Mixing
Permutation of ordered pairs
$begingroup$
$DeclareMathOperator*{maxi}{maximize}$
Let $Z$ be a set of n ordered pairs, defined as
$Z = {(a_{k}, b_{k}) | a_{k}, b_{k} in mathbb{R}, k in [1, n]}$.
We define a permutation $pi$ of the n ordered pairs in the set $Z$, $pi = bigl( begin{smallmatrix} (a_{1}, b_{1}) & (a_{2}, b_{2}) & cdots & (a_{n}, b_{n})end{smallmatrix} bigr) $.
Is it true that $max_{1 leq i leq n} ( (sum_{j = 1}^{i} a_{j}) + b_{i} )$ for a permutation $pi$ is minimum if and only if the permutation satsifies the condition $b_{1} geq b_{2} geq cdots geq b_{n}$?
permutations ordered-groups
edited Jan 27 at 23:25
Andrés E. Caicedo
65.8k8160251
asked Jan 27 at 17:19
Nikhil BalwaniNikhil Balwani
83
$endgroup$
add a comment |
$begingroup$
$DeclareMathOperator*{maxi}{maximize}$
Let $Z$ be a set of n ordered pairs, defined as
$Z = {(a_{k}, b_{k}) | a_{k}, b_{k} in mathbb{R}, k in [1, n]}$.
We define a permutation $pi$ of the n ordered pairs in the set $Z$, $pi = bigl( begin{smallmatrix} (a_{1}, b_{1}) & (a_{2}, b_{2}) & cdots & (a_{n}, b_{n})end{smallmatrix} bigr) $.
Is it true that $max_{1 leq i leq n} ( (sum_{j = 1}^{i} a_{j}) + b_{i} )$ for a permutation $pi$ is minimum if and only if the permutation satsifies the condition $b_{1} geq b_{2} geq cdots geq b_{n}$?
permutations ordered-groups
edited Jan 27 at 23:25
Andrés E. Caicedo
65.8k8160251
asked Jan 27 at 17:19
Nikhil BalwaniNikhil Balwani
83
$endgroup$
$begingroup$
I don't think so cause if you set the order by this, and then change any 2 elements one with each other, when non of them is maximum element in b, you are still with the same minimum I guess but the order isn't right..
$endgroup$
– Shaq
Jan 27 at 17:29
add a comment |
$begingroup$
$DeclareMathOperator*{maxi}{maximize}$
Let $Z$ be a set of n ordered pairs, defined as
$Z = {(a_{k}, b_{k}) | a_{k}, b_{k} in mathbb{R}, k in [1, n]}$.
We define a permutation $pi$ of the n ordered pairs in the set $Z$, $pi = bigl( begin{smallmatrix} (a_{1}, b_{1}) & (a_{2}, b_{2}) & cdots & (a_{n}, b_{n})end{smallmatrix} bigr) $.
Is it true that $max_{1 leq i leq n} ( (sum_{j = 1}^{i} a_{j}) + b_{i} )$ for a permutation $pi$ is minimum if and only if the permutation satsifies the condition $b_{1} geq b_{2} geq cdots geq b_{n}$?
permutations ordered-groups
edited Jan 27 at 23:25
Andrés E. Caicedo
65.8k8160251
asked Jan 27 at 17:19
Nikhil BalwaniNikhil Balwani
83
$endgroup$
$DeclareMathOperator*{maxi}{maximize}$
Let $Z$ be a set of n ordered pairs, defined as
$Z = {(a_{k}, b_{k}) | a_{k}, b_{k} in mathbb{R}, k in [1, n]}$.
We define a permutation $pi$ of the n ordered pairs in the set $Z$, $pi = bigl( begin{smallmatrix} (a_{1}, b_{1}) & (a_{2}, b_{2}) & cdots & (a_{n}, b_{n})end{smallmatrix} bigr) $.
Is it true that $max_{1 leq i leq n} ( (sum_{j = 1}^{i} a_{j}) + b_{i} )$ for a permutation $pi$ is minimum if and only if the permutation satsifies the condition $b_{1} geq b_{2} geq cdots geq b_{n}$?
permutations ordered-groups
permutations ordered-groups
edited Jan 27 at 23:25
Andrés E. Caicedo
65.8k8160251
asked Jan 27 at 17:19
Nikhil BalwaniNikhil Balwani
83
edited Jan 27 at 23:25
Andrés E. Caicedo
65.8k8160251
asked Jan 27 at 17:19
Nikhil BalwaniNikhil Balwani
83
edited Jan 27 at 23:25
Andrés E. Caicedo
65.8k8160251
edited Jan 27 at 23:25
Andrés E. Caicedo
65.8k8160251
edited Jan 27 at 23:25
Andrés E. Caicedo
65.8k8160251
65.8k8160251
asked Jan 27 at 17:19
Nikhil BalwaniNikhil Balwani
83
asked Jan 27 at 17:19
Nikhil BalwaniNikhil Balwani
83
asked Jan 27 at 17:19
Nikhil BalwaniNikhil Balwani
83
83
$begingroup$
I don't think so cause if you set the order by this, and then change any 2 elements one with each other, when non of them is maximum element in b, you are still with the same minimum I guess but the order isn't right..
$endgroup$
– Shaq
Jan 27 at 17:29
add a comment |
$begingroup$
I don't think so cause if you set the order by this, and then change any 2 elements one with each other, when non of them is maximum element in b, you are still with the same minimum I guess but the order isn't right..
$endgroup$
– Shaq
Jan 27 at 17:29
$begingroup$
I don't think so cause if you set the order by this, and then change any 2 elements one with each other, when non of them is maximum element in b, you are still with the same minimum I guess but the order isn't right..
$endgroup$
– Shaq
Jan 27 at 17:29
$begingroup$
I don't think so cause if you set the order by this, and then change any 2 elements one with each other, when non of them is maximum element in b, you are still with the same minimum I guess but the order isn't right..
$endgroup$
– Shaq
Jan 27 at 17:29
add a comment |
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$begingroup$
I don't think so cause if you set the order by this, and then change any 2 elements one with each other, when non of them is maximum element in b, you are still with the same minimum I guess but the order isn't right..
$endgroup$
– Shaq
Jan 27 at 17:29