Families of ordered set partitions with disjoint blocks
$begingroup$
Let $C_1,dots, C_m$ be a family of ordered set partitions of $[n]$ with exactly $k$ blocks.
Write $C_i = {B_{i1}, dots, B_{ik}}$ for $i=1,dots, m$ where $B_{ij}$ are the blocks of the ordered set partition $C_i$.
Suppose this family also has the property that for each $j=1,dots, k$
$$B_{1j} cup cdots cup B_{mj}$$
is also a partition of $[n]$
Can one determine the maximal number of members in such a family $m$, or at least a decent upper bound on $m$?
Edit:
It might also be worth noting that if we take $k=n$, then $m=n$ since this would be equivalent to the existence of a latin square. I am in particular interested in the case $k=2$.
co.combinatorics partitions
edited 7 hours ago
darij grinberg
18.4k373188
asked 10 hours ago
user94267user94267
1006
$endgroup$
add a comment |
$begingroup$
Let $C_1,dots, C_m$ be a family of ordered set partitions of $[n]$ with exactly $k$ blocks.
Write $C_i = {B_{i1}, dots, B_{ik}}$ for $i=1,dots, m$ where $B_{ij}$ are the blocks of the ordered set partition $C_i$.
Suppose this family also has the property that for each $j=1,dots, k$
$$B_{1j} cup cdots cup B_{mj}$$
is also a partition of $[n]$
Can one determine the maximal number of members in such a family $m$, or at least a decent upper bound on $m$?
Edit:
It might also be worth noting that if we take $k=n$, then $m=n$ since this would be equivalent to the existence of a latin square. I am in particular interested in the case $k=2$.
co.combinatorics partitions
edited 7 hours ago
darij grinberg
18.4k373188
asked 10 hours ago
user94267user94267
1006
$endgroup$
add a comment |
$begingroup$
Let $C_1,dots, C_m$ be a family of ordered set partitions of $[n]$ with exactly $k$ blocks.
Write $C_i = {B_{i1}, dots, B_{ik}}$ for $i=1,dots, m$ where $B_{ij}$ are the blocks of the ordered set partition $C_i$.
Suppose this family also has the property that for each $j=1,dots, k$
$$B_{1j} cup cdots cup B_{mj}$$
is also a partition of $[n]$
Can one determine the maximal number of members in such a family $m$, or at least a decent upper bound on $m$?
Edit:
It might also be worth noting that if we take $k=n$, then $m=n$ since this would be equivalent to the existence of a latin square. I am in particular interested in the case $k=2$.
co.combinatorics partitions
edited 7 hours ago
darij grinberg
18.4k373188
asked 10 hours ago
user94267user94267
1006
$endgroup$
Let $C_1,dots, C_m$ be a family of ordered set partitions of $[n]$ with exactly $k$ blocks.
Write $C_i = {B_{i1}, dots, B_{ik}}$ for $i=1,dots, m$ where $B_{ij}$ are the blocks of the ordered set partition $C_i$.
Suppose this family also has the property that for each $j=1,dots, k$
$$B_{1j} cup cdots cup B_{mj}$$
is also a partition of $[n]$
Can one determine the maximal number of members in such a family $m$, or at least a decent upper bound on $m$?
Edit:
It might also be worth noting that if we take $k=n$, then $m=n$ since this would be equivalent to the existence of a latin square. I am in particular interested in the case $k=2$.
co.combinatorics partitions
co.combinatorics partitions
edited 7 hours ago
darij grinberg
18.4k373188
asked 10 hours ago
user94267user94267
1006
edited 7 hours ago
darij grinberg
18.4k373188
asked 10 hours ago
user94267user94267
1006
edited 7 hours ago
darij grinberg
18.4k373188
edited 7 hours ago
darij grinberg
18.4k373188
edited 7 hours ago
darij grinberg
18.4k373188
18.4k373188
asked 10 hours ago
user94267user94267
1006
asked 10 hours ago
user94267user94267
1006
asked 10 hours ago
user94267user94267
1006
1006
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
We have $$mn=sum_isum_j |B_{ij}|=sum_jsum_i |B_{ij}|=kn,$$
thus $m=k$.
answered 9 hours ago
Fedor PetrovFedor Petrov
52.1k6122239
$endgroup$
add a comment |
$begingroup$
Answer: $m=k$.
Put indeed your blocks $B_{ij}$ in a $mtimes k$ array and then "read" this array:
-- row-wise: any element of $[n]$ appears then $m$ times.
