Find the Recurrence relation for $q_n$ given the following condition:
$begingroup$
Let $q_n$ denote the number of strings of length $n$ (formed from digits 0,1,2,3) which have even number of $2$'s. set up a recurrence relation for $q_n$.
discrete-mathematics recurrence-relations generating-functions
asked Dec 10 '18 at 8:54
CosmicCosmic
8810
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$begingroup$
Let $q_n$ denote the number of strings of length $n$ (formed from digits 0,1,2,3) which have even number of $2$'s. set up a recurrence relation for $q_n$.
discrete-mathematics recurrence-relations generating-functions
asked Dec 10 '18 at 8:54
CosmicCosmic
8810
$endgroup$
add a comment |
$begingroup$
Let $q_n$ denote the number of strings of length $n$ (formed from digits 0,1,2,3) which have even number of $2$'s. set up a recurrence relation for $q_n$.
discrete-mathematics recurrence-relations generating-functions
asked Dec 10 '18 at 8:54
CosmicCosmic
8810
$endgroup$
Let $q_n$ denote the number of strings of length $n$ (formed from digits 0,1,2,3) which have even number of $2$'s. set up a recurrence relation for $q_n$.
discrete-mathematics recurrence-relations generating-functions
discrete-mathematics recurrence-relations generating-functions
asked Dec 10 '18 at 8:54
CosmicCosmic
8810
asked Dec 10 '18 at 8:54
CosmicCosmic
8810
edited Dec 10 '18 at 9:19
Cosmic
asked Dec 10 '18 at 8:54
CosmicCosmic
8810
asked Dec 10 '18 at 8:54
CosmicCosmic
8810
asked Dec 10 '18 at 8:54
CosmicCosmic
8810
8810
add a comment |
add a comment |
1 Answer
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Indeed it can be defined by two recurrence. Suppose $p_n$ is the number of strings which have odd 2's. Now we have:
$$q_n = 3q_{n-1} + p_{n-1}$$
$$p_n = q_{n-1} + 3 p_{n-1}$$
Now we can unified them by matrix operation by $T_n = begin{bmatrix}q_n \ p_n end{bmatrix}$:
$$T_n=begin{bmatrix}3 & 0 \ 0 & 1 end{bmatrix}T_{n-1}+begin{bmatrix}0 & 1 \ 3 & 0 end{bmatrix}T_{n-1}=begin{bmatrix}3 & 1 \ 3 & 1 end{bmatrix}T_{n-1}.$$
Now we can say $T_n = begin{bmatrix}q_n \ p_n end{bmatrix} = begin{bmatrix}3 & 1 \ 3 & 1 end{bmatrix}^{n-1}T_1$. We can compute that $T_1 = begin{bmatrix}3 \ 1 end{bmatrix}$, and
$$T_n = begin{bmatrix}3 & 1 \ 3 & 1 end{bmatrix}^{n-1}begin{bmatrix}3 \ 1 end{bmatrix}$$
answered Dec 10 '18 at 9:49
OmGOmG
2,502822
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1 Answer
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1 Answer
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$begingroup$
Indeed it can be defined by two recurrence. Suppose $p_n$ is the number of strings which have odd 2's. Now we have:
$$q_n = 3q_{n-1} + p_{n-1}$$
$$p_n = q_{n-1} + 3 p_{n-1}$$
Now we can unified them by matrix operation by $T_n = begin{bmatrix}q_n \ p_n end{bmatrix}$:
$$T_n=begin{bmatrix}3 & 0 \ 0 & 1 end{bmatrix}T_{n-1}+begin{bmatrix}0 & 1 \ 3 & 0 end{bmatrix}T_{n-1}=begin{bmatrix}3 & 1 \ 3 & 1 end{bmatrix}T_{n-1}.$$
