Changes

MyWikiBiz, Author Your Legacy — Wednesday July 03, 2024
Jump to navigationJump to search
Line 624: Line 624:  
|-
 
|-
 
|
 
|
<math>\begin{array}{c|cc}
+
<math>\begin{array}{c|cc|cc|cc|cc|cc|c}
t & u & v \\
+
t & u & v & du & dv & d^2 u & d^2 v & d^3 u & d^3 v & d^4 u & d^4 v & \ldots \\
 
\\
 
\\
0 &  1 &  1 \\
+
0 &  1 &  1 &  0 &  0 &    0 &    0 &    0 &    0 &    0 &    0 & \ldots \\
1 &  1 &  1 \\
+
1 &  1 &  1 &  0 &  0 &    0 &    0 &    0 &    0 &    0 &    0 & \ldots \\
2 & '' & '' \\
+
4 & '' & '' & '' & '' &    '' &    '' &    '' &    '' &    '' &    '' & \ldots \\
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
   −
<pre>
+
To be honest, I have never thought of trying to hack the problem in such a brute-force way until just now, and so I know enough to expect a not inappreciable probability of error about all that I've taken the risk to write out here, but let me forge ahead and see what I can see.
Incipit 2.  <u, v> = <1, 1>
  −
o-----o-----o-----o-----o
  −
| d d | d d | d d | d d |
  −
| 0 0 | 1 1 | 2 2 | 3 3 |
  −
| u v | u v | u v | u v |
  −
o-----o-----o-----o-----o
  −
| 1 1 | 0 0 | 0 0 | = = |
  −
| 1 1 | 0 0 | 0 0 | = = |
  −
| " " | " " | " " | " " |
  −
o-----o-----o-----o-----o
  −
</pre>
  −
 
  −
To be honest, I have never thought of trying to hack the problem in such a brute-force way until just now, and so I know enough to expect a not unappreciable probability of error about all that I've taken the risk to write out here, but let me forge ahead and see what I can see.
      
What we are looking for is &mdash; one rule to rule them all, a rule that applies to every state and works every time.
 
What we are looking for is &mdash; one rule to rule them all, a rule that applies to every state and works every time.
12,080

edits

Navigation menu