2014_10_15_lecture_23

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EGR 115 Introduction to Computing for
Engineers
Loops and Vectorization – Part 2
Wednesday 15 Oct 2014
EGR 115 Introduction to Computing for Engineers
Lecture Outline
• Testing the Random Number Generator “rand”
• Vectorization
Wednesday 15 Oct 2014
EGR 115 Introduction to Computing for Engineers
Slide 2 of 18
Testing the Random Num Generator “rand”
• Problem: How can we test the “randomness” of the
random number generator “rand” in MATLAB?
• Solution: Develop a histogram of the random
numbers generated!!
 Generate N total random integers each between 1 and 10
 Count how many times each of the numbers appear
 Plot the result as a bar chart
o
Scale the count to total 100%
 Plot the result as a pie chart
Wednesday 15 Oct 2014
EGR 115 Introduction to Computing for Engineers
Slide 3 of 18
Testing the Random Num Generator “rand”
• PseudoCode:
 Prompt the user for N = # of random nums to be generated
 Initialize a histogram to store the 10 sums
 In a loop: 1:N
o
o
Generate a random integer (1 to 10)
Update the histogram (i.e., appropriate sum)
 Scale the histogram to total 100%
 Plot the result as a bar chart
 Plot the result as a pie chart
o
Explode the most frequently occurring integer
RandTest.m courtesy of Hilken, Tanner R. <[email protected]>
Wednesday 15 Oct 2014
EGR 115 Introduction to Computing for Engineers
Slide 4 of 18
Testing the Random Num Generator “rand”
• Results obtained for N = 500 points
Random Number Generator Test: N = 500 points
Random Number Generator Test: N = 500 points
12
8%
10%
9%
10
Freequency of Occurance (%)
12%
8
10%
6
9%
4
9%
11%
2
11%
0
11%
1
2
3
4
5
6
7
Value Generated
Wednesday 15 Oct 2014
8
9
10
EGR 115 Introduction to Computing for Engineers
Slide 5 of 18
1
2
3
4
5
6
7
8
9
10
Loops and Vectorization
Preallocation of Arrays
• MATLAB allows arrays to grow dynamically.
a = 1:4;
for i = 1:6
a(i) = i^2;
end
a=
1
2
3
4
Loop 1: i =1  a(1) = 1^2 = 1
a=
1
2
3
4
Loop 2: i =2  a(2) = 2^2 = 4
a=
1
4
3
4
Loop 3: i =3  a(3) = 3^2 = 9
a=
1
4
9
4
Loop 4: i =4  a(4) = 4^2 = 16
a=
1
4
9
16
Loop 5: i =5  a(5) = 5^2 = 25
a=
1
4
9
16
25
Loop 6: i =6  a(6) = 6^2 = 36
a=
1
4
9
16
25
Wednesday 15 Oct 2014
EGR 115 Introduction to Computing for Engineers
36
Slide 6 of 18
Loops and Vectorization
Preallocation of Arrays
• Dynamically resizing arrays is VERY slow!!
 Pre-allocate array sizes appropriately
N = 1000;
a = zeros(1, N);
for i = 1:N
a(i) = i^2;
end
Initialization pre-allocates
the size of the array “a”
Non-Preallocated Array Case
Preallocated Array Case
for i = 1:10000000
x(i) = i;
end
x=1:10000000;
Elapsed time = 1.9668 sec
Elapsed time = 0.023375 sec
Wednesday 15 Oct 2014
EGR 115 Introduction to Computing for Engineers
84 Times
FASTER!!
Slide 7 of 18
Loops and Vectorization
Vectorization
• MATLAB natively “understands” arrays
 Operating on vector objects is fast
 E.g., Compute: a  Sin  3t  t 2 , t  0 :106 :1
Non-Vectorized Array Case
N =
a =
i =
for
1000000;
zeros(1, N);
0;
t = 0:1/N:1
i = i+1;
a(i) = sin(3*t)*t^2;
Vectorized Array Case
N = 1000000;
t = 0:1/N:1;
a = sin(3*t).*t.^2;
198 Times
FASTER!!
end
Elapsed time = 1.249 sec
Wednesday 15 Oct 2014
Elapsed time = 0.0063078 sec
EGR 115 Introduction to Computing for Engineers
Slide 8 of 18
Loops and Vectorization
The break and continue statements
• What if you want to exit a for or while loop or skip
back to the top of the loop?
• break -> exits the loop to statement after end
 If the break statement is in a nested loop – control jumps
the next loop level.
 E.g., break will not kick you all the way out of nested loops
– just the current level.
• continue -> jumps to end and loops again (unless
finished).
 Statements between continue and end are not executed!
Wednesday 15 Oct 2014
EGR 115 Introduction to Computing for Engineers
Slide 9 of 18
Loops and Vectorization
The break and continue statements
• The following code uses both continue and
break.
for i=1:10
if (i > 3) && (i <= 5)
continue
elseif i == 8
break
end
fprintf('This is loop %d\n',i);
end
Loop 1: This is loop 1
Loop 2: This is loop 2
Loop 3: This is loop 3
Loop 4: - continue Loop 5: - continue Loop 6: This is loop 6
Output:
This is loop 1
This is loop 2
This is loop 3
This is loop 6
This is loop 7
Wednesday 15 Oct 2014
Loop 7: This is loop 7
Loop 8: - break -
EGR 115 Introduction to Computing for Engineers
Slide 10 of 18
Loops and Vectorization
Quiz
• What is the result from executing the following?
1.
for k = 8:10
fprintf('k = %g \n',k);
end
2.
for j = 8:-1:10
fprintf(' j = %g \n',j);
end
3.
k=8
k=9
k = 10
NOTHING!!
for l = 1:10:10
fprintf(' l = %g \n',l);
end
i=1
4.
for i = -10:3:-7
fprintf(' i = %g \n',i);
end
i = -10
i = -7
5.
for m = [0 2 -3]
fprintf(' m = %g \n',m);
end
m=0
m=2
m = -3
Wednesday 15 Oct 2014
EGR 115 Introduction to Computing for Engineers
Slide 11 of 18
Loops and Vectorization
Quiz
• What is the value of “x” at the end of each loop?
1.
2.
3.
x = 0;
for m = 1:5
x = x + 1;
end
x = 0;
for m = 1:5
x = x + m;
end
x=5
x = 15
x = 1;
for m = 1:5
if m == 2
continue;
elseif x > 8
break;
end
x = x + m;
fprintf(' m = %g & x = %g \n', m, x);
end
fprintf(' End: x = %g \n', x);
Wednesday 15 Oct 2014
EGR 115 Introduction to Computing for Engineers
m=1&x=2
m=3&x=5
m=4&x=9
End: x = 9
Slide 12 of 18
Loops and Vectorization
An Example - Fitting noisy data to a line
• Example: Fitting noisy data to a line
 Eqn. of a Line: y = m x + c
 Fit the following measured data to a line
Noisy Data
12
10
noisy data
8
y
Speed of a DC
motor as the
voltage was
increased
6
4
Least
Squares Fit?
2
0
0
1
2
3
4
5
6
7
8
9
10
11
x
Wednesday 15 Oct 2014
EGR 115 Introduction to Computing for Engineers
Slide 13 of 18
Loops and Vectorization
An Example - Fitting noisy data to a line
• Problem:
 Given ‘n’ noisy data points [𝑥𝑖 ] and [𝑦𝑖 ], determine the “best”
choice of 𝑚 and 𝑐 such that the error between 𝑦𝑖 = 𝑚𝑥𝑖 + 𝑐
and [𝑦𝑖 ] is minimized.
e    yi  yˆi     yi  m xi  c 
2
e


