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digitalmars.D.learn - Best way to get ceil(log2(x)) of a BigInt?

reply pineapple <meapineapple gmail.com> writes:
I'm trying to do some math stuff with std.bigint and realized 
there's no obvious way to calculate the ceil of log2 of a bigint. 
Help?
Nov 02 2016
parent reply Andrea Fontana <nospam example.com> writes:
On Wednesday, 2 November 2016 at 14:05:50 UTC, pineapple wrote:

 I'm trying to do some math stuff with std.bigint and realized 
 there's no obvious way to calculate the ceil of log2 of a 
 bigint. Help?


How big are your bigints?
Nov 02 2016
parent reply pineapple <meapineapple gmail.com> writes:
On Wednesday, 2 November 2016 at 14:24:42 UTC, Andrea Fontana 
wrote:

 On Wednesday, 2 November 2016 at 14:05:50 UTC, pineapple wrote:

 I'm trying to do some math stuff with std.bigint and realized 
 there's no obvious way to calculate the ceil of log2 of a 
 bigint. Help?


 How big are your bigints?


I think they'll generally stay between 0 and 2^200

I came up with this, I guess it'll work?

    auto clog2(BigInt number) in{
         assert(number > 0);
     }body{
         uint log;
         auto n = number - 1;
         while(n > 0){
             log++; n >>= 1;
         }
         return log;
     }
Nov 02 2016
parent reply Andrea Fontana <nospam example.com> writes:
On Wednesday, 2 November 2016 at 14:49:08 UTC, pineapple wrote:

 On Wednesday, 2 November 2016 at 14:24:42 UTC, Andrea Fontana 
 wrote:

 On Wednesday, 2 November 2016 at 14:05:50 UTC, pineapple wrote:

 I'm trying to do some math stuff with std.bigint and realized 
 there's no obvious way to calculate the ceil of log2 of a 
 bigint. Help?


 How big are your bigints?


 I think they'll generally stay between 0 and 2^200

 I came up with this, I guess it'll work?

    auto clog2(BigInt number) in{
         assert(number > 0);
     }body{
         uint log;
         auto n = number - 1;
         while(n > 0){
             log++; n >>= 1;
         }
         return log;
     }


Why don't you perform a binary search over 200 power of 2?
Something like:

BigInt powers[] = [BigInt(2), BigInt(4), BigInt(8), ...];

And I think you can reduce your search in a smaller interval. 
Something like:

    string number = "71459266416693160362545788781600";
    BigInt i = number;
    ulong l = number.length;

    ulong approxMin = cast(ulong)((l-1)/0.30103);
    ulong approxMax = cast(ulong)((l)/0.301029);

So you can search on powers[approxMin .. approxMax], if i'm right.
Nov 02 2016
parent Andrea Fontana <nospam example.com> writes:
On Wednesday, 2 November 2016 at 15:15:18 UTC, Andrea Fontana 
wrote:

 Why don't you perform a binary search over 200 power of 2?


Something like: http://paste.ofcode.org/scMD5JbmLMZkrv3bWRmPPT

I wonder if a simple binary search on whole array is faster than 
search for limits as in this example

PS: If you need ceil, just use the else branch with <= instead of 
<.

Andrea
Nov 02 2016