Write a script that asks for an integer (n) and then computes the following based on the value of the integer: While the value of n is greater than 1, replace the integer with half of its value (n/2) if the integer is even. Otherwise, replace the integer with three times its value, plus 1 (3*n + 1).
Make provision to count the number of values in (or the length of) the sequence that results.
Example calculation: If n = 10, the sequence of integers is 5, 16, 8, 4, 2, 1 and so the length is 6.
((Make a plot of the length of the sequence that occurs as a function of the integers from 2 to 30. For example, when n = 10, the length is 6 while for n = 15, the length is 17. Is there any pattern? Is there any integer for which the sequence does not terminate?))
Hi,
how I can plot this if I have a function(calculate) and the output is the length
for i=2:30
p = calculate(i)
plot(i,p)
end
is that correct ??
You should really just run it and see, but you have options to make it work:
Store each p generated in the loop, and plot after the loop.
Plot each point (not connected with lines) inside the loop with plot and hold on.
This is fundamental. Read the plot documentation, please.
Related
i wrote this code for randomize and round numbers
x=3+5*rand(3,4);
for n=1:3
for m=1:4
y(n,m)=round(x(n,m));
end
end
y
Using the randperm() function may be an option. The first argument of randperm() sets the range. In this case randperm(6,4) will generate 4 numbers that are within the range 1 to 6 (a random permuatation of integers in this case permutations of 6). If we add 2 to this result we can generate an array of length 4 that will have values ranging from 3 to 8. Here we can use one for-loop and generate the rows upon each iteration.
Array = zeros(3,4);
for Row = 1: 3
Array(Row,:) = randperm(6,4) + 2;
end
Array
First of all the rand() function returns numbers between 0 and 1, so it probably doesn't make sense to use this function and then round the numbers off. If you're looking for random integers use randi() instead.
With this in mind, the following code produces a 3 by 4 matrix filled with random integers and no repeats:
maxInteger = 12; %change to any number greater than 3x4 = 12
y = randi(maxInteger, 3, 4);
used = [];
for i = 1:numel(y)
while sum(find(used == y(i)))>0
y(i) = randi(maxInteger);
end
used = [used, y(i)];
end
If the while loop takes too long (as might happen with large matrices) consider filling a matrix by pulling and removing elements from a predecided list of integers.
I want to only use "randi" this function to produce the 6 different number randomly in matlab ,and the range of these 6 number is 1 ~ 12.
number=randi([1,12],1,6)
c=0;
for n=1:6%when "n" is 1 to 6
while c <= 6 %while c is less equal than 6,do the program below
c = c + 1; %c=c+1
if number(n) == number(c) %when the nth element is equal to cth element
number(n) = randi(12); %produce a random integer in the nth element
c = 0; %the reason why i set c=0 again is because i want to check again whether the new random integer is the same as cth element or not
end
end
end
final_number=number
but the result still show me like
1 "2" 6 11 "2" 3
5 "8" "8" 12 3 1
How do i improve my code to produce 6 different numbers.i don't want to always rely on the convenient matlab instruction too much,so my tags will also write c.hoping someone can help me to improve this
If you're trying to reproduce randsample (or randperm), why not just reproduce the algorithm MATLAB uses? (As far as we can tell...)
This is the Fisher-Yates shuffle. If you have a vector v, each iteration selects a random, previously unused element and puts it at the end of the unselected elements. If you do k iterations, the last k elements of the list are your random sample. If k equals the number of elements in v, you've shuffled the entire array.
function sample = fisher_yates_sample(v, k)
% Select k random elements without replacement from vector v
% if k == numel(v), this is simply a fisher-yates shuffle
for n = 0:k-1
randnum = randi(numel(v)-n); % choose from unused values
% swap elements v(end-n) and v(randnum)
v([end-n, randnum]) = v([randnum, end-n]);
end
sample = v(end-k+1:end);
end
Unlike MATLAB's version, mine requires a vector as input, so to get 6 random values in the range 1:12 you'd call the function like this:
>> fisher_yates_sample(1:12,6)
ans =
5 11 6 10 8 4
Since you're re-selecting single random numbers, when there is one occuring multiple times, why not just re-selecting all numbers at once?
% Initial selecting of random numbers.
number = randi([1, 12], 1, 6)
% While the amount of unique elements in numbers is less than 6:
while (numel(unique(number)) < 6)
% Re-select random numbers.
number = randi([1, 12], 1, 6)
end
And since you wrote, you specifically want to use the randi method, I guess there is a reason, you don't want to use randperm(12, 6)!?
