# binary value

how to obtain binary of following an integer.as the object is very large 189866136719308462018271159242437168532L

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(I tried repeated times to post the answer, got internal error, that i should report...)

bin( 189866136719308462018271159242437168532L )

( 2018-04-07 15:54:04 +0200 )edit

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One can convert this "long" to a "Sage integer", and use the method "bits".

sage: a = 189866136719308462018271159242437168532L
sage: b = ZZ(a).bits()


How many bits do we have:

sage: len(b)
128


Show the first eight bits

sage: b[:8]
[0, 0, 1, 0, 1, 0, 0, 1]


and the last eight bits

sage: b[-8:]
[0, 1, 1, 1, 0, 0, 0, 1]

more

Alternatively, if we want the binary representation as a p h o n e number:

sage: a = 189866136719308462018271159242437168532L
sage: bin(a)
'0b10001110110101101110001101000111101110110011101001111001010101010010100111110111010100011010100111001000110100000000110110010100'

sage: type(bin(a))
<type 'str'>


This is similar to

sage: hex(a)
'0x8ed6e347bb3a795529f751a9c8d00d94L'
sage: oct(a)
'02166556150756635171252247672432471064006624L'


We can check that this matches the bits solution from the answer of slelievre:

sage: ZZ(a).bits() == [ int(s) for s in bin(a)[:1:-1] ]
True


Of course.

Note: Repeated times i tried to post versions of this answer. (The solution was to take a shower, then i wrote the word phone in an "unexpected" way. It seems that we cannot post the string pho.. number followed by an obvious such number here.)

more

@dan_fulea -- regarding pho.. ..mber, yes, this is one of the strings that cause questions or answers to be rejected automatically, as it frequently signals spam. It would be nice if that could be bypassed for users with sufficient karma, but I'm not sure the Askbot engine allows that.

( 2018-04-07 21:14:15 +0200 )edit