1. pTrees predicate Tree technologies
provide fast, accurate horizontal processing of
compressed, data-mining-ready, vertical data structures.
Applications:
PINE Podium Incremental Neighborhood Evaluator
uses pTrees for Closed k Nearest Neighbor Classification.
FAUST Fast Accurate Unsupervised, Supervised Treemining
uses pTtrees for classification and clustering of spatial data. 13 12 1
document 2 3 4 5
course
Text
person
Enroll
Buy
MYRRH ManY-Relationship-Rule Harvester
uses pTrees for association rule mining of multiple relationships.
PGP-D Pretty Good Protection of Data
protects vertical pTree data.
5,54 | 7,539 | 87,3 | 209,126 | 25,896 | 888,23 | ...
key=array(offset,pad)
ConCur Concurrency Control
uses pTrees for ROCC and ROLL concurrency control.
DOVE DOmain VEctors
Uses pTrees for database query processing.

2. MYRRH pTree-based ManY-Relationship-Rule Harvester
RoloDex Model:
2 Entities many relationships
MYRRH pTree-based ManY-Relationship-Rule Harvester
uses pTrees for ARM of multiple relationships.
 Customer 1 2 3 4
Item 1
customer rates movie as 5 card
cust item card 1 5
people 2 3 4
items
terms
DataCube Model for 3 entities, items, people and terms. 1 2 3 4
Author 5 6 7
People 
movie 2 3 1 5 4
customer rates movie card 2 3 4 5 PI 2 3 4 5 PI 4 3 2 1
Course
Enrollments 1
Doc
termdoc card
authordoc card 1 3 2
Doc 1 2 3 4
Gene
genegene card (ppi)
docdoc
People 
term  7 6 5 4 3
Gene 1 2 3 4 G 5 6 7 6 5 4 3 2 t 1
termterm card (share stem?) 1 3
Exp
expPI card
expgene card
genegene card (ppi)
Items: i1 i2 i3 i4 i5
|0 001|0 |0 11|
|1 001|0 |1 01|
|2 010|1 |0 10|
People: p1 p p3 p4
|0 100|A|M|
|1 001|T|M|
|2 010|S|F|
|3 011|B|F|
|4 100|C|M|
Terms: t1 t2 t3 t4 t5 t6
|1 010|1 101|2 11|
|2 001|0 000|3 11|
|3 011|1 001|3 11|
|4 011|3 001|0 00|
Relationship: p1 i1 t1
|0 0| 1
|0 1| 1
|1 0| 1
|2 0| 2
|3 0| 2
|4 1| 2
|5 1|_2
Relational Model:

3. The "hop" approach to MYRRH2 3 4 5 F
A hop is a relationship, R, hopping from one entity, E, to another entity, F. Standard ARM finds strong (frequent, confident) 1-hop, 1-entity (A,CE) rules, AC, s.t. 4 1 3 1 2 1 1 1
ct(&eACRe)  mnsp
ct(&eARe &eCRe) / ct(&eARe)  mncf E
R(E,F)
The 1-hop, transitive, F-focused rule, A(E)C(F), is strong if: ( focused on refers to where the counts are taken)
ct(&eARe)  mnsp (frequent)
ct(&eARe &RC) / ct(&eARe)  mncf (confident)
1-hop, transitive, F-focused APRIORI can mine strong rules efficiently by
1. (antecedent downward closure)
If A is frequent, all of its subsets are frequent. Or, if A is infrequent, then so are all of its supersets.
Since frequency involves only A, we can mine for all qualifying antecedents efficiently using downward closure.
2. (consequent upward closure) If AC is non-confident, then so is AD for all subsets, D, of C.
So  frequent antecedent, A, use upward closure to mine for all of its' confident consequents.
The theorem we demonstrate throughout this section is: For transitive (a+c)-hop Apriori strong rule mining with a focus entity which is a hops from the antecedent and c hops from the consequent, if a/c is odd/even then one can use downward/upward closure on that step in the mining of strong (frequent and confident) rules.
In this case A is 1-hop from F (odd, use downward closure).
C is 0-hops from F (even, use upward closure).
We will be checking more examples to see if the Odddownward Evenupward theorem seems to hold.
1-hop, transitive, E-focused rule, AC is strong if:
|A|=ct(RA)  mnsp
ct(RA&fCRf) / ct(RA)  mncf
1. (antecedent upward closure) If A is infrequent, then so are all of its subsets.
2. (consequent downward closure) If AC is non-confident, then so is AD for all supersets, D, of C.
In this case A is 0-hops from E (even, use upward closure).
C is 1-hop from E (odd, use downward closure).

