Re: [algogeeks] Re: validate bit pattern : MS question
@Ashgoel: My solution worked when the pattern of ones followed by zeros
followed by ones is right-justified in a larger integer, so that there are
zeros to the left of the leftmost one. E.g., my algorithm will say that
00110011 in an 8-bit integer is valid.
Dave
On Friday, May 18, 2012 1:23:28 AM UTC-5, ashgoel wrote:
i changed the last step to return (n == ~0);
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Fri, May 18, 2012 at 11:27 AM, Vishal Thanki vishaltha...@gmail.comwrote:
On Sat, Apr 7, 2012 at 11:24 AM, Dave dave_and_da...@juno.com wrote:
@Ashgoel: Try this:
bool OnesZerosOnes(unsigned int n)
{
if( !(n 1) || !(n = n+1) ) return 0;
n |= n-1;
return !(n (n+1));
}
Here is how it works:
!(n 1) is true if the number has trailing zeros.
If the number has trailing ones, n = n+1 replaces the trailing ones
with
zeros.
!(n = n+1) is true if there are only trailing ones, i.e., the original
number was zeros followed by ones.
n |= n-1 replaces trailing zeros with ones. Thus, if the original
number is
ones followed by zeros followed by ones, the zeros have been changed to
ones.
(n (n+1)) replaces the trailing ones with zeros. If the number is now
zero, the number is valid, otherwise the number is invalid.
Dave
Superb, Dave!!
On Tuesday, April 3, 2012 7:00:36 PM UTC-5, ashgoel wrote:
verify that the bits of a number are in format 1s followed by 0s
followed
by 1s like 1110001 is valid but 100100100 is not
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
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Re: [algogeeks] Re: validate bit pattern : MS question
i changed the last step to return (n == ~0);
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Fri, May 18, 2012 at 11:27 AM, Vishal Thanki vishaltha...@gmail.comwrote:
On Sat, Apr 7, 2012 at 11:24 AM, Dave dave_and_da...@juno.com wrote:
@Ashgoel: Try this:
bool OnesZerosOnes(unsigned int n)
{
if( !(n 1) || !(n = n+1) ) return 0;
n |= n-1;
return !(n (n+1));
}
Here is how it works:
!(n 1) is true if the number has trailing zeros.
If the number has trailing ones, n = n+1 replaces the trailing ones with
zeros.
!(n = n+1) is true if there are only trailing ones, i.e., the original
number was zeros followed by ones.
n |= n-1 replaces trailing zeros with ones. Thus, if the original number
is
ones followed by zeros followed by ones, the zeros have been changed to
ones.
(n (n+1)) replaces the trailing ones with zeros. If the number is now
zero, the number is valid, otherwise the number is invalid.
Dave
Superb, Dave!!
On Tuesday, April 3, 2012 7:00:36 PM UTC-5, ashgoel wrote:
verify that the bits of a number are in format 1s followed by 0s
followed
by 1s like 1110001 is valid but 100100100 is not
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
--
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Re: [algogeeks] Re: validate bit pattern : MS question
Hello Dave,
Was trying this
bool OnesZerosOnes(unsigned int n)
{
if( !(n 1) || !(n = n+1) ) return 0; /*step1*/
n |= n-1;/*step 2*/
return !(n (n+1)); /*step 3*/
}
for 1110011
after step 1 it becomes 1110010100=111
after step2 it becomes 111|110=111(all ones)
on step 3 we return~(111000) = 1
for 1010011
after step 1 it becomes 10100111010100=101
after step2 it becomes 101|100=101
on step 3 we return~(101110) = 0
what are we trying to do with step 3 here
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Sat, Apr 7, 2012 at 11:24 AM, Dave dave_and_da...@juno.com wrote:
bool OnesZerosOnes(unsigned int n)
{
if( !(n 1) || !(n = n+1) ) return 0;
n |= n-1;
return !(n (n+1));
}
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Re: [algogeeks] Re: validate bit pattern : MS question
got it...super...
