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the algorithm of decimal<->binary conversion

I'm writing a program for listing all binary numbers of the same length
with the same number of 1s.

e.g.
0011
0101
0110
1001
1010
1100

my algorithm is as follows...
0. use a formula to calculate the number of numbers
1. temp=1; outputed=0;
2. count from 1 to 2 to the power of the required length
2.1 convert temp to a binary number in a char[]
2.1.1 temp/2 get remainder
2.1.2 blah blah blah.........
2.1.2.1 if no. of 1s == required no., output+1, printf
2.2 temp+1;
2.3 if output>=the value calculated, break

it works perfectly...
but it seems too slow even when calculating binary numbers of length 20...
(as this is a program for the demo online acm contest, time limit is set...)

how can I make it more efficient????
Thx~~~~

Mars.
Nov 14 '05 #1
16 2434
Mars wrote:

I'm writing a program for listing all binary numbers of the same length
with the same number of 1s.
<snip>
it works perfectly...
but it seems too slow even when calculating binary numbers of length 20...
(as this is a program for the demo online acm contest, time limit is set...)

how can I make it more efficient????


Make sure you are using an efficient algorithm.
Use a profiler to identify the principal bottleneck.
Rewrite the bottleneck code.
Iterate until fast enough.
Nov 14 '05 #2
infobahn mentioned:
Mars wrote:
I'm writing a program for listing all binary numbers of the same length
with the same number of 1s.


<snip>
it works perfectly...
but it seems too slow even when calculating binary numbers of length 20...
(as this is a program for the demo online acm contest, time limit is set...)

how can I make it more efficient????

Make sure you are using an efficient algorithm.
Use a profiler to identify the principal bottleneck.
Rewrite the bottleneck code.
Iterate until fast enough.


ummm....
is this good enough??

typedef struct
{
char bits[2000];
int pointer;
int length;

}conv;

conv convert(long long input)
{
int pass;
conv rconv;

rconv=init();

while (1)
{
if ((input!=1)&&(input!=0))
{
pass=input%2;
input/=2;
}
else
{
rconv.bits[rconv.pointer]=input+48;
rconv.pointer++;
if (input==1)
rconv.length++;
break;
}

if (pass==1)
{
rconv.bits[rconv.pointer]='1';
rconv.pointer++;
rconv.length++;
}
else
{
rconv.bits[rconv.pointer]='0';
rconv.pointer++;
}
}
Nov 14 '05 #3
On Mon, 21 Feb 2005 01:18:35 +0800, Mars
<Mars@Mars> wrote:
I'm writing a program for listing all binary numbers of the same length
with the same number of 1s.
That's not what the subject line says.
it works perfectly...
but it seems too slow even when calculating binary numbers of length 20...
Yes, that's a million (and a bit) tests.
(as this is a program for the demo online acm contest, time limit is set...)
Ah. Might it be that you are looking to use someone else's work in a
contest? Might that be considered just a tad unethical?
how can I make it more efficient????


Change your algorithm to a more efficient one. Hint: instead of an
exponential time with the number of bits, it can be done O(M*N) (where M
is the number of bits set and N is the total number of bits). In fact
it can be done better than that. Think shifts...

Chris C
Nov 14 '05 #4
Chris Croughton wrote:
On Mon, 21 Feb 2005 01:18:35 +0800, Mars
<Mars@Mars> wrote:

[Request for an improvement of program/algorithm to output all
numbers representable with N bits where M bits are set]
how can I make it more efficient????


Change your algorithm to a more efficient one. Hint: instead of an
exponential time with the number of bits, it can be done O(M*N) (where M
is the number of bits set and N is the total number of bits). In fact
it can be done better than that. Think shifts...


Hmmm, I cannot follow:
The number of numbers is
/ N \
| |
\ M /
with a maximum at M = N/2.
If I use M = 3 and consider N=n and N=2*n, I get first
n*(n-1)*(n-2)/6
and then
4*(2n-1)*n*(n-1)/6 > 8*n*(n-1)*(n-2)/6
which is by no means linear in N. We have to output
considerably more than O(M*N) numbers -- how do you plan
to do that in at most quadratic time?

