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Programming`s Greatest Hits of the
60s and 70s
Michelle Brush Engineering Director, Cerner Corporation
Chapter Leader, Kansas City Girl Develop It Conference Organizer, Midwest.io
Where Are We?
(41.911582, -‐87.634803)
Where Are We?
(41.911582, -‐87.634803)
Latitude: Y value
Where Are We?
(41.911582, -‐87.634803)
Latitude: Y value Longitude: X value
Assuming we have a data store of the locations of all
bars in the world…
https://www.flickr.com/photos/shutterrunner/4334879403/“Today. The Green Door Tavern.” Shutter Runner CC BY 2.0
Let’s find a fun place
nearbyto have a beer.
Interview Question:
Write an API to support searching for the nearest bars.
How would you do it?
Right Answer
Google Places API
Sometimes our job is to find ways to not write code.
Let’s rephrase the question.
Interview Question:
Write an API to support searching for the nearest bars
from scratch.
How would you do it?
Binary Search2 4 7 9 11 13 34 37 92 1050 1 2 3 4 5 6 7 8 9
?
?
?Found It!
Everyone used to know binary search.
They just didn’t know what it was called.
(cc) http://www.recyclethis.co.uk(and picked inefficient pivot points)
SOME SORTING REQUIRED.
INSERT SORT ALGORITHM NOW.
Sort by X, then Y!
Well, that breaks.
Which is closer?
United Lily Growers
in Smith River California
~1800 miles
The Green Door Tavern
not that far in Chicago
1.26 miles
or
X gets preferential treatment over Y.
It doesn’t deserve it.
How about brute force?
1. Iterate over every bar. 2. Calculate the distance. 3. Sort by the distance.
1630 bars in Chicago alone.
Is there a right answer?
Was I just a mean interviewer?
Some people got it right.
Physics Majors
In particular…
Physics majors that care about particle collisions.
Divide & Conquer the Space
Divide & Conquer the Space
Divide & Conquer the Space
Divide & Conquer the Space
Rules
Quadrants are not always equal in space.
They can be equal in capacity.
Capacity = Number of Objects
Center Point
(X, Y)
Or Bounding Box
( X1, Y1 )
( X2, Y2 )
Create a Tree of Nested Quadrants
Follow the Rectangles Down
Follow the Rectangles Down
Follow the Rectangles Down
Follow the Rectangles Down
1974
Jon L. Bentley & Raphael Finkel
Quad-Trees
1974
Jon L. Bentley (includes buggy implementation of binary search)
Quad-tree is a very good answer.
It can be generalized to more dimensions.
Octrees
WhiteTimberwolf 2010, Schematic drawing of an octree, a data structure of computer science.
37
37
37
37
37
37
37
37
Binary Search2 4 7 9 11 13 34 37 92 1050 1 2 3 4 5 6 7 8 9
?
?
?Found It!
SOME SORTING REQUIRED.
INSERT SORT ALGORITHM NOW.
I have 2 values. I need 1.
Bit Concatenation
(x, y)
(x3x2x1x0, y3y2y1y0)
x3x2x1x0 y3y2y1y0
(9, 3)
(1001, 0011)
10010011
Bit Interleaving
(x, y)
(x3x2x1x0, y3y2y1y0)
x3y3x2y2x1y1x0y0
(9, 3)
(1001, 0011)
10000111
1966
G.M. Morton
Z Order
3 (0,3) (1,3) (2,3) (3,3)
2 (0,2) (1,2) (2,2) (3,2)
1 (0,1) (1,1) (2,1) (3,1)
0 (0,0) (1,0) (2,0) (3,0)
0 1 2 3
3 1010 10
1011 11
1110 14
1111 15
2 1000 8
1001 9
1100 12
1101 13
1 0010 2
0011 3
0110 6
0111 7
0 0000 0
0001 1
0100 4
0101 5
0 1 2 3
3 1010 10
1011 11
1110 14
1111 15
2 1000 8
1001 9
1100 12
1101 13
1 0010 2
0011 3
0110 6
0111 7
0 0000 0
0001 1
0100 4
0101 5
0 1 2 3
Newly created number determines order.
