FAQ
Is there a function that takes a number with binary numeral a1...an to the
number with binary numeral b1...bn, where each bi is 1 if ai is 0, and vice
versa? (For example, the function's value at 18 [binary 10010] would be 13
[binary 01101].) I thought this was what the tilde operator (~) did, but when I
went to try it I found out that wasn't the case. I discovered by experiment (and
verified by looking at the documentation) that the tilde operator takes n
to -(n+1). I can't imagine what that has to do with binary numerals. Can anyone
shed some light on that? (In case you're curious, I'm writing a script that will
play Nim, just as a way of familiarizing myself with bitwise operators. Good
thing, too: I thought I understood them, but apparently I don't.)

Muchas gracias for any and all helps and hints.

Peace,
EJ

## Search Discussions

• at Aug 17, 2003 at 5:28 am ⇧ Elaine Jackson wrote:

Is there a function that takes a number with binary numeral a1...an
to the number with binary numeral b1...bn, where each bi is 1 if ai
is 0, and vice versa? (For example, the function's value at 18
[binary 10010] would be 13 [binary 01101].) I thought this was what
the tilde operator (~) did, but when I went to try it I found out
that wasn't the case. I discovered by experiment (and verified by
looking at the documentation) that the tilde operator takes n to
-(n+1). I can't imagine what that has to do with binary numerals.
Can anyone shed some light on that? (In case you're curious, I'm
writing a script that will play Nim, just as a way of familiarizing
myself with bitwise operators. Good thing, too: I thought I
understood them, but apparently I don't.)
I think this is because it's also inverting the leading zero bits.
That is, 18 isn't really 10010, it's 000...0010010 with a bunch of 0s
at the front. (I'm not sure, but I think the number of bits in the
integer type can change with different implementations.) So when you
bitwise-not it, you get 111...1101101, which is interpreted as -19.

--
--OKB (not okblacke)
no path, and leave a trail."
--author unknown
• at Aug 17, 2003 at 5:35 am ⇧ Elaine Jackson wrote:
Is there a function that takes a number with binary numeral a1...an to the
number with binary numeral b1...bn, where each bi is 1 if ai is 0, and vice
versa? (For example, the function's value at 18 [binary 10010] would be 13
[binary 01101].) I thought this was what the tilde operator (~) did, but when I
went to try it I found out that wasn't the case. I discovered by experiment (and
verified by looking at the documentation) that the tilde operator takes n
to -(n+1). I can't imagine what that has to do with binary numerals.
It has a lot to do with binary! Google for "two's complement".

In the meantime, try this:
~18 & 31
13

The '~' operator cannot care about precision -- that is, how many bits
you're operating on, or expecting in your result. In your example, you
represent decimal 18 as '10010', but '000000010010' is also correct,
right?

In two's complement math, both inverses, '01101' and '111111101101'
respectively, are equivalent to decimal -19.

And-ing with a mask that is the of length 'n' will ensure that you only
get the least significant n bits -- and this is what you're looking for.
Since you're operating on five bits in your example, I chose decimal 31,
or '11111'.

-- Graham
• at Aug 17, 2003 at 5:35 am ⇧ [Elaine Jackson]
Is there a function that takes a number with binary numeral a1...an
to the number with binary numeral b1...bn, where each bi is 1 if ai
is 0, and vice versa? (For example, the function's value at 18
[binary 10010] would be 13 [binary 01101].) I thought this was what
the tilde operator (~) did,
Surprise: that is what ~ does.
but when I went to try it I found out that wasn't the case.
Please give a specific example. To understand your example above, note that
binary 10010 actually has an unbounded number of 0 bits "to the left":

...00000010010 = 13

The bitwise inversion of that therefore has an unbounded number of 1 bits
"to the left":

...11111101101 = -19

This is why you get -19:
n = 18
~n
-19
>>>

Any answer other than that is wishing away the 0 bits at the left in the
input. Python can't *guess* how many bits you want to keep. If you only
want to keep the rightmost 5 bits, then explicitly mask off the rightmost 5
bits:
(~18) & 0x1f
13
>>>

So that's the 13 "you expected".
I discovered by experiment (and verified by looking at the
documentation) that the tilde operator takes n to -(n+1). I can't
imagine what that has to do with binary numerals. Can anyone
shed some light on that?
Python represents negative integers in unbounded 2's-complement form (well,
it doesn't really under the covers, but it maintains the illusion of doing
so in all arithmetic and logical operators). The 2's-complement of an
integer is equal to 1 plus its 1's-complement form, and 1's-complement is
identical to bitwise inversion. So

-n = 1 + (~n)

Then

~n = -(n+1)

follows from that via rearrangement.
(In case you're curious, I'm writing a script that will play Nim,
just as a way of familiarizing myself with bitwise operators. Good
thing, too: I thought I understood them, but apparently I don't.)
You're only overlooking the consequences of an infinite amount of
information <wink>.
• at Aug 17, 2003 at 6:14 am ⇧ Tim Peters wrote:

(In case you're curious, I'm writing a script that will play Nim,
just as a way of familiarizing myself with bitwise operators. Good
thing, too: I thought I understood them, but apparently I don't.)
You're only overlooking the consequences of an infinite amount of
information <wink>.
Excellent line! I may borrow that one, Tim; I'll use it on the boss,
next time that he asks why my application doesn't work the way he
expected it to. Or why we ran out of coffee, or why the printer isn't
working... It may need to be accompanied by a smoldering, wide-eyed
Rasputin-ish stare, however, or it might lack the necessary mystique to
supress any further questions.

