🍲 🧔🏼 👈 あなたが数学者ならクリスマスツリーを作る方法＃2 🧔🏻 🤯 🙍🏿

以下のfYolkaに関する昨日の記事の続き。

基本機能

台形

$y = \左| x-4 \右| + \左| x + 2 \右| -5.5$

ここで、数値の係数が2回適用され、係数と減算値の下の定数が変更されます。このセグメントの定数値yとy自体の値でセグメントの長さを調整できます。この機能は、後でドリフトやバケットに役立ちます。

代替楕円

$\ sqrt {\左（x-1 \右）^ {2} +1.9 \左（y-2 \右）^ {2}}-1.3 = 0$

代替の楕円表記。ブラケット内の定数は楕円の中心の座標を担当し、ブラケットの前の定数は軸に沿った圧縮率を担当し、ルートの後ろの数字は半径です。

2点の楕円

$\ sqrt {\左（x-1 \右）^ {2} + \左（y-2 \右）^ {2}} + \ sqrt {\左（x-0.1 \右）^ {2} + \左（y + 1 \右）^ {2}}-3.3 = 0$

, - , (A B ) . .

, .

$\ max \ left（\ left | x \ right | -1、\ left | y \ right | -1 \ right）\ le0$

- :

$\ max \ left（\ left | x \ right |、\ left | y \ right | \ right）\ le1$

-

$s_{1}=\sqrt{\left(x-10\right)^{2}+1.1\left(y-3.85\right)^{2}}-0.55$ $s_{2}=\sqrt{\left(x-10\right)^{2}+1.1\left(y-2.7\right)^{2}}-0.85$ $s_{3}=\sqrt{\left(x-10\right)^{2}+1.2\left(y-1.05\right)^{2}}-1.15$

- min .

$\min\left(s_{1},\ s_{2},s_{3}\right)\le0$

!

-

$-\left|x-1\right|-\left|x+1\right|-y\ge0$

- ,

$2-1.9\left|x-0.3\right|-1.9\left|x+0.3\right|-y\ge0$

, - 2 , - .

-

, .

$x=\frac{\left(\left|y\right|+y\right)}{2}$

$x=\frac{100\left(\left|y\right|-y\right)}{2}$

$x=\left(\frac{\left(\left|y\right|+y\right)}{2}+\frac{100\left(\left|y\right|-y\right)}{2}\right)$

$2-1.9\left|x-0.3\right|-1.9\left|x+0.3\right|-\left(\frac{\left(\left|y\right|+y\right)}{2}+\frac{100\left(\left|y\right|-y\right)}{2}\right)\ge0$

-

$2-1.9\left|x-9.7\right|-1.9\left|x-10.3\right|-\left(\frac{\left(\left|y-4\right|+y-4\right)}{2}+\frac{100\left(\left|y-4\right|-y+4\right)}{2}\right)\ge0$

$s_{4}=2-1.9\left|x-9.7\right|-1.9\left|x-10.3\right|-\left(\frac{\left(\left|y-4\right|+y-4\right)}{2}+\frac{100\left(\left|y-4\right|-y+4\right)}{2}\right)$ $\min\left(s_{1},\ s_{2},s_{3},-s_{4}\right)\le0$

s₄ , >0, <0, .

-

- x = 10, , , .

$h_{1}=\sqrt{\left(\left|x-10\right|\ -\ 0.8\right)^{2}+\left(y-2.7\right)^{2}}+\sqrt{\left(\left|x-10\right|\ -\ 2.8\right)^{2}+\left(y-2.5\right)^{2}}-2.015\\h_1\le0$

, = 10, = 2.55

$h_{2}=\sqrt{\left(\left|x-10\right|\ -\ 1.9\right)^{2}+\left(y-2.55\right)^{2}}+\sqrt{\left(\left|x-10\right|\ -\ 2.3\right)^{2}+\left(\left|y-2.55\right|-0.3\right)^{2}}-0.51\\h_2\le0$

