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$begingroup$
Let an automaton $A$ sit on point $O$ $(0,0)$ and turned to the North. That automaton can execute only any combination of three different commands in each step:
- Move one unit forward
- Turn 90 degrees clockwise
- Turn 90 degrees counterclockwise
Let $Sigma = {F, R, L}$ represent the mentioned commands. Now, $w$ is a word in $Sigma^*$ with a routine that the automaton will execute indefinitely.
Example: $w =$ 'F'
at step 0, A is on (0,0) pointing to North
at step 1, A is on (0,1) stil pointing to North
...
$w =$ 'FL'
at step 0, A is on (0,0) pointing to North
at step 1, A is on (0,1) pointing to West
at step 2, A is on (-1,1) pointing to South
at step 3, A is on (-1,0) pointing to East
at step 4, A is on (0,0) pointing to North (again)
Question: Given $w$, decide if the $A$ will be stuck in a cycle.
I can intuitively answer that question, but I cannot formalize:
If after step 1, $A$ is on $(0,0)$ or pointing to West, South, or East, then $A$ is in a cycle.
Suppose that after step 1, for any $w$, $A$ is at $(a, b)$. Let $S$ a sequence of positions that $A$ takes in each step, $S = {s_0, s_1, s_2, s_4, ldots }$, $s_0 = (0,0)$, $s_1 =(a,b)$. Let $u$ be the vector $(a,b)$ and $f$ be a function of rotation relative to initial orientation to final orientation after $w$. Finally, let $w = f(u)$. Then
begin{array}
ss_1 = s_0 + u, \
s_2 = s_0 + u + w, \
s_3 = s_0 + u + w + f(f(u)) , \
s_4 = s_0 + u + w + f(f(u)) + f(f(w)),
end{array}
If $f$ is a 90 degrees rotation (clockwise or counterclockwise) then $f(f(u)) = -u$. If $f$ is a 180 degrees rotation then $f(u) = -u$. In such cases,
begin{array}
ss_4 = s_0
end{array}
therefore, $A$ only "escapes" if:
f is the identity rotation
u != (0,0)
To verify whether $f$ is the identity rotation, the number of 'L' commands in $w$ should be equal to the number of 'R' commands.
linear-algebra geometry automata
asked Mar 20 at 11:49
rdllopesrdllopes
1063
$endgroup$
add a comment |
$begingroup$
Let an automaton $A$ sit on point $O$ $(0,0)$ and turned to the North. That automaton can execute only any combination of three different commands in each step:
- Move one unit forward
- Turn 90 degrees clockwise
- Turn 90 degrees counterclockwise
Let $Sigma = {F, R, L}$ represent the mentioned commands. Now, $w$ is a word in $Sigma^*$ with a routine that the automaton will execute indefinitely.
Example: $w =$ 'F'
at step 0, A is on (0,0) pointing to North
at step 1, A is on (0,1) stil pointing to North
...
$w =$ 'FL'
at step 0, A is on (0,0) pointing to North
at step 1, A is on (0,1) pointing to West
at step 2, A is on (-1,1) pointing to South
at step 3, A is on (-1,0) pointing to East
at step 4, A is on (0,0) pointing to North (again)
Question: Given $w$, decide if the $A$ will be stuck in a cycle.
I can intuitively answer that question, but I cannot formalize:
If after step 1, $A$ is on $(0,0)$ or pointing to West, South, or East, then $A$ is in a cycle.
Suppose that after step 1, for any $w$, $A$ is at $(a, b)$. Let $S$ a sequence of positions that $A$ takes in each step, $S = {s_0, s_1, s_2, s_4, ldots }$, $s_0 = (0,0)$, $s_1 =(a,b)$. Let $u$ be the vector $(a,b)$ and $f$ be a function of rotation relative to initial orientation to final orientation after $w$. Finally, let $w = f(u)$. Then
begin{array}
ss_1 = s_0 + u, \
s_2 = s_0 + u + w, \
s_3 = s_0 + u + w + f(f(u)) , \
s_4 = s_0 + u + w + f(f(u)) + f(f(w)),
end{array}
If $f$ is a 90 degrees rotation (clockwise or counterclockwise) then $f(f(u)) = -u$. If $f$ is a 180 degrees rotation then $f(u) = -u$. In such cases,
begin{array}
ss_4 = s_0
end{array}
therefore, $A$ only "escapes" if:
f is the identity rotation
u != (0,0)
To verify whether $f$ is the identity rotation, the number of 'L' commands in $w$ should be equal to the number of 'R' commands.