4. ct(&eARe &gCSg) / ct(&eARe)  mncf
2-hop transitive F-focused (focus on middle entity, F) C G
S(F,G) 1 4 1 3
AC strong if:
ct(&eARe)  mnsp
ct(&eARe &gCSg) / ct(&eARe)  mncf 1 2 1 1
1. (antecedent downward closure) If A is infrequent, then so are all of its supersets. 2 3 4 5 F
2. (consequent downward closure) If AC is non-confident, so is AD for all supersets, D. 4 1
3. Apriori for 2-hops: Find all freq antecedents, A, using downward closure. For each:
find C1G, the set of g's s.t. A{g} is confident.
Find C2G, the set of C1G pairs that are confident consequents for antecedent, A.
Find C3G, the set of triples (from C2G) s.t. all subpairs are in C2G (ala Apriori), etc. 3 1 2 1 1 1
A  E
R(E,F)
1,1 odd so down, down correct.
ct(&flist&eAReSf)mnsp
 mncf
ct(&flist&eAReSf & PC)
/ &flist&eAReSf
2-hop trans G-foc
ct(&fl&eAReSf)mnsp
 mncf
ct(&f&eAReSf & PC)
/ &f&eAReSf
1. (antecedent upward closure) If A is infrequent, then so for are all subsets.
2,0 even so up,up is correct.
2. (consequent upward closure) If AC is non-confident, so is AD for all subsets, D.
From here onward we will not include the "list". ANDing over a pTree means ANDing over its' list form.
2-hop trans E-foc
ct(PA)mnsp
 mncf
ct(PA&f&gCSgRf )
/ ct(PA)
1. (antecedent upward closure) If A is infrequent, then so for are all subsets.
2. (consequent upward closure) If AC is non-confident, so is AD for all subsets, D.
0,2 even so up,up is correct.
Does standard ARM give the same info after collapsing R and S into T via, 1 2 3 4 G
T(E,G) E
A  C
T(e,g) = 1 iff (&eARe)g = 1
NO! Standard ARM has A and C both subsets of E (or G)
If G=E and S(F,E)=transposeR(E,F) then it is standard ARM?
What if G=E=F and T is collapsed? Think about the facebook case (where G=E=people, R and S are directional friendships...?

5. ct(&eARe ct(&eARe &g&hCThSg) / ct(&eARe ct(&f&eAReSf) &hCTh)
3-hop
Collapse T: TC≡ {gG|T(g,h) hC} That's just 2-hop case w TCG replacing C. ( can be replaced by  or any other quantifier. The choice of quantifier should match that intended for C.).
Collapse T and S: STC≡{fF |S(f,g) gTC} Then it's 1-hop w STC replacing C.
 mnsup
ct(&eARe
S(F,G)
R(E,F) 1 2 3 4 E F 5 G A C
T(G,H) H
Focus on F
 mncnf
ct(&eARe
&g&hCThSg)
/ ct(&eARe
antecedent downward closure: A infreq. implies supersets infreq. A 1-hop from F (down
consequent upward closure: AC noncnf implies AD noncnf. DC. C 2-hops (up
 mnsp
ct(&f&eAReSf)
Focus on G
 mncnf
&hCTh)
ct(&f&eAReSf
/ ct(&f&eAReSf)
antecedent upward closure: A infreq. implies all subsets infreq. A 2-hop from G (up)
consequent downward closure: AC noncnf impl AD noncnf. DC. C 1-hops (down)
ct( &g=1,3,4 Sg ) /ct(1001)
ct( &1001&1000&1100) / 2
ct( ) / = 1/2
Focus on F Are they different? Yes, because the confidences can be different numbers Focus on G.
ct(&eARe
&glist&hCThSg )
/ct(&eARe
&hCTh)
ct(&flist&eAReSf
/ ct(&flist&eAReSf)
ct(&f=2,5Sf
&1101 )
/ ct(&f=2,5Sf
ct(1101 & &
/ ct(1101 & )
ct(0001 )
/ ct(0001) = 1/1 =1
ct(PA & Rf)
f&g&hCThSg
/ ct(PA)  mncnf
 mnsup
ct(PA)
Focus on E
antecedent upward closure: A infreq. implies subsets infreq. A 0-hops from E (up)
consequent downward closure: AC noncnf implies AD noncnf. DC. C 3-hops (down)
Focus on H
antecedent downward closure: A infreq. implies all subsets infreq. A 3-hops from G (down)
consequent upward closure: AC noncnf impl AD noncnf. DC. C 0-hops (up)
ct(& Tg & PC)
g&f&eAReSf
mncnf
/ct(& Tg)
ct(& Tg)
mnsp

