how to identify odd 1's in a binary number using regexp
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From: mallikarjunarao on 5 Aug 2010 07:44 hi , how to identify odd 1's in a binary number my binary numbers is 110010001010101111110110 11010101010101111110110 11101101010101100011011 11111100110101010111101
From: mallikarjunarao on 5 Aug 2010 08:04 On Aug 5, 4:44 pm, mallikarjunarao <malli....(a)gmail.com> wrote: > hi , > how to identify odd 1's in a binary number my binary numbers is > 110010001010101111110110 > 11010101010101111110110 > 11101101010101100011011 > 11111100110101010111101 for better understanding, identify 1 in the odd position of the binary number
From: Andreas Leitgeb on 5 Aug 2010 08:47 mallikarjunarao <malli.kv2(a)gmail.com> wrote: > On Aug 5, 4:44 pm, mallikarjunarao <malli....(a)gmail.com> wrote: >> hi , >> how to identify odd 1's in a binary number my binary numbers is >> 110010001010101111110110 >> 11010101010101111110110 >> 11101101010101100011011 >> 11111100110101010111101 > for better understanding, identify 1 in the odd position of the binary > number {^(..)*.1} will match up to the first odd-positioned "1". For further hits, remove an even number of digits from start including the just found "1". another approach: set binNumber 110010001010101111110110 regsub -all {(..)} $binNumber {\1:} tmp foreach h [regexp -all -indices -inline {1:} $tmp] { set realIndex [expr {[lindex $h 0]/3*2+1}] puts "pos $realIndex: [string index $binNumber $realIndex]" } If I mixed up odd and even (that is, if positions are meant as 1-based), then change the regexp (in foreach) to {1.:} and remove the "+1" in the expr.
From: Uwe Klein on 5 Aug 2010 10:12 Andreas Leitgeb wrote: > mallikarjunarao <malli.kv2(a)gmail.com> wrote: > >>On Aug 5, 4:44 pm, mallikarjunarao <malli....(a)gmail.com> wrote: >> >>>hi , >>>how to identify odd 1's in a binary number my binary numbers is >>>110010001010101111110110 >>>11010101010101111110110 >>>11101101010101100011011 >>>11111100110101010111101 >> >>for better understanding, identify 1 in the odd position of the binary >>number > > > {^(..)*.1} will match up to the first odd-positioned "1". > For further hits, remove an even number of digits from start including > the just found "1". > > another approach: > > set binNumber 110010001010101111110110 > regsub -all {(..)} $binNumber {\1:} tmp > foreach h [regexp -all -indices -inline {1:} $tmp] { > set realIndex [expr {[lindex $h 0]/3*2+1}] > puts "pos $realIndex: [string index $binNumber $realIndex]" > } > > If I mixed up odd and even (that is, if positions are meant as 1-based), > then change the regexp (in foreach) to {1.:} and remove the "+1" in the > expr. > if {[string length $binNumber] % 2} { .... } uwe
From: Bezoar on 5 Aug 2010 13:27
On Aug 5, 6:44 am, mallikarjunarao <malli....(a)gmail.com> wrote: > hi , > how to identify odd 1's in a binary number my binary numbers is > 110010001010101111110110 > 11010101010101111110110 > 11101101010101100011011 > 11111100110101010111101 Are you saying you want to identify the binary numbers where there is a 1 in and odd position(e.g 001, 1 is in position 3 (if count starts at 1 ) ), the resuting number is odd, the count of ones showing up in the binary number is odd? for the second just check to see if last digit is 1 proc isOddNum { num } { return [expr ([string index $num end ] == "1") ? 1 : 0 ] } for the latter : set num 11111100110101010111101 proc isCountOfOnesOdd { num } { # compress 0's then apply modulus 2 to resulting string return [expr {[string length [string match {"0" ""} $num ] ] % 2 }]; } |