Binary Numbers what is the solution V2 [closed]

问题

Does anyone know how I can solve this problem ? any help would be great...... i cant seem to get my head around it.

As you know Binary can only be either 1 or 0

Say you had a 8 digit Binary number like a byte >>>>>> 0001 1000

Im trying to figure out an equation or what would be the maximum amount of combinations you could get from an 8 digit number

What i mean is.... say you had a two digit binary number,the maximum Bnary combinations that you could have are either

11
Therefore total maximum combinations from a 2 digit Binary number = 4

Example 2

If you had a 3 digit number , maximum Binary Combinations would be

000

001

010

100

101

111

110

011
Therefore total maximum Binary combinations from a 3 digit number = 8

Example 3

If it were a 4 digit number, maximum binary combinations that you could have are either

0000

0001

0010

0100

1000

0111

0110

1111

1110

1101

1011

1001 Total maximum combination = 12

I Recently Asked this Question and it was answered thank you manu-fatto and zgnilec they were kind enough to let me know it was a simple equation the answer/equation is 2^digit size.

I guess my next problem is how do i write a small program that can show these combinations in Xcode or NSLog. I'm good with objective C an output I can view like NSLog would be great.

All I know is it would look something like:

int DigitSize=8
int CombinationTotal = 2^8
CombinationSize = NSMutableArray  ArraywithCapacity 8;

Output

NSString  Combination1 =@"0000 0000";
NSString  Combination2 =@"0000 0001";
NSString  Combination3 =@"0000 0010";

Nslog @"combination 1 = %@  ,Combination1";
Nslog @"combination 2 = %@  ,Combination2";
Nslog @"combination 3 = %@  ,Combination3";

……

Nslog @"combination 256 = ???? ???? ";

Sorry for the vague language I only started learning programming 3 months ago and I still have a lot of tutorials to go through.

**Im trying to build a data compression algorithm...

basically data compression is about reducing the number of bits ... the lesser the bits the smaller a file is

A file with 700bits is smaller than a file with 900bits

8 bits = 1 byte

1024bytes = 1kb

1024kb = 1 mb

i donno if its even possible but i just thought what if you had an algorithm that could read 1024 bits at a time ...with the equation thats = 2^1024 = math error :( == total number of bit combinations possible

Once you have the total number of combinations you set each combination to a symbol like eg 000101010010101011001011011010101010140010101101000000001110100101100001010100000......0011010 = symbol #

So from now on whenever the computer sees the symbol # it recognises it is equal to the binary number 000101010010101011001011011010101010140010101101000000001110100101100001010100000......0011010

to better understand it ...just think of number plates on a car/vehicle, they are only a few characters but wen you punch them into police database or any car data base more information comes out its the same principle....

basically the symbols are a key to more data

i dont know if it make sense but... in theory if you could read 8388608 bits at a time

8388608 bits = 1megabyte ......

ten symbols could mean 10mb...you could create digital media 2d barcodes its just a thought i had watching starGate lol :)**

回答1:

2 to the power of 8, where 8 is number of digits.

Edit- only read first question :)

create function that will display an integer as binary

for (i = 0; i < pow(2,n), i++)
{
    displayBits(i);
}

回答2:

A quick implementation

#import <Foundation/Foundation.h>

int main(int argc, const char * argv[])
{

    @autoreleasepool {
        NSUInteger length = 8;  // number of digits
        NSUInteger n = pow(2, length); // number of possible values
        for (int i = 0; i < n; i++) {
            NSString *repr = @"" ;
            for (int j = 0; j < length; ++j) {
                if([repr length] % 5 == 0)
                    repr = [@" " stringByAppendingString:repr]; // add a blank after evey 4th digit
                int x =( i >> j) &1;
                repr = [[NSString stringWithFormat:@"%u",x] stringByAppendingString:repr];
            }
            NSLog(@"%@", repr);                
        }
    }
    return 0;
}

Output

The core of this program is this:

for (int i = 0; i < n; i++) {
    //…
    for (int j = 0; j < length; ++j) {
        int x =( i >> j) &1;
        //…
    }
}

this will run for i =0 bis (2^n)-1 and in the inner for-loop for every j of the n bits, to check, if the least bit is 1, and append it to the representation string.

As you are a beginner, you probably dont know, what this means: int x =( i >> j) & 1;

>> shifts the bits of the left side integer by as many decimal places as the right side defines. and & 1 performs a bit wise addition

so for i == 3 and n == 8

       3 as binary                         string representation
j = 0: 00000011 >> 0 ->  0000 0011 
                        &0000 0001
                        -----------
                        00000 0001 -> 1     repr = 1

j = 1: 00000011 >> 1 ->   000 0001 
                        &0000 0001
                        -----------
                        00000 0001 -> 1     repr = 11

j = 2: 00000011 >> 2 ->    00 0000
                        &0000 0001
                        -----------
                        00000 0000 -> 0     repr = 011

j = 3: 00000011 >> 3 ->     0 0000 
                        &0000 0001
                        -----------
                        00000 0000 -> 0     repr = 0011

  (the same till j = 7)                     repr = 0000 0011

来源：https://stackoverflow.com/questions/14927833/binary-numbers-what-is-the-solution-v2

标签

objective-c

binary

binary-data

equation