A small red unbordered square. That way we get each of the four properties (size, shape, colour, borderedness) being split 3 and 3 in the objects, i.e. three red, three green; three with a border, three without a border…
We also form a pretty shape in the 4d (size, shape, colour, borderedness) space, though I’m not sure what to call it. Is it an octahedron?

I first came to the same answer as Paul, Oscar, and VDegeler. And then I thought: is this a set, or is it a sequence?

If it is a set, add me to the list. Small, red, unbordered square.

But if this is a sequence, notice that two properties stay the same and two change as we move to the next item.

1. small red bordered circle
2. large red bordered square
3. ????? ??? ???????? ??????
4. large grn bordered circle
5. small grn unbordrd circle
6. large grn unbordrd square

There are four objects that could complete this sequence:
a) large red unbordrd circle or
b) large grn unbordrd square or
c) small red bordered circle or
d) small grn bordered square

Since b and c already exist, aesthetically I am drawn to a or b, large red unbordrd circle or small grn bordrd square

Alternatively, a full stop. The first two shapes represent a question with a question mark. The green shapes are the answer, unfortunately missing the appropriate grammar.

A copy of object No. 4: big green circle with a border! This is the only way to make each of the properties (shape, size, border type and color) represented an even number of times. This is of course complementary to the answer by Paul and others (replace “even” by “odd”). If you think of objects as vectors in F_2^4, my answer amounts to make the sum equal to 0.

Small green bordered square. The items are in pairs, and between first and second members they a) retain colour, b) retain border state, c) change shape.

A small green square with a border would complete the pair, if the answer is meant to complete the three pairs.
If it is meant to be two complementary triplets, then a large red circle without a border is okay.
Either way, the question mark is the odd one out.

By the way, if the las two (green unbordered) figures were brown, then nothing could be missing, because the green bordered circle could implicate an operator instead of a figure in the sequence.

Still, I would prefer a small green square with a border, because the first two and last two figures are in the same sequence – and that would indicate three pairs.
Within these pairs there is only change in shape, at least that is what the first and the last pair suggest.

But again, is it 3 x 2 (small green square border), 2 x 3 (large red circle borderless), or even 1 x 6 (where small red borderless square would be right)?

@Jonathan: (“aesthetically I am drawn to a or b, large red unbordrd circle or small grn bordrd square”)
If there is no reason to choose one over the other than aesthetics, then that rules out the puzzle being about sequence.
@Merlign: The problem with thinking about it as 2 triplets or 3 pairs is that two of anything is not enough to establish a pattern.
We already have 3 large, 3 circles, 3 borders, 3 green. We also have 2 small+1?, 2 squares+1?, 2 borderless+1?, 2 red+1?
All the characteristics clearly occur 3 times; nothing suggests one should leave a characteristic only happening twice by making its opposite happen 4 times: The ? is a small, red, borderless square.

The most holistic is the 1 x 6 option (your choice), where the choice is based on the sum of all specifics of all elements.

Two of anything (elements) is enough to establish a pattern, if:
1. there are more than one group of two elements, or
2. If the two elements within one group are comparable in their specifics. E.g. a group consisting of two equal dots is uniform – that is a pattern, or a group consisting of two green but different elements, that is a pattern as well.
Even a group consisting of two totally different elements creates a pattern of high contrast.
So nr 2. is always valid, but only if you acknowledge that the group consists of the two elements in question.

There is a good reason to say that there are two elements per group:
a. the repetition in pairs (first two and last two elements)
b. alternate large and small elements
a. and b. would suggest a triple pairing sequence, consisting of 3 x 2 elements (3 groups of 2 elements).
And with that comes the acknowledgement that one group should consist of two elements.
I think it’s even preferable, because both a. and b. form a premise in the puzzle.

I am, however, making the following assumptions about the pattern:
– that there should be the same number of large and small objects, and that these alternate small, large, small, large, etc.
– that there should be the same number of green and red objects, with reds on one side and greens on the other

The last two assumptions (or maybe one large assumption) I’m having a hard time phrasing. But it’s basically that the missing shape should be such that , when the set is looked at as subsets of any one attribute, the two subsets are reversed mirrors of the other in all other traits.

