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For whatever it's worth, I know offhand from my own personal experience with defender clones that I tend to start off by stationing a full million of them to be able to handle everything (in one fight, no rematches) if they need to, and usually by the time I reach 95% damage reduction and... I forget how many levels in Might's "clones on planet+"... But usually I can cut it down to 100k or maybe even 50k defenders and still be safe for every UB attack that comes in. Honestly, less than that could probably be fine but I haven't bothered to try and push the limit.
I think in the same topic with the "power theory" guy it was determined that the wiki's suggested numbers (800k for ITRTG) were probably based on level 50 Might, even though the wiki seemed to make it sound like suggestions were derived based on having no levels in Might. Even without looking at damage reduction on defender clones the suggestion is already kind of flexible depending on that Might level.
Found that topic again (he actually called it battle theory), and a rundown of the basic idea (not getting into the excessive specifics with the number of turns that it takes for a fight to end) would be something like...
* your clone army has a "power value" equal or proportional to their attack, their defense, and the number squared. PV = n*(n+1)*Attack*Defense
* at zero Might we could treat it as PV = n*(n+1)*1*1 or just PV = n*(n+1)
* at 100 Might we could treat it as PV = n*(n+1)*2*2 or PV = 2n*(2n+1), ie lvl 100 Might granting +100% attack and defense is roughly equivalent to having twice as many clones without Might (both ways calculate 4 times as much power value)
* if any two forces (the guy said his theory could apply to armies in RTS games as well) with power values of A and B fight each other and A is larger than B then side A will have AB power value remaining
So in a hypothetical scenario if you had 100k clones and level 123 Might and they were defending with 45% damage reduction (ie 55% damage taken) that'd be roughly PV = (100k)^2 * 2.23 * (2.23 * (1/0.55)) = (100k)^2 * 9.0416 = (300,693.14)^2 = power equivalent to almost 301k clones without any bonuses. And if the opponent you were facing had a (I'm making up this number) power value of (200k)^2, then you should win with a remaining power value of (300,693.14)^2  (200,000)^2 = (224,537.90)^2 which you then divide out that 9.0416 from earlier (which was the product of attack and defense) to get (74,673.44)^2 which means that you should have about 74,673 clones remaining.
Of course, this is an approximation. Don't try to use it as a precision tool. Err on the side of caution and round up and see how it goes. (Meaning that while you could theoretically tinker with the numbers in that hypothetical example to find a number of clones that'd have equal power value as the opponent, you'd still want to use a higher number than that to leave some room for error.)
Edit: N clones with 200 Might and 95 damage reduction would be roughly PV = N^2 * 3 * (3 * (1/0.05)) = N^2 * 180. I don't know what we might assume to be ITRTG's power value, but you'd want a number of clones, N, for which 180 * N^2 results in a larger power value than ITRTG's value. But 800k clones with those bonuses would be equivalent to about 10.7 million clones without any bonuses.
Hmmm.... Maybe if we...? Okay, so with 800k at whatever lower bonuses being an amount that's known to stand a chance against ITRTG... (800k)^2 / 180 = (59,628.48)^2, so roughly 59,628 clones with the 200 Might and 95% DR should be comparable in power to that? I might have goofed that one but I'm tired of it at this point. But if that was a reasonable way to handle it, then you'd want to round up to 100k to leave room for error and see how well they do when ITRTG attacks. (I suppose when I mentioned eventually lowering my number of defenders to 50k that might have been with Might over 250, not 200.)
There's definitly some randomness in the fights (if you check the defense logs you'll see you lose different amounts of clones in the same fight) so it's probably better to have a bit more than the minimum, just in case.