$\min\left(s_{1},\ s_{2},s_{3},-s_{4},h_{1},h_{2}\right)\le0$

- 2

$100\left(\left|x-10\right|-0.2\right)^{2}+100\left(y-3.95\right)^{2}\le1$

-

$\left(300\left(\left|x-10\right|-0.03-0.-\left(y-3.6\right)\right)^{2}+3000\left(y-3.6\right)^{2}\right)\le1$

desmos

s_{1}=\sqrt{\left(x-10\right)^{2}+1.1\left(y-2.7\right)^{2}}-0.85

s_{2}=\sqrt{\left(x-10\right)^{2}+1.2\left(y-1.05\right)^{2}}-1.15

s_{3}=\sqrt{\left(x-10\right)^{2}+1.1\left(y-3.85\right)^{2}}-0.55

s_{4}=2-1.9\left|x-9.7\right|-1.9\left|x-10.3\right|-\left(\frac{\left(\left|y-4\right|+y-4\right)}{2}+\frac{100\left(\left|y-4\right|-y+4\right)}{2}\right)

h_{1}=\sqrt{\left(\left|x-10\right|\ -\ 0.8\right)^{2}+\left(y-2.7\right)^{2}}+\sqrt{\left(\left|x-10\right|\ -\ 2.8\right)^{2}+\left(y-2.5\right)^{2}}-2.015

h_{2}=\sqrt{\left(\left|x-10\right|\ -\ 1.9\right)^{2}+\left(y-2.55\right)^{2}}+\sqrt{\left(\left|x-10\right|\ -\ 2.3\right)^{2}+\left(\left|y-2.55\right|-0.3\right)^{2}}-0.51

\min\left(s_{1},\ s_{2},s_{3},-s_{4},h_{1},h_{2}\right)\le0

100\left(\left|x-10\right|-0.2\right)^{2}+100\left(y-3.95\right)^{2}\le1

\left(300\left(\left|x-10\right|-0.03-0.-\left(y-3.6\right)\right)^{2}+3000\left(y-3.6\right)^{2}\right)\le1

- . .

$d_{1}=-\left|x+7\right|-\left|x-14\right|+22\\d_{2}=\left|x+2.7\right|+\left|x-2.7\right|-6.35\\d_{3}=\left|x-9\right|+\left|x-11\right|-2.8$

$d=d_{1}+\left|d_{1}\right|+d_{2}-\left|d_{2}\right|+d_{3}-\left|d_{3}\right|$

$0.3d\left|\sin\left(13x\right)\right|$

d_{1}=-\left|x+7\right|-\left|x-14\right|+22

d_{2}=\left|x+2.7\right|+\left|x-2.7\right|-6.35

d_{3}=\left|x-9\right|+\left|x-11\right|-2.8

d=d_{1}+\left|d_{1}\right|+d_{2}-\left|d_{2}\right|+d_{3}-\left|d_{3}\right|

0.3d\left|\sin\left(13x\right)\right|

. - " ", , , x .

$\sqrt{\left|x\right|}+\sqrt{\left|y\right|}-0.45\le0$

mod,

$\left|\operatorname{mod}\left(x,2\right)-1\right|$

$f_{1}=\sqrt{\left|\operatorname{mod}\left(x,2\right)-1\right|}+\sqrt{\left|\operatorname{mod}\left(y,2\right)-1\right|}-0.45\\f_1\le0$

, .

$f_{2}=2xx+\left(y-6\right)^{2}-40\\f_{3}=2\left(x-10\right)^{2}+\left(y-2.5\right)^{2}-10\\f_2\le0\\f_3\le0$

$\min\left(-f_{1},f_{2},f_{3}\right)\ge0$

f_{1}=\sqrt{\left|\operatorname{mod}\left(x,2\right)-1\right|}+\sqrt{\left|\operatorname{mod}\left(y,2\right)-1\right|}-0.45

f_{2}=2xx+\left(y-6\right)^{2}-40

f_{3}=2\left(x-10\right)^{2}+\left(y-2.5\right)^{2}-10

\min\left(-f_{1},f_{2},f_{3}\right)\ge0

- . , |x| .

$j_{1}=\left|0.9\left|\left|x\right|-2.1\right|\right|-\left(y-1\right)-0.2\\j_1\le0$

$j_{2}=\left|\left|x\right|-2.1\right|^{2}-\left(y-1\right)^{2}-0.05\\j_2\ge0$

$j_{3}=0.2\left|\left|x\right|-2.1\right|^{2}+0.2\left(y-1\right)^{2}-0.1\\j_3\le0$

$j_{4}=\left(0.5\left|\left|x\right|-2.1\right|\right)^{2}+\left(y-1\right)^{2}-0.02\\j_4\le0$

:

$j_1j_4\le0$

$\max\left(j_{1}j_{4},\ -j_{2},\ j_{3}\right)\le0$

j_{1}=\left|0.9\left|\left|x\right|-2.1\right|\right|-\left(y-1\right)-0.2

j_{2}=\left|\left|x\right|-2.1\right|^{2}-\left(y-1\right)^{2}-0.05

j_{3}=0.2\left|\left|x\right|-2.1\right|^{2}+0.2\left(y-1\right)^{2}-0.1

j_{4}=\left(0.5\left|\left|x\right|-2.1\right|\right)^{2}+\left(y-1\right)^{2}-0.02

\max\left(j_{1}j_{4},\ -j_{2},\ j_{3}\right)\le0

, , .

