91–100 of 120 posts
Set family {{1,2,3,4}, {1}, {2}, {3}, {4}} has all 4 elements appearing in exactly 2 sets out of the 5, that is 40%. What am I missing?
"A collection, or family, of sets is considered “union-closed” if the union of any two sets in the family equals any existing set in the family."
EDIT: In your hypothetical collection of sets, the union of "2" and "3" would produce the set "2,3" which is not an enumerated set in your list.
Set family {{1,2,3,4}, {1}, {2}, {3}, {4}} has all 4 elements appearing in exactly 2 sets out of the 5, that is 40%. What am I missing?
Always cool to see unexpected applications of information theory, especially outside of probabilistic contexts. A related toy example that comes to mind is the puzzle where you're given 12 metal balls and told that one of them is either heavier or lighter than the rest. You're then given a balance/scale and instructed to figure out which ball is different and whether it is heavier or lighter using the balance a maxim…
I once made a cool solution to this problem using information theory, in particular coding theory. The usual solution uses the fact there are 3 places to put items: left scale, off scale, and right scale. Then those solutions do if/then cases to juggle items around to find the mis-weighted ball. My solution numbered the balls in base 3, then did four weighings with no if/thens,. The weightings each gave a base 3 resu…
Impressive result however.
Earlier quoted context omitted.
Just took a stab at recreating my older solution. It goes something like this Number the balls 7 8 9 10 11 12 14 15 16 17 18 19 Then do three weighings according to rows like this: 0 1 -1 0 1 -1 1 -1 0 1 -1 0 1 1 -1 -1 -1 0 0 1 1 1 -1 -1 -1 -1 0 0 0 0 0 0 0 0 1 1 Where -1 means index on left side of of balance, 0 means not on balance, 1 means on right side of balance. Note each row has the same number of -1 and 1. If…
I like how simple the deciding process becomes but I’m afraid there might be a problem. This solution works when the odd ball is always heavier, but it could either be heavier or lighter, and you don’t know which. For example, say ball 7 is heavier. From top to bottom, the weightings would result in: 120. However, if ball 19 was lighter you would obtain the same weighing sequence 120. I believe that there is another…
Earlier quoted context omitted.
Just took a stab at recreating my older solution. It goes something like this Number the balls 7 8 9 10 11 12 14 15 16 17 18 19 Then do three weighings according to rows like this: 0 1 -1 0 1 -1 1 -1 0 1 -1 0 1 1 -1 -1 -1 0 0 1 1 1 -1 -1 -1 -1 0 0 0 0 0 0 0 0 1 1 Where -1 means index on left side of of balance, 0 means not on balance, 1 means on right side of balance. Note each row has the same number of -1 and 1. If…
I like how simple the deciding process becomes but I’m afraid there might be a problem. This solution works when the odd ball is always heavier, but it could either be heavier or lighter, and you don’t know which. For example, say ball 7 is heavier. From top to bottom, the weightings would result in: 120. However, if ball 19 was lighter you would obtain the same weighing sequence 120. I believe that there is another…
The idea works. When I solved it before I cider it up to test
I’m glade to know that “did a PhD in maths six years ago and work in applied mathematics” is now being long out of math. I see clickbaity titles are still doing fine in 2023. Impressive result however.
It is possible for someone who is not a full-time mathematician by still working on mathematics in their spare time or in a mathy-field in their day job but very very few real examples of that can be found.
So I don't find the title clickbaity. It is really really hard to come up with results like this without working on mathematics research (not AI research) 10 hours a day!
Earlier quoted context omitted.
I once made a cool solution to this problem using information theory, in particular coding theory. The usual solution uses the fact there are 3 places to put items: left scale, off scale, and right scale. Then those solutions do if/then cases to juggle items around to find the mis-weighted ball. My solution numbered the balls in base 3, then did four weighings with no if/thens,. The weightings each gave a base 3 resu…
Just took a stab at recreating my older solution. It goes something like this Number the balls 7 8 9 10 11 12 14 15 16 17 18 19 Then do three weighings according to rows like this: 0 1 -1 0 1 -1 1 -1 0 1 -1 0 1 1 -1 -1 -1 0 0 1 1 1 -1 -1 -1 -1 0 0 0 0 0 0 0 0 1 1 Where -1 means index on left side of of balance, 0 means not on balance, 1 means on right side of balance. Note each row has the same number of -1 and 1. If…
I’m glade to know that “did a PhD in maths six years ago and work in applied mathematics” is now being long out of math. I see clickbaity titles are still doing fine in 2023. Impressive result however.
To be fair, results like this are very improbable to come from someone who is not a full-time mathematician. And no that does not mean someone working in machine learning or someone who did their PhD 6 years ago. How many such results do you see from someone who is not a full-time mathematician? It is possible for someone who is not a full-time mathematician by still working on mathematics in their spare time or in a…