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Determining combination of bits

Say I have a dictionary like the following

{1:'one',2:'two',4:'three',8:'four',16:'five', etc...}

and I am given some numbers, say 22, 25, and 9. I want to determine the
keys, powers of 2, that comprise the number.

Ex. 22 = 16+4+2
25 = 16+8+1
9 = 8+1
....etc...

How do I get these keys?
Jul 18 '05 #1
25 1823
Sean Berry <sean <at> buildingonline.com> writes:

I want to determine the keys, powers of 2, that comprise the number.

Ex. 22 = 16+4+2
25 = 16+8+1
9 = 8+1
...etc...


This sounds suspiciously like a homework, so I'm just going to give a hint here:
import math
math.log(22, 2) 4.4594316186372973 2**int(math.log(22, 2)) 16 2**int(math.log(22 - 16, 2)) 4 2**int(math.log(22 - 16 - 4 - 2, 2)) 2


Hope this is helpful.

Steve

Jul 18 '05 #2
You didn't say anything about how large the
numbers might get. If they stay small (e.g. <thousands)
you could just look them up directly from a dictionary.
If they can get arbitrarily large you must use bit
shifing and boolean & (and).

Sounds a lot like a homework assignment, but I'll
give you some "hints".

1) Use bit shifting operator (>>) and boolean &
operator to isolate each bit in the integer.

2) It will be either zero or one. Build up a
list of these which will represent the power
of two that each one bit represents.

Larry Bates

Sean Berry wrote:
Say I have a dictionary like the following

{1:'one',2:'two',4:'three',8:'four',16:'five', etc...}

and I am given some numbers, say 22, 25, and 9. I want to determine the
keys, powers of 2, that comprise the number.

Ex. 22 = 16+4+2
25 = 16+8+1
9 = 8+1
...etc...

How do I get these keys?

Jul 18 '05 #3

"Sean Berry" <se**@buildingonline.com> wrote in message
news:x3Qjd.121786$hj.41260@fed1read07...
and I am given some numbers, say 22, 25, and 9. I want to determine the
keys, powers of 2, that comprise the number.
How do I get these keys?


if n%2: print 'has a positive one bit'

n//2 == n>>1 deletes that bit

keep track of divisions/shifts and stop when n == 0

Terry J. Reedy

Jul 18 '05 #4
Just to set everyone's mind at ease... I haven't had a homework assignment
for about four years now.

I am using two database tables. One will have some options and a code
(power of 2).

Then, someone will check off checkboxes and submit. The number will be
added and saved in a cookie. Then, later, I want to be able to redisplay
their choices by reading the value from the cookie.

I expect the values will get no bigger than 2^32 = 4294967296. Is this
getting too big???

"Terry Reedy" <tj*****@udel.edu> wrote in message
news:ma**************************************@pyth on.org...

"Sean Berry" <se**@buildingonline.com> wrote in message
news:x3Qjd.121786$hj.41260@fed1read07...
and I am given some numbers, say 22, 25, and 9. I want to determine the
keys, powers of 2, that comprise the number.
How do I get these keys?


if n%2: print 'has a positive one bit'

n//2 == n>>1 deletes that bit

keep track of divisions/shifts and stop when n == 0

Terry J. Reedy

Jul 18 '05 #5
Sean Berry <sean <at> buildingonline.com> writes:

I am using two database tables. One will have some options and a code
(power of 2).

Then, someone will check off checkboxes and submit. The number will be
added and saved in a cookie. Then, later, I want to be able to redisplay
their choices by reading the value from the cookie.


You might look at the number_in_base thread:

http://groups.google.com/groups?thre...%40news.oz.net

This gives a solution something like:
def number_in_base(n, b=10, digits='0123456789ABCDEF'): .... return '-'[:n<0]+''.join(reversed(list(
.... digits[r]
.... for q in [abs(n)]
.... for q, r in iter(lambda: divmod(q, b), (0, 0))))) or digits[0]
.... digits = number_in_base(22, 2)
[power_of_2

.... for power_of_2, digit
.... in zip(reversed([2**x for x in range(len(digits))]), digits)
.... if digit == '1']
[16, 4, 2]
Steve

Jul 18 '05 #6
Sean Berry wrote:
Just to set everyone's mind at ease... I haven't had a homework assignment
for about four years now.

