Equal Sub Set Sum Problem.how many ways we can choose 3 and more consecutive number from set of N numbersSet of ten distinct two-digit natural numbersDivide set of numbers into twosub sets with equal totalsDetermine the number of subsets of $1,2,3,4,…,50$ whose sum of elements is larger than or equal to $638$.Sum of absolute difference between each pair of element in a setSum of combinations of setsA homework problem about set theoryBritish Maths Olympiad (BMO) 2003 Round 1 Question 4 what is wrong with this approach, can it be fixed?How many labeled trees are there if there're $4$ leaves that are set?Is there a cleverer way to find the largest Sidon set in 1,…,N?
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Equal Sub Set Sum Problem.
how many ways we can choose 3 and more consecutive number from set of N numbersSet of ten distinct two-digit natural numbersDivide set of numbers into twosub sets with equal totalsDetermine the number of subsets of $1,2,3,4,…,50$ whose sum of elements is larger than or equal to $638$.Sum of absolute difference between each pair of element in a setSum of combinations of setsA homework problem about set theoryBritish Maths Olympiad (BMO) 2003 Round 1 Question 4 what is wrong with this approach, can it be fixed?How many labeled trees are there if there're $4$ leaves that are set?Is there a cleverer way to find the largest Sidon set in 1,…,N?
$begingroup$
I am trying to solve the equal subset sum problem , where we have to tell if its possible to generate to disjoint subsets whose (non-zero) sum is equal and you don't have to consider all the elements. The numbers will be unique and in the range of 1-99.
I used a recursive approach of considering each element to be part of first/second/neither set. I am looking to optimize the code.
The second approach is to generate all possible sums and if any sum occurs more than once it would return true. The problem with the second approach is I am not able to prove that the sum generated is from disjoint sets.
Consider sets A = 1,2,3,4 and B = 9,1. The sum of both the sets is 10 and they are disjoint. Is it safe to say that if we have a set of non disjoint sets whose sum is equal, can be converted into two disjoint sets of equal sum by removing the common elements. If not can somebody give me a counter example . Like in this case we will have 2,3,4 and 9.
combinatorics
edited Mar 28 at 17:03
Andrés E. Caicedo
65.8k8160252
asked Mar 28 at 14:31
Syam KumarSyam Kumar
388
$endgroup$
add a comment |
$begingroup$
I am trying to solve the equal subset sum problem , where we have to tell if its possible to generate to disjoint subsets whose (non-zero) sum is equal and you don't have to consider all the elements. The numbers will be unique and in the range of 1-99.
I used a recursive approach of considering each element to be part of first/second/neither set. I am looking to optimize the code.
The second approach is to generate all possible sums and if any sum occurs more than once it would return true. The problem with the second approach is I am not able to prove that the sum generated is from disjoint sets.
Consider sets A = 1,2,3,4 and B = 9,1. The sum of both the sets is 10 and they are disjoint. Is it safe to say that if we have a set of non disjoint sets whose sum is equal, can be converted into two disjoint sets of equal sum by removing the common elements. If not can somebody give me a counter example . Like in this case we will have 2,3,4 and 9.
combinatorics
edited Mar 28 at 17:03
Andrés E. Caicedo
65.8k8160252
asked Mar 28 at 14:31
Syam KumarSyam Kumar
388
$endgroup$
1
$begingroup$
It is quite safe to say that!
$endgroup$
– Mike Earnest
Mar 28 at 16:03
add a comment |
$begingroup$
I am trying to solve the equal subset sum problem , where we have to tell if its possible to generate to disjoint subsets whose (non-zero) sum is equal and you don't have to consider all the elements. The numbers will be unique and in the range of 1-99.
I used a recursive approach of considering each element to be part of first/second/neither set. I am looking to optimize the code.
The second approach is to generate all possible sums and if any sum occurs more than once it would return true. The problem with the second approach is I am not able to prove that the sum generated is from disjoint sets.