I would say a large red circle with no outline. I am basising this on the fact that there is no red circle in the pattern and it only seems logical. Then again, I may be missing something completely.

@ Zach
Yes, I also thought about the operator thing.
But as it is in the picture, I see no consequent application of the potential operator:
a. border + border operator becomes: no border
The result is different from the elements as well as the operator itself
b. red + green operator becomes: green
The result is different from the elements but not different from the operator.

But if you take a self styled pattern then this goes as well:

2 and 4 have the border in common
4 and 6 have the colour in common
so:
1 and 3 have the border in common
3 and 5 have the colour in common
hence:
a small green bordered square.

I like Edward Shaw’s “answer” even if slightly tongue in cheek. Our first question, or at least a question that should be asked before we get into red/green or large/small etc., should be “What is the set of objects from which we can pick?” Our universe can be safely defined at the 16 shapes with each variation of the four characteristics and split into the two subsets of the 5 defined members and 11 non members.

Counting the commanalities beween each 5 members and 11 of the non-members, a curious fact is observed. There are 8 non-members which have at least 3 common characteristics with at least 1 member, there are 2 non-members with exactly 2 characterisics shared with every member of the set, and one that shares 2 characteristics with every member save the big, green, outlined, circle, with which it shares no characteristics.

The two non-members that share two characteristics with all members are of course diametric opposites. And, as you will guess, there are 4 choose 2 = 6 elements of our universe that share exactly 2 characteristics with the opposites noted. In fact, of the 8 diametric opposites only the two noted above have all 5 of the members fit into the set of elements which match exactly 2 characteristics of the opposites. There is a sixth member, of course, the small red square with no outline.

If we assign a binary function to their qualities of the objects we can see that the first two and the last two are direct opposites of each other, therefor the 3rd shape must be directly opposite to the 4th shape

It’s another trick question! ALL the possible combinations of red/green, square/circle large/small and border/non-border are “correct”, depending on the parameters and logic you use. The “properties” of the object can be large or small, red or green, square or circle, and bordered or unbordered. The permutations are:
small red unbordered circle
small red unbordered square
small red bordered circle
small red unbordered square
small green unbordered circle
small green unbordered square
small green bordered circle
small green bordered square
large red unbordered circle
large red unbordered square
large red bordered circle
large red unbordered square
large green unbordered circle
large green unbordered square
large green bordered circle
large green bordered square

You can come up with “logical” solutions to ALL these permutations, though some follow very simple logic, others rather convoluted. This reminds me of the 11+ exam I sat many years ago, when questions like this were common. You had to be careful to use the MOST logical method to arrive at the solution – which wan’t really fair! In this case, taken as a set, and assuming that the sequence is less relevant than the set itself (is that the best logic?) The solution is this: there are already 3 bordered objects, so the missing object should be unbordered; there are three green objects, so the missing object should be red, there are 3 circles, so the missing object should be square, and there are 3 large objects, so the missing object should be small.
A small, red, unbordered square is the solution. But ONLY inamuch as this is the “simplest” solution!
Great little puzzle, got any more?

Oops! I made a little mistake with the list of permutations, here is the correct list:
small red unbordered circle
small red unbordered square
small red bordered circle
small red bordered square
small green unbordered circle
small green unbordered square
small green bordered circle
small green bordered square
large red unbordered circle
large red unbordered square
large red bordered circle
large red bordered square
large green unbordered circle
large green unbordered square
large green bordered circle
large green bordered square
Doug π

Silly me! I forgot to remove the objects already in the puzzle from the list! There are no repeated objects, so the final list is actually:
small red unbordered circle
small red unbordered square
small red bordered square
small green unbordered square
small green bordered circle
small green bordered square
large red unbordered circle
large red unbordered square
large red bordered circle
large green unbordered circle
large green bordered square
….although I’m quite sure a government statistician would find ways to “solve” the puzzle for any of these (already included) objects!
Doug

