[Tutor] Help needed

[C Smith]

Just glancing at your work, I see you have curly braces around what looks
like it should be a list. If you are concerned with the order of your
output, dictionaries do not have a concept of order.


On Sat, Apr 26, 2014 at 3:16 PM, Suhana Vidyarthi <suhanavidyarthi at gmail.com
> wrote:

> Hi Danny,
>
> Let me give you a high level brief of what I am doing:
> I am working on doing "disaster aware routing" considering the 24-node US
> network where I will be setting up connection between two any two nodes (I
> will select the source and destination nodes randomly). Also I have some
> links whose "probability of failure" is mentioned in the attached file.
> Other links, which are not mentioned in the file - we suppose their
> "probability of failure" is zero. So between the source-destination nodes,
> there will be multiple paths and I will select the one which has "least
> probability of failure".
>
> Now to setup the connection between two nodes, I have to select a path
> whose "probability of failure" is least. To do that first I will calculate
> the risk of each path from the attached file and then select the path with
> least risk value. Did you get this part? I know it can be a bit confusing.
>
> Now I break the problem into parts:
>
> 1. I have to topology of the 24-node map
> 2. I have the link values of each link - where risk values are the
> "probability of failure"
> 3. I calculate the total "probability of failure" of each path (a path may
> have multiple links): Suppose my source node is "a" and destination node is
> "b". I can setup a path between a to b via c or via d (a-c-b or a-d-c):
> Here I will check the risk values of a-c and c-b; also risk values of a-d
> and d-c. If the total risk valure of a-c-b is lower that risk value of
> a-d-c, then I select the path a-c-d to setup the connection. (again risk
> value = probability of failure)
>
> Now, I will first calculate the "total probability of failure" of each
> link (using the file.txt) and since some links are repeated their values
> will be added. The probabilities get added if a link is mentioned twice
> or thrice. For example:  link 3—5 is repeated 3 times: in line one, it
> has a probability of failure as 0.03, in line two it is 0.11 and in line
> three it is 0.04. So the probability of failure for link 3—5 is
> 0.03+0.11+0.04 = 0.18
>
> The length of each array will be same. You see the code I wrote: here is
> the output for it:
>
> Links ->
>
> [('10', '13'), ('14', '18'), ('7', '8'), ('15', '20'), ('5', '8'), ('5',
> '4'), ('11', '9'), ('21', '22'), ('12', '13'), ('21', '20'), ('17', '13'),
> ('20', '21'), ('21', '16'), ('14', '10'), ('11', '12'), ('11', '19'),
> ('14', '13'), ('3', '5'), ('11', '6'), ('19', '20')]
>
>
> Probability ->
>
> [0.04, 0.06, 0.04, 0.24, 0.08, 0.15, 0.08, 0.27, 0.04, 0.29, 0.08, 0.27,
> 0.27, 0.04, 0.08, 0.08, 0.08, 0.18000000000000002, 0.08, 0.24]
>
>
> It means that link [10,13] has a "probability of failure" as [0.04] and
> since the link [3-5] is repeated thrice with probability of 0.03, 0.11 and
> 0.04, its "probability of failure" is [0.18] (third last element in the
> Probability array). For some reason instead of 0.18 it is showing
> 0.180000000002, which I cannot figure to why.
>
>
> Please see the attached code. If you see the file.txt and my output: the
> output is not displayed in sequence and that is what I need help with. I
> want this to display :
>
>
> Links = { [3,5] [5,4] [5,8] [7,8] [14,10] [14,13] [17,13] [14,18] [10,13]
> [14,13] [17,13] [12,13] [11,6] [11,9] [11,12] [11,19] [19,20] [15,20]
> [21,20] [20,21] [21,16] [21,22] }
>
>
> Prob = {[0.18] [0.15] [0.08] [0.04] [0.04] [0.04] [0.08] [0.04] [0.08]
> [0.08] [0.08] [0.08] [0.24] [0.24] [0.34] [0.27] [0.27]}
>
>
> If you can figure why the output is not generated in same sequence as in
> the file.txt for me, it will be very helpful.
>
>
> let me know if I explained correctly, and if you have any questions or
> doubts?
>
>
> On Sat, Apr 26, 2014 at 11:41 AM, Danny Yoo <dyoo at hashcollision.org>wrote:
>
>> >>> I want to create two arrays using the above file (Links array and Prob
>> >>> array) that should give following output:
