how to find all integers between two integers
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Write a function called int_col that has one input argument, a positive integer n that is greater than 1, and one output argument v that is a column vector of length n containing all the positive integers smaller than or equal to n, arranged in such a way that no element of the vector equals its own index. In other words, v(k) is not equal to k for any valid index k.
2 Comments
John D'Errico
on 2 May 2015
Ok, so you managed to type in your homework assignment. Good for you. Of maybe you just did a copy and paste.
Why not make an effort? After all if you cannot do so, you will never learn MATLAB. Learning involves using the language. Making an effort, even if you don't succeed, you learn by trying. This is what homework is for. If we do it for you, then all you learn is how to con someone else into doing your work for you. And while that may be laudable for some people, it is not so here.
So try it out. Start writing. And when you DO have a question when you get stuck, THEN ask it, and show what you tried, and explain what failed. You will probably get some help then.
Accepted Answer
John D'Errico
on 2 May 2015
Ok, so your solution works when n is an even integer. For an odd integer, the middle element of that string will cause it to fail, because it stays unchanged in the permutation you have chosen.
You might also choose a random permutation. Something like this:
vec = 1:n;
ind = 1:n;
while any(vec == ind)
vec = vec(randperm(n));
end
The problem is this might take a while. Have you ever seen the birthday paradox? This is actually a variation of that paradox, in that you will get hits surprisingly often for even moderately sized values of n. (A nice question is how large does n need be so that you will see a problem in more than 50% of the random samplings that you take? What value of n assures a hit over 99% of the time?)
But here is a simple solution that will never fail. Suppose you took the last number, and put it first? So, for example,
[6 1 2 3 4 5]
or
[3 1 2]
You get the idea. Just do a circular shift on the vector. Is there a function in MATLAB that does such a thing? Try this:
lookfor circular
Does it show any functions in MATLAB that will do a circular shift?
Of course, this is simple enough to do without a built-in function. So you might try this:
[n,1:(n-1)]
3 Comments
Adwaith G
on 16 Jun 2022
function v = int_col(n)
v = circshift((1:n)',1); % Transposed it , to make it a column vector and then shifted by 1
Is this correct or not? (I am very knew to MATLAB , so i am sorry for any dumb comments or doubts
More Answers (5)
Image Analyst
on 2 May 2015
Hint: read up on the colon operator, fliplr(), the == operator, and the ' (transpose) operator (to turn row vectors into column vectors). Also, read this: http://www.mathworks.com/matlabcentral/answers/8626-how-do-i-get-help-on-homework-questions-on-matlab-answers The rem() function may come in handy to determine if n is even or odd.
0 Comments
Muhammad Usman Saleem
on 9 May 2015
Edited: Muhammad Usman Saleem
on 9 May 2015
@alisha ali , take my codes and guide where is the need for correction.. thanks in advance for assistance
function v=int_col(n)
v=1:n;
for i=1:n
if v(i)==i
v(i)=[];
end
end
end
end
1 Comment
John D'Errico
on 9 May 2015
This code will not actually satisfy the requirement that all n integers will be in the list, just in a permuted order.
In fact, what this code will do is delete the FIRST element, then leave the rest of them unchanged in order, but shifted. (Look carefully at what it does. Think about it.) Therefore, your vector of integers will never have the number 1 in it.
Remember that the requirement is for a vector that contains "all the positive integers smaller than or equal to n". So deleting integers that fail is not the correct idea. Look instead at my answer.
srinivas
on 11 May 2015
function [v]=int_col(n)
if n<3
v=[2;1];
else
v=1:n;
x=v(3:n);
x1=transpose(x);
k1=[1;2];
y=cat(1,x1,k1);
v=y;
end
end
Thomas Nguyen
on 5 Apr 2018
function[v] = int_col(n)
%Const:
%n=some_int_here;
v=size(n);
for i=1:n-1
v(i+1)=i;
end
end
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