Of course, after a good night's sleep, maybe it won't seem like such a
good idea. ;-)

nocturnally yours,

-- Graham
• at Aug 17, 2003 at 8:00 pm ⇧ Graham Fawcett <fawcett at teksavvy.com> wrote in message news:<mailman.1061100930.7013.python-list at python.org>...
Tim Peters wrote:
You're only overlooking the consequences of an infinite amount of
information <wink>.
Excellent line! [...]
It may need to be accompanied by a smoldering, wide-eyed
Rasputin-ish stare, however, or it might lack the necessary mystique to
supress any further questions.
Somehow, I associated it with a smoldering, wide-eyed Tom Baker-ish
stare. To each his own.

--
Tim Lesher
tim at lesher.ws
• at Aug 17, 2003 at 7:01 am ⇧ "Elaine Jackson" <elainejackson7355 at home.com> schrieb im Newsbeitrag
news:1YD%a.747879\$3C2.17342633 at news3.calgary.shaw.ca...

...
I'm writing a script that will
play Nim, just as a way of familiarizing myself with bitwise operators. Good
thing, too: I thought I understood them, but apparently I don't.)
Hi Elaine,
You have described a general misconception you seem to be not the only one
to live with.
The enlightening answers having been posted might sufficwe, but a schuld
liek to add some more "enlightenment":
Bit complements have a lot to do with set complents and the aritmetic
negation (sometimes called two's complement for obvious reasons). Consider
the set of of "red" of "blue". Now whats the complement? "green" and
"yellow" is obviously the wrong answer. You in fact cannot give any answer
befor you define the total set you are dealing with. The same applies to
logical bit operations. Generally you take a "processor" word or a part of
it to be defined. Some high level languages are more flexible; and even some
computers ("vector processors") are.

The only rule is, that ~(~x) == x

The same situation with numbers: What is the negation of +5. You have to
think very hard! This is a trick question and you probably will give a
"trick answer": -5. You should be aware that this is just a trick. "-5"
contains no other information as that it is some "complement" of 5. (same
with complex "imaginary" numbers: 5j (in Python) just says it is some fancy
5.)

Now we define a transformation between positive numbers and bit patterns 5 LoL. Note that 5 == ...000005 or LoL == ....ooooLoL does not help any
understanding so you generally skip this part.

Now you do some arithmetic "inversion": 5 -> -5 This however can (and
should) stay a secret of the processor! By no means should you be interested
in how the machine represents "-5". If you are courious then know that
there had been times when computers represente -5 as ...LLLoLo. Yes it
worked! And you had two diffrent "zeros" then: +0 and -0 !!!!

Most computers do not distinguish between the representation of negativ
numbers and complemented sets (let alone note a special "total set" the
complemt was referring to). Thus the "secret" of modern two's-complement
computern arithmetic is always disclosed to you.

Note that there is no use in something like "masking" the MSB, i.e. that
bits-complements only work on 31 bits. This will lead to ~5 == ~LoL == ~
LL..LLLLoL == 2,147,483,643 Not much improvement, eh!?

Kindly
Michael P
• at Aug 17, 2003 at 1:05 pm ⇧ "Tim Peters" <tim.one at comcast.net> wrote in message
news:mailman.1061098645.13097.python-list at python.org...

<snip>
To understand your example above, note that
binary 10010 actually has an unbounded number of 0 bits "to the left":

...00000010010 = 13

The bitwise inversion of that therefore has an unbounded number of 1 bits
"to the left":

...11111101101 = -19
** What you're saying, the way I read it, is that 5+S=(-19), where S is the
(divergent) sum of all the powers 2^k of 2 with k>=5. I'm still at sea, in other
words.

<snip>
Python can't *guess* how many bits you want to keep.
** But it could if someone had told it that the leftmost nonzero digit is the
place to start. I just assumed somebody had told it that.

<snip>
Python represents negative integers in unbounded 2's-complement form
** Now we're getting someplace. That was the missing piece in my puzzle.

Thanks for the help.

Peace,
EJ
• at Aug 17, 2003 at 1:54 pm ⇧ "Elaine Jackson" <elainejackson7355 at home.com> schrieb im Newsbeitrag
news:d4L%a.781431\$Vi5.17573533 at news1.calgary.shaw.ca...
"Tim Peters" <tim.one at comcast.net> wrote in message
news:mailman.1061098645.13097.python-list at python.org...

