Families of ordered set partitions with disjoint blocks Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Deligne-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
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Deligne-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 Apr 9 at 18:59
darij grinberg
18.4k373189
asked Apr 9 at 16:42
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 Apr 9 at 18:59
darij grinberg
18.4k373189
asked Apr 9 at 16:42
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 Apr 9 at 18:59
darij grinberg
18.4k373189
asked Apr 9 at 16:42
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 Apr 9 at 18:59
darij grinberg
18.4k373189
asked Apr 9 at 16:42
user94267user94267
1007
edited Apr 9 at 18:59
darij grinberg
18.4k373189
asked Apr 9 at 16:42
user94267user94267
1007
edited Apr 9 at 18:59
darij grinberg
18.4k373189
edited Apr 9 at 18:59
darij grinberg
18.4k373189
edited Apr 9 at 18:59
darij grinberg
18.4k373189
18.4k373189
asked Apr 9 at 16:42
user94267user94267
1007
asked Apr 9 at 16:42
user94267user94267
1007
asked Apr 9 at 16:42
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 Apr 9 at 17:06
Fedor PetrovFedor Petrov
52.3k6122240
$endgroup$
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 Apr 9 at 17:07
Teo BanicaTeo Banica
478528
$endgroup$
$begingroup$
26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
$endgroup$
– Teo Banica
Apr 9 at 17:09
$begingroup$
you actually multiply 1 by $m$ and by $k$ :)
$endgroup$
– Fedor Petrov
Apr 9 at 17:13
add a comment |
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2 Answers
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2 Answers
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$begingroup$
We have $$mn=sum_isum_j |B_ij|=sum_jsum_i |B_ij|=kn,$$
thus $m=k$.
answered Apr 9 at 17:06
Fedor PetrovFedor Petrov
52.3k6122240
$endgroup$
add a comment |
$begingroup$
We have $$mn=sum_isum_j |B_ij|=sum_jsum_i |B_ij|=kn,$$
thus $m=k$.
answered Apr 9 at 17:06
Fedor PetrovFedor Petrov
52.3k6122240
$endgroup$
add a comment |
$begingroup$
We have $$mn=sum_isum_j |B_ij|=sum_jsum_i |B_ij|=kn,$$
thus $m=k$.
answered Apr 9 at 17:06
Fedor PetrovFedor Petrov
52.3k6122240
$endgroup$
We have $$mn=sum_isum_j |B_ij|=sum_jsum_i |B_ij|=kn,$$
thus $m=k$.
answered Apr 9 at 17:06
Fedor PetrovFedor Petrov
52.3k6122240
answered Apr 9 at 17:06
Fedor PetrovFedor Petrov
52.3k6122240
answered Apr 9 at 17:06
Fedor PetrovFedor Petrov
52.3k6122240
answered Apr 9 at 17:06
Fedor PetrovFedor Petrov
52.3k6122240
52.3k6122240
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 Apr 9 at 17:07
Teo BanicaTeo Banica
478528
$endgroup$
$begingroup$
26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
$endgroup$
– Teo Banica
Apr 9 at 17:09
$begingroup$
you actually multiply 1 by $m$ and by $k$ :)
$endgroup$
– Fedor Petrov
Apr 9 at 17:13
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 Apr 9 at 17:07
Teo BanicaTeo Banica
478528
$endgroup$
$begingroup$
26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
$endgroup$
– Teo Banica
Apr 9 at 17:09
$begingroup$
you actually multiply 1 by $m$ and by $k$ :)
$endgroup$
– Fedor Petrov
Apr 9 at 17:13
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 Apr 9 at 17:07
Teo BanicaTeo Banica
478528
$endgroup$
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 Apr 9 at 17:07
Teo BanicaTeo Banica
478528
answered Apr 9 at 17:07
Teo BanicaTeo Banica
478528
answered Apr 9 at 17:07
Teo BanicaTeo Banica
478528
answered Apr 9 at 17:07
Teo BanicaTeo Banica
478528
478528
$begingroup$
26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
$endgroup$
– Teo Banica
Apr 9 at 17:09
$begingroup$
you actually multiply 1 by $m$ and by $k$ :)
$endgroup$
– Fedor Petrov
Apr 9 at 17:13
add a comment |
$begingroup$
26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
$endgroup$
– Teo Banica
Apr 9 at 17:09
$begingroup$
you actually multiply 1 by $m$ and by $k$ :)
$endgroup$
– Fedor Petrov
Apr 9 at 17:13
$begingroup$
26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
$endgroup$
– Teo Banica
Apr 9 at 17:09
$begingroup$
26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
$endgroup$
– Teo Banica
Apr 9 at 17:09
$begingroup$
you actually multiply 1 by $m$ and by $k$ :)
$endgroup$
– Fedor Petrov
Apr 9 at 17:13
$begingroup$
you actually multiply 1 by $m$ and by $k$ :)
$endgroup$
– Fedor Petrov
Apr 9 at 17:13
add a comment |
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