What you are looking for is randperm. It produces a random permutation of a range of integers, so that if you select the first k numbers, you are sure that you get k unique integers in the range [1;n].
In your case, simply call:
randperm(12,6)
So I'm trying to figure out why my code doesn't seem to be displaying the properly uniformed quantized image into 4 levels.
Q1 =uint8(zeros(ROWS, COLS, CHANNELS));
for band = 1 : CHANNELS,
for x = 1 : ROWS,
for y = 1 : COLS,
Q1(ROWS,COLS,CHANNELS) = uint8(double(I1(ROWS,COLS,CHANNELS) / 2^4)*2^4);
end
end
end
No5 = figure;
imshow(Q1);
title('Part D: K = 4');
It is because you are not quantifying. You divide a double by 16, then multiply again by 16, then convert it to uint8. The right way to quantize is to divide by 16, throw away any decimals, then multiply by 16:
Q1 = uint8(floor(I1 / 16) * 16);
In the code snippet above, I assume I1 is a double. Convert it to double if its not: I1=double(I1).
Note that you don't need the loops, MATLAB will apply the operation to each element in the matrix.
Note also that if I1 is an integer type, you can do something like this:
Q1 = (uint8(I1) / 16) * 16;
but this is actually equivalent to replacing the floor by round in the first example. This means you get an uneven distribution of values: 0-7 are mapped to 0, 8-23 are mapped to 16, etc. and 248-255 are all mapped to 255 (not a multiple of 16!). That is, 8 numbers are mapped to 0, and 8 are mapped to 255, instead of mapping 16 numbers to each possible multiple of 16 as the floor code does.
The 16 in the code above means that there will be 256/16=16 different grey levels in the output. If you want a different number, say n, use 256/n instead of 16.
It's because you are using ROWS, COLS, CHANNELS as your index, it should be x,y,band. Also, the final multiplication of 2^4 has be after the uint8 cast otherwise no rounding ever takes place.
In practice you should avoid the for loops in Matlab since matrix operations are much faster. Replace your code with
Q1=uint8(double(I1/2^4))*2^4
No5 = figure;
imshow(Q1);
title('Part D: K = 4');
I want to find the smallest integer P, such that the number of primes in the set {1,2,..., P} is less than P/6.
I think have the answer via (long) trial and error but would like to know how to verify this through MATLAB.
You can use isprime to check if any value in an array is a prime number. If we want to check all integers up until the integer N we can do
% You can change this to the maximum number that you'd like to consider for P
N = 2000;
possible_P_values = 2:N; % We omit 1 here since it's not a prime number
primes = isprime(possible_P_values);
To determine how many primes have occured up to a given integer N we can use cumsum of this logical matrix (the cumulative sum)
nPrimes_less_than_or_equal_to_P = cumsum(primes);
Then we can divide possible_P_values by 6 and check where the number of primes up to a certain point is less than that number.
is_less_than_P_over_6 = nPrimes_less_than_or_equal_to_P < (possible_P_values ./ 6);
Then we can identify the first occurance with find
possible_P_values(find(is_less_than_P_over_6, 1, 'first'))
% 1081
How can I get the numeric values in a column (let's say column 10) when the numeric values in another column (let's say column 9) are equal to a specific number and plot this in a graph.
e.g., When values of column 9 == 4, get the corresponding value of column 10 and plot. I am using row index number as a marker for time.
I am plotting all of column 10 to get a waveform then I want to use the data of column 9 to add markers to my waveform that are representative of a command occurring at a certain point in time.
Here is my code:
E = csvread('Experiment_at_10_45_1.csv');
[signal_rows, signal_columns] = size(E);
t=(1:signal_rows)/128; %128 samples per second
%% SNR plot for down frequency
plot(t,E(:,13),'k')
I hope my explanation is clear, as I have attempted to use a minimum working example of my code for the first time.
You'll want to use logical indexing to do this. You want to first create an array of 0 (false) and 1 (true) values where column 9 is equal to the value you want.
bool = E(:,9) == 4;
Then you'll want to use this 0 and 1 array as the row index. This will grab only the rows where column 9 was equal to 4. This is referred to as logical indexing.
E(bool, 10)
Then you can plot this
plot(t(bool), E(bool, 10))
As pointed out though, it is possible that the values aren't exactly to 4 due to floating point representation. To get around this, you just want to check if they are "close enough" using a very small epsilon.
bool = abs(E(:,9) - 4) < 1e-12;