<snip>
To understand your example above, note that
binary 10010 actually has an unbounded number of 0 bits "to the left":

...00000010010 = 13
The bitwise inversion of that therefore has an unbounded number of 1
bits
"to the left":

...11111101101 = -19
** What you're saying, the way I read it, is that 5+S=(-19), where S is the
(divergent) sum of all the powers 2^k of 2 with k>=5. I'm still at sea, in other
words.
I think not. Tim seems to point out that there is no natural final
"representation" of negative numbers unless we invent some convention. The
interpretation of 1101101 or 11111101101 or ......1111101101 as -19 is a
convention and has nothing to do with the more natural interpretaton of
101101 as 18. The main reason is that binary adders used in computers are
very simple too be (ab-) used for subtraction in that way....

Kindly
Michael P
• at Aug 17, 2003 at 6:03 pm ⇧ On Sun, 17 Aug 2003 13:05:13 GMT, "Elaine Jackson" wrote:
"Tim Peters" <tim.one at comcast.net> wrote in message
news:mailman.1061098645.13097.python-list at python.org...

<snip>
To understand your example above, note that
binary 10010 actually has an unbounded number of 0 bits "to the left":

...00000010010 = 13

The bitwise inversion of that therefore has an unbounded number of 1 bits
"to the left":

...11111101101 = -19
** What you're saying, the way I read it, is that 5+S=(-19), where S is the
(divergent) sum of all the powers 2^k of 2 with k>=5. I'm still at sea, in other
words.
Since all the sign bits extend infinitely, we can chop them off any place above
the first of the repeating bits to get N bits with some number of repetitions at the
top. We will see that the interpreted numeric value does not depend on how far
to the left we chop the bits, so long as we keep at least the first of
the repeating bits.

This representation of a signed integer with N total bits b(i) little-endianly numbered
with i going 0 to N-1 and each bit having a value b(i) of 0 or 1 is interpreted as having
the (N-1)th bit as the sign bit, and the rest positive, i.e.,

___ i=N-2
\
a = -b(N-1)*2**(N-1)i + > b(i)*2**i
/__
i=0

or in python, operating on a little-ending list of N bits,
def bitsvalue(b): #b is bitlist, to follow convention above
... N=len(b)
... sum = -b[N-1]*2**(N-1)
... for i in xrange(N-1):
... sum += b[i]*2**i
... return sum
...

(Remember this is little-endian: least significant bit to the left, sign at the right)
bitsvalue([0,1,0,0,1,0])
18
bitsvalue([1,0,1,1,0,1])
-19

It doesn't matter how far you extend a copy of the top bit, the value remains the same:
bitsvalue([1,0,1,1,0,1,1,1,1,1,1,1])
-19

You can verify it from the math of the sum, if you separate it into two sums, one with
the top repeating sign bits b(N-r) to b(N-1) and the other with the rest, which has the
same constant value. I'll leave it as an exercise.
bitsvalue([1,1,1,1,1,1])
-1
bitsvalue([1,1,1,1])
-1
bitsvalue([1,1,1,0])
7
<snip>
Python can't *guess* how many bits you want to keep.
** But it could if someone had told it that the leftmost nonzero digit is the
place to start. I just assumed somebody had told it that.
Or the leftmost bit that is either the only bit of all or different
from the one to its right is the minimum sign bit to copy arbitrarily leftwards.
<snip>
Python represents negative integers in unbounded 2's-complement form
** Now we're getting someplace. That was the missing piece in my puzzle.
Ok, I guess you don't need the demo prog after all ;-)

Regards,
Bengt Richter
• at Aug 17, 2003 at 10:08 pm ⇧ Elaine Jackson wrote:
<snip>
Python can't *guess* how many bits you want to keep.
** But it could if someone had told it that the leftmost nonzero
digit is the place to start. I just assumed somebody had told it
that.

And if someone had done that, it would violate the invariant:

~(~x) == x

In fact, by repeatedly applying ~ you'd eventually zero all the bits.

--
CARL BANKS http://www.aerojockey.com/software
"You don't run Microsoft Windows. Microsoft Windows runs you."
• Irmen de Jong at Aug 17, 2003 at 3:21 pm ⇧ While others explained how the ~ operator works, let me suggest
another possibility: the bitwise exclusive or.
def bin(i):
... l = ['0000', '0001', '0010', '0011', '0100', '0101', '0110', '0111',
... '1000', '1001', '1010', '1011', '1100', '1101', '1110', '1111']
... s = ''.join(map(lambda x, l=l: l[int(x, 16)], hex(i)[2:]))
... if s == '1' and i > 0:
... s = '0000' + s
... return s
...
bin(18)
'00010010'
~18
-19
bin(~18) # tricky...
'11111111111111111111111111101101'
~18 & 0x1f
13
bin(~18 & 0x1f)
'00001101'
18 ^ 0x1f # XOR!
13
bin(18 ^ 0x1f) # XOR
'00001101'
>>>

You still have to think about the number of bits you want to invert.
x ^ 0x1f inverts the 5 least significant bits of x.
x ^ 0xff inverts the 8 least significant bits of x, and so on.