6. * ct(&iCUi) ) ct(&f&eAReSf) ct( &f&eAReSf &h&iCUiTh )
4-hop
S(F,G)
R(E,F) 1 2 3 4 E F 5 G A C
T(G,H) H
U(H,I) I
Focus on G? Replace C by UC; A by RA as above (not different from 2 hop?)
Focus on H (RA for A, use 3-hop) or focus on F (UC for C, use 3-hop).
Another focus on G (the main way)
 mnsup
ct(&f&eAReSf)
 mncnf
ct( &f&eAReSf
&h&iCUiTh )
/ ct(&f&eAReSf)
...
R(E,G) 1 2 3 4 E G 5 A
Sn(G,G)
S1(G,G)
U(G,I) C I
F=G=H=genes and S,T=gene-gene intereactions.
More than 3, S1, ..., Sn?
&iCUi))+
(ct(S1(&eARe
mncnf
/ ( (ct(&eARe))n
* ct(&iCUi) )
&iCUi))+...
ct(S2(&eARe
&iCUi)) )
ct(Sn(&eARe
If the S cube can be implemented so counts can be can be made of the 3-rectangle in blue directly, calculation of confidence would be fast.
4-hop APRIORI focus on G:
 mnsup
ct(&f&eAReSf)
 mncnf
ct(&f&eAReSf
&h&iCUiTh)
/ ct(&f&eAReSf)
1. (antecedent upward closure) If A is infrequent, then so are all of its subsets (the "list" will be larger, so the AND over the list will produce fewer ones) Frequency involves only A, so mine all qualifying antecedents using upward closure.
2. (consequent upward closure) If AC is non-confident, then so is AD for all subsets, D, of C (the "list" will be larger, so the AND over the list will produce fewer ones) So  frequent antecedent, A, use upward closure to mine out all confident consequents, C.

7. ct(&f&eAReSf) ct( &f&eAReSf &h(& )UiTh ) / ct(&f&eAReSf)
S(F,G)
R(E,F) 1 2 3 4 E F 5 G A C
T(G,H) H
U(H,I) I
V(I,J) J
5-hop
Focus on G:
 mnsup
ct(&f&eAReSf)
ct( &f&eAReSf
&h(& )UiTh ) /
ct(&f&eAReSf)
 mncnf
i(&jCVj)
5-hop APRIORI focus on G:
1. (antecedent upward closure) If A is infrequent, then so are all of its subsets (the "list" will be larger, so the AND over the list will produce fewer ones) Frequency involves only A, so mine all qualifying antecedents using upward closure.
2. (consequent downward closure) If AC is non-confident, then so is AD for all supersets, D, of C. So  frequent antecedent, A, use downward closure to mine out all confident consequents, C.

8. ct( &f(& )ReSf) ct( &f(& )ReSf &h(& )UiTh) / ct( &f(& )ReSf )
6-hop
S(F,G)
R(E,F) 1 2 3 4 E F 5 G A C
T(G,H) H
U(H,I) I
V(I,J) J D
Q(D,E)
The conclusion we have demonstrated (but not proven) is: for (a+c)-hop transitive Apriori ARM with focus the entity which is a hops from the antecedent and c hops from the consequent, if a/c is odd/even use downward/upward closure on that step in the mining of strong (frequent and confident) rules.
Focus on G:
ct(
&f(& )ReSf)
 mnsup
e(&dDQd)
ct(
&f(& )ReSf
&h(& )UiTh) /
e(&dDQd)
i(&jCVj)
ct(
&f(& )ReSf )
 mncnf
e(&dDQd)
6-hop APRIORI:
1. (antecedent downward closure) If A is infrequent, then so are all of its supersetsbsets. Frequency involves only A, so mine all qualifying antecedents using downward closure.
2. (consequent downward closure) If AC is non-confident, then so is AD for all supersets, D, of C. So  frequent antecedent, A, use downward closure to mine out all confident consequents, C.

