1
$\begingroup$

Find the rule that is all correct on the left side, but none on the right side.

correction in correction

Raw text: (Orange, Yellow, Green, Cyan)

Left:

OGYCO
C
YYGG
GCOOOYGC
CCOCO
YCGCOCY
CCC
COCOCYY
CGCCCG
GG

Right:

O
GYCY
CGCGCO
CG
OOOCCC
YOOCCOYC
GGCCCCG
GC
YCGOOY
CCCYYY
Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

I think it’s simply that:

There are an even total number of orange, green and yellow. Or, equivalently, leaving out cyan leaves an even number of squares.

How I got to this:

The single cyan and single orange showed that one of those two colours was important in the rule. Looking, there was a lot of cyan, so I decided to see what would happen if I ignored it, giving this answer.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ But the 7th (from top) on the right side has 4 greens? $\endgroup$
    – Peter Wirdemo
    Commented Oct 1, 2018 at 8:12
  • $\begingroup$ @u_ndefined still, 4th from top has zero non-cyan. Zero is even too. $\endgroup$
    – Peter Wirdemo
    Commented Oct 1, 2018 at 10:35
  • $\begingroup$ @Plarsen there isn’t much contrast between the two colours but the fourth and seventh from the top on the right side have one and three greens respectively. $\endgroup$
    – boboquack
    Commented Oct 2, 2018 at 8:07
  • $\begingroup$ yes, sure, after the edit that is true. Not what it looked like yesterday (see revisions): i.sstatic.net/lI6LJ.png $\endgroup$
    – Peter Wirdemo
    Commented Oct 2, 2018 at 11:36

Not the answer you're looking for? Browse other questions tagged or ask your own question.