Alan, I think the point of this puzzle is that there isn’t a definitive “solution”. As I said above, logic can arrive at ANY of the possible outcomes, and logic isn’t subjective (or, it’s not meant to be!) Perhaps this puzzle helps to demonstrate that indeed logic IS subjective – as mathematicians have been realising more and more.
My analogy to government statisticians, whilst slightly tongue-in-cheek, is also a reminder that our brains WILL square the circle when required. In other words, we will adapt our perceptions of the Universe to fit our current needs. Generally, we prefer the most elegant, or simplest solutions to problems. This puzzle has been designed to offer multiple solutions, and the choice of which solution is “best”, “most elegant”, or “simplest” is entirely subjective.
Would you believe it – an entire philosophy in a simple puzzle!
π

I agree with you Doug. In reading this problem I wasn’t expecting a definite solution, I was looking for other peoples ways of proposing a solution. As with all “what comes next” puzzles its open to every ones individual logic, and therein lies the interest.

Sorry if I came across as aggressive in my post I really wasn’t trying to be, and if this was a conversation I doubt my tone of voice would convey that if this was a conversation. What I was really trying to say was ‘how would you solve this?’

sorry if I came across as aggressive in my post I really wasnβt trying to be, and if this was a conversation I doubt my tone of voice would convey that. What I was really trying to say was βhow would you solve this?β

“what object is missing?”
errrm.. from where?
my bedroom?
may car?
my life?
nothing is really specified!!..
but some standard coloured, standard shapes are presented!!
..is someone trying to tell me something here?

I think they related by 3 things and whats missing we add them and create the missing object, so thats what came up :
3 lined Objects
3 Circles
3 Green Objects
3 Big Objects
2 Red Objects
2 Unlined Objects
2 Small Objects
2 Squares

So whats missing will be “Red Unlined small square”

For the sake of completeness I would say small red unbordered square, for in this case there are 3 squares, 3 circles, 3 big shapes, 3 small shapes, 3 bordered shapes, 3 unbordered shapes, 3 red figures and 3 green figures.

On the other hand, searching for some antisymmetry, if we think that the “opposite” of red is green, that of square is circle, that of big is small and that of bordered is unbordered… we can think that the missing figure is small, red, unbordered square. In this case the three first figures are the symmetric-opposite of the other three.

Since there are two different, rather logical, criteria to choose small, red, unbordered square I am forced to choose it.

Still no comment from the authors…
@Gary
Nifty solution, but how do you know what has to be 1, and what has to be 0?
E.g why is Small 1, and Large 0?
These self appointed binary numbers are not about what is actually shown in the picture.
It looks a bit like styling your own pattern so that something beautiful can be made of it.

@Doug 15 June 2010, 5:37 pm
Well, in a philosophical sense subjectivity is the only thing we know.
In a concrete sense, logic can be subjective but can also be objective.
Simple arithmetics show this objectivity, as well as conclusions based upon concrete premises.
It begins to get harder if the matter involves multiple factors, then concrete premises have to be found and weighed against each other.
Then you have a coherent set of premises, which emphasizes certain aspects and disregards others.
Here we have to complete a picture that consists of coherent and concrete patterns.
We must not forget that we can only use what the picture shows us if we want to make a valid conclusion.
Therefore we must ask ourselves which patterns are dominant.
It is well known that the shape outweighs all other factors in importance: if we draw a shape of a tree and colour it blue, if we make it small or large, we still call it a tree. It is the shape that determines the type of element.
Here in the picture, there is even a consistency in pairs of elements.
Size and colour are below the shape, and are more or less equal in weight (depending on the circumstances).
Here, the colours could be each others opposites, but the red colour could also be seen as a signaling colour for instance, to emphasize that the first two elements are a pair – which would be very dominant in the pattern in that case, as well as the repetition of this pair in green on the right side.
The borders are of the least importance because these do not change the essential aspects of each element.
Alan’s solution with the binary sytem is the only way the binary system can be applied without bias in what should be considered 1 or 0, and in which order you put the factors shape, size, colour an border.

