f



MISSING NUMBER

* suppose you have a set of unsorted numbers from 1 to N (a big
number, say 1 million). One number is missing, how to find that
missing number most efficiently?

Adding the number then subtract N(N+1)/2 may not be possible since the
number could be too big to store.
0
1/10/2009 6:55:56 PM
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"Tagore" <c.lang.myself@gmail.com> wrote in message 
news:2d6e8332-d116-4d55-a6e2-ca0e204a63c5@n33g2000pri.googlegroups.com...
> * suppose you have a set of unsorted numbers from 1 to N (a big
> number, say 1 million). One number is missing, how to find that
> missing number most efficiently?
>
> Adding the number then subtract N(N+1)/2 may not be possible since the
> number could be too big to store.

How do you want to calculate the effeciency?
Time? Memory? (Both?)
If you may use O(n) memory then there is an easy O(n) algorithm to find the 
number.
There may still be such an algorithm if you may not use so much memory but I 
can't think of one right now. 

0
1/10/2009 7:20:50 PM
On Sat, 10 Jan 2009 10:55:56 -0800 (PST), Tagore
<c.lang.myself@gmail.com> wrote:

>* suppose you have a set of unsorted numbers from 1 to N (a big
>number, say 1 million). One number is missing, how to find that
>missing number most efficiently?
>
>Adding the number then subtract N(N+1)/2 may not be possible since the
>number could be too big to store.

The obvious way is to add the numbers modulo N.  There is a
little trick.  If N is odd sum(1..N) modulo N = 0; if N is even
it is N/2.  For example, if N is 1000000 and the missing number
is 489733 the sum in module 1000000 arithmetic is 10267.

You can also precompute what the result of xor'ing all of the
numbers from 1 to N would be and then xor that with xor'ing of
all of the given numbers.  The precomputing can be done on a bit
by bit basis but I'm too lazy to spell out the details.



Richard Harter, cri@tiac.net
http://home.tiac.net/~cri, http://www.varinoma.com
Save the Earth now!!
It's the only planet with chocolate.
0
cri (1432)
1/10/2009 7:42:49 PM
On Jan 10, 12:55=EF=BF=BDpm, Tagore <c.lang.mys...@gmail.com> wrote:
> * suppose you have a set of unsorted numbers from 1 to N (a big
> number, say 1 million). One number is missing, how to find that
> missing number most efficiently?
>
> Adding the number then subtract N(N+1)/2 may not be possible since the
> number could be too big to store.

Use a math library where "too big to store" isn't
an issue.
0
mensanator (1210)
1/10/2009 8:06:58 PM
Richard Harter wrote:
> You can also precompute what the result of xor'ing all of the
> numbers from 1 to N would be

  Out of curiosity: Is there any faster way of doing that other than
making a loop from 1 to N and xorring the loop counter?

  I was thinking something along the lines of: If N modulo 4 is 1 or 2,
you know the least-significant bit of the xor will be 1, else it will be
0. Could similar rules be developed for all the other bits as well?
0
nospam270 (2948)
1/10/2009 8:12:43 PM
Mensanator wrote:
> On Jan 10, 12:55�pm, Tagore <c.lang.mys...@gmail.com> wrote:
>> * suppose you have a set of unsorted numbers from 1 to N (a big
>> number, say 1 million). One number is missing, how to find that
>> missing number most efficiently?
>>
>> Adding the number then subtract N(N+1)/2 may not be possible since the
>> number could be too big to store.
> 
> Use a math library where "too big to store" isn't
> an issue.

  He wanted the most efficient way of doing that, and using a library
for numbers of unlimited size is certainly *not* the most efficient way
of doing what he wants.
0
nospam270 (2948)
1/10/2009 8:14:04 PM
On Sat, 10 Jan 2009 12:06:58 -0800 (PST), Mensanator
<mensanator@aol.com> wrote:

>On Jan 10, 12:55=EF=BF=BDpm, Tagore <c.lang.mys...@gmail.com> wrote:
>> * suppose you have a set of unsorted numbers from 1 to N (a big
>> number, say 1 million). One number is missing, how to find that
>> missing number most efficiently?
>>
>> Adding the number then subtract N(N+1)/2 may not be possible since the
>> number could be too big to store.
>
>Use a math library where "too big to store" isn't
>an issue.