linear-algebra geometry automata
asked Mar 20 at 11:49
rdllopesrdllopes
1063
$endgroup$
1
$begingroup$
UseSigmafor Big_Sigma. You should have a look through math.meta.stackexchange.com/questions/5020/… for a quick reference on formatting mathematical expressions in MathJax, which is the norm here.
$endgroup$
– amd
Mar 20 at 18:21
add a comment |
$begingroup$
Let an automaton $A$ sit on point $O$ $(0,0)$ and turned to the North. That automaton can execute only any combination of three different commands in each step:
- Move one unit forward
- Turn 90 degrees clockwise
- Turn 90 degrees counterclockwise
Let $Sigma = {F, R, L}$ represent the mentioned commands. Now, $w$ is a word in $Sigma^*$ with a routine that the automaton will execute indefinitely.
Example: $w =$ 'F'
at step 0, A is on (0,0) pointing to North
at step 1, A is on (0,1) stil pointing to North
...
$w =$ 'FL'
at step 0, A is on (0,0) pointing to North
at step 1, A is on (0,1) pointing to West
at step 2, A is on (-1,1) pointing to South
at step 3, A is on (-1,0) pointing to East
at step 4, A is on (0,0) pointing to North (again)
Question: Given $w$, decide if the $A$ will be stuck in a cycle.
I can intuitively answer that question, but I cannot formalize:
If after step 1, $A$ is on $(0,0)$ or pointing to West, South, or East, then $A$ is in a cycle.
Suppose that after step 1, for any $w$, $A$ is at $(a, b)$. Let $S$ a sequence of positions that $A$ takes in each step, $S = {s_0, s_1, s_2, s_4, ldots }$, $s_0 = (0,0)$, $s_1 =(a,b)$. Let $u$ be the vector $(a,b)$ and $f$ be a function of rotation relative to initial orientation to final orientation after $w$. Finally, let $w = f(u)$. Then
begin{array}
ss_1 = s_0 + u, \
s_2 = s_0 + u + w, \
s_3 = s_0 + u + w + f(f(u)) , \
s_4 = s_0 + u + w + f(f(u)) + f(f(w)),
end{array}
If $f$ is a 90 degrees rotation (clockwise or counterclockwise) then $f(f(u)) = -u$. If $f$ is a 180 degrees rotation then $f(u) = -u$. In such cases,
begin{array}
ss_4 = s_0
end{array}
therefore, $A$ only "escapes" if:
f is the identity rotation
u != (0,0)
To verify whether $f$ is the identity rotation, the number of 'L' commands in $w$ should be equal to the number of 'R' commands.
linear-algebra geometry automata
asked Mar 20 at 11:49
rdllopesrdllopes
1063
$endgroup$
Let an automaton $A$ sit on point $O$ $(0,0)$ and turned to the North. That automaton can execute only any combination of three different commands in each step:
- Move one unit forward
- Turn 90 degrees clockwise
- Turn 90 degrees counterclockwise
Let $Sigma = {F, R, L}$ represent the mentioned commands. Now, $w$ is a word in $Sigma^*$ with a routine that the automaton will execute indefinitely.
Example: $w =$ 'F'
at step 0, A is on (0,0) pointing to North
at step 1, A is on (0,1) stil pointing to North
...
$w =$ 'FL'
at step 0, A is on (0,0) pointing to North
at step 1, A is on (0,1) pointing to West
at step 2, A is on (-1,1) pointing to South
at step 3, A is on (-1,0) pointing to East
at step 4, A is on (0,0) pointing to North (again)
Question: Given $w$, decide if the $A$ will be stuck in a cycle.
