Question

I have 4 binary bits

Bit 3  Bit 2  Bit 1  Bit 0 

Normally the answer is simple: 2^4, or 16 different combinations; and it would looks something like the following:

0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111

However, The LSB (Bit 0) changes state every iteration.

I need an algorithm where the state of a bit only changes once through all the iterations; i.e, I need all my bits to act like the MSB (Bit 3).

How can I do this?

Edit

It seems that most people are converging to there being only 5 possible solutions. However this assumes there is a starting point for the value and an ending point. This doesn’t matter so I’m going to give a real world scenario to better explain.

Suppose I have an digital alarm clock that gives me 4 outputs. Each output can be programmed to go on at a certain time and off at a certain time and are programmed independent of each other, eg. I can program output 1 to go on at 1 am and off at 3 am, while I can program output 2 to go on at 7 pm and off at 2 am. There are no restrictions to how long each output can stay on.

Now I want to hook this alarm clock to a computer and get as close as possible to the current correct time. i.e If the clock says the time is 2:15 pm, my computer knows that the alarm is within the 12 pm to 6 pm range for example. I want to be able to get the smallest possible range. Whats the smallest possible range I can get?

Answer 1

1. There are 4 bits.
2. Each bit may change state only once.
3. For each new value, at least one of the bits must have changed state from the previous value.

Therefore, you can have at most 4 state changes, and at most 5 different values.

Example:

0000 -> 0001 -> 0011 -> 0111 -> 1111 

Edit:

Very well, let’s restate from what you mean rather than from what you say.

1. There are 4 bits.
2. Each bit may change state only twice. (once from 0 to 1, and once from 1 to 0)
3. For each new value, at least one of the bits must have changed state from the previous value.

Therefore, you can have at most 8 state changes, and at most 8 different values (since the last state change necessarily brings all bits back to their initial state)

Example:

0000 -> 0001 -> 0011 -> 0111 -> 1111 -> 1110 -> 1100 -> 1000 -> 0000 

So, by setting the outputs for: 3 AM – 3 PM, 6 AM – 6 PM, 9 AM – 9 PM and noon – midnight, you can determine which 3-hour period it is from the outputs. I’d suggest plugging wires into the visual output instead.