A regular expression for strings containing at most two $texttt{0}$s?
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Let $L_1$ be the set of strings in $Sigma^{*}$ that contain at most two $0s$. A regular expression for $L_1$ is
$$R_1 = 1^{∗} + 1^{∗}01^{∗} + 1^{∗}01^{∗}01^{∗}$$
Informally, our regular expression is correct (that is, it denotes $L_1$), because the first term, $1^∗$
,
denotes the set of strings that contain no $0$s at all; the second term, $1^{∗}01^{∗}$
, denotes the set
of strings that contain exactly one $0$; and the third term, $1^{∗}01^{∗}01^{∗}$
, denotes the set of strings
that contain exactly two $0$s. This example is so simple that a more detailed proof is not really
needed; this informal explanation is convincing enough to most people.
What I'm trying to do :
Let $L_1$ be the set of strings in $Sigma^{*}$ that contain at most four $0s$. A regular expression for $L_1$ is
$$R_1 = 1^{∗} + 1^{∗}01^{∗} + 1^{∗}01^{∗}01^{∗} + 1^*01^*01^*01^* + 1^*01^*01^*01^*01^*$$
Would this be right?
computer-science
edited Nov 23 at 14:37
dkaeae
302311
asked Nov 5 at 9:47
shah
384
add a comment |
up vote
0
down vote
favorite
Let $L_1$ be the set of strings in $Sigma^{*}$ that contain at most two $0s$. A regular expression for $L_1$ is
$$R_1 = 1^{∗} + 1^{∗}01^{∗} + 1^{∗}01^{∗}01^{∗}$$
Informally, our regular expression is correct (that is, it denotes $L_1$), because the first term, $1^∗$
,
denotes the set of strings that contain no $0$s at all; the second term, $1^{∗}01^{∗}$
, denotes the set
of strings that contain exactly one $0$; and the third term, $1^{∗}01^{∗}01^{∗}$
, denotes the set of strings
that contain exactly two $0$s. This example is so simple that a more detailed proof is not really
needed; this informal explanation is convincing enough to most people.
What I'm trying to do :
Let $L_1$ be the set of strings in $Sigma^{*}$ that contain at most four $0s$. A regular expression for $L_1$ is
$$R_1 = 1^{∗} + 1^{∗}01^{∗} + 1^{∗}01^{∗}01^{∗} + 1^*01^*01^*01^* + 1^*01^*01^*01^*01^*$$
Would this be right?
computer-science
edited Nov 23 at 14:37
dkaeae
302311
asked Nov 5 at 9:47
shah
384
It's fine. You could do this with $ngeq 0$ number of 0's in general terms.
– Wuestenfux
Nov 5 at 9:49
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $L_1$ be the set of strings in $Sigma^{*}$ that contain at most two $0s$. A regular expression for $L_1$ is
$$R_1 = 1^{∗} + 1^{∗}01^{∗} + 1^{∗}01^{∗}01^{∗}$$
Informally, our regular expression is correct (that is, it denotes $L_1$), because the first term, $1^∗$
,
denotes the set of strings that contain no $0$s at all; the second term, $1^{∗}01^{∗}$
, denotes the set
of strings that contain exactly one $0$; and the third term, $1^{∗}01^{∗}01^{∗}$
, denotes the set of strings
that contain exactly two $0$s. This example is so simple that a more detailed proof is not really
needed; this informal explanation is convincing enough to most people.
What I'm trying to do :
Let $L_1$ be the set of strings in $Sigma^{*}$ that contain at most four $0s$. A regular expression for $L_1$ is
$$R_1 = 1^{∗} + 1^{∗}01^{∗} + 1^{∗}01^{∗}01^{∗} + 1^*01^*01^*01^* + 1^*01^*01^*01^*01^*$$
Would this be right?
computer-science
edited Nov 23 at 14:37
dkaeae
302311
asked Nov 5 at 9:47
shah
384
Let $L_1$ be the set of strings in $Sigma^{*}$ that contain at most two $0s$. A regular expression for $L_1$ is
$$R_1 = 1^{∗} + 1^{∗}01^{∗} + 1^{∗}01^{∗}01^{∗}$$
Informally, our regular expression is correct (that is, it denotes $L_1$), because the first term, $1^∗$
,
denotes the set of strings that contain no $0$s at all; the second term, $1^{∗}01^{∗}$
, denotes the set
of strings that contain exactly one $0$; and the third term, $1^{∗}01^{∗}01^{∗}$
, denotes the set of strings
that contain exactly two $0$s. This example is so simple that a more detailed proof is not really
needed; this informal explanation is convincing enough to most people.
What I'm trying to do :
Let $L_1$ be the set of strings in $Sigma^{*}$ that contain at most four $0s$. A regular expression for $L_1$ is
$$R_1 = 1^{∗} + 1^{∗}01^{∗} + 1^{∗}01^{∗}01^{∗} + 1^*01^*01^*01^* + 1^*01^*01^*01^*01^*$$
Would this be right?
computer-science
computer-science
edited Nov 23 at 14:37
dkaeae
302311
asked Nov 5 at 9:47
shah
384
edited Nov 23 at 14:37
dkaeae
302311
asked Nov 5 at 9:47
shah
384
edited Nov 23 at 14:37
dkaeae
302311
edited Nov 23 at 14:37
dkaeae
302311
edited Nov 23 at 14:37
dkaeae
302311
302311
asked Nov 5 at 9:47
shah
384
asked Nov 5 at 9:47
shah
384
asked Nov 5 at 9:47
shah
384
384
It's fine. You could do this with $ngeq 0$ number of 0's in general terms.
– Wuestenfux
Nov 5 at 9:49
add a comment |
It's fine. You could do this with $ngeq 0$ number of 0's in general terms.
– Wuestenfux
Nov 5 at 9:49
It's fine. You could do this with $ngeq 0$ number of 0's in general terms.
– Wuestenfux
Nov 5 at 9:49
It's fine. You could do this with $ngeq 0$ number of 0's in general terms.
– Wuestenfux
Nov 5 at 9:49
add a comment |
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It's fine. You could do this with $ngeq 0$ number of 0's in general terms.
– Wuestenfux
Nov 5 at 9:49