finding if number(iteration) is the sum of 2 numbers inside same array - matlab

hello,

im trying to write a code that will check every number inside a 1 dimention array ( loop),

and check if the number can be summed by 2 numbers inside the array.

thanks in advance.

[EDITED] ill try to clarify.

given an array = [ 20 40 50 60 80 100]

if one of the element is the sum of 2 other element, print 1/ true

if it does not, than print 0/false

in this instance it should be true for 2 occurnaces, (20+40 = 60) and (20 + 80 = 100)

thanks in advance

[Sam Chak]

@Oren Savir

Can you define what the number cannot be summed by 2 numbers inside the array?

More specifically, is the problem related to the Fibonacci numbers?

If true, then it becomes a problem of checking whether each element in the array is a Finonacci number or not...

Answer 1

[Sam Chak]

Hi @Oren Savir

This is just a very simple example. You can turn it into a loop to check each element in the array.

% The array
X = [1 1 2; 3 5 8; 13 21 34]
X = transpose(X);

% Reshape array into vector
A = reshape(X, 1, [])

% Generate fibonacci sequence
B = fibonacci(1:length(A))

% Check true/false (1 means true, 0 means false)
tf = isequal(A, B)

X =

1 1 2

3 5 8

13 21 34

A =

1 1 2 3 5 8 13 21 34

B =

1 1 2 3 5 8 13 21 34

tf =

logical

1

Answer 2

[Jan]

x = [20 40 50 60 80 100];

n      = numel(x);          % Number of elements
index  = nchoosek(1:n, 2);  % Pairs of elements
sumx   = sum(x(index), 2);  % Sum of pairs
sumx   = unique(sumx);      % Cheaper to ignore multiple values
result = ismember(x, sumx)  % TRUE if element is part of sumx

result = 1×6 logical array
   0   0   0   1   1   1

Answer 3

[Voss]

Here's something that may be close to what you want to do:

X = [20 40 50 60 80 100];

ismember(X,sum(X(nchoosek(1:numel(X),2)),2))

ans = 1×6 logical array
   0   0   0   1   1   1

That tells you that 20, 40, and 50 cannot be made by adding two distinct elements in X, but 60, 80, and 100 can.

Breaking down that expression, step-by-step:

disp(nchoosek(1:numel(X),2)) % generate all pairs of distinct indexes into X

     1     2
     1     3
     1     4
     1     5
     1     6
     2     3
     2     4
     2     5
     2     6
     3     4
     3     5
     3     6
     4     5
     4     6
     5     6

disp(X(nchoosek(1:numel(X),2))) % index into X using those indexes, i.e., get the elements from X

    20    40
    20    50
    20    60
    20    80
    20   100
    40    50
    40    60
    40    80
    40   100
    50    60
    50    80
    50   100
    60    80
    60   100
    80   100

disp(sum(X(nchoosek(1:numel(X),2)),2)) % add each pair of elements

    60
    70
    80
   100
   120
    90
   100
   120
   140
   110
   130
   150
   140
   160
   180

disp(ismember(X,sum(X(nchoosek(1:numel(X),2)),2))) % see which elements of X are in the summed-pairs array

   0   0   0   1   1   1

Note that this method is not "the sum of two other elements"; it is "the sum of any two elements" (really, the sum of elements at any two indices). So if you had a zero in X, then every non-zero element would be counted because it is itself+0.

X = [20 40 0 60 80 100];
disp(ismember(X,sum(X(nchoosek(1:numel(X),2)),2)))

   1   1   0   1   1   1

and if you had more than one 0 element, then all elements would always be counted:

X = [20 40 0 60 80 100 0];
disp(ismember(X,sum(X(nchoosek(1:numel(X),2)),2)))

   1   1   1   1   1   1   1

(I point it out because it's not clear what your expected result would be in those cases, based on one example with no repeated elements and no zero elements.)