--Irmen de Jong
• at Aug 17, 2003 at 9:41 pm ⇧ "Dennis Lee Bieber" <wlfraed at ix.netcom.com> schrieb im Newsbeitrag
news:c84511-ol3.ln1 at beastie.ix.netcom.com...
Elaine Jackson fed this fish to the penguins on Saturday 16 August 2003
09:58 pm:

Is there a function that takes a number with binary numeral a1...an to
the number with binary numeral b1...bn, where each bi is 1 if ai is 0,
and vice versa? (For example, the function's value at 18 [binary
10010] would be 13
[binary 01101].) I thought this was what the tilde operator (~) did,
[but when I
went to try it I found out that wasn't the case. I discovered by
experiment (and verified by looking at the documentation) that the
tilde operator takes n to -(n+1). I can't imagine what that has to do
with binary numerals.
[..]
You've had lots of answers at the moment though I haven't seen anyone
explain away the "+1" part...

Most computers use twos-complement arithmetic to avoid the problem of
having two valid values for integer 0, which is what appears in ones
complement arithmetic.

For argument, assume an 8-bit integer. The value of "5" would be
represented as 00000101. The one's complement negative would be
11111010. So far there isn't any problem... But consider the value of
0, represented as 00000000. A one's complement negative would become
11111111 -- But mathematically, +0 = -0; in one's complement math, this
does not hold true.

So a little trick is played, to create twos complement... To negate a
number, we take the ones complement, and then add 1 to the result. The
"5" then goes through: 00000101 -> 11111010 + 1 -> 11111011... Looks
strange, doesn't it... But watch what happens to that 8-bit 0: 00000000
-> 11111111 + 1 -> (overflows) 00000000.... Negative 0 is the same as
positive 0.
[..]

I have the impression (may be wrong) that you are working under the
misconception that there can be a "natural" binary represensation of
negative numbers!?
Three conventions have commonly been used so far:
1- Complement
2-Complement
Sign tag plus absolut binary values

All of them have their pros and cons. For a mixture of very technical
reasons (you mentioned the +0/-0 conflict, I might add the use of binary
adders for subtraction) most modern computers use 2-complement, and this now

Kindly
Michael P
• at Aug 17, 2003 at 11:26 pm ⇧ "Michael Peuser" <mpeuser at web.de> wrote in message
news:bhosrr\$u57\$06\$1 at news.t-online.com...
I have the impression (may be wrong) that you are working under the
misconception that there can be a "natural" binary represensation of
negative numbers!?
Three conventions have commonly been used so far:
1- Complement
2-Complement
Sign tag plus absolut binary values
The last alternative sounds like what I was assuming. If it is, I would argue
that it's pretty darn natural. Here's a little function to illustrate what I
mean:

def matilda(n): ## "my tilde"
if 0<=n<pow(2,29):
for i in range(1,31):
iOnes=pow(2,i)-1
if n<=iOnes:
return iOnes-n
else:
raise
• at Aug 18, 2003 at 12:18 am ⇧ [Michael Peuser]
I have the impression (may be wrong) that you are working under the
misconception that there can be a "natural" binary represensation of
negative numbers!? Three conventions have commonly been used so far:
1- Complement
2-Complement
Sign tag plus absolut binary values
[Elaine Jackson]
The last alternative sounds like what I was assuming. If it is, I
would argue that it's pretty darn natural. Here's a little function to
illustrate what I mean:

def matilda(n): ## "my tilde"
if 0<=n<pow(2,29):
for i in range(1,31):
iOnes=pow(2,i)-1
if n<=iOnes:
return iOnes-n
else:
raise
As Carl Banks pointed out, under this meaning the fundamental identity

~~x == x

almost never holds. For example, matilda(18) is 13, but matilda(13) is 2.
~ is its own inverse under Python's meaning. You can see that formally by
algrebraic manipulation:

~~n = ~(-(n+1)) = -(-(n+1)+1) == n

or more easily by noting that ~x in Python acts the same as xor'ing x with
an infinite string of 1 bits.
• at Aug 18, 2003 at 3:12 am ⇧ In article <bhosrr\$u57\$06\$1 at news.t-online.com>, Michael Peuser wrote:

I have the impression (may be wrong) that you are working under the
misconception that there can be a "natural" binary represensation of
negative numbers!?
Three conventions have commonly been used so far:
1- Complement
2- Complement
Sign tag plus absolut binary values