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Thu, May 17, 2012 at 8:21 PM, Ashish Goel ashg...@gmail.com wrote:
Hello Dave,
Was trying this
bool OnesZerosOnes(unsigned int n)
{
if( !(n 1) || !(n = n+1) ) return 0; /*step1*/
n |= n-1;/*step 2*/
return !(n (n+1)); /*step 3*/
}
for 1110011
after step 1 it becomes 1110010100=111
after step2 it becomes 111|110=111(all ones)
on step 3 we return~(111000) = 1
for 1010011
after step 1 it becomes 10100111010100=101
after step2 it becomes 101|100=101
on step 3 we return~(101110) = 0
what are we trying to do with step 3 here
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Sat, Apr 7, 2012 at 11:24 AM, Dave dave_and_da...@juno.com wrote:
bool OnesZerosOnes(unsigned int n)
{
if( !(n 1) || !(n = n+1) ) return 0;
n |= n-1;
return !(n (n+1));
}
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Re: [algogeeks] Re: validate bit pattern : MS question
Hii Can anyone explain to me what this code is doing??
On 17 May 2012 20:36, Ashish Goel ashg...@gmail.com wrote:
got it...super...
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Thu, May 17, 2012 at 8:21 PM, Ashish Goel ashg...@gmail.com wrote:
Hello Dave,
Was trying this
bool OnesZerosOnes(unsigned int n)
{
if( !(n 1) || !(n = n+1) ) return 0; /*step1*/
n |= n-1;/*step 2*/
return !(n (n+1)); /*step 3*/
}
for 1110011
after step 1 it becomes 1110010100=111
after step2 it becomes 111|110=111(all ones)
on step 3 we return~(111000) = 1
for 1010011
after step 1 it becomes 10100111010100=101
after step2 it becomes 101|100=101
on step 3 we return~(101110) = 0
what are we trying to do with step 3 here
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Sat, Apr 7, 2012 at 11:24 AM, Dave dave_and_da...@juno.com wrote:
bool OnesZerosOnes(unsigned int n)
{
if( !(n 1) || !(n = n+1) ) return 0;
n |= n-1;
return !(n (n+1));
}
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Re: [algogeeks] Re: validate bit pattern : MS question
On Sat, Apr 7, 2012 at 11:24 AM, Dave dave_and_da...@juno.com wrote:
@Ashgoel: Try this:
bool OnesZerosOnes(unsigned int n)
{
if( !(n 1) || !(n = n+1) ) return 0;
n |= n-1;
return !(n (n+1));
}
Here is how it works:
!(n 1) is true if the number has trailing zeros.
If the number has trailing ones, n = n+1 replaces the trailing ones with
zeros.
!(n = n+1) is true if there are only trailing ones, i.e., the original
number was zeros followed by ones.
n |= n-1 replaces trailing zeros with ones. Thus, if the original number is
ones followed by zeros followed by ones, the zeros have been changed to
ones.
(n (n+1)) replaces the trailing ones with zeros. If the number is now
zero, the number is valid, otherwise the number is invalid.
Dave
Superb, Dave!!
On Tuesday, April 3, 2012 7:00:36 PM UTC-5, ashgoel wrote:
verify that the bits of a number are in format 1s followed by 0s followed
by 1s like 1110001 is valid but 100100100 is not
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
--
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[algogeeks] Re: validate bit pattern : MS question
Hello Ashish, I would do something like here - http://www.evaluzio.net/library/verify-1-0-bits-in-num.htm Basically you just count the number of 01 and 10 binary sequences using (n 3 == 1) or (n 3 == 2) Regards, Ilya On Tuesday, 3 April 2012 17:00:36 UTC-7, ashgoel wrote: verify that the bits of a number are in format 1s followed by 0s followed by 1s like 1110001 is valid but 100100100 is not Best Regards Ashish Goel Think positive and find fuel in failure +919985813081 +919966006652 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/DFNF8rFO62UJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: validate bit pattern : MS question
@Ashgoel: Try this:
bool OnesZerosOnes(unsigned int n)
{
if( !(n 1) || !(n = n+1) ) return 0;
n |= n-1;
return !(n (n+1));
}
Here is how it works:
!(n 1) is true if the number has trailing zeros.
If the number has trailing ones, n = n+1 replaces the trailing ones with
zeros.
!(n = n+1) is true if there are only trailing ones, i.e., the original
number was zeros followed by ones.
n |= n-1 replaces trailing zeros with ones. Thus, if the original number is
ones followed by zeros followed by ones, the zeros have been changed to
ones.
(n (n+1)) replaces the trailing ones with zeros. If the number is now
zero, the number is valid, otherwise the number is invalid.
Dave
On Tuesday, April 3, 2012 7:00:36 PM UTC-5, ashgoel wrote:
verify that the bits of a number are in format 1s followed by 0s followed
by 1s like 1110001 is valid but 100100100 is not
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
--
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