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Problem 1
Reconsider the US air pollution data set:
Loading required package: tools
── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
✔ dplyr 1.1.4 ✔ readr 2.1.5
✔ forcats 1.0.0 ✔ stringr 1.5.1
✔ ggplot2 3.5.2 ✔ tibble 3.3.0
✔ lubridate 1.9.4 ✔ tidyr 1.3.1
✔ purrr 1.1.0
── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
✖ dplyr::filter() masks stats::filter()
✖ dplyr::lag() masks stats::lag()
ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errors
library (dplyr)data (USairpollution)
A)
Perform singular value decomposition of this data matrix. Then create the matrix \(D\) . Describe what this matrix looks like.
<- svd (USairpollution)<- components$ d<- matrix (diag (D),nrow= 7 ,ncol= 7 )
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] 7051.949 0.0000 0.000 0.00000 0.00000 0.0000 0.0000
[2,] 0.000 931.1211 0.000 0.00000 0.00000 0.0000 0.0000
[3,] 0.000 0.0000 540.463 0.00000 0.00000 0.0000 0.0000
[4,] 0.000 0.0000 0.000 92.70909 0.00000 0.0000 0.0000
[5,] 0.000 0.0000 0.000 0.00000 85.23724 0.0000 0.0000
[6,] 0.000 0.0000 0.000 0.00000 0.00000 52.9465 0.0000
[7,] 0.000 0.0000 0.000 0.00000 0.00000 0.0000 10.1409
The matrix looks like a diagonal matrix with all values aligned across the diagonal line and 0 filling the rest of the space.
B)
Verify that \(X=UDV^T\) by plotting all the entries of \(X\) versus all the entries of \(UDV^T\) with the 0/1 line.
<- 7 <- components$ d<- components$ u<- components$ v<- (U %*% diag (D) %*% t (V)) %>% round (1 )<- U[,1 : k] %*% diag (D[1 : k]) %*% t (V[,1 : k])) %>% round (1 )
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] 46 47.6 44 116 8.8 33.4 135
[2,] 11 56.8 46 244 8.9 7.8 58
[3,] 24 61.5 368 497 9.1 48.3 115
[4,] 47 55.0 625 905 9.6 41.3 111
[5,] 11 47.1 391 463 12.4 36.1 166
[6,] 31 55.2 35 71 6.5 40.7 148
[7,] 110 50.6 3344 3369 10.4 34.4 122
[8,] 23 54.0 462 453 7.1 39.0 132
[9,] 65 49.7 1007 751 10.9 35.0 155
[10,] 26 51.5 266 540 8.6 37.0 134
[11,] 9 66.2 641 844 10.9 35.9 78
[12,] 17 51.9 454 515 9.0 12.9 86
[13,] 17 49.0 104 201 11.2 30.8 103
[14,] 35 49.9 1064 1513 10.1 31.0 129
[15,] 56 49.1 412 158 9.0 43.4 127
[16,] 10 68.9 721 1233 10.8 48.2 103
[17,] 28 52.3 361 746 9.7 38.7 121
[18,] 14 68.4 136 529 8.8 54.5 116
[19,] 14 54.5 381 507 10.0 37.0 99
[20,] 13 61.0 91 132 8.2 48.5 100
[21,] 30 55.6 291 593 8.3 43.1 123
[22,] 10 61.6 337 624 9.2 49.1 105
[23,] 10 75.5 207 335 9.0 59.8 128
[24,] 16 45.7 569 717 11.8 29.1 123
[25,] 29 43.5 699 744 10.6 25.9 137
[26,] 18 59.4 275 448 7.9 46.0 119
[27,] 9 68.3 204 361 8.4 56.8 113
[28,] 31 59.3 96 308 10.6 44.7 116
[29,] 14 51.5 181 347 10.9 30.2 98
[30,] 69 54.6 1692 1950 9.6 39.9 115
[31,] 10 70.3 213 582 6.0 7.1 36
[32,] 61 50.4 347 520 9.4 36.2 147
[33,] 94 50.0 343 179 10.6 42.7 125
[34,] 26 57.8 197 299 7.6 42.6 115
[35,] 28 51.0 137 176 8.7 15.2 89
[36,] 12 56.7 453 716 8.7 20.7 67
[37,] 29 51.1 379 531 9.4 38.8 164
[38,] 56 55.9 775 622 9.5 35.9 105
[39,] 29 57.3 434 757 9.3 38.9 111
[40,] 8 56.6 125 277 12.7 30.6 82
[41,] 36 54.0 80 80 9.0 40.2 114
plot (X_back, Xtilde);abline (0 ,1 )cor (as.vector (X_back), as.vector (Xtilde))
C)
Consider low-dimensional approximations of the data matrix. What is the fewest number of dimensions required to yield a correlation between the entries of \(X\) and \(\tilde X\) of at least 0.9?
