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The Post Office Problem

k-d trees, k-nn search, and the Johnson-
Lindenstrauss lemma
Who am I?
   Jeremy Holland
   Senior lead developer at
    Centresource
   Math and algorithms nerd
   @awebneck,
    github.com/awebneck, freenode:
    awebneck, (you get the idea)
If you like the talk...
   I like scotch. Just putting it out there.
What is the Post Office Problem?
   Don Knuth, professional CS badass.
   TAOCP, vol. 3
   Otherwise known as ”Nearest Neighbor search”
   Let's say you've...
Just moved to Denmark!
But you need to mail a letter!
      Which post office do you go to?
Finding free images of post
offices is hard, so...
        We'll just reduce it to this:




              q
Naive implementation
   Calculate distance to all points, find smallest
min = INFINITY
P = <points to be searched>
K = <dimensionality of points, e.g. 2>
q = <query point>
best = nil
for p in P do
  dimDistSum = 0
  for k in K do
    dimDistSum += (q[k]-p[k])**2
  dist = dimDistSum.sqrt
  if dist < min
    min = dist
    best = p
return best
With a little preprocessing...
   But that takes      time! - can we do better?
   You bet!
   k-d tree
   Binary tree (each node has at most two
    children)
   Each node represents a single point in the set
    to be searched
Each node looks like...
   Domain: the vector describing the point (i.e.
    [p[0], p[1], … p[k-1]])
   Range: Some identifying characteristic (e.g. PK
    in a database)
   Split: A chosen dimension from 0 ≤ split < k
   Left: The left child (left.domain[split] <
    self.domain[split])
   Right: The right child (right.domain[split] ≥
    self.domain[split])
Let's build a k-d tree!
             Point 1: [20,10]
Let's build a k-d tree!
          Let's split on the x axis
Let's build a k-d tree!
          Add a new point: [10,5]
Let's build a k-d tree!
 The new point is the Left Child of the first point
Let's build a k-d tree!
        Let's split him on the y axis
Let's build a k-d tree!
         And add a 3rd point: [25,3]
Let's build a k-d tree!
 The new point is the Right Child of the first point
Let's build a k-d tree!
           So on and so forth...
Let's build a k-d tree!
            Giving you a tree:
How do we search it?
 Step 1: Find the best bin (where the query point
           would otherwise be inserted)


              q

                               root
How do we search it?
 NOTE: There is no node for this bin – just the
     space a node would be if existed!


             q

                              root
How do we search it?
 Step 2: Make the current leaf node the current
                 ”best guess”




                     Best guess
How do we search it?
  … and set the ”best guess radius” to be the
  distance between the query and that point




                      Best guess radius
How do we search it?
      Step 3: Back up the tree 1 node




                    Current node
How do we search it?
 If the distance between the query and the new
     node is less than the best guess radius...
How do we search it?
   Then set the best guess radius to the new
 distance, and make the current node the best
How do we search it?
Step 4: If the hypersphere described by the best
    guess radius crosses the current split...



                        Oh nooooh!
How do we search it?
 And the current node has a child on the other
                    side...




                      Oh snap!
How do we search it?
 … then make that node the current node, and
                   repeat:
How do we search it?
Here, the distance is not less than the best guess
                     radius...
How do we search it?
  … and the hypersphere neither crosses the
                  split ...
                             Whew, missed it!
How do we search it?
… nor does the current node have any children ...

                               Whew, missed it!
How do we search it?
So we can eliminate it and back up the tree again!
How do we search it?
We've already compared this node, so let's keep
            going back up the tree
How do we search it?
 Again, the radius is bigger than the best guess,
   and there is no crossing – back up again!
How do we search it?
           ...and again...
How do we search it?
       All the way back to the root!
How do we search it?
And you have your nearest neighbor, with a good
         case of        running time!


