Case 1
- Number of distinct values (N)
- 1000
10bit
Open with these valuesBits Needed Calculator
Change any value to update the result.
10bit
Bits needed: 10 bitExamples
How the result moves
Number of distinct values (N) → Bits needed (bit)Your current valid inputs, with one value varied across the range.
How the result moves
| Number of distinct values (N) | Bits needed (bit) |
|---|---|
| 800 | 10 |
| 1,000Your value | 10 |
| 1,200 | 11 |
Bits = ⌈log₂N⌉, always rounded up, because there is no such thing as 9.97 of a bit. Labelling 1000 values takes 10 bits: 9 bits cover only 512 patterns, 10 bits cover 1024. One single value takes 0 bits.
Worked examples
Case 2
- Number of distinct values (N)
- 256
8bit
Open with these valuesCase 3
- Number of distinct values (N)
- 1025
11bit
Open with these valuesHow it's calculated
bits = ⌈log₂N⌉
- StepEnter how many different values you need to tell apart.Number of distinct values (N)=1,000
- StepThe base-2 logarithm says how many bits that would take.
- ResultThe result is rounded up, because bits come whole.10bit
What this number means
The rounding up is the whole calculation
log₂1000 comes out at 9.97, and 0.97 of a bit does not exist. The answer is 10, because 9 bits label only 512 patterns.
n bits hold 2ⁿ values
One bit covers 2 values, 8 bits cover 256 and 10 bits cover 1024. Counting from zero, the largest value n bits can carry is 2ⁿ − 1.
N is the count of values, not the biggest one
The integers 0 through 999 are 1000 values, which needs 10 bits. If you only know the largest value M and count from zero, enter M + 1.
One value too many costs a whole bit
1024 values still fit in 10 bits, but 1025 need 11. The jump happens immediately after every power of two.
Commonly misread
MisconceptionRound 9.97 bits to whichever whole number is closer.
CorrectionAlways upwards. Rounding down would leave part of the values without a pattern of their own.
MisconceptionA single value still needs one bit.
CorrectionIt needs none, since log₂1 is 0 and there is nothing to tell it apart from. Bits start being necessary at two values.
MisconceptionTwice as many values costs twice as many bits.
CorrectionDoubling costs exactly one extra bit. 512 values take 9 bits, 1024 take 10.
Reference table
| Values (N) | Nearest power of two | Bits |
|---|---|---|
| 1 | nothing to distinguish | 0 |
| 2 | exactly 2¹ | 1 |
| 3 | past 2¹ | 2 |
| 16 | exactly 2⁴ | 4 |
| 100 | past 2⁶ | 7 |
| 256 | exactly 2⁸ | 8 |
| 1000 | past 2⁹ | 10 |
| 1024 | exactly 2¹⁰ | 10 |
| 1025 | past 2¹⁰ | 11 |
Questions
How many bits do I need to represent N values?
Take the base-2 logarithm of N and round up: bits = ⌈log₂N⌉. Labelling 1000 distinct values needs ⌈9.97⌉ = 10 bits, because 9 bits cover only 512 patterns while 10 bits cover 1024.
Why round up instead of down?
Bits come in whole units — you cannot store 9.97 of a bit. Rounding down would leave some values without a unique pattern. Only when N is an exact power of two does the logarithm come out whole and no rounding is needed.
How many values can n bits represent?
Exactly 2ⁿ. One bit covers 2 values, 8 bits cover 256, and 10 bits cover 1024. Every extra bit doubles the count, which is why bit counts grow slowly even as the number of values explodes.
Why does 1 value need 0 bits?
With one possible value there is nothing to distinguish it from, and log₂1 is 0. You only start needing bits once there are at least two values to tell apart.
Is N the count of values or the largest number?
The count of distinct values. To store the integers 0 through 999 you have 1000 values, so ⌈log₂1000⌉ = 10 bits. If you know the largest value M and count from 0, enter M + 1.