<- 2 <- U[,1 : k2] %*% diag (D[1 : k2]) %*% t (V[,1 : k2])) %>% round (1 )
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] 3.5 23.8 17.2 151.5 3.6 15.2 42.1
[2,] 5.9 35.8 45.7 244.5 5.5 22.8 63.7
[3,] 17.3 37.8 352.3 516.8 6.2 24.6 73.1
[4,] 30.4 63.4 629.6 900.0 10.5 41.3 123.5
[5,] 17.1 32.9 363.0 498.0 5.5 21.5 64.8
[6,] 2.4 20.0 0.5 115.4 3.0 12.7 35.1
[7,] 131.5 80.9 3371.0 3334.4 17.1 56.6 208.0
[8,] 18.2 20.7 433.8 488.4 3.8 13.8 44.9
[9,] 34.5 -2.1 962.8 807.5 1.1 0.1 15.2
[10,] 16.2 56.9 260.0 548.2 9.0 36.6 104.9
[11,] 29.3 50.8 643.7 839.4 8.6 33.3 101.6
[12,] 19.2 26.4 443.0 528.5 4.6 17.4 54.9
[13,] 6.3 26.2 86.1 223.8 4.1 16.8 47.7
[14,] 50.9 96.7 1088.6 1481.9 16.2 63.2 190.8
[15,] 11.3 -15.3 362.9 221.1 -1.8 -9.2 -19.7
[16,] 38.8 98.3 746.5 1199.8 16.0 63.6 186.9
[17,] 22.1 73.6 368.8 736.7 11.7 47.4 136.3
[18,] 13.5 71.1 138.3 526.0 11.0 45.4 127.3
[19,] 17.7 35.8 369.9 520.6 6.0 23.4 70.1
[20,] 4.6 18.8 65.6 163.8 2.9 12.1 34.2
[21,] 17.7 60.9 289.1 595.9 9.6 39.2 112.5
[22,] 19.2 58.8 336.5 624.1 9.4 37.9 109.7
[23,] 11.0 36.3 183.1 364.5 5.8 23.4 67.3
[24,] 25.5 43.8 559.9 728.1 7.4 28.7 87.7
[25,] 28.5 32.6 680.9 766.8 5.9 21.8 70.5
[26,] 14.5 42.3 261.2 465.3 6.8 27.3 79.3
[27,] 11.4 38.9 186.7 382.2 6.2 25.0 71.8
[28,] 8.3 44.1 83.6 324.4 6.8 28.2 78.9
[29,] 10.6 37.2 171.2 359.4 5.9 23.9 68.6
[30,] 71.6 81.4 1712.1 1924.8 14.8 54.4 176.3
[31,] 15.9 62.2 233.5 555.9 9.8 39.9 113.6
[32,] 17.4 45.1 331.6 541.4 7.3 29.2 85.6
[33,] 10.4 -4.0 301.8 233.9 -0.2 -2.1 -1.1
[34,] 10.0 30.3 177.0 324.6 4.8 19.5 56.6
[35,] 6.4 17.3 118.4 200.1 2.8 11.2 32.7
[36,] 23.2 53.9 462.5 703.5 8.8 35.0 103.6
[37,] 18.2 44.1 358.0 558.0 7.2 28.6 84.3
[38,] 27.5 4.6 745.0 660.4 1.8 4.0 22.6
[39,] 23.7 65.8 438.8 751.3 10.6 42.5 123.9
[40,] 8.1 33.2 115.3 288.9 5.2 21.3 60.5
[41,] 3.4 13.3 49.9 118.9 2.1 8.5 24.3
plot (X_back, Xtilde);abline (0 ,1 )cor (as.vector (X_back), as.vector (Xtilde))
The fewest number of dimensions to yield a correlation of at least 0.9 is 2 dimensions.
