96
$\begingroup$

The website of my university posted a riddle that goes something like this:

Riddle

There are three men named 1,2 and 3 and each one has two colored dots on his forehead. Possible colors are black and red. No color is used more than four times. The men can see the colors on the other mens head, but not the ones on their own. A game-master asks the men in the order 1,2,3,1,2,3,... whether they know the colors of their dots. If one says "no", he proceeds with the next man. If one says "yes", he has to state his guess. If he is correct, all men win the game, if he is wrong then all men lose. The men were not given any time to make up a strategy.

As it turnes out the men answer "no", "no", "no", "no", "yes". What is the color on the head of man 2?

I was going to solve this riddle systematically by trying to reason what color combinations make sense and what information each man can deduce from the others saying "no".

However, I came to a much more direct solution, that completely ignores most of the riddle and only works by the assumption that the riddle is solvable.

Solution: Man 2 must have a red and a black dot on his forehead.

Reason: I (from the meta perspective) only have the information about the answer-pattern and no actual colors. Since the game is symmetric w.r.t. the colors red and black, the game would have the same answer-pattern if we switch the colors. This means, any reasoning that explains that man 2 has e.g. two red dots, will also work to reason that he has two black dots. If any of these two would be the solution, I would not be able to conclusively find it. Hence the solution must be "one red and one black dot". Note that I cannot answer how man 2 came to his conclusion, but this was not part of the question.

For me, this kind of solving the riddle was very interesting and I wondered if there are more such riddles out there which (intentionally or unintentionally) can be solved by such meta-assumptions.

Question: Are there more such riddles which have a surprisingly easy solution by meta-assumptions like e.g. "the riddle has a solution" or "the riddle is solvable with reasonable effort" etc.


Update

Because it was asked in the comments, and maybe needs a clarification, here is what I am looking for in other words:

I ask for riddles or math-problems

  • which are formulated in a way that suggests some specific meta-assumption (e.g. the riddle must have a solution which I as a reader can find).
  • the meta-information is not obviously presented to the reader as something that should be used to solve the riddle (preferably might not even be needed).
  • the meta-information turns out to be unexpectedly helpful.

I am not specifically asking about the assumptions of "existence of a unique solution", but also other meta-assumptions, like e.g. the context of the question, the number of answers (if it is multiple-choice), the expected time-frame of solution, maybe even the color of the paper the question was asked on, etc.

Some type of examples which came up a lot (and which I liked) are riddles/problems where the meta-information was the absence of some information. The absence implied that the solution will probably not depend on this missing information and hence we can choose a very simple instance of the problem.

One other example that came to my mind (but it does not fit so perfectly) is the graph-problem presented in this video of 3Blue1Brown. It can (and must) be solved by the meta-information that it is presented on a cup and not on a piece of paper. Unfortunatly the riddle can only be solved by recognizing that it is on a cup. I would prefer that the meta-way is not the intended solution.

$\endgroup$
  • 7
    $\begingroup$ @dEmigOd I do not take this into account. You are right that by seeing four dots of the same color he would have concluded that he had two of the same color. But we as an "external solver of the riddle" had no way to conclude whether he saw four red or four black dots because the riddle has no further information for us which is asymmetric w.r.t. the colors. A real situation could be staged in such a way that the second man concludes e.g. red-red, but in the riddle there is no fixed distribution of colors. $\endgroup$ – M. Winter Feb 12 '18 at 12:55
  • 15
    $\begingroup$ Plenty of standardized, quantitative multiple choice questions (SAT, GRE general and subject, etc) can be attacked with meta-reasoning... "I know the answer is one of the listed choices, so ...," for example. This is one reason why such tests aren't great: you can get the right answer for the "wrong" reason (wrong in the sense that no such meta-reasoning is available if you had to answer an open-ended question with no answer choices). $\endgroup$ – symplectomorphic Feb 12 '18 at 14:07
  • 4
    $\begingroup$ Relevant and possibly interesting: An exam question that asks about property of a group of a certain order, but without saying what the group is, On an exam one can assume it doesn't matter which group it is, take the group to be whichever one happens to make the answer easiest to find, and then write down the answer for that particular group. I said: "Reasoning from the premise that the question is well-posed is fair game on a written exam, but is not legitimate mathematics." $\endgroup$ – MJD Feb 12 '18 at 16:06
  • 7
    $\begingroup$ In contrast, here's a puzzle that many people get wrong, because they assume (incorrectly) that the puzzle is solvable. The puzzle has a unique solution, but only under the (possibly false) assumption that it has a unique solution. $\endgroup$ – MJD Feb 12 '18 at 16:15
  • 2
    $\begingroup$ Do let us know whether this belongs to Puzzling.SE $\endgroup$ – ABcDexter Feb 13 '18 at 5:44

11 Answers 11

44
$\begingroup$

Here's a riddle that fits right into this category. Hope you like it:

Begin with a sphere. Drill a circularly-cylindrical hole directly through the center of the sphere [i.e the result is still a 'solid of revolution'], and remove the drilled material. Measure the height of the hole from one circular edge through the body of the sphere to the other circular edge; the measured height is 10 centimeters.