            I'm the answer!
But that was a pretty good case...
   We barely had to backtrack at all – best case is

   Worst case (lots of backtracking – examining
    almost every node) can get up to
   Amount of backtracking is directly proportional
    to k!
   If k is small (say 2, as in this example) and n is
    large, we see a huge improvement over linear
    search
   As k becomes large, the benefits of this over a
    naive implementation virtually disappear!
The Curse of Dimensionality
   Curse you, dimensionality!
   High-dimensional vector spaces are darned
    hard to search!
   Why? Too many dimensions! Why are there so
    many dimensions!?!
   What can we do about it?
   Get rid of the extra weight!
   Enter Mssrs. Johnson and Lindenstrauss
It turns out...
   Your vectors have a high dimension
   Absolute distance and precise location versus
    relative distance between points
   Relative distance can be largely preserved by a
    lower dimensional space
   Reduce k dimensions to kproj dimensions,
     kproj << k
Example: 2d to 1d

                    11.180
   5.0
       00




                               8.246
 7.28




                                       17.
                                           464
     0




                                                  7.
                                  6                  0
                               .29                       71
                             10
            3.16




                                         13.34
                                              2
                2




                                       14.000
Example: 2d to 1d, 1st attempt

                    11.180                        Projection Plane
   5.0
       00




                               8.246
 7.28




                                       17.
                                           464
     0




                                                      7.
                                  6                      0
                               .29                           71
                             10
            3.16




                                         13.34
                                              2
                2




                                       14.000
Example: 2d to 1d, 1st attempt

                    11.180
   5.0
       00




                               8.246
 7.28




                                        17.
                                            464
     0




                                                       7.
                                  6                       0
                               .29                            71
                             10
            3.16




                                           13.34
                                                2
                2




                                         14.000




                             Finished 1-d Projection
Example: 2d to 1d, 2nd attempt

                    11.180                        Projection Plane
   5.0
       00




                               8.246
 7.28




                                       17.
                                           464
     0




                                                      7.
                                  6                      0
                               .29                           71
                             10
            3.16




                                         13.34
                                              2
                2




                                       14.000
Example: 2d to 1d, 2nd attempt

                    11.180
   5.0
       00




                               8.246
 7.28




                                        17.
                                            464
     0




                                                       7.
                                  6                       0
                               .29                            71
                             10
            3.16




                                           13.34
                                                2
                2




                                         14.000




                             Finished 1-d Projection
Example: 2d to 1d, 3rd attempt

                    11.180                        Projection Plane
   5.0
       00




                               8.246
 7.28




                                       17.
                                           464
     0




                                                      7.
                                  6                      0
                               .29                           71
                             10
            3.16




                                         13.34
                                              2
                2




                                       14.000
Example: 2d to 1d, 3rd attempt

                    11.180                             Projection Plane
   5.0
       00




                               8.246
 7.28




                                        17.
                                            464
     0




                                                           7.
                                  6                           0
                               .29                                71
                             10
            3.16




                                           13.34
                                                2
                2




                                        14.000




                             Finished 1-d Projection
It turns out...




   Relative distance can be largely but not
    completely preserved by a lower dimensional
    space
   Every projection will have errors
   How do you choose one with the fewest?
   Trick question: Let fate decide!
Multiple random projection
   Choose the projections radomly
   Multiple projections
   Exchange cost in resources for cost in accuracy
       More projections = greater resource cost = greater
        accuracy
       Fewer projections = lesser resource cost = lesser
        accuracy
   Trivially parallelizable
   Learn to be happy with ”good enough”
Multiple random projections
 Get the nearest from each projection, then run a
       naive nearest on the results thereof.
Nns = []
P = <projections>
q = <query point>
for p in P do
  pq = <project q to the same plane as p>
  nns << <nearest neighbor to pq from projection>
<execute naive nearest on nns to find nearest of result>
return nn



                         Et voilá!
Multiple random projection
   Experiments yield > 98% accuracy when
    multiple nearest neighbors are selected from
    each projection and d is reduced from 256 to
    15, with approximately 30% of the calculation.
    (see credits)
   Additional experiments yielded similar results,
    as did my own
   That's pretty darn-tootin' good
Stuff to watch out for
   Balancing is vitally important (assuming uniform
    distribution of points): careful attention must be
    paid to selection of nodes (node with median
    coordinate for split axis)
   Cycle through axes for each level of the tree –
    root should split on 0, lvl 1 on 1, lvl 2 on 2, etc.
Stuff to watch out for
   Building the trees still takes some time
       Building the projections is effectively matrix
        multiplication, time in            (Strassen's
        algorithm)
       Building the (balanced) trees from the projections
        takes time in approximately
   Solution: build the trees ahead of time and
    store them for later querying (i.e. index those
    bad boys!)
Thanks!
   Credits:
       Based in large part on research conducted by
        Yousuf Ahmed, NYU: http://bit.ly/NZ7ZHo
       K-d trees: J. L. Bentley, Stanford U.:
        http://bit.ly/Mpy05p
       Dimensionality reduction: W. B. Johnson and J.
        Lindenstrauss: http://bit.ly/m9SGPN
       Research Fuel: Ardbeg Uigeadail:
        http://bit.ly/fcag0E

