linear algebra in a computational setting alan kaylor cline deans scholars seminar january 22, 2014
DESCRIPTION
After examining the code you believe that the running time depends entirely upon some input parameter n and … a good model for the running time is Time(n) = a + b·log 2 (n) + c·n + d·n·log 2 (n) where a, b, c, and d are constants but currently unknown.TRANSCRIPT
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Linear Algebrain a
Computational Setting
Alan Kaylor Cline
Dean’s Scholars SeminarJanuary 22, 2014
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How long does it take for this code to run?
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After examining the code you believe that the running time depends entirely upon some input
parameter n and …
a good model for the running time is
Time(n) = a + b·log2(n) + c·n + d·n·log2(n)
where a, b, c, and d are constants but currently unknown.
![Page 4: Linear Algebra in a Computational Setting Alan Kaylor Cline Deans Scholars Seminar January 22, 2014](https://reader033.vdocuments.us/reader033/viewer/2022060206/5a4d1ba37f8b9ab0599c8473/html5/thumbnails/4.jpg)
So you time the code for 4 values of n, namely n = 10, 100, 500, and 1000
and you get the times
According to the model you then have 4 equations in the 4 unknowns a, b, c, and d:
a + b·log2(10) + c·10 + d·10·log2(10) = 0.685a + b·log2(100) + c·100 + d·100·log2(100) = 7.247a + b·log2(500) + c·5000 + d·500·log2(500) = 38.511a + b·log2(1000) + c·1000+ d·1000·log2(1000) = 79.134
Time(10) = 0.685 ms.Time(100) = 7.247ms.Time(500) = 38.511ms.Time(1000) = 79.134 ms.
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These equations are linear in the unknowns a, b, c, and d.
a + b·log2(10) + c·10 + d·10·log2(10) = 0.685a + b·log2(100) + c·100 + d·100·log2(100) = 7.247a + b·log2(500) + c·5000 + d·500·log2(500) = 38.511a + b·log2(1000) + c·1000+ d·1000·log2(1000) = 79.134
We solve them and obtain: a = 6.5
b = 10.3 c = 57.1
d = 2.2
So the final model for the running time is
Time(n) = 6.5 + 10.3·log2(n) + 57.1·n + 2.2·n·log2(n)
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and now we may apply the model Time(n) = 6.5 + 10.3·log2(n) + 57.1·n + 2.2·n·log2(n)
for a particular value of n (for example, n = 10,000)to estimate a running time of
Time(10,000) = 6.5 + 10.3·log2(10,000) + 57.1· 10,000 + 2.2· 10,000 ·log2(10,000) = 863.47 ms.
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Picture from MarsOct. 1, 1976
One approach to the coloring of the images was to adjust manually the colors in the various patches.
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Picture from MarsOct. 1, 1976
One approach to the coloring of the images was to adjust manually the colors in the various patches.
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Picture from MarsOct. 1, 1976
One approach to the coloring of the images was to adjust manually the colors in the various patches. This is what resulted:
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Picture from MarsOct. 1, 1976
The other approach was to solve 6 x 6 linear systems for each pixel. The systems were based on the filters being used.
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Picture from MarsOct. 1, 1976
The other approach was to solve 6 x 6 linear systems for each pixel. The systems were based on the filters being used.This is what resulted:
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Picture from MarsOct. 1, 1976
Compare the two:
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But have we ignored something?
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0 100 200 300 400 500 600 700 800 9000
2
4
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10
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18x 10
5
So you time the code for 30 values of n, and you get these times {(ni,ti)}
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If the model was perfect and there were no errors in the timings
then for some values a, b, c, d, and e:
a + b·log2(ni) + c·ni + d·ni·log2(ni) +e·ni2 = ti
for i =1,…,30
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But the model was not perfect and there were error in the timings
So we do not expect to get any values a, b, c, d, and e so that:
a + b·log2(ni) + c·ni + d·ni·log2(ni) +e·ni2 = ti
for i =1,…,30
We will settle for values a, b, c, d, and e so that:
a + b·log2(ni) + c·ni + d·ni·log2(ni) +e·ni2 ti
for i =1,…,30
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Our sense of
a + b·log2(ni) + c·ni + d·ni·log2(ni) +e·ni2 ti
for i =1,…,30
Will be to get a, b, c, d, and e so that sum of squares of all of the differences
(a + b·log2(ni) + c·ni + d·ni·log2(ni) +e·ni2 - ti)2
is minimized over all possible choices of a, b, c, d, and e
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After solving the least squares system to get the best values of a, b, c, d, and e, we plot
a + b·log2(n) + c·n + d·n·log2(n) + e·n2
0 100 200 300 400 500 600 700 800 900 10000
0.5
1
1.5
2
2.5x 10
6
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What’s a “good” solutionwhen we don’t have the exact
solution?
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What’s a “good” solutionwhen we don’t have the exact
solution?
“Hey. That’s not a question that was discussed in other math classes.”
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What’s a “good” solutionwhen we don’t have the exact
solution?
.780 𝑥+ .563 𝑦= .217
.913 𝑥+ .659 𝑦=.254
Consider the two equations:
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Consider two approximate solution pairs:
.780 𝑥+ .563 𝑦= .217
.913 𝑥+ .659 𝑦=.254
and these two equations:
𝑥1𝑦 1
¿¿.999−1.001
𝑥2𝑦 2
¿¿.341−0.087
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Consider two approximate solution pairs:
.780 𝑥+ .563 𝑦= .217
.913 𝑥+ .659 𝑦=.254
and these two equations:
𝑥1𝑦 1
¿¿.999−1.001
𝑥2𝑦 2
¿¿.341−0.087
Which pair of these two is better?