All of them have their pros and cons. For a mixture of very technical
reasons (you mentioned the +0/-0 conflict, I might add the use of binary
The latter is _far_ more important than the former. Being able
to use a simple binary adder to do operations on either signed
or unsigned values is a huge savings in CPU and ISA design. I
doubt that anybody really cares about the +0 vs. -0 issue very
much (IEEE FP has zeros of both signs, and nobody seems to
care).
most modern computers use 2-complement, and this now leads to
those funny speculations in this thread. ;-)
--
Grant Edwards grante Yow! An Italian is COMBING
at his hair in suburban DES
visi.com MOINES!
• at Aug 18, 2003 at 4:11 am ⇧ On Sun, 17 Aug 2003 22:18:39 GMT, Dennis Lee Bieber wrote:
Michael Peuser fed this fish to the penguins on Sunday 17 August 2003
02:41 pm:
I have the impression (may be wrong) that you are working under the
misconception that there can be a "natural" binary represensation of
negative numbers!?
Apologies if I gave that impression... the +/- 0 technical affair is
the main reason I went into the whole thing...
Three conventions have commonly been used so far:
1- Complement
2-Complement
Sign tag plus absolut binary values

All of them have their pros and cons. For a mixture of very technical
reasons (you mentioned the +0/-0 conflict, I might add the use of
binary adders for subtraction) most modern computers use 2-complement,
and this now leads to those funny speculations in this thread. ;-)
most "natural"; after all, what is -19 in human terms but a "pure" 19
prefaced with a negation tag marker... (I believe my college
mainframe's BCD hardware unit actually put the sign marker in the
nibble representing the decimal point location -- but it has been 25
years since I had to know what a Xerox Sigma did for COBOL packed
decimal <G>).

ie, 00010011 vs -00010011 <G>

1s complement is electrically easy; just "not" each bit.

2s complement is mathematically cleaner as 0 is 0, but requires an
adder to the 1s complement circuit... Though both complement styles
lead to the ambiguity of signed vs unsigned values
Everyone says "two's complement" and then usually starts talking about numbers
that are bigger than two. I'll add another interpretation, which is what I thought
when I first heard of it w.r.t. a cpu that was designed on the basis that all its
"integer" numbers were fixed point fractions up to 0.9999.. to whatever precision
the binary fractional bits provided. There was no units bit. And if you took one
of these fractional values 0.xxxx and subtracted it from 2.0, you would have a
complementary number with respect to two. Well, for addition and subtraction, that turns
out to work just like the "two's complement" integers we are used to. But since the
value of fractional bits were all in negative powers of two, squaring e.g., .5 had
to result in a consistent representation of 0.25 -- i.e. in binary squaring 0.1
resulted in 0.01 -- which is shifted one bit from what you get looking at the numbers
as integers with the lsb at the bottom of the registers and the result.

I.e., a 32-bit positive integer n in the fractional world was n*2**-31. If you square
that for 64 bits, you get n**2, but in the fractional world that looks like (n**2)*2**-63,
where it's supposed to be (n*2**-31)**2 => (n**2)*2**-62 with respect to the binary point.
The fractional model preserved an extra bit of precision in multiplies.

So on that machine we used to count bits from the left instead of the right, and place imaginary
binary points in the representations, so a binary 0.101 could be read as "5 at 3" or "2.5 at 2"
or "10 at 4" etc. And the multiplying rule was x at xbit times y at ybit => x*y at xbit+ybit.

You can do the same counting the bit positions leftwards from lsb at 0, as we usually do now,
of course, to play with fixed point fractions. A 5 at 0 is then 1.25 at 2 ;-)

Anyway, my point is that there was a "two's complement" implementation that really meant
a numeric value complement with respect to the value two ;-)

Regards,
Bengt Richter
• at Aug 18, 2003 at 7:42 am ⇧ "Bengt Richter" <bokr at oz.net> schrieb im Newsbeitrag
news:bhpjm1\$g01\$0 at 216.39.172.122...
On Sun, 17 Aug 2003 22:18:39 GMT, Dennis Lee Bieber
wrote:
Michael Peuser fed this fish to the penguins on Sunday 17 August 2003
02:41 pm:
I have the impression (may be wrong) that you are working under the
misconception that there can be a "natural" binary represensation of
negative numbers!?
Apologies if I gave that impression... the +/- 0 technical affair
is
the main reason I went into the whole thing...
Three conventions have commonly been used so far:
1- Complement
2-Complement
Sign tag plus absolut binary values

All of them have their pros and cons. For a mixture of very technical
reasons (you mentioned the +0/-0 conflict, I might add the use of
binary adders for subtraction) most modern computers use 2-complement,
and this now leads to those funny speculations in this thread. ;-)
the
most "natural"; after all, what is -19 in human terms but a "pure" 19
prefaced with a negation tag marker... (I believe my college
mainframe's BCD hardware unit actually put the sign marker in the
nibble representing the decimal point location -- but it has been 25
years since I had to know what a Xerox Sigma did for COBOL packed
decimal <G>).

ie, 00010011 vs -00010011 <G>

1s complement is electrically easy; just "not" each bit.