m m
e 

c c
   y  mx  c  
2
i
i
2
 2  yi  mxi  c  xi  0
c   xi   m   xi2    yi xi
   y  mx  c  
2
i
i
 2  yi  mxi  c   0
nc  m   xi    yi
Wednesday 15 Oct 2014
EGR 115 Introduction to Computing for Engineers
Slide 14 of 18
Loops and Vectorization
An Example - Fitting noisy data to a line
nc  m   xi    yi
 n n xxi icc  yyii  
x x 2 mmx y 
i    i i 
 i
 c   yi 
M 


x
y
m
   i i
c   xi   m   xi2    yi xi
c  n
 m    x
   i
m
1
xi   yi 
xi2  xi yi 
xi yi  nxi yi
 xi 
Wednesday 15 Oct 2014
2
 nx
2
i
c
M
1
 yi 
 x y 
 i i
xi2 yi  xi xi yi
nx   xi 
2
i
EGR 115 Introduction to Computing for Engineers
2
Slide 15 of 18
Loops and Vectorization
An Example - Fitting noisy data to a line
• Given the data:
x = [ 0.17, 1.02, 2.13, 3.13, 4.22, 5.12, 6.10, 7.17, 8.19, 9.13, 10.09];
y = [ 0.95, 3.34, 2.79, 4.10, 7.55, 6.70, 7.96, 10.07, 9.89, 12.16, 11.74];
c  n
 m    x
   i
1
xi   yi 
xi2  xi yi 
1
1.29 
56.47   77.25 
 11
56.47 400.29  519.74   1.12 



 

yˆi  m xi  c
yˆi  1.29 xi  1.12
Wednesday 15 Oct 2014
EGR 115 Introduction to Computing for Engineers
Slide 16 of 18
Loops and Vectorization
An Example - Fitting noisy data to a line
• Least Square fit of “noisy” data to a line
Least-Square fit of Noisy Data to a Line
12
10
noisy data
LS fit
LS_line_fit.m
y
8
6
4
2
0
0
1
2
3
4
5
6
7
8
9
10
11
x
Wednesday 15 Oct 2014
EGR 115 Introduction to Computing for Engineers
Slide 17 of 18
Next Lecture
• Nested Loops
• Profiling
Wednesday 15 Oct 2014
EGR 115 Introduction to Computing for Engineers
Slide 18 of 18
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