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First part of the algorithm may fail, if we have started it on the wrong foot. For example $2,3,4,5,6,7,8,9,10,11,12,13,15$ has the basic solution $2,3,4,6$ which you get if you start algorithm from 6. However we can start our algorithm from 7, since there is nothing in step 4. that would say not to, and lock ourselves in, the algorithm cannot end properly. The reason is that 7 is part of another incorrect solution $5,4,3,3$ 7=5+2 and 7=4+3 fromand 4 and 3 are in the first solution. So locked algorithm does not always mean that there is no solution, just to try again with lower initial value. In that case, some ideas about the possible values are hidden within remaining $S_{ij}$. That is why we suggested starting from there in case of failure.

First part of the algorithm may fail, if we have started it on the wrong foot. For example $2,3,4,5,6,7,8,9,10,11,12,13,15$ has the basic solution $2,3,4,6$ which you get if you start algorithm from 6. However we can start our algorithm from 7, since there is nothing in step 4. that would say not to, and lock ourselves in, the algorithm cannot end properly. The reason is that 7 is part of another incorrect solution $5,4,3,3$ 7=5+2 and 7=4+3 from the first solution. So locked algorithm does not always mean that there is no solution, just to try again with lower initial value. In that case, some ideas about the possible values are hidden within remaining $S_{ij}$. That is why we suggested starting from there in case of failure.

First part of the algorithm may fail, if we have started it on the wrong foot. For example $2,3,4,5,6,7,8,9,10,11,12,13,15$ has the basic solution $2,3,4,6$ which you get if you start algorithm from 6. However we can start our algorithm from 7, since there is nothing in step 4. that would say not to, and lock ourselves in, the algorithm cannot end properly. The reason is that 7 is 7=4+3 and 4 and 3 are in the solution. So locked algorithm does not always mean that there is no solution, just to try again with lower initial value. In that case, some ideas about the possible values are hidden within remaining $S_{ij}$. That is why we suggested starting from there in case of failure.

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First part of the algorithm may fail, if we have started it on the wrong foot. For example $2,3,4,5,6,7,8,9,10,11,12,13,15$ has the basic solution $2,3,4,6$ which you get if you start algorithm from 6. However we can start our algorithm from 7, since there is nothing in step 4. that would say not to, and lock ourselves in, the algorithm cannot end properly. The reason is that 7 is part of another incorrect solution $5,3,3,2,2$$5,4,3,3$ 7=5+2 and 7=4+3 from the first solution. So locked algorithm does not always mean that there is no solution, just to try again with lower initial value. In that case, some ideas about the possible values are hidden within remaining $S_{ij}$. That is why we suggested starting from there in case of failure.

First part of the algorithm may fail, if we have started it on the wrong foot. For example $2,3,4,5,6,7,8,9,10,11,12,13,15$ has the basic solution $2,3,4,6$ which you get if you start algorithm from 6. However we can start our algorithm from 7, since there is nothing in step 4. that would say not to, and lock ourselves in, the algorithm cannot end properly. The reason is that 7 is part of another solution $5,3,3,2,2$ 7=5+2 and 7=4+3 from the first solution. So locked algorithm does not always mean that there is no solution, just to try again with lower initial value. In that case, some ideas about the possible values are hidden within remaining $S_{ij}$. That is why we suggested starting from there in case of failure.

First part of the algorithm may fail, if we have started it on the wrong foot. For example $2,3,4,5,6,7,8,9,10,11,12,13,15$ has the basic solution $2,3,4,6$ which you get if you start algorithm from 6. However we can start our algorithm from 7, since there is nothing in step 4. that would say not to, and lock ourselves in, the algorithm cannot end properly. The reason is that 7 is part of another incorrect solution $5,4,3,3$ 7=5+2 and 7=4+3 from the first solution. So locked algorithm does not always mean that there is no solution, just to try again with lower initial value. In that case, some ideas about the possible values are hidden within remaining $S_{ij}$. That is why we suggested starting from there in case of failure.

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  1. Find maximal element $a_{m}$ from the sum (multi)set. $P$, the potential minimal (multi)set is initially empty.

  2. Unless there is only one group, represent $a_{m}$ in all possible ways as a pair of sums that add up to $a_{m}$, $S_{ij}=\{(a_{i},a_{j})|a_{i}+a_{j}=a_{m}\}$

  3. Check that all elements from the set of sums are included.

  4. Find maximal element $a_{s}$ from all $S_{ij}$ (meaning together) with the following property: for each $S_{ij}$, $a_{s}$ is either in $S_{ij}$, or we can find $a_{p}$ from the set of sums so that $a_{p}+a_{s}$ is in $S_{ij}$.

  5. If it is the case that $S_{ij}$ does not contain $a_{s}$, just the sum $a_{s}+a_{p}$, remove $a_{p}+a_{s}$ from $S_{ij}$ (or just set a mark to ignore it) and insert $a_{p}$ and $a_{s}$ in $S_{ij}$ instead.

  6. If an element is present in every $S_{ij}$ remove it from all $S_{ij}$ once (or just set a mark to ignore it and not to touch it any longer) and add it to the list of elements of potential minimal set $P$.

  7. Repeat until all $S_{ij}$ are empty

  8. If some of $S_{ij}$ remains non-empty and we cannot continue, try again the entire procedure fromwith the largest elementmaximum value from all remaining $S_{ij}$.

