<P>看来你的方法是层先法。下面举两个你的问题的典型例子,括号内的公式主要用于顶层棱块问题,但下面是套用于你的问题。第一图公式本身是同层两棱翻色互换,第二图公式本身是同层三棱块轮换,其中两个棱块翻色,一个棱块不翻色。这两式可以举一反三地变换出多个派生公式。</P>
<P> </P>
<applet code="RevengePlayer.class" codebase=4 width="300" height="300">
<param name="scrptLanguage" value="SupersetENG">
<param name="scrpt" value=" U L CF CR' (ML' U2 ML' U2 ML' U2 ML' U2 ML' U MF' U' F2 U MF U' F2 )CR CF' L' U'">
<param name="stickersFront" value="6,6,0,6,0,0,0,0,0,0,0,0,0,0,0,0">
<param name="stickersRight" value="6,6,6,6,1,1,1,1,5,1,1,1,1,1,1,1">
<param name="stickersBack" value="6,6,6,6,3,3,3,3,3,3,3,3,3,3,3,3">
<param name="stickersLeft" value="6,6,6,6,4,4,4,4,4,4,4,4,4,4,4,4">
<param name="stickersUp" value="6,6,6,6,6,5,5,6,6,5,5,6,6,6,1,6">
</applet>
<P> </P>
<applet code="RevengePlayer.class" codebase=4 width="300" height="300">
<param name="scrptLanguage" value="SupersetENG">
<param name="scrpt" value=" CR ( TF D MR D' R2 D MR' D' R2 TF' ) CR' ">
<param name="stickersFront" value="6,6,0,6,0,0,0,5,0,0,0,1,0,0,0,0">
<param name="stickersRight" value="6,6,6,6,0,1,1,1,0,1,1,1,1,1,1,1">
<param name="stickersBack" value="6,6,6,6,3,3,3,3,3,3,3,3,3,3,3,3">
<param name="stickersLeft" value="6,6,6,6,4,4,4,4,4,4,4,4,4,4,4,4">
<param name="stickersUp" value="6,6,6,6,6,5,5,6,6,5,5,6,6,6,1,6">
</applet> |