Code breaker needed

[bluedot]

I believe there must be "correct" answers based on the fact that they're obviously correct. Like imatfaal's. Otherwise all of us are capable of making some made-up answers and start philosophizing to make others believe they're the answers.

Well, I am convinced some questions are just badly composed (like Q2 and Q3) so you must start philosophizing to make sense of them.

Anyway, here's take on Q4.

Process of elimination. "D" can't be becouse it is just one figure. "A" can't be becouse its figures have rounded corners.

For the clarity of explanation let's enumerate boxes in the pic as follows:

1 4 7

2 5 8

3 6 9

Observation a: all the boxes containing two figures that differ in number of sides have one figure bold-sided and other figure thin-sided while all the boxes containing figures with equal number of sides are with both figures thin sided. Therefore, "B" eliminated.

Observation b: In the 1st column (1,2,3) 1 has outer figure bold-sided and inner figure thin while 3 has outer figure thin and inner figure bold. In the 3rd column (7,8,9) 8 has outer figure bold-sided and inner figure thin sided while 9 has outer one thin-sided and inner bold-sided. This would indicate that 2nd column got to have two boxes with inverted thickness of inner and outer figure. Therefore, "C" eliminated.

Correct answer:E

 

Do you agree with this philosophy ?

[tar]

Bluedot,

 

I agree with your answer to 4.

 

I think the code is:

Put a symbol inside another and if one symbol has more sides than the other, bold it.

 

E follows the code. And in addition there are three boxes with symbols of the same number of sides, three with the bolded greater number of sides on the inside symbol, and only two with the bolded, greater number of sides being the outside symbol. So we need a combination of symbols where the outside symbol has more sides than the inside, and the outside symbol is bolded. E is just such a combination.

 

I will look at your answer to number five again, to see if I understand what you are saying, but I think a guide to anwering the questions, is to forget completely about the answers given, and see if you can see the answer without being given any posibilities. Once you are sure of what the answer is going to look like...only one will be like that...and that will be the answer.

 

Regards, TAR

 

Three to go.

Bluedot,

 

OK, I see your 5,3 reasoning for number 5, but there is something about it that seems obscure or fishy, so that I am not really sure if you have the code. Your code works and you might have broken the code, in this case, but it is not as satisfying as Imatfaal's break of number one, and I am holding out hope that there might be a more satisfying answer for 5.

 

So, still three to go. (or at least 2)

 

Regards, TAR

[bluedot]

 

OK, I see your 5,3 reasoning for number 5, but there is something about it that seems obscure or fishy, so that I am not really sure if you have the code. Your code works and you might have broken the code, in this case, but it is not as satisfying as Imatfaal's break of number one, and I am holding out hope that there might be a more satisfying answer for 5.

 

So, still three to go. (or at least 2)

 

Regards, TAR

Well, let me put it like this: if I'm 99 % sure "E" is the correct answer for Q4, I'm 99.9 % sure "C" is the correct answer for Q5

It's not just that two simple eqs presented perfectly match situation in the picture, but interplay of numbers 3 and 5 there.

This is a 3x3=9 boxes puzzle, each dotted box has 5x5=25 dots in total, and solution to the problem is (m,n)=(3,5).

Accident? I don't thinks so. This is a very nice puzzle and whoever composed it knew what he/she was doing.

 

BTW, any idea for Q3?

[tar]

Bluedot,

 

On 2 you were thinking "the question may simply ask "which one is the most like others?".

 

I do not think in general that is what the code maker is asking in the 5 questions.

I think it is more like "do you see what I am doing here?" or in the spirit of code breaking "I am thinking something here, and you will know what it is, as soon as you see it, as soon as you break the code"

 

Regards TAR

Bluedot,

 

Well since you know exactly what the code maker was thinking in five, why did he/she put the symbols in the northwest and southeast corners in the other two boxes and not in the answer?

 

I guess you did not like my initial figuring for 3, or you would not be asking for something reasonable. I initially was thinking the answer had to be a new symbol, since all the others were different, and just tried to figure why the background and symbol shading would be a certain way.