D)
Find \(\Sigma\) , the covariance matrix of this data set. Then perform eigen-decomposition of this matrix. Verify that
The eigenvectors of \(\Sigma\) equal the columns of \(V\)
The eigenvalues of \(\Sigma\) equal the diagonals of \(D^2/(n-1)\)
Attaching package: 'MASS'
The following object is masked from 'package:dplyr':
select
<- (USairpollution%>% scale (center= TRUE , scale = FALSE )t (US_df)%*% US_df/ (41-1 )
SO2 temp manu popul wind
SO2 550.947561 -73.560671 8527.7201 6711.9945 3.1753049
temp -73.560671 52.239878 -773.9713 -262.3496 -3.6113537
manu 8527.720122 -773.971341 317502.8902 311718.8140 191.5481098
popul 6711.994512 -262.349634 311718.8140 335371.8939 175.9300610
wind 3.175305 -3.611354 191.5481 175.9301 2.0410244
precip 15.001799 32.862988 -215.0199 -178.0529 -0.2185311
predays 229.929878 -82.426159 1968.9598 645.9860 6.2143902
precip predays
SO2 15.0017988 229.92988
temp 32.8629884 -82.42616
manu -215.0199024 1968.95976
popul -178.0528902 645.98598
wind -0.2185311 6.21439
precip 138.5693840 154.79290
predays 154.7929024 702.59024
SO2 temp manu popul wind
SO2 550.947561 -73.560671 8527.7201 6711.9945 3.1753049
temp -73.560671 52.239878 -773.9713 -262.3496 -3.6113537
manu 8527.720122 -773.971341 317502.8902 311718.8140 191.5481098
popul 6711.994512 -262.349634 311718.8140 335371.8939 175.9300610
wind 3.175305 -3.611354 191.5481 175.9301 2.0410244
precip 15.001799 32.862988 -215.0199 -178.0529 -0.2185311
predays 229.929878 -82.426159 1968.9598 645.9860 6.2143902
precip predays
SO2 15.0017988 229.92988
temp 32.8629884 -82.42616
manu -215.0199024 1968.95976
popul -178.0528902 645.98598
wind -0.2185311 6.21439
precip 138.5693840 154.79290
predays 154.7929024 702.59024
<- svd (US_df)<- eigen (cov (US_df))$ vectors
[,1] [,2] [,3] [,4] [,5]
[1,] 0.0168607518 -0.099835625 0.208775573 0.95883106 -0.152191203
[2,] -0.0011417794 0.025814390 -0.071600745 -0.11014784 -0.477854201
[3,] 0.6968327936 -0.710249079 -0.067182201 -0.07319788 -0.009643654
[4,] 0.7170284512 0.692912523 0.056666935 0.04906669 0.010735457
[5,] 0.0004067530 -0.001011680 0.005386606 -0.01506609 0.025401917
[6,] -0.0004336922 0.001225937 0.265807619 -0.16261712 -0.832729325
[7,] 0.0028836950 -0.069155051 0.934279828 -0.18459052 0.232812295
[,6] [,7]
[1,] -0.053952911 -2.704138e-02
[2,] -0.852534945 -1.640507e-01
[3,] -0.002153023 1.136208e-03
[4,] 0.002915751 -5.682006e-05
[5,] 0.176541256 -9.838347e-01
[6,] 0.453206155 6.376788e-02
[7,] -0.183568735 -1.891454e-02
[,1] [,2] [,3] [,4] [,5]
[1,] -0.0168607518 0.099835625 0.208775573 -0.95883106 0.152191203
[2,] 0.0011417794 -0.025814390 -0.071600745 0.11014784 0.477854201
[3,] -0.6968327936 0.710249079 -0.067182201 0.07319788 0.009643654
[4,] -0.7170284512 -0.692912523 0.056666935 -0.04906669 -0.010735457
[5,] -0.0004067530 0.001011680 0.005386606 0.01506609 -0.025401917
[6,] 0.0004336922 -0.001225937 0.265807619 0.16261712 0.832729325
[7,] -0.0028836950 0.069155051 0.934279828 0.18459052 -0.232812295
[,6] [,7]
[1,] -0.053952911 -2.704138e-02
[2,] -0.852534945 -1.640507e-01
[3,] -0.002153023 1.136208e-03
[4,] 0.002915751 -5.682006e-05
[5,] 0.176541256 -9.838347e-01
[6,] 0.453206155 6.376788e-02
[7,] -0.183568735 -1.891454e-02
[1] 6.384720e+05 1.481204e+04 7.019599e+02 2.050019e+02 1.167047e+02
[6] 1.205705e+01 1.448704e+00
[1] 6.384720e+05 1.481204e+04 7.019599e+02 2.050019e+02 1.167047e+02
[6] 1.205705e+01 1.448704e+00
Problem 2
In this problem we explore how “high” a low-dimensional SVD approximation of an image has to be before you can recognize it.