OK, then here's a great little bit of education:

n & -n == least significant bit of n

So, for example:

def bits(n):
assert n >= 0 # This is an infinite loop if n < 0
while n:
lsb = n & -b
yield lsb
n ^= lsb

-Scott David Daniels
Sc***********@Acm.Org
Jul 18 '05 #7
Scott David Daniels wrote:
Sean Berry wrote:
Just to set everyone's mind at ease... I haven't had a
homework assignment for about four years now.

OK, then here's a great little bit of education:

n & -n == least significant bit of n


A lapse of mind?
n = 6
n & (-n)

2

You probably meant n&1 or perhaps n%2.
Jul 18 '05 #8
> A lapse of mind?
n = 6
n & (-n)

2

You probably meant n&1 or perhaps n%2.


No, the exact right thing: 6 is binary

110

with the least significant bit beeing

10

thus the decimal value of 2.

Albeit I have to admit that I didn't know the trick.
--
Regards,

Diez B. Roggisch
Jul 18 '05 #9
On 2004-11-08, Diez B. Roggisch <de*********@web.de> wrote:
A lapse of mind?
> n = 6
> n & (-n)

2

You probably meant n&1 or perhaps n%2.


No, the exact right thing: 6 is binary

110

with the least significant bit beeing

10


The least significant bit of 110 is this one here -----+
^ |
| |
+-----------------------+

It's a 0 (zero).

What I think you're trying to say is something like the value
of the rightmost 1.

--
Grant Edwards grante Yow! The PINK SOCKS were
at ORIGINALLY from 1952!! But
visi.com they went to MARS around
1953!!
Jul 18 '05 #10
Diez B. Roggisch wrote:
A lapse of mind?
> n = 6
> n & (-n)

2

You probably meant n&1 or perhaps n%2.


No, the exact right thing: 6 is binary

110

with the least significant bit beeing

10

thus the decimal value of 2.


Wow. I always thought "least significant bit" meant, in the common big-endian bit notation, the last bit - with emphasis on the bit
word, i.e. either 1 or 0.
Have I got a real mess in my head (probably not, I just googled for definitions and the onse I checked are consistent with what I
wrote, http://www.google.com/search?hl=en&l...ificant+bit%22) or is the meaning of the phrase poorly defined?
Jul 18 '05 #11
I wrote:
def bits(n):
assert n >= 0 # This is an infinite loop if n < 0
while n:
lsb = n & -b
yield lsb
n ^= lsb
Diez B. Roggisch wrote:
A lapse of mind?
>n = 6
>n & (-n)

2


Stupid typo, should be:
def bits(n):
assert n >= 0 # This is an infinite loop if n < 0
while n:
lsb = n & -n ### Here was ... & -b, which is clearly wrong
yield lsb
n ^= lsb

list(bits(6)) == [2, 4]
sum(bits(6)) == 6
list(bits(5000)) == [8, 128, 256, 512, 4096]
sum(bits(5000)) == 5000
Albeit I have to admit that I didn't know the trick.

