July 15th, 2010, 06:54 PM  #1 
Newbie Joined: Jul 2010 Posts: 3 Thanks: 0  boolean and modulo
Please do you know if (a&b)modp(where & stands for bitwise boolean AND operator)is the same as (amodp)&(bmodp)?Or can you direct me to somewhere i can find more information on this?Thanks.

July 15th, 2010, 07:08 PM  #2 
Senior Member Joined: Aug 2008 From: Blacksburg VA USA Posts: 344 Thanks: 6 Math Focus: primes of course  Re: boolean and modulo
ok, via wolframalpha, input box: (45 BitAnd 69) mod 4 returns 1 45mod4=1, 69mod4=1 1BitAnd1 gives 1, so on the first test check it worked. You may try others now ... 
July 16th, 2010, 07:18 AM  #3 
Senior Member Joined: Apr 2010 Posts: 215 Thanks: 0  Re: boolean and modulo
Your statement is valid if (for some natural integer n). Any other p will not work. 
July 16th, 2010, 03:18 PM  #4 
Senior Member Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3  Re: boolean and modulo
If a and b are less than p, then the statements are equivalent, as a&b can't exceed min{a,b}. Otherwise (if p is not a power of 2) the situation is complicated. I have plotted results for (a,b) using modulo 3 and modulo 17. A white dot means equivalence, a black dot nonequivalence. 
July 16th, 2010, 04:30 PM  #5  
Senior Member Joined: Apr 2010 Posts: 215 Thanks: 0  Re: boolean and modulo Quote:
They are obviously symmetrical because the ANDoperation is communitative: (a&b)%p == (b&a)%p (a%p)&(b%p) == (b%p)&(a%p)  
July 17th, 2010, 02:37 AM  #6 
Newbie Joined: Jul 2010 Posts: 3 Thanks: 0  Re: boolean and modulo
Thank you guys so much for replying.I feel so at home here on this forum.So many on point posts.I sorta suspected this about powers of two,but for the when a and b are less than p case that was a revelation.Thank you so much


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