What is the volume of the remaining solid?

$\endgroup$
  • 4
    $\begingroup$ I know this from a VSauce video :D. Despite from concluding that the the volume seems to be independent of the exact sphere-radius and drill-hole-radius (as long as the height is 10cm), can we somehow trivially conclude the exact volume? $\endgroup$ – M. Winter Feb 12 '18 at 14:58
  • 14
    $\begingroup$ @M.Winter - Consider the case when the drill-hole radius is $0$. If there is a solution, it has to work for this case (or at least for cases approximating this one to any resolution). $\endgroup$ – Paul Sinclair Feb 12 '18 at 15:24
  • 3
    $\begingroup$ This was the puzzle I was thinking of. I got it from one of Martin Gardner's books of mathematical puzzles from the 1960-80s (don't know exact year, book may have been old when I read it. Not sure if its original to him, or came from elsewhere before him. $\endgroup$ – Stilez Feb 13 '18 at 16:28
  • 6
    $\begingroup$ @Shufflepants It definitely is. The assumption tells us that the answer must be a number. If the hole radius is 0, the sphere radius is 5 and the volume is 4/3*PI*5^3. As the "question must be solvable" assumption tells us it does not depend on the hole/sphere radius, the value must be 4/3*PI*5^3 for any hole/sphere radius values. If the question were not phrased the way it was. I would probably resort to calculus. $\endgroup$ – kaine Feb 13 '18 at 17:33
  • 3
    $\begingroup$ A similar problem: suppose you draw two concentric circles such that a chord of the larger circle is tangent to the smaller circle and has length 1. What is the area of the region between the two circles? $\endgroup$ – Acccumulation Feb 13 '18 at 21:11
55
$\begingroup$

Having solved my share of logic puzzles, I can report that logic puzzles can be solved using the meta-assumption that the logic puzzle is solvable surprisingly often (in fact, I had the occasion of 'solving' the very puzzle you mention just a few months ago in exactly the way you did). I use 'solving' since of course you are not really solving the riddle at all, but merely showing that 'if the riddle has a solution, then it has to be ...'. Here is another example:

In the Sudoku-solving (and other grid-filling puzzles) community the meta-assumption that a Sudoku puzzle has a unique solution is called the Uniqueness assumption and can be used to great effect. For example, suppose you know that you have a row where there are two open cells left (say in columns $2$ and $3$) which must contain the numbers $1$ and $4$. Suppose you have another row with three open cells left (say columns $2$, $3$, and $5$), which must contain the numbers $1$, $4$, and $7$. Then using the Uniqueness assumption we can immediately say that the cell in that second row in column $5$ does not contain a $7$, since if it did, you could place the remaining $1$'s and $4$'s in two different ways, and clearly this has no bearing on the solvability of the rest of the puzzle.

$\endgroup$
  • $\begingroup$ Thanks for adding the piece about Sudoku's. I considered posting it as an isolated example, but felt that I wouldn't have too much to say about it. $\endgroup$ – Jyrki Lahtonen Feb 15 '18 at 17:53
  • $\begingroup$ Do you have a reference? I've solved some Sudoku previously, but I've not heard of this, and it would have made some seemingly impossible puzzles solvable if I'm not misunderstanding you. $\endgroup$ – Ryan The Leach Feb 16 '18 at 6:50
  • $\begingroup$ I have actually seen Sudoku's before that have more than 1 solution. I don't think this would work on them, would it? $\endgroup$ – Belle-Sophie Feb 16 '18 at 11:23
  • 1
    $\begingroup$ @Belle Correct! Applying the Uniqueness Assumption to a puzzle that does not have a unique solution can quickly lead to trouble. Take the situation as described in my post, for example. Suppose that particular Sudoku actually has exactly two solutions, with one solution having the one row having a $1$ in column $2$, and a $4$ in column $3$, and the other row having a $4$ in column $2$, and a $1$ in column $3$, while the second solution has those $1$'s and $4$'s swapped. Then both solutions do have a $7$ in column $5$ of that second row, but the Uniqueness Assumption said there is not a 7! $\endgroup$ – Bram28 Feb 16 '18 at 14:32
25
$\begingroup$

Here's one example.

The Riddle

Assume that the Earth is a perfectly round sphere and that you are constrained to the surface of the Earth. Is there a point P on Earth such that if you start at P, walk one mile North, then walk one mile East, and then walk one mile South, you will end up exactly where you started, at point P?

The Solution

The South Pole is one solution. Any direction away from the South Pole is North if you are standing at the South Pole. Just walk along a longitude, then a latitude, then a longitude and you will be back at the South Pole.

It turns out that there are infinitely many other solutions. Any point on a certain circle near the North Pole will work. But besides that, the number of circles which will work is also infinite. All of them have the North Pole as their center.

