Such a nice place AoCZone is <3One could, provided it is really impossible. But how does one know that?
You know what, I think I'll give a rigorous answer to that. Hold on to your seats!
[OneMillion doing his thing.]
Such a nice place AoCZone is <3One could, provided it is really impossible. But how does one know that?
You know what, I think I'll give a rigorous answer to that. Hold on to your seats!
[OneMillion doing his thing.]
One could, provided it is really impossible. But how does one know that?
You know what, I think I'll give a rigorous answer to that. Hold on to your seats!
Theorem: A task is optimised if, and only if, the resources left over upon execution is zero.
Proof: Define an object which we will call a task s. A task resides in the set of all tasks, which is a vector space. Then we can ....bla math things bla bla....
One could, provided it is really impossible. But how does one know that?
You know what, I think I'll give a rigorous answer to that. Hold on to your seats!
Theorem: A task is optimised if, and only if, the resources left over upon execution is zero.
Proof: Define an object which we will call a task s. A task resides in the set of all tasks, which is a vector space. Then we can define a function f(s) which associates with each task a four-vector specifying the resources required to execute that task. We can define a further function g(f(s),t) which adds the cost of any auxiliary spending, such as villager production. Define an auxiliary function h(r,t), which returns the resources generated in a given time t in which a town centre stops producing villagers temporarily at time r. The partial derivatives of this function with respect to r and t are positive. Suppose also r and t obey a relation t-r>=d, for some d. Next, define a fourth function w(r,s,t)=h(r,t)-g(f(s),t). Then for a given r,s and t, if w(r,s,t)>0, we can take a t-dt and s-ds to get w(r,s,t)-(6w/6t)dt-(6w/6s)ds<w(r,s,t). When r',s,t' are found such that w(r',s,t')=0, we can consider a t'-dt and s'+ds such that w(r',s,t')-(6w/6t)dt+(6w/6s)ds=0 subject to t-r>=d. When t-r=d, it is impossible to lower t without also lowering r, and bringing w below zero, accomplishing a task before the resources necessary have been gathered. Therefore, w(r'',s,t'')=0 and t'' is the optimised time.
Corollary: Goths produce 40 champions faster ab initio than Byzantines.
Proof: Consider the task s of producing 40 champions ab initio. Then let a be the optimised time for the Koreans. Let r and t be the optimised parameters. Then w(r,s,t)=0. The Byzantines have a one third discount on the imperial age, which saves 600 resources. This yields w(r,s',t) with trace equal to 600. The Goths meanwhile save 1120 resources on swordsmen, so w(r,s'',t) has trace 1120. Optimise with w(r,s',t) to yield w(r',s',t')=0. Then the trace of w(r',s'',t') is 520. Finally, we have r''<r', t'<t'' such that w(r'',s'',t'')=0.
I have skipped over some subtleties, but in plain English, it means, if you have resources left over, you can do it faster, and if it costs less, you can do it faster. I can't see anyone other than Goths coming out on top in this one!
Im getting close to min 25 :P
You disappoint me. I went to that effort just for you!tl;dr lol as I said after thinking about it, it is obvious that goths are faster. I was referring to how you can prove that it is impossible to be under 25min 11
I'm at 24 minutes.
I knew it! 1111 one million :lol:
I knew it! 1111 one million :lol:
If someone submits a recording doing it in under 25 minutes as Byzantines, I'll admit it was possible after all. I wouldn't have thought Goths would be able to go that low (24:xx we don't know yet how many seconds), but it doesn't prove Byzantines can be done in less than 25 minutes.
One could, provided it is really impossible. But how does one know that?
You know what, I think I'll give a rigorous answer to that. Hold on to your seats!
Theorem: A task is optimised if, and only if, the resources left over upon execution is zero.
Proof: Define an object which we will call a task s. A task resides in the set of all tasks, which is a vector space. Then we can define a function f(s) which associates with each task a four-vector specifying the resources required to execute that task. We can define a further function g(f(s),t) which adds the cost of any auxiliary spending, such as villager production. Define an auxiliary function h(r,t), which returns the resources generated in a given time t in which a town centre stops producing villagers temporarily at time r. The partial derivatives of this function with respect to r and t are positive. Suppose also r and t obey a relation t-r>=d, for some d. Next, define a fourth function w(r,s,t)=h(r,t)-g(f(s),t). Then for a given r,s and t, if w(r,s,t)>0, we can take a t-dt and s-ds to get w(r,s,t)-(6w/6t)dt-(6w/6s)ds<w(r,s,t). When r',s,t' are found such that w(r',s,t')=0, we can consider a t'-dt and s'+ds such that w(r',s,t')-(6w/6t)dt+(6w/6s)ds=0 subject to t-r>=d. When t-r=d, it is impossible to lower t without also lowering r, and bringing w below zero, accomplishing a task before the resources necessary have been gathered. Therefore, w(r'',s,t'')=0 and t'' is the optimised time.
I'm at 24 minutes.
100% that byzantines can't. Fayt is just trolling and saying he got 24:xx time. Don't think goths can do it either, or atleast I'm nowhere close 11.
no recording submitted yet. I guess people might think i am going to give away their strategy and are waiting for the last days to do so?
Hey Mr.OneMillion, I would probably have no interpretation about this if i didn't take "Optimization Theory" this semester . This conclusion comes from KKT, namely Karush–Kuhn–Tucker conditions right ? I respect your adaptation to AoC related problem. I remember reading some of your cool mathematical approach into other stuff. What do you do for living if i may ask you ?
I knew it! 1111 one million :lol:
If someone submits a recording doing it in under 25 minutes as Byzantines, I'll admit it was possible after all. I wouldn't have thought Goths would be able to go that low (24:xx we don't know yet how many seconds), but it doesn't prove Byzantines can be done in less than 25 minutes.
100% that byzantines can't. Fayt is just trolling and saying he got 24:xx time. Don't think goths can do it either, or atleast I'm nowhere close 11.