Undoubtedly there is strong argument for a small red square without border, if you balance out the 1s and 0s (three of each), or if you draw a line in the middle where the left three elements are in every way the opposite of the right three elements, a kind of mirror that shows the inverse: border vs. borderless, red vs. green, large vs. small, and round vs. angular (not circle vs. square, because a square is not the opposite of a circle – then a triangle would be the opposite of a circle as well).
The odd thing about this ‘mirror’ solution is that it ignores the concrete shapes themselves, but instead handles the characteristics of these shapes: round versus angular as opposites.

Still, I prefer the small green bordered square, because it answers to the repeating patterns in pairs, which stand on themselves in the sequence, and have no dependence on the element which is asked for.
In other words: these patterns are self evident regardless of the question mark.

small green square, with black outline… it’s obvious! π

The color doesn’t change between the two members of a pair, neither does the outline. The first member of the pair is always small. And, if the first member is a square, the second member of the pair is a circle, and vice-versa. Therefore, a large green outlined circle must be preceded by a small green outlined square.

## Leo B.:

I’ll say, a large non-outlined red circle.

9 January 2010, 11:37 pm## Gregory Marton:

Big red borderless circle?

10 January 2010, 1:07 am## Sana:

A cat

10 January 2010, 1:59 am## Paul:

A small, red square without a border. Because then there are an equal number of either shape, size, border type and colour.

10 January 2010, 3:44 am## Oscar Cunningham:

A small red unbordered square. That way we get each of the four properties (size, shape, colour, borderedness) being split 3 and 3 in the objects, i.e. three red, three green; three with a border, three without a border…

10 January 2010, 4:17 amWe also form a pretty shape in the 4d (size, shape, colour, borderedness) space, though I’m not sure what to call it. Is it an octahedron?

## Eka:

A Red dot,perhaps? I thought the sequence changes color after three points..

10 January 2010, 5:40 am## VDegeler:

A big red square.

So we have:

10 January 2010, 6:04 am3 circles and 3 squares

3 big items and 3 small items

3 red items and 3 green items.

## VDegeler:

(My first comment had mistake: should be small instead of big π

A small red square.

So we have:

10 January 2010, 6:05 am3 circles and 3 squares

3 big items and 3 small items

3 red items and 3 green items.

## p:

I’d say a large, red, borderless circle is missing since every object is matched with another that has the inverse colour and border width.

10 January 2010, 7:15 am## Jonathan:

I first came to the same answer as Paul, Oscar, and VDegeler. And then I thought: is this a set, or is it a sequence?

If it is a set, add me to the list. Small, red, unbordered square.

But if this is a sequence, notice that two properties stay the same and two change as we move to the next item.

1. small red bordered circle

2. large red bordered square

3. ????? ??? ???????? ??????

4. large grn bordered circle

5. small grn unbordrd circle

6. large grn unbordrd square

There are four objects that could complete this sequence:

a) large red unbordrd circle or

b) large grn unbordrd square or

c) small red bordered circle or

d) small grn bordered square

Since b and c already exist, aesthetically I am drawn to a or b, large red unbordrd circle or small grn bordrd square

10 January 2010, 10:24 am## Edward Shaw:

10 question marks. With 4 attributes, this makes a total of 16 different possible shapes. We have 6 here, there must be 10 others.

10 January 2010, 11:02 am## Edward Shaw:

Alternatively, a full stop. The first two shapes represent a question with a question mark. The green shapes are the answer, unfortunately missing the appropriate grammar.

10 January 2010, 11:05 am## Andrei Zelevinsky:

A copy of object No. 4: big green circle with a border! This is the only way to make each of the properties (shape, size, border type and color) represented an even number of times. This is of course complementary to the answer by Paul and others (replace “even” by “odd”). If you think of objects as vectors in F_2^4, my answer amounts to make the sum equal to 0.

10 January 2010, 11:44 am## Neil F:

Small green bordered square. The items are in pairs, and between first and second members they a) retain colour, b) retain border state, c) change shape.

10 January 2010, 6:17 pm## Merlijn van Veen:

A small green square with a border would complete the pair, if the answer is meant to complete the three pairs.