3 1010 10
1011 11
1110 14
1111 15
2 1000 8
1001 9
1100 12
1101 13
1 0010 2
0011 3
0110 6
0111 7
0 0000 0
0001 1
0100 4
0101 5
0 1 2 3
Newly created number determines order.
start
3 1010 10
1011 11
1110 14
1111 15
2 1000 8
1001 9
1100 12
1101 13
1 0010 2
0011 3
0110 6
0111 7
0 0000 0
0001 1
0100 4
0101 5
0 1 2 3
Newly created number determines order.
start
finish
3101010
101111
111014
111115
210008
10019
110012
110113
1 00102
00113
01106
01117
0 00000
00011
01004
01015
0 1 2 3
3101010
101111
111014
111115
210008
10019
110012
110113
1 00102
00113
01106
01117
0 00000
00011
01004
01015
0 1 2 3
3101010
101111
111014
111115
210008
10019
110012
110113
1 00102
00113
01106
01117
0 00000
00011
01004
01015
0 1 2 3
3101010
101111
111014
111115
210008
10019
110012
110113
1 00102
00113
01106
01117
0 00000
00011
01004
01015
0 1 2 3
3101010
101111
111014
111115
210008
10019
110012
110113
1 00102
00113
01106
01117
0 00000
00011
01004
01015
0 1 2 3
3101010
101111
111014
111115
210008
10019
110012
110113
1 00102
00113
01106
01117
0 00000
00011
01004
01015
0 1 2 3
3101010
101111
111014
111115
210008
10019
110012
110113
1 00102
00113
01106
01117
0 00000
00011
01004
01015
0 1 2 3
3101010
101111
111014
111115
210008
10019
110012
110113
1 00102
00113
01106
01117
0 00000
00011
01004
01015
0 1 2 3
3101010
101111
111014
111115
210008
10019
110012
110113
1 00102
00113
01106
01117
0 00000
00011
01004
01015
0 1 2 3
3101010
101111
111014
111115
210008
10019
110012
110113
1 00102
00113
01106
01117
0 00000
00011
01004
01015
0 1 2 3
3101010
101111
111014
111115
210008
10019
110012
110113
1 00102
00113
01106
01117
0 00000
00011
01004
01015
0 1 2 3
3101010
101111
111014
111115
210008
10019
110012
110113
1 00102
00113
01106
01117
0 00000
00011
01004
01015
0 1 2 3
3101010
101111
111014
111115
210008
10019
110012
110113
1 00102
00113
01106
01117
0 00000
00011
01004
01015
0 1 2 3
3101010
101111
111014
111115
210008
10019
110012
110113
1 00102
00113
01106
01117
0 00000
00011
01004
01015
0 1 2 3
3101010
101111
111014
111115
210008
10019
110012
110113
1 00102
00113
01106
01117
0 00000
00011
01004
01015
0 1 2 3
3101010
101111
111014
111115
210008
10019
110012
110113
1 00102
00113
01106
01117
0 00000
00011
01004
01015
0 1 2 3
(0,0)
(1,1)
space-filling curve
Bit-interleaving can be extended to
multiple dimensions.
from space to graphs…
lost in suburbia
Blue Spruce Dr.
Arborwood Ct.
Elmwood St.
Blue Fir Ln.
Go right.
Go right.
