$n/2.$$ Replacing $$U$$ by $$-U$$ if necessary, we may suppose $$\dim E_{+} \geq n/2.$$ Since $$|V - H| < n/2,$$ we can find an eigenvector $$x \in E_{+}$$ such that $$x_i = 0$$ when $$v_i \notin H.$$ Scaling $$x$$ if necessary, we may assume that $$\|x\|_\infty = 1$$ and that there is a $$j$$ with $$x_j = 1.$$ Then $$1 = |x_j| = \Big|\sum_{k=1}^n w(v_j,v_k)x_k\Big| \leq \sum_{k=1}^n |w(v_j,v_k)||x_k| \leq \Vert U\Vert_\infty |\{ k \mid w(v_j,v_k) \neq 0, x_k \neq 0\}|.$$ Now $$v_j \in H$$ since $$x_j \neq 0$$ and since $$\{ k \mid w(v_j,v_k) \neq 0, x_k \neq 0\} \subseteq \{ k \mid (v_j,v_k) \in E, v_k \in H \}$$ we see that $$v_j$$ is connected to at least $$1/\Vert U \Vert_\infty$$ vertices in $$H.$$ Very recently, Hao Huang proved the Sensitivity Conjecture, which had been open for 30 years or so. Huang's proof is surprisingly short and easy. Here is Huang's preprint, a discussion on Scott Aaronson's blog, and a one-page streamlined proof by Don Knuth. Huang's proof relies on the existence of a $$2^n \times 2^n$$ Hermitian unitary matrix $$U_n$$ ($$= A_n/\sqrt n$$, where $$A_n$$ is from Huang's paper) such that, for some enumeration $$v_1,v_2,\ldots,v_{2^n}$$ of the vertices of the hypercube graph $$Q_n$$, we have that the $$(i,j)$$th entry of $$U_n$$ is $$0$$ unless $$(v_i,v_j)$$ is an edge of $$Q_n$$. (Huang's matrix is actually symmetric but the more general Hermitian case also handles the skew-symmetric Klee-Minty matrix described by Knuth in a footnote.) We can view this Hermitian unitary matrix as a (complex-valued) weighting on the edges of the graph. In the general case, for a graph $$G = (V,E)$$, lets say that a weighting $$w:E \to \mathbb{C}$$ is Hermitian unitary if the matrix $$U$$ whose entries are $$u_{ij} = \begin{cases} w(u_i,u_j) & \text{if (u_i,u_j) \in E,} \\ 0 & \text{otherwise,} \end{cases}$$ is Hermitian unitary for any (and hence all) enumerations $$v_1,v_2,\ldots,v_n$$ of $$V$$. Theorem. Suppose $$G = (V,E)$$ has a Hermitian unitary weighting $$w$$. Then for every set $$H \subseteq V$$ with $$|H| > n/2$$, there is a $$v \in H$$ which is connected to at least $$1/\Vert U\Vert_{\infty}$$ vertices inside $$H$$, where $$\|U\|_\infty$$ is the maximum absolute value $$|w(u,v)|$$ ranging over edges $$uv$$ of $$G$$. (In the case of Huang's proof, $$U = \frac{1}{\sqrt{n}}A_n$$ has all entries in $$\{0,\pm1/\sqrt{n}\}$$ and the result follows immediately. In the Klee-Minty case the entries of $$U = \frac{i}{\sqrt n}\widehat{A}_n$$ are all in $$\{0,\pm i/\sqrt{n}\},$$ using notation from Knuth.) It is surprising that this was not noticed for 30 years until Huang put the pieces together. (The Klee-Minty cube has been around since 1973!) Why this was never noticed is interesting but, even after this fact, there are still some follow-up questions: Which graphs have Hermitian unitary weightings as above? Is this rare, common, neither? Is there an algorithm to build such edge weights, given that one exists? How hard is it to compute the minimum value of $$\Vert U \Vert_\infty,$$ given that a Hermitian unitary weighting exists? And probably many more related questions... Since there is strong evidence that connections between existing literature are missing, this is an opportune time to bridge these gaps. Since the theorem above is somewhat more general than what Huang proved, here is a quick proof: Since $$U$$ is Hermitian, it is unitarily diagonalizable with real eigenvalues. Since $$U$$ is unitary, the only possible eigenvalues are $$\pm1.$$ One of the two eigenspaces $$E_{\pm} = \{ x : Ux = \pm x \}$$ has dimension at least $$n/2.$$ Replacing $$U$$ by $$-U$$ if necessary, we may suppose $$\dim E_{+} \geq n/2.$$ Since $$|V - H| < n/2,$$ we can find an eigenvector $$x \in E_{+}$$ such that $$x_i = 0$$ when $$v_i \notin H.$$ Scaling $$x$$ if necessary, we may assume that $$\|x\|_\infty = 1$$ and that there is a $$j$$ with $$x_j = 1.$$ Then $$1 = |x_j| = \Big|\sum_{k=1}^n w(v_j,v_k)x_k\Big| \leq \sum_{k=1}^n |w(v_j,v_k)||x_k| \leq \Vert U\Vert_\infty |\{ k \mid w(v_j,v_k) \neq 0, x_k \neq 0\}|.$$ Now $$v_j \in H$$ since $$x_j \neq 0$$ and since $$\{ k \mid w(v_j,v_k) \neq 0, x_k \neq 0\} \subseteq \{ k \mid (v_j,v_k) \in E, v_k \in H \}$$ we see that $$v_j$$ is connected to at least $$1/\Vert U \Vert_\infty$$ vertices in $$H.$$ 6 edited tags | link edited Aug 5 at 14:28 François G. Dorais♦ 38k55 gold badges123123 silver badges205205 bronze badges 5 added proof edited Aug 5 at 14:04 François G. Dorais♦ 38k55 gold badges123123 silver badges205205 bronze badges 4 typo edited Aug 5 at 11:28 François G. Dorais♦ 38k55 gold badges123123 silver badges205205 bronze badges Post Undeleted by François G. Dorais♦ occurred Aug 5 at 5:18 3 edited body edited Aug 5 at 5:18 François G. Dorais♦ 38k55 gold badges123123 silver badges205205 bronze badges Post Deleted by François G. Dorais♦ occurred Aug 5 at 5:12 2 added 7 characters in body edited Aug 5 at 5:06 François G. Dorais♦ 38k55 gold badges123123 silver badges205205 bronze badges Post Made Community Wiki by François G. Dorais♦ occurred Aug 5 at 5:00 1 asked Aug 5 at 5:00 François G. Dorais♦ 38k55 gold badges123123 silver badges205205 bronze badges
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