I can intuitively answer that question, but I cannot formalize:
If after step 1, $A$ is on $(0,0)$ or pointing to West, South, or East, then $A$ is in a cycle.
Suppose that after step 1, for any $w$, $A$ is at $(a, b)$. Let $S$ a sequence of positions that $A$ takes in each step, $S = {s_0, s_1, s_2, s_4, ldots }$, $s_0 = (0,0)$, $s_1 =(a,b)$. Let $u$ be the vector $(a,b)$ and $f$ be a function of rotation relative to initial orientation to final orientation after $w$. Finally, let $w = f(u)$. Then
begin{array}
ss_1 = s_0 + u, \
s_2 = s_0 + u + w, \
s_3 = s_0 + u + w + f(f(u)) , \
s_4 = s_0 + u + w + f(f(u)) + f(f(w)),
end{array}
If $f$ is a 90 degrees rotation (clockwise or counterclockwise) then $f(f(u)) = -u$. If $f$ is a 180 degrees rotation then $f(u) = -u$. In such cases,
begin{array}
ss_4 = s_0
end{array}
therefore, $A$ only "escapes" if:
f is the identity rotation
u != (0,0)
To verify whether $f$ is the identity rotation, the number of 'L' commands in $w$ should be equal to the number of 'R' commands.
linear-algebra geometry automata
linear-algebra geometry automata
asked Mar 20 at 11:49
rdllopesrdllopes
1063
asked Mar 20 at 11:49
rdllopesrdllopes
1063
edited Mar 21 at 10:00
rdllopes
asked Mar 20 at 11:49
rdllopesrdllopes
1063
asked Mar 20 at 11:49
rdllopesrdllopes
1063
asked Mar 20 at 11:49
rdllopesrdllopes
1063
1063
1
$begingroup$
UseSigmafor Big_Sigma. You should have a look through math.meta.stackexchange.com/questions/5020/… for a quick reference on formatting mathematical expressions in MathJax, which is the norm here.
$endgroup$
– amd
Mar 20 at 18:21
add a comment |
1
$begingroup$
UseSigmafor Big_Sigma. You should have a look through math.meta.stackexchange.com/questions/5020/… for a quick reference on formatting mathematical expressions in MathJax, which is the norm here.
$endgroup$
– amd
Mar 20 at 18:21
1
1
$begingroup$
Use
Sigma for Big_Sigma. You should have a look through math.meta.stackexchange.com/questions/5020/… for a quick reference on formatting mathematical expressions in MathJax, which is the norm here.$endgroup$
– amd
Mar 20 at 18:21
$begingroup$
Use
Sigma for Big_Sigma. You should have a look through math.meta.stackexchange.com/questions/5020/… for a quick reference on formatting mathematical expressions in MathJax, which is the norm here.$endgroup$
– amd
Mar 20 at 18:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let initial state be $P_0=(0,0,N)$ and $P_1=(a,b,X)$ be the state after step 1. Then:
If $P_1=(0,0,X)$ then $A$ will stay forever at its starting point.
If $P_1=(a,b,E)$ then $A$ underwent a counterclockwise rotation of $90°$ about center
$displaystyle C=left({a-bover2},{a+bover2}right)$. Thus after 4 repetitions $A$ is back to start.If $P_1=(a,b,W)$ then $A$ underwent a clockwise rotation of $90°$ about center
$displaystyle C=left({a+bover2},{a-bover2}right)$. Thus after 4 repetitions $A$ is back to start.If $P_1=(a,b,S)$ then $A$ underwent a rotation of $180°$ about center
$displaystyle C=left({aover2},{bover2}right)$. Thus after 2 repetitions $A$ is back to start.If $P_1=(a,b,N)$ then $A$ underwent a translation by vector $(a,b)$. Thus it will never return to starting point, unless $(a,b)=(0,0)$.
answered Mar 20 at 16:15
AretinoAretino
25.8k31545
$endgroup$
$begingroup$
That solution just ignores the path performed by $A$ and assume an arc between $P_0$ and $P_1$.