19th CenturyCharles Pierre Trémaux
depth-first search
BREADTH-FIRST SEARCH
BREADTH-FIRST SEARCH1950s
BREADTH-FIRST SEARCH1950s
A
B2
C3
2
D4 2
E
4
1
4
My Example Graph
Open List
A
B2
C3
2
D4 2
E
4
1
4
My Example Graph
Open List
A
B2
C3
2
D4 2
E
4
1
4
My Example Graph
sum of all costs to get here
Open List Closed List
A
B2
C3
2
D4 2
E
4
1
4
My Example Graph
sum of all costs to get here
Open List Closed List
A 0
A
Open List Closed List
A 0
A
B2
C3
D4
Open List Closed List
A 0
A
B2
C3
D4
B 2
C 3
D 4
Open List Closed List
B 2
C 3
D 4
A 0
A
B2
C3
2
D4
E
4
My Example Graph
Open List Closed List
C 3
D 4
E 6
A 0
B 2
A
B2
C3
2
D4
E
4
My Example Graph
Open List Closed List
C 3
D 4
E 6
A 0
B 2
A
B2
C3
2
D4 2
E1
4
My Example Graph
Open List Closed List
D 4
E 4
A 0
B 2
C 3
A
B2
C3
2
D4 2
E1
4
My Example Graph
Open List Closed List
E 4 A 0
B 2
C 3D 4A
B2
C3
2
D4 2
E
4
1
4
My Example Graph
Open List Closed List
E 4 A 0
B 2
C 3D 4A
B2
C3
2
D4 2
E
4
1
4
My Example Graph
1956
Edsger W. Djikstra
Djikstra’s Algorithm
Hamilton Richards - manuscripts of Edsger W. Dijkstra, University Texas at Austin
1956
Edsger W. Djikstra
Djikstra’s Algorithm
Hamilton Richards - manuscripts of Edsger W. Dijkstra, University Texas at Austin
1968
Peter Hart, Nils Nilsson, & Bertram Raphael
A*A
Open List Closed List
A
B2
C3
2
D4 2
E
4
1
4
My Example Graph
sum cost to get here plus distance to goal
0
2
2
13
Open List Closed List
A 3
A
B2
C3
D4
My Example Graph
2
2
13
Open List Closed List
B 4
C 4
D 6
A 3
A
B2
C3
D4
My Example Graph
2
2
13
Open List Closed List
A 3
A
B2
C3
2
D4
E
4
My Example Graph
0
2
2
13
B 4
C 4
D 6
Open List Closed List
A 3
B 4
A
B2
C3
2
D4
E
4
My Example Graph
0
2
2
13
C 4
D 6
E 6
Open List Closed List
A
B2
C3
2
D4 2
E1
4
My Example Graph
0
2
2
13
A 3
B 4
C 4
D 6
E 6
Open List Closed List
A
B2
C3
2
D4 2
E1
4
My Example Graph
0
2
2
13
A 3
B 4
C 4
E 4
D 6
Open List Closed List
A
B2
C3
2
D4 2
E1
4
My Example Graph
0
2
2
13
A 3
B 4
C 4
E 4
D 6
We didn’t have to close D.
Breadth-first Search
Breadth-first Search
Djikstra’s Algorithm
Breadth-first Search
Djikstra’s Algorithm
A*
The power of heuristics.
Breadth-first Search
Djikstra’s Algorithm
A*
Are all edge weights equal?
What’s your most expensive operation?
Do you have insights about your target?
Breadth-First
Djikstra
A*
yes
no
visitshe
uristic
s
?
?
?
breadth-first search f(x) = c
djikstra’s algorithm f(x) = g(x) + c
a* f(x) = g(x) + h(x) + c
The others were specializations.
and we can do more than search with graphs…
1974
Charles L. Forgy
Rete
Premise 1
Premise 2
Premise 3
Conclusion
Rules
Premise 1
Premise 2
Premise 3
Take Action
Rules
(Premise 1, Premise 2, Premise 3)
(Action)
=>
(Premise 1, Premise 2, Premise 3)
(Action)
=>Left Hand Side
(Premise 1, Premise 2, Premise 3)
(Action)
=>Left Hand Side
Right Hand Side
A
B
C
D
ERules are a
directed acyclic graph.
