均分田地，田埂最短问题

mathe

模仿圆形结果，正方形n=9可以得到类似结果
3.96402976129625， 有效率90.4%,第一个超过90%

        A(0.91187069637516,0.50000000000000)
        B(0.72210436901679,0.72268643503705)
        C(0.55268911505887,0.64573925360558)
        D(0.37858788414340,0.74457369504822)
        E(0.26841856989262,0.66672036190947)
        F(0.26841856989262,0.33327963809053)
        G(0.37858788414340,0.25542630495178)
        H(0.55268911505887,0.35426074639442)
        I(0.72210436901679,0.27731356496295)
        J(1.00000000000000,0.50000000000000)
        K(0.74423827925252,1.00000000000000)
        L(0.37222716160246,1.00000000000000)
        M(0.00000000000000,0.71976551259378)
        N(0.00000000000000,0.28023448740622)
        O(0.37222716160247,0.00000000000000)
        P(0.74423827925252,0.00000000000000)
        Theta(B=>A)=-0.18215276949528
        Theta(B=>C)=0.06202961194765
        Theta(C=>H)=0.03523401271244
        Theta(D=>C)=-0.04251423927608
        Theta(E=>D)=-0.04179512196850
        Theta(F=>E)=0.13338470977671
        Theta(G=>F)=-0.04179512196850
        Theta(H=>G)=-0.04251423927608
        Theta(H=>I)=0.06202961194765
        Theta(I=>A)=0.18215276949528
        Theta(A=>J)=0.00000000000000
        Theta(P=>I)=0.07964661830387
        Theta(O=>G)=-0.02489723291985
        Theta(N=>F)=0.19510703291080
        Theta(M=>E)=-0.19510703291080
        Theta(L=>D)=0.02489723291985
        Theta(K=>B)=-0.07964661830387

另外我们可以注意到除了n=5意外，现在所有其它情况我们找到的最佳结果圆形和方形都使用了相同的“点线结构”

mathe

方形n=9的情况还是中心“八边形”的险胜，再次体现了对称的优势
3.94963096319756

        A(0.67780660170043, 0.63671810616270)=: (a, b)
        B(b, a), C(1-b, a), D(1-a, b), E(1-a, 1-b), F(1-b, 1-a), G(b, 1-a), H(a, 1-b)
        I(1, 0.68315675108767)=: (1, c),
        J(c, 1), K(1-c, 1), L(0, c), M(0, 1-c), N(c, 0), P(1-c,0), Q(1, 1-c)
        角弧AB=CD=EF=GH=-0.02449461204714
        角弧BC=DE=FG=HA=-0.23730477575201
        角弧AI=-BJ=CK=-DL=EM=-FN=GP=-HQ=-0.14314699992314

zeroieme

mathe 发表于 2019-6-29 22:37
另外圆形情况在n充分大时，所有区域都近似边长为a的六边形，可以得出田埂总长近似$3sqrt{{2pi n}/{3sqrt{3} ...

提个小意见
\(3 \sqrt{\frac{2 \pi  n}{3 \sqrt{3}}}=\sqrt{2 \pi  \sqrt{3} n}\)

数学星空

关于104#进一步结果：





下面构型I，\(N=12\sim 20\)情形

\(N= 12, R = 0.290764378774303, x = 0.290764378774303, L = 9.62851830604856\)

\(N= 13, R = 0.240136028738117, x = 0.279171182175020, L = 10.4113552667414\)

\(N= 14, R = 0.202976820173302, x = 0.268866694777500, L = 11.2070447597888\)

\(N= 15, R = 0.174681928762035, x = 0.259627942927826, L = 12.0138870043366\)

\(N= 16, R = 0.152510229545802, x = 0.251282616850686, L = 12.8304949585247\)

\(N= 17, R = 0.134732092094370, x = 0.243695101808107, L = 13.6557247145566\)

\(N= 18, R = 0.120204704362928, x = 0.236757049610299, L = 14.4886225525779\)

\(N= 19, R = 0.108144170032553, x = 0.230380646126255, L = 15.3283862126107\)

\(N= 20, R = 0.0979958664296144, x = 0.224493865257003, L = 16.1743348171956\)

通过渐近分析可以得到：\(N=n+1\)

