Brilliant series! You have a wonderful delivery, and it is a pleasure to listen and watch. Your grasp of the material is made clear by your ability to break down the information into understandable chunks, without losing anything in the process. AAA!
Thanks you so much! After struggling much with no possible exhaustive content available online that I was aware of, your video was like a saving grace!
@Ben1994, When I tried to generate a random composition function over a set of five items, I started by using shuffled playing cards to generate five different possible permutations of five items. Then I tried to generate a composition function from those. But I immediately ran into a failure to get closure. Applying my first permutation to itself resulted in a permutation that wasn't among the five random permutations I selected for use at the start of the algorithm. Have I misunderstand your algorithm?
You need to make sure you use the full set of possible permutations of your underlying set. Choosing an arbitrary subset of the permutations will not generate a group, for the exact reason you mentioned. For a set of n elements there are n! permutations, so you would need to use all 120 permutations in your case. It just so happened that in the example in the video the underlying set had the same number of elements as the set of permutations (because 2! = 2). If you want to try another example, try using a set of 3 elements. You should get 6 permutations total, meaning 6 elements in your group. By the way, if you're interested in understanding why you can't use an arbitrary subset of the permutations of a given set to generate a group, look into subgroups.
Bro @10:53 the other way to fill the table: Take for example i composed with i. Say, ioi(a)=i(i(a))=i(a)=a. So we started with ioi(a)-->a which is identity. Similarly, if we want to find io(tau)(a)=I(tau(a))=I(b)=b. So it went from a to b so it is tau. Like that we can complete the table
That's pretty much the same method, except the author of this video did this more visually, which is less tedious and less error-prone, and calculates multiple paths in parallel.
Do composition functions (over a finite number of elements) exist that are not homomorphic to modular arithmetic? If so, I would want my algorithm for generating composition functions to include them.
@@waynemv Using permutations of sets isn't "complete" either (ie not all groups are isomorphic to a symmetric group). If you want an algorithm to enumerate finite groups, just enumerate and test all possible composition tables.
@@waynemv Yes, modular arithmetic is commutative and there are non-commutative groups. Composing permutations don't generally commute, for example. But this is not meant to be an exhaustive approach for constructing groups, just an example of one way of doing it.
@@khoavo5758 Which becomes very tedious, because the number of possible ways of building a Roman square grows combinatorially with the table's size. Even with a computer it takes hours for tables with only ~10…20 elements, and quickly becomes infeasible for bigger tables. Then you also have to check the associativity for each of them to see which ones are groups and which ones aren't, and after that, you would have to check every pair of them for isomorphism to avoid counting isomorphic tables more than once. There HAS to be a better way.