In what way is this "most efficiently"?  Not, I fancy, in terms
of efficiency of computer space or time.  Not even in terms of
programmer time if you take into account the work of linkin in a
library?  Now if only you had said, use a language with bignums,
you could make a case for efficient use of programmer effort, but
as it is ...

Be that as it may, this is a programming puzzle rather than a
real problem as such.  Traditionally, "cheating" is a legitimate
response, where "cheating" means an answer that relies on a
lawyerly reading of the terms of the puzzle.  But in the context
of answering puzzles, even cheating must be fair.



Richard Harter, cri@tiac.net
http://home.tiac.net/~cri, http://www.varinoma.com
Save the Earth now!!
It's the only planet with chocolate.
0
cri (1432)
1/10/2009 8:31:01 PM
On Jan 10, 10:55=A0am, Tagore <c.lang.mys...@gmail.com> wrote:
> * suppose you have a set of unsorted numbers from 1 to N (a big
> number, say 1 million). One number is missing, how to find that
> missing number most efficiently?
>
> Adding the number then subtract N(N+1)/2 may not be possible since the
> number could be too big to store.

You only require that you calculate the sum modulo K, where K > N.
I.e., if the number is too big, go ahead and let it wrap around.
There are complicated ways of computing N(N+1)/2 doing this through
direct calculation, but why bother:

    numSum =3D 0;
    inputSum =3D 0;
    for (i=3D0; i < N-1; i++) {
        numSum +=3D i;
        inputSum +=3D array[i];
    }
    return (numSum + N-1) - inputSum;

In any 2s complement system the above will worth without issue,
regardless of overflow issues.  Another way is to use a xor sum rather
than a direct sum:

    numSum =3D 0;
    inputSum =3D 0;
    for (i=3D0; i < N-1; i++) {
        numSum ^=3D i;
        inputSum ^=3D array[i];
    }
    return numSum ^ (N-1) ^ inputSum;

--
Paul Hsieh
http://www.pobox.com/~qed/

0
websnarf (1153)
1/10/2009 9:59:07 PM
cri@tiac.net (Richard Harter) writes:

> On Sat, 10 Jan 2009 10:55:56 -0800 (PST), Tagore
> <c.lang.myself@gmail.com> wrote:
>
>>* suppose you have a set of unsorted numbers from 1 to N (a big
>>number, say 1 million). One number is missing, how to find that
>>missing number most efficiently?
>>
>>Adding the number then subtract N(N+1)/2 may not be possible since the
>>number could be too big to store.
>
> The obvious way is to add the numbers modulo N.  There is a
> little trick.  If N is odd sum(1..N) modulo N = 0; if N is even
> it is N/2.  For example, if N is 1000000 and the missing number
> is 489733 the sum in module 1000000 arithmetic is 10267.
>
> You can also precompute what the result of xor'ing all of the
> numbers from 1 to N would be and then xor that with xor'ing of
> all of the given numbers.  The precomputing can be done on a bit
> by bit basis but I'm too lazy to spell out the details.

That's smart.

But anyways, the memory size needed to store N(N+1)/2 is insignificant
compared to the memory size needed to store the numbers from 1 to N
(minus one).  For example, for N=1e6, the size to store N(N+1)/2 is 2
millionth of the size to store the other numbers.

-- 
__Pascal Bourguignon__
0
pjb (7869)
1/10/2009 10:12:28 PM
"Juha Nieminen" <nospam@thanks.invalid> wrote in message 
news:%i7al.210$nD1.180@read4.inet.fi...
> Richard Harter wrote:
>> You can also precompute what the result of xor'ing all of the
>> numbers from 1 to N would be
>
>  Out of curiosity: Is there any faster way of doing that other than
> making a loop from 1 to N and xorring the loop counter?
>
>  I was thinking something along the lines of: If N modulo 4 is 1 or 2,
> you know the least-significant bit of the xor will be 1, else it will be
> 0. Could similar rules be developed for all the other bits as well?