2s complement is mathematically cleaner as 0 is 0, but requires
an
adder to the 1s complement circuit... Though both complement styles
lead to the ambiguity of signed vs unsigned values
Everyone says "two's complement" and then usually starts talking about numbers
that are bigger than two. I'll add another interpretation, which is what I thought
when I first heard of it w.r.t. a cpu that was designed on the basis that all its
"integer" numbers were fixed point fractions up to 0.9999.. to whatever precision
the binary fractional bits provided. There was no units bit. And if you took one
of these fractional values 0.xxxx and subtracted it from 2.0, you would have a
complementary number with respect to two. Well, for addition and
subtraction, that turns
out to work just like the "two's complement" integers we are used to. But since the
value of fractional bits were all in negative powers of two, squaring
to result in a consistent representation of 0.25 -- i.e. in binary
squaring 0.1
resulted in 0.01 -- which is shifted one bit from what you get looking at
the numbers
as integers with the lsb at the bottom of the registers and the result.

I.e., a 32-bit positive integer n in the fractional world was n*2**-31. If
you square
that for 64 bits, you get n**2, but in the fractional world that looks
like (n**2)*2**-63,
where it's supposed to be (n*2**-31)**2 => (n**2)*2**-62 with respect to
the binary point.
The fractional model preserved an extra bit of precision in multiplies.

So on that machine we used to count bits from the left instead of the
right, and place imaginary
binary points in the representations, so a binary 0.101 could be read as
"5 at 3" or "2.5 at 2"
or "10 at 4" etc. And the multiplying rule was x at xbit times y at ybit
=> x*y at xbit+ybit.
You can do the same counting the bit positions leftwards from lsb at 0, as
we usually do now,
of course, to play with fixed point fractions. A 5 at 0 is then 1.25 at 2 ;-)
Anyway, my point is that there was a "two's complement" implementation
that really meant
a numeric value complement with respect to the value two ;-)

Regards,
Bengt Richter

A very good point! I might add that this is my no means an exotic feature.
Mathematically speaking there is great charme in computing just inside the
invervall (-1,+1). And if you have no FPU you can do *a lot* of pseudo real
operations. You have get track of the scale of course - it is a little bit
like working with sliding rules if anyone can remember those tools ;-)

Even modern chips have support for this format, e.g. there is the 5\$ Atmel
Mega AVR which has two kinds of multiplication instructions: one for the
integer multiplication and one which automatically adds a left shift after
the multiplication! I leave it as an exercise to find out why this is
necessary when multiplying fractional numbers ;-)

Negative numbers are formed according to the same rule for fractionals and
integers:
Take the maximum positive number: 2**32-1 or 0.999999
Add one bit: 2*32 or 1
Double it: 2*33 or 2
Subtract the number in question

Kindly Michael P
• at Aug 18, 2003 at 1:11 pm ⇧ In article <bhq01j\$tge\$03\$1 at news.t-online.com>, Michael Peuser wrote:

A very good point! I might add that this is my no means an exotic feature.
Mathematically speaking there is great charme in computing just inside the
invervall (-1,+1). And if you have no FPU you can do *a lot* of pseudo real
operations. You have get track of the scale of course - it is a little bit
like working with sliding rules if anyone can remember those tools ;-)
Sure. I've got two sitting at home. :)

FWIW, it used to be fairly common for process-control systems
to define operations only over the interval (-1,+1). This made
implimentation easy, and the input and output devices
(temp/pressure sensors, valves, whatnot) all had pre-defined
ranges that mapped logically to the (-1,+1) interval.

--
Grant Edwards grante Yow! The SAME WAVE keeps
at coming in and COLLAPSING
visi.com like a rayon MUU-MUU...
• at Aug 18, 2003 at 2:58 pm ⇧ "Grant Edwards" <grante at visi.com> schrieb im Newsbeitrag
In article <bhq01j\$tge\$03\$1 at news.t-online.com>, Michael Peuser wrote:
A very good point! I might add that this is my no means an exotic
feature.
Mathematically speaking there is great charme in computing just inside
the
invervall (-1,+1). And if you have no FPU you can do *a lot* of pseudo
real
operations. You have get track of the scale of course - it is a little
bit
like working with sliding rules if anyone can remember those tools ;-)
Sure. I've got two sitting at home. :)

FWIW, it used to be fairly common for process-control systems
to define operations only over the interval (-1,+1). This made
implimentation easy, and the input and output devices
(temp/pressure sensors, valves, whatnot) all had pre-defined
ranges that mapped logically to the (-1,+1) interval.

--
Yes it simplifies a lot of matters, even when using full floating point
numbers. Take OpenGL e.g. The colour space is a 1x1x1 cube. Very fine! No
magic numbers near 256 ;-)

Kindly
Michael P
• at Aug 17, 2003 at 11:35 pm ⇧ In case any fellow newbies have been following this thread, here is the finished
Nim script. It doesn't use bit-manipulation as much as I thought it would.
(Still, I got the basics straight. Thanks again to everyone who helped out!) For
a readable account of the optimal strategy for Nim, see Hardy & Wright's
"Introduction to the Theory of Numbers". Peace.

