See attachment

x.12

S1: <P1,P2,P5,P2,P7,P3>
S2: <P1,P4,P5,P1,P6,P7>
S3: <P1,P6,P1,P4>
S4: <P5,P4,P1,P6,P7>
S5: <P5,P3>
S6: <P1,P2,P7,P3>
S7: <P2,P7>
S8: <P1,P2,P4,P1,P2,P6,P7,P3>


REAL:

S1: <P1,P2,P5>
S2: <P1,P2,P7,P3>
S3: <P1,P4,P5,P1,P6,P7>
S4: <P1,P6,P1,P4,P5,P4,P1,P6,P7>
S5: <P5,P3,P1,P2,P7,P3>
S6: <P2,P7>
S7: <P1,P2,P4,P1,P2,P6,P7,P3>

Mcr: The counts of real seesions contained in constructed sessions / all the real sessions.
Mcrs : The counts of real seesions contained in constructed sessions and share the same start page / all the real sessions.  
Mcre : The counts of real seesions contained in constructed sessions and share the same end page / all the real sessions. 
Mcrse : The counts of real seesions contained in constructed sessions and share the same start and end pages / all the real sessions.  
Mo : 1) Find Maximum of (Count of certain real session intersect one reconstructed session /count of the certain real session) for all possible reconstructed sessions.
     2) Find the avg of result of  1) for all the real sessions.  
Ms : 1) Find Maximum of (Count of certain real session intersect one reconstructed session /count of the certain real session union one reconstructed session ) for all possible reconstructed sessions.
     2) Find the avg of result of  1) for all the real sessions. 



Mcr = (1/*from RealS1*/ + 1/*from RealS2*/ + 1/*from RealS3*/ + 3/*from RealS6*/ + 1/*from RealS7*/) / 7 = 1
Mcrs = (1/*from RealS1*/ + 1/*from RealS2*/ + 1/*from RealS3*/ + 1/*from RealS6*/ + 1/*from RealS7*/ / 7 = 5/7
Mcre = (1/*from RealS1*/ + 1/*from RealS2*/ + 1/*from RealS3*/ + 1/*from RealS6*/ + 1/*from RealS7*/ / 7 = 5/7
Mcrse = (1/*from RealS1*/ + 1/*from RealS2*/ + 1/*from RealS3*/ + 1/*from RealS6*/ + 1/*from RealS7*/ / 7 = 5/7

Mo = (1/*from RealS1*/ + 1/*from RealS2*/ + 1/*from RealS3*/ + 4/9/*from RealS4*/ + 2/6/*from RealS5*/ + 1/*from RealS6*/ + 1/*from RealS7*/) / 7 = 0.82
Ms = (3/5/*from RealS1*/ + 1/*from RealS2*/ + 1/*from RealS3*/ + 4/9/*from RealS4*/ + 2/6/*from RealS5*/ + 1/*from RealS6*/ + 1/*from RealS7*/) / 7 = 0.76

batcat wrote:

See attachment

I thought that the sentence for the M_crse is kind of misleading, you said:

Of these 5 sessions only S2, S3, S6, and S7 share the first and last element with the constructed session

I thought it's suppose to be:

Of these 5 sessions only S2, S3, S6, and S7 are identical with the constructed session

and I still don't understand the Largest Common Intersection.. For example:

Real Session, S = <1, 2, 3>

Constructed Session, C = <3, 2, 6, 1>

what is the LCI? is it 3 or 1?

if it's 1, then how about

C = <2, 8>, I thought that it's suppose to be 0..

HelloWorld wrote:

batcat wrote:

See attachment

I thought that the sentence for the M_crse is kind of misleading, you said:

Of these 5 sessions only S2, S3, S6, and S7 share the first and last element with the constructed session

I thought it's suppose to be:

Of these 5 sessions only S2, S3, S6, and S7 are identical with the constructed session

I was referring to a defintion that states

A real session and constructed session are identical if and only if the following 2 conditions hold:

1) The real session is completely reconstructed by the constructed session

2) The 2 session have the same start and end element

But that is actually making things more complicated than is needed. It is easier and better just to check which ones look indentical

and I still don't understand the Largest Common Intersection.. For example:

Real Session, S = <1, 2, 3>

Constructed Session, C = <3, 2, 6, 1>

what is the LCI? is it 3 or 1?

if it's 1, then how about

C = <2, 8>, I thought that it's suppose to be 0..

Remember, with containment the entire sequence(maintaining order and adjacency) of the real session had to be in the constructed session.

Example:

Real session <P1, P2, P5> is contained in constructed session <P1, P2, P5, P2, P7, P3>

However, with common fragments and CLI we can also have small or broken up pieces of the real session sequence. (The indiviual fragments must still maintain order and adjacency)

Example:

Real session <P1, P2, P5> is not contained in constructed session <P1, P2, P8, P5> (interrupted by P8)

However, there are a couple common fragments here namely <P1, P2> and <P5>

<P1, P2> is the longest common fragement so it would be LCI

Another example:

Real session <P1, P2, P5> is not contained in constructed session <P1, P2>

but they do share LCI of <P1, P2>

In your example

Real Session, S = <1, 2, 3>

Constructed Session, C = <3, 2, 6, 1>

LCI could be fragment 1 or 2 or 3 because they are all have length 1.

