Number of bit strings of length 19 that contain either 000 or 11.
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I have a bit string of length 19. How many bit strings of length 19 contain three consecutive 0s or 2 consecutive 1s?
combinatorics
asked Dec 5 '18 at 0:49
NdjsjsNdjsjs
1
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add a comment |
$begingroup$
I have a bit string of length 19. How many bit strings of length 19 contain three consecutive 0s or 2 consecutive 1s?
combinatorics
asked Dec 5 '18 at 0:49
NdjsjsNdjsjs
1
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add a comment |
$begingroup$
I have a bit string of length 19. How many bit strings of length 19 contain three consecutive 0s or 2 consecutive 1s?
combinatorics
asked Dec 5 '18 at 0:49
NdjsjsNdjsjs
1
$endgroup$
I have a bit string of length 19. How many bit strings of length 19 contain three consecutive 0s or 2 consecutive 1s?
combinatorics
combinatorics
asked Dec 5 '18 at 0:49
NdjsjsNdjsjs
1
asked Dec 5 '18 at 0:49
NdjsjsNdjsjs
1
asked Dec 5 '18 at 0:49
NdjsjsNdjsjs
1
asked Dec 5 '18 at 0:49
NdjsjsNdjsjs
1
asked Dec 5 '18 at 0:49
NdjsjsNdjsjs
1
1
add a comment |
add a comment |
1 Answer
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$begingroup$
Let $a_n$ be the number of strings of length $n$ which contain neither $000$ nor $11$.
Let $b_n$ be the number of strings which satisfy these conditions, and also end with a $0$.
By conditioning on whether the string of length $n$ ends with $10$ or $100$, you can show
$$
b_n = b_{n-2}+b_{n-3}
$$
The above formula gives you a way to recursively compute $b_n$. For $n=19$, this is doable with pen and paper. You can also get an analytical result by solving this recurrence. However, the expression you get is not nice.
To find $a_n$, use the fact that $$a_n=b_n+b_{n-1}.tag{$nge 1$}$$ Proving this is left to the reader.
answered Dec 5 '18 at 6:40
Mike EarnestMike Earnest
22.3k12051
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1 Answer
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$begingroup$
Let $a_n$ be the number of strings of length $n$ which contain neither $000$ nor $11$.
Let $b_n$ be the number of strings which satisfy these conditions, and also end with a $0$.
By conditioning on whether the string of length $n$ ends with $10$ or $100$, you can show
$$
b_n = b_{n-2}+b_{n-3}
$$
The above formula gives you a way to recursively compute $b_n$. For $n=19$, this is doable with pen and paper. You can also get an analytical result by solving this recurrence. However, the expression you get is not nice.
To find $a_n$, use the fact that $$a_n=b_n+b_{n-1}.tag{$nge 1$}$$ Proving this is left to the reader.
answered Dec 5 '18 at 6:40
Mike EarnestMike Earnest
22.3k12051
$endgroup$
add a comment |
$begingroup$
Let $a_n$ be the number of strings of length $n$ which contain neither $000$ nor $11$.
Let $b_n$ be the number of strings which satisfy these conditions, and also end with a $0$.
By conditioning on whether the string of length $n$ ends with $10$ or $100$, you can show
$$
b_n = b_{n-2}+b_{n-3}
$$
The above formula gives you a way to recursively compute $b_n$. For $n=19$, this is doable with pen and paper. You can also get an analytical result by solving this recurrence. However, the expression you get is not nice.
To find $a_n$, use the fact that $$a_n=b_n+b_{n-1}.tag{$nge 1$}$$ Proving this is left to the reader.
answered Dec 5 '18 at 6:40
Mike EarnestMike Earnest
22.3k12051
$endgroup$
add a comment |
$begingroup$
Let $a_n$ be the number of strings of length $n$ which contain neither $000$ nor $11$.
Let $b_n$ be the number of strings which satisfy these conditions, and also end with a $0$.
By conditioning on whether the string of length $n$ ends with $10$ or $100$, you can show
$$
b_n = b_{n-2}+b_{n-3}
$$
The above formula gives you a way to recursively compute $b_n$. For $n=19$, this is doable with pen and paper. You can also get an analytical result by solving this recurrence. However, the expression you get is not nice.
To find $a_n$, use the fact that $$a_n=b_n+b_{n-1}.tag{$nge 1$}$$ Proving this is left to the reader.
answered Dec 5 '18 at 6:40
Mike EarnestMike Earnest
22.3k12051
$endgroup$
Let $a_n$ be the number of strings of length $n$ which contain neither $000$ nor $11$.
Let $b_n$ be the number of strings which satisfy these conditions, and also end with a $0$.
By conditioning on whether the string of length $n$ ends with $10$ or $100$, you can show
$$
b_n = b_{n-2}+b_{n-3}
$$
The above formula gives you a way to recursively compute $b_n$. For $n=19$, this is doable with pen and paper. You can also get an analytical result by solving this recurrence. However, the expression you get is not nice.
To find $a_n$, use the fact that $$a_n=b_n+b_{n-1}.tag{$nge 1$}$$ Proving this is left to the reader.
answered Dec 5 '18 at 6:40
Mike EarnestMike Earnest
22.3k12051
edited Dec 5 '18 at 7:22
answered Dec 5 '18 at 6:40
Mike EarnestMike Earnest
22.3k12051
answered Dec 5 '18 at 6:40
Mike EarnestMike Earnest
22.3k12051
answered Dec 5 '18 at 6:40
Mike EarnestMike Earnest
22.3k12051
22.3k12051
add a comment |
add a comment |
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