1. Efficient Equal Interval Neighborhood Ring (P-trees technology is patented by NDSU)

2. OUTLINE Review: HOBBit Metric Equal Interval Neighborhood Ring (EINring) –Prototype Problem –Definition of Range Mask –Propositions –Definition of EINring –Calculation of EINring Using P-trees Summary

3. Review: HOBBit Similarity Metric Let X and Y be two values, the HOBBit similarity between X and Y is defined by where x i and y i are the bits of X and Y respectively,  denotes XOR. In another word, it is the left most position at which X and Y differ.

4. Review: HOBBit Distance & Ring The HOBBit distance between two tuples X and Y is defined by HOBBit Ring: The HOBBit ring of radii, r 1 and r 2, centered at c is defined as R(c, r 1, r 2 ) = {x X | r 1  d(c,x) < r 2 }, where d(c,x) is HOBBit distance.

5. Diagram of HOBBit Ring

6. Example of HOBBit Ring HOBBit RingBinary RangeDecimal R(22,7,8)00010110-0001011122-23 R(22,6,8)00010100-0001011120-23 R(22,5,8)00010000-0001011116-23 R(22,4,8)00010000-0001111116-31 R(22,3,8)00000000-000111110-31 R(22,2,8)00000000-001111110-63 R(22,1,8)00000000-000101110-127 R(22,0,8)00000000-000101110-255

7. Summary of HOBBit Metric The HOBBit metric is based on the most significant matching bit positions starting from the left. HOBBit ring is a geometric ring whose diameter increases exponentially. HOBBit ring is eccentric ring.

8. Equal Interval Neighborhood Ring (EINring)

9. Outline Prototype Problem Definition of Range Mask Propositions and Theorem Definition of EINring Calculation of EINring using P-trees

10. Prototype Problem Problem: x > (4) 10 > (100) 2 Conjecture: Px>(100) 2 = P3(P2P1) 7 7 7 7 5 5 1 1 7 7 7 7 1 1 1 1 5 5 7 7 4 4 1 1 5 7 7 7 4 5 5 1 6 6 6 6 3 3 0 0 6 6 6 6 0 0 0 0 2 2 6 6 3 3 0 0 2 6 6 6 3 3 3 0 8x8 data set

11. Walk Through: Peano Trees

12. Result of Crude Method

13. Result of Conjecture

14. Definition of Range Mask Range Mask The Range Mask is the P-tree mask that calculates any data point, x, that satisfies range inequality, i.e., x  c 1, x > c 1, x  c 2, etc., where c 1, c 2 are integers. Example: P x>100 is a P-tree mask that calculates any data point greater than 100.

15. Proposition 1 Let m be the number of binary bit of j th attribute of data point x, P j,m, P j,m-1, … P j,0 be the basic P-trees of i th bit of j th attribute, and integers c=b m b m-1 … b 0, where b i is i th binary bit value of c. Let P xjr be the Range Mask that satisfies inequality x j  r, then P xjr = P j, m … P j,i op j,i … P j,0, s.t. 1) Op j,j is  if b i =1, 2) Op j,i is  if b i =0, 3) right binding within each attribute. Example: P xj  (70) 10 = P xj  (01000110)2 = P7(P6(P5(P4( P3( P2P1P0))))

16. Proof Sketch Without loss of generality, assume data point x has one attribute. Let c= b m …b i …b 0, where b i is i th bit value of c. P xj  c is the range mask that satisfy x  c. If b i =1, the ith bit of x should be set 1 when x and c have the same bit value from position m th to i th position, e.i., P xj  c =P m …P i  …P 0. (Partially done!) If b i =0, there are two cases that satisfy x  c, one is to set i th bit of x, x i =1, another is to set x i =0. Thus P xj  c = (P m … P i )  (P m …P i ’  P i-1 …P 0 ). = P xj  c =(P m …(P i  (P i-1 …P 0 )). Done!

