I had a perl script that answerred a similar question hanging arround. I added a line or two for you. A little overkill but here is a sample output:
perl bits.pl 3
0 000 000 zero=3, one=0, longest=3 (0)
1 001 001 zero=2, one=1, longest=2 (0)
2 002 010 zero=2, one=1, longest=1 (0)
3 003 011 zero=1, one=2, longest=2 (1)
4 004 100 zero=2, one=1, longest=2 (0)
5 005 101 zero=1, one=2, longest=1 (1)
6 006 110 zero=1, one=2, longest=2 (1)
7 007 111 zero=0, one=3, longest=3 (1)
and:
perl bits.pl 8 | grep "e=4" | sort -k 6 | tail -4
15 00F 00001111 zero=4, one=4, longest=4 (1)
30 01E 00011110 zero=4, one=4, longest=4 (1)
60 03C 00111100 zero=4, one=4, longest=4 (1)
120 078 01111000 zero=4, one=4, longest=4 (1)
Script below.
Enjoy...
Hein.
---- bits.pl ---
$w = shift or die "How many bits wide?";
while ($i < (1<<$w)) {
$zero = $one = $long = $longest = $n = 0;
$last = 2;
$tmp = $i;
while ($n++ < $w) {
$bit = $tmp & 1;
$tmp >>= 1;
if ($bit) { $one++ } else { $zero++ }
$long++ if ($bit == $last);
if ($long > $longest) { $longest = $long; $long_last = $last};
if ($bit != $last) { $last = $bit; $long = 1 };
# print "bit=$bit, last=$last, zero=$zero, one=$one, longest=$longest, ll=$long_last\n";
}
printf ("%4d %03X %.*b zero=%d, one=%d, longest=%d (%d)\n",
$i,$i,$w,$i,$zero,$one,$longest,$long_last);
$i++;
}