The modulo 4 pattern is as such:

0, 1, 3, 0,
4, 1, 7, 0,
8, 1, B, 0,
C, 1, F, 0
10, 1, 13, 0
14, 1, 17, 0
18, 1, 1B, 0
etc

Looks like a closed-form formula shouldn't be too hard to find 

0
1/10/2009 10:40:21 PM
Richard Harter wrote:
> Tagore <c.lang.myself@gmail.com> wrote:
> 
>> suppose you have a set of unsorted numbers from 1 to N (a big
>> number, say 1 million). One number is missing, how to find that
>> missing number most efficiently?
>>
>> Adding the number then subtract N(N+1)/2 may not be possible
>> since the number could be too big to store.
> 
> The obvious way is to add the numbers modulo N.  There is a
> little trick.  If N is odd sum(1..N) modulo N = 0; if N is even
> it is N/2.  For example, if N is 1000000 and the missing number
> is 489733 the sum in module 1000000 arithmetic is 10267.
> 
> You can also precompute what the result of xor'ing all of the
> numbers from 1 to N would be and then xor that with xor'ing of
> all of the given numbers.  The precomputing can be done on a bit
> by bit basis but I'm too lazy to spell out the details.

That's ingenious.  

-- 
 [mail]: Chuck F (cbfalconer at maineline dot net) 
 [page]: <http://cbfalconer.home.att.net>
            Try the download section.
0
cbfalconer (19194)
1/11/2009 1:02:31 AM
On Jan 10, 2:31=EF=BF=BDpm, c...@tiac.net (Richard Harter) wrote:
> On Sat, 10 Jan 2009 12:06:58 -0800 (PST), Mensanator
>
> <mensana...@aol.com> wrote:
> >On Jan 10, 12:55=3DEF=3DBF=3DBDpm, Tagore <c.lang.mys...@gmail.com> wrot=
e:
> >> * suppose you have a set of unsorted numbers from 1 to N (a big
> >> number, say 1 million). One number is missing, how to find that
> >> missing number most efficiently?
>
> >> Adding the number then subtract N(N+1)/2 may not be possible since the
> >> number could be too big to store.
>
> >Use a math library where "too big to store" isn't
> >an issue.
>
> In what way is this "most efficiently"?

The OP suggested an algorithm, but claimed that it
could be too big to store. I was addressing whether
said algorithm could work at all, not whether it
was most efficient.

>=EF=BF=BDNot, I fancy, in terms of efficiency of computer space or time. =
=EF=BF=BD

Is there an algorithm that DOESN'T require storing the
999999 numbers?

> Not even in terms of
> programmer time if you take into account the work of linkin in a
> library? =EF=BF=BD

Come on! How hard can that be? Actually, I already know.
For your benefit, its:

  #include <gmp.h>

Lotta work, that is, eh?

> Now if only you had said, use a language with bignums,

Oh, like Python? Won't even have to do an import.
But technically, you're correct. But how much effort
are you going to save when you optimze 0.001% of the
problem?

> you could make a case for efficient use of programmer effort, but
> as it is ...

It's pretty hard to see how ANYTHING could be
more efficient in terms of programmer effort.

>>> big =3D range(1,1000001)
>>> n =3D random.choice(big)
>>> big.pop(n)
>>> s =3D sum(big)
>>> soi =3D (1000000**2 + 1000000)/2
>>> soi-s
151072L

Note, the first three lines are the given, the solution
only takes the last three lines.

>
> Be that as it may, this is a programming puzzle rather than a
> real problem as such. =EF=BF=BD

So? It was the OP's algorithm, not mine. All I'm
doing is pointing out that there is no impediment
to said algorithm.

> Traditionally, "cheating" is a legitimate
> response, where "cheating" means an answer that relies on a
> lawyerly reading of the terms of the puzzle. =EF=BF=BDBut in the context
> of answering puzzles, even cheating must be fair.

Fine, when it's posted that only native C code, with
it's 32-bit integers are allowed. But that wasn't said.

>
> Richard Harter, c...@tiac.nethttp://home.tiac.net/~cri,http://www.varinom=
a.com
> Save the Earth now!!
> It's the only planet with chocolate.

0
mensanator (1210)
1/11/2009 4:00:34 AM
On Jan 10, 2:14=EF=BF=BDpm, Juha Nieminen <nos...@thanks.invalid> wrote:
> Mensanator wrote:
> > On Jan 10, 12:55 pm, Tagore <c.lang.mys...@gmail.com> wrote:
> >> * suppose you have a set of unsorted numbers from 1 to N (a big
> >> number, say 1 million). One number is missing, how to find that
> >> missing number most efficiently?
>
> >> Adding the number then subtract N(N+1)/2 may not be possible since the
> >> number could be too big to store.
>
> > Use a math library where "too big to store" isn't
> > an issue.
>
> =EF=BF=BD He wanted the most efficient way of doing that, and using a lib=
rary
> for numbers of unlimited size is certainly *not* the most efficient way
> of doing what he wants.

