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# HEELELELEEP

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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?

Jun 19, 2019

#1
+26222
+4

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.

Jun 19, 2019
#3
+209
+1

Great Job Heureka!

NoobGuest  Jun 19, 2019
edited by Guest  Jun 19, 2019
#4
+23
+1

How do you do the box in latex?

ProffesorNobody  Jun 19, 2019
#5
+209
-1

\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...

$$\begin{array}{|rcrcrcrc|} \hline & &f(x+5,y+5) & & f(x-4,y) & & f(x,y-4) \\ \hline f(1,1) &\rightarrow & f(6,6) &\rightarrow & f(2,6) & \rightarrow & f(2,2) \\ &\rightarrow & f(7,7) &\rightarrow & f(3,7) & \rightarrow & f(3,3) & \\ &\rightarrow & f(8,8) &\rightarrow & f(4,8) & \rightarrow & f(4,4) & \\ &\rightarrow & f(9,9) &\rightarrow & f(5,9) \\ & & &\rightarrow & f(1,9) & \rightarrow & f(1,5) & \\ & & & & & \rightarrow & f(1,1) \\ \hline \end{array}$$

NoobGuest  Jun 19, 2019
edited by NoobGuest  Jun 19, 2019
#6
+115391
+2

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!

Melody  Jun 19, 2019
edited by Melody  Jun 19, 2019
#7
+26222
+3

Thank you NoobGuest !

heureka  Jun 19, 2019
#8
+26222
+3

Thank you, Melody !

heureka  Jun 19, 2019
#9
+209
0

np!!!!

NoobGuest  Jun 19, 2019