Mixing
Number of triplets in array
$begingroup$
We have array $a_0,a_1,...,a_n.$
What is number of triplets $(a[i],a[j],a[k])$ where $0le i,j,k le n$ and $a[i]&a[j]&a[k]=0$?
& is bitwise and
combinatorics
edited Jan 27 at 6:18
saulspatz
17.1k31435
asked Jan 27 at 3:26
JimJim
112
$endgroup$
add a comment |
$begingroup$
We have array $a_0,a_1,...,a_n.$
What is number of triplets $(a[i],a[j],a[k])$ where $0le i,j,k le n$ and $a[i]&a[j]&a[k]=0$?
& is bitwise and
combinatorics
edited Jan 27 at 6:18
saulspatz
17.1k31435
asked Jan 27 at 3:26
JimJim
112
$endgroup$
$begingroup$
This can only be answered with an algorithm, since the answer depends on the array. That's not really appropriate for this site.
$endgroup$
– Matt Samuel
Jan 27 at 4:18
add a comment |
$begingroup$
We have array $a_0,a_1,...,a_n.$
What is number of triplets $(a[i],a[j],a[k])$ where $0le i,j,k le n$ and $a[i]&a[j]&a[k]=0$?
& is bitwise and
combinatorics
edited Jan 27 at 6:18
saulspatz
17.1k31435
asked Jan 27 at 3:26
JimJim
112
$endgroup$
We have array $a_0,a_1,...,a_n.$
What is number of triplets $(a[i],a[j],a[k])$ where $0le i,j,k le n$ and $a[i]&a[j]&a[k]=0$?
& is bitwise and
combinatorics
combinatorics
edited Jan 27 at 6:18
saulspatz
17.1k31435
asked Jan 27 at 3:26
JimJim
112
edited Jan 27 at 6:18
saulspatz
17.1k31435
asked Jan 27 at 3:26
JimJim
112
edited Jan 27 at 6:18
saulspatz
17.1k31435
edited Jan 27 at 6:18
saulspatz
17.1k31435
edited Jan 27 at 6:18
saulspatz
17.1k31435
17.1k31435
asked Jan 27 at 3:26
JimJim
112
asked Jan 27 at 3:26
JimJim
112
asked Jan 27 at 3:26
JimJim
112
112
$begingroup$
This can only be answered with an algorithm, since the answer depends on the array. That's not really appropriate for this site.
$endgroup$
– Matt Samuel
Jan 27 at 4:18
add a comment |
$begingroup$
This can only be answered with an algorithm, since the answer depends on the array. That's not really appropriate for this site.
$endgroup$
– Matt Samuel
Jan 27 at 4:18
$begingroup$
This can only be answered with an algorithm, since the answer depends on the array. That's not really appropriate for this site.
$endgroup$
– Matt Samuel
Jan 27 at 4:18
$begingroup$
This can only be answered with an algorithm, since the answer depends on the array. That's not really appropriate for this site.
$endgroup$
– Matt Samuel
Jan 27 at 4:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You basically have to use the principle of inclusion exclusion. Count the number of triples, $(n+1)^3$. From those, for each bit, subtract the triples which are all one in that bit. Letting $b_i$ be the number of triplets whose $i^{th}$ bit is one, we this leaves
$$
(n+1)^3-b_1^3-b_2^2-dots-b_k^3,
$$
where $k$ is the largest values for which some $a_i$ has a nonzero $k^{th}$ bit. But we are not done yet... triplets for which $a[i],&,a[j],&,a[k]$ is one in two places have been subtracted twice. We need to add these back in. And then subtract the triplets which whose bitwise and is one in three places, and so on. The final result is
$$
sum_{S} (-1)^{|S|} b_S^3,
$$
where $S$ ranges over subsets of bits, and $b_S$ is the number of $a_i$ which are one for all bits in $S$.