\(L=n-\sqrt{n}-1+\frac{\frac{3}{2}+\frac{\sqrt{3}\pi}{2}+\frac{\pi^2}{3}}{\sqrt{n}}+\frac{\frac{\pi^4}{18}+\frac{\sqrt{3}\pi^3}{6}-\frac{47\pi^2}{24}+\frac{7\pi\sqrt{3}}{6}-\frac{7}{8}}{n^{\frac{3}{2}}}+\dots\)

数学星空

关于mathe 在124#提到的构型III,N=8~11的图形





对于N=13,根据对称性，我们可以构造如下：

第一排都是全对称的，第二排调整了一条到圆周的触角长度，使其相邻的两条触角（人字形）长度不相等~



很可惜，计算结果表明N=13以上对称的构型III好像都不存在？？？

mathe

n=13对称图还是比较容易构造的
9.97723132549458

        A(0.26412405041761,0.15249209160808)
        B(-0.00000000000002,0.30498418321613)
        C(-0.26412405041763,0.15249209160805)
        D(-0.26412405041761,-0.15249209160808)
        E(0.00000000000002,-0.30498418321613)
        F(0.26412405041763,-0.15249209160805)
        G(0.53929396519804,0.31136151597946)
        H(-0.00000000000003,0.62272303195890)
        I(-0.53929396519807,0.31136151597941)
        J(-0.53929396519806,-0.31136151597947)
        K(0.00000000000003,-0.62272303195887)
        L(0.53929396519810,-0.31136151597943)
        M(0.85065548117751,-0.00000000000000)
        N(0.42532774058869,0.73668925656823)
        O(-0.42532774058877,0.73668925656819)
        P(-0.85065548117751,-0.00000000000006)
        Q(-0.42532774058875,-0.73668925656820)
        R(0.42532774058882,-0.73668925656816)
        S(1.00000000000000,0.00000000000000)
        T(0.49999999999993,0.86602540378448)
        U(-0.50000000000001,0.86602540378443)
        V(-1.00000000000000,-0.00000000000007)
        W(-0.49999999999999,-0.86602540378444)
        X(0.50000000000008,-0.86602540378439)
        Theta(A=>B)=0.00000000000000
        Theta(A=>G)=-0.00000000000000
        Theta(B=>C)=-0.00000000000000
        Theta(C=>D)=0.00000000000000
        Theta(D=>E)=0.00000000000000
        Theta(E=>F)=-0.00000000000000
        Theta(F=>A)=-0.00000000000000
        Theta(G=>N)=0.26179938779917
        Theta(H=>B)=-0.00000000000000
        Theta(H=>O)=0.26179938779914
        Theta(I=>C)=0.00000000000000
        Theta(I=>P)=0.26179938779916
        Theta(J=>D)=0.00000000000000
        Theta(J=>Q)=0.26179938779913
        Theta(K=>E)=-0.00000000000000
        Theta(K=>R)=0.26179938779917
        Theta(L=>F)=0.00000000000000
        Theta(L=>M)=0.26179938779913
        Theta(M=>G)=0.26179938779917
        Theta(N=>H)=0.26179938779913
        Theta(O=>I)=0.26179938779916
        Theta(P=>J)=0.26179938779914
        Theta(Q=>K)=0.26179938779917
        Theta(R=>L)=0.26179938779913
        Theta(M=>S)=-0.00000000000000
        Theta(X=>R)=-0.00000000000000
        Theta(W=>Q)=0.00000000000000
        Theta(V=>P)=-0.00000000000000
        Theta(U=>O)=0.00000000000000
        Theta(T=>N)=-0.00000000000000

倪举鹏

用四种等量互不相容的油，同时滴在水面，等摊开，边缘线就是了

mathe

现在二维问题已经解决的相当不错了，但是相应的还有三维问题。比如上海进行垃圾分类，所有垃圾被分为四类。已知现在有一些直径为1米高也为1米的垃圾桶，要求将之用垂直方向的隔板分割为四个体积相同的部分，而且使用的材料最少，那么这个问题等价于本题n=4的问题。考虑到各种垃圾产生量的不同，可以要求四个部分有不同的体积，比如1:2:3:4。
更进一步，我们可以去除隔板为垂直方向的条件，但是要求这四个部分在垃圾桶开口方向(向上)具有相同的开口面积(即使体积不同)，请问该使用什么形状的隔板才使得隔板面积最小

aimisiyou

mathe 发表于 2019-6-30 14:53
方形n=9的情况还是中心“八边形”的险胜，再次体现了对称的优势
3.94963096319756

9等分不是九宫格形状么？

mathe

132#的结果有点接近九宫格