In fact I wouldn't know it if I hadn't worked on ancient machines
with "exposed guts". One machine (a PDP-8, if I remember correctly)
did not have a "negate" operation. Instead you did, "ones complement
(OCA) and increment (INA)". Note that a ones complement turns all of
the least significant zeros to ones, and the least significant one to
a zero. When you increment that the carry propagates back to the 0 for
the least one. The carry stops there. This will exactly match all the
low bits, and none of the bits above the least significant 1. For the
least significant 1 bit, the calculation is 1 == 1 & 1,
for the others it is one of 0 == 0&1 == 1&0 == 0&0

Sample:
123456 = ...000 011 110 001 001 000 000
~123456 = ...111 100 001 110 110 111 111
1+~123456 = ...111 100 001 110 111 000 000

123456 = ...000 011 110 001 001 000 000
-123456 = ...111 100 001 110 111 000 000
123456^-123456 = ...000 000 000 000 001 000 000
Jul 18 '05 #12

Grant Edwards <gr****@visi.com> wrote:

On 2004-11-08, Diez B. Roggisch <de*********@web.de> wrote:
A lapse of mind?
>> n = 6
>> n & (-n)
2

You probably meant n&1 or perhaps n%2.


No, the exact right thing: 6 is binary

110

with the least significant bit beeing

10


The least significant bit of 110 is this one here -----+
^ |
| |
+-----------------------+

It's a 0 (zero).

What I think you're trying to say is something like the value
of the rightmost 1.


From what I remember of high school chemistry (7 years ago), we used to
talk about 'significant figures' quite often. If the teacher asked for
some number to 5 significant figures, it had better have them...

sig_fig(83737256,5) -> 83737000
sig_fig(1,5) -> 1.0000

etc.

Now, in the case of 'least significant bit of n', that can be
interpreted as either n&1, or the rightmost bit that is significant
(nonzero).

The n&-n produces the rightmost bit that is nonzero, which is certainly
a valid interpretation of 'least significant bit of n'.

- Josiah

Jul 18 '05 #13
>
The least significant bit of 110 is this one here -----+
^ |
| |
+-----------------------+

It's a 0 (zero).

What I think you're trying to say is something like the value
of the rightmost 1.


Yup.

--
Regards,

Diez B. Roggisch
Jul 18 '05 #14
> Wow. I always thought "least significant bit" meant, in the common
big-endian bit notation, the last bit - with emphasis on the bit word,
i.e. either 1 or 0. Have I got a real mess in my head (probably not, I
just googled for definitions and the onse I checked are consistent with
what I wrote,
http://www.google.com/search?hl=en&l...ificant+bit%22) or
is the meaning of the phrase poorly defined?


No, seems to be well defined - I was the confused one. I have to admit that
I was somewhat impressed by that neat trick that I simply extended the
meaning of LSB in my faulty wet-ware.

I've done my share of assembler coding, but never happened to have used or
seen that one.

--
Regards,

Diez B. Roggisch
Jul 18 '05 #15
On 2004-11-08, Josiah Carlson <jc******@uci.edu> wrote:
The least significant bit of 110 is this one here -----+
^ |
| |
+-----------------------+

It's a 0 (zero).

What I think you're trying to say is something like the value
of the rightmost 1.


From what I remember of high school chemistry (7 years ago), we used to
talk about 'significant figures' quite often. If the teacher asked for
some number to 5 significant figures, it had better have them...

sig_fig(83737256,5) -> 83737000
sig_fig(1,5) -> 1.0000

etc.

Now, in the case of 'least significant bit of n', that can be
interpreted as either n&1, or the rightmost bit that is significant
(nonzero).

The n&-n produces the rightmost bit that is nonzero, which is certainly
a valid interpretation of 'least significant bit of n'.


Perhaps it's "valid", but in 25+ years of doing hardware and
low-level software this is the first time I've ever heard the
phrase "least significant bit" refer to anything other than the
"rightmost" bit (the one with a weighting of 1).

--
Grant Edwards grante Yow! When you get your
at PH.D. will you get able to
visi.com work at BURGER KING?
Jul 18 '05 #16

"Scott David Daniels" <Sc***********@Acm.Org> wrote in message
news:41********@nntp0.pdx.net...
Sean Berry wrote:
Just to set everyone's mind at ease... I haven't had a homework
assignment for about four years now. OK, then here's a great little bit of education:


Better than the education I got at UCSB...
n & -n == least significant bit of n

So, for example:

def bits(n):
assert n >= 0 # This is an infinite loop if n < 0
while n:
lsb = n & -b
yield lsb
n ^= lsb

Perfect. This not only does exactly what I wanted... but I have just
started to learn about generators, and this does great.

