For which integers $k$ there exists an $(k-1)$-$(k, k, 1)$ orthogonal array?
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This is what I've found out so far:
Since the first $k-1$ columns must have all the possible $k^{k-1}$ strings of length $k-1$ in exactly one row and the rows can be reorganized without changing the "orthogonality" of the array, I can think of this $k^{k-1}times k$ array instead:
$$
begin{array}{ccccc}
0 &0 &cdots&0 & a_{00cdots0} \
0 &0 &cdots & 1 &a_{00cdots1} \
vdots & vdots &vdots &vdots & vdots \
k-1 & k-1&cdots& k-1 & a_{(k-1)(k-1)cdots(k-1)}
end{array}
$$
Where the alphabet is ${0, 1, cdots, k-1}$ and each element $a$ in the last column is indexed by the corresponding row in the smaller $k^{k-1}times (k-1)$ array.
Now I can conclude that looking at the last $k-1$ columns, there must be the string $00cdots00$ somewhere, as well as $00cdots01$ and so on up to $00cdots0(k-1)$. So the set of elements ${a_{00cdots00}, a_{10cdots00}, cdots, a_{(k-1)0cdots00}}$ has the property that all elements of it must be different (otherwise I would certainly get exactly the same row twice somewhere. And we can repeat this process for all sets of $k-1$ columns and all strings.
But I don't really know how to use this to prove anything. Maybe this is useless, but I am completely stuck right now and I wanted to show you why. How can I solve this problem?
If it helps, I can show the process I was describing for the $3$-$(3,3,1)$ orthogonal array. But I wanted to keep this short.
orthogonality latin-square
asked Nov 24 at 10:41
SlowerPhoton
408111
add a comment |
up vote
0
down vote
favorite
This is what I've found out so far:
Since the first $k-1$ columns must have all the possible $k^{k-1}$ strings of length $k-1$ in exactly one row and the rows can be reorganized without changing the "orthogonality" of the array, I can think of this $k^{k-1}times k$ array instead:
$$
begin{array}{ccccc}
0 &0 &cdots&0 & a_{00cdots0} \
0 &0 &cdots & 1 &a_{00cdots1} \
vdots & vdots &vdots &vdots & vdots \
k-1 & k-1&cdots& k-1 & a_{(k-1)(k-1)cdots(k-1)}
end{array}
$$
Where the alphabet is ${0, 1, cdots, k-1}$ and each element $a$ in the last column is indexed by the corresponding row in the smaller $k^{k-1}times (k-1)$ array.
Now I can conclude that looking at the last $k-1$ columns, there must be the string $00cdots00$ somewhere, as well as $00cdots01$ and so on up to $00cdots0(k-1)$. So the set of elements ${a_{00cdots00}, a_{10cdots00}, cdots, a_{(k-1)0cdots00}}$ has the property that all elements of it must be different (otherwise I would certainly get exactly the same row twice somewhere. And we can repeat this process for all sets of $k-1$ columns and all strings.
But I don't really know how to use this to prove anything. Maybe this is useless, but I am completely stuck right now and I wanted to show you why. How can I solve this problem?
If it helps, I can show the process I was describing for the $3$-$(3,3,1)$ orthogonal array. But I wanted to keep this short.
orthogonality latin-square
asked Nov 24 at 10:41
SlowerPhoton
408111
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This is what I've found out so far:
Since the first $k-1$ columns must have all the possible $k^{k-1}$ strings of length $k-1$ in exactly one row and the rows can be reorganized without changing the "orthogonality" of the array, I can think of this $k^{k-1}times k$ array instead:
$$
begin{array}{ccccc}
0 &0 &cdots&0 & a_{00cdots0} \
0 &0 &cdots & 1 &a_{00cdots1} \
vdots & vdots &vdots &vdots & vdots \
k-1 & k-1&cdots& k-1 & a_{(k-1)(k-1)cdots(k-1)}
end{array}
$$
Where the alphabet is ${0, 1, cdots, k-1}$ and each element $a$ in the last column is indexed by the corresponding row in the smaller $k^{k-1}times (k-1)$ array.
Now I can conclude that looking at the last $k-1$ columns, there must be the string $00cdots00$ somewhere, as well as $00cdots01$ and so on up to $00cdots0(k-1)$. So the set of elements ${a_{00cdots00}, a_{10cdots00}, cdots, a_{(k-1)0cdots00}}$ has the property that all elements of it must be different (otherwise I would certainly get exactly the same row twice somewhere. And we can repeat this process for all sets of $k-1$ columns and all strings.
But I don't really know how to use this to prove anything. Maybe this is useless, but I am completely stuck right now and I wanted to show you why. How can I solve this problem?
If it helps, I can show the process I was describing for the $3$-$(3,3,1)$ orthogonal array. But I wanted to keep this short.
orthogonality latin-square
asked Nov 24 at 10:41
SlowerPhoton
408111
This is what I've found out so far:
Since the first $k-1$ columns must have all the possible $k^{k-1}$ strings of length $k-1$ in exactly one row and the rows can be reorganized without changing the "orthogonality" of the array, I can think of this $k^{k-1}times k$ array instead:
$$
begin{array}{ccccc}
0 &0 &cdots&0 & a_{00cdots0} \
0 &0 &cdots & 1 &a_{00cdots1} \
vdots & vdots &vdots &vdots & vdots \
k-1 & k-1&cdots& k-1 & a_{(k-1)(k-1)cdots(k-1)}
end{array}
$$
Where the alphabet is ${0, 1, cdots, k-1}$ and each element $a$ in the last column is indexed by the corresponding row in the smaller $k^{k-1}times (k-1)$ array.
Now I can conclude that looking at the last $k-1$ columns, there must be the string $00cdots00$ somewhere, as well as $00cdots01$ and so on up to $00cdots0(k-1)$. So the set of elements ${a_{00cdots00}, a_{10cdots00}, cdots, a_{(k-1)0cdots00}}$ has the property that all elements of it must be different (otherwise I would certainly get exactly the same row twice somewhere. And we can repeat this process for all sets of $k-1$ columns and all strings.
But I don't really know how to use this to prove anything. Maybe this is useless, but I am completely stuck right now and I wanted to show you why. How can I solve this problem?
If it helps, I can show the process I was describing for the $3$-$(3,3,1)$ orthogonal array. But I wanted to keep this short.
orthogonality latin-square
orthogonality latin-square
asked Nov 24 at 10:41
SlowerPhoton
408111
asked Nov 24 at 10:41
SlowerPhoton
408111
asked Nov 24 at 10:41
SlowerPhoton
408111
asked Nov 24 at 10:41
SlowerPhoton
408111
asked Nov 24 at 10:41
SlowerPhoton
408111
408111
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