Topic-Sensitive PageRank
Reference: Taher H. Haveliwala, “Topic-Sensitive PageRank: A
context-sensitive ranking algorithm for websearch”, IEEE Trans. On
Knowledge and Data Engineering, vol. 15, No. 4, PP. 784-796.
The original PageRank:
• purely based on the hyperlinks of web pages.
• Contents are not considered.
• A vector of PageRank is used for all web pages.
Topic-Sensitive PageRank
• For each topic, a vector of PageRank is created.
• Each page has several PageRank values. One for each
topic.
Creating a Page Rank vector for each topic
• How to select topics?
– Using a small set of topics is important for low
computation cost and quick response time.
– Open Directory: http://www.dmoz.org
• 16 top level topics
Original PageRank
• Rank=MRank
– Rank is a vector, one element for each web
page.
– M is a nn matric
• If there is a link from page j to page I, then
Mi,j =1/Nj, where Nj is the number of out-links of
page j.
Another Version
• Let n be the total number of web pages.
P=[1/n] n1 be a vector.
• d is a n1 matrix.
– di=1 if page i has no out links. Otherwise, di=0.
• D=ppT, and E=p[1] 1n
• M’=(1-)(M+D) +E.
• Rank=M’Rank=(1-)(M+D) Rank +P.
Topic sensitive Page Rank
• Let Tj be the set of URLs in the ODP category
c j.
• P=vj, where
– Vj,i =1/|Tj| if page j points to page i. Otherwisr vj,i=0.
• The pageRank vector for topic cj is PR(, vj).
• Compute the pageRAnk for all pages related to
topic cj as if for the original PageRank by
considering Tj.
The Retrieval Score
• Let r j, d be the PageRank of document d given
by the rank PR( , vj).
• Sq,d=  j P(cj|q)•r j,d ,
• P(cj|q) is the score that topic is related to q.
Similarity Measures for Induced
Rankings Another Version
• Let 1 and 2 be two rankings of documents.
• OSim(1,2) indicates the degree of overlap
between the top k URLs of the two rankings.
• OSim (1,2) =|AB|/k.
• KSimn (1,2) =
|(u, v): 1 and 2 agree on order of (u, v)|/(|U|)(|U|-1)
Let  be the true ranking given by user. To
compare 1 and 2, we can use OSim (1,2)
or KSimn (1,2) .
Weighted PageRank
• Assign larger rank values to more important pages.
• Each outlink page gets its value proportional to it
popularity.
• W in (v, u) is the weight of link(v, u) calculated based on
the number of inlinks of page u and the number of
inlinks of all reference pages of page v.
w in (v, u) =Iu/pR(v) Ip,
Iu—number of inlinks of page u.
R(v)-the set of all pages that v points to.
Weighted PageRank
• w out (v, u) =Ou/pR(v) Op,
Ou—number of outlinks of page u.
Let B(u) be the set of pages that points to v.
PR(u)=(1-d) +d  vB(u) PR(v) w in (v, u) w out (v, u)
Reference: Wenpu Xing and Ali Ghorbani, Weighted
PageRank Algorithm, Proceedings of the 2nd Annual
Conference on Communication Networks and Services
Research (CNSR’04), 2004.
Choices of Search Engines
• Many search engines exist to compete for
users
– The results are not necessarily the same
– Different users prefer different search engines
– Search results may, in the future, be biased
towards paid advertisements.
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MetaSearch Engine
• Metasearch Engines are designed to
increase the coverage of web by forwarding
users’ queries to multiple search engines
– Users’ requests are sent to multiple search
engines such as AlltheWeb, Google, MSN.
• Then the results from the individual search
engine are combined into a single result set
to present to users.
Longest common subsequence
• Definition 1: Given a sequence X=x1x2...xm,
another sequence Z=z1z2...zk is a
subsequence of X if there exists a strictly
increasing sequence i1i2...ik of indices of X
such that for all j=1,2,...k, we have xij=zj.