$x_{1}=x\\y_{1}=y$

2021 MMXXI,

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t2,

$t_{2}=\max\left(\left|\left|x_{1}\right|-1\right|,\left|y_{1}-0.89\right|\right)-0.95\\t_{2}\le0$

"",

$\max\left(\left|1.2\left|x_{1}\right|-1.2\right|,\left|y_{1}-0.9\right|\right)-1\ge0$

V-

$\min\left(\left|2\left|x_{1}\right|-2\right|-y_{1},-\left|2\left|x_{1}\right|-2\right|+y_{1}+0.2,-t_{2}\right)\ge0$

$\max\left(\min\left(-t_{2},\max\left(\left|1.2\left|x_{1}\right|-1.2\right|,\left|y_{1}-0.9\right|\right)-1\right),\min\left(\left|2\left|x_{1}\right|-2\right|-y_{1},-\left|2\left|x_{1}\right|-2\right|+y_{1}+0.2,-t_{2}\right)\right)\ge0$

""

$\max\left(\left|\left|x_{1}\right|-1.05\right|,\left|y_{1}-0.9\right|\right)-1\le0$

$\min\left(\left|\left|x_{1}\right|-1.05\right|-\left|y_{1}-0.9\right|,\ -\left|\left|x_{1}\right|-1.05\right|+\left|y_{1}-0.9\right|+0.15\right)\ge0$

$\max\left(-\min\left(\left|\left|x_{1}\right|-1.05\right|-\left|y_{1}-0.9\right|,\ -\left|\left|x_{1}\right|-1.05\right|+\left|y_{1}-0.9\right|+0.15\right),\max\left(\left|\left|x_{1}\right|-1.05\right|,\left|y_{1}-0.9\right|\right)-1\right)\le0$

4.1 ,

$\max\left(-\min\left(\left|\left|x_{1}-4.1\right|-1.05\right|-\left|y_{1}-0.9\right|,\ -\left|\left|x_{1}-4.1\right|-1.05\right|+\left|y_{1}-0.9\right|+0.15\right),\max\left(\left|\left|x_{1}-4.1\right|-1.05\right|,\left|y_{1}-0.9\right|\right)-1\right)\le0$

"I"

, "I"

$\max\left(\left|x_{1}-6.4\right|-0.06,\left|y_{1}-0.9\right|-01\right)\le0$

t_{2}=\max\left(\left|\left|x_{1}\right|-1\right|,\left|y_{1}-0.89\right|\right)-0.95

\max\left(\min\left(-t_{2},\max\left(\left|1.2\left|x_{1}\right|-1.2\right|,\left|y_{1}-0.9\right|\right)-1\right),\min\left(\left|2\left|x_{1}\right|-2\right|-y_{1},-\left|2\left|x_{1}\right|-2\right|+y_{1}+0.2,-t_{2}\right)\right)\ge0

\max\left(-\min\left(\left|\left|x_{1}-4.1\right|-1.05\right|-\left|y_{1}-0.9\right|,\ -\left|\left|x_{1}-4.1\right|-1.05\right|+\left|y_{1}-0.9\right|+0.15\right),\max\left(\left|\left|x_{1}-4.1\right|-1.05\right|,\left|y_{1}-0.9\right|\right)-1\right)\le0

$x_ {1} = \左（x \ -8 \右）\ cdot1.3 \\ y_ {1} = \左（y-9.3 \右）\ cdot1.3$

, - , , , sin(x), x∈(-5, 5). .

$min = \ frac {f + g} {2}-\ left | \ frac {fg} {2} \ right | \\ max = \ frac {f + g} {2} + \ left | \ frac {fg} {2} \右|$

したがって、図の式でmin関数とmax関数を使用することは、このタスクでは合法です。

あなたが数学者ならクリスマスツリーを作る方法＃2

基本機能

台形

代替楕円

2点の楕円

-

!

-

- ,

-

-

-

- 2

-

desmos

:

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