首页> 外文期刊>Automatic Control and Computer Sciences >n -Dimensional Simplices Satisfying Inclusions $$S subset {{[0,1]}^{n}} subset nS$$ ]]>
【24h】

n -Dimensional Simplices Satisfying Inclusions $$S subset {{[0,1]}^{n}} subset nS$$ ]]>

机译: $$ s subset {{[0,1]} ^ {n}} subset ns $$ <! - autcont1807019nevskiii-m1 - > ]]>

获取原文
获取原文并翻译 | 示例
           
-->

    摘要

    Let $$n in mathbb{N}$$ , $${{Q}_{n}} = {{[0,1]}^{n}}.$$ For a nondegenerate simplex $$S subset {{mathbb{R}}^{n}}$$ , by $$sigma S$$ we denote the homothetic image of $$S$$ with center of homothety in the center of gravity of $$S$$ and ratio of homothety $$sigma $$ . By $${{d}_{i}}(S)$$ we mean the $$i$$ th axial diameter of $$S$$ , i. e. the maximum length of a line segment in $$S$$ parallel to the $$i$$ th coordinate axis. Let $$xi (S) = min{ext{{ }}sigma geqslant 1:{{Q}_{n}} subset sigma S{ext{} }},$$ $${{xi }_{n}} = min{ext{{ }}xi (S):S subset {{Q}_{n}}{ext{} }}.$$ By $$lpha (S)$$ we denote the minimal $$sigma > 0$$ such that $${{Q}_{n}}$$ is contained in a translate of simplex $$sigma S$$ . Consider $$(n + 1) imes (n + 1)$$ -matrix $${mathbf{A}}$$ with the rows containing coordinates of vertices of $$S$$ ; last column of $${mathbf{A}}$$ consists of 1’s. Put $${{{mathbf{A}}}^{{ - 1}}}$$ $$ = ({{l}_{{ij}}})$$ . Denote by $${{lambda }_{j}}$$ linear function on $${{mathbb{R}}^{n}}$$ with coefficients from the $$j$$ th column of $${{{mathbf{A}}}^{{ - 1}}}$$ , i.e. $${{lambda }_{j}}(x) = {{l}_{{1j}}}{{x}_{1}} + ldots + {{l}_{{nj}}}{{x}_{n}} + {{l}_{{n + 1,j}}}.$$ Earlier the first author proved the equalities $$frac{1}{{{{d}_{i}}(S)}} = frac{1}{2}sumolimits_{j = 1}^{n + 1} left| {{{l}_{{ij}}}} ight|,lpha (S) = sumolimits_{i = 1}^n frac{1}{{{{d}_{i}}(S)}}.$$ In the present paper we consider the case $$S subset {{Q}_{n}}$$ . Then all the $${{d}_{i}}(S) leqslant 1$$ , therefore, $$n leqslant lpha (S) leqslant xi (S).$$ If for some simplex $$S{kern 1pt} {{'}} subset {{Q}_{n}}$$ holds $$xi (S{kern 1pt} {{'}}) = n,$$ then $${{xi }_{n}} = n$$ , $$xi (S{kern 1pt} {{'}}) = lpha (S{kern 1pt} {{'}})$$ , and $${{d}_{i}}(S{kern 1pt} {{'}}) = 1$$ . However, such the simplices S? ' exist not for all the dimensions $$n$$ . The first value of $$n$$ with such a property is equal to $$2$$ . For each 2-dimensional simplex, $$xi (S) geqslant {{xi }_{2}} = 1 + frac{{3sqrt 5 }}{5} = 2.34 ldots > 2$$ . We have an estimate $$n leqslant {{xi }_{n}} < n + 1$$ . The equality $${{xi }_{n}} = n$$ takes place if there exist an Hadamard matrix of order $$n + 1$$ . Further investigation showed that $${{xi }_{n}} = n$$ also for some other $$n$$ . In