## NIM
from random import random
piles=[0,0,0]

########################################

def take(n,pileNum):
if piles[pileNum]>=n:
piles[pileNum]=piles[pileNum]-n
print piles
else:
print "illegal move"

########################################

def newGame():
for i in range(3):
piles[i]=int(9*random())+1
print piles

########################################

def indexOfMax():
returnValue=0
for i in range(1,3):
if piles[i]>piles[returnValue]:
returnValue=i
return returnValue

########################################

def leftmostBitIndex(n):
if 0<n<pow(2,29):
for i in range(1,31):
if pow(2,i)>n:
return (i-1)
else:
raise

########################################

def yourMove():
magicNum=piles^piles^piles
if magicNum==0:
i=indexOfMax()
piles[i]=piles[i]-1
else:
magicIndex=leftmostBitIndex(magicNum)
targetIndex=(-1)
for i in range(3):
if (piles[i]>>magicIndex)%2==1:
targetNum=piles[i]
targetIndex=i
break
replacement=0
for i in range(magicIndex):
magicDigit=(magicNum>>i)%2
targetDigit=(piles[targetIndex]>>i)%2
if magicDigit==1:
replacementDigit=(magicDigit-targetDigit)
else:
replacementDigit=targetDigit
replacement=replacement+replacementDigit*pow(2,i)
newNum=targetNum-(targetNum%pow(2,magicIndex+1))+replacement
piles[targetIndex]=newNum
print piles

############ not used in this script:

def matilda(n):
if 0<=n<pow(2,29):
for i in range(1,31):
iOnes=pow(2,i)-1
if n<=iOnes:
return iOnes-n
else:
raise
• at Aug 18, 2003 at 1:11 am ⇧ [Elaine Jackson]
In case any fellow newbies have been following this thread, here is
the finished Nim script. It doesn't use bit-manipulation as much as I
thought it would. (Still, I got the basics straight. Thanks again to
everyone who helped out!) For a readable account of the optimal
strategy for Nim, see Hardy & Wright's "Introduction to the Theory of
Numbers". Peace.

## NIM
from random import random
piles=[0,0,0]

...

def yourMove():
magicNum=piles^piles^piles
if magicNum==0:
i=indexOfMax()
piles[i]=piles[i]-1
else:
magicIndex=leftmostBitIndex(magicNum)
targetIndex=(-1)
for i in range(3):
if (piles[i]>>magicIndex)%2==1:
targetNum=piles[i]
targetIndex=i
break
replacement=0
for i in range(magicIndex):
magicDigit=(magicNum>>i)%2
targetDigit=(piles[targetIndex]>>i)%2
if magicDigit==1:
replacementDigit=(magicDigit-targetDigit)
else:
replacementDigit=targetDigit
replacement=replacement+replacementDigit*pow(2,i)
newNum=targetNum-(targetNum%pow(2,magicIndex+1))+replacement
piles[targetIndex]=newNum
print piles
OK, what you're trying to do is force the xor of the pile counts to 0.
There may be more than one way to do that. For example, if piles is

[2, 6, 7]

you can force a win by doing any of these:

remove 1 from the 2 pile, leaving [1, 6, 7]
remove 1 from the 6 pile, leaving [2, 5, 7]
remove 3 from the 7 pile, leaving [2, 6, 4]

because 1^6^7 == 2^5^7 == 2^6^4 == 0. Now the educational <wink> thing is
that it's possible to find all these solutions easily *without* picking
apart any bits:

def myMove(piles):
all = 0
for count in piles:
all ^= count
for i, count in enumerate(piles): # enumerate() new in 2.3
all_without_count = all ^ count
if all_without_count < count:
print "leave %d in pile #%d" % (all_without_count, i)

For example,
myMove([2, 6, 7])
leave 1 in pile #0
leave 5 in pile #1
leave 4 in pile #2
>>>

Note that the function works for any number of piles, and doesn't care how
big each pile may be; for example,
myMove([2**50, 3**67, 5**12, 76**30])
leave 92709463147897838211662074651978 in pile #3
2**50 ^ 3**67 ^ 5**12 ^ 92709463147897838211662074651978
0L
>>>

I think you'll enjoy figuring out why that function works; proving that it
finds all solutions is a bit delicate, but not truly hard.
• at Aug 18, 2003 at 4:50 pm ⇧ Cool. Thanks a bunch. I find the documentation about 'enumerate' a little
sketchy, but it seems like, as a first approximation, I can think of
enumerate(piles) as the list of all pairs [ i, piles[i] ] with 0 <= i <
len(piles). Is that right?

"Tim Peters" <tim.one at comcast.net> wrote in message
news:mailman.1061169151.30094.python-list at python.org...
[Elaine Jackson]
In case any fellow newbies have been following this thread, here is
the finished Nim script. It doesn't use bit-manipulation as much as I
thought it would. (Still, I got the basics straight. Thanks again to
everyone who helped out!) For a readable account of the optimal
strategy for Nim, see Hardy & Wright's "Introduction to the Theory of
Numbers". Peace.