-- column-wise: any element of $[n]$ appears then $k$ times.
answered 9 hours ago
Teo BanicaTeo Banica
568528
$endgroup$
$begingroup$
26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
$endgroup$
– Teo Banica
9 hours ago
$begingroup$
you actually multiply 1 by $m$ and by $k$ :)
$endgroup$
– Fedor Petrov
9 hours ago
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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oldest
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$begingroup$
We have $$mn=sum_isum_j |B_{ij}|=sum_jsum_i |B_{ij}|=kn,$$
thus $m=k$.
answered 9 hours ago
Fedor PetrovFedor Petrov
52.1k6122239
$endgroup$
add a comment |
$begingroup$
We have $$mn=sum_isum_j |B_{ij}|=sum_jsum_i |B_{ij}|=kn,$$
thus $m=k$.
answered 9 hours ago
Fedor PetrovFedor Petrov
52.1k6122239
$endgroup$
add a comment |
$begingroup$
We have $$mn=sum_isum_j |B_{ij}|=sum_jsum_i |B_{ij}|=kn,$$
thus $m=k$.
answered 9 hours ago
Fedor PetrovFedor Petrov
52.1k6122239
$endgroup$
We have $$mn=sum_isum_j |B_{ij}|=sum_jsum_i |B_{ij}|=kn,$$
thus $m=k$.
answered 9 hours ago
Fedor PetrovFedor Petrov
52.1k6122239
answered 9 hours ago
Fedor PetrovFedor Petrov
52.1k6122239
answered 9 hours ago
Fedor PetrovFedor Petrov
52.1k6122239
answered 9 hours ago
Fedor PetrovFedor Petrov
52.1k6122239
52.1k6122239
add a comment |
add a comment |
$begingroup$
Answer: $m=k$.
Put indeed your blocks $B_{ij}$ in a $mtimes k$ array and then "read" this array:
-- row-wise: any element of $[n]$ appears then $m$ times.
-- column-wise: any element of $[n]$ appears then $k$ times.
answered 9 hours ago
Teo BanicaTeo Banica
568528
$endgroup$
$begingroup$
26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
$endgroup$
– Teo Banica
9 hours ago
$begingroup$
you actually multiply 1 by $m$ and by $k$ :)
$endgroup$
– Fedor Petrov
9 hours ago
add a comment |
$begingroup$
Answer: $m=k$.
Put indeed your blocks $B_{ij}$ in a $mtimes k$ array and then "read" this array:
-- row-wise: any element of $[n]$ appears then $m$ times.
-- column-wise: any element of $[n]$ appears then $k$ times.
answered 9 hours ago
Teo BanicaTeo Banica
568528
$endgroup$
$begingroup$
26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
$endgroup$
– Teo Banica
9 hours ago
$begingroup$
you actually multiply 1 by $m$ and by $k$ :)
$endgroup$
– Fedor Petrov
9 hours ago
add a comment |
$begingroup$
Answer: $m=k$.
Put indeed your blocks $B_{ij}$ in a $mtimes k$ array and then "read" this array:
-- row-wise: any element of $[n]$ appears then $m$ times.
-- column-wise: any element of $[n]$ appears then $k$ times.
answered 9 hours ago
Teo BanicaTeo Banica
568528
$endgroup$
Answer: $m=k$.
Put indeed your blocks $B_{ij}$ in a $mtimes k$ array and then "read" this array:
-- row-wise: any element of $[n]$ appears then $m$ times.
-- column-wise: any element of $[n]$ appears then $k$ times.
answered 9 hours ago
Teo BanicaTeo Banica
568528
answered 9 hours ago
Teo BanicaTeo Banica
568528
answered 9 hours ago
Teo BanicaTeo Banica
568528
answered 9 hours ago
Teo BanicaTeo Banica
568528
568528
$begingroup$
26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
$endgroup$
– Teo Banica
9 hours ago
$begingroup$
you actually multiply 1 by $m$ and by $k$ :)
$endgroup$
– Fedor Petrov
9 hours ago
add a comment |
$begingroup$
26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
$endgroup$
– Teo Banica
9 hours ago
$begingroup$
you actually multiply 1 by $m$ and by $k$ :)
$endgroup$
– Fedor Petrov
9 hours ago
$begingroup$
26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
$endgroup$
– Teo Banica
9 hours ago
$begingroup$
26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
$endgroup$
– Teo Banica
9 hours ago
$begingroup$
you actually multiply 1 by $m$ and by $k$ :)
$endgroup$
– Fedor Petrov
9 hours ago
$begingroup$
you actually multiply 1 by $m$ and by $k$ :)
$endgroup$
– Fedor Petrov
9 hours ago
add a comment |
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