二阶肓拧教程

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x下载了！第一次见二阶的盲拧

乌木

<P>文中这两个图最好互换一下，因为一般人习惯把公式前的图看作公式要解决的初态。</P>
<P> </P>
<P>&nbsp;</P>

QQ406263

二阶~绝对简单~~

乌木

<SPAN lang=EN-US style="FONT-SIZE: 12pt; COLOR: blue; FONT-FAMILY: 'Times New Roman'; mso-fareast-font-family: 宋体; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA"><SPAN lang=EN-US style="FONT-SIZE: 12pt; COLOR: blue; FONT-FAMILY: 'Times New Roman'; mso-fareast-font-family: 宋体; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">
<P class=MsoNormal style="MARGIN: 0cm 0cm 0pt"><SPAN style="FONT-SIZE: 12pt; FONT-FAMILY: 宋体; mso-font-kerning: 0pt; mso-ascii-font-family: 'Times New Roman'; mso-hansi-font-family: 'Times New Roman'"><FONT color=#000000>教程中讲位置调整法的时候，前面举例为（46574），后面的具体操作变为47865。最好前后一致。</FONT></SPAN></P></SPAN></SPAN>

kexin_xiao

我也和农民探讨过后面调整位置部分的问题，建议农民修改的再通俗一点。

乌木

<P>有人要看看java 演示的例子。下面用三阶的角块代替二阶。第一图为初态，已经整体翻滚得3号角（红－绿－黄）处于复原态。色向编码为1101&nbsp; 1221 ，位置编码及将做的循环解码为 （282）（47654）。</P>
<P>&nbsp;</P>
<applet code="RubikPlayer.class" codebase=3 width="250" height="250">
<param name="ColorTable" value="0xf8f8f8,0x00732f,0xff4400,0xffd200,0x003373,0x8c000f,0x858585">
  <param name="scrptLanguage" value="SupersetENG">
  <param name="beta" value="29">
  <param name="stickersFront" value="4,6,2,6,6,6,5,6,3">
  <param name="stickersRight" value="3,6,1,6,6,6,5,6,2">
  <param name="stickersDown" value="1,6,4,6,6,6,0,6,4">
  <param name="stickersBack" value="3,6,2,6,6,6,3,6,2">
  <param name="stickersLeft" value="1,6,5,6,6,6,4,6,0">
  <param name="stickersUp" value="0,6,5,6,6,6,0,6,1">
</applet>

<applet code="RubikPlayer.class" codebase=3 width="300" height="400">
<param name="ColorTable" value="0xf8f8f8,0x00732f,0xff4400,0xffd200,0x003373,0x8c000f,0x858585">
  <param name="scrptLanguage" value="SupersetENG">
  <param name="scrpt" value="(R U R' U R U2 R' L' U' L U' L' U2 L)  \n R2 B2 (R U R' U R U2 R' L' U' L U' L' U2 L)B2 R2 \n D2 B2 (R U R' U R U2 R' L' U' L U' L' U2 L)B2 D2 \n D' B2 (L' U2 L  U L' U L R U2 R' U' R U' R' )B2 D \n B2 (L' U2 L  U L' U L R U2 R' U' R U' R' )B2 \n D B2 (R U R' U R U2 R' L' U' L U' L' U2 L)B2 D' ">
  <param name="beta" value="29">
  <param name="stickersFront" value="4,6,2,6,6,6,5,6,3">
  <param name="stickersRight" value="3,6,1,6,6,6,5,6,2">
  <param name="stickersDown" value="1,6,4,6,6,6,0,6,4">
  <param name="stickersBack" value="3,6,2,6,6,6,3,6,2">
  <param name="stickersLeft" value="1,6,5,6,6,6,4,6,0">
  <param name="stickersUp" value="0,6,5,6,6,6,0,6,1">
</applet>
<P>&nbsp;</P>   
未完

乌木

17楼的复原态取上红前白右绿……

乌木

<P>继续调位置，1号角无循环，3号角初始即处于复原态，解码：（282）（47654）。</P>
<P>&nbsp;</P>
<applet code="RubikPlayer.class" codebase=3 width="300" height="420">
<param name="ColorTable" value="0xf8f8f8,0x00732f,0xff4400,0xffd200,0x003373,0x8c000f,0x858585">
  <param name="scrptLanguage" value="SupersetENG">
  <param name="scrpt" value="( R U' L' U R' U2 L U' L' U2 L) \n D B2 ( R U' L' U R' U2 L U' L' U2 L)B2 D' \n ( R U' L' U R' U2 L U' L' U2 L)\n R2 B2( R U' L' U R' U2 L U' L' U2 L) B2 R2 \n B2 ( R U' L' U R' U2 L U' L' U2 L)B2 \n D' B2( R U' L' U R' U2 L U' L' U2 L) B2 D \n D2 B2 ( R U' L' U R' U2 L U' L' U2 L)B2 D2 \n R2 B2( R U' L' U R' U2 L U' L' U2 L) B2 R2 ">
  <param name="beta" value="29">
  <param name="stickersFront" value="0,6,3,6,6,6,0,6,4">
  <param name="stickersRight" value="1,6,1,6,6,6,3,6,3">
  <param name="stickersDown" value="5,6,5,6,6,6,2,6,2">
  <param name="stickersBack" value="3,6,1,6,6,6,4,6,4">
  <param name="stickersLeft" value="0,6,4,6,6,6,0,6,1">
  <param name="stickersUp" value="2,6,5,6,6,6,5,6,2">
</applet>

kexin_xiao

乌木老师的JAVA很直观，顶起来！

kexin_xiao

乌木老师的JAVA很直观，顶起来！