9. Given any 1-hop labeled relationship (e. g
Given any 1-hop labeled relationship (e.g., cells have values from {1,2,…,n} then there is a natural n-hop transitive relationship, A implies D, by alternating entities for each specific label value relationship.
E.g., in netflix, Rating(Cust,Movie) has label set {0,1,2,3,4,5} so that generates a bonafide 6-hop transitive relationship:
E.g., equity trading on a given day, QuantityBought(Cust,Stock) w labels {0,1,2,3,4,5} (where n means n thousand shares) so that generates a bonafide 6-hop transitive relationship:
R3(C,M)
R2(M,C) 1 2 3 4 M C 5 A D
R4(M,C)
R5(C,M)
R0(M,C)
R1(C,M)
E.g., equity trading - moved similarly, (define moved similarly on a day then define relationship StockStock(#DaysMovedSimilarlyOfLast10)
E.g., equity trading - moved similarly2, (define moved similarly to mean that stock2 moved similarly to what stock1 did the previous day.Define relationship StockStock(#DaysMovedSimilarlyOfLast10)
E.g., Gene-Experiment, Label values could be "expression level". Intervalize and go!
Has Strong Transitive Rule Mining (STRM) been done? Are their downward and upward closure theorems already for it?
Is it useful? That is, are there good examples of use: stocks, gene-experiment, MBR, Netflix predictor,...

10. MYRRH APPENDIX 2e_2r pARM is MYRRH_2e_1r )
e.g., Rate5(Cust,Book) or R5(C,B), Purchase(Book,Cust) or P(B,C)
pre-computed 
BpTtreec 1-counts 3 2 1 2 3 1
BpTtreeb 1-cts  1
P(B,C)
(S(E,F)) 1
If cust, c, rates book, b as 5, then c purchase b. For bB, {c| rate5(b,c)=y}{c| purchase(c,b)=y}
ct(R5pTreei & PpTreei) / ct(R5pTreei)  mncnf ct(R5pTreei) / sz(R5pTreei)  mnsp 1 1 4 3
Speed of AND: R5pTreeSet & PpTreeSet? (Compute each ct(R5pTreeb&PpTreeb).)
Slice counts, bB, ct(R5pTreeb & PpTreeb) w AND? 2 3 4 5 C
(E) 2 1 1
pre-comR5pTtreeb 1-cts  1
R5(C,B)
(R(E,F)) B
(F) 1 1
Given eE, If R(e,f), then S(e,f)
ct(Re & Se)/ct(Re)mncnf, ct(Re)/sz(Re)mnsp 1
If eA R(e,f), then eB S(e,f)
R5pTtreec 1-cts 1 2
ct( &eARe &eBSe) / ct(&eARe)  mncnf. ...
Schema: size(C)=size(R5pTreeb)=size(BpTreeb)=4
size(B)= size(R5pTreec)=size(BpTreec)=4
If eA R(e,f), then eB S(e,f)
ct( &eARe OReBSe) / ct(&eARe)  mncnf. ...
If eA R(e,f), then eB S(e,f)
ct( OReARe &eBSe) / ct(OReARe)  mncnf. ... 1
R5(C,B) 1
R5pTtreeb&PpTreeb
1-counts 1
P(B,C)
If eA R(e,f), then eB S(e,f)
ct( OReARe OReBSe) / ct(OReARe)  mncnf. ...
Consder 2 Customer classes, Class1={C=2|3} and Class2={C=4|5}. Then P(B,C) is TrainingSet:
C\B
Book=4 is very discriminative of Class1 and Class2,
e.g., Class1=salary>$100K
Then the DiffSup table is:
B=1 B=2 B=3 B=4
P1={B=1|2} P2={B=3|4} C C DS
P1 [and P2, B=2 and B=3] is somewhat discriminative of the classes, whereas B=1 is not..
Are "Discriminative Patterns" covered by ARM? E.g., does the same information come out of strong rule mining?
Does "DP" yield information across multiple relationships? E.g., determining the classes via the other relationship?