I didn't say it was. See my reply to RH.
0
mensanator (1210)
1/11/2009 4:02:45 AM
"Tagore" <c.lang.myself@gmail.com> wrote in message
news:2d6e8332-d116-4d55-a6e2-ca0e204a63c5@n33g2000pri.googlegroups.com...
> * suppose you have a set of unsorted numbers from 1 to N (a big
> number, say 1 million). One number is missing, how to find that
> missing number most efficiently?
>
> Adding the number then subtract N(N+1)/2 may not be possible since the
> number could be too big to store.

Use double precsion arithmetic.


0
robin_v (2737)
1/12/2009 1:31:22 AM
Mensanator wrote:
> Juha Nieminen <nos...@thanks.invalid> wrote:
>> Mensanator wrote:
>>> Tagore <c.lang.mys...@gmail.com> wrote:
>>>
>>>> suppose you have a set of unsorted numbers from 1 to N (a big
>>>> number, say 1 million). One number is missing, how to find
>>>> that missing number most efficiently?
>>>>
>>>> Adding the number then subtract N(N+1)/2 may not be possible
>>>> since the number could be too big to store.
>>>
>>> Use a math library where "too big to store" isn't an issue.
>>
>> He wanted the most efficient way of doing that, and using a
>> library for numbers of unlimited size is certainly *not* the
>> most efficient way of doing what he wants.
> 
> I didn't say it was. See my reply to RH.

Actually, for 10e6 numbers, you don't need 10e6 4 byte number
storage, you only need 10e6 bits, which will always fit in 125k
bytes (or less, for bigger bytes).  All you have to record is seen
vs not-seen.  At completion, scan the bit array to find the missing
beasts.  You can also do some ingenious optimizations.  Shouldn't
take much longer than the file reading time.

-- 
 [mail]: Chuck F (cbfalconer at maineline dot net) 
 [page]: <http://cbfalconer.home.att.net>
            Try the download section.
0
cbfalconer (19194)
1/12/2009 1:38:16 AM
robin wrote:
> "Tagore" <c.lang.myself@gmail.com> wrote:
>
>> suppose you have a set of unsorted numbers from 1 to N (a big
>> number, say 1 million). One number is missing, how to find that
>> missing number most efficiently?
>>
>> Adding the number then subtract N(N+1)/2 may not be possible
>> since the number could be too big to store.
> 
> Use double precsion arithmetic.

To see what you need, compute (N * N + N + 1) and divide it by 2. 
If N is roughly 10e6 the result will be roughly 10e12.

-- 
 [mail]: Chuck F (cbfalconer at maineline dot net) 
 [page]: <http://cbfalconer.home.att.net>
            Try the download section.
0
cbfalconer (19194)
1/12/2009 3:56:09 AM
On Sat, 10 Jan 2009 23:12:28 +0100, pjb@informatimago.com (Pascal
J. Bourguignon) wrote:

>cri@tiac.net (Richard Harter) writes:
>
>> On Sat, 10 Jan 2009 10:55:56 -0800 (PST), Tagore
>> <c.lang.myself@gmail.com> wrote:
>>
>>>* suppose you have a set of unsorted numbers from 1 to N (a big
>>>number, say 1 million). One number is missing, how to find that
>>>missing number most efficiently?
>>>
>>>Adding the number then subtract N(N+1)/2 may not be possible since the
>>>number could be too big to store.
>>
>> The obvious way is to add the numbers modulo N.  There is a
>> little trick.  If N is odd sum(1..N) modulo N = 0; if N is even
>> it is N/2.  For example, if N is 1000000 and the missing number
>> is 489733 the sum in module 1000000 arithmetic is 10267.
>>
>> You can also precompute what the result of xor'ing all of the
>> numbers from 1 to N would be and then xor that with xor'ing of
>> all of the given numbers.  The precomputing can be done on a bit
>> by bit basis but I'm too lazy to spell out the details.
>
>That's smart.
>
>But anyways, the memory size needed to store N(N+1)/2 is insignificant
>compared to the memory size needed to store the numbers from 1 to N
>(minus one).  For example, for N=1e6, the size to store N(N+1)/2 is 2
>millionth of the size to store the other numbers.

If you are reading the numbers from a file you don't need to
store the numbers.  