11 January 2010, 3:32 pmIf it is meant to be two complementary triplets, then a large red circle without a border is okay.

Either way, the question mark is the odd one out.

## Merlijn van Veen:

By the way, if the las two (green unbordered) figures were brown, then nothing could be missing, because the green bordered circle could implicate an operator instead of a figure in the sequence.

Still, I would prefer a small green square with a border, because the first two and last two figures are in the same sequence – and that would indicate three pairs.

Within these pairs there is only change in shape, at least that is what the first and the last pair suggest.

But again, is it 3 x 2 (small green square border), 2 x 3 (large red circle borderless), or even 1 x 6 (where small red borderless square would be right)?

11 January 2010, 3:58 pm## sympath:

@Jonathan: (“aesthetically I am drawn to a or b, large red unbordrd circle or small grn bordrd square”)

12 January 2010, 12:23 amIf there is no reason to choose one over the other than aesthetics, then that rules out the puzzle being about sequence.

@Merlign: The problem with thinking about it as 2 triplets or 3 pairs is that two of anything is not enough to establish a pattern.

We already have 3 large, 3 circles, 3 borders, 3 green. We also have 2 small+1?, 2 squares+1?, 2 borderless+1?, 2 red+1?

All the characteristics clearly occur 3 times; nothing suggests one should leave a characteristic only happening twice by making its opposite happen 4 times: The ? is a small, red, borderless square.

## Merlijn van Veen:

@sympath

The most holistic is the 1 x 6 option (your choice), where the choice is based on the sum of all specifics of all elements.

Two of anything (elements) is enough to establish a pattern, if:

1. there are more than one group of two elements, or

2. If the two elements within one group are comparable in their specifics. E.g. a group consisting of two equal dots is uniform – that is a pattern, or a group consisting of two green but different elements, that is a pattern as well.

Even a group consisting of two totally different elements creates a pattern of high contrast.

So nr 2. is always valid, but only if you acknowledge that the group consists of the two elements in question.

There is a good reason to say that there are two elements per group:

12 January 2010, 3:10 pma. the repetition in pairs (first two and last two elements)

b. alternate large and small elements

a. and b. would suggest a triple pairing sequence, consisting of 3 x 2 elements (3 groups of 2 elements).

And with that comes the acknowledgement that one group should consist of two elements.

I think it’s even preferable, because both a. and b. form a premise in the puzzle.

## Noelle:

A small, red, non-outlined square.

I am, however, making the following assumptions about the pattern:

– that there should be the same number of large and small objects, and that these alternate small, large, small, large, etc.

– that there should be the same number of green and red objects, with reds on one side and greens on the other

The last two assumptions (or maybe one large assumption) I’m having a hard time phrasing. But it’s basically that the missing shape should be such that , when the set is looked at as subsets of any one attribute, the two subsets are reversed mirrors of the other in all other traits.

I hope that’s not too odd/hard to understand.

12 January 2010, 9:40 pm## Zach:

There is nothing missing:

We have:

[1] A Nice Red Circle & Square on the left side each with a border

[2] A Nice Green Circle & Square on the right side each wiht out a border

[3] Nice Green Circle With A Border In The Middle To That Show The Changes Between [1] and [2] (color and border)

13 January 2010, 4:35 pm## Joey:

I would say a large red circle with no outline. I am basising this on the fact that there is no red circle in the pattern and it only seems logical. Then again, I may be missing something completely.

13 January 2010, 5:13 pm## Merlijn van Veen:

@ Zach

13 January 2010, 10:36 pmYes, I also thought about the operator thing.

But as it is in the picture, I see no consequent application of the potential operator:

a. border + border operator becomes: no border

The result is different from the elements as well as the operator itself

b. red + green operator becomes: green

The result is different from the elements but not different from the operator.

## misha:

………………………………

14 January 2010, 12:07 amand that embittered mind which boils

in empty deeds and futile toils.

