Return true if the elements of the input vector increase monotonically (i.e. each element is larger than the previous). Return false otherwise.

Return true if the elements of the input vector increase monotonically (i.e. each element is larger than the previous). Return false otherwise. this is my code

function tf = mono_increase(x)
% we need to check every number 
L=length(x);
% we nust use for loop 
% then we must use if statement 
if L==1 
    tf=true;   % adding a condition for single element 
    return %we must close the case for faster operation
end 
for ii = 1:(L-1)
    if x(ii+1) > x(ii)  % we have problems in the -ve numbers
        tf= true ;
    else
        tf=false  ;  
    end 
end

it doesn't work for negative numbers like : input x= [ -3 -4 1 2 4] the output must be false but I get true I think the wrong is in the indexing

Answer 1

[Jeffrey Eymer]

for ii = 1:(L-1)
    if x(ii+1) > x(ii)  % we have problems in the -ve numbers
        tf= true ;
    else
        tf=false  ;  
    end 
end

Look at this portion of the code. For each ii in 1:(L-1), this portion of code gets called. This means that the value of tf is being updated 'L' number of times (i.e., when checking every pair of elements).

For example, the last two elements in [ -3 -4 1 2 4] are 2 and 4, which are monotonically increasing. Therefore, on the last iteration the value of tf is set to true and is not affected by the rest of the input.

You should try rewriting your code in a way where the value of tf is only updated to false when an issue is found.

[Ahmed Diaa]

get it ..thanks I edit return to fix the problem

[D. Plotnick]

You really want to try to avoid loops, especially for relatively simple operations like this. In this case, you can use a break to stop at the first non-monotonic jump, but the code is way clunkier and a bit slower.

 % Set up spaces
x1 = linspace(0,1,1000001).'; % Column it
x2 = rand(1000001,1);
% Using any
tic;
dx1 = diff(x1);
tf1 = ~any(dx1<=0)
dx2 = diff(x2);
tf2 = ~any(dx2<=0)
toc

 % Versus loops:
 tic
tf1 = true;
tf2 = true;
for ii = 1:length(x1)-1
    if x1(ii) >= x1(ii+1)
        tf1 = false;
        break
    end
end
tf1
for ii = 1:length(x2)-1
    if x2(ii) >= x2(ii+1)
        tf2 = false;
        break
    end
end
tf2
toc

The any method requires just 2 lines to execute.

Note too that using any you can also return the indices of the bad numbers

 function [tf,inds] = ismonotonic(x)
dx = diff(x);
inds = dx<=0;
tf = ~any(inds);
end

Answer 2

[D. Plotnick]

One idea: dx = diff(x); tf = ~any(dx<=0);

Note this will also work on matrices, if you want. E.g. to look at each row

x = {some numbers}; dx = diff(x,2); tf = ~any(dx<=0,2);

Here, tf will return a column of logicals.