如果我们仅仅查看126#中第二和第三条方程，由于三个变量两条方程，我们还需要额外的极值条件。
如果选择a=0.2072,使用第二条方程，可以解得t2=1.0823532480783912061963768584031196565,
代入第三条方程，对应行列式为关于t1的一个函数，图像如下
可以看出，这时没有t1使得第三条方程成立。
而如果我们选择a=0.2071,使用第二条方程，可以解得t2=1.0583754560589007360690340217022844133
代入第三条方程，对应行列式的图像如下
所以这时可以解得t1=0.86322743127987940258188507393774721629或t1=1.2064235503364808008993124412531140525
类似如果我们修改a=0.20715,t2=1.0694919889194442972821067666160149949,
那么可以求得t1=0.94777682413999966409749131019961311368或1.1319161809545791216932267155646217128
对应26边正方形的图（中间两个正方形没有做出）
这个过程必然有一个最大的a,使得得出a,t2代入第三条方程后，行列式对应的图像和x轴相切，这代表这时这个解应该同时满足第三条方程关于t1的偏导数。
但是为什么软件求解没有找到a在0.20715和0.2072之间的解呢？
然后马上发现其实这时中间小正方形不需要使用不同的倾斜方向，所以我们可以有下图更好的结果
边长0.20731795816007595663838595129040459670
本帖最后由 uk702 于 2020-12-4 08:57 编辑
mathe 发表于 2020-12-4 07:50
如果我们仅仅查看126#中第二和第三条方程，由于三个变量两条方程，我们还需要额外的极值条件。
如果选择a= ...
n=26 才能做到 a=0.2073... 吗？我不太懂 n=26 时需要的全部约束，但如果只是求行列式的解的话，现在
a = 0.263411195766482410000000000000000000; t2 = 0.674388195727422620000000000000000000;t1 = 0.549563195778975530000000000000000000;
求得行列式误差为 7.054261726841471997705696027840869*10^-16
另外，谁能告诉我 Mathematica 如何进行更更精度的计算，现在精确到/显示到小数点后16位后再也无法进一步提高精度，无论怎么样，后面显示的全是 0，
以下的语句，包括设置 $, $MinPrecision = 1000, $MaxPrecision = 1000，都不管用。
Print, NumberForm, NumberForm, NumberForm]
实在搞不定 Mathematica 的精度，弄得每次计算出来的结果不可重复！
本帖最后由 uk702 于 2020-12-4 17:15 编辑
mathe 发表于 2020-12-4 07:50
如果我们仅仅查看126#中第二和第三条方程，由于三个变量两条方程，我们还需要额外的极值条件。
如果选择a= ...
只要是这种构图方式，a = 0.207317958160075956638385951290... 应该是到头了，不管 25 个正方形，还是 26 个正方形。
显然 GK =2 *a <= GH = Sqrt[(-1 + 3 a)^2 + (-6 a + Sqrt)^2]，当等号成立时，解得 a =0.207317958160075956638385951290...
本帖最后由 dlpg070 于 2020-12-5 10:26 编辑
uk702 发表于 2020-12-4 17:07
只要是这种构图方式，a = 0.207317958160075956638385951290... 应该是到头了，不管 25 个正方形，还是 ...
分析有新意,但画图有些粗糙,斜放的正方形位置都不对
为方便你精确画图,把由a26.GGB提取的点,线的数据提供给你
说明:
1 GGB 定义了2个slider,只显示a,b未显示
a=0.20731795816007595
b=1.0694919889194443
2 GGB point--->MMA Point
GGB segment--->MMA
3 初步分析,待完善
用提取数据重画的图形:
a26ggb提取.png
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本帖最后由 uk702 于 2020-12-5 10:51 编辑
dlpg070 发表于 2020-12-5 10:14
分析有新意,但画图有些粗糙,斜放的正方形位置都不对
为方便你精确画图,把由a26.GGB提取的点,线的数据 ...
