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

Asked by Ahmed Diaa

### Ahmed Diaa (view profile)

on 28 Jun 2018
Latest activity Edited by D. Plotnick

### D. Plotnick (view profile)

on 28 Jun 2018
Accepted Answer by Jeffrey Eymer

### Jeffrey Eymer (view profile)

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

### Products

Answer by Jeffrey Eymer

### Jeffrey Eymer (view profile)

on 28 Jun 2018

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

### Ahmed Diaa (view profile)

on 28 Jun 2018
get it ..thanks I edit return to fix the problem
D. Plotnick

### D. Plotnick (view profile)

on 28 Jun 2018
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 by D. Plotnick

### D. Plotnick (view profile)

on 28 Jun 2018
Edited by D. Plotnick

### D. Plotnick (view profile)

on 28 Jun 2018

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.