Re: Binary number digits <- > Decimal number digits

On Dec 9, 3:44 pm, Adam <n...@xxxxxxxx> wrote:
On Sun, 9 Dec 2007 09:28:13 -0800 (PST), mensana...@xxxxxxxxxxx wrote:
On Dec 9, 10:50?am, Jeremy Boden <jer...@xxxxxxxxxxxxxxx> wrote:
On Sun, 09 Dec 2007 16:58:11 +0100, Helmut Richter wrote:
On Sun, 9 Dec 2007, fc wrote:

For example, what calculus was made to arrive that a decimal number of
617 digits must have 2048 digits in binary? (Obviously, without make
the conversion of the decimal number 999999...99999 [617 9 numbers] to
binary) And: how to generalize for any decimal number of any length?

The number of digits of x in some base b is, not regarding rounding to
integers, the logarithm of x with respect to base b, which is ?log x /
log b (log to arbitrary base).

So the number of decimal digits of x is ?log 2 / log 10 = 0.30103 .

For a rough calculation, 2^10 = 1024,
So 10 bits is approximately 3 digits.

How come you didn't round up?

Rounding up is appropriate in this case -- for the value 1024.

It is not always appropriate -- for an arbitrary n-bit value.

But that wasn't asked.


The following is an incomplete question:
"How many decimal digits are available from an n-bit binary number?"

What was asked was how many bits are required
for an n-digit decimal number.


What does the questioner mean by "How many?"

Do they mean "at _most_ 'x' digits"? If so, we always round up.

It was _exactly_ 617 decimal digits, with each
digit a 9.


Or do they mean "at _least_ 'x' digits"? If so, we might round down.

Not applicable.


Let's take a closer look at a 10-bit integer.

1023 is an unsigned 10-bit integer in which all bits are set:

11 1111 1111 (binary) = 1023 (decimal) -- four decimal digits.

10 * log(2)/log(10) = 3.01

Rounding up may suggest that we can always get 4 decimal digits from a
10-bit integer when the highest-order bit is in use.

If you have 3.x gallons of water in a tank a
1 gallon bucket to transfer it with, you're going
to make 4 goes regardless of how small x is as long
as x>0.


But such is not the case.

Because you're doing it backwards?


512 is also an unsigned 10-bit integer -- it requires the use of the
highest order bit in a 10-bit integer, which is 2^9:

10 0000 0000 (binary) = 512 (decimal) -- three decimal digits.

The following are true statements:

1. We can get _at most_ 4 decimal digits from a 10-bit unsigned
integer.

2. We can get _at least_ 3 decimal digits from a 10-bit unsigned
integer when the highest-order bit is in use.

Whether we round the digit-conversion calculation up or down may
depend on:
1. What we're after when we ask "how many digits"?
2. The width of the binary representation -- which determines the
fractional part in the result of n * log 2 / log 10. It may be that
when this fractional part is "large enough", we round up for either
case.

On Sun, 9 Dec 2007 07:39:24 -0800 (PST), fc wrote:
And: how to generalize for any decimal number of any length?

Consider the following two 5-digit decimal numbers:

expression decimal number bits required
2^16 - 1 65,535 16
2^16 + 2^14 81,920 17

5 * log(10) / log(2) = 16.6

Round up or down comes into play again.

Nope. A 617 digit decimal display will _require_
a 2050-bit register if you plan to shown _every_
possible 617-digit decimal number.

There may be some 617-digit numbers (such as RSA
challenge) that can get away with fewer bits, but
the OP specifically said all 9's, so rounding
down doesn't enter into it.


Adam

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