n=26 时我个人觉得已经没什么招了，想改为尝试改善 n=25 的情况。
这时 GH 要能够排列 3 个小正方形，假设 H 的坐标为 {1-a, 3a-m*a}，因此，必须有 EuclideanDistance >= 3a，Solve==3*a]，解得 m→0.19233210366681976732550681，代入，得 H 坐标为
{0.79268204183992404336161404871,0.582079975459390737912459267}
其倾角为：
a=0.207317958160075956638385951290;h=Sqrt; G={a, h-3*a};
H={0.7926820418399240433616140487, 0.582079975459390737912459267};
t=ArcTan[(h-3*a-0.582079975459390737912459267)/(0.7926820418399240433616140487 - a)]; 求得：t=0.34472149047858860338673920
J 点的坐标为：H - a*{Sin, Cos}
= {0.722622137695431896370256847,0.38695861423277470900471656}
由于
J2=H-2*a*{Sin,Cos}={0.652562233550939749378899646,0.19183725300615868009697386}
所以，直接拉2个单位的话，就会与底下的那排正方形冲突。
K 点为过 J2 作 GH 的平行线与直线 y=a 的交点。
l1=InfiniteLine[{J2,J2-{Cos,-Sin}}];l2=InfiniteLine[{{0,a},{a,a}}];
K=RegionIntersection[]
解得：K= {0.6094474691323290055078487,0.20731795816007595663838595129}
L 点的坐标为：L=K+3*a*{- Cos,Sin}
解得：L= {0.0240833854524809187846206,0.417497670593552397612457555}
16个正方形除了69#的排列方式，还可以
两者边长相同，但是这个结果和70#中17个正方形的方案更加接近
本帖最后由 uk702 于 2020-12-6 10:52 编辑
uk702 发表于 2020-12-5 10:49
n=26 时我个人觉得已经没什么招了，想改为尝试改善 n=25 的情况。
这时 GH 要能够排列 3 个小正方形 ...
@dlpg070 是这样的，中间那张图不是程序生成的，而是手工编辑出来的，Mathematica 支持双击某个图元，然后复制、删除、拖动、旋转（刚发现也能旋转，当保存成 nb 文件时，还能将修改后的图形保存下来，这就方便了），就这样改出来，我觉得这样探测空隙是否有足够空间的一个好办法。
本帖最后由 uk702 于 2020-12-9 21:29 编辑
n=26 时，不知这种情况下大伙是否计算过能否提供 a > 0.207317958160075956638385951... 的边长，从图上来看可能在毫厘之间。(* a262start，尝试放 26 个 *)
(* a262start，尝试放 26 个 *)
a262 := Module[{$MinPrecision = 50, $MaxPrecision = 50, w, h, f, a, esp, sqs, ra, rb, r1, r2, r3},
w = 1.0; h = Sqrt;
f = Rectangle[{0, 0}, {1, h}];
(* a=0.211520823989986052222250797555493157307027099075156414441346; *)
a=0.207317958160075956638385951290;
esp = 10.0^-10;
total = 0.0;
(* 各个矩形 *)
sqs = {}; intersections={};
(* 底下第一排 *)
sqs = Append];
sqs = Append];
sqs = Append];
sqs = Append];
(* 底下第二排 *)
sqs = Append];
(* sqs = Append]; *)
sqs = Append];
sqs = Append];
(* 底下第三排 *)
sqs = Append];
sqs = Append];
sqs = Append];
(* 第四排 *)
sqs = Append];
sqs = Append];
sqs = Append];
(* sqs = Append]; *)
(* 第五排 *)
sqs = Append];
sqs = Append];
sqs = Append];
sqs = Append];
(* 第六排 *)
sqs = Append];
sqs = Append];
