Solve iteration algorithm in ruby

Question

So this is hurting my head, I am not very good with programming obviously. I have,

LetterArray = [a,b,c,d,e,f,g]
NumArray = [1,2,3,4,5,6,7,8,9,10]
ListOfLetters = []

and I want to take an element from NumArray and, starting on LetterArray[0], go up var x amount of times in LetterArray, and add that element (say var y to the array. Then starting on y go up the next number in NumArray, and so on. Then print the ListOfLetters to console.

My goal is for the output to be like this: [a, c, f, c, a, f, e, e, f, a].

I am drawing a blank on how to go about this in code.

It's not really clear what you want, could you explain it better? — nikhil
– nikhil, Commented Jun 6, 2012 at 15:20
Looks like a job/ interview quiz. Interesting question though. Please post all details. — Anil
– Anil, Commented Jun 6, 2012 at 15:41
Nah i just made it up while i was playing guitar...decided i wanted to use the fib sequence to generate a sequence of notes in E mixolydian... — Dustin
– Dustin, Commented Jun 6, 2012 at 16:02
For one, that isn't an array of letters, it's an array of references that are the variables a through g. You probably want strings and thus need to quote them (e.g. "a"). — Andrew Marshall
– Andrew Marshall, Commented Jun 6, 2012 at 16:35

Victor Moroz · Accepted Answer · 2012-06-06 16:23:46Z

2

Something like this (if I get your requirements right of course)?

letter_array = %w[a b c d e f g]
number_array = [1,2,3,4,5,6,7,8,9,10]
list_of_letters = []

number_array.inject(0) do |offset, delta| 
  list_of_letters << letter_array[offset] 
  (offset + delta) % letter_array.size
end

p list_of_letters #=> ["a", "b", "d", "g", "d", "b", "a", "a", "b", "d"]

answered Jun 6, 2012 at 16:23

Victor Moroz

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Comments

Michael Kohl · Accepted Answer · 2012-06-06 15:54:35Z

0

Either I don't understand your problem description, or the example output you showed is wrong from a certain point onwards. Anyway, maybe this gets you started:

letter_array = [*?a..?g]
number_array = *1..10
list_of_letters = []

number_array.inject(0) do |s, n| 
  i = s + n
  list_of_letters << letter_array[i % letter_array.size - 1] 
  i 
end

This produces the output ["a", "c", "f", "c", "a", "g", "g", "a", "c", "f"].

Alternatively you can also first create the indices and then use them (this doesn't require a pre-initialized list_of_letters):

indices = number_array.inject([]) { |a, n| a << (a.last || 0) + n ; a})
list_of_letters = indices.map { |i| letter_array[i%letter_array.size-1] }

edited Jun 6, 2012 at 15:54

answered Jun 6, 2012 at 15:46

Michael Kohl

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2 Comments

Dustin Over a year ago

I think my example output is wrong. But letter_array = ['E', 'F#', 'G#','A', 'B', 'C#', 'D'] number_array = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597] list_of_letters = [] number_array.inject(0) do |s, n| i = s + n list_of_letters << letter_array[i % letter_array.size - 1] i end p list_of_letters is what i was going for...though, the output gives me ["E", "F#", "A", "D", "B", "C#", "B", "B", "A", "G#", "E", "B", "D", "C#", "D", "D", "E"]

Dustin Over a year ago

but when I hand count it, I got E, F#, G#, B, E, B, B....etc. I want to start on E, go up 1 to F#, then go up 1 to G#, then up 2, to B....i hope this makes this more clear?

steenslag · Accepted Answer · 2012-06-06 16:00:41Z

0

ar = ('a'..'g').to_a.cycle #keeps on cycling 
res = []
p 10.times.map do |n|
  n.times{ar.next} #cycle one time too short (the first time n is 0)
  res << ar.next #cycle once more and store
end
p res #=>["a", "c", "f", "c", "a", "g", "g", "a", "c", "f"]

answered Jun 6, 2012 at 16:00

steenslag

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