Families of ordered set partitions with disjoint blocks The 2019 Stack Overflow Developer Survey Results Are InDeligne-Simpson problem in the symmetric groupProcreation with several gendersI am searching for the name of a partition (if it already exists)Existence problem for a generalisation of Latin squares (matrices with fixed row and column sets)Simple lower bounds for Bell numbers (number of set partitions)?Can a partition free family in $2^[n]$ always be enlarged to one of size $2^n-1$?Looking for N-dimensional spheres in the configuration space of the colorful Tverberg problemBalanced partitions of vector setsCan we cover a set by a particular family of sets?genus zero permutation and noncrossing partition
Families of ordered set partitions with disjoint blocks
The 2019 Stack Overflow Developer Survey Results Are InDeligne-Simpson problem in the symmetric groupProcreation with several gendersI am searching for the name of a partition (if it already exists)Existence problem for a generalisation of Latin squares (matrices with fixed row and column sets)Simple lower bounds for Bell numbers (number of set partitions)?Can a partition free family in $2^[n]$ always be enlarged to one of size $2^n-1$?Looking for N-dimensional spheres in the configuration space of the colorful Tverberg problemBalanced partitions of vector setsCan we cover a set by a particular family of sets?genus zero permutation and noncrossing partition
$begingroup$
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 11 hours ago
darij grinberg
18.4k373188
asked 14 hours ago
user94267user94267
1007
$endgroup$
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 11 hours ago
darij grinberg
18.4k373188
asked 14 hours ago
user94267user94267
1007
$endgroup$
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 11 hours ago
darij grinberg
18.4k373188
asked 14 hours ago
user94267user94267
1007
$endgroup$
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 11 hours ago
darij grinberg
18.4k373188
asked 14 hours ago
user94267user94267
1007
edited 11 hours ago
darij grinberg
18.4k373188
asked 14 hours ago
user94267user94267
1007
edited 11 hours ago
darij grinberg
18.4k373188
edited 11 hours ago
darij grinberg
18.4k373188
edited 11 hours ago
darij grinberg
18.4k373188
18.4k373188
asked 14 hours ago
user94267user94267
1007
asked 14 hours ago
user94267user94267
1007
asked 14 hours ago
user94267user94267
1007
1007
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We have $$mn=sum_isum_j |B_ij|=sum_jsum_i |B_ij|=kn,$$
thus $m=k$.
answered 13 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 13 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
13 hours ago
$begingroup$
you actually multiply 1 by $m$ and by $k$ :)
$endgroup$
– Fedor Petrov
13 hours ago
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
We have $$mn=sum_isum_j |B_ij|=sum_jsum_i |B_ij|=kn,$$
thus $m=k$.
answered 13 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 13 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 13 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 13 hours ago
Fedor PetrovFedor Petrov
52.1k6122239
answered 13 hours ago
Fedor PetrovFedor Petrov
52.1k6122239
answered 13 hours ago
Fedor PetrovFedor Petrov
52.1k6122239
answered 13 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 13 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
13 hours ago
$begingroup$
you actually multiply 1 by $m$ and by $k$ :)
$endgroup$
– Fedor Petrov
13 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 13 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
13 hours ago
$begingroup$
you actually multiply 1 by $m$ and by $k$ :)
$endgroup$
– Fedor Petrov
13 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 13 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 13 hours ago
Teo BanicaTeo Banica
568528
answered 13 hours ago
Teo BanicaTeo Banica
568528
answered 13 hours ago
Teo BanicaTeo Banica
568528
answered 13 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
13 hours ago
$begingroup$
you actually multiply 1 by $m$ and by $k$ :)
$endgroup$
– Fedor Petrov
13 hours ago
add a comment |
$begingroup$
26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
$endgroup$
– Teo Banica
13 hours ago
$begingroup$
you actually multiply 1 by $m$ and by $k$ :)
$endgroup$
– Fedor Petrov
13 hours ago
$begingroup$
26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
$endgroup$
– Teo Banica
13 hours ago
$begingroup$
26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
$endgroup$
– Teo Banica
13 hours ago
$begingroup$
you actually multiply 1 by $m$ and by $k$ :)
$endgroup$
– Fedor Petrov
13 hours ago
$begingroup$
you actually multiply 1 by $m$ and by $k$ :)
$endgroup$
– Fedor Petrov
13 hours ago
add a comment |
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