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Topic: Reverse a number (Read 1283 times) |
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u07103
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Reverse a number
« on: Nov 2nd, 2008, 11:42am » |
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How do you reverse a number using bitwise operator only...cud not find a related thread!!!!
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towr
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Re: Reverse a number
« Reply #1 on: Nov 2nd, 2008, 12:31pm » |
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It's here
Basically split the number in parts using bitmasks, and then use shifts to reorder the parts.
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u07103
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Re: Reverse a number
« Reply #2 on: Nov 2nd, 2008, 9:58pm » |
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dis wud just reverse the binary representation...question says reverse the number...i.e. rev(1234) = 4321
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towr
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Re: Reverse a number
« Reply #3 on: Nov 3rd, 2008, 2:11am » |
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That would just reverse the decimal representation. There is no such thing as a reversed number. 
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R
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Re: Reverse a number
« Reply #4 on: Apr 16th, 2009, 6:34pm » |
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on Nov 3rd, 2008, 2:11am, towr wrote:| That would just reverse the decimal representation. |
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I didn't get it. How will reversing the binary representation of a number reverse the number in decimal too. For example 12 = (1100) -> reversed to 0011 = 3.
Quote:| There is no such thing as a reversed number. |
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What did you mean?
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| « Last Edit: Apr 16th, 2009, 6:35pm by R » |
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towr
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Re: Reverse a number
« Reply #5 on: Apr 17th, 2009, 12:08am » |
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on Apr 16th, 2009, 6:34pm, R wrote:| I didn't get it. How will reversing the binary representation of a number reverse the number in decimal too. |
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It won't. Not generally, at least.
Quote:There is a difference between a number and its representation. A number is not something that can be reversed. At least I don't see how. It's not that type of concept.
Confusing a number with any of its representations, whether binary or decimal is just, well, confused. It's like saying the reverse of walking north is htron gniklaw, rather than walking south. "Walking north" is just a representation in words of an action; you don't get the reverse of the action by reversing its representation. Likewise you don't get the reverse of a number by reversing any of its representations.
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R
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Re: Reverse a number
« Reply #6 on: Apr 17th, 2009, 12:20am » |
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on Apr 17th, 2009, 12:08am, towr wrote:
There is a difference between a number and its representation. A number is not something that can be reversed. At least I don't see how. It's not that type of concept.
Confusing a number with any of its representations, whether binary or decimal is just, well, confused. It's like saying the reverse of walking north is htron gniklaw, rather than walking south. "Walking north" is just a representation in words of an action; you don't get the reverse of the action by reversing its representation. Likewise you don't get the reverse of a number by reversing any of its representations. |
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Ok I get your point about definition of "Reverse". But I think, this question is just about calculating a number which contains digits of the given number N, in the reverse order. And it is not about the mathematical or ideal definition of REVERSE.
It is just one of the number manipulating programming exercise. Isn't it?
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towr
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Re: Reverse a number
« Reply #7 on: Apr 17th, 2009, 1:58am » |
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on Apr 17th, 2009, 12:20am, R wrote:| It is just one of the number manipulating programming exercise. Isn't it? |
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Yeah. But if it isn't specified, reversing the binary representation makes more sense to me than reversing the decimal one. Especially if you're using only bitwise operators.
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R
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Re: Reverse a number
« Reply #8 on: Apr 17th, 2009, 8:26am » |
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on Apr 17th, 2009, 1:58am, towr wrote:
Yeah. But if it isn't specified,
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In Reply#2, the original poster of this questions has specified the meaning.
Quote:
reversing the binary representation makes more sense to me than reversing the decimal one. Especially if you're using only bitwise operators.
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That is a question on the correctness of this prblem. May be the original poster mis-interpreted the question from where he heard it.
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oldspy
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Re: Reverse a number
« Reply #9 on: May 15th, 2009, 5:16pm » |
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For dicimal version, it's simply. Get one digit a time from right and push it to the result from right:
int reverseDicimal( int n )
{
int res = 0;
int factor = 1;
int digit = 0;
while ( n != 0 )
{
digit = n % 10;
res = (res + digit ) * factor;
factor *= 10;
n = ( n - digit ) / 10;
}
return res;
}
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towr
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Re: Reverse a number
« Reply #10 on: May 16th, 2009, 2:39am » |
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on May 15th, 2009, 5:16pm, oldspy wrote:| For dicimal version, it's simply. Get one digit a time from right and push it to the result from right: |
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The question requires you to use only bitwise operators, you're not using them at all and are instead using only arithmetic operators which indeed makes it very simple.
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Grimbal
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Re: Reverse a number
« Reply #11 on: May 16th, 2009, 6:27pm » |
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There is still the possibility that the number is supposed to be encoded in EBCDIC.
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oldspy
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Re: Reverse a number
« Reply #12 on: May 16th, 2009, 7:07pm » |
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For binary reverse, here are the steps:
Write a loop to swap all adjacent bit blocks where bit block span from 1 to half of the total bits of the number.
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oldspy
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Re: Reverse a number
« Reply #13 on: May 16th, 2009, 7:29pm » |
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Here is the code:
int reverseBits( int n )
{
int mask = 1;
int i = 1;
int bits = 8 * sizeof n;
while ( i < bits )
{
int mask2 = mask;
int tmp = 0;
int k = 0;
// swap all adjacent blocks with span of i bits
do
{
tmp += ((n & mask2) << i ) + ((n & (mask2 << i)) >> i );
mask2 = mask2 << (2*i);
k += 2*i;
}
while ( k < bits );
n = tmp;
i = i << 1;
mask = (1 << i) - 1;
}
return n;
}
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Grimbal
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Re: Reverse a number
« Reply #14 on: May 18th, 2009, 1:15am » |
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That looks like O(#bits) to me. In the first outer loop you are swapping every pair of bits. You might just as well swap them one by one to where they belong.
int reverseBits( int n )
{
int rev = 0;
unsigned mask1 = 1;
unsigned mask2 = ~(~0U>>1);
while( mask1<mask2 ){
if( mask1 & n ) rev |= mask2;
if( mask2 & n ) rev |= mask1;
mask1 >>= 1;
mask2 <<= 1;
}
return rev;
}
int bits = 8 * sizeof n;
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oldspy
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Re: Reverse a number
« Reply #15 on: May 18th, 2009, 1:33am » |
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Yeah. It's bits - 1. But the algorithm can be improved to get to O(log (bits)).
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