## NIM
from random import random
piles=[0,0,0]

...

def yourMove():
magicNum=piles^piles^piles
if magicNum==0:
i=indexOfMax()
piles[i]=piles[i]-1
else:
magicIndex=leftmostBitIndex(magicNum)
targetIndex=(-1)
for i in range(3):
if (piles[i]>>magicIndex)%2==1:
targetNum=piles[i]
targetIndex=i
break
replacement=0
for i in range(magicIndex):
magicDigit=(magicNum>>i)%2
targetDigit=(piles[targetIndex]>>i)%2
if magicDigit==1:
replacementDigit=(magicDigit-targetDigit)
else:
replacementDigit=targetDigit
replacement=replacement+replacementDigit*pow(2,i)
newNum=targetNum-(targetNum%pow(2,magicIndex+1))+replacement
piles[targetIndex]=newNum
print piles
OK, what you're trying to do is force the xor of the pile counts to 0.
There may be more than one way to do that. For example, if piles is

[2, 6, 7]

you can force a win by doing any of these:

remove 1 from the 2 pile, leaving [1, 6, 7]
remove 1 from the 6 pile, leaving [2, 5, 7]
remove 3 from the 7 pile, leaving [2, 6, 4]

because 1^6^7 == 2^5^7 == 2^6^4 == 0. Now the educational <wink> thing is
that it's possible to find all these solutions easily *without* picking
apart any bits:

def myMove(piles):
all = 0
for count in piles:
all ^= count
for i, count in enumerate(piles): # enumerate() new in 2.3
all_without_count = all ^ count
if all_without_count < count:
print "leave %d in pile #%d" % (all_without_count, i)

For example,
myMove([2, 6, 7])
leave 1 in pile #0
leave 5 in pile #1
leave 4 in pile #2
Note that the function works for any number of piles, and doesn't care how
big each pile may be; for example,
myMove([2**50, 3**67, 5**12, 76**30])
leave 92709463147897838211662074651978 in pile #3
2**50 ^ 3**67 ^ 5**12 ^ 92709463147897838211662074651978
0L
I think you'll enjoy figuring out why that function works; proving that it
finds all solutions is a bit delicate, but not truly hard.
• at Aug 19, 2003 at 4:43 am ⇧ In case anybody is still following this thread, the correctness and completeness
of the 'myMove' function (see below) are easy consequences of the following
lemmas:
1) xor is commutative and associative
2) zero is the identity element for xor
3) every number is its own inverse wrt xor

"Tim Peters" <tim.one at comcast.net> wrote in message
news:mailman.1061169151.30094.python-list at python.org...
[Elaine Jackson]
In case any fellow newbies have been following this thread, here is
the finished Nim script. It doesn't use bit-manipulation as much as I
thought it would. (Still, I got the basics straight. Thanks again to
everyone who helped out!) For a readable account of the optimal
strategy for Nim, see Hardy & Wright's "Introduction to the Theory of
Numbers". Peace.

## NIM
from random import random
piles=[0,0,0]

...

def yourMove():
magicNum=piles^piles^piles
if magicNum==0:
i=indexOfMax()
piles[i]=piles[i]-1
else:
magicIndex=leftmostBitIndex(magicNum)
targetIndex=(-1)
for i in range(3):
if (piles[i]>>magicIndex)%2==1:
targetNum=piles[i]
targetIndex=i
break
replacement=0
for i in range(magicIndex):
magicDigit=(magicNum>>i)%2
targetDigit=(piles[targetIndex]>>i)%2
if magicDigit==1:
replacementDigit=(magicDigit-targetDigit)
else:
replacementDigit=targetDigit
replacement=replacement+replacementDigit*pow(2,i)
newNum=targetNum-(targetNum%pow(2,magicIndex+1))+replacement
piles[targetIndex]=newNum
print piles
OK, what you're trying to do is force the xor of the pile counts to 0.
There may be more than one way to do that. For example, if piles is

[2, 6, 7]

you can force a win by doing any of these:

remove 1 from the 2 pile, leaving [1, 6, 7]
remove 1 from the 6 pile, leaving [2, 5, 7]
remove 3 from the 7 pile, leaving [2, 6, 4]

because 1^6^7 == 2^5^7 == 2^6^4 == 0. Now the educational <wink> thing is
that it's possible to find all these solutions easily *without* picking
apart any bits:

def myMove(piles):
all = 0
for count in piles:
all ^= count
for i, count in enumerate(piles): # enumerate() new in 2.3
all_without_count = all ^ count
if all_without_count < count:
print "leave %d in pile #%d" % (all_without_count, i)

For example,
myMove([2, 6, 7])
leave 1 in pile #0
leave 5 in pile #1
leave 4 in pile #2
Note that the function works for any number of piles, and doesn't care how
big each pile may be; for example,
myMove([2**50, 3**67, 5**12, 76**30])
leave 92709463147897838211662074651978 in pile #3
2**50 ^ 3**67 ^ 5**12 ^ 92709463147897838211662074651978
0L
I think you'll enjoy figuring out why that function works; proving that it
finds all solutions is a bit delicate, but not truly hard.