$endgroup$
– rdllopes
Mar 21 at 9:58
1
$begingroup$
A rotation is a transformation of points in the plane, no arc implied. I used those rotations because not only they carry $P_0$ to $P_1$, but they also give the correct transformation from the orientation at $P_0$ to the orientation at $P_1$.
$endgroup$
– Aretino
Mar 21 at 13:49
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let initial state be $P_0=(0,0,N)$ and $P_1=(a,b,X)$ be the state after step 1. Then:
If $P_1=(0,0,X)$ then $A$ will stay forever at its starting point.
If $P_1=(a,b,E)$ then $A$ underwent a counterclockwise rotation of $90°$ about center
$displaystyle C=left({a-bover2},{a+bover2}right)$. Thus after 4 repetitions $A$ is back to start.If $P_1=(a,b,W)$ then $A$ underwent a clockwise rotation of $90°$ about center
$displaystyle C=left({a+bover2},{a-bover2}right)$. Thus after 4 repetitions $A$ is back to start.If $P_1=(a,b,S)$ then $A$ underwent a rotation of $180°$ about center
$displaystyle C=left({aover2},{bover2}right)$. Thus after 2 repetitions $A$ is back to start.If $P_1=(a,b,N)$ then $A$ underwent a translation by vector $(a,b)$. Thus it will never return to starting point, unless $(a,b)=(0,0)$.
answered Mar 20 at 16:15
AretinoAretino
25.8k31545
$endgroup$
$begingroup$
That solution just ignores the path performed by $A$ and assume an arc between $P_0$ and $P_1$.
$endgroup$
– rdllopes
Mar 21 at 9:58
1
$begingroup$
A rotation is a transformation of points in the plane, no arc implied. I used those rotations because not only they carry $P_0$ to $P_1$, but they also give the correct transformation from the orientation at $P_0$ to the orientation at $P_1$.
$endgroup$
– Aretino
Mar 21 at 13:49
add a comment |
$begingroup$
Let initial state be $P_0=(0,0,N)$ and $P_1=(a,b,X)$ be the state after step 1. Then:
If $P_1=(0,0,X)$ then $A$ will stay forever at its starting point.
If $P_1=(a,b,E)$ then $A$ underwent a counterclockwise rotation of $90°$ about center
$displaystyle C=left({a-bover2},{a+bover2}right)$. Thus after 4 repetitions $A$ is back to start.If $P_1=(a,b,W)$ then $A$ underwent a clockwise rotation of $90°$ about center
$displaystyle C=left({a+bover2},{a-bover2}right)$. Thus after 4 repetitions $A$ is back to start.If $P_1=(a,b,S)$ then $A$ underwent a rotation of $180°$ about center
$displaystyle C=left({aover2},{bover2}right)$. Thus after 2 repetitions $A$ is back to start.If $P_1=(a,b,N)$ then $A$ underwent a translation by vector $(a,b)$. Thus it will never return to starting point, unless $(a,b)=(0,0)$.
answered Mar 20 at 16:15
AretinoAretino
25.8k31545
$endgroup$
$begingroup$
That solution just ignores the path performed by $A$ and assume an arc between $P_0$ and $P_1$.
$endgroup$
– rdllopes
Mar 21 at 9:58
1
$begingroup$
A rotation is a transformation of points in the plane, no arc implied. I used those rotations because not only they carry $P_0$ to $P_1$, but they also give the correct transformation from the orientation at $P_0$ to the orientation at $P_1$.
$endgroup$
– Aretino
Mar 21 at 13:49
add a comment |
$begingroup$
Let initial state be $P_0=(0,0,N)$ and $P_1=(a,b,X)$ be the state after step 1. Then:
If $P_1=(0,0,X)$ then $A$ will stay forever at its starting point.
