Do bitwise operators (other than shifts) make any mathematical sense in base-10

bit-manipulationmath

According to wiki shifts can be used to calculate powers of 2:

A left arithmetic shift by n is
equivalent to multiplying by 2^n
(provided the value does not
overflow), while a right arithmetic
shift by n of a two's complement value
is equivalent to dividing by 2^n and
rounding toward negative infinity.

I was always wondering if any other bitwise operators (~,|,&,^) make any mathematical sense when applied to base-10? I understand how they work, but do results of such operations can be used to calculate anything useful in decimal world?

Best Answer

"yep base-10 is what I mean"

In that case, yes, they can be extended to base-10 in several ways, though they aren't nearly as useful as in binary.

One idea is that &, |, etc. are the same as doing arithmetic mod-2 to the individual binary digits. If a and b are single binary-digits, then

a & b = a * b (mod 2)
a ^ b = a + b (mod 2)
   ~a = 1-a   (mod 2)
a | b = ~(~a & ~b) = 1 - (1-a)*(1-b) (mod 2)

The equivalents in base-10 would be (note again these are applied per-digit, not to the whole number)

a & b = a * b (mod 10)
a ^ b = a + b (mod 10)
   ~a = 9-a   (mod 10)
a | b = ~(~a & ~b) = 9 - (9-a)*(9-b) (mod 10)

The first three are useful when designing circuits which use BCD (~a being the 9's complement), such as non-graphing calculators, though we just use * and + rather than & and ^ when writing the equations. The first is also apparently used in some old ciphers.

Related Solutions

Bit shift when there is no … bit shift operator

In two's-complement arithmetic, the only impact that negative values have occurs when dividing by 2 to shift right: the intended right shift will take place, but it will also introduce a new 1-bit in the most significant bit (MSB) position to "keep the value negative" -- unless the original value was -1, in which case all bits become 0. So to correct for this, try the following pseudocode:

rightshift(x) {
    if x >= 0 return x / 2;
    if x < -1 return x / 2 - MINVAL;    # Strip out sign bit
    # x must be -1, i.e. "all bits on"
    return x - MINVAL;
}

MINVAL should be the value whose representation consists of just the MSB on and all other bits off, which is -32768 for 16 bits. (So named because it will be the most negative representable number using two's-complement.) Interestingly, adding MINVAL works just as well as subtracting it in the above pseudocode, since in two's-complement arithmetic, x - y = x + NOT(y) + 1, and MINVAL == NOT(MINVAL) + 1.

Left shifts using multiply-by-2 work for negative numbers just as well as they do for positive ones.

R – How many concurrent outbound HttpWebRequest calls can be made in ASP.NET / IIS7

By default, an HTTP/1.1 server is limited to two connection, and a HTTP/1.0 server is limited to four connections. So, your ASP.NEt app will have serious throughput problems if you are trying to issue more than two outstanding requests to an HTTP/1.1 server, for eg.

You will need to increase the connection limit, either per server, or globally.

For eg, globally:

ServicePointManager.DefaultConnectionLimit = 10; // allow 10 outstanding connections

Hope this helps.

Best Answer

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Bit shift when there is no … bit shift operator

R – How many concurrent outbound HttpWebRequest calls can be made in ASP.NET / IIS7

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