Thank you very much for the help everyone.

-Scott David Daniels
Sc***********@Acm.Org

Jul 18 '05 #17
On Mon, 8 Nov 2004 12:33:48 -0800, "Sean Berry"
<se**@buildingonline.com> declaimed the following in comp.lang.python:
Then, someone will check off checkboxes and submit. The number will be
added and saved in a cookie. Then, later, I want to be able to redisplay
their choices by reading the value from the cookie.
I'd just save the raw number, and build the list of choices by
ANDing against the number.

I expect the values will get no bigger than 2^32 = 4294967296. Is this
getting too big???
Well, it will go to Long int in Python as I recall... 2^31 is
largest signed int...
My first cut though, was:

-=-=-=-=-=-=-=-=-

# powers of two in integer value

val = int(raw_input("Enter the integer to be evaluated> "))
val = abs(val)
sval = val

pwrs = []
bit = 0

while val:
if val & 1:
pwrs.append(bit)
val = val >> 1
bit = bit + 1

pwrs.reverse()

print sval, "=",

sequence = 0
for p in pwrs:
if sequence:
print "+",
else:
sequence = 1
print 2**p,

print

-=-=-=-=-=-=-=-=-=-
22 = 16 + 4 + 2
25 = 16 + 8 + 1
9 = 8 + 1

Note: 2^1 = 2, so your dictionary is already in error...

Now -- on the concept of rebuilding a list of choices....

CheckBoxes = { "FirstChoice" : 1,
"SecondChoice" : 2,
"ThirdChoice" : 4,
"FourthChoice": 8,
"FifthChoice" : 16,
"SixthChoice" : 32 }
for num in [22, 25, 9]:
for k,i in CheckBoxes.items():
if num & i:
print k,
print
FifthChoice SecondChoice ThirdChoice
FirstChoice FifthChoice FourthChoice
FirstChoice FourthChoice
-- ================================================== ============ <
wl*****@ix.netcom.com | Wulfraed Dennis Lee Bieber KD6MOG <
wu******@dm.net | Bestiaria Support Staff <
================================================== ============ <
Home Page: <http://www.dm.net/~wulfraed/> <
Overflow Page: <http://wlfraed.home.netcom.com/> <

Jul 18 '05 #18
On Mon, 08 Nov 2004 15:33:01 -0800, Scott David Daniels
<Sc***********@Acm.Org> declaimed the following in comp.lang.python:

(OCA) and increment (INA)". Note that a ones complement turns all of
the least significant zeros to ones, and the least significant one to
a zero. When you increment that the carry propagates back to the 0 for
Ones complement turns ALL 0 bits to 1, and ALL 1 bits to 0.

00000000
1C 11111111 ones complement has a "negative zero"
+1 00000000 twos complement "overflows" back to single zero

-- ================================================== ============ <
wl*****@ix.netcom.com | Wulfraed Dennis Lee Bieber KD6MOG <
wu******@dm.net | Bestiaria Support Staff <
================================================== ============ <
Home Page: <http://www.dm.net/~wulfraed/> <
Overflow Page: <http://wlfraed.home.netcom.com/> <

Jul 18 '05 #19
> Note: 2^1 = 2, so your dictionary is already in error...


The dictionary was filled with arbitrary values, not
{ x : 2^x } values like you might have thought.

It is actually more like {1:123, 2:664, 4:323, 8:990, 16:221... etc}

Jul 18 '05 #20
Larry Bates <lb****@syscononline.com> wrote:
Sounds a lot like a homework assignment
Indeed!
but I'll give you some "hints".

1) Use bit shifting operator (>>) and boolean &
operator to isolate each bit in the integer.
Shifting isn't necessary - D.keys() contains a list of all possible
(for this problem) binary numbers.
2) It will be either zero or one. Build up a
list of these which will represent the power
of two that each one bit represents.