## Sunil:

A big red borderless circle

14 January 2010, 12:57 am## DeepButi:

1 and 6 have nothing in common

2 and 5 neither

3 and 4 neither, so: Small red unbordered square.

Or maybe a yellow medium dash-bordered triangle π

14 January 2010, 3:15 pm## Merlijn van Veen:

@ DeepButi

But if you take a self styled pattern then this goes as well:

2 and 4 have the border in common

4 and 6 have the colour in common

so:

1 and 3 have the border in common

3 and 5 have the colour in common

hence:

a small green bordered square.

…it is not valid this way.

16 January 2010, 2:12 pm## colorblind:

I like Edward Shaw’s “answer” even if slightly tongue in cheek. Our first question, or at least a question that should be asked before we get into red/green or large/small etc., should be “What is the set of objects from which we can pick?” Our universe can be safely defined at the 16 shapes with each variation of the four characteristics and split into the two subsets of the 5 defined members and 11 non members.

Counting the commanalities beween each 5 members and 11 of the non-members, a curious fact is observed. There are 8 non-members which have at least 3 common characteristics with at least 1 member, there are 2 non-members with exactly 2 characterisics shared with every member of the set, and one that shares 2 characteristics with every member save the big, green, outlined, circle, with which it shares no characteristics.

The two non-members that share two characteristics with all members are of course diametric opposites. And, as you will guess, there are 4 choose 2 = 6 elements of our universe that share exactly 2 characteristics with the opposites noted. In fact, of the 8 diametric opposites only the two noted above have all 5 of the members fit into the set of elements which match exactly 2 characteristics of the opposites. There is a sixth member, of course, the small red square with no outline.

At least, that’s how I’d work it.

17 January 2010, 1:54 am## Anoush:

Consider the first pair of objects. First they both change their colour and shape. Then they change their shape and border.

19 January 2010, 6:10 amA green bordered square.

## Merlin:

The authors are somewhat silent…

6 May 2010, 8:18 pm## Alan:

square 0 1 ? 0 0 1

red 1 1 ? 0 0 0

border 1 1 ? 1 0 0

Big 0 1 ? 1 0 1

If we assign a binary function to their qualities of the objects we can see that the first two and the last two are direct opposites of each other, therefor the 3rd shape must be directly opposite to the 4th shape

square 0 1 1 0 0 1

red 1 1 1 0 0 0

border 1 1 0 1 0 0

Big 0 1 0 1 0 1

therefor the shape we are looking for is a small red square without a border

9 June 2010, 8:17 am## Bob Ress:

a baby’s arm, holding an appple

9 June 2010, 9:04 am## Bob Ress:

A small red square, no outline.

9 June 2010, 9:10 am## Bill Mitchell:

Think paralax – large red circle, no border

11 June 2010, 9:28 pm## Doug Roberts:

It’s another trick question! ALL the possible combinations of red/green, square/circle large/small and border/non-border are “correct”, depending on the parameters and logic you use. The “properties” of the object can be large or small, red or green, square or circle, and bordered or unbordered. The permutations are:

small red unbordered circle

small red unbordered square

small red bordered circle

small red unbordered square

small green unbordered circle

small green unbordered square

small green bordered circle

small green bordered square

large red unbordered circle

large red unbordered square

large red bordered circle

large red unbordered square

large green unbordered circle

large green unbordered square

large green bordered circle

large green bordered square

You can come up with “logical” solutions to ALL these permutations, though some follow very simple logic, others rather convoluted. This reminds me of the 11+ exam I sat many years ago, when questions like this were common. You had to be careful to use the MOST logical method to arrive at the solution – which wan’t really fair! In this case, taken as a set, and assuming that the sequence is less relevant than the set itself (is that the best logic?) The solution is this: there are already 3 bordered objects, so the missing object should be unbordered; there are three green objects, so the missing object should be red, there are 3 circles, so the missing object should be square, and there are 3 large objects, so the missing object should be small.

15 June 2010, 11:28 amA small, red, unbordered square is the solution. But ONLY inamuch as this is the “simplest” solution!

Great little puzzle, got any more?