Now we can say $T_n = begin{bmatrix}q_n \ p_n end{bmatrix} = begin{bmatrix}3 & 1 \ 3 & 1 end{bmatrix}^{n-1}T_1$. We can compute that $T_1 = begin{bmatrix}3 \ 1 end{bmatrix}$, and
$$T_n = begin{bmatrix}3 & 1 \ 3 & 1 end{bmatrix}^{n-1}begin{bmatrix}3 \ 1 end{bmatrix}$$
answered Dec 10 '18 at 9:49
OmGOmG
2,502822
$endgroup$
add a comment |
$begingroup$
Indeed it can be defined by two recurrence. Suppose $p_n$ is the number of strings which have odd 2's. Now we have:
$$q_n = 3q_{n-1} + p_{n-1}$$
$$p_n = q_{n-1} + 3 p_{n-1}$$
Now we can unified them by matrix operation by $T_n = begin{bmatrix}q_n \ p_n end{bmatrix}$:
$$T_n=begin{bmatrix}3 & 0 \ 0 & 1 end{bmatrix}T_{n-1}+begin{bmatrix}0 & 1 \ 3 & 0 end{bmatrix}T_{n-1}=begin{bmatrix}3 & 1 \ 3 & 1 end{bmatrix}T_{n-1}.$$
Now we can say $T_n = begin{bmatrix}q_n \ p_n end{bmatrix} = begin{bmatrix}3 & 1 \ 3 & 1 end{bmatrix}^{n-1}T_1$. We can compute that $T_1 = begin{bmatrix}3 \ 1 end{bmatrix}$, and
$$T_n = begin{bmatrix}3 & 1 \ 3 & 1 end{bmatrix}^{n-1}begin{bmatrix}3 \ 1 end{bmatrix}$$
answered Dec 10 '18 at 9:49
OmGOmG
2,502822
$endgroup$
add a comment |
$begingroup$
Indeed it can be defined by two recurrence. Suppose $p_n$ is the number of strings which have odd 2's. Now we have:
$$q_n = 3q_{n-1} + p_{n-1}$$
$$p_n = q_{n-1} + 3 p_{n-1}$$
Now we can unified them by matrix operation by $T_n = begin{bmatrix}q_n \ p_n end{bmatrix}$:
$$T_n=begin{bmatrix}3 & 0 \ 0 & 1 end{bmatrix}T_{n-1}+begin{bmatrix}0 & 1 \ 3 & 0 end{bmatrix}T_{n-1}=begin{bmatrix}3 & 1 \ 3 & 1 end{bmatrix}T_{n-1}.$$
Now we can say $T_n = begin{bmatrix}q_n \ p_n end{bmatrix} = begin{bmatrix}3 & 1 \ 3 & 1 end{bmatrix}^{n-1}T_1$. We can compute that $T_1 = begin{bmatrix}3 \ 1 end{bmatrix}$, and
$$T_n = begin{bmatrix}3 & 1 \ 3 & 1 end{bmatrix}^{n-1}begin{bmatrix}3 \ 1 end{bmatrix}$$
answered Dec 10 '18 at 9:49
OmGOmG
2,502822
$endgroup$
Indeed it can be defined by two recurrence. Suppose $p_n$ is the number of strings which have odd 2's. Now we have:
$$q_n = 3q_{n-1} + p_{n-1}$$
$$p_n = q_{n-1} + 3 p_{n-1}$$
Now we can unified them by matrix operation by $T_n = begin{bmatrix}q_n \ p_n end{bmatrix}$:
$$T_n=begin{bmatrix}3 & 0 \ 0 & 1 end{bmatrix}T_{n-1}+begin{bmatrix}0 & 1 \ 3 & 0 end{bmatrix}T_{n-1}=begin{bmatrix}3 & 1 \ 3 & 1 end{bmatrix}T_{n-1}.$$
Now we can say $T_n = begin{bmatrix}q_n \ p_n end{bmatrix} = begin{bmatrix}3 & 1 \ 3 & 1 end{bmatrix}^{n-1}T_1$. We can compute that $T_1 = begin{bmatrix}3 \ 1 end{bmatrix}$, and
$$T_n = begin{bmatrix}3 & 1 \ 3 & 1 end{bmatrix}^{n-1}begin{bmatrix}3 \ 1 end{bmatrix}$$
answered Dec 10 '18 at 9:49
OmGOmG
2,502822
edited Dec 10 '18 at 9:57
answered Dec 10 '18 at 9:49
OmGOmG
2,502822
answered Dec 10 '18 at 9:49
OmGOmG
2,502822
answered Dec 10 '18 at 9:49
OmGOmG
2,502822
2,502822
add a comment |
add a comment |
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