# Ranges in Boolean Expressions

• Jul 12, 2006, 09:14
WoW_15
Ranges in Boolean Expressions
Can someone explain how the state machine works for a range in a boolean expression? I'm trying to figure out WHY the block below would give a value for 17 rather than nil (and also, why wouldn't 15 pass the i%3 == 0 test?)

Code:

```a = (11..20).collect {|i| (i%4 == 0)..(i%3 == 0) ? i : nil}  a  » [nil, 12, nil, nil, nil, 16, 17, 18, nil, 20]```

And what is happening internally with the example below?

Code:

```a = (11..20).collect {|i| (i%4 == 0)...(i%3 == 0) ? i : nil}  a  » [nil, 12, 13, 14, 15, 16, 17, 18, nil, 20]```
• Jul 13, 2006, 08:56
WoW_15
I think I figured it out myself... If anyone cares to hear my explanation, I'd be glad to share it.
• Jul 13, 2006, 09:44
Fenrir2
I do!
• Jul 13, 2006, 14:59
WoW_15
This is how I *think* it works:

Ruby maintains the true/false state of the boolean expression when using a range... it switches back and forth between the start and end value of the range with each state change. I think it's easiest to explain by stepping through the process using the example below:

Code:

```a = (11..20).collect {|i| (i%4 == 0)...(i%3 == 0) ? i : nil}  a  » [nil, 12, 13, 14, 15, 16, 17, 18, nil, 20]```
The initial state of the boolean expression is false.
We start by passing the value 11 into the code block:

11 is passed in, (11%4 == 0) is False, so the state does not change... it remains false, so nil is returned.
12 is passed in, (12%4 == 0) is True, so the state CHANGES to true.. the value of 'i' is returned

Since the state has now changed, ruby switches to the end value in our range when evaluating the boolean expression

13 is passed in, now we use the end expression of the range: (i%3 == 0). Since (13%3 == 0) is False, no change is made to the state. However, Ruby remembers that the state was ALREADY true, so it STILL treats it as true, and returns the value of 'i'

14 is passed in, (14%3 == 0) is false, so again, no state change. The boolean is still considered true, and the value of i is returned

15 is passed in, (15%3 == 0) is True, so the state changes (since it was already true, it now changes to false). Ruby still returns the value of i before completing the loop. However, on the next iteration, it switches back to the first expression in the range - (i%4 == 0)

16 is passed in, (16%4 == 0) is True also, so the state changes yet again (it was false, so now it's back to true!) the value of i is returned and again we switch to the end value of our range when evaluating the next iteration

and so it goes... flipping back and forth between the start and end value of the range with each state change

When you use the 2 dot range in a boolean expression, ruby will test the start and end value of the range during the SAME iteration IF the first expression of the range returns true... phew... so in effect it changes state twice in one iteration, while the 3 dot range I used above can only change state once with each iteration.

Does that make sense? And is that how it really works??
• Jul 14, 2006, 01:46
Fenrir2
That does make sense, but I don't know if that's how it really works. Thanks for the explanation!