Prove the maximum sum of products algorithm
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We are given two equal sized sets of positive integers (n integers). We can multiply any two numbers from different sets, each number being used once. All multiplied pairs are added. Prove that maximum sum pairs is when we sort the sets by highest to lowest numbers and multiply highest elements in both sets then 2nd highest and so on. I can prove for n=2 but can't seem to extend that for greater n
sorting multisets
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We are given two equal sized sets of positive integers (n integers). We can multiply any two numbers from different sets, each number being used once. All multiplied pairs are added. Prove that maximum sum pairs is when we sort the sets by highest to lowest numbers and multiply highest elements in both sets then 2nd highest and so on. I can prove for n=2 but can't seem to extend that for greater n
sorting multisets
1
Often such problems yield to an attack where you use the case $n=2$ to make improvements to the arrangement of a larger number of pairs.
â hardmath
Jul 31 at 14:48
2
Please search for 'rearrangement inequality' on wiki.
â J.Guldan
Jul 31 at 14:53
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
We are given two equal sized sets of positive integers (n integers). We can multiply any two numbers from different sets, each number being used once. All multiplied pairs are added. Prove that maximum sum pairs is when we sort the sets by highest to lowest numbers and multiply highest elements in both sets then 2nd highest and so on. I can prove for n=2 but can't seem to extend that for greater n
sorting multisets
We are given two equal sized sets of positive integers (n integers). We can multiply any two numbers from different sets, each number being used once. All multiplied pairs are added. Prove that maximum sum pairs is when we sort the sets by highest to lowest numbers and multiply highest elements in both sets then 2nd highest and so on. I can prove for n=2 but can't seem to extend that for greater n
sorting multisets
asked Jul 31 at 14:39
jatin
1011
1011
1
Often such problems yield to an attack where you use the case $n=2$ to make improvements to the arrangement of a larger number of pairs.
â hardmath
Jul 31 at 14:48
2
Please search for 'rearrangement inequality' on wiki.
â J.Guldan
Jul 31 at 14:53
add a comment |Â
1
Often such problems yield to an attack where you use the case $n=2$ to make improvements to the arrangement of a larger number of pairs.
â hardmath
Jul 31 at 14:48
2
Please search for 'rearrangement inequality' on wiki.
â J.Guldan
Jul 31 at 14:53
1
1
Often such problems yield to an attack where you use the case $n=2$ to make improvements to the arrangement of a larger number of pairs.
â hardmath
Jul 31 at 14:48
Often such problems yield to an attack where you use the case $n=2$ to make improvements to the arrangement of a larger number of pairs.
â hardmath
Jul 31 at 14:48
2
2
Please search for 'rearrangement inequality' on wiki.
â J.Guldan
Jul 31 at 14:53
Please search for 'rearrangement inequality' on wiki.
â J.Guldan
Jul 31 at 14:53
add a comment |Â
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1
Often such problems yield to an attack where you use the case $n=2$ to make improvements to the arrangement of a larger number of pairs.
â hardmath
Jul 31 at 14:48
2
Please search for 'rearrangement inequality' on wiki.
â J.Guldan
Jul 31 at 14:53