11. MYRRH_2e_3r Rate1(Cust,Book) or R5(C,B), Purchase(Book,Cust) or P(B,C)
Sell(Cust,Book) or S(B,C)
Cust,c. Rates book,b as 1, and c Purchases b, likely c Sells b at term end
For bB, {c| R1(c,b)=y & P(c,b)=y}  {c| S(c,b)=y}
ct(R1pTreeb & PpTreeb & SpTreeb) / ct(R1pTreeb & PpTreeb)  minconf 1
R1(C,B) 2 3 4
P(B,C) B C 5
S(B,C) 1
R5(S,C) 2 3
PHC(B,S) B S 4 C 5
Rate5(Student,Course), PurchHardCov(Book,Stu)
If a student, s, rates any course as 5,
then s Purchases a HardCover book.
3e_2r
3e_3r
Students who buy b and courses using b, student enrolls in the course? {(s,c)| Buy(s,b)=y & Text(b,c)=y){(s,c)|Enroll(s,c)=y}. cnt(EpTreeSubSet(BpTreeb×TpTreeb))/(cnt(BpTreeb)*(cnt(TpTreeb)>mncf 13 12 1
book 2 3 4 5
course
Text
student
Enroll
Buy 13 12 1
book 2 3 4 5
course
Text
student
Enroll
Buy
offering
Location
If s enrolls in c, And c is Offered at L And L uses Text=b, Then s Buys b
4e_4r
Any 2 adjacent relationships can be collapsed into 1: R(c,b) and P(b,e) iff RP(c,e).
By doing so, we have a whole new relationship to analyze
Given c, {b|R(c,b)} is List(PR.c)
For b in List(PR,c), {eC|P(b,e)} is List(PP,b)
Therefore {e|RP(c,e)}=ORbListPR,cPP,b 1
R(C,B) 2 3 4
P(B,C) B C 5 1
RP(C,C) 2 3 4 5 C

12. PTc = ORbListTcPb PEc = ORsListEcPs ETs = ORcListEsTc
PT=PURCHASE_TEXT(S,C) 1
E=ENROLL(S,C)
S=STUDENT 2 3 4 5
C=COURSE
PTc = ORbListTcPb
also PTs = ORbListPsTb
P=PURCHASE(S,B) 1
E=ENROLL(S,C) 2 3 4
B=BOOK
S=STUDENT 5
C=COURSE
T=TEXT(C,B)
Let Tc=C-pTree of T for C=c
with list={b|T(c,b)}
PE=PURCHASE_ENROLL(C,B) 1 2 3 4
B=BOOK
C=COURSE
T=TEXT(C,B)
PEc = ORsListEcPs
also PEb = ORsListPbEs
P=PURCHASE(S,B) 1
ET=ENROLL_TEXT(S,B) 2 3 4
B=BOOK
S=STUDENT 5
ETs = ORcListEsTc
also ETb=ORcListTbEc
Interesting 2-hop examples?
Good 2-hop transitive relationships to study include:
WebpageWasVisitedByCustomerPurchasedItem.
(earlier rendition of 3 that may or may not be useful?): If {g} is the consequent of no confident rules, then no superset C of g is either. So it would make sense to consider singleton C's first. g we can replicate Sg and AND it into R making T. If ct(&eATe)=ctTA is small and the conf is high, ct(&eARe) is small also. If ctTA is small, then ctTB is smaller BA. So we can start with A=E and work down. As soon as we find A too small, then we need not look at any of its subsets...
In general, how can we speed up 2-hop Apriori? PTreeSet ANDs? Better algorithm?

