Relation and Function problems
Clash Royale CLAN TAG#URR8PPP
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Trying to solve these problems for a whole day, but just can't figure out where to start.
let $Sigma=a,b$ and $L = w inSigma^*: 3midtextlength(w)$
Define R â ã
* àã
* as follows: (w, w'
) â R if there is a v â ã
*
such
that: either wv â L and w
'v â L, or wv â L and w
'v â L.
Define $SsubseteqSigma^*timesSigma^*$ as the complement of $R$. That is $(w, w') in S$ iff $(w, w')notin R$.
- State a simple rule for determining if $(w, w') in S$
- show that $S$ is an equivalence relation
- How many equivalence classes does $S$ have?
Any tips or direction will be much appreciated.
relations
add a comment |Â
up vote
1
down vote
favorite
Trying to solve these problems for a whole day, but just can't figure out where to start.
let $Sigma=a,b$ and $L = w inSigma^*: 3midtextlength(w)$
Define R â ã
* àã
* as follows: (w, w'
) â R if there is a v â ã
*
such
that: either wv â L and w
'v â L, or wv â L and w
'v â L.
Define $SsubseteqSigma^*timesSigma^*$ as the complement of $R$. That is $(w, w') in S$ iff $(w, w')notin R$.
- State a simple rule for determining if $(w, w') in S$
- show that $S$ is an equivalence relation
- How many equivalence classes does $S$ have?
Any tips or direction will be much appreciated.
relations
2
What is the definition of R in this?
â Q the Platypus
Aug 24 at 7:25
there is another part of the question before this 'Define R â é* x é* as follows: (w, w') â R if there is a v â é* such that: either vw â L and w'v â L, or wv â L and w'v â L. this wat the first part of the question, I tought they are not related, or are they?
â DSt_FTW
Aug 24 at 8:08
@AlexDSt The info in your comment is essential and must be a part of your question. Check my edit.
â drhab
Aug 24 at 8:29
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Trying to solve these problems for a whole day, but just can't figure out where to start.
let $Sigma=a,b$ and $L = w inSigma^*: 3midtextlength(w)$
Define R â ã
* àã
* as follows: (w, w'
) â R if there is a v â ã
*
such
that: either wv â L and w
'v â L, or wv â L and w
'v â L.
Define $SsubseteqSigma^*timesSigma^*$ as the complement of $R$. That is $(w, w') in S$ iff $(w, w')notin R$.
- State a simple rule for determining if $(w, w') in S$
- show that $S$ is an equivalence relation
- How many equivalence classes does $S$ have?
Any tips or direction will be much appreciated.
relations
Trying to solve these problems for a whole day, but just can't figure out where to start.
let $Sigma=a,b$ and $L = w inSigma^*: 3midtextlength(w)$
Define R â ã
* àã
* as follows: (w, w'
) â R if there is a v â ã
*
such
that: either wv â L and w
'v â L, or wv â L and w
'v â L.
Define $SsubseteqSigma^*timesSigma^*$ as the complement of $R$. That is $(w, w') in S$ iff $(w, w')notin R$.
- State a simple rule for determining if $(w, w') in S$
- show that $S$ is an equivalence relation
- How many equivalence classes does $S$ have?
Any tips or direction will be much appreciated.
relations
edited Aug 24 at 12:38
asked Aug 24 at 6:46
DSt_FTW
226
226
2
What is the definition of R in this?
â Q the Platypus
Aug 24 at 7:25
there is another part of the question before this 'Define R â é* x é* as follows: (w, w') â R if there is a v â é* such that: either vw â L and w'v â L, or wv â L and w'v â L. this wat the first part of the question, I tought they are not related, or are they?
â DSt_FTW
Aug 24 at 8:08
@AlexDSt The info in your comment is essential and must be a part of your question. Check my edit.
â drhab
Aug 24 at 8:29
add a comment |Â
2
What is the definition of R in this?
â Q the Platypus
Aug 24 at 7:25
there is another part of the question before this 'Define R â é* x é* as follows: (w, w') â R if there is a v â é* such that: either vw â L and w'v â L, or wv â L and w'v â L. this wat the first part of the question, I tought they are not related, or are they?
â DSt_FTW
Aug 24 at 8:08
@AlexDSt The info in your comment is essential and must be a part of your question. Check my edit.
â drhab
Aug 24 at 8:29
2
2
What is the definition of R in this?
â Q the Platypus
Aug 24 at 7:25
What is the definition of R in this?
â Q the Platypus
Aug 24 at 7:25
there is another part of the question before this 'Define R â é* x é* as follows: (w, w') â R if there is a v â é* such that: either vw â L and w'v â L, or wv â L and w'v â L. this wat the first part of the question, I tought they are not related, or are they?
â DSt_FTW
Aug 24 at 8:08
there is another part of the question before this 'Define R â é* x é* as follows: (w, w') â R if there is a v â é* such that: either vw â L and w'v â L, or wv â L and w'v â L. this wat the first part of the question, I tought they are not related, or are they?
