I HAVE A KEY WHICH IS 1*16 MATRIX. HOW CAN I DETERMINE IT'S BIT SIZE?

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KEY=[0 0 1 0 3 12 8 7 7 8 12 3 0 1 0 0]

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KSSV
KSSV el 16 de Ag. de 2022
KEY=[0 0 1 0 3 12 8 7 7 8 12 3 0 1 0 0] ;
a = single(KEY) ;
iwant = whos('a')
iwant = struct with fields:
name: 'a' size: [1 16] bytes: 64 class: 'single' global: 0 sparse: 0 complex: 0 nesting: [1×1 struct] persistent: 0
  3 comentarios
Walter Roberson
Walter Roberson el 16 de Ag. de 2022
Editada: Walter Roberson el 16 de Ag. de 2022
Will this key prevent brute force attack
NO it will not prevent brute force attack.
There is no known way of preventing brute force attacks.
There are some famous cases where particular encryption challenges were broken by way of contests that organized thousands of computers on the Internet to keep trying sequential possibilities.
You have 4 bits per entry and 16 entries, for a total of 64 bits. It is accepted that the NSA can brute force 64 bit DES encryption keys, using their custom-built hardware.
sabitri
sabitri el 16 de Ag. de 2022
Okay...It’s clear to me now..Thanks

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Más respuestas (2)

Abderrahim. B
Abderrahim. B el 16 de Ag. de 2022
Editada: Abderrahim. B el 16 de Ag. de 2022
Hi!
Try this:
KEY = [0 0 1 0 3 12 8 7 7 8 12 3 0 1 0 0] ;
KEY = single(KEY) ;
% WHOS returns a structure
S = whos("KEY") ;
S.class
ans = 'single'
% Bit is an eighth of a byte
bitSize = S.bytes * 8
bitSize = 512
Hope this helps

Walter Roberson
Walter Roberson el 16 de Ag. de 2022
KEY=[0 0 1 0 3 12 8 7 7 8 12 3 0 1 0 0]
KEY = 1×16
0 0 1 0 3 12 8 7 7 8 12 3 0 1 0 0
bits_required_per_entry = max( ceil(log2(KEY)) )
bits_required_per_entry = 4
However the calculation changes if any entry might be negative.

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