Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Looking for an example of a question that would seem not to have sufficient information for an answer, or a question that the solution would not require (or maybe even maybe hindered ) by the extra provided (relevant) information.

Is it possible to have questions were a solution would depend on lack of some certain information?

For example, a variant of Hat Puzzle, can be answered correctly by realizing the hesitation of other participants to conclusively answer and including it into the deduction. But that problem includes a temporal aspect into the riddle.

share|improve this question

4 Answers 4

up vote 1 down vote accepted

Lack of information is often useful in terms of generality, that is, if you don't know the details, the same way should work for a wide class of objects.

For example, suppose you have a function $f : \mathbb{N} \to \mathbb{N}$. It might be any kind of function, e.g. $f_1(n) = n$ or $f_2(n) = n+1$, etc. (what I mean is that if you know that your input is a natural number, then you can modify it with any operation that works with natural numbers).

However, suppose you have a function $f : \alpha \to \alpha$ for any $\alpha$ (that is a single function that provided with an object of type $\alpha$ returns an object of the same type; I know this is not formal, I am just sketching the idea), then there is not much this function could do. It has to be the identity, the reason being, this function doesn't know how to manipulate an object of some unknown, given type (which might not yet exists when creating the definition of $f$).

In other words, if there is a lack of information, then the thing (whatever is the object of consideration) must treat all the given inputs in some way uniformly, and this constraint might be just big enough to pinpoint how that thing works.

If you are interested, look at this excellent blog post by Dan Piponi (informal but easy to understand): Reverse Engineering Machines with the Yoneda Lemma and this great paper by Philip Wadler (formal but less approachable): Theorems for free!

I hope I didn't make it too vague, but if I were to put it formally, we would all drown in unreadable notation and it wouldn't be then fun at all...

share|improve this answer
    
+1, very impressive observation "Lack of information is often useful in terms of generality", would have +10'ed if I could. –  Arjang Jan 10 '13 at 21:23

A ring is created by drilling a hole straight through the center of a solid sphere. The ring is six inches tall after it has been drilled out. What is the volume of the ring?

That's all the info you need to solve the problem, even though you have no idea whether you took a sphere slightly bigger than 6" around and drilled a skinny hole through it, or started out with a much larger sphere and bored a much fatter hole through it. No matter what size sphere you started with, the volume after drilling is the same as the volume of a sphere with a diameter of six inches, i.e. 36*pi.

share|improve this answer

What is the sum of the roots of the following quadratic polynomial?

$$ x^2 - 18 x + \text{<illegible smudge>} $$


An easy way to hinder a solution is to simply add in extra irrelevant information to give people false paths to waste time on, or better yet, misleading information. For example:

Many numbers can be written as the sum of squares of three distinct integers, such as $29 = 2^2 + 3^2 + 4^2$ or $75 = 1^2 + 5^2 + 7^2$.

What is the smallest number that can be written in such a way?

The answer is $2 = (-1)^2 + 0^2 + 1^2$. The examples are crafted to plant/reinforce the idea of positive integer in the reader's head, making it more difficult to think of using negative numbers. I've seen some very well done examples of this trick, but none come to mind at the moment.

(I would have given $12$ as an example of a number that can't, because $2^2 + 2^2 +2^2$ doesn't count -- but I'm not sure if that would throw people off even further, or help them think of the idea if considering using both $x^2$ and $(-x)^2$)


Special cases can sometimes be unhelpful. For example, at one time in my life, I found the sum

$$ 1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + 4 \cdot 2^4 + \cdots + 1000 \cdot 2^{1000} $$

much harder to compute than the sum

$$ \sum_{n=1}^{1000} n x^n $$

because seeing the special case prevented me from thinking of general methods.

share|improve this answer
    
The most honest answer I've ever encountered on SE. People nowadays are more prone to act like a fox who would hide his trace whenever it's possible. –  Metta World Peace Jan 10 '13 at 13:09
    
Thank you Hurkyl, "special case prevented me from thinking of general methods" is food for thought, wonder if the was used in teaching of mathematics, poor kids wouldn't have to solve the same problems 100 times over in name of learning math. –  Arjang Jan 10 '13 at 21:27

Perhaps the following is what you are looking for. Suppose I have some distribution over the set $[0,1]$ which I use in order to pick two numbers $x,y$. The only thing you know about the distribution is that the probability that $x=y$ is zero. Now, I show you $x$ and I ask you to guess whether $y>x$ or $y<x$. Clearly, if you always guess that $x<y$ then your chances of success are 50%. It would seem that it is impossible to devise a strategy whereby the chances of success will be greater than 50% but, surprisingly, it is possible.

share|improve this answer
    
Well, does this strategy have strictly greater probability than $1/2$ for any distribution? Let $A$ be the set of $x$ that would make you answer $x < y$ by your strategy. Then setting some distribution with $P(x \in A) = 1$ would make your chance of success $1/2$. This would mean that your set of $x$-es would need to be non-measurable and it doesn't kind of fit the "devise a strategy" part. I am confused. –  dtldarek Jan 10 '13 at 13:31
    
the solution is very straightforward and feasible yielding a strategy with strictly greater than 1/2 probability of success. –  Ittay Weiss Jan 10 '13 at 21:23
    
Well, strategy such like "if $x < 1/2$ then $x < y$ else $x > y$" works for distributions which has non-zero measure at both sides of $1/2$. However, the strategy I sketched up in previous comment makes you answer $x < y$ in all cases (non-deterministic strategies could be solved so that my distribution is dependent on the random variables used in the strategy; in fact it is all defined in the same probability space). Does the strategy works in such case (i.e. when all your answers are $x < y$)? You have blown my mind :-P –  dtldarek Jan 10 '13 at 22:55
    
The strategy works for any setting as given in my answer. –  Ittay Weiss Jan 10 '13 at 22:57
    
You have blown my mind twice :-) –  dtldarek Jan 10 '13 at 23:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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