Consider sets A = 1,2,3,4 and B = 9,1. The sum of both the sets is 10 and they are disjoint. Is it safe to say that if we have a set of non disjoint sets whose sum is equal, can be converted into two disjoint sets of equal sum by removing the common elements. If not can somebody give me a counter example . Like in this case we will have 2,3,4 and 9.
combinatorics
edited Mar 28 at 17:03
Andrés E. Caicedo
65.8k8160252
asked Mar 28 at 14:31
Syam KumarSyam Kumar
388
$endgroup$
I am trying to solve the equal subset sum problem , where we have to tell if its possible to generate to disjoint subsets whose (non-zero) sum is equal and you don't have to consider all the elements. The numbers will be unique and in the range of 1-99.
I used a recursive approach of considering each element to be part of first/second/neither set. I am looking to optimize the code.
The second approach is to generate all possible sums and if any sum occurs more than once it would return true. The problem with the second approach is I am not able to prove that the sum generated is from disjoint sets.
Consider sets A = 1,2,3,4 and B = 9,1. The sum of both the sets is 10 and they are disjoint. Is it safe to say that if we have a set of non disjoint sets whose sum is equal, can be converted into two disjoint sets of equal sum by removing the common elements. If not can somebody give me a counter example . Like in this case we will have 2,3,4 and 9.
combinatorics
combinatorics
edited Mar 28 at 17:03
Andrés E. Caicedo
65.8k8160252
asked Mar 28 at 14:31
Syam KumarSyam Kumar
388
edited Mar 28 at 17:03
Andrés E. Caicedo
65.8k8160252
asked Mar 28 at 14:31
Syam KumarSyam Kumar
388
edited Mar 28 at 17:03
Andrés E. Caicedo
65.8k8160252
edited Mar 28 at 17:03
Andrés E. Caicedo
65.8k8160252
edited Mar 28 at 17:03
Andrés E. Caicedo
65.8k8160252
65.8k8160252
asked Mar 28 at 14:31
Syam KumarSyam Kumar
388
asked Mar 28 at 14:31
Syam KumarSyam Kumar
388
asked Mar 28 at 14:31
Syam KumarSyam Kumar
388
388
1
$begingroup$
It is quite safe to say that!
$endgroup$
– Mike Earnest
Mar 28 at 16:03
add a comment |
1
$begingroup$
It is quite safe to say that!
$endgroup$
– Mike Earnest
Mar 28 at 16:03
1
1
$begingroup$
It is quite safe to say that!
$endgroup$
– Mike Earnest
Mar 28 at 16:03
$begingroup$
It is quite safe to say that!
$endgroup$
– Mike Earnest
Mar 28 at 16:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Ok, first of all: If you don't have to use all numbers, may I suggest to take $A = B = emptyset$?
Next: We can write
$$sum_x in A x = sum_x in A cap B x + sum_x in A backslash B x$$ and the same for $B$. This should already finish the proof as long as $A neq B$ (if they are equal you will get the empty set mentioned above).
answered Mar 28 at 14:34
DirkDirk
4,478219
$endgroup$
$begingroup$
Updated the question. I need non-zero sum.
$endgroup$
– Syam Kumar
Mar 28 at 14:38
$begingroup$
My answer still holds. As soon as you have two different sets $A,B$, you can cut out the intersection while still keeping the same sum, using the formula above. This sum will then be over a non-empty set and therefore positive, as you only consider positive numbers.
$endgroup$
– Dirk
Mar 28 at 14:52
add a comment |
Your Answer
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1 Answer
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active
oldest
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1 Answer
1
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oldest
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votes
$begingroup$
Ok, first of all: If you don't have to use all numbers, may I suggest to take $A = B = emptyset$?
Next: We can write
$$sum_x in A x = sum_x in A cap B x + sum_x in A backslash B x$$ and the same for $B$. This should already finish the proof as long as $A neq B$ (if they are equal you will get the empty set mentioned above).
answered Mar 28 at 14:34
DirkDirk
4,478219
$endgroup$
$begingroup$
Updated the question. I need non-zero sum.