Could not really come up with anything. Made me wonder if the puzzle was originally in color and the shading did not give the proper clues. Anyway, that was before Imatfaal cracked the first one. Now I am thinking he/she is doing something obvious, and I am not yet seeing it. Let me look at it a bit, without looking at the answers and see if I can figure what should be where the question mark is.

 

Regards, TAR

Edited September 27, 2014 by tar

[bluedot]

Well since you know exactly what the code maker was thinking in five, why did he/she put the symbols in the northwest and southeast corners in the other two boxes and not in the answer?

 

 

Quite likely becouse order of operations is irrelevant for multiplication (m*n=n*m) and for addition (m+n=n+m). IOW, solutions (3,5) and (5,3) are equally valid. If the position of polygons in boxes were of any importance, in the set of offered answers he/she would offer two boxes with changed positions of same polygons, wouldn't he/she?. Well, there is no such pair of boxes among offered answers A,B,C,D,E...

As concerns Q3, I'm totally clueless about it. Maybe it is a very tricky or very stupid one, or maybe it lost colors as you said...

Edited September 27, 2014 by bluedot

[tar]

Bluedot,

 

Ok. So Imatfaal got number one. You got number 5. We agree on number 4, But number two is still up for grabs, Endy0816 sees some rhombus triangle thing resulting in E as the answer, you see some five line thing resulting in C, and I see some left to right progression of keep and change that points to A. And number 3 has not fallen into place for anybody yet.

 

So, two to go.

 

Regards, TAR

Bluedot,

 

Just noticed these are not questions 1 through 5, as I had first assumed.

 

QNS is a clinical laboratory abbreviation for Quantity Not Sufficient.

 

Don't know if that observation helps, but it seems to indicate that to "get" the answer, you have to bring something to the party. Sort of a BYOS (specimen) party.

 

Regards, TAR

[bluedot]

Tar, few posts above is my best take on Q2: #22

Therefore, my opinion is the best choice here is "B" (not "C"). But that is just my opinion and I've already changed my mind twice about that Q.

What interests me the most is Q3 however

[tar]

Bluedot,

 

Well, I can not pick up any pattern or progression or notice any "other" way to look at the boxes that would guide one to know what the question mark box in number 3 should be.

 

My only reasoning is that it seems more likely a figure that has not been yet presented would be the answer, than one already presented. The previous box has a figure with eight sides, the following box has a figure with eight sides, so perhaps the answer should have eight sides. There are two answers with this new eight sided figure, so maybe one of them. However I don't get what is happening with the background and forground black, grey and white, so just guess the one I originally guessed.

 

But this whole reasoning is exactly the reasoning I had before Imatfaal got number one. Number one was so "good" and clear and satisfying that I expect the same kind of code to be "hidden" in number 3. Nobody has seen it yet, but I think its there. Hardly ill concieved or lame the answer will be. We already know what the maker of the challenge is capable of.

 

Regards, TAR

[tar]

bluedot,

 

OK, on three.

 

If the code is that 10 sided figures should be rendered filled with dark grey on a dark grey background.

8 sided figures should be filled with white on a dark grey background,

and 6 sided figures should be filled with light grey on a white background,

then there is only one correct answer that fits the above rules.

 

I won't mention it, because there is only one that fits the above code.

 

Regards, TAR

Edited October 11, 2014 by tar

[bluedot]

bluedot,

 

OK, on three.

 

If the code is that 10 sided figures should be rendered filled with dark grey on a dark grey background.

8 sided figures should be filled with white on a dark grey background,

and 6 sided figures should be filled with light grey on a white background,

then there is only one correct answer that fits the above rules.

 

I won't mention it, because there is only one that fits the above code.

 

Regards, TAR

I agree, this make sense. Due to very limited number of figures given in the sequence we can't be sure, but you have my vote for "C"

[tar]

Bluedot,

 

Not very happy with 3 though. Just can not come up with anything better. At least it makes sense, although it is not as satisfyingly obvious as Imatfaal's solution to number one.

 

Regards, TAR

 

Still looking for 2's answer. If we can come up with an agreeable answer for 2 we can ask the OP for an assessment of how we are doing, since we have "conditional" agreement on the other 4.