answered Jan 28 at 17:30
Mike EarnestMike Earnest
25.6k22151
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You basically have to use the principle of inclusion exclusion. Count the number of triples, $(n+1)^3$. From those, for each bit, subtract the triples which are all one in that bit. Letting $b_i$ be the number of triplets whose $i^{th}$ bit is one, we this leaves
$$
(n+1)^3-b_1^3-b_2^2-dots-b_k^3,
$$
where $k$ is the largest values for which some $a_i$ has a nonzero $k^{th}$ bit. But we are not done yet... triplets for which $a[i],&,a[j],&,a[k]$ is one in two places have been subtracted twice. We need to add these back in. And then subtract the triplets which whose bitwise and is one in three places, and so on. The final result is
$$
sum_{S} (-1)^{|S|} b_S^3,
$$
where $S$ ranges over subsets of bits, and $b_S$ is the number of $a_i$ which are one for all bits in $S$.
answered Jan 28 at 17:30
Mike EarnestMike Earnest
25.6k22151
$endgroup$
add a comment |
$begingroup$
You basically have to use the principle of inclusion exclusion. Count the number of triples, $(n+1)^3$. From those, for each bit, subtract the triples which are all one in that bit. Letting $b_i$ be the number of triplets whose $i^{th}$ bit is one, we this leaves
$$
(n+1)^3-b_1^3-b_2^2-dots-b_k^3,
$$
where $k$ is the largest values for which some $a_i$ has a nonzero $k^{th}$ bit. But we are not done yet... triplets for which $a[i],&,a[j],&,a[k]$ is one in two places have been subtracted twice. We need to add these back in. And then subtract the triplets which whose bitwise and is one in three places, and so on. The final result is
$$
sum_{S} (-1)^{|S|} b_S^3,
$$
where $S$ ranges over subsets of bits, and $b_S$ is the number of $a_i$ which are one for all bits in $S$.
answered Jan 28 at 17:30
Mike EarnestMike Earnest
25.6k22151
$endgroup$
add a comment |
$begingroup$
You basically have to use the principle of inclusion exclusion. Count the number of triples, $(n+1)^3$. From those, for each bit, subtract the triples which are all one in that bit. Letting $b_i$ be the number of triplets whose $i^{th}$ bit is one, we this leaves
$$
(n+1)^3-b_1^3-b_2^2-dots-b_k^3,
$$
where $k$ is the largest values for which some $a_i$ has a nonzero $k^{th}$ bit. But we are not done yet... triplets for which $a[i],&,a[j],&,a[k]$ is one in two places have been subtracted twice. We need to add these back in. And then subtract the triplets which whose bitwise and is one in three places, and so on. The final result is
$$
sum_{S} (-1)^{|S|} b_S^3,
$$
where $S$ ranges over subsets of bits, and $b_S$ is the number of $a_i$ which are one for all bits in $S$.
answered Jan 28 at 17:30
Mike EarnestMike Earnest
25.6k22151
$endgroup$
You basically have to use the principle of inclusion exclusion. Count the number of triples, $(n+1)^3$. From those, for each bit, subtract the triples which are all one in that bit. Letting $b_i$ be the number of triplets whose $i^{th}$ bit is one, we this leaves
$$
(n+1)^3-b_1^3-b_2^2-dots-b_k^3,
$$
where $k$ is the largest values for which some $a_i$ has a nonzero $k^{th}$ bit. But we are not done yet... triplets for which $a[i],&,a[j],&,a[k]$ is one in two places have been subtracted twice. We need to add these back in. And then subtract the triplets which whose bitwise and is one in three places, and so on. The final result is
$$
sum_{S} (-1)^{|S|} b_S^3,
$$
where $S$ ranges over subsets of bits, and $b_S$ is the number of $a_i$ which are one for all bits in $S$.
answered Jan 28 at 17:30
Mike EarnestMike Earnest
25.6k22151
answered Jan 28 at 17:30
Mike EarnestMike Earnest
25.6k22151
answered Jan 28 at 17:30
Mike EarnestMike Earnest
25.6k22151
answered Jan 28 at 17:30
Mike EarnestMike Earnest
25.6k22151
25.6k22151
add a comment |
add a comment |
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$begingroup$
This can only be answered with an algorithm, since the answer depends on the array. That's not really appropriate for this site.
$endgroup$
– Matt Samuel
Jan 27 at 4:18