Loop queue algorithm issue - arrays

Question

I'm trying to do the following, and can't figure out how exactly do It without crashing or infinite looping:
I have to create a queue in which I have to distribute different tasks a different number of times each, alternatively with this kind of info:
Task X: [NextOne,LastOne]
Task 1: [30,32]
Task 2: [76,81]
Task 3: [2,2]
Task 4: [5,8]
Meaning that "Task X" will be made "LastOne - NextOne" times, and if both are equal, it won't be enqueued, and they enter the queue in X order.
With this example, the queue should look like:
FIRST
Task1[30]
Task2[76]
Task4[5]
Task1[31]
Task2[77]
Task4[6]
Task1[32]
Task2[78]
Task4[7]
Task2[79]
Task4[8]
Task2[80]
Task2[81]
LAST
It's not a language issue, it's more of an algorithm issue I have here. Using PHP I've made the following:
$tasks = array(
'Task1' => array(30,32),
'Task2' => array(76,81),
'Task3' => array(2,2),
'Task4' => array(5,8)
);
$aux = array();
$i=0;
foreach($tasks as $s=>$n) {
$aux[$i]['task'] = $s;
$aux[$i]['times'] = $n[1]-$n[0];
$aux[$i]['first'] = $n[0];
$i++;
}
But as you imagine this actually does nothing, just change the shape of the information. I'm really stuck here I don't know why, this actually shouldn't be hard to figure out. I'd appreciate any help.

Answer 1

In python (I may be misinterpreting your "it's not a language issue" comment - forgive me):
tasks = [
("Task1", 30, 32),
("Task2", 76, 81),
("Task3", 2, 2),
("Task4", 5, 8) ]
while not tasks == []:
# Pop first task off the current list
(n, s, e) = tasks[0]
tasks = tasks[1:]
print n, s
if s != e:
tasks.append( (n, s+1, e) )
Sorry it's not in php - it's not my forte, but perhaps this'll help? Output:
Task1 30
Task2 76
Task3 2
Task4 5
Task1 31
Task2 77
Task4 6
Task1 32
Task2 78
Task4 7
Task2 79
Task4 8
Task2 80
Task2 81

Answer 2

I guess you can use $s as a key for the hash(or array equivalent) and just increase the value by 1 whenever it encounters the element with the same key. The default value would be 0 in this case.
For example,
Task1 => array(30,32)
will be come in order like
Task1[30] (default value to 0)
...
...
Task1[31] (add 1 which becomes 1)
...
Task1[32] (add 1 which becomes 2)
This means that Task1 has appeared 3 times overall, and the final times value should be 2.
I think you can use array_key_exists helper function to check if certain task had appeared previously.

Answer 3

In C#
The result is as you wish.
I added a number as a flag: 1 = Not to be enqueued, 2 = Last record of the task.
Not efficient but works!
private static void Main()
{
var tasks = new Dictionary<string, int[]>
{
{"Task1", new[] {30, 32, 0}},
{"Task2", new[] {76, 81, 0}},
{"Task3", new[] {2, 2, 0}},
{"Task4", new[] {5, 8, 0}}
};
int loopCounter = 0;
Console.WriteLine("FIRST");
while (loopCounter < tasks.Count)
{
foreach (var task in tasks)
{
if (task.Value[0] == task.Value[1])
{
if (task.Value[2] == 2)
{
loopCounter++;
Console.WriteLine(task.Key + "[" + task.Value[0] + "]");
task.Value[2] = 1;
}
else if (task.Value[2] == 0)
{
loopCounter++;
task.Value[2] = 1;
}
}
else
{
Console.WriteLine(task.Key + "[" + task.Value[0] + "]");
task.Value[0]++;
if (task.Value[0] == task.Value[1])
task.Value[2] = 2;
}
}
}
Console.WriteLine("LAST");
Console.ReadLine();
}
Output:
FIRST
Task1[30]
Task2[76]
Task4[5]
Task1[31]
Task2[77]
Task4[6]
Task1[32]
Task2[78]
Task4[7]
Task2[79]
Task4[8]
Task2[80]
Task2[81]
LAST
Hope this helps.