# Binary Counter / Flipper

Miling Shah

Greenhorn

Posts: 6

posted 3 years ago

- 0

My task is to print an n-bit(n is accepted as input from user) binary sequence as :

For n=4,

Output:

0000

0001

0010

0011

0100

0101

0110

0111

1000

1001

1010

1011

1100

1101

1110

1111

I found three ways to do this:

Algo 1. Run a loop for i=0 to 2^(n), step i by 1

print(Dec2Bin(i,n));

where Dec2Bin is a function that converts a decimal i into a String of size n (if possible without data value damage/loss )

Code for this algorithm:

Algo 2. Create a bit array

Check bits from right (LSB) to left (MSB).

if bitvalue @ index == 0, flip bit @ index to 1

else if bitvalue @ index == 1, recursively check for left bits,

as terminal case.. if bit @ index=0 is 0 n all bits to the right are 1

flip bit at index=0 to 1 and flip all the bits to the right to 0

I tried implementing this algo, but it didnt work well,, it gives me the same output for n>4, as that given by n=4.

I would be glad if someone could help me with debugging/fixing this piece of code.

My Code :

Algo 3. I just thought that maybe every bit 0 or 1 can be calculated dynamically and directly displayed..

using the Bin2Dec single bit conversion in simpler steps as shown in this code snippet, but unfortunately even this doesnt work:

Any help is appreciated... I want to fix the code, atleast there is a silly logical error in Algo 3.

Algo 2 seems a little too weird.. But still i guess it does work if coded well.

For n=4,

Output:

0000

0001

0010

0011

0100

0101

0110

0111

1000

1001

1010

1011

1100

1101

1110

1111

I found three ways to do this:

Algo 1. Run a loop for i=0 to 2^(n), step i by 1

print(Dec2Bin(i,n));

where Dec2Bin is a function that converts a decimal i into a String of size n (if possible without data value damage/loss )

Code for this algorithm:

Algo 2. Create a bit array

Check bits from right (LSB) to left (MSB).

if bitvalue @ index == 0, flip bit @ index to 1

else if bitvalue @ index == 1, recursively check for left bits,

as terminal case.. if bit @ index=0 is 0 n all bits to the right are 1

flip bit at index=0 to 1 and flip all the bits to the right to 0

I tried implementing this algo, but it didnt work well,, it gives me the same output for n>4, as that given by n=4.

I would be glad if someone could help me with debugging/fixing this piece of code.

My Code :

Algo 3. I just thought that maybe every bit 0 or 1 can be calculated dynamically and directly displayed..

using the Bin2Dec single bit conversion in simpler steps as shown in this code snippet, but unfortunately even this doesnt work:

Any help is appreciated... I want to fix the code, atleast there is a silly logical error in Algo 3.

Algo 2 seems a little too weird.. But still i guess it does work if coded well.

Campbell Ritchie

Sheriff

Posts: 47244

52

posted 3 years ago

- 0

I would have thought you can reduce those algorithms to about ten lines, including lines with only a { on. I would use a StringBuilder rather than toBinaryString. The + operator for String catenation is possible, but less efficient if used in several statements.

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