17. Proposition 2 Let m be the number of binary bit of j th attribute of data point x, P ’ j,m, P ’ j,m-1, … P ’ j,0 be the complement P-trees of i th bit of j th attribute, and integers c=b m b m- 1 … b 0, where b i is i th binary bit value of c. Let P xj  c be the Range Mask that satisfies x j  c, then P xj  r = P ’ j,m … P ’ j,i op j,i … P ’ j,0 s.t. 1) Op j,i is  if b i =0, 2) Op j,i is  if b i =1, 3) right binding within each attribute Example: P xj  (198)10 = P xj  (11000101)2 = P7 ’  (P6 ’  (P5 ’ P4 ’ P3 ’ (P2 ’  (P1 ’ P0 ’ )))

18. Proposition 3 Let m be the number of binary bit of j th attribute of data point x, P j,m, P j,m-1, … P j,0 be the basic P-trees of i th bit of j th attribute, and integers c=b m b m- 1 … b 0, where b i is c ’ s i th binary bit value. Let P xj>c be the Range Mask that satisfies inequality x j > c, then P xj>c = P j,m … P i,j op i,j … P j,k, s.t. 1) op i,j is  if b i =1, 2) op i,j is  if b i =0, 3) right binding within each attribute 4) b k =0, and b j =1 j<k. Example: P xj > (72)10 = P xj > (01001000)2 = P7  (P6  (P5  P4  P3))

19. Proposition 4 Let m be the number of binary bit of j th attribute of data point x, P ’ j,m, P ’ j,m-1, … P ’ j,0 be the complement P-trees of i th bit of j th attribute, and integer c=b m b m- 1 … b 0, where b i is c ’ s i th binary bit value. Let P xj<c be the Range Mask that satisfies x j < c, then P xj<c = P ’ j,m … P ’ j,i op j,i … P ’ j,k, s.t. 1) op i,j is  if b i =0, 2) op i,j is  if b i =1, 3) right binding within each attribute, 4) b k =0, and bj=1 j<k. Example: P xj < (72)10 = P xj < (01001000)2 = P7 ’ P6 ’  (P5 ’ P4 ’ P3 ’ )

20. More Examples c=(70) 10 =(01001000) 2 P x<c =P7 ’ P6 ’  (P5 ’ P4 ’ P3 ’ ) c=(72) 10 =(01001000) 2 P x>c =P7  P6 (P5  P4  P3) c=(198) 10 =(11000101) 2, P x<=c = P7 ’  (P6 ’  (P5 ’ P4 ’ P3 ’ (P2 ’  (P1 ’ P0 ’ ))) Let c=(198) 10 =(11000101) 2, P x=c =P7P6(P5  (P4  (P3 (P2(P1 P0)))))

21. Theorem – Range Mask Complement Rule Theorem Range Mask Complement Rule Let P xj c be the Range Mask of j th attribute of any data point x, where c is integer, then P xj  c = P’ xj c hold.

22. Definition of Neighborhood Ring Neighborhood Ring: The Neighborhood ring of radii, r 1 and r 2, centered at c is defined as R(c, r 1, r 2 ) = {x X | r 1 < abs(x-c)  r 2 }, where abs(x-c) is absolute length between x and c.

23. Definition of Equal Interval Neighborhood Ring (EINring) The Equal Interval Neighborhood ring of radii, r 1 and r 2, centered at c is defined as R(c, r 1, r 2 ) = {x X | r 1 <abs(x-c)  r 2 }, and abs(r 2 -r 1 )=2k, where k=1,2, …, abs(x-c) is absolute length between x and c, and is interval

24. Diagram of EINring

25. Example of EINring HOBBit RingBinary RangeDecimal R 16 (86,1)01000110-0110011070-102 R 16 (86,2)00110110-1111011054-118 R 16 (86,3)00100110-1000011038-134 R 16 (86,4)00001110-1001011022-150 ……… R 16 (86,10)00000000-111001100-230 R 16 (86,11)00000000-11110110 0-246 R 16 (86,12)00000000-11111111 0-255

26. Neighbor Count within EINring For any data point, x, let x = (x 1,x 2, … x m ), where x,j is x ’ s j th attribute column. Let r be vectors with m elements, we define the range mask P x>c+r as P x>c-r = P x1>c-r1  P x2>c-r2  ….  P xj>c-rj and define the range mask P xc+r as P xc+r = P x1c+r1  P x2c+r2  ….  P xjc+rj where c is a constant.

27. Neighbor Count within EINring: The range mask for any data points x within the neighborhood ring, R(c, 0, r), are calculated by P c,r = P x>c-r  P xc+r The neighbor count for x within the neighborhood ring, R(c, 0, r), are calculated by NC (c,0,r) = RC(P c,r ) where RC is the root count of P-tree.

28. Neighbor Count within EINring The Neighbor Count NC(c, r 1, r 2 ) of c within EINring R(c, r 1, r 2 ) is calculated as NC(c, r 1,r 2 ) =RC(P c,r2 )-RC(P c,r1 )

29. Summary Equal Interval Neighborhood Ring (EINring) is much finer than HOBBit ring. Calculation of EINring using P-trees is efficient, comparable to that of HOBBit ring.