.Rdata objects are essentially R workspace memory snapshots that, when loaded, load any type of R object that you want into your global environment. The command below, when executed, will load three objects into your memory: mysteryU4, mysteryD4, and mysteryV4. These are the first \(k\) vectors and singular values of an SVD I performed on a 700-pixels-tall \(\times\) 600-pixels-wide image of a well-known villain.
load ('Data/mystery_person_k4.Rdata' )
A)
Write a function that takes SVD ingredients u, d and v and renders the \(700 \times 600\) image produced by this approximation using functions from the magick package. Use your function to determine whether a 4-dimensional approximation to this image is enough for you to tell who the mystery villain is. Recall that you will likely need to rescale your recomposed approximation so that all pixels are in [0,1].
Linking to ImageMagick 6.9.13.29
Enabled features: cairo, fontconfig, freetype, heic, lcms, pango, raw, rsvg, webp
Disabled features: fftw, ghostscript, x11
<- mysteryU4<- mysteryD4<- mysteryV4<- 4 <- U2[,1 : k3] %*% diag (D2[1 : k3]) %*% t (V2[,1 : k3])<- max (recompose_mystery)- min (recompose_mystery)<- (recompose_mystery- min (recompose_mystery))/ R%>% as.raster%>% image_read
B)
I’m giving you slightly higher-dimensional approximations (\(k=10\) and \(k=50\) , respectively) in the objects below:
load ('Data/mystery_person_k10.Rdata' )load ('Data/mystery_person_k50.Rdata' )
Create both of the images produced by these approximations. At what point can you tell who the mystery villain is?
library (magick)<- mysteryU10<- mysteryD10<- mysteryV10<- 10 <- U10[,1 : k4] %*% diag (D10[1 : k4]) %*% t (V10[,1 : k4])<- max (recompose_mystery2)- min (recompose_mystery2)<- (recompose_mystery2- min (recompose_mystery2))/ R2%>% as.raster%>% image_read
library (magick)<- mysteryU50<- mysteryD50<- mysteryV50<- 50 <- U50[,1 : k5] %*% diag (D50[1 : k5]) %*% t (V50[,1 : k5])<- max (recompose_mystery3)- min (recompose_mystery3)<- (recompose_mystery3- min (recompose_mystery3))/ R3%>% as.raster%>% image_read
I could make a very good inference at 10 dimensions but the 50 dimensions is a much better image and confirms what I was thinking at 10 dimensions.
C)
How many numbers need to be stored in memory for each of the following:
A full \(700\times 600\) image?
\((700\times 600 = 420000)\)
A 4-dimensional approximation?
\(k = 4\) \((700\times 4 + 4 + 4^2 = 2820)\)
A 10-dimensional approximation?
\(k = 10\) \((700\times 10 + 10 + 10^2 = 7110)\)
A 50-dimensional approximation?
\(k = 50\) \((700\times 50 + 50 + 50^2 = 37550)\)