sqs = Append];
sqs = Append];
G = {2*a, h-4*a}; H = {1-2*a, 2*a};
(* 过 w1 作 GH 的平行线 l1，过 G 点作 l1 的垂线 l2 *)
w1 = {(2*a + 1 - a)/2, (h-3*a + 2*a)/2};
p1 = w1 + a*(G-H)/2/EuclideanDistance;
p2 = w1 - a*(G-H)/2/EuclideanDistance;
v = RotationTransform;
p3 = p1 + v;
p4 = p2 + v;
r0 = Polygon[{p1, p3, p4, p2}];
r1 = TransformedRegion]; (* 暂时手工写一个 *)
r2 = TransformedRegion];
r3 = TransformedRegion];
r4 = TransformedRegion];
r5 = TransformedRegion];
(* 语句后面不能加 ; ，否则相当于静默不显示图形 *)
Graphics[{EdgeForm[{Thin, RGBColor["#807F7D"]}], White, f,
EdgeForm[{Thin, RGBColor["#E79967"]}], RGBColor["#FBF0E8"],
sqs, r1, r2, r3, r4, r5
(* Yellow, r1, *)
(* Blue, l1, l2 *)
(* Red, PointSize, Point[{w1}] *)
}]
]; a262
(* a262end *)
上面方案在正方形倾斜角度为0.73793551389460718773965984811536761268时，达到边长最大上限0.20721104897365918253946967525090920670。
过(a,a)的正方形一条边的方程为$y-a = -ctan(t)(x-a)$, 于是平移后得到右上方倾斜正方形最上方边方程为$y-a = -ctan(t)(x-a)+\frac{5a}{\sin(t)}$
将x=1代入得到倾斜正方形和最右边边界接触点坐标为$(1,a-ctan(t)(1-a)+\frac{5a}{\sin(t)})$.
这个点和坐标$(4a, \sqrt(2)-2a)$连线在$(-\sin(t),\cos(t))$方向投影为a得到方程$(1-4a)\sin(t)+( \sqrt(2)-2a-a+\frac{1-a}{\tan(t)}-\frac{5a}{\sin(t)})\cos(t)=a$
由此得出上面的最大上限
本帖最后由 dlpg070 于 2020-12-10 21:39 编辑
uk702 发表于 2020-12-9 21:03
n=26 时，不知这种情况下大伙是否计算过能否提供 a > 0.207317958160075956638385951... 的边长，从图上来 ...
利用mathe 139#的计算结果画图,验证
t=0.73793551389460718773965984811536761268;
a =0.20721104897365918253946967525090920670;
是第2优的解
画出了3个破坏的正方形,供参考
从图形看最上面斜放的正方形还可以提高一行,a可稍有有提高,大于0.2072110489,但可能仍小于0.207317958
(*uk702 138# dlpg070修改*)
(*a262start，尝试放 26 个*)
Clear["Global`*"];
t=0.73793551389460718773965984811536761268;
a =0.20721104897365918253946967525090920670;
n=26;u=4;v=6;d=3;f=5;
sub=0;
b=1-u a;
c=Sqrt-v a;
td=t*180/Pi;
Print["n= ",n," \na= ",a,"\nb= ",b,"\nc= ",c,"\nt= ",t,"\ntd= ",td];
am=
(Sin +Sqrt Cos+Cos/Tan)/(4Sin+3Cos + Cos/Tan+5 Cos/Sin+ 1 )
xt=b Sin;yt=b Cos;(* 公式*)
xs=c Cos;ys=c Sin; (* 公式*)
(* a=b Sin+ c Cos*) (* a *);
pBt={a+xt Sin,a-xt Cos};
pAt=pBt+{-aSin,a Cos};
pCt=pBt+{a Cos,a Sin};
pDt=pCt+{-aSin,a Cos};
Print[" xt= ",xt,"\nn= ",n," \na= ",a," \nt= ",t,"\npAt= ",N,"\npBt= ",N,"\nPCt= ",N,"\npDt= ",N];