13. ct(&bAPb &cDEc) / ct(&bAPb)  minconf
2-hop transitive rules
(specific examples)
E(S,C)
P(B,S) 1 2 3 4 B S 5 C
A 
 D
AD: If bA P(b,s), then cD E(s,c) is a strong rule if:
ct(&bAPb)  minsupp
ct(&bAPb &cDEc) / ct(&bAPb)  minconf
2-hop
Enroll
Book
If a student Purchases every book in A, then that student is likely to enroll in every course in D, and lots of students purchase every book in A. In short, P(A,s) E(s,D) is confident and P(A,s) is frequent
PJ(C,I)
PD(I,C) 1 2 3 4 I C 5
A 
 D
2-hop
Purchase
Dec/Jan
AD: If iA PD(i,c), then iD PJ(c,i) is a strong rule if:
ct(&iAPDi)  minsupp
ct(&iAPDi &iDPJi) / ct(&iAPDi)  minconf
If a customer Purchases every item in A in December, then that customer is likely to purchase every item in D in January, and lots of customers purchase every item in A in December: PD(A,c)PJ(c,D) conf and PD(A,c) freq.
B(P,I)
O(E,P) 1 2 3 4 E P 5 I
A 
 D
2-hop
Event
Buy
AD: If eA O(e,p), then iD B(p,i) is a strong rule if:
ct(&eAOe)  minsupp
ct(&eAOe &iDBi) / ct(&eAOe)  minconf
If every Event in A occurred in a person's life last year, then that person is likely to buy every item in D this year, and lots of people had every Event in A occur last year: O(A,p)B(p,D) conf and O(A,p) freq.

14. ct(&eAOe &mDTm) / ct(&eAOe)  minconf
2-hop
stock
trading
T(S,M)
O(E,S) 1 2 3 4 E S 5 M
A 
 D
AD: If eA O(e,s), then mD T(s,m) is a strong rule if:
ct(&eAOe)  minsupp
ct(&eAOe &mDTm) / ct(&eAOe)  minconf
If every Event in A occurs for a company in time period 1, then the price of that stock experienced every move in D time period 2, and lots of companies had every Event in A occur in period 1: O(A,s)T(s,D) conf and O(A,s) freq. (T=True; e.g., m=1 down a lot, m=2 down a little, m=3 up a little, m=4 up a lot.)
T(C,M)
O(E,C) 1 2 3 4 E C 5 M
A 
 D
AD: If eA O(e,c), then mD T(c,m) is a strong rule if:
2-hop
commodity
trading
ct(&eAOe)  minsupp
ct(&eAOe &mDTm) / ct(&eAOe)  minconf
If every Event in A occurs for a commodity in time period 1, then the price of that commodity experienced every move in D time period 2, and lots of commodities had every Event in A occur in period 1: O(A,c)T(c,D) conf and O(A,c) freq.
B(P,I)
F(P,P) 1 2 3 4 P 5 I
A 
 D
AD: If pA P(p,q), then iD B(p,i) is a strong rule if:
2-hop
facebook
friends
buying
ct(&pAFp)  minsupp
ct(&pAFp &iDBi) / ct(&pAFp)  minconf
F(p,q)=1 iff q is a facebook friend of p. B(p,i)=1 iff p buys item i.
People befriended by everyone in A (= &pAFp denoted FA for short ) likely buy everything in D. And FA is large.
So every time a new person appears in FA that person is sent ads for items in D.

