112
Starfield has "fewest bugs that any game from Bethesda has ever shipped with", Microsoft says
(www.eurogamer.net)
this post was submitted on 13 Jun 2023
112 points (100.0% liked)
Gaming
30573 readers
88 users here now
From video gaming to card games and stuff in between, if it's gaming you can probably discuss it here!
Please Note: Gaming memes are permitted to be posted on Meme Mondays, but will otherwise be removed in an effort to allow other discussions to take place.
See also Gaming's sister community Tabletop Gaming.
This community's icon was made by Aaron Schneider, under the CC-BY-NC-SA 4.0 license.
founded 2 years ago
MODERATORS
you are viewing a single comment's thread
view the rest of the comments
view the rest of the comments
Yes! There's actually two facets to consider:
Infinities can be countable or uncountable:
The set of integers is a countable infinity. This is pretty obvious, since you can easily count from one member to the next.
The set of irrational numbers is an uncountable infinity. This is because if I give you one member, you can't give me an objectively "next" one. There's infinitely many choices.
Example: I say what's the next member of the set of irrational numbers after 1.05? Well, there's 1.050001, 1.056, etc.
Can a member of an infinite set be mapped to a corresponding member of another infinite set? And if so, how?
Spoiler, there are three different ways: surjective, injective, and bijective.
In this situation, the sets are both countable. QA can open bug #1, bug #2, etc. It's also - for now - at least a surjective mapping of Starfield bugs -> Skyrim bugs. Because they're both countable, for each bug in Starfield you can find at least one bug in Skyrim (because it's a known bigger set at the moment).
But we don't know more than that right now.
I love that this comment represents more work into the issue of bugs than Bethesda bothers with.
I'd just like to chip in that it isn't necessary for a countably infinite set to have an obvious method of counting. Listing all of the rationals in numerical order isn't possible (what's the smallest fraction above 0?) but it is nevertheless possible to create a bijection with the naturals.
Great point! It's been a while since my degree (and I don't use it), so I knew I'd probably get something wrong.