Richard Harter, cri@tiac.net
http://home.tiac.net/~cri, http://www.varinoma.com
Save the Earth now!!
It's the only planet with chocolate.
0
cri (1432)
1/12/2009 4:52:08 AM
On Sat, 10 Jan 2009 20:12:43 GMT, Juha Nieminen
<nospam@thanks.invalid> wrote:

>Richard Harter wrote:
>> You can also precompute what the result of xor'ing all of the
>> numbers from 1 to N would be
>
>  Out of curiosity: Is there any faster way of doing that other than
>making a loop from 1 to N and xorring the loop counter?
>
>  I was thinking something along the lines of: If N modulo 4 is 1 or 2,
>you know the least-significant bit of the xor will be 1, else it will be
>0. Could similar rules be developed for all the other bits as well?

I'm going to let you work out the formula for yourself.  The key
observations are (a) if the difference between the number of 0's
and 1's is even the final xor is 0, otherwise it is odd, (b) in
bit position k (starting the position number with 0) there are
alternating sequences of 2^k 0's and 2^k 1's in the sequence of
integers, (c) in consequence of (b) we only need to look at N
modulo 2^(k+1), and (d) we only need to know whether N modulo
2^(k+1) is odd or even.  

 

Richard Harter, cri@tiac.net
http://home.tiac.net/~cri, http://www.varinoma.com
Save the Earth now!!
It's the only planet with chocolate.
0
cri (1432)
1/12/2009 5:00:11 AM
On Mon, 12 Jan 2009 05:00:11 GMT, cri@tiac.net (Richard Harter)
wrote:

>On Sat, 10 Jan 2009 20:12:43 GMT, Juha Nieminen
><nospam@thanks.invalid> wrote:
>
>>Richard Harter wrote:
>>> You can also precompute what the result of xor'ing all of the
>>> numbers from 1 to N would be
>>
>>  Out of curiosity: Is there any faster way of doing that other than
>>making a loop from 1 to N and xorring the loop counter?
>>
>>  I was thinking something along the lines of: If N modulo 4 is 1 or 2,
>>you know the least-significant bit of the xor will be 1, else it will be
>>0. Could similar rules be developed for all the other bits as well?
>
>I'm going to let you work out the formula for yourself.  The key
>observations are (a) if the difference between the number of 0's
>and 1's is even the final xor is 0, otherwise it is odd, (b) in
>bit position k (starting the position number with 0) there are
>alternating sequences of 2^k 0's and 2^k 1's in the sequence of
>integers, (c) in consequence of (b) we only need to look at N
>modulo 2^(k+1), and (d) we only need to know whether N modulo
>2^(k+1) is odd or even.  


Mea culpa.  The above is wrong.  The key observation is that if
we xor a mixed sequence of 0's and 1's the result is 0 if the
number of 1's is even and 1 if it is odd.  Bit  0 of the xor is
the complement of bit 0.  For bit k of the xor set m = N mod
2^(k+1).  If m < 2^k bit k of the xor is 0.  Otherwise it is the
complement of bit 0 of m.

I hope I haven't screwed this up yet again.  :-)

 
Richard Harter, cri@tiac.net
http://home.tiac.net/~cri, http://www.varinoma.com
Save the Earth now!!
It's the only planet with chocolate.
0
cri (1432)
1/12/2009 12:16:00 PM
On Jan 12, 4:16=A0am, c...@tiac.net (Richard Harter) wrote:
> On Mon, 12 Jan 2009 05:00:11 GMT, c...@tiac.net (Richard Harter)
> wrote:
>
> >On Sat, 10 Jan 2009 20:12:43 GMT, Juha Nieminen
> ><nos...@thanks.invalid> wrote:
>
> >>Richard Harter wrote:
> >>> You can also precompute what the result of xor'ing all of the
> >>> numbers from 1 to N would be
>
> >> =A0Out of curiosity: Is there any faster way of doing that other than
> >>making a loop from 1 to N and xorring the loop counter?
>
> >> =A0I was thinking something along the lines of: If N modulo 4 is 1 or =
2,
> >>you know the least-significant bit of the xor will be 1, else it will b=
e
> >>0. Could similar rules be developed for all the other bits as well?
>
> >I'm going to let you work out the formula for yourself. =A0The key
> >observations are (a) if the difference between the number of 0's
> >and 1's is even the final xor is 0, otherwise it is odd, (b) in
> >bit position k (starting the position number with 0) there are
> >alternating sequences of 2^k 0's and 2^k 1's in the sequence of
> >integers, (c) in consequence of (b) we only need to look at N
> >modulo 2^(k+1), and (d) we only need to know whether N modulo
> >2^(k+1) is odd or even. =A0
>
> Mea culpa. =A0The above is wrong. =A0The key observation is that if
> we xor a mixed sequence of 0's and 1's the result is 0 if the
> number of 1's is even and 1 if it is odd. =A0Bit =A00 of the xor is
> the complement of bit 0. =A0For bit k of the xor set m =3D N mod
> 2^(k+1). =A0If m < 2^k bit k of the xor is 0. =A0Otherwise it is the
> complement of bit 0 of m.
>
> I hope I haven't screwed this up yet again. =A0:-)

