
Post by stamos65 on Jun 8, 2018 1:03:47 GMT 5
Let's have some fun.How many ways can you fit weapons on the four slots on a patton such that no adjacent slot has the same weapon?Adjacent slot means for example the two side weapons on the patton or the top right and bottom right weapon slots.



Post by stamos65 on Jun 8, 2018 1:06:33 GMT 5
You can also post your War Robots math questions here



Post by windcaster on Jun 8, 2018 1:46:16 GMT 5
Ok, someone go for his legs, I got the top.


Deleted
Deleted Member
Posts: 0
Karma:

Post by Deleted on Jun 8, 2018 2:27:50 GMT 5
Lots, using Terry Pratchett's counting for trolls.



Post by Pulse Hadron on Jun 8, 2018 2:42:37 GMT 5
Figure out the number of light weapons which I’ll label N. Then I think it’s N*(N1)^3
This is a first guess, I’ll ponder it some more later. Btw, are you asking or do you already know?
edit: oh wait, need clarification what you mean by adjacent? Are you only talking about adjacency between the weapons on either side but not the adjacency of the 2 weapons on top? Are the top and bottom weapons the same on both sides?....



Post by eco on Jun 8, 2018 5:56:07 GMT 5
it is M where M is the proper number of non adjacent linear escalations. I did not consider hyperextrapolation solutions since they would be located on the zaxis. anyhow you could easily formulate the solution of these as M+L. All statements are made with the following assumtions (to get a reproducable result) 20 degree celsius, 0m above sea level. 40% humidity, 50% full moon cycle, lastly but not least I assume the bot stands still.



Post by stamos65 on Jun 8, 2018 9:41:59 GMT 5
Figure out the number of light weapons which I’ll label N. Then I think it’s N*(N1)^3 This is a first guess, I’ll ponder it some more later. Btw, are you asking or do you already know? edit: oh wait, need clarification what you mean by adjacent? Are you only talking about adjacency between the weapons on either side but not the adjacency of the 2 weapons on top? Are the top and bottom weapons the same on both sides?.... Think of it as a square where the four corners are where the light weapons are supposed to go Edit:yeah it was for fun,I already know the answer.



Post by Pulse Hadron on Jun 8, 2018 21:00:42 GMT 5
Ah, ok. n*(n1)^3 works for a segment, the first slot you have n choices and for every subsequent slot there’s (n1) choices.
But now considering a ring where the last slot of the square is adjacent to the first, some of those (n1) choices of the last slot are the same as the first slot and they need to be subtracted. To find that amount I drew out a tree for 4 types and reasoned about the counts (because my permutation knowledge is rusty). So given types ABCD (n=4), slot 1 has A which branches to B C and D in slot 2, which branch to slot 3 as B>ACD, C>ABD, D>ABC, and those branch into slot 4 resulting in 27 paths down the tree. But 6 of those paths end in A.
Looking at the tree I could see that A only appears in slot 4 once for every node in slot 3 that is not A. That is, the number of terminal A equals the number of nonA in slot 3. To get the number of nonA in slot 3 note that every node in slot 2 is necessarily not A and there are (n1) of them. Each of those (n1) nodes in slot 2 has (n1) branches to slot 3, but one of those branches goes to A. So there are (n2) branches from slot 2 that go to nonA in slot 3. Therefore, the number of terminal A to subtract is the number of nodes in slot 2 (n1) times the number of branches that don’t go to A (n2).
The counts of that tree are for one type in slot 1. Multiply by n to get the count for all types in slot 1 and yields the expression...
n * ( (n1)^3  (n1)*(n2) )
(n1)^3 gives the total paths from a single starting type, subtract (n1)*(n2) to remove terminals with the same starting type, and multiply by n for every starting type.
Trying to simplify I foiled that to n * (n^3  4*n^2 + 6*n  3) which isn’t much nicer. Betting there’s a cleaner way to write this.
Anyways, there are 12 light weapons and plugging that in gives 14652 possibilities.
While looking up the number of light weapons I noticed a complication, a 13th light slot item that is not a weapon and can’t be mounted on the top 2 slots, the god damned Écu shield. I will ponder this later. I’m also curious for a general expression of nonadjacency on any ring size.



Post by Pulse Hadron on Jun 8, 2018 21:32:30 GMT 5
Factoring in Ecu was pretty easy. It can only by mounted in 1 slot since it has to be bottom left or bottom right which are adjacent. Considering mounted on bottom left there are n choices for top left (n is still 12) and (n1) choices for both top right and bottom right. That’s 1452 possibilities with Ecu on the bottom left, multiply by 2 for it also mounted bottom right for a total of 2904 possibilities with Ecu. Add that to the possibilities without Ecu, 14652, for a grand total of 17556 possibilities of mounting light items on a Patton with no adjacency repetition.



Post by Browncoats4ever on Jun 8, 2018 23:46:51 GMT 5
You lost me at math.



Post by zombiewolf907 on Jun 8, 2018 23:53:54 GMT 5
Three  wait  that is how many licks it takes to get to the center of a Tootsie Pop! (Showing my age)



Post by Browncoats4ever on Jun 9, 2018 0:27:04 GMT 5
37?!



Post by HarvesterOsorrow on Jun 9, 2018 9:33:41 GMT 5
Is adjacency really a word? Huh, learn something new every day
Next thread is grammar and English within War Robots
Good times LOL



Post by Pulse Hadron on Jun 9, 2018 21:27:24 GMT 5
I already know the answer. So am I right?



Post by stamos65 on Jun 11, 2018 9:06:55 GMT 5
I already know the answer. So am I right? Actually,my answer was 12210.How? Here's the formula(X is the number of weapons) {(X1)*((X2)^3)}{(X1)*(X2)(X3)})[C1]{(1*X*(X1)^2)*2}[C2] Where C stands for case.The case 1 bit is to calculate the number of builds without the ecu,that's why it's X1.The case 2 bit was to calculate the builds with the ecu.I multiplied it by 2 cuz,you know,the builds with ecu on the left can be reflected to the right.



Post by Pulse Hadron on Jun 11, 2018 22:09:21 GMT 5
{(X1)*((X2)^3)}{(X1)*(X2)(X3)})[C1]{(1*X*(X1)^2)*2}[C2] Not being sure where all of your expression comes from (there’s an extra closing parentheses before C1) and not reproducing your result value from it I noticed there’s only 13^4 = 28561 total permutations so wrote a bit of code to brute force test each one. dim arr(3), i0, i1, i2, i3, count As integer
for i0 = 0 to 12 arr(0) = i0 for i1 = 0 to 12 arr(1) = i1 for i2 = 0 to 12 arr(2) = i2 for i3 = 0 to 12 arr(3) = i3
if hasNoAdjacencyRing(arr) then count = count + 1
next next next next
resultField.Text = Str(count)
Function hasNoAdjacencyRing(arr() As integer) As Boolean //a value of 0 in the array is considered an Ecu //index 0=bottomleft 1=topleft 2=topright 3=bottomright dim i As integer
for i = 0 to 2 //Test adjacency on ring if arr(i) = arr(i+1) then return False next if arr(0) = arr(3) then return False
if arr(1) = 0 or arr(2) = 0 then return False //filter Ecu on top
return True //no adjacency on ring and no misplaced Ecu
End Function
The result is 17556. I’ll list them out later

