how can i check if n2 is a sub number of n1? (HW qst.)

for ex.:

n1 = 123; n2 = 12; (n1>=n2)

n2 is a sub number of n1

i can't make any use of strings or arrays

[Image Analyst]

What is the class of n1 and n2? Strings, number??? What does this say:

whos n1

I'm not sure you can do it if you can use strings or arrays. Is a scalar (single number but not an array of 2 or more number) allowed? Also define subnumber. Is 2 a subnumber of 4 because it's a factor of 4? Or do the digits have to be in order in the larger number, like 22 and 42 are subnumbers of 4422, even though they may not be integer factors of it.

[Zaza]

the class of n1 and n2 is double

scalar is allowed

the digits have to be in order in the larger number, like 22 and 42 are subnumbers of 4422, even though they may not be integer factors of it.

[Sean de Wolski]

Not using strings, this is not trivial (at least I haven't come up with a trivial way to do it yet, maybe a conversion to binary and comparison? or dynamic programming with mod, log10?).

[Zaza]

i suppose it's connected somehow with mod fun.

[Zaza]

suppose i have

n1 = 12345;

how can i omit the first (in the highest power of 10) digit to get

n1 = 2345;

and so on

in this way if n2 is a sub number of n1 it'll match...

[Azzi Abdelmalek]

The highest power of 10 correspond to 1 not 5

[Azzi Abdelmalek]

check this

2345-fix(2345/10^3)*10^3

then

345-fix(345/10^2)*10^2

and so on

Answer 1

[Sean de Wolski]

Here is a hint:

a = 131245;
b = 12;
op = @(x)hidden_for_you_to_figure_out(x);
pa = op(a);
pb = op(b);
isASub = false; %guilty until proven innocent
for ii = pb:pa
  amii = mod(a,(10^ii));
  if amii == b
      isASub = true;
      return;
  else
      for jj = 1:(ii-pb)            
          afjj = floor(amii./(10^jj));            
          if b == afjj
              isASub = true;
              return;
          end
      end
  end
end

All you have to do it figure out the contents of op().

[Zaza]

op() == num_of_digits() ???

Answer 2

[Azzi Abdelmalek]

n1=12456;
n2=245
a=num2str(n1);
b=num2str(n2);
c=all(ismember(b,a))

if c is equal to 1 that means n2 is a sub number of n1

[Zaza]

i'm not sure about num2str because strings are not allowed in this problem

[Azzi Abdelmalek]

Why?

[Zaza]

these are the requirements

[Azzi Abdelmalek]

I will give you hints:

a=124 ;a1=1, a2=2, a3=4

to find a3:

 a3=fix(a/10^2); %(an=fix(a/10^(n-1))

In your previous question the number of digit in an integer number is

int=-123456789
digit_number=max(ceil(log10(abs(int))),1)

[Zaza]

can i use floor instead of fix?

[Azzi Abdelmalek]

Yes you can

Answer 3

[Walter Roberson]

Create a function that calculates the decimal expansion of a number in reverse order. mod() 10 gives the next digit to be output, integer division by 10, if the result is non-zero keep going.

Apply that to both numbers.

Now match the reversed n2 to the reversed n1.