â DSt_FTW
Aug 24 at 8:08
@AlexDSt The info in your comment is essential and must be a part of your question. Check my edit.
â drhab
Aug 24 at 8:29
@AlexDSt The info in your comment is essential and must be a part of your question. Check my edit.
â drhab
Aug 24 at 8:29
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
If I understand well then $wSw'$ iff $$textlength(w)equivtextlength(w')mod 3$$or equivlently:$$3midtextlength(w)-textlength(w')$$
This can also be expressed by: $$f(w)=f(w')$$where $f:Sigma^*to0,1,2$ is the function prescribed by $$wmapstotextlength(w)-3lfloorfrac13cdottextlength(w)rfloor$$
On base of $wSw'iff f(w)=f(w')$ it is easy to prove that $S$ is an equivalence relation:
- $f(w)=f(w)$ for all $winSigma^*$ so reflexivity.
- $f(w)=f(v)implies f(v)=f(w)$ so symmetry.
- $f(w)=f(v)wedge f(v)=f(u)implies f(w)=f(u)$ so transitivity.
The equivalence classes are the sets:
- $f^-1(0)=L$
- $f^-1(1)=vwmid vinSigmawedge win L$
- $f^-1(2)=uvwmid u,vinSigmawedge win L$
took me a while to get through the first huddle of understanding the first 3, but what is this means? wâ¦length(w)âÂÂ3âÂÂ1/3â length(w)âÂÂ, if length of w = 0, wont this means 0-3*0 = 0, and if length(w)= 3, then 3-3*3/3 = 0?
â DSt_FTW
Aug 24 at 11:08
Yes. It means that $0mapsto0$,$1mapsto1$,$2mapsto 2$,$3mapsto0$,$4mapsto1$,$5mapsto2$,$6mapsto0$,$7mapsto1$,... et cetera.
â drhab
Aug 24 at 11:26
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If I understand well then $wSw'$ iff $$textlength(w)equivtextlength(w')mod 3$$or equivlently:$$3midtextlength(w)-textlength(w')$$
This can also be expressed by: $$f(w)=f(w')$$where $f:Sigma^*to0,1,2$ is the function prescribed by $$wmapstotextlength(w)-3lfloorfrac13cdottextlength(w)rfloor$$
On base of $wSw'iff f(w)=f(w')$ it is easy to prove that $S$ is an equivalence relation:
- $f(w)=f(w)$ for all $winSigma^*$ so reflexivity.
- $f(w)=f(v)implies f(v)=f(w)$ so symmetry.
- $f(w)=f(v)wedge f(v)=f(u)implies f(w)=f(u)$ so transitivity.
The equivalence classes are the sets:
- $f^-1(0)=L$
- $f^-1(1)=vwmid vinSigmawedge win L$
- $f^-1(2)=uvwmid u,vinSigmawedge win L$
took me a while to get through the first huddle of understanding the first 3, but what is this means? wâ¦length(w)âÂÂ3âÂÂ1/3â length(w)âÂÂ, if length of w = 0, wont this means 0-3*0 = 0, and if length(w)= 3, then 3-3*3/3 = 0?
â DSt_FTW
Aug 24 at 11:08
Yes. It means that $0mapsto0$,$1mapsto1$,$2mapsto 2$,$3mapsto0$,$4mapsto1$,$5mapsto2$,$6mapsto0$,$7mapsto1$,... et cetera.
â drhab
Aug 24 at 11:26
add a comment |Â
up vote
2
down vote
accepted
If I understand well then $wSw'$ iff $$textlength(w)equivtextlength(w')mod 3$$or equivlently:$$3midtextlength(w)-textlength(w')$$
This can also be expressed by: $$f(w)=f(w')$$where $f:Sigma^*to0,1,2$ is the function prescribed by $$wmapstotextlength(w)-3lfloorfrac13cdottextlength(w)rfloor$$
On base of $wSw'iff f(w)=f(w')$ it is easy to prove that $S$ is an equivalence relation:
- $f(w)=f(w)$ for all $winSigma^*$ so reflexivity.
- $f(w)=f(v)implies f(v)=f(w)$ so symmetry.
- $f(w)=f(v)wedge f(v)=f(u)implies f(w)=f(u)$ so transitivity.
The equivalence classes are the sets:
- $f^-1(0)=L$
- $f^-1(1)=vwmid vinSigmawedge win L$
- $f^-1(2)=uvwmid u,vinSigmawedge win L$
took me a while to get through the first huddle of understanding the first 3, but what is this means? wâ¦length(w)âÂÂ3âÂÂ1/3â length(w)âÂÂ, if length of w = 0, wont this means 0-3*0 = 0, and if length(w)= 3, then 3-3*3/3 = 0?
â DSt_FTW
Aug 24 at 11:08
Yes. It means that $0mapsto0$,$1mapsto1$,$2mapsto 2$,$3mapsto0$,$4mapsto1$,$5mapsto2$,$6mapsto0$,$7mapsto1$,... et cetera.
