assigning rows and columns - swift

Question

I am trying to plot data to a grid that is made up of hexagons. Because of this, the row lengths alternate between two different values.
ie: a grid would look like this with row lengths 4 and 5
0 0 0 0 0
0 0 0 0
0 0 0 0 0
0 0 0 0
Does any one know a clever way to approach this? I thought about using flags to tell you which row you are in, but feel like there can be a more elegant solution

Answer 1

If all you're trying to do is figure out the row length, it's quite simple: Just use modulus of 2.
In your example, assume the top row has an index of 0, and the index increases as you go down.
rowLength = rowIndex % 2 == 0 ? 5 : 4;
0 % 2 == 0 --> 5
1 % 2 == 1 --> 4
2 % 2 == 0 --> 5
3 % 2 == 1 --> 4
Alternatively, you can have the rows always have a length of five, and on every other row store a value (such as null) that indicates the value should be skipped:
0 0 0 0 0
0 0 0 0 (null)
0 0 0 0 0
0 0 0 0 (null)

Answer 2

One solution is to write a "generic" pattern generator:
Takes a pattern, in your case "0 ". This specific pattern uses 2 characters.
Given number of rows, say nR, and number of characters per row, say nC (in your case, nC=5*2-1=9): repeat the pattern into one string that contains this pattern, however, without proper line breaks.
Using number of characters per row nC: insert a line break exactly after nC characters continually throughout your string.
This can be done in a somewhat "swifty functional" manner:
func plotGrid(numRows: Int, numCharsPerRow: Int, myPattern: String) {
let totalNumChars = numRows*numCharsPerRow
let numCharsInPattern = myPattern.characters.count
let gridTemplate = [String](count:totalNumChars/numCharsInPattern + totalNumChars%numCharsInPattern, repeatedValue: myPattern).reduce("", combine: +)
let grid = 0.stride(to: totalNumChars, by: numCharsPerRow)
.map {
i -> String in
let a = gridTemplate.startIndex.advancedBy(i)
let b = a.advancedBy(numCharsPerRow, limit: gridTemplate.endIndex)
return gridTemplate[a..<b] + "\n"
}.reduce("", combine: +)
print(grid)
}
With the results (for your example and another 4-character pattern example, respectively)
plotGrid(5, numCharsPerRow: 9, myPattern: "0 ")
//0 0 0 0 0
// 0 0 0 0
//0 0 0 0 0
// 0 0 0 0
//0 0 0 0 0
plotGrid(4, numCharsPerRow: 10, myPattern: "X O ")
//X O X O X
//O X O X O
//X O X O X
//O X O X O
Note, however, that in the let grid assignment, you create an array that is immediately reduced, so there will be an unnecessary overhead here, especially for large grids. The imperative approach below should be faster, but the above, perhaps, neater.
One possible (among many) classic imperative approach:
let numRows = 5
let numCharsPerRow = 9
let myPattern = "0 "
let totalNumChars = numRows*numCharsPerRow
var grid = "0"
var addZero = false
for i in 2...totalNumChars {
grid += addZero ? "0" : " "
addZero = !addZero
if i%numCharsPerRow == 0 {
grid += "\n"
}
}
print(grid)
//0 0 0 0 0
// 0 0 0 0
//0 0 0 0 0
// 0 0 0 0
//0 0 0 0 0