Yes its confusing to me too because I am used to taking unions, intersections, and cardinality with sets not session reconstructions with

duplicates. For the Ms I decided to take union to just mean all the elements including duplicates of both sessions. But your way of taking

union makes more sense and accomplishes the goal of Ms which is take c as well as r into account.

xieguahu wrote:

In the paper, they said "The “degree of similarity” between a real session and a constructed session is the number
of elements in their largest continuous intersection divided by the total number of elements in the two sessions"

So you need to consider all the element in the union and not just pick the duplicates of one set.

HelloWorld wrote:

Please check my answer..

As far as the Union thing.. I thought that it makes more sense to pick the duplicates of one set, but not both.. So for example:

S1 = <P1, P1, P2, P6>

S1' = <P1, P7>

OR

S1 = <P1, P7>

 

S1' = <P1, P1, P2, P6>

 

then S1 Union S1' = <P1, P1, P2, P6, P7>

Notice that there are 2 P1's coming from S1

x is a member of the union if it is an element present in set A or in set B, or both.

 

GX is right, I missed that part..

So the only different of my answer and batcat is

S6: deg(r, c) = 2/2, 1/2, 0/2, 1/2, 0/2, 2/2, 2/2, 1/2 max = 1

S7: deg(r, c) = 2/8, 2/8, 1/8, 2/8, 1/8, 2/8, 1/8, 8/8 max = 1

for the Degree of Overlap, though, it won't change the final answer..

HelloWorld wrote:

Hmmm.. I also feel that there's something strange about this question, for example about the Union between C and R

Suppose,

S1 = <P1, P2, P1>

and 

S1' = <P1, P6>

What is S1 Union S1'? is it <P1, P2, P1, P6>? or <P1, P2, P6>? I just know that Union should ignore duplicates, right?

As per my understanding of # Union of Sessions... If you have Real Session S = <P1, P2, P1> and Constructed Session C = <P1, P6> then | S U C | = 5. If I am wrong then let me know, please...

jadiagaurang wrote:

HelloWorld wrote:

Hmmm.. I also feel that there's something strange about this question, for example about the Union between C and R

Suppose,

S1 = <P1, P2, P1>

and 

S1' = <P1, P6>

What is S1 Union S1'? is it <P1, P2, P1, P6>? or <P1, P2, P6>? I just know that Union should ignore duplicates, right?

As per my understanding of # Union of Sessions... If you have Real Session S = <P1, P2, P1> and Constructed Session C = <P1, P6> then | S U C | = 5. If I am wrong then let me know, please...

Yes you are correct and that is how I did it in my solution at the top of the page

HelloWorld wrote:

xieguahu wrote:

In the paper, they said "The “degree of similarity” between a real session and a constructed session is the number
of elements in their largest continuous intersection divided by the total number of elements in the two sessions"

So you need to consider all the element in the union and not just pick the duplicates of one set.

HelloWorld wrote:

Please check my answer..

As far as the Union thing.. I thought that it makes more sense to pick the duplicates of one set, but not both.. So for example:

S1 = <P1, P1, P2, P6>

S1' = <P1, P7>

OR

S1 = <P1, P7>

 

S1' = <P1, P1, P2, P6>

 

then S1 Union S1' = <P1, P1, P2, P6, P7>

Notice that there are 2 P1's coming from S1

x is a member of the union if it is an element present in set A or in set B, or both.

 

GX is right, I missed that part..

So the only different of my answer and batcat is

S6: deg(r, c) = 2/2, 1/2, 0/2, 1/2, 0/2, 2/2, 2/2, 1/2 max = 1

S7: deg(r, c) = 2/8, 2/8, 1/8, 2/8, 1/8, 2/8, 1/8, 8/8 max = 1

for the Degree of Overlap, though, it won't change the final answer..

Yes in my solution at the top of page 3,

For S6: I forgot to type the 1/2 at the end

For S7: I miscounted the number of elements in the real session which should be 8, not 7

But it doesn't change any of the max values and therefore doesn't change the final answer

Regarding M_s, in DEFINITION 9 of the paper, the text says "... divided by the total number of elements in the two sessions" (as quote by Guanhua earlier), but the equation shows |r U c|, which means the size of the union, not the sum of the sizes. I believe the equation is correct, i.e. the text is wrong, because otherwise the best similarity we can get is 0.5 (when two sessions are identical), but that contradicts Figure 1 and 2 which show M_s that are way above 0.6.

Due to this ambiguity in the paper, when I grade the exercises/exams, I'll take both interpretation as correct.