New character?

[paupal244]

Yay, hooray, the black kitten is back. Good job

[The MAZZTer]

Hit New Post instead of Reply? :X

[Jelly]

I hope the black kitty stays.

[Dhokarena56]

DCS very well might develop him. But he could turn into another Locke, just sort of discussed from time to time but not too important anymore.

[Jelly]

Ya true. Hey, is he the same one from before when Millie was trying to make Ozy have bad luck?

[Tom Flapwell]

I'm not sure if Dhokarena was talking about Stevie or Mu-nan.

[Dhokarena56]

Actually, I would say that for both.

[Svix]

Hit New Post instead of Reply? :X

Relax, 71% of paupal's posts are new topics.

[Sage]

black kitty might become lyke Emily teh skunk

[A dude named Vince]

Ya true. Hey, is he the same one from before when Millie was trying to make Ozy have bad luck?

Most probaly.

[Extreme-Speed]

Ya true. Hey, is he the same one from before when Millie was trying to make Ozy have bad luck?

Most probaly.

Yep.

[Jelly]

Hit New Post instead of Reply? :X

Don't worry if you make a mistake like that you don't have to t ype stuff in the box you can just close the windown.

[Muninn]

Also moved this topic to the Ozy and Millie forum, Current Strips are for discussing the comics.

[KJ Fellie]

The black cat is Stevie. We have seen him before. I think he even wore the same shirt.

Also, the little puzzle in Extreme-Speed's signature:
To make this sentence true, the number of times we see 1 is _____, 2 is _____, 3 is _____, 4 is _____, and 5 is _____.
It can be answered with "one, one, one, one, and one," because if you spell it out, you only see "1" one time, "2" one time, "3" one time, "4" one time, and "5" one time. If you try to use the actual digits, it gets tricky.

[Doc Sigma]

Also, the little puzzle in Extreme-Speed's signature:
To make this sentence true, the number of times we see 1 is _____, 2 is _____, 3 is _____, 4 is _____, and 5 is _____.
It can be answered with "one, one, one, one, and one," because if you spell it out, you only see "1" one time, "2" one time, "3" one time, "4" one time, and "5" one time. If you try to use the actual digits, it gets tricky.

3, 2, 3, 1, 1. Wrote a dumb brute-force Perl script to solve it.

Code: Select all

my $n = 6; for my $one (1..$n) { for my $two (1..$n) { for my $three (1..$n) { for my $four (1..$n) { for my $five (1..$n) { my @t = (0, 1, 1, 1, 1, 1); $t[$_] += 1 for (0, $one, $two, $three, $four, $five); if (($t[1] == $one) && ($t[2] == $two) && ($t[3] == $three) && ($t[4] == $four) && ($t[5] == $five)) { print "$one, $two, $three, $four, $five\n"; exit(0); } # if } # for } # for } # for } # for } # for print "Unsolvable for n(1,$n)\n"; exit(1);