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Spoiler wrote:There are 6 Z-Bots and 3 rows, and since 2 Z-Bots are moved, that means 6 divided by 3 equals 2, right?
And there are 120 Z-Bots and maybe 15 rows, 120 divided by 15 is 8.
Indeed.StinkerSquad01 wrote:I was expecting this.
The tennis adventures aren't all that impressive compared to people's main adventures.Sammy_P wrote:You could've atleast featured the tennis WAE levels, just saying.
Anyone who's interested may PM me. 
15 rows of z-bots, and it will take 45 moves (I think...)
45 moves is not very far from my solution, but I can do it with less moves.
I was able to get it in 41 move and I'm not going to try any further. EDIT: Marinus has beaten me, not only solving it in 40 moves but also finding out it's the minimum number of moves.
Marinus, is you solution similar?It involves some math, but can there be proof without it, right?
To find minimal number of z-bots to move, we must minimize area (special "z-bot" area) of three triangles that are outside of overlapping zone of two triangles with side 15 (it's easy to find, as number of z-bots is sum of arithmetical progression, so n(n+1)/2=120 and n=15 (also -16, but obviously it can't be negative)), when one of them is turned upside down.
First of all, it's obvious that biggest coverage we'll receive when at least one height of that triangles coincide (we'll definitely got more bots outside if shift it aside). So we'll have three small triangles who's area we need to calculate, and we already have formula for z-bots count in them: n(n+1)/2.
We'll have two same symmetric triangles on bottom - left and right. And we'll have one triangle on top. Another observation: we have to increase size of top triangle each time by 2 to have bottom triangles same. On one edge we'll have one huge triangle with side 14, and two small were reduced to 0, and on other edge - two with side 7, and that that was big - reduced to 0 too. It's easy to see that if size of bottom triangle is n, then the size of top triangle is m=k-2n. (k=14-constant restriction based on original triangle size) Now as we have n and m we can easily calculate what is number of z-bots outside overlapping zone:
s=2n(n+1)/2+m(m+1)/2=n(n+1)+(k-2n)(k-2n+1)/2=3n^2-2kn+(k^2+k)/2.
This is quadratic function (on n) and it will have it's minimum in point k/3.
When this point isn't integer, we'll have to round it to closest integer. Function is parabolic, so it we be okay (I can also prove this is someone don't believe, but I'm too lazy to do it now).
When k=14:
s=3n^2-28n+105 and minimum is 14/3=4.(6), and closest integer point is 5. And in point 5 there will be minimal value of function: s=40. This value consists of three triangles, two with side 5 (15 bots each) and one with side 4 (10 bots). Substituting other close points, we'll receive other solutions listed up here:
n=p=4, m=6, s=41
n=p=6, m=2, s=45
p.s. After thinking a bit more, it's not so obvious that heights have to coincide. Let's prove that too. So we can have three different triangles with the rule that restricts their sized: m+n+p=k and we have to minimize function m(m+1)/2+n(n+1)/2+p(p+1)/2. To do calculations we can drop divisions by two as they don't impact the point of minimum (function is always positive), and also drop +1s after opening brackets as they will always gave us k in sum, that also don't change the result. So the function is equivalent to the m^2+n^2+p^2. It will have absolute minimum when all of the m n and p are same and equal to k/3 (suspicious similar result, huh?:)). But when k/3 isn't integer, we need to find closes integer point that satisfies minimum condition. Let's assume that in solution none of heights coincide, so we have three different small triangles m!=n, m!=p and n!=p. Function is parabolic, so the further we go from optimum point, the bigger difference is, so we can assume that difference between m, n and p will not exceed 1. The only possible case when all that inequalities for this is when m+1=n=p-1 (or similar with different order). But in this case k=m+n+p=3n, so m+1=n=p-1 isn't optimal solution as m=n=p is better. As a result we have that at least two of the m, n and p coincide, so overlapping is symmetrical and this means that one height of triangles must coincide.
No, actually my solution is by trying things out: https://docs.google.com/spreadsheet/ccc ... l=nl#gid=1Marinus, is you solution similar?
Thank you, actually I'm not completely new, as I had around dozen post time ago. But it seems that now I have one againStinkerSquad01 wrote:Welcome to the forums, professor_k.