Is there a concise description of what ‘special’ damage is?

Hello RPG Stack this may be somewhat a worthless question as I’ve looked diligently without success. I don’t believe the information is physically present but I thought I’d give it a shot. In my games things have temporarily side stepped in to the sword and planet genre (lol don’t ask) and I find myself in need of modelling a famous character from the genre. I was looking at page 146 of the Dungeon Master’s Guide under the Futuristic Weapons heading as it states:

‘Futuristic weapons are like other ranged projectile weapons, though the type of damage they deal is special.’

My question, is there anywhere in D&D 3.5 rules that defines concisely what special damage is?

How to make my code for batch label for graphs more readable and concise?

I’d like to label my graphs at once, so I write following codes.

chemgraph = CompleteGraph /@ Range[3]; labels := {1/2, 1/3, 1/4}; Table[chemgraph[[i]] //    Labeled[Graph[#, ImageSize -> 100, VertexLabels -> None],      Style[#,         20] & /@ (ToString[#, StandardForm] & /@ labels)[[i]]] &, {i,    1, labels // Length}] 

It can do that, but I think it is not easy to read and not beautiful here with using Table.
Is there any way to improve? enter image description here

Concise CSS Tips

What are some good coding practices to keep CSS code concise so that CSS code is effective and saving you time.

Could you share tips that you have picked up on the way and lessons learned?

I have picked up a few things along to way.

  • Using styles for things that require specific formatting on large websites as opposed to targeting elements individually which will produce long repeated code.
  • Grouping elements that are in the same section to save typing same properties over and…

Concise CSS Tips

Concise way of “updating” element in nested collections

Imagine simple game:

type Combatant = {     hp : int     attack : int }  type CombatantGroup = Combatant list type CombatantGroups = CombatantGroup list  type Battle = {     combatantGroups : CombatantGroups }  

there is some battle, in which some amount groups participate. In each group there is several combatants. Now I want to implement a function for one combatant to attack another. Due to immutability, I have to replace whole attacked target, and whole group and whole battle. If the battle was just between two participants, I could write code like this:

if target = battle.combatant1 then     { battle with combatant1 = { battle.combatant1 with hp = battle.combatant1.hp - attacker.attack }} else     { battle with combatant2 = { battle.combatant2 with hp = battle.combatant2.hp - attacker.attack }} 

which is also terrible code, I would appreciate any advice on it. But with nested collections it gets even more complicated, and event doesn’t look functional to me anymore:

let containsTarget = List.contains target         let transformCombatant combatant = if combatant = target then { combatant with hp = combatant.hp - attacker.attack } else combatant         let transformGroup = List.map transformCombatant                 let checkGroup group =     if group |> containsTarget then         group |> transformGroup     else         group         let transformGroups = List.map checkGroup  { battle with combatantGroups = transformGroups battle.combatantGroups } 

Can you please give me advise on how can I use features of F# and functional programming, to make given code more pretty and concise?

Concise description for a sequence with a specified fractal property

I have a sequence A with the fractal pattern that the innermost values x, where length(x) = length(A)/2, has sum(x) = sum(A)/2. Also this is a fractal pattern so that for sequence x with the fractal pattern that the innermost values y, where length(y) = length(x)/2, has sum(y) = sum(x)/2. This continues as long as the innermost values length is divisible by 2. Example:

A = 31,18,37,24,27,46,33,36,23,-6,-3,16,19,38,41,12,-1,2,-11,8,11,-2,17,4 x = 33,36,23,-6,-3,16,19,38,41,12,-1,2 y = -6,-3,16,19,38,41

A(length)=24 A(sum)=420 x(length)=12 x(sum)=210 y(length)=6 y(sum)=105

I have a list of other constraints to aid in this: for a sequence of integers z that is 48 values in length, for values x and y, at positions a and b, x+y=35, where a=48-(b-1), partitioning z into 16 sections each with 3 consecutive values from z, and summing the pieces giving 3 final values, M,N,O, ie M is a sum of values at position 1,4,7,10,13,…46, where:

M < N < O, and O – N = N – M, and N -M is a power of 2. How many possible integer sequences z exist?

z=27, -18, 1, 4, 23, 26, 13, 32, 19, 22, 41, 44, 31, 18, 37, 24, 27, 46, 33, 36, 23, -6, -3, 16, 19, 38, 41, 12, -1, 2, -11, 8, 11, -2, 17, 4, -9, -6, 13, 16, 3, 22, 9, 12, 31, 34, 53, 8. Also A is a subset of z.

In this example: M=216,N=280,O=344,N-M=64=2^6

What is a concise way to describe this property? Thanks!

Can I make this Dictionary sort (smallest to biggest) algorithm more concise? Swift

I have the bellow sorting algorithm which takes an array of dictionary values as declared below.

 guard var imageUrlString = anyImage.value as? [String:AnyObject] else { return } 

I then loop through it adding the values with the smallest Int key to a new array to be used afterward. Thus the values in this array of AnyObjects would go from 1 to n.

I was wondering if I could make it more concise.

        var values = [AnyObject]()         var keys = [String]()         var Done = false         var j = 1           while !Done {             for i in imageUrlString {                 let key = Int(String(i.key.last!))                  if j == key {                     values.append(i.value)                     keys.append(i.key)                     print(i, " This is teh i for in if ")                     if imageUrlString.count == j {                         print("Done yet: yes", values[0], " ", values[3])                         Done = true                         break;                     }                     j+=1                 } else {                     print("No,,.")                 }             }         }