• Example 1: If X=abcdefg, Z=abdg is a
subsequence of X.
X=abcdefg,
Z=ab d g
• Definition 2: Given two sequences X
and Y, a sequence Z is a common
subsequence of X and Y if Z is a
subsequence of both X and Y.
• Example 2: X=abcdefg and Y=aaadgfd.
Z=adf is a common subsequence of X
and Y.
X=abc defg
Y=aaaadgfd
Z=a d f
• Definition 3: A longest common
subsequence of X and Y is a common
subsequence of X and Y with the
longest length. (The length of a
sequence is the number of letters in the
seuqence.)
• Longest common subsequence may
not be unique.
• Example: abcd
acbd
Both acd and abd are LCS.
Longest common subsequence problem
• Input:
Two sequences X=x1x2...xm, and
Y=y1y2...yn.
• Output: a longest common subsequence of X and Y.
• A brute-force approach
Suppose that mn. Try all subsequence of X
(There are 2m subsequence of X), test if such a
subsequence is also a subsequence of Y, and
select the one with the longest length.
Charactering a longest common
subsequence
• Theorem (Optimal substructure of an LCS)
• Let X=x1x2...xm, and Y=y1y2...yn be two
sequences, and
• Z=z1z2...zk be any LCS of X and Y.
• 1. If xm=yn, then zk=xm=yn and Z[1..k-1] is an
LCS of X[1..m-1] and Y[1..n-1].
• 2. If xm yn, then zkxm implies that Z is an LCS
of X[1..m-1] and Y.
• 2. If xm yn, then zkyn implies that Z is an LCS
of X and Y[1..n-1].
The recursive equation
• Let c[i,j] be the length of an LCS of X[1...i] and
X[1...j].
• c[i,j] can be computed as follows:
0
if i=0 or j=0,
c[i,j]= c[i-1,j-1]+1
if i,j>0 and xi=yj,
max{c[i,j-1],c[i-1,j]} if i,j>0 and xiyj.
Computing the length of an LCS
• There are nm c[i,j]’s. So we can compute them
in a specific order.
The algorithm to compute an LCS
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
1. for i=1 to m do
2.
c[i,0]=0;
3. for j=0 to n do
4.
c[0,j]=0;
5. for i=1 to m do
6.
for j=1 to n do
7.
{
8.
if x[I] ==y[j] then
9.
c[i,j]=c[i-1,j-1]=1;
10
b[i,j]=1;
11.
else
if c[i-1,j]>=c[i,j-1] then
12.
c[i,j]=c[i-1,j]
13.
b[i,j]=2;
14.
else c[i,j]=c[i,j-1]
15.
b[i,j]=3;
14
}
Example 3: X=BDCABA and Y=ABCBDAB.
Constructing an LCS (back-tracking)
• We can find an LCS using b[i,j]’s.
• We start with b[n,m] and track back to some cell b[0,i] or
b[i,0].
• The algorithm to construct an LCS
1.
2.
3.
4.
i=m
j=n;
if i==0 or j==0 then exit;
if b[i,j]==1 then
{
i=i-1;
j=j-1;
print “xi”;
}
5. if b[i,j]==2
i=i-1
6. if b[i,j]==3
j=j-1
7. Goto Step 3.
• The time complexity: O(nm).
Shortest common supersequence
• Definition: Let X and Y be two sequences. A
sequence Z is a supersequence of X and Y if
both X and Y are subsequence of Z.
• Shortest common supersequence problem:
Input: Two sequences X and Y.
Output: a shortest common supersequence of X and Y.
• Example: X=abc and Y=abb. Both abbc and
abcb are the shortest common supersequences
for X and Y.
Recursive Equation:
• Let c[i,j] be the length of an LCS of
X[1...i] and X[1...j].
• c[i,j] can be computed as follows:
c[i,j]=
j
if i=0
i
if j=0,
c[i-1,j-1]+1
if i,j>0 and xi=yj,
min{c[i,j-1]+1,c[i-1,j]+1} if i,j>0 and xiyj.