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The Post Office Problem

  • 1. The Post Office Problem k-d trees, k-nn search, and the Johnson- Lindenstrauss lemma
  • 2. Who am I?  Jeremy Holland  Senior lead developer at Centresource  Math and algorithms nerd  @awebneck, github.com/awebneck, freenode: awebneck, (you get the idea)
  • 3. If you like the talk...  I like scotch. Just putting it out there.
  • 4. What is the Post Office Problem?  Don Knuth, professional CS badass.  TAOCP, vol. 3  Otherwise known as ”Nearest Neighbor search”  Let's say you've...
  • 5. Just moved to Denmark!
  • 6. But you need to mail a letter! Which post office do you go to?
  • 7. Finding free images of post offices is hard, so... We'll just reduce it to this: q
  • 8. Naive implementation Calculate distance to all points, find smallest min = INFINITY P = <points to be searched> K = <dimensionality of points, e.g. 2> q = <query point> best = nil for p in P do dimDistSum = 0 for k in K do dimDistSum += (q[k]-p[k])**2 dist = dimDistSum.sqrt if dist < min min = dist best = p return best
  • 9. With a little preprocessing...  But that takes time! - can we do better?  You bet!  k-d tree  Binary tree (each node has at most two children)  Each node represents a single point in the set to be searched
  • 10. Each node looks like...  Domain: the vector describing the point (i.e. [p[0], p[1], … p[k-1]])  Range: Some identifying characteristic (e.g. PK in a database)  Split: A chosen dimension from 0 ≤ split < k  Left: The left child (left.domain[split] < self.domain[split])  Right: The right child (right.domain[split] ≥ self.domain[split])
  • 11. Let's build a k-d tree! Point 1: [20,10]
  • 12. Let's build a k-d tree! Let's split on the x axis
  • 13. Let's build a k-d tree! Add a new point: [10,5]
  • 14. Let's build a k-d tree! The new point is the Left Child of the first point
  • 15. Let's build a k-d tree! Let's split him on the y axis
  • 16. Let's build a k-d tree! And add a 3rd point: [25,3]
  • 17. Let's build a k-d tree! The new point is the Right Child of the first point
  • 18. Let's build a k-d tree! So on and so forth...
  • 19. Let's build a k-d tree! Giving you a tree:
  • 20. How do we search it? Step 1: Find the best bin (where the query point would otherwise be inserted) q root
  • 21. How do we search it? NOTE: There is no node for this bin – just the space a node would be if existed! q root
  • 22. How do we search it? Step 2: Make the current leaf node the current ”best guess” Best guess
  • 23. How do we search it? … and set the ”best guess radius” to be the distance between the query and that point Best guess radius
  • 24. How do we search it? Step 3: Back up the tree 1 node Current node
  • 25. How do we search it? If the distance between the query and the new node is less than the best guess radius...
  • 26. How do we search it? Then set the best guess radius to the new distance, and make the current node the best
  • 27. How do we search it? Step 4: If the hypersphere described by the best guess radius crosses the current split... Oh nooooh!
  • 28. How do we search it? And the current node has a child on the other side... Oh snap!
  • 29. How do we search it? … then make that node the current node, and repeat:
  • 30. How do we search it? Here, the distance is not less than the best guess radius...
  • 31. How do we search it? … and the hypersphere neither crosses the split ... Whew, missed it!
  • 32. How do we search it? … nor does the current node have any children ... Whew, missed it!
  • 33. How do we search it? So we can eliminate it and back up the tree again!
  • 34. How do we search it? We've already compared this node, so let's keep going back up the tree
  • 35. How do we search it? Again, the radius is bigger than the best guess, and there is no crossing – back up again!
  • 36. How do we search it? ...and again...
  • 37. How do we search it? All the way back to the root!
  • 38. How do we search it? And you have your nearest neighbor, with a good case of running time! I'm the answer!