â drhab
Aug 24 at 11:26
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If I understand well then $wSw'$ iff $$textlength(w)equivtextlength(w')mod 3$$or equivlently:$$3midtextlength(w)-textlength(w')$$
This can also be expressed by: $$f(w)=f(w')$$where $f:Sigma^*to0,1,2$ is the function prescribed by $$wmapstotextlength(w)-3lfloorfrac13cdottextlength(w)rfloor$$
On base of $wSw'iff f(w)=f(w')$ it is easy to prove that $S$ is an equivalence relation:
- $f(w)=f(w)$ for all $winSigma^*$ so reflexivity.
- $f(w)=f(v)implies f(v)=f(w)$ so symmetry.
- $f(w)=f(v)wedge f(v)=f(u)implies f(w)=f(u)$ so transitivity.
The equivalence classes are the sets:
- $f^-1(0)=L$
- $f^-1(1)=vwmid vinSigmawedge win L$
- $f^-1(2)=uvwmid u,vinSigmawedge win L$
If I understand well then $wSw'$ iff $$textlength(w)equivtextlength(w')mod 3$$or equivlently:$$3midtextlength(w)-textlength(w')$$
This can also be expressed by: $$f(w)=f(w')$$where $f:Sigma^*to0,1,2$ is the function prescribed by $$wmapstotextlength(w)-3lfloorfrac13cdottextlength(w)rfloor$$
On base of $wSw'iff f(w)=f(w')$ it is easy to prove that $S$ is an equivalence relation:
- $f(w)=f(w)$ for all $winSigma^*$ so reflexivity.
- $f(w)=f(v)implies f(v)=f(w)$ so symmetry.
- $f(w)=f(v)wedge f(v)=f(u)implies f(w)=f(u)$ so transitivity.
The equivalence classes are the sets:
- $f^-1(0)=L$
- $f^-1(1)=vwmid vinSigmawedge win L$
- $f^-1(2)=uvwmid u,vinSigmawedge win L$
answered Aug 24 at 8:59
drhab
88.3k541120
88.3k541120
took me a while to get through the first huddle of understanding the first 3, but what is this means? wâ¦length(w)âÂÂ3âÂÂ1/3â length(w)âÂÂ, if length of w = 0, wont this means 0-3*0 = 0, and if length(w)= 3, then 3-3*3/3 = 0?
â DSt_FTW
Aug 24 at 11:08
Yes. It means that $0mapsto0$,$1mapsto1$,$2mapsto 2$,$3mapsto0$,$4mapsto1$,$5mapsto2$,$6mapsto0$,$7mapsto1$,... et cetera.
â drhab
Aug 24 at 11:26
add a comment |Â
took me a while to get through the first huddle of understanding the first 3, but what is this means? wâ¦length(w)âÂÂ3âÂÂ1/3â length(w)âÂÂ, if length of w = 0, wont this means 0-3*0 = 0, and if length(w)= 3, then 3-3*3/3 = 0?
â DSt_FTW
Aug 24 at 11:08
Yes. It means that $0mapsto0$,$1mapsto1$,$2mapsto 2$,$3mapsto0$,$4mapsto1$,$5mapsto2$,$6mapsto0$,$7mapsto1$,... et cetera.
â drhab
Aug 24 at 11:26
took me a while to get through the first huddle of understanding the first 3, but what is this means? wâ¦length(w)âÂÂ3âÂÂ1/3â length(w)âÂÂ, if length of w = 0, wont this means 0-3*0 = 0, and if length(w)= 3, then 3-3*3/3 = 0?
â DSt_FTW
Aug 24 at 11:08
took me a while to get through the first huddle of understanding the first 3, but what is this means? wâ¦length(w)âÂÂ3âÂÂ1/3â length(w)âÂÂ, if length of w = 0, wont this means 0-3*0 = 0, and if length(w)= 3, then 3-3*3/3 = 0?
â DSt_FTW
Aug 24 at 11:08
Yes. It means that $0mapsto0$,$1mapsto1$,$2mapsto 2$,$3mapsto0$,$4mapsto1$,$5mapsto2$,$6mapsto0$,$7mapsto1$,... et cetera.
â drhab
Aug 24 at 11:26
Yes. It means that $0mapsto0$,$1mapsto1$,$2mapsto 2$,$3mapsto0$,$4mapsto1$,$5mapsto2$,$6mapsto0$,$7mapsto1$,... et cetera.
â drhab
Aug 24 at 11:26
add a comment |Â
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2
What is the definition of R in this?
â Q the Platypus
Aug 24 at 7:25
there is another part of the question before this 'Define R â é* x é* as follows: (w, w') â R if there is a v â é* such that: either vw â L and w'v â L, or wv â L and w'v â L. this wat the first part of the question, I tought they are not related, or are they?
â DSt_FTW
Aug 24 at 8:08
@AlexDSt The info in your comment is essential and must be a part of your question. Check my edit.
â drhab
Aug 24 at 8:29