Well ... I don't know about these hints.  Remember that he was
originally looking at N (mod 4) which is not a bad place to start in
figuring out the xor-sum formula.  If you don't have the math insight,
it takes very little effort to figure it out by brute force: Just
print out the first 100 results in 4 columns and 50% of the pattern
will be blatantly obvious, while each of the remaining 25% will
probably be clear enough.  Then its a very simple matter of proving
it.

--
Paul Hsieh
http://www.pobox.com/~qed/
http://bstring.sf.net/
0
websnarf (1153)
1/13/2009 2:22:44 AM
On Mon, 12 Jan 2009 18:22:44 -0800 (PST), Paul Hsieh
<websnarf@gmail.com> wrote:

>On Jan 12, 4:16=A0am, c...@tiac.net (Richard Harter) wrote:
>> On Mon, 12 Jan 2009 05:00:11 GMT, c...@tiac.net (Richard Harter)
>> wrote:
>>
>> >On Sat, 10 Jan 2009 20:12:43 GMT, Juha Nieminen
>> ><nos...@thanks.invalid> wrote:
>>
>> >>Richard Harter wrote:
>> >>> You can also precompute what the result of xor'ing all of the
>> >>> numbers from 1 to N would be
>>
>> >> =A0Out of curiosity: Is there any faster way of doing that other than
>> >>making a loop from 1 to N and xorring the loop counter?
>>
>> >> =A0I was thinking something along the lines of: If N modulo 4 is 1 or =
>2,
>> >>you know the least-significant bit of the xor will be 1, else it will b=
>e
>> >>0. Could similar rules be developed for all the other bits as well?
>>
>> >I'm going to let you work out the formula for yourself. =A0The key
>> >observations are (a) if the difference between the number of 0's
>> >and 1's is even the final xor is 0, otherwise it is odd, (b) in
>> >bit position k (starting the position number with 0) there are
>> >alternating sequences of 2^k 0's and 2^k 1's in the sequence of
>> >integers, (c) in consequence of (b) we only need to look at N
>> >modulo 2^(k+1), and (d) we only need to know whether N modulo
>> >2^(k+1) is odd or even. =A0
>>
>> Mea culpa. =A0The above is wrong. =A0The key observation is that if
>> we xor a mixed sequence of 0's and 1's the result is 0 if the
>> number of 1's is even and 1 if it is odd. =A0Bit =A00 of the xor is
>> the complement of bit 0. =A0For bit k of the xor set m =3D N mod
>> 2^(k+1). =A0If m < 2^k bit k of the xor is 0. =A0Otherwise it is the
>> complement of bit 0 of m.
>>
>> I hope I haven't screwed this up yet again. =A0:-)
>
>Well ... I don't know about these hints.  Remember that he was
>originally looking at N (mod 4) which is not a bad place to start in
>figuring out the xor-sum formula.  If you don't have the math insight,
>it takes very little effort to figure it out by brute force: Just
>print out the first 100 results in 4 columns and 50% of the pattern
>will be blatantly obvious, while each of the remaining 25% will
>probably be clear enough.  Then its a very simple matter of proving
>it.

I worked out an entirely different approach; I wasn't familiar
with the formula you are talking about, which is a much simpler
approach.