  • 39. But that was a pretty good case...  We barely had to backtrack at all – best case is  Worst case (lots of backtracking – examining almost every node) can get up to  Amount of backtracking is directly proportional to k!  If k is small (say 2, as in this example) and n is large, we see a huge improvement over linear search  As k becomes large, the benefits of this over a naive implementation virtually disappear!
  • 40. The Curse of Dimensionality  Curse you, dimensionality!  High-dimensional vector spaces are darned hard to search!  Why? Too many dimensions! Why are there so many dimensions!?!  What can we do about it?  Get rid of the extra weight!  Enter Mssrs. Johnson and Lindenstrauss
  • 41. It turns out...  Your vectors have a high dimension  Absolute distance and precise location versus relative distance between points  Relative distance can be largely preserved by a lower dimensional space  Reduce k dimensions to kproj dimensions, kproj << k
  • 42. Example: 2d to 1d 11.180 5.0 00 8.246 7.28 17. 464 0 7. 6 0 .29 71 10 3.16 13.34 2 2 14.000
  • 43. Example: 2d to 1d, 1st attempt 11.180 Projection Plane 5.0 00 8.246 7.28 17. 464 0 7. 6 0 .29 71 10 3.16 13.34 2 2 14.000
  • 44. Example: 2d to 1d, 1st attempt 11.180 5.0 00 8.246 7.28 17. 464 0 7. 6 0 .29 71 10 3.16 13.34 2 2 14.000 Finished 1-d Projection
  • 45. Example: 2d to 1d, 2nd attempt 11.180 Projection Plane 5.0 00 8.246 7.28 17. 464 0 7. 6 0 .29 71 10 3.16 13.34 2 2 14.000
  • 46. Example: 2d to 1d, 2nd attempt 11.180 5.0 00 8.246 7.28 17. 464 0 7. 6 0 .29 71 10 3.16 13.34 2 2 14.000 Finished 1-d Projection
  • 47. Example: 2d to 1d, 3rd attempt 11.180 Projection Plane 5.0 00 8.246 7.28 17. 464 0 7. 6 0 .29 71 10 3.16 13.34 2 2 14.000
  • 48. Example: 2d to 1d, 3rd attempt 11.180 Projection Plane 5.0 00 8.246 7.28 17. 464 0 7. 6 0 .29 71 10 3.16 13.34 2 2 14.000 Finished 1-d Projection
  • 49. It turns out...  Relative distance can be largely but not completely preserved by a lower dimensional space  Every projection will have errors  How do you choose one with the fewest?  Trick question: Let fate decide!
  • 50. Multiple random projection  Choose the projections radomly  Multiple projections  Exchange cost in resources for cost in accuracy  More projections = greater resource cost = greater accuracy  Fewer projections = lesser resource cost = lesser accuracy  Trivially parallelizable  Learn to be happy with ”good enough”
  • 51. Multiple random projections Get the nearest from each projection, then run a naive nearest on the results thereof. Nns = [] P = <projections> q = <query point> for p in P do pq = <project q to the same plane as p> nns << <nearest neighbor to pq from projection> <execute naive nearest on nns to find nearest of result> return nn Et voilá!
  • 52. Multiple random projection  Experiments yield > 98% accuracy when multiple nearest neighbors are selected from each projection and d is reduced from 256 to 15, with approximately 30% of the calculation. (see credits)  Additional experiments yielded similar results, as did my own  That's pretty darn-tootin' good
  • 53. Stuff to watch out for  Balancing is vitally important (assuming uniform distribution of points): careful attention must be paid to selection of nodes (node with median coordinate for split axis)  Cycle through axes for each level of the tree – root should split on 0, lvl 1 on 1, lvl 2 on 2, etc.
  • 54. Stuff to watch out for  Building the trees still takes some time  Building the projections is effectively matrix multiplication, time in (Strassen's algorithm)  Building the (balanced) trees from the projections takes time in approximately  Solution: build the trees ahead of time and store them for later querying (i.e. index those bad boys!)
  • 55. Thanks!  Credits:  Based in large part on research conducted by Yousuf Ahmed, NYU: http://bit.ly/NZ7ZHo  K-d trees: J. L. Bentley, Stanford U.: http://bit.ly/Mpy05p  Dimensionality reduction: W. B. Johnson and J. Lindenstrauss: http://bit.ly/m9SGPN  Research Fuel: Ardbeg Uigeadail: http://bit.ly/fcag0E