15. ct(&aAAOa&gCB'g)/ct(&aAAOa)mncf ct( AOa=4)mnsp
How do we construct interesting 2-hop examples? Method-1: Use a feature attribute of a 1-hop entity.
Start with a 1-hop, e.g., customers buy items, stocks have prices or people befriend people then focus on one feature attribute of one of the entities. The relationship is the projection of that entity table onto the feature attribute and the entity id attribute (key) e.g. Age, Gender, Income Level, Ethnicity, Weight, Height... of people or customer entity
These are not bonafide 2-hop transitive relationships since they are many-to-one relationships, not a many-to-many (because the original entity is the primary key of its feature table). Thus, we don't get a fully transitive relationship since collapsing the original entity leaves nearly the same information as the transitive situation was intended to add. Here is an example.
If, from the new transitive relationship, AgeIsAgeOfCustomerPurchasedItem, Customer is collapsed we have AgePurchaseItem and the Customer-to-Age info is still available to us in the Cust table. The relationship between Customers and Items is lost, but presumably, the reason for mining, AgeIsAgeOfCustomerPurchaseItem is to find AgePurchaseItem rules independent of the Customers involved. Then when a high confidence Age implies Item rule is found, the Customers who are of that age can be looked up from the Customer feature table and sent a flyer for that item. Also, in CustomerPurchaseItem, the antecedent, A, could have been chosen to be an age-group. So most AgePurchaseItem info would come out of CustomerPurchaseItem directly.
Given a 1-hop relationship, R(E,F) and a feature attribute, A of E, if there is a pertinent way to raise E up the semantic hierarchy (cluster it) producing E', then the relationship between A and E ' is many-to-many, e.g., cluster Customers by Income Level, IL. Then AgeIsAgeOfIL is a many-to-many relationship. Note, what we're really doing here is using the many-to-many relationship between two feature attributes in one of the entity tables and then replacing the entity by the second feature. E.g., if B(C,I) is a relationship, and IL is a feature attribute in the entity table C(A,G,IL,E,W,H), then clustering (Classifying) C by IL produces a relationship, B'(IL,I), given by B'(il,i)=1 iff B(c,i)=1 for  50% of cil, which is many-to-many provided IL is not a candidate key. So from the 1-hop relationship, CB(C,I)I, we get a bonafide 2-hop relationship, AAO(A,IL)ILB'(IL,I)I.
ct(&aAAOe)mnsp
ct(&aAAOa&gCB'g)/ct(&aAAOa)mncf I
B(C,I) 1 C A IL 2 3 4 5 1 2 3 4 I
B'(IL,I) IL
ct( AOa=4)mnsp
ct(AOa=4 &g=3,4B'g )/ct(AOa= )mncf C
ct( )mnsp
ct( &100&110)/ ct(010) mncf
1 mnsp
0 / mncf
Bil=2 = Bc=3OR Bc=5
Bil=1 = Bc=2
Bil=3 = Bc=4 4 5 I
AO(A,IL) 1
ct(&cC(A)Bc)mnsp
ct(&cC(A)Bc &IC)/ct(&cC(A)Bc)mncf
ct( Bc= )mnsp
ct(Bc= &0011)/ct(Bc= )mncf A
ct( )mnsp
ct( &0011)/ct( )mncf
So these are different rules.
2 mnsp
1 / mncf

16. ct(&aALa&cCPc)/ct(&aALa)  mncf
stride=10 level-1 val SL SW PL PW
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SL SW PL rnd(PW/10)
Method-2: Use 3 feature attributes of an entity. Start with an Entity (e.g., IRIS(SL,SW,PL,PW). Take 3 attributes SL,PL,PW; form 2 many-to-many relationships L(SL,PL) and P(PL,PW) in which a cell value is 1 iff there is a IRIS sample with those values (Could also cluster IRIS on SW first then add PL and PW, so that the key, IRIS-ID is involved.)
Let ASL be {6,7} and CPW be {1,2}
ct(&aALa) mnsp
ct(&aALa&cCPc)/ct(&aALa)  mncf
ct( )mnsp
ct( & )/ct( )mncf
1  mnsp
1 / 1  mncf
PW=
PL=
So, with very high confidence, SL[55,74] PL PW[5,24], however the support of this rule is very low. What about the 1-hop version?
(aA)(SLPWaC) sup conf=ct(ORaASLPWa&pTreeC) / ct(ORaASLPWa)
PW=
SL=
conf=ct( & ) / ct( )
conf= 2/2 = Supp=2
PL=
SL=
What's the difference?
The 1-hop says "if SLA then PW will likely C.
2-hop says "If all SL in A are related to a PL then that PL is likely to be related to all PW in C."
These are different. 2-hop seems convoluted. Why?
2-hops from one table may always make less sense than the corresponding direct 1-hop.
How do you mine out all confident (or all strong) rules?

