Test 1, Gov02  7/01/02

9.  Make up your own rule for a cellular automaton and state the rule.  Then work out its binary coding.  Don't make the rule trivial (e.g., don't have it turning on every cell if all three are off), but you can make it pretty simple if you wish.  Show how the rule works for the first 5 transitions.

Then make up a complex rule--one you think won't give you a definite pattern--and trace out the first 10 steps in its development.  Give the number for your rule.

9) "If only the cell to the left or right is on but not the cell itself then the cell is on. If the cell and its neighbor to the right but not its left is on, then the cell is on."

The binary coding for this rule, using the standard 8-digit coding rule, would be 01011000 or

0*2^7+1*2^6+0*2^5+1*2^4+1*2^3+0*2^2+0*2^1+0*2^0=88.    

"The cell is on if the cell and its neighbor to the left but not to its right is on.  The cell is also on if only the cell itself is on, or one or both of its neighbors is on but the cell itself is off."

The binary coding for this rule is 01101110 or 0*2^7+1*2^6+1*2^5+0*2^4+1*2^3+1*2^2+1*2^1+0*2^0= 110. 
To find the binary coding, we use the rule stated and the cell possibilities (cells that are on have an o in them) :

Using the rule that is given, look at the cells above.  If the cell that is in question is turned on by any of the above combinations, you put a 1 by the cell combination.  If the cell is not turned on by a combination above, you put a 0 by the combination.  As stated above, the binary code for this cellular automaton would be 0 (the first set would not turn it on), 1 (the 2nd set would turn it on), 1 (the 3rd set would turn it on), 0 (the 4th set would not turn it on), 1110.  To get the number of the binary code, take the binary code and for each digit multiply it by 2^x in descending order, starting with 2^7.