Whenever you think "build up a list" you should be thinking list
comprehension. A conditional add to a list should make you think of
the if clause of a list comprehension.
D={1:'one',2:'two',4:'three',8:'four',16:'five'}
def f(n): return [D[i] for i in D.keys() if XXXXX] .... f(9) ['four', 'one'] f(22) ['two', 'three', 'five'] f(25)

['four', 'one', 'five']

I've left XXXXX as an excercise - see 1) above for a hint ;-)

--
Nick Craig-Wood <ni**@craig-wood.com> -- http://www.craig-wood.com/nick
Jul 18 '05 #21
On Mon, 8 Nov 2004 11:48:22 -0800, "Sean Berry" <se**@buildingonline.com> wrote:
Say I have a dictionary like the following

{1:'one',2:'two',4:'three',8:'four',16:'five', etc...}

and I am given some numbers, say 22, 25, and 9. I want to determine the
keys, powers of 2, that comprise the number.

Ex. 22 = 16+4+2
25 = 16+8+1
9 = 8+1
...etc...

How do I get these keys?

Since it's not homework, and I seem to have read your question a little
differently from the others, maybe this will be useful...
bitdict = {1:'one',2:'two',4:'three',8:'four',16:'five'}
def bitnames(n): ... return [t[1] for t in sorted([kv for kv in bitdict.items() if n&kv[0]])][::-1]
... bitnames(11) ['four', 'two', 'one'] bitnames(15) ['four', 'three', 'two', 'one'] bitnames(31) ['five', 'four', 'three', 'two', 'one'] bitnames(9) ['four', 'one'] bitnames(20) ['five', 'three']

If you don't care about the order, you can leave out the sorting and reversal.

I gather you want to put in something other than strings 'one','two', etc. as bit definitions,
otherwise you could define your dict from names in a single list, which makes the numbers less
typo-prone (since you're not typing them ;-) e.g.
namelist = 'one two three four five'.split()
bitdict = dict((2**i,name) for i,name in enumerate(namelist))
bitdict {8: 'four', 1: 'one', 2: 'two', 4: 'three', 16: 'five'}

This uses python 2.4b1 BTW, so you will have to change sorted and put [] around the
dict generator expression argument above.

You could also use the list instead of a dict, since you know you have an ordered
compact set of values corresponding to the bits, and since the order is still there
you don't need to sort. E.g.,
def bitnames2(n): ... return [name for i, name in enumerate(namelist) if n&2**i][::-1]
... bitnames2(11) ['four', 'two', 'one'] bitnames2(9) ['four', 'one'] bitnames2(31)

['five', 'four', 'three', 'two', 'one']

HTH

Regards,
Bengt Richter
Jul 18 '05 #22
On Mon, 8 Nov 2004 21:18:36 -0800, "news.west.cox.net"
<se*********@cox.net> declaimed the following in comp.lang.python:
Note: 2^1 = 2, so your dictionary is already in error...

The dictionary was filled with arbitrary values, not
{ x : 2^x } values like you might have thought.


Well, you had stated "powers of two"... If all you wanted is a
bit mapping you could probably drop the dictionary and just use a list
of the values, indexed by the bit position, and my first attempt
logic...

It is actually more like {1:123, 2:664, 4:323, 8:990, 16:221... etc}


CheckBoxes = [ "FirstChoice",
"SecondChoice",
"ThirdChoice",
"FourthChoice",
"FifthChoice",
"SixthChoice" ]
for num in [22, 25, 9]:
bit = 0
while num:
if num & 1:
print CheckBoxes[bit],
bit = bit + 1
num = num >> 1
print

SecondChoice ThirdChoice FifthChoice
FirstChoice FourthChoice FifthChoice
FirstChoice FourthChoice

where "num" is the sum of the checkbox index values (or whatever
selection mechanism is used), assuming /they/ were set up in 2^(n+1)
scheme (n = bit position, starting with 0)...

