How do I execute binary operations in Mathematica? I want to to multiply say 1010101011 (binary) to 1111101110 (binary) and getting the result 10100111101111111010? Then I want to add the result to say 10101010000111 (binary)? Assume the all binary numbers are given as lists such as {1,0,1...}. Thanks!
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2 Answers
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There isn't really such a thing as binary arithmetic (at least in Mathematica). Numbers can be represented in any base, and this user-visible representation is completely independent from how arithmetic is done.
Try this:
BaseForm[(2^^1010101011)*(2^^1111101110), 2]
Things to look up:
answered Mar 28, 2014 at 16:19
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You can easily convert between decimal representation and a representation in terms of binary lists with IntegerDigits and FromDigits as in the following example:
IntegerDigits[57, 2]
{1, 1, 1, 0, 0, 1}
FromDigits[%, 2]
57
It is then straight forward to write functions for arithmetic like for example:
Multiply[a_, b_] := IntegerDigits[FromDigits[a] FromDigits[b], 2];
To get a multiplication operator symbol you can use for example:
MakeExpression[RowBox[{x_, "x", y_}], StandardForm] :=
MakeExpression[RowBox[{"Multiply", "[", x, ",", y, "]"}], StandardForm]
This then make to the following evaluation possible:
{1, 0, 0} x {1, 0, 1}
{1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0}
Maybe it would be better to use another symbol than "x"...
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$\begingroup$ I'm looking for something like {1,1,1,0,1} x {1,1,1,0,1} = {1, 0, 1, 0, 0, 1, 1, 0, 1, 1} $\endgroup$– GiorgioMar 28, 2014 at 16:14
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$\begingroup$ I see. I edited the answer accordingly. You might also find mathematica.stackexchange.com/questions/31375/… interesting. $\endgroup$– cgogolinMar 28, 2014 at 16:27
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$\begingroup$ Hi @Cgogolin, thanks! What I'm trying to do is to find the fastest way to multiply a binary number say b1 (with is list with one million components of zeros and ones) times b2 also a list with one million components of zeros and ones. Are you familiar with the Karatsuba Multiplication method? It claims to be much faster than conventional multiplication. The result should be a list with 2 million bits. I'll try your suggestion above and will see what happens, specially when I feed it with 2 lists of one million components each. $\endgroup$– GiorgioMar 29, 2014 at 0:51
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$\begingroup$ I wasn't aware that the numbers were that large. I was trying to come up with a convenient method, rather than an efficient one. No, I am not familiar with Karatsuba Multiplication. $\endgroup$– cgogolinMar 31, 2014 at 8:10
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$\begingroup$ @Giorgio If the purpose is to obtain "faster" integer multiplication, this approach is not going to get you there. $\endgroup$ Mar 31, 2014 at 22:07
ListConvolve[l1,l2,,{1,-1},Modulus->2]. $\endgroup$