-Michael
--
E-Mail: Mine is an /at/ gmx /dot/ de address.
Nov 14 '05 #5
In article <sl******************@ccserver.keris.net>,
Chris Croughton <ch***@keristor.net> wrote:
:On Mon, 21 Feb 2005 01:18:35 +0800, Mars
: <Mars@Mars> wrote:

:> I'm writing a program for listing all binary numbers of the same length
:> with the same number of 1s.

:Hint: instead of an
:exponential time with the number of bits, it can be done O(M*N) (where M
:is the number of bits set and N is the total number of bits). In fact
:it can be done better than that. Think shifts...

M bits set out of N is (N choose N), which will be N! / (M! (N-M)!)
That's a bit bigger than O(M*N). For example, 8 bits 3 set, you
predict O(8*3), but instead it is O((8*7*6)/(1*2*3)) = O(56):

00000111 00001011 00001101 00001110 00010011 00010101 00010110 00011001
00011010 00011100 00100011 00100101 00100110 00101001 00101010 00101100
00110001 00110010 00110100 00111000 01000011 01000101 01000110 01001001
01001010 01001100 01010001 01010010 01010100 01011000 01100001 01100010
01100100 01101000 01110000 10000011 10000101 10000110 10001001 10001010
10001100 10010001 10010010 10010100 10011000 10100001 10100010 10100100
10101000 10110000 11000001 11000010 11000100 11001000 11010000 11100000
--
Pity the poor electron, floating around minding its own business for
billions of years; and then suddenly Bam!! -- annihilated just so
you could read this posting.
Nov 14 '05 #6
Mars wrote:
I'm writing a program for listing all binary numbers of the same length with the same number of 1s.

e.g.
0011
0101
0110
1001
1010
1100


Use bit manipulations. Given a valid number, add the last bit to
the original, then put enough 1s in the low end to keep the total
1-bits consistent.

% type nmo.c
long strtol(const char*,char**,int);void q1(unsigned q2,int
q3){unsigned q4=(q3>0)? 1u<<(q3-1):-1u/2 +1;for(;q4;q4>>=1)
putchar('0'+!!(q2&q4));}int q5(unsigned q2){int q6;for(q6=0
;q2;q2&= q2-1,q6++);return q6 ;}int main(int q7,char **q8){
long q3,q4;unsigned q2,q9, q10;if(q7!=3)return 0;q3=strtol(
q8[1],0,0);q4=strtol(q8[2],0,0);if(q3<1||q4<1||q3<q4||q3>q5
(-1u))return 0;q2=(1u<<(q4-1)<<1)-1;q9=q2<<(q3-q4);for(;;){
q1(q2,q3);putchar('\n');if(q2==q9)break;q10=q2&-q2;q2+=q10+
((((q2+q10)&-(q2+q10))/q10)-1)/2;}return 0;}

% gcc -ansi -pedantic nmo.c -o nmo.exe

% nmo 4 2
0011
0101
0110
1001
1010
1100

%

--
Peter

Nov 14 '05 #7
Peter Nilsson wrote:
Mars wrote:
I'm writing a program for listing all binary numbers of the same

length
with the same number of 1s.

e.g.
0011
0101
0110
1001
1010
1100


Use bit manipulations. Given a valid number, add the last bit to
the original, then put enough 1s in the low end to keep the total
1-bits consistent.


Alternatively, if you want to deal with strings...

http://groups-beta.google.com/group/...c0260bb56e83f2

--
Peter

Nov 14 '05 #8
Mars wrote:
ummm....
is this good enough??

conv convert(long long input)
{
int pass;
conv rconv;

rconv=init();

while (1)
{
if ((input!=1)&&(input!=0))
{
pass=input%2;
input/=2;


Division is slow (modulus is essentially division as well.) Also, a
note on code comments for Usenet: Since lines can be permanently
wrapped, you are best to put all comments in /* */ style if you are
going to post. This allows people to copy and paste directly and still
have it compile.
But I say this only because you have none.