like a trie
C
I OH
I DT
C EY
A K
G E
O N
Root Node
Root Node
Type Nodes
Root Node
Selection Nodes
Type Nodes
Root Node
Negation Nodes
Selection Nodes
Type Nodes
Root Node
Join Nodes
Negation Nodes
Selection Nodes
Type Nodes
Root Node
Terminal Nodes
Join Nodes
Negation Nodes
Selection Nodes
Type Nodes
Root Node
Alpha Memory
Terminal Nodes
Join Nodes
Negation Nodes
Selection Nodes
Type Nodes
Root Node
Beta Memory
Alpha Memory
Terminal Nodes
Join Nodes
Negation Nodes
Selection Nodes
Type Nodes
recommend-diet-changes
Hypertension (severity == moderate)
Demographics (age > 60)
prescribe-bp-meds
Hypertension
HypertensionDemographics
Hypertension (severity == moderate)
Ryan Brush “Retaking Rules for Developers”
Hypertension
Demographics Hypertension (severity == moderate)
Demographics (age > 60)
recommend-diet-changes
prescribe-bp-meds
Ryan Brush “Retaking Rules for Developers”
Sometimes our job is to find ways to not write code.
Let’s not build an expert system!
Let’s not build an expert system!
Let’s not build an expert system!
(-‐>clara)
Let’s not build an expert system!
(-‐>clara)
+
=
(-‐>clara)
+
=
Triskel99 - Own work AMC Pacer, photographed during August 2009.
Efficacy
Cost
Convenience
import java.util.Comparator;
public class ChoiceComparator implements Comparator<Choice> {
@Override public int compare(Choice o1, Choice o2) { if( o2.getEfficacy() == o1.getEfficacy) { if( o2.getCost() == o1.getCost) { return o2.getConvenience() -‐ o1.getConvenience(); } return o2.getCost() -‐ o1.getCost(); } return o2.getEfficacy() -‐ o2.getEfficacy(); } }
import java.util.Comparator;
public class ChoiceComparator implements Comparator<Choice> {
@Override public int compare(Choice o1, Choice o2) { if( o2.getEfficacy() == o1.getEfficacy) { if( o2.getCost() == o1.getCost) { return o2.getConvenience() -‐ o1.getConvenience(); } return o2.getCost() -‐ o1.getCost(); } return o2.getEfficacy() -‐ o2.getEfficacy(); } }
public class EfficacyComparator implements Comparator<Choice>
public class CostComparator implements Comparator<Choice>
public class ConvenienceComparator implements Comparator<Choice>
1.SORT 2.TAKE 10 3.GOTO 1
public class EfficacyComparator implements Comparator<Choice>
public class CostComparator implements Comparator<Choice>
public class ConvenienceComparator implements Comparator<Choice>
1.SORT 2.TAKE 10 3.GOTO 1
1974
Jon L. Bentley & Raphael Finkel
Quad-Trees
1966
G.M. Morton
Z Order
for(int i = 0; i < count1; ++i) { for(int j = 0; j < count2; ++j) { for(int k = 0; k < count3; ++k) { for(int l = 0; l < count4; ++l) { for(int m = 0; m < count5; ++m) { for(int n = 0; n < count6; ++n) { for(int o = 0; o < count7; ++o) { for(int p = 0; p < count8; ++p) { for(int q = 0; q < count9; ++q) { for(int r = 0; r < count10; ++r) { for(int s = 0; s < count 11 for(int t = 0; t < count for(int u = 0; for(int v = 0 for(int w for(i fo
for-nado?
You are using the wrong algorithm.
Crack open an algorithms book.
Do a Depth-First Search of Wikipedia
Do a Depth-First Search of Wikipedia
Your solution is out there.
BLOOM FILTER
trie
Gale Shapley
A*
R-Tree
TREAP
Dijkstra’s
Floyd Warshall
D* Hu
ffm
anID
3
CARTRet
e
And to be honest…
It is probably from the 60s or 70s.
1979Engineering Director, Cerner Corporation Chapter Leader, Kansas City Girl Develop It Conference Organizer, Midwest.io
@michellebrush
Michelle Brush