## Doug Roberts:

Oops! I made a little mistake with the list of permutations, here is the correct list:

15 June 2010, 11:38 amsmall red unbordered circle

small red unbordered square

small red bordered circle

small red bordered square

small green unbordered circle

small green unbordered square

small green bordered circle

small green bordered square

large red unbordered circle

large red unbordered square

large red bordered circle

large red bordered square

large green unbordered circle

large green unbordered square

large green bordered circle

large green bordered square

Doug π

## Doug Roberts:

Silly me! I forgot to remove the objects already in the puzzle from the list! There are no repeated objects, so the final list is actually:

15 June 2010, 11:53 amsmall red unbordered circle

small red unbordered square

small red bordered square

small green unbordered square

small green bordered circle

small green bordered square

large red unbordered circle

large red unbordered square

large red bordered circle

large green unbordered circle

large green bordered square

….although I’m quite sure a government statistician would find ways to “solve” the puzzle for any of these (already included) objects!

Doug

## Alan:

I guess the point of this puzzle was to see your logic in proposing a solution, saying it could be anything isn’t really a solution.

15 June 2010, 12:42 pm## Doug Roberts:

Alan, I think the point of this puzzle is that there isn’t a definitive “solution”. As I said above, logic can arrive at ANY of the possible outcomes, and logic isn’t subjective (or, it’s not meant to be!) Perhaps this puzzle helps to demonstrate that indeed logic IS subjective – as mathematicians have been realising more and more.

15 June 2010, 5:37 pmMy analogy to government statisticians, whilst slightly tongue-in-cheek, is also a reminder that our brains WILL square the circle when required. In other words, we will adapt our perceptions of the Universe to fit our current needs. Generally, we prefer the most elegant, or simplest solutions to problems. This puzzle has been designed to offer multiple solutions, and the choice of which solution is “best”, “most elegant”, or “simplest” is entirely subjective.

Would you believe it – an entire philosophy in a simple puzzle!

π

## Alan:

I agree with you Doug. In reading this problem I wasn’t expecting a definite solution, I was looking for other peoples ways of proposing a solution. As with all “what comes next” puzzles its open to every ones individual logic, and therein lies the interest.

Sorry if I came across as aggressive in my post I really wasn’t trying to be, and if this was a conversation I doubt my tone of voice would convey that if this was a conversation. What I was really trying to say was ‘how would you solve this?’

18 June 2010, 6:00 am## Alan:

sorry if I came across as aggressive in my post I really wasnβt trying to be, and if this was a conversation I doubt my tone of voice would convey that. What I was really trying to say was βhow would you solve this?β

edit (sorry its early)

18 June 2010, 6:01 am## Van Helmont:

Breasts? (hey I’m a Freudian)

21 June 2010, 12:53 pm## Lucas Brasher:

The only one missing is a large unbordered Red circle.

26 June 2010, 1:57 am## AnotherBrokenToaster:

Small green border less circle definitely!

13 July 2010, 6:39 pm## AnotherBrokenToaster:

Wait no not circle small green bordered square.

13 July 2010, 6:40 pm## person:

something

16 July 2010, 7:54 am## aamir:

“what object is missing?”

2 August 2010, 6:20 amerrrm.. from where?

my bedroom?

may car?

my life?

nothing is really specified!!..

but some standard coloured, standard shapes are presented!!

..is someone trying to tell me something here?

## Blooshi:

I think they related by 3 things and whats missing we add them and create the missing object, so thats what came up :

3 lined Objects

3 Circles

3 Green Objects

3 Big Objects

2 Red Objects

2 Unlined Objects

2 Small Objects

2 Squares

So whats missing will be “Red Unlined small square”

5 August 2010, 1:28 am## Antonio:

For the sake of completeness I would say small red unbordered square, for in this case there are 3 squares, 3 circles, 3 big shapes, 3 small shapes, 3 bordered shapes, 3 unbordered shapes, 3 red figures and 3 green figures.

On the other hand, searching for some antisymmetry, if we think that the “opposite” of red is green, that of square is circle, that of big is small and that of bordered is unbordered… we can think that the missing figure is small, red, unbordered square. In this case the three first figures are the symmetric-opposite of the other three.