2 >> 1 = 1 //division by two
4 >> 1 = 2
6 >> 1 = 3
8 >> 1 = 4
8 >> 2 = 2 //division by four

1 & 1 = 1 //modulus of 2
2 & 1 = 0
3 & 1 = 1
4 & 1 = 0
4 & 2 = 0 //modulus of 4

-Chris

Nov 14 '05 #9
On Sun, 20 Feb 2005 21:37:18 +0100, Michael Mair
<Mi**********@invalid.invalid> wrote:
Chris Croughton wrote:
On Mon, 21 Feb 2005 01:18:35 +0800, Mars
<Mars@Mars> wrote:

[Request for an improvement of program/algorithm to output all
numbers representable with N bits where M bits are set]
how can I make it more efficient????


Change your algorithm to a more efficient one. Hint: instead of an
exponential time with the number of bits, it can be done O(M*N) (where M
is the number of bits set and N is the total number of bits). In fact
it can be done better than that. Think shifts...


Hmmm, I cannot follow:
The number of numbers is
/ N \
| |
\ M /
with a maximum at M = N/2.
If I use M = 3 and consider N=n and N=2*n, I get first
n*(n-1)*(n-2)/6
and then
4*(2n-1)*n*(n-1)/6 > 8*n*(n-1)*(n-2)/6
which is by no means linear in N. We have to output
considerably more than O(M*N) numbers -- how do you plan
to do that in at most quadratic time?


Yes, you're right, it's worse than I was thinking, it's something better
than O(N^M) (O(N^m) for any given M=m). Which is still better than
O(2^N), it's permutation not exponential, for N=20 and M=2 it will be
n(n-1)/2 = 190, for N=40 M=3 it will be n(n-1)(n-2)/6 = 9880 rather than
1,048,576 and 1,099,511,627,776 respectively.

(Using ^ as exponentiation operator above, just in case anyone was
wondering why N xor M would give meaningful result!)

(My O(M*N) was right for at least one case, M=1 <g>...)

Chris C
Nov 14 '05 #10
Chris Croughton mentioned:
On Mon, 21 Feb 2005 01:18:35 +0800, Mars
<Mars@Mars> wrote:

I'm writing a program for listing all binary numbers of the same length
with the same number of 1s.

That's not what the subject line says.

it works perfectly...
but it seems too slow even when calculating binary numbers of length 20...

Yes, that's a million (and a bit) tests.

(as this is a program for the demo online acm contest, time limit is set...)

Ah. Might it be that you are looking to use someone else's work in a
contest? Might that be considered just a tad unethical?


Rest assured.
That's just a DEMO contest system, all questions are past questions used
in the real ACM Contest.
Everyone can go register and try them.

http://acm.uva.es/problemset/
Nov 14 '05 #11
Chris Williams wrote:
Mars wrote:
ummm....
is this good enough??

conv convert(long long input)
{
int pass;
conv rconv;

rconv=init();

while (1)
{
if ((input!=1)&&(input!=0))
{
pass=input%2;
input/=2;

Division is slow (modulus is essentially division as well.)


Not on all systems.
Also, a
note on code comments for Usenet: Since lines can be permanently
wrapped, you are best to put all comments in /* */ style if you are
going to post. This allows people to copy and paste directly and still
have it compile.
That is true.
But I say this only because you have none.

2 >> 1 = 1 //division by two
4 >> 1 = 2
6 >> 1 = 3
8 >> 1 = 4
8 >> 2 = 2 //division by four

1 & 1 = 1 //modulus of 2
2 & 1 = 0
3 & 1 = 1
4 & 1 = 0
4 & 2 = 0 //modulus of 4


Any decent optimising compiler will do this optimisation when dealing
with division by a constant. So it is generally better to write what you
mean rather than making the code harder to read.
--
Flash Gordon
Living in interesting times.
Although my email address says spam, it is real and I read it.
Nov 14 '05 #12
Peter Nilsson mentioned:

