棱块三置换之二:利用中层
**** Hidden Message *****</p>[此贴子已经被作者于2006-9-16 1:33:47编辑过]
<p>超短循环,在成对解决棱块时效率很高。</p><p><applet codebase="3" height="300" width="300" code="RubikPlayer.class"><param value="SupersetENG" name="scrptLanguage"/><param value="MF' R2 MF R2 " name="scrpt"/></applet></p><p></p> <p>这里所列出的公式均以 UR 棱块为出发点,</p><p>抛离这一局限,利用其原理用于更自由的棱块还原就更美妙了 :0</p><p>角块或许也有类似的解决方案?</p> <p>Hi!</p><p></p> <p>这个如何:</p><p> <applet codebase="3" height="200" width="200" code="RubikPlayer.class"><param value="SupersetENG" name="scrptLanguage"/><param value="R U' R' U R U' R' U " name="scrpt"/><param value="6,0,6,0,0,0,6,0,6" name="stickersFront"/><param value="6,1,6,1,1,1,6,1,6" name="stickersRight"/><param value="6,2,6,2,2,2,6,2,6" name="stickersDown"/><param value="6,3,6,3,3,3,6,3,6" name="stickersBack"/><param value="6,4,6,4,4,4,6,4,6" name="stickersLeft"/><param value="6,5,6,5,5,5,6,5,6" name="stickersUp"/></applet></p> 顶顶顶!收藏!!! <div class="msgheader">QUOTE:</div><div class="msgborder"><b>以下是引用<i>乌木</i>在2006-9-16 10:52:27的发言:</b><br/><p>这个如何:</p><p> <applet codebase="3" height="200" width="200" code="RubikPlayer.class"><param value="SupersetENG" name="scrptLanguage"/><param value="R U' R' U R U' R' U " name="scrpt"/><param value="6,0,6,0,0,0,6,0,6" name="stickersFront"/><param value="6,1,6,1,1,1,6,1,6" name="stickersRight"/><param value="6,2,6,2,2,2,6,2,6" name="stickersDown"/><param value="6,3,6,3,3,3,6,3,6" name="stickersBack"/><param value="6,4,6,4,4,4,6,4,6" name="stickersLeft"/><param value="6,5,6,5,5,5,6,5,6" name="stickersUp"/></applet></p></div><p>乌木先生这个会影响角块,盲拧的时候会比较复杂。</p> 看看 <p></p><p> 呵呵,支持一下!<br/></p>