Richard Harter, cri@tiac.net
http://home.tiac.net/~cri, http://www.varinoma.com
Save the Earth now!!
It's the only planet with chocolate.
0
cri (1432)
1/16/2009 6:35:58 AM
On Jan 10, 3:59=A0pm, Paul Hsieh <websn...@gmail.com> wrote:
> On Jan 10, 10:55=A0am, Tagore <c.lang.mys...@gmail.com> wrote:
>
> > * suppose you have a set of unsorted numbers from 1 to N (a big
> > number, say 1 million). One number is missing, how to find that
> > missing number most efficiently?
>
> > Adding the number then subtract N(N+1)/2 may not be possible since the
> > number could be too big to store.
>
> You only require that you calculate the sum modulo K, where K > N.
> I.e., if the number is too big, go ahead and let it wrap around.
> There are complicated ways of computing N(N+1)/2 doing this through
> direct calculation, but why bother:
>
> =A0 =A0 numSum =3D 0;
> =A0 =A0 inputSum =3D 0;
> =A0 =A0 for (i=3D0; i < N-1; i++) {
> =A0 =A0 =A0 =A0 numSum +=3D i;
> =A0 =A0 =A0 =A0 inputSum +=3D array[i];
> =A0 =A0 }
> =A0 =A0 return (numSum + N-1) - inputSum;


It's not even necessary to keep two separate sums:

     Sum =3D N-1;
     for (i=3D0; i < N-1; i++)
         Sum +=3D i - array[i];
     return Sum;


0
robertwessel2 (1674)
1/16/2009 6:53:06 AM
"robertwessel2@yahoo.com" <robertwessel2@yahoo.com> writes:

> On Jan 10, 3:59�pm, Paul Hsieh <websn...@gmail.com> wrote:
>> On Jan 10, 10:55�am, Tagore <c.lang.mys...@gmail.com> wrote:
>>
>> > * suppose you have a set of unsorted numbers from 1 to N (a big
>> > number, say 1 million). One number is missing, how to find that
>> > missing number most efficiently?
>>
>> > Adding the number then subtract N(N+1)/2 may not be possible since the
>> > number could be too big to store.
>>
>> You only require that you calculate the sum modulo K, where K > N.
>> I.e., if the number is too big, go ahead and let it wrap around.
>> There are complicated ways of computing N(N+1)/2 doing this through
>> direct calculation, but why bother:

Computing the XOR of all the numbers works too.
It takes less memory since you only need the number of bits of the biggest number, N.

(defun test (n)
  (let ((missing (random n)))
    (loop
       :for i :from 1 :to n
       :for xor = (if (= i missing) 0 i) :then (if (= i missing) xor (logxor xor i))
       :finally (setf xor (logxor n xor))
                (return (values xor missing (= xor missing))))))

(loop :repeat 10 :do (print (multiple-value-list  (test 1000000))))

(340555 340555 T) 
(228983 228983 T) 
(854712 854712 T) 
(662113 662113 T) 
(371568 371568 T) 
(907540 907540 T) 
(889816 889816 T) 
(670949 670949 T) 
(109844 109844 T) 
(544193 544193 T) 


(But I don't understand why we need to xor N again to get the result...)


-- 
__Pascal Bourguignon__
0
pjb (7869)
1/16/2009 9:46:40 AM
In article <2d6e8332-d116-4d55-a6e2-ca0e204a63c5
@n33g2000pri.googlegroups.com>, c.lang.myself@gmail.com says...
> * suppose you have a set of unsorted numbers from 1 to N (a big
> number, say 1 million). One number is missing, how to find that
> missing number most efficiently?
> 
> Adding the number then subtract N(N+1)/2 may not be possible since the
> number could be too big to store.

Well, a little modulo arithmetic should solve this particular homework 
problem.

- Gerry Quinn
0
gerryq2 (435)
1/23/2009 4:18:09 AM
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Missing nurse declared major crime
POLICE hold serious concerns for a nurse who was abducted from her remote home in South Australia’s Far North, with her disappearance declared ...

'Precariat' generation missing out on Australian lifestyle
The gap between generations is growing with young people behind in material wealth and mental wellbeing, a new book reveals.

Body found in search for missing SA nurse Gayle Woodford
The body of a woman believed to be that of missing health worker Gayle Woodford has been found in South Australia.

Police find body believed to be that of missing South Australian nurse Gayle Woodford
Police locate a body believed to be that of missing outback SA nurse Gayle Woodford.

Family in central Ohio prays missing couple is found safe after Brussels attacks
Lesa Stone's niece and her husband have been missing since the attacks this week.

Resources last updated: 3/26/2016 4:52:01 PM