Since there are two different, rather logical, criteria to choose small, red, unbordered square I am forced to choose it.

13 September 2010, 1:02 pm## Gary:

The Solution is Large, Green Square with Boarder

I created the following scheme to solve it

Small = 1 Red = 1 Square = 1 Boarder = 1

Large = 0 Green = 0 Circle = 0 No Boarder = 0

Now set out the qualities in line as a binary number

This 1101 represents the first figure is Small, Red, Circle, with Boarder

and our sequence is

1101 = 13

0101 = 5

xxxx = ?

0001 = 1

1000 = 8

0010 = 2

This is the Fibonacci sequence out of order so

xxxx = 3

In binary

= 0011

using my scheme this is

Large, Green, Square, Boarder

Gary

PS this was a great puzzle thanks Tanya and Gregory

7 December 2010, 11:06 pm## Merlin:

Still no comment from the authors…

30 March 2012, 7:52 pm@Gary

Nifty solution, but how do you know what has to be 1, and what has to be 0?

E.g why is Small 1, and Large 0?

These self appointed binary numbers are not about what is actually shown in the picture.

It looks a bit like styling your own pattern so that something beautiful can be made of it.

## Merlin:

@Doug 15 June 2010, 5:37 pm

Well, in a philosophical sense subjectivity is the only thing we know.

In a concrete sense, logic can be subjective but can also be objective.

Simple arithmetics show this objectivity, as well as conclusions based upon concrete premises.

It begins to get harder if the matter involves multiple factors, then concrete premises have to be found and weighed against each other.

Then you have a coherent set of premises, which emphasizes certain aspects and disregards others.

Here we have to complete a picture that consists of coherent and concrete patterns.

We must not forget that we can only use what the picture shows us if we want to make a valid conclusion.

Therefore we must ask ourselves which patterns are dominant.

It is well known that the shape outweighs all other factors in importance: if we draw a shape of a tree and colour it blue, if we make it small or large, we still call it a tree. It is the shape that determines the type of element.

Here in the picture, there is even a consistency in pairs of elements.

Size and colour are below the shape, and are more or less equal in weight (depending on the circumstances).

Here, the colours could be each others opposites, but the red colour could also be seen as a signaling colour for instance, to emphasize that the first two elements are a pair – which would be very dominant in the pattern in that case, as well as the repetition of this pair in green on the right side.

The borders are of the least importance because these do not change the essential aspects of each element.

Alan’s solution with the binary sytem is the only way the binary system can be applied without bias in what should be considered 1 or 0, and in which order you put the factors shape, size, colour an border.

Undoubtedly there is strong argument for a small red square without border, if you balance out the 1s and 0s (three of each), or if you draw a line in the middle where the left three elements are in every way the opposite of the right three elements, a kind of mirror that shows the inverse: border vs. borderless, red vs. green, large vs. small, and round vs. angular (not circle vs. square, because a square is not the opposite of a circle – then a triangle would be the opposite of a circle as well).

The odd thing about this ‘mirror’ solution is that it ignores the concrete shapes themselves, but instead handles the characteristics of these shapes: round versus angular as opposites.

Still, I prefer the small green bordered square, because it answers to the repeating patterns in pairs, which stand on themselves in the sequence, and have no dependence on the element which is asked for.

17 April 2012, 1:15 pmIn other words: these patterns are self evident regardless of the question mark.

## ObiJohn:

small green square, with black outline… it’s obvious! π

The color doesn’t change between the two members of a pair, neither does the outline. The first member of the pair is always small. And, if the first member is a square, the second member of the pair is a circle, and vice-versa. Therefore, a large green outlined circle must be preceded by a small green outlined square.

24 April 2012, 6:43 am## JRG:

I would say a does-not-equal sign” i.e. a “β ”. A red circle and a red square does not equal two green circles and a green square.

8 May 2012, 5:27 pm## aristogeit:

I’d say a triangle ther is no fun inthere

6 February 2014, 9:38 pm