-- ================================================== ============ <
wl*****@ix.netcom.com | Wulfraed Dennis Lee Bieber KD6MOG <
wu******@dm.net | Bestiaria Support Staff <
================================================== ============ <
Home Page: <http://www.dm.net/~wulfraed/> <
Overflow Page: <http://wlfraed.home.netcom.com/> <

Jul 18 '05 #23
"Sean Berry" <se**@buildingonline.com> wrote in message news:<x3Qjd.121786$hj.41260@fed1read07>...
Say I have a dictionary like the following

{1:'one',2:'two',4:'three',8:'four',16:'five', etc...}
Dont know exactly what your dictionary should represent.
Since 2**0 = 1
2**1 = 2
2**2 = 4
etc.
So I would have expected something like:
{1:'zero',2:'one',4:'two',8:'three',16:'four',32:' five', etc...}


and I am given some numbers, say 22, 25, and 9. I want to determine the
keys, powers of 2, that comprise the number.

Ex. 22 = 16+4+2
25 = 16+8+1
9 = 8+1
...etc...

How do I get these keys?


Solution No. XXXXXXXX:
def fn(n): .... number=n
.... if n<=0:
.... return 'n must be greater 0'
.... keys=[]
.... i=1
.... while n:
.... if n&1:
.... keys.append(i)
.... n=n>>1
.... i*=2
.... keys.reverse()
.... l=map(str,keys)
.... print '%d = %s' % (number,'+'.join(l))
.... return keys
....
print fn(22) 22 = 16+4+2
[16, 4, 2] print fn(25) 25 = 16+8+1
[16, 8, 1] print fn(255) 255 = 128+64+32+16+8+4+2+1
[128, 64, 32, 16, 8, 4, 2, 1]


Sorry I have only Python 2.2. and though I'm a one-liner-fan my
solution should be clear.

Regards Peter
Jul 18 '05 #24
Grant Edwards wrote:
Perhaps it's "valid", but in 25+ years of doing hardware and
low-level software this is the first time I've ever heard the
phrase "least significant bit" refer to anything other than the
"rightmost" bit (the one with a weighting of 1).

I tried to say, 'least significant one bit' (or 'on bit), but may
missed it in at least a sentence or two.

--Scott David Daniels
Sc***********@Acm.Org
Jul 18 '05 #25
Dennis Lee Bieber wrote:
On Mon, 08 Nov 2004 15:33:01 -0800, Scott David Daniels
<Sc***********@Acm.Org> declaimed the following in comp.lang.python:
(OCA) and increment (INA)". Note that a ones complement turns all of
the least significant zeros to ones, and the least significant one to
a zero. When you increment that the carry propagates back to the 0 for Ones complement turns ALL 0 bits to 1, and ALL 1 bits to 0.

Right. In particular, all of the lowest order zeroes turn to 1s,
the one directly before them turns to zero. Those bits are the only
bits where I care what the value is, all others are simply inverted
(and it doesn't matter what values they have).
00000000
1C 11111111 ones complement has a "negative zero"
+1 00000000 twos complement "overflows" back to single zero


Correct, and this isoloates the least significant one bit for this value
as well (inasmuch as it doesn't exist).

--Scott David Daniels
Sc***********@Acm.Org

Jul 18 '05 #26

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Hi Guys, I have a domain whose name is BytesLimited.com, and I want to sell it. Does anyone know about platforms that allow me to list my domain in auction for free. Thank you
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by: Faith0G | last post by:
I am starting a new it consulting business and it's been a while since I setup a new website. Is wordpress still the best web based software for hosting a 5 page website? The webpages will be...
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isladogs
by: isladogs | last post by:
The next Access Europe User Group meeting will be on Wednesday 3 Apr 2024 starting at 18:00 UK time (6PM UTC+1) and finishing by 19:30 (7.30PM). In this session, we are pleased to welcome former...

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