The function f(x,y) accepts an ordered pair as input and gives another ordered pair as output. It is defined according to the following rules: If x>4,f(x,y) = (x-4,y) . If x<= 4 but y>4, f(x,y) = (x,y-4). Otherwise,(x+5,y+6) . A robot starts by moving to the point (1,1). Every time it arrives at a point (x,y), it applies f to that point and then moves to f(x,y). If the robot runs forever, how many different points will it visit?
+26393 The function f(x,y) accepts an ordered pair as input and gives another ordered pair as output.
It is defined according to the following rules: If x>4,f(x,y) = (x-4,y) . If x<= 4 but y>4, f(x,y) = (x,y-4). Otherwise,(x+5,y+6) .
A robot starts by moving to the point (1,1).
Every time it arrives at a point (x,y), it applies f to that point and then moves to f(x,y).
If the robot runs forever, how many different points will it visit?
\( f(x,y) =\begin{cases} f(x-4,y), & \text{if } x>4 \\\\ f(x,y-4), & \text{if } x\le4 \text{ and } y > 4 \\\\ f(x+5,y+5), & \text{if } x\le4 \text{ and } y \le 4 \\ \end{cases}\)
\(\begin{array}{|rcrcrcrc|} \hline & &f(x+5,y+5) & & f(x-4,y) & & f(x,y-4) \\ \hline f(1,1) &\rightarrow & f(6,6) &\rightarrow & f(2,6) & \rightarrow & f(2,2) \\ &\rightarrow & f(7,7) &\rightarrow & f(3,7) & \rightarrow & f(3,3) & \\ &\rightarrow & f(8,8) &\rightarrow & f(4,8) & \rightarrow & f(4,4) & \\ &\rightarrow & f(9,9) &\rightarrow & f(5,9) \\ & & &\rightarrow & f(1,9) & \rightarrow & f(1,5) & \\ & & & & & \rightarrow & f(1,1) \\ \hline \end{array} \)
The robot will 14 different points visit.
+209 \begin{array}{|rcrcrcrc|} \hline & &f(x+5,y+5) & & f(x-4,y) & & f(x,y-4) \\ \hline f(1,1) &\rightarrow & f(6,6) &\rightarrow & f(2,6) & \rightarrow & f(2,2) \\ &\rightarrow & f(7,7) &\rightarrow & f(3,7) & \rightarrow & f(3,3) & \\ &\rightarrow & f(8,8) &\rightarrow & f(4,8) & \rightarrow & f(4,4) & \\ &\rightarrow & f(9,9) &\rightarrow & f(5,9) \\ & & &\rightarrow & f(1,9) & \rightarrow & f(1,5) & \\ & & & & & \rightarrow & f(1,1) \\ \hline \end{array just add a } at the end of that code...
\(\begin{array}{|rcrcrcrc|} \hline & &f(x+5,y+5) & & f(x-4,y) & & f(x,y-4) \\ \hline f(1,1) &\rightarrow & f(6,6) &\rightarrow & f(2,6) & \rightarrow & f(2,2) \\ &\rightarrow & f(7,7) &\rightarrow & f(3,7) & \rightarrow & f(3,3) & \\ &\rightarrow & f(8,8) &\rightarrow & f(4,8) & \rightarrow & f(4,4) & \\ &\rightarrow & f(9,9) &\rightarrow & f(5,9) \\ & & &\rightarrow & f(1,9) & \rightarrow & f(1,5) & \\ & & & & & \rightarrow & f(1,1) \\ \hline \end{array}\)
+118673 Hi ProfesorNobody, (I hope you realize that professor has 2 esses. 
)
If you right click on the latex box you will bring up this
Now, clik on Tex Commands
and then you will be able to see the code.
Highlight the code and press [ctrl] C to copy it.
Then you can paste with [cnrl] V it where ever you want and then you can start makeing sense of it.
I will do some of that for you.
I have copied and pasted the code here.
\begin{array}{|rcrcrcrc|} \hline & &f(x+5,y+5) & & f(x-4,y) & & f(x,y-4) \\ \hline f(1,1) &\rightarrow & f(6,6) &\rightarrow & f(2,6) & \rightarrow & f(2,2) \\ &\rightarrow & f(7,7) &\rightarrow & f(3,7) & \rightarrow & f(3,3) & \\ &\rightarrow & f(8,8) &\rightarrow & f(4,8) & \rightarrow & f(4,4) & \\ &\rightarrow & f(9,9) &\rightarrow & f(5,9) \\ & & &\rightarrow & f(1,9) & \rightarrow & f(1,5) & \\ & & & & & \rightarrow & f(1,1) \\ \hline \end{array}
Now I can reoganise it so it makes more sense.
\begin{array}{|rcrcrcrc|} \hline & &f(x+5,y+5) & & f(x-4,y) & & f(x,y-4) \\
\hline f(1,1) &\rightarrow & f(6,6) &\rightarrow & f(2,6) & \rightarrow & f(2,2) \\
&\rightarrow & f(7,7) &\rightarrow & f(3,7) & \rightarrow & f(3,3) & \\
&\rightarrow & f(8,8) &\rightarrow & f(4,8) & \rightarrow & f(4,4) & \\ &\rightarrow & f(9,9) &\rightarrow & f(5,9) \\
& & &\rightarrow & f(1,9) & \rightarrow & f(1,5) & \\
& & & & & \rightarrow & f(1,1) \\
\hline \end{array}
the bars at the ends of this |rcrcrcrc| will give the vertical lines and \hline gives the horizontal lines.
It is quite complicated. Heueka is excellent with Latex!