a262:=Module[{$MinPrecision=50,$MaxPrecision=50,w,h,f,esp,sqs,ra,rb,r1,r2,r3,r4,r5},w=1;h=N,20];(*注意h的写法,确保精度*)
Print["h= ",h];
fr=Rectangle[{0,0},{1,h}];
(*a=0.211520823989986052222250797555493157307027099075156414441346;*)(*a=0.207317958160075956638385951290;*)
esp=10.0^-10;
total=0.0;
(*各个矩形*)sqs={};intersections={};
(*底下第一排*)sqs=Append];
sqs=Append];
sqs=Append];
sqs=Append];
(*底下第二排*)sqs=Append];
(*sqs=Append];*)sqs=Append];
sqs=Append];
(*底下第三排*)sqs=Append];
sqs=Append];
sqs=Append];
(*第四排*)sqs=Append];
sqs=Append];
sqs=Append];
(*sqs=Append];*)(*第五排*)sqs=Append];
sqs=Append];
sqs=Append];
sqs=Append];
(*第六排*)sqs=Append];
sqs=Append];
sqs=Append];
sqs=Append];
sqsd={};(*破坏的正方形 sqsd *)
pgnd=Polygon[{
{1 a, (1+0)a+c},
{(1+1) a, (1+0) a+c},
{(1+1) a, (1+1) a+c},
{1 a,(1+1) a+c}
}];
AppendTo;
pgnd=Polygon[{
{2 a, (2+0)a+c},
{(2+1) a, (2+0) a+c},
{(2+1) a, (2+1) a+c},
{2 a,(2+1) a+c}
}];
AppendTo;
pgnd=Polygon[{
{3 a, (3+0)a+c},
{(3+1) a, (3+0) a+c},
{(3+1) a, (3+1) a+c},
{3 a,(3+1) a+c}
}];
AppendTo;
b=1-4 a;
c=Sqrt-6 a;
M={1,a-c Tan (1-a) + (5 a)/Sin -3.62a};
M={1,a-b Tan (1-a) + (5 a)/Sin -3.62a};(* c?b? 3.62? *)
Nn={4a,Sqrt-2 a};
G={2*a,h-4*a};H={1-2*a,2*a};
(*过 w1 作 GH 的平行线 l1，过 G 点作 l1 的垂线 l2*)w1={(2*a+1-a)/2,(h-3*a+2*a)/2};
p1=w1+a*(G-H)/2/EuclideanDistance;
p2=w1-a*(G-H)/2/EuclideanDistance;
v=RotationTransform;
v=(M-pCt)/4;
Print["v=",v];
p3=p1+v;
p4=p2+v;
r0=Polygon[{p1,p3,p4,p2}];
(*r1=TransformedRegion];*)
r1=Polygon[{pAt,pBt,pCt,pDt}];(*暂时手工写一个*)
r2=Polygon[{pAt+v,pBt+v,pCt+v,pDt+v} ];
r3=Polygon[{pAt+2v,pBt+2 v,pCt+2 v,pDt+2 v}];
r4=Polygon[{pAt+3v,pBt+3v,pCt+3v,pDt+3v}];
r5=Polygon[{pAt+4v,pBt+4v,pCt+4v,pDt+4v}];
(*
r2=TransformedRegion];
r3=TransformedRegion];
r4=TransformedRegion];
r5=TransformedRegion];
*)
(*语句后面不能加;，否则相当于静默不显示图形*)Graphics[{EdgeForm[{Thin,RGBColor["#807F7D"]}],White,fr,Opacity,EdgeForm[{Thin,RGBColor["#a099FF"]}],RGBColor["#FBE0E0"],sqs,r1,r2,r3,r4,r5,
LightGray,sqsd,
(*Yellow,r0,*)
(*Blue,rt,*)
Text,M+{0.03,0}],
Text,Nn+{0.03,0}],
Text,G+{0.03,0}],
Text,H+{0.03,0}],
Text,p1+{0.03,0}],
Text,p2+{0.03,0}],
Red,PointSize,Point[{w1}]}
,PlotLabel->Style<>"_"<>ToString<>"a"<>ToString]],10,Blue,Background-> RGBColor["#FBE0E0"]]]];a262
(*a262end*)