What is the maximum number of some unit B that some unit A can kill, fighting that number of unit B all at once, killing the weakest first, without micromanagement or reinforcement?
I do *not* currently have time beyond reasonable doubt to explain the derivation behind this formula (although I intend to within the near future), but, this *is* beyond reasonable doubt the correct formula for applicable units:
Let
ƒ( n ) = ⌈n⌉( n − ⌊n⌋ ) + (⌈n⌉^2)/2 − ⌈n⌉/2
Where "⌈ ⌉" denotes rounding up to the nearest integer and "⌊ ⌋" denotes rounding down to the nearest integer.
LetR(A) denote the *singular* unit ratio for some unit A, given an opposing unit B (as defined here in my previous post). Then, for n that is *not* an integer such that
R(A) = ƒ( n )
n is the ratio expressing the maximum number of some unit B that some unit A can kill, fighting that number of unit B all at once, killing the weakest first, without micromanagement or reinforcement.
So, for example, given a unit ratio of 0.63, we observe that
0.63 = ⌈0.63⌉( 0.63 − ⌊0.63⌋ ) + (⌈0.63⌉^2)/2 − ⌈0.63⌉/2
0.63 = (1)(0.63 − 0) + (1^2)/2 − 1/2
0.63 = 0.63 + 1/2 − 1/2
0.63 = 0.63
Thus, in this case,n = R(A) = 0.63.
So, for example, given a unit ratio of 2.86, we observe that
2.86 = ⌈1.93⌉( 1.93 − ⌊1.93⌋ ) + (⌈1.93⌉^2)/2 − ⌈1.93⌉/2
2.86 = (2)(1.93 − 1) + (2^2)/2 − 2/2
2.86 = 2(0.93) + 4/2 − 1
2.86 = 1.86 + 2 − 1
2.86 = 1.86 + 1
2.86 = 2.86
Thus, in this case,n = 1.93.
I do *not* currently have time beyond reasonable doubt to explain the derivation behind this formula (although I intend to within the near future), but, this *is* beyond reasonable doubt the correct formula for applicable units:
Let
Where "⌈
Let
So, for example, given a unit ratio of 0.63, we observe that
0.63 = ⌈
0.63 = (1)(0.63 −
0.63 = 0.63 + 1/2 −
0.63 = 0.63
Thus, in this case,
So, for example, given a unit ratio of 2.86, we observe that
2.86 = ⌈
2.86 = (2)(1.93 −
2.86 = 2(0.93) + 4/2 −
2.86 = 1.86 + 2 −
2.86 = 1.86 + 1
2.86 = 2.86
Thus, in this case,
[This message has been edited by Sagacious (edited 10-15-2017 @ 07:15 PM).]