% type nmo.c
long strtol(const char*,char**,int);void q1(unsigned q2,int
q3){unsigned q4=(q3>0)? 1u<<(q3-1):-1u/2 +1;for(;q4;q4>>=1)
putchar('0'+!!(q2&q4));}int q5(unsigned q2){int q6;for(q6=0
;q2;q2&= q2-1,q6++);return q6 ;}int main(int q7,char **q8){
long q3,q4;unsigned q2,q9, q10;if(q7!=3)return 0;q3=strtol(
q8[1],0,0);q4=strtol(q8[2],0,0);if(q3<1||q4<1||q3<q4||q3>q5
(-1u))return 0;q2=(1u<<(q4-1)<<1)-1;q9=q2<<(q3-q4);for(;;){
q1(q2,q3);putchar('\n');if(q2==q9)break;q10=q2&-q2;q2+=q10+
((((q2+q10)&-(q2+q10))/q10)-1)/2;}return 0;}



Thx for your answer~
......but I don't quite understand....

What does this mean??

unsigned q4=(q3>0)? 1u<<(q3-1):-1u/2 +1;

Thx again.



Mars.
Nov 14 '05 #13
Peter Nilsson mentioned:
Peter Nilsson wrote:


And what's that "1u" mean in your program??

e.g.
q2=(1u<<(q4-1)<<1)-1;

Thx~

Mars
Nov 14 '05 #14


Mars wrote:
Peter Nilsson mentioned:

% type nmo.c
long strtol(const char*,char**,int);void q1(unsigned q2,int
q3){unsigned q4=(q3>0)? 1u<<(q3-1):-1u/2 +1;for(;q4;q4>>=1)
putchar('0'+!!(q2&q4));}int q5(unsigned q2){int q6;for(q6=0
;q2;q2&= q2-1,q6++);return q6 ;}int main(int q7,char **q8){
long q3,q4;unsigned q2,q9, q10;if(q7!=3)return 0;q3=strtol(
q8[1],0,0);q4=strtol(q8[2],0,0);if(q3<1||q4<1||q3<q4||q3>q5
(-1u))return 0;q2=(1u<<(q4-1)<<1)-1;q9=q2<<(q3-q4);for(;;){
q1(q2,q3);putchar('\n');if(q2==q9)break;q10=q2&-q2;q2+=q10+
((((q2+q10)&-(q2+q10))/q10)-1)/2;}return 0;}
Thx for your answer~
.....but I don't quite understand....


This may be a good part of the intention.

What does this mean??

unsigned q4=(q3>0)? 1u<<(q3-1):-1u/2 +1;


Shift left by q3, then rotate right by 1. Store the result in q4.
Cheers
Michael
--
E-Mail: Mine is a gmx dot de address.

Nov 14 '05 #15


Mars wrote:
Peter Nilsson mentioned:
Peter Nilsson wrote:


And what's that "1u" mean in your program??

e.g.
q2=(1u<<(q4-1)<<1)-1;


You were not there when they were talking about constants...

1 (int)
1u, 1U (unsigned int)
1l, 1L (long)
1UL, 1ul, 1LU, 1lu ....
1.0 (double)
1.0F,1.0f (float)
1.0L,1.0l (long double)
-Michael
--
E-Mail: Mine is a gmx dot de address.

Nov 14 '05 #16
Flash Gordon wrote:
Chris Williams wrote:
2 >> 1 = 1 //division by two
4 >> 1 = 2
6 >> 1 = 3
8 >> 1 = 4
8 >> 2 = 2 //division by four

1 & 1 = 1 //modulus of 2
2 & 1 = 0
3 & 1 = 1
4 & 1 = 0
4 & 2 = 0 //modulus of 4
Any decent optimising compiler will do this optimisation when dealing

with division by a constant. So it is generally better to write what you mean rather than making the code harder to read.


True. It's a two second test, but indeed there might be no change if
your compiler was spitting your code out using shifts and ands in
secret.

Though I was never big on trusting the compiler to do anything past
predetermining all math between constants--e.g. time = y + 1000 * 60;
-> time = y + 60000; Just the general idea that between knowing how to
do it myself or having the best optimizing compiler, I'd rather know
what I'm doing. Though I would want one for any large project to deal
with the 99% of code that won't be used intensively.

-Chris

Nov 14 '05 #17

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