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

2 次查看（过去 30 天）
sabitri
sabitri 2022-8-16
评论： sabitri 2022-8-16
KEY=[0 0 1 0 3 12 8 7 7 8 12 3 0 1 0 0]

请先登录，再回答此问题。

采纳的回答

KSSV
KSSV 2022-8-16
Ran in:
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 个评论
显示 1更早的评论
Walter Roberson
Walter Roberson 2022-8-16
编辑：Walter Roberson 2022-8-16
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.

请先登录，再进行评论。

更多回答（2 个）

Abderrahim. B
Abderrahim. B 2022-8-16
编辑：Abderrahim. B 2022-8-16
Ran in:
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 2022-8-16
Ran in:
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.

类别

在 Help Center 和 File Exchange 中查找有关 Matrices and Arrays 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by