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Important fact to consider:
𝑥=1𝑦=−1
The exact solution is:
𝑥1𝑦 1
¿¿.999−1.001
𝑥2𝑦 2
¿¿.341−0.087
Which pair of these two is better?
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Consider two approximate solution pairs:
.780 𝑥+ .563 𝑦= .217
.913 𝑥+ .659 𝑦=.254
and these two equations:
𝑥1𝑦 1
¿¿.999−1.001
𝑥2𝑦 2
¿¿.341−0.087
Which pair of these two is better?
![Page 26: Linear Algebra in a Computational Setting Alan Kaylor Cline Deans Scholars Seminar January 22, 2014](https://reader033.vdocuments.us/reader033/viewer/2022060206/5a4d1ba37f8b9ab0599c8473/html5/thumbnails/26.jpg)
Important fact to consider:
.780 𝑥+ .563 𝑦= .217
.913 𝑥+ .659 𝑦=.254
𝑥1𝑦 1
¿¿.999−1.001
𝑥2𝑦 2
¿¿.341−0.087
Which pair of these two is better?
.780 𝑥1+.563 𝑦1=.215757
.913 𝑥1+.659 𝑦 1= .252428
.780 𝑥2+.563 𝑦 2=.216999.913𝑥 2+.659 𝑦 2=.254
Recall we are trying to solve:
For the first pair, we have:
For the second pair, we have:
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Important fact to consider:
𝑥1𝑦 1
¿¿.999−1.001
𝑥2𝑦 2
¿¿.341−0.087
Which pair of these two is better?
![Page 28: Linear Algebra in a Computational Setting Alan Kaylor Cline Deans Scholars Seminar January 22, 2014](https://reader033.vdocuments.us/reader033/viewer/2022060206/5a4d1ba37f8b9ab0599c8473/html5/thumbnails/28.jpg)
Student: “Is there something funny about
that problem?”
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Student: “Is there something funny about
that problem?”
Professor: “You bet your life. It looks innocent but it
is very strange. The problem is knowing when you have a strange case on
your hands.”
CLINE
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Professor: “Geometrically, solving equations is like
finding the intersections of lines.”CLINE
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here’s the intersection?
When lines have no thickness …
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where’s the intersection?
… but when lines have thickness …
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25.96 miles
Galveston Island
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25.96 miles
Galveston Island
Where’s the intersection?
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London Olympics Swimming
• http://www.youtube.com/watch?v=fFiV4ymEDfY&feature=related
• 1:19
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How do you transform this image …
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How do you transform this image …
into the coordinate system of another image?
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and in greater generality, transform
3-dimensional objects
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The $25 Billion Eigenvector
How does Google do Pagerank?
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The $25 Billion Eigenvector
How did Google do Pagerank?
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The Imaginary Web Surfer:
• Starts at any page,• Randomly goes to a page linked from the
current page,• Randomly goes to any web page from a
dangling page,• … except sometimes (e.g. 15% of the time),
goes to a purely random page.
![Page 42: Linear Algebra in a Computational Setting Alan Kaylor Cline Deans Scholars Seminar January 22, 2014](https://reader033.vdocuments.us/reader033/viewer/2022060206/5a4d1ba37f8b9ab0599c8473/html5/thumbnails/42.jpg)
A tiny web: who should get the highest rank?
J A B
I C
DH
G F E
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The associated stochastic matrix:
0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.4400 0.0150 0.0150 0.2983 0.4400 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.2983 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.2983 0.8650 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.4400 0.0150 0.0150 0.8650 0.0150 0.8650 0.0150 0.0150 0.0150 0.0150 0.0150 0.2983 0.0150 0.0150 0.8650 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.8650 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.8650 0.2983 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.2983 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.4400 0.0150 0.0150 0.0150
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We seek to find a vector x so that
A x = x
One way is to start with some initial x0,and then:
for k = 1, 2, 3,…xk = A xk-1
This converges to an x so that A x = x
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1 20
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0.35
Start with equal components
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1 20
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One iteration
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1 20
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Two iterations
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1 20
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Three iterations
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1 20
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Four iterations
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1 20
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Five iterations
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1 20
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Six iterations
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1 20
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Seven iterations
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1 20
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Eight iterations
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1 20
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Nine iterations
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1 20
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Ten iterations
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The Eigenvector
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[U,G] = surfer ('http://cns.utexas.edu/honors/honors-programs-center/deans-scholars', 100)
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0
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40
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70
80
90
100
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Pagerank Power Iteration the limit
0 20 40 60 80 100 1200
0.02
0.04
0.06
0.08
0.1
0.12
0.14Page Rank
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And the winners are…
'http://www.utexas.edu''http://www.utexas.edu/maps''http://www''http://m.utexas.edu''http://www.utexas.edu/emergency''http://www.lib.utexas.edu' 'http://www.utexas.edu/parking/transportation/shuttle''http://healthyhorns.utexas.edu''http://www.utexas.edu/know/feed''http://www.utexas.edu/know''http://cns.utexas.edu'
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How much storage to hold this array?
• Current estimate of indexed WWW:4.7 · 1010 web pages
• If placed into an array this would have2.21 · 1021 elements
• If each element is stored in 4 bytes, this would be 8.8 · 1022 bytes
• Current estimate of world’s data storage capacity is 3.0 · 1018 bytes (.003% of necessary space)
http://www.smartplanet.com/blog/thinking-tech/what-is-the-worlds-data-storage-capacity/6256
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How much time to do one power step?
• Current estimate of indexed WWW:4.7 · 1010 web pages
• If placed into an array this would have2.21 · 1021 elements
• Fastest current machine does 33.86 · 1015
operations per second• One step of y = Ay takes 3.68 days
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Go Home Now.