$endgroup$
– Syam Kumar
Mar 28 at 14:38
$begingroup$
My answer still holds. As soon as you have two different sets $A,B$, you can cut out the intersection while still keeping the same sum, using the formula above. This sum will then be over a non-empty set and therefore positive, as you only consider positive numbers.
$endgroup$
– Dirk
Mar 28 at 14:52
add a comment |
$begingroup$
Ok, first of all: If you don't have to use all numbers, may I suggest to take $A = B = emptyset$?
Next: We can write
$$sum_x in A x = sum_x in A cap B x + sum_x in A backslash B x$$ and the same for $B$. This should already finish the proof as long as $A neq B$ (if they are equal you will get the empty set mentioned above).
answered Mar 28 at 14:34
DirkDirk
4,478219
$endgroup$
$begingroup$
Updated the question. I need non-zero sum.
$endgroup$
– Syam Kumar
Mar 28 at 14:38
$begingroup$
My answer still holds. As soon as you have two different sets $A,B$, you can cut out the intersection while still keeping the same sum, using the formula above. This sum will then be over a non-empty set and therefore positive, as you only consider positive numbers.
$endgroup$
– Dirk
Mar 28 at 14:52
add a comment |
$begingroup$
Ok, first of all: If you don't have to use all numbers, may I suggest to take $A = B = emptyset$?
Next: We can write
$$sum_x in A x = sum_x in A cap B x + sum_x in A backslash B x$$ and the same for $B$. This should already finish the proof as long as $A neq B$ (if they are equal you will get the empty set mentioned above).
answered Mar 28 at 14:34
DirkDirk
4,478219
$endgroup$
Ok, first of all: If you don't have to use all numbers, may I suggest to take $A = B = emptyset$?
Next: We can write
$$sum_x in A x = sum_x in A cap B x + sum_x in A backslash B x$$ and the same for $B$. This should already finish the proof as long as $A neq B$ (if they are equal you will get the empty set mentioned above).
answered Mar 28 at 14:34
DirkDirk
4,478219
answered Mar 28 at 14:34
DirkDirk
4,478219
answered Mar 28 at 14:34
DirkDirk
4,478219
answered Mar 28 at 14:34
DirkDirk
4,478219
4,478219
$begingroup$
Updated the question. I need non-zero sum.
$endgroup$
– Syam Kumar
Mar 28 at 14:38
$begingroup$
My answer still holds. As soon as you have two different sets $A,B$, you can cut out the intersection while still keeping the same sum, using the formula above. This sum will then be over a non-empty set and therefore positive, as you only consider positive numbers.
$endgroup$
– Dirk
Mar 28 at 14:52
add a comment |
$begingroup$
Updated the question. I need non-zero sum.
$endgroup$
– Syam Kumar
Mar 28 at 14:38
$begingroup$
My answer still holds. As soon as you have two different sets $A,B$, you can cut out the intersection while still keeping the same sum, using the formula above. This sum will then be over a non-empty set and therefore positive, as you only consider positive numbers.
$endgroup$
– Dirk
Mar 28 at 14:52
$begingroup$
Updated the question. I need non-zero sum.
$endgroup$
– Syam Kumar
Mar 28 at 14:38
$begingroup$
Updated the question. I need non-zero sum.
$endgroup$
– Syam Kumar
Mar 28 at 14:38
$begingroup$
My answer still holds. As soon as you have two different sets $A,B$, you can cut out the intersection while still keeping the same sum, using the formula above. This sum will then be over a non-empty set and therefore positive, as you only consider positive numbers.
$endgroup$
– Dirk
Mar 28 at 14:52
$begingroup$
My answer still holds. As soon as you have two different sets $A,B$, you can cut out the intersection while still keeping the same sum, using the formula above. This sum will then be over a non-empty set and therefore positive, as you only consider positive numbers.
$endgroup$
– Dirk
Mar 28 at 14:52
add a comment |
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$begingroup$
It is quite safe to say that!
$endgroup$
– Mike Earnest
Mar 28 at 16:03