17. / ct(&pw&swARswSpw)
4-hop using attributes of
IRIS(Cls,SL,SW,PL,PW)
stride=10 level-1 val SL SW PL PW
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SL SW PL rnd(PW/10) SW PW 1 2 3 4 5 6 7 S R PL SL 1 2 3 4 5 6 7 U T
CLS se ve vi
AC confident?
C={se}
ct( &pw&swARswSpw
&sl&clsCUclsTsl )
/ ct(&pw&swARswSpw)
= 1/2
pl={1,2}
pl={1}

18. ct(RA&cls{se}Rcls) / ct(RA)
1-hop: IRIS(Cls,SL,SW,PL,PW)
stride=10 level-1 val SL SW PL PW
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SL SW PL rnd(PW/10) SW
CLS se ve vi
C={se} R
1-hop AC is more confident:
ct(RA&cls{se}Rcls) / ct(RA)
= 1
sw=
{3,4}
sw=
{3,4}
sw=
{3,4}
But what about just taking R{class}? Gives {3,4}se {2,3}ve {3}vi This is not very differentiating of class. Include the other three? SL
CLS se ve vi
{4,5}se {5,6}ve {6,7}vi
These rules were derived from the binary relationships only.
A minimal Decision Tree Classifier suggested by the rules:
/ \
PW= else |
se PL{3,4} & SW=2 & SL=5 else ve
2 of 3 of: else
PL{3,4,5} |
SW={2,3} vi
SL={5,6} SW
CLS se ve vi
{3,4}se {2,3}ve {3}vi PL
CLS se ve vi
{1,2}se {3,4,5}ve {5,6}vi PW
CLS se ve vi
{0}se {1,2}ve {1,2}vi
I was hoping for a "Look at that!" but it didn't happen ;-)

19. =1 ct(ORplATpl &clsCUcls) / ct(ORplATpl)
2-hop
stride=10 level-1 val SL SW PL PW
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SL SW PL rnd(PW/10) PL SL 1 2 3 4 5 6 7 U T
CLS se ve vi
C={se}
ct(ORplATpl &clsCUcls) / ct(ORplATpl) =1
Mine out all confident se-rules with minsup = 3/4:
sl={4,5}
Closure: If A{se} is nonconfident and AUse then B{se} is nonconfident for all B  A. So starting with singleton A's:
ct(Tpl=1 & Use) / ct(Tpl=1) = 2/2 yes.
A= {1,3} {1,4} {1,5} or {1,6} will yield nonconfidence and AUse so all supersets will yield nonconfidence.
ct(Tpl=2 & Use) / ct(Tpl=2) = 1/1 yes.
A= {1,2} will yield confidence.
ct(Tpl=3 & Use) / ct(Tpl=3) = 0/1 no.
A= {2,3} {2,4} {2,5} or {2,6} will yield nonconfidence but the closure property does not apply.
ct(Tpl=4 & Use) / ct(Tpl=4) = 0/1 no.
ct(Tpl=5 & Use) / ct(Tpl=5) = 1/2 no.
ct(Tpl=6 & Use) / ct(Tpl=6) = 0/1 no. etc.
I conclude that this closure property is just too weak to be useful. And also it appears from this example that trying to use myrrh to do classification (at least in this way) does not appear to be productive.

20. 2012_02_04 We are looking for real-life examples, e.g.,
“friends of friends of those who buy X, buy Y”. 2 3 4 5
J=Person
AreGivenBy=U(H,I)
I=Customers C 1 4 1 1 1 3 1
6-hop myrrh example (from Damian).
Market agency targeting advertising to friends of customers:
Entities:
1. advertisements markets merchants products
5. reviews customers 7. friends
The hop logic would be:
1. Advertisements target specific markets.
2. Markets have particular merchants
3. Merchants sell individual products.
4. Products have reviews.
5. Reviews are provided by customers.
6. Customers have friends. 1 2 1 1 1 1 1
IsAFriend?(I,J) 2 3 4 5
H=rating
Sell=S(F,G)
G=products 1 4 1 1 1 3 1 1 2 1 1 1 1 1
RatedAt(G,H) 2 3 4 5
F=merchants
targetsQ(D,E)
E=Markets 1 4 1 1 3 1 1 2 1 1 1 1
IsUsedBy= R(E,F) 2 3 4 5
D=Ads A
Another example (from Ya).
In 779 slide "03vertical_audio1.ppt", 2 techniques to address the curse of cardinality and dimensionality are: Parallelizing the processing engine. ... Parallelize the greyware engine on clusters of people  (i.e., enable visualization and use the web...).
There is a good example of the "visualization and use the web": Crystal structure of a monomeric retroviral protease solved by protein folding game players
Although this is a protein structure problem, not a curse of cardinality or curse of dimensionality problem.