particular, simplices with the condition $$S subset {{Q}_{n}} subset nS$$ were built for any odd $$n$$ in the interval $$1 leqslant n leqslant 11$$ . In the first part of the paper we present some new results concerning simplices with such a condition. If $$S subset {{Q}_{n}} subset nS$$ , then center of gravity of $$S$$ coincide with center of $${{Q}_{n}}$$ . We prove that $$sumolimits_{j = 1}^{n + 1} left| {{{l}_{{ij}}}} ight| = 2,(1 leqslant i leqslant n),sumolimits_{i = 1}^n left| {{{l}_{{ij}}}} ight| = frac{{2n}}{{n + 1}}(1 leqslant j leqslant n + 1).$$ Also we give some corollaries. In the second part of the paper we consider the following conjecture. Let for simplex $$S subset {{Q}_{n}}$$ an equality $$xi (S) = {{xi }_{n}}$$ holds. Then $$(n - 1)$$ -dimensional hyperplanes containing the faces of $$S$$ cut off from the cube $${{Q}_{n}}$$ the equal-sized parts. Though it is true for $$n = 2$$ and $$n = 3$$ , in general case this conjecture is not valid .
    机译:让$$ n in mathbb {n} $$,$$ {{q} _ {n}} = {{{[0,1]} ^ {n}}。$$对于nondegenerate simplex $$ s 子集{{ mathbb {r}} ^ {n}} $$,由$$ sigma s $$,我们表示$$ s $$中心的$$的同性恋形象$和homothety $$ sigma $$的比例。 $$ {{d} _ {i}}(s)$$我们的意思是$$我$$的轴向直径$$ e。 $$ S $$平行于$$ I $$ TH坐标轴的最大线段的最大长度。让$$ xi(s)= min { text { {}} sigma geqslant 1:{{q} _ {n}} subset sigma s { text {}}},$$ $ $ {{ xi} _ {n}} = min { text { {}} xi(s):s subset {{q} _ {n}} {n}} { text {}}}。$$通过$$ alpha(s)$$我们表示最小的$$ sigma> 0 $$,这样可以包含在SIMPLEX $$ SIGMA S $的翻译中$。考虑$$(n + 1) times(n + 1)$$ -matrix $$ { mathbf {a}} $$包含包含$$ s $$顶点的坐标的行; $$ { mathbf {a}} $$的最后一列由1的组成。 $$ {{{ mathbf {a}}} ^ {{ - 1}} $$ $$ =({{l} _ {{ij}})$$。在$$ {{ lambda} _ {j} _ {j}} $$ {{ mathbb {r}} ^ {n}} $$上来自$$ J $$的系数{{{ mathbf {a}}} ^ {{ - 1}} $$,即$$ {{ lambda} _ {j}}(x)= {{l} _ {{1j}}} { {x} _ {1}} + ldots + {{l} _ {{nj}} {{x} _ {n}} + {{l} _ {{n + 1,j}}}。$ $前面的第一作者证明了平等$$ tfrac {1} {{{d} _ {i}}(s)}}}}}} = tfrac {1} {2} sum nolimits_ {j = 1} ^ {n + 1} 左手| {{{l} _ {{ij}}} light |, alpha(s)= sum nolimits_ {i = 1} ^ n tfrac {1} {{{{d} _ {i}} (s)}}。$$在本文中,我们考虑案例$$ s subset {{q} _ {n}} $$。然后所有$$ {{d} _ {i}}(s) leqslant 1 $$,因此,$$ n leqslant alpha(s) leqslant xi(s)。$$是否为somethex $ $ s { kern 1pt} {{'}} {{'}} subset {{q} _ {n}} $$ holds $$ xi(s { kern 1pt} {{'}})= n,$$然后是$ $ {{ xi} _ {n}} = n $$,$$,$$ xi(s { kern 1pt} {{'}})= alpha(s { kern 1pt} {{'}})$ $,$$ {{d} _ {i}}(s { kern 1pt} {{'}})= 1 $$。但是,这样的简单s? '不存在所有维度$$ n $$。 $$ n $$的第一个值等于$ 2 $$。对于每2维单纯x,$$ xi(s) geqslant {{ xi} _ {2}} = 1 + tfrac {{3 sqrt 5}} {5} = 2.34 ldots> 2 $$ 。我们有一个估计$$ n leqslant {{ xi} _ {n}}

    著录项

    相似文献

    • 外文文献
    • 中文文献
    • 专利