Timeline for Set with distinct subset sums
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 23, 2014 at 23:54 | audit | Reopen votes | |||
| Dec 23, 2014 at 23:55 | |||||
| Dec 20, 2014 at 11:39 | vote | accept | ZeroG | ||
| Dec 19, 2014 at 18:14 | answer | added | Asinomás | timeline score: 7 | |
| Dec 17, 2014 at 18:40 | history | edited | ZeroG | CC BY-SA 3.0 | added 32 characters in body |
| Dec 17, 2014 at 18:40 | comment | added | ZeroG | Yes, B also consists of positive integers. | |
| Dec 17, 2014 at 18:08 | comment | added | Asinomás | does $B$ also have to be of positive integers? | |
| Dec 17, 2014 at 9:20 | history | edited | ZeroG | CC BY-SA 3.0 | added 9 characters in body |
| Dec 17, 2014 at 9:19 | comment | added | ZeroG | Yes. All elements of A are strictly positive. Let me add that to the problem statement. | |
| Dec 17, 2014 at 7:44 | comment | added | JimmyK4542 | Are all the elements of $A$ non-negative? | |
| Dec 17, 2014 at 7:26 | history | edited | Alex Ravsky | edited tags | |
| Dec 16, 2014 at 16:56 | comment | added | ZeroG | The problem is the removal/replacement strategy. How do we replace? For e.g., A = 1,2,3,4 , set B = 1,2,3,4. Now since 4 = 1+3, we remove 4. But the only solution for this case is B = 1,2,4. So, removing 4 is not an option. | |
| Dec 16, 2014 at 16:36 | comment | added | AlexR | Algorithmically you could start with $B=A$ and remove elements until 3. is met, making sure that 2. always holds. Some work would be required that if 2. holds and $\sum B_1 = \sum B_2$ for some $B_1\ne B_2\subset B$ that we can remove (or replace) at least one element from either $B_1$ or $B_2$ such that 1. is not violated. WLOG we may assume that $B_1\cap B_2 = \emptyset$ | |
| Dec 16, 2014 at 16:30 | history | asked | ZeroG | CC BY-SA 3.0 |