Index in position 1 exceeds array bound. Index must not exceeds 1.?

I'm finding this error in Line 124 : dN=[dN dist(v(i,:),A(r,:))]; of my code. Can anyone tell me how to solve this problem.

clc;
clear;
close all;
tic


% Budget Constraint
% The price of each sensor is C = $3000
% The price of each monitor is C' = $122000
% The total available budget is P = $313000

CS = 3000;
CM = 122000;
P = 313000;
ln = 2;      % the minimum no. of monitors


% Population Data
Pop = [44252	74524	85060	66989	94922
    23631	50185	74016	80964	86887
    40666	69841	65646	29660	15504
    29549	21068	9487	2984	2260 
    4267	    2293	2042	2393	1711];

add = sum(sum(Pop)); 
Pop=Pop/add*100;    % Population Percentage

% Emission PM2.5 Data
e = [0.293849555	0.497287887	0.700726218	0.665801844	0.630877469
    0.199782061	0.310923724	0.422065388	0.393195231	0.364325075
    0.105714567	0.124559562	0.143404557	0.120588619	0.09777268
    0.056276604	0.085150863	0.114025122	0.14243416	0.170843197
    0.006838642	0.045742165	0.084645688	0.164279701	0.243913714
    ];

add1 = sum(sum(e)); 
e=e/add1*100;    % Emission PM2.5 Percentage


% The empty set of Grids of point of interest (V)
v=[0 0];  
A=[1 5];
s=[0 0];
B=[0 0];
H=[0 0];
t=size(Pop);
% The set of V Grids and their locations
for i=1:t(1)        
    for j = 1:t(2)
        v(end+1,:)=[i j];
        s(end+1,:)=[i j];
    end
end
% This is to remove first row of v & s
v(1,:)=[];
s(1,:)=[];
B(1,:)=[];
H(1,:)=[];
% For A = empty set, d = infinite, g(d) =0

% This for loop is for averaging population and emission percentage
% for each grid 
m=zeros(25,1);
for i = 1:length(v)
    b=v(i,1:2);
    m(i,:) = m(i,:) + (Pop(b(1),b(2))+e(b(1),b(2)))/2;
end
m;
M =[4.6314 7.8192 10.0010 8.7974 9.9390
    2.8197 5.0719 7.1852 7.3060 7.3746
    2.9277 4.5674 4.5058 2.4869 1.5808
    1.9613 1.7624 1.4054 1.3036 1.4963
    0.2728 0.4867 0.7884 1.4500 2.0590];


% For A = empty set, d = infinite, g(d) =0
% This nested for loop is to find the 1st position 
% where sensor can be deployed

f1=zeros(25,1);
for j = 1:length(s)
    for i = 1:length(v)
        ds=dist(v(i,:),s(j,:));
        dA=dist(v(i,:),A(1,:));
        d=min([ds dA]);
        b=v(i,1:2);
        f1(j,:) = f1(j,:) + exp(-d/1)*M(b(1),b(2)) - exp(-dA/1)*M(b(1),b(2));
    end 
end

Unable to perform assignment because the size of the left side is 1-by-1 and the size of the right side is 1-by-2.

f1   % f(A) matrix
[fmax,grid_p] = max(f1)   % find the maximum of f(A)
B = [B;2 3]; 

% This nested for loop is to find the 2nd position 
% where sensor can be deployed

f2=zeros(25,1);
for j = 1:length(s)
    for i = 1:length(v)
        ds=dist(v(i,:),s(j,:));
        dA1=dist(v(i,:),A(1,:));
        dB1=dist(v(i,:),B(1,:));
        d=min([ds dA1 dB1]);
        dA=min([dA1 dB1]);
        b=v(i,1:2);
        f2(j,:) = f2(j,:) + exp(-d/1)*M(b(1),b(2)) - exp(-dA/1)*M(b(1),b(2));
    end
end
f2     % f(A) matrix
[fmax,grid_p] = max(f2)   % find the maximum of f(A)
B = [B;1 2];

% To find next sensors locations


for ii=1:23
    f=zeros(25,1);
    if (CS*size(A) + CM*size(B)) <= P
        for j=1:length(s)
            for i=1:length(v)
                dN=[];
                ds=dist(v(i,:),s(j,:));
                for r =1:11
                    dN=[dN dist(v(i,:),A(r,:))];
                end
                dB1=dist(v(i,:),B(1,:));
                dB2=dist(v(i,:),B(2,:));
                d=min([ds dN dB1 dB2 dA1]);
                dA=min([dN dB1 dB2 dA1]);
                b=v(i,1:2);
                f(j,:) = f(j,:) + exp(-d/1)*M(b(1),b(2)) - exp(-dA/1)*M(b(1),b(2));
            end
        end
        f    % f(A) matrix
        [fmax,grid_p] = max(f)   % find the maximum of f(A)
        
        if mod(grid_p,5)==0
            A = [A;quorem(sym(grid_p), sym(5)) 5];
        else
            A = [A;quorem(sym(grid_p), sym(5))+1 mod(grid_p,5)];
        end
    else 
        break
    end
    f =[];
end

[Dyuman Joshi]

There's another error occuring before that, see above.

As for the error, the message clearly states what's wrong.

As for the code, what are you trying to do? What is the objective?

I don't understand what you are trying to do. For e.g -

You define B and H like this

B=[0 0];
H=[0 0];

and then you delete them

B(1,:)=[];
H(1,:)=[];

What's the point of initializing then?

[Nishant Ajnoti]

Objective is to find the location where sensors and monitors can be deployed and the code works on greedy algorithm. "B" is defined separately because it will store the locations for monitors and "A" for location of sensors. H should be in comments, it was for different purpose but I deleted them, didn't delete from initializing.

Answer 1

[Pooja Kumari]

Hi Nishant,

It is my understanding that you are facing “Index in position 1 exceeds array bounds. Index must not exceed 1.” error in line 124 of the provided code.

As Dyuman Joshi pointed out, another error occurring before line 124 at line 84 which is “Unable to perform assignment because the size of the left side is 1-by-1 and the size of the right side is 1-by-2.” happening due to the size of ‘dA’ which is of size 1 x 2. This error can be resolved by taking one value of ‘dA’. I have replaced ‘dA’ with ‘dA(2)’ in line 84.

Updated code of line 84 is given below:

f1(j,:) = f1(j,:) + exp(-d/1)*M(b(1),b(2)) - exp(-dA/1)*M(b(1),b(2)); 

The error occurring in line 124 is due to value of ‘r’. “Index in position 1 exceeds array bound as the value of ‘r’ is 11 and ‘A’ is 1 dimensional row vector with 2 values. The error indicates that the index ‘r’ exceeds the size of array ‘A’.

Corrected code for error in line 124 is given below:

for r =1:2
    dN=[dN dist(v(i,:),A(:,r))];  % A is a row vector with 2 column
end

A(r,:) is the error because ‘A’ is a row vector with 2 values, either value of ‘r’ can be reduced to 2 or vector ‘A’ can be redefined with 11 values.

I hope this helps!