  9. Recreate the recursive steps without removals and continue with power set coverage algorithm over $P$. (Before this, you can make a safe-check that $P$ includes all elements that cannot be represented as a sum of two elements so they must be in underlying set for sure. For example, the minimal element must be in $P$.)

First part of the algorithm may fail, if we have started it on the wrong foot. For example $2,3,4,5,6,7,8,9,10,11,12,13,15$ has the basic solution $2,3,4,6$ which you get if you start algorithm from 6. However we can start our algorithm from 7, since there is nothing in step 4. that would say not to, and lock ourselves in, the algorithm cannot end properly. ThisThe reason is that 7 is part of another solution $5,3,3,2,2$ 7=5+2 and 7=4+3 from the first solution. So locked algorithm does not always mean that there is no solution, just to try again with lower initial value. In that case, some ideas about the possible values are hidden within remaining $S_{ij}$. That is why we suggested starting from there in case of failure.

  1. Find maximal element $a_{m}$ from the sum (multi)set. $P$, the potential minimal (multi)set is initially empty.

  2. Unless there is only one group, represent $a_{m}$ in all possible ways as a pair of sums that add up to $a_{m}$, $S_{ij}=\{(a_{i},a_{j})|a_{i}+a_{j}=a_{m}\}$

  3. Check that all elements from the set of sums are included.

  4. Find maximal element $a_{s}$ from all $S_{ij}$ (meaning together) with the following property: for each $S_{ij}$, $a_{s}$ is either in $S_{ij}$, or we can find $a_{p}$ from the set of sums so that $a_{p}+a_{s}$ is in $S_{ij}$.

  5. If it is the case that $S_{ij}$ does not contain $a_{s}$, just the sum $a_{s}+a_{p}$, remove $a_{p}+a_{s}$ from $S_{ij}$ (or just set a mark to ignore it) and insert $a_{p}$ and $a_{s}$ in $S_{ij}$ instead.

  6. If an element is present in every $S_{ij}$ remove it from all $S_{ij}$ once (or just set a mark to ignore it and not to touch it any longer) and add it to the list of elements of potential minimal set $P$.

  7. Repeat until all $S_{ij}$ are empty

  8. If some of $S_{ij}$ remains non-empty and we cannot continue, try again the entire procedure from the largest element from all remaining $S_{ij}$.

  9. Recreate the recursive steps without removals and continue with power set coverage algorithm over $P$. (Before this, you can make a safe-check that $P$ includes all elements that cannot be represented as a sum of two elements so they must be in underlying set for sure. For example, the minimal element must be in $P$.)

First part of the algorithm may fail, if we have started it on the wrong foot. For example $2,3,4,5,6,7,8,9,10,11,12,13,15$ has the basic solution $2,3,4,6$ which you get if you start algorithm from 6. However we can start our algorithm from 7, since there is nothing in step 4. that would say not to, and lock ourselves in, the algorithm cannot end properly. This does not always mean that there is no solution, just to try again with lower initial value. In that case, some ideas about the possible values are hidden within remaining $S_{ij}$. That is why we suggested starting from there in case of failure.

  1. Find maximal element $a_{m}$ from the sum (multi)set. $P$, the potential minimal (multi)set is initially empty.

  2. Unless there is only one group, represent $a_{m}$ in all possible ways as a pair of sums that add up to $a_{m}$, $S_{ij}=\{(a_{i},a_{j})|a_{i}+a_{j}=a_{m}\}$

  3. Check that all elements from the set of sums are included.

  4. Find maximal element $a_{s}$ from all $S_{ij}$ (meaning together) with the following property: for each $S_{ij}$, $a_{s}$ is either in $S_{ij}$, or we can find $a_{p}$ from the set of sums so that $a_{p}+a_{s}$ is in $S_{ij}$.

  5. If it is the case that $S_{ij}$ does not contain $a_{s}$, just the sum $a_{s}+a_{p}$, remove $a_{p}+a_{s}$ from $S_{ij}$ (or just set a mark to ignore it) and insert $a_{p}$ and $a_{s}$ in $S_{ij}$ instead.

  6. If an element is present in every $S_{ij}$ remove it from all $S_{ij}$ once (or just set a mark to ignore it and not to touch it any longer) and add it to the list of elements of potential minimal set $P$.

  7. Repeat until all $S_{ij}$ are empty

  8. If some of $S_{ij}$ remains non-empty and we cannot continue, try again with the maximum value from all $S_{ij}$.

  9. Recreate the recursive steps without removals and continue with power set coverage algorithm over $P$. (Before this, you can make a safe-check that $P$ includes all elements that cannot be represented as a sum of two elements so they must be in underlying set for sure. For example, the minimal element must be in $P$.)

First part of the algorithm may fail, if we have started it on the wrong foot. For example $2,3,4,5,6,7,8,9,10,11,12,13,15$ has the basic solution $2,3,4,6$ which you get if you start algorithm from 6. However we can start our algorithm from 7, since there is nothing in step 4. that would say not to, and lock ourselves in, the algorithm cannot end properly. The reason is that 7 is part of another solution $5,3,3,2,2$ 7=5+2 and 7=4+3 from the first solution. So locked algorithm does not always mean that there is no solution, just to try again with lower initial value. In that case, some ideas about the possible values are hidden within remaining $S_{ij}$. That is why we suggested starting from there in case of failure.

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