>> >>>
>> >>> *Links *= { [3,5] [5,4] [5,8] [7,8] [14,10] [14,13] [17,13] [14,18]
>> >>> [10,13] [14,13] [17,13] [12,13] [11,6] [11,9][11,12] [11,19] [19,20]
>> >>> [15,20] [21,20] [20,21] [21,16] [21,22] }
>> >>>
>> >>> *Prob *= {[0.28] [0.15] [0.08] [0.04] [0.04] [0.04] [0.08] [0.04]
>> [0.08]
>> >>> [0.08] [0.08] [0.08] [0.24] [0.24] [0.34] [0.27] [0.27]}
>> >>
>> >>
>> >> I don't understand how you develop this? The first list has 22 items
>> the
>> >> second 17. I would have expected them to be the same?
>> >
>> >
>> > In the "Prob" array the elements are less because if you read the note
>> > below: I said the links that are repeating for example [3,5] their
>> > probabilities  get added and stored as a single value in the "Prob"
>> array.
>>
>>
>> But what will you plan to do with these values afterwards?  I think
>> Alan's point here is that if there's no direct relationship between
>> the elements in Links and the elements in Probs, those values aren't
>> going to be very useful to solve the rest of the problem.
>>
>>
>> One way to look at this problem is to simplify or normalize the input;
>> the original structure in the file is slightly weird to process, since
>> a single line of the input represents several link/failure pairs.
>>
>> One concrete example is:
>>
>>     4,10,13,14,13,17,13,12,13,0.04
>>
>> where all these numbers are uninterpreted.
>>
>> You can imagine something that takes the line above, and breaks it
>> down into a series of LinkFailure items.
>>
>> ######
>> class LinkFailure(object):
>>     """Represents a link and the probability of failure."""
>>     def __init__(self, start, end, failure):
>>         self.start = start
>>         self.end = end
>>         self.failure = failure
>> ######
>>
>> which represent a link and failure structure.  If we have a structure
>> like this, then it explicitly represents a relationship between a link
>> and its failure, and the string line:
>>
>>     4,10,13,14,13,17,13,12,13,0.04
>>
>> can be distilled and represented as a collection of LinkFailure instances:
>>
>>     [LinkFailure(10, 13, 0.04), LinkFailure(14, 13, 0.04),
>> LinkFailure(17, 13, 0.04), LinkFailure(12, 13, 0.04)]
>>
>> Then the relationship is explicit.
>>
>>
>> >> Also why do you want these two lists? What do you plan on doing with
>> them?
>> >> I would have thought a mapping of link to probability would be much
>> more
>> >> useful? (mapping => dictionary)
>> >
>> >
>> > I want these two lists because using the links and probs array, I will
>> check
>> > which link has the lowest probability. Here lowest probability means the
>> > risk of failure for that link. So based on which link has least
>> probability
>> > of failure, I will use it to setup a connection (I have a source and
>> > destination and the links mentioned above are the paths between them) I
>> want
>> > to select the path which has least failure probability.
>> >
>> > Did it make sense?
>>
>>
>> Unfortunately, I'm still confused.  If you just have the Links and the
>> Probs lists of unequal length, unless there's some additional
>> information that you're represented, then I don't see the necessary
>> connection between the two lists that lets you go any further in the
>> problem.  There's no one-to-one-ness: given a link in Links, which
>> Probs do you want to look at?  If you don't represent that linkage in
>> some way, I don't understand yet where you go next.
>>
>>
>> >>> So the first element in Links array is [3,5] and its probability of
>> >>> failure is the first element in Prob array i.e. 0.28
>>
>> But if the two lists are different lengths, what probability of
>> failure is associates with the last element in the Prob array?
>>
>>
>> The representation of data is important: if you choose an awkward
>> representation, it makes solving this problem more difficult than it
>> needs be.  That is, if you're already compressing multiple elements in
>> Prob that correspond to the same link, you also need some way to
>> figure out what link that a compressed probability refer to.
>> Otherwise, you don't have enough information to solve the problem
>> anymore.
>>
>
>
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