If $P_1=(a,b,E)$ then $A$ underwent a counterclockwise rotation of $90°$ about center
$displaystyle C=left({a-bover2},{a+bover2}right)$. Thus after 4 repetitions $A$ is back to start.If $P_1=(a,b,W)$ then $A$ underwent a clockwise rotation of $90°$ about center
$displaystyle C=left({a+bover2},{a-bover2}right)$. Thus after 4 repetitions $A$ is back to start.If $P_1=(a,b,S)$ then $A$ underwent a rotation of $180°$ about center
$displaystyle C=left({aover2},{bover2}right)$. Thus after 2 repetitions $A$ is back to start.If $P_1=(a,b,N)$ then $A$ underwent a translation by vector $(a,b)$. Thus it will never return to starting point, unless $(a,b)=(0,0)$.
answered Mar 20 at 16:15
AretinoAretino
25.8k31545
$endgroup$
Let initial state be $P_0=(0,0,N)$ and $P_1=(a,b,X)$ be the state after step 1. Then:
If $P_1=(0,0,X)$ then $A$ will stay forever at its starting point.
If $P_1=(a,b,E)$ then $A$ underwent a counterclockwise rotation of $90°$ about center
$displaystyle C=left({a-bover2},{a+bover2}right)$. Thus after 4 repetitions $A$ is back to start.If $P_1=(a,b,W)$ then $A$ underwent a clockwise rotation of $90°$ about center
$displaystyle C=left({a+bover2},{a-bover2}right)$. Thus after 4 repetitions $A$ is back to start.If $P_1=(a,b,S)$ then $A$ underwent a rotation of $180°$ about center
$displaystyle C=left({aover2},{bover2}right)$. Thus after 2 repetitions $A$ is back to start.If $P_1=(a,b,N)$ then $A$ underwent a translation by vector $(a,b)$. Thus it will never return to starting point, unless $(a,b)=(0,0)$.
answered Mar 20 at 16:15
AretinoAretino
25.8k31545
answered Mar 20 at 16:15
AretinoAretino
25.8k31545
answered Mar 20 at 16:15
AretinoAretino
25.8k31545
answered Mar 20 at 16:15
AretinoAretino
25.8k31545
25.8k31545
$begingroup$
That solution just ignores the path performed by $A$ and assume an arc between $P_0$ and $P_1$.
$endgroup$
– rdllopes
Mar 21 at 9:58
1
$begingroup$
A rotation is a transformation of points in the plane, no arc implied. I used those rotations because not only they carry $P_0$ to $P_1$, but they also give the correct transformation from the orientation at $P_0$ to the orientation at $P_1$.
$endgroup$
– Aretino
Mar 21 at 13:49
add a comment |
$begingroup$
That solution just ignores the path performed by $A$ and assume an arc between $P_0$ and $P_1$.
$endgroup$
– rdllopes
Mar 21 at 9:58
1
$begingroup$
A rotation is a transformation of points in the plane, no arc implied. I used those rotations because not only they carry $P_0$ to $P_1$, but they also give the correct transformation from the orientation at $P_0$ to the orientation at $P_1$.
$endgroup$
– Aretino
Mar 21 at 13:49
$begingroup$
That solution just ignores the path performed by $A$ and assume an arc between $P_0$ and $P_1$.
$endgroup$
– rdllopes
Mar 21 at 9:58
$begingroup$
That solution just ignores the path performed by $A$ and assume an arc between $P_0$ and $P_1$.
$endgroup$
– rdllopes
Mar 21 at 9:58
1
1
$begingroup$
A rotation is a transformation of points in the plane, no arc implied. I used those rotations because not only they carry $P_0$ to $P_1$, but they also give the correct transformation from the orientation at $P_0$ to the orientation at $P_1$.
$endgroup$
– Aretino
Mar 21 at 13:49
$begingroup$
A rotation is a transformation of points in the plane, no arc implied. I used those rotations because not only they carry $P_0$ to $P_1$, but they also give the correct transformation from the orientation at $P_0$ to the orientation at $P_1$.
$endgroup$
– Aretino
Mar 21 at 13:49
add a comment |
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$begingroup$
Use
Sigmafor Big_Sigma. You should have a look through math.meta.stackexchange.com/questions/5020/… for a quick reference on formatting mathematical expressions in MathJax, which is the norm here.$endgroup$
– amd
Mar 20 at 18:21