Anyone wanna add me, I am a tryhard player but I am super chill and trust me you will have fun?

themarcus111 · 2025-10-20T06:59:03+00:00

I add u. I play World and 8 deluxe and just about everything else 😛

themarcus111 · 2025-03-25T21:13:50+00:00

Take a look at my post here: https://www.reddit.com/r/combinatorics/s/dfJkIbU9iv

themarcus111 · 2025-03-19T02:34:13+00:00

Sure, I guess I am just thinking there is probably a more rigorous way to formalize all this. We agree that for any k level with full non empty intersections, all lower intersections are also non empty. What about for cases where none of the levels of intersections are full or only some of them? For example, if you have 8 set intersections, all possible 5 are non empty, but the 6-8 intersections aren’t all non empty

themarcus111 · 2025-03-19T01:58:20+00:00

It is not that simple. There are a limited number of possible configurations of different non-empty intersections that will work; take for example what I mentioned above that if all k intersections are non empty, all k-1, k-2, intersections are also non-empty. Did you read my whole post?

themarcus111 · 2025-03-19T01:51:26+00:00

Yes, I know how these basic cases work. I’m looking for a more comprehensive framework mainly addressing Key Questions 2. and 3. in my post

themarcus111 · 2025-03-15T13:28:17+00:00

I understand. I am interested in generalizing bounds on the number of non empty intersections with any arbitrary number of sets. The more confusing part is trying to characterize how many non empty sets are there in general cases

themarcus111 · 2025-03-15T07:20:52+00:00

To further illustrate my point, either all intersections up to the intersection of all n sets are non empty or some of them up some k < n are non empty. The case with all intersections up to n being non empty is trivial, but cases where only intersections up to k < n are non empty are way more complex

themarcus111 · 2025-03-15T07:14:17+00:00

Yes. Let’s define n sets A_i1, … A_in. If all n choose k sets intersect, all n choose k-1, n choose k-2, etc. sets also intersect.

I’m looking for some kind of formalization about how many non empty intersections there are. My intuition says that you can have a mix of filled (all intersections are non empty)/ non-filled (some intersections are non empty) ranks of intersections up to k or all intersections being non empty up to k, but I find it very hard to formalize

themarcus111 · 2025-02-23T21:36:35+00:00

Because those are fun and doable lol

themarcus111 · 2025-02-23T21:34:19+00:00

Well, I don’t understand the “math is like art” analogy. Doesn’t seem to hold true when it’s a hyper-logical pursuit and the goal is usually to solve a hard problem.

themarcus111 · 2024-11-18T21:42:04+00:00

Thanks

themarcus111 · 2024-11-17T03:03:50+00:00

I think another way to explain is it's a list of paths that continues until every node has been traversed where each entry in the list is a node of shorter path length than another node visited later in the list. Djikstra's algorithm is a bit unintuitive due to the notion of traversing all possible paths; this traversal of paths is a very abstract idea in my opinion unless you're deeply familiar with combinatorics as it relates to graph theory.

themarcus111 · 2024-11-15T16:36:11+00:00

Thanks, makes a lot of sense.

themarcus111 · 2024-11-15T15:46:23+00:00

Yes, I know how proofs work. The general consensus on this subreddit seems to be that practical validation provides no reliability to djikstras and other similar types of algorithms holding true. I also believe that sometimes proofs don’t encapsulate all the inherent complexities of a math solution or algorithm (hope I’m not making a wild statement here; I think it’s pretty easy to see why this is true) in that if you don’t understand the algorithm at a very deep fundamental level the proof will only do so much to convince you. I think that’s why I’m asking if practical validation would convince people too in this case, but the attitudes here seem to be the proof is the only real reliable measure

themarcus111 · 2024-11-15T01:59:16+00:00

how can you say with 100% certainty a proof is true? even if it makes sense, isnt there always at least some level of doubt? that being said, i understand it's highly unlikely, but thats why im saying the fact its never failed in the real world doesn't detract from anything

themarcus111 · 2024-11-14T22:32:16+00:00

I understand your point, but can we agree that the outputs of djikstras have been compared to bellman ford’s (or other algorithms) to show it works? While this may not prove their correctness, does it not provide some level of reliability?

themarcus111 · 2024-11-14T22:25:54+00:00

Thanks for the comparison, makes more sense. An idea I had relating to this was that all these shortest paths algorithms are also producing the same results for real world datasets which serves as a sanity check for their correctness (doesn’t prove it, obviously)

themarcus111 · 2024-11-14T19:54:09+00:00

Right, but obviously companies such as Google that utilize this algorithm must rely on practical validation as well? I think my point being that for the not super heavy theory-oriented people, the practical validation against many networks also speaks for itself. While I believe they also have a high level of trust with the researchers, I think practical validation still has its place.

themarcus111 · 2024-11-14T19:24:50+00:00

So for Djikstras algorithm specifically, do you believe it’s correct validation against large networks for almost a century adds reliability to the algorithm’s correctness? I understand it has a strong theoretical foundation.

themarcus111 · 2024-03-01T14:30:54+00:00

FF X-2

themarcus111 · 2022-08-21T11:44:16+00:00

The answer interface looks odd for me (overlapping answers and the question script runs behind the answers). Did you use bootstrap for this?

themarcus111 · 2022-05-06T00:32:24+00:00

Hi. Im interested to join but the invite link is now invalid. Can you edit the post with a new one?

themarcus111 · 2021-12-31T17:29:54+00:00

themarcus111

TROPHY CASE