Solving the Riddle Just by Assuming that a Solution Exists

I asked this question to a fellow mathematician. She immediately assumed that a solution exists and (being a good mathematician) considered an extreme case. She was talking out loud, rapid-fire comments and questions so I know what she was thinking. She said it must be one of the poles and figured out in two seconds that it must be the South Pole.

Then I told her, "what if I told you that there is another entirely different solution". She thought about it for a few seconds and asked me if the number of solutions is finite or infinite. I thought about it and replied, "I can't tell you. Because even that would give away too much information." She then assumed that it is probably an infinite number of solutions because of the symmetry of the sphere. If (besides the poles) one point will work then there should be another that works just as well, and another, and another, and then maybe the locus of points forms a circle on the sphere and so on.

This all took about sixty seconds.

It's cool to realize that here, not just some meta-information helped her solve the riddle, but rather some meta-meta-information helped her as well. The fact that me telling her the number of solutions would give her too much information, gave her some information too.

Explaining the Complete Solution

Consider the following picture. enter image description here

In this picture, start on the black dot, go North for one mile along the red arc. Then go East for one mile and you will have traversed the entire red circle. Then go one mile South, which is along the red arc again, and you will be at the black dot again. Therefore the black dot is another solution to the riddle (as the riddle is stated). In fact, any point on the blue circle is a valid solution.

Besides these solutions, there is another red circle, centered at the North Pole, such that its circumference is 1/2 mile so that going East on that circle will make you go around twice. Therefore you simply start a mile below it.

Similarly there is another smaller red circle with circumference 1/3 mile and you can start a mile below it. And so on.

There is a blue circle for every natural integer. So there are countably many (blue) circles centered around the North Pole, converging on the circle one mile South of the North Pole, which are all valid solutions for this riddle. Some of them are shown here.

enter image description here

$\endgroup$
  • 3
    $\begingroup$ That's a cool example, because I know this riddle but never thought about othern solutions than the south pole! :D Shame on me. $\endgroup$ – M. Winter Feb 12 '18 at 20:00
  • 9
    $\begingroup$ @CiaPan I think you misunderstand. You walk one mile north, till you're far enough from the northpole that walking one mile east is a circle with circumference one mile. (1/2pi mile) Then you walk one mile south. You never go cross the northpole. (basically you start at 1 mile+ 1/2pi mile from the northpole.) $\endgroup$ – Pieter B Feb 13 '18 at 14:09
  • 1
    $\begingroup$ @PieterB Actually, you could start at a point that leaves you on a circle of 1/2 mile (walking East, you'll go round twice) circumference or 1/3 mile circumference, or 1/4 mile etc etc etc. $\endgroup$ – JeremyP Feb 13 '18 at 16:33
  • $\begingroup$ @CiaPan PieterB and JeremyP are both correct. I have added a picture in the post to show what the solutions are. $\endgroup$ – Fixed Point Feb 13 '18 at 19:37
  • $\begingroup$ This one was also mentioned by Martin Gardner, IIRC. $\endgroup$ – PM 2Ring Feb 14 '18 at 14:05
19
$\begingroup$

In the spirit of providing more examples, I see this come up a lot in the puzzle Slant from Simon Tatham's Portable Puzzle Collection. The rules of this puzzle are simple:

Fill in a diagonal line in every grid square so that there are no loops in the grid, and so that every numbered point has that many lines meeting at it.

The meta-information is that each grid square can be filled in unambiguously.

You can use this information to deduce that the upper left corner must be the NW-SE ("\") diagonal, because otherwise it could form a loop somewhere (in fact, it must otherwise form a loop somewhere, otherwise the corner would be ambiguous.)

Example from Slant

Without this meta-information, you would have to wait until later in the game to place this piece.

$\endgroup$
13
$\begingroup$

I was taking (decades ago) an advanced topics math course in high school. We got a test with several problems, one of which was "Prove that ...". I forget what we were expected to prove, but at the time I had no clue how to approach it. I went on to the other problems, solved them all, and came back to this one. I still had no clue. When the hour was just about up, I wrote "If it weren't true, you wouldn't ask us to prove it." I don't remember what score I got, but all future questions were "Prove or disprove that ..."

$\endgroup$
  • 2
    $\begingroup$ I'd hope that you got a zero score for that proof. It is only probably true that he wouldn't ask you to prove it if it weren't true. For one, the Teacher could have been wrong or he could just be devious. So, not only is your proof wrong, it's not even a proof in a mathematical sense. It's only circumstantial evidence pointing toward a likely conclusion. $\endgroup$ – Shufflepants Feb 13 '18 at 16:41
  • 4
    $\begingroup$ @Shufflepants and yet, if Ross had disproven it thoroughly and rigorously, he would not have followed the instructions on the exam paper, would he? The response is brilliant and shows thinking outside the box. $\endgroup$ – Wildcard Feb 13 '18 at 23:09
  • 1
    $\begingroup$ @Wildcard And yet if he had correctly disproven it, he probably would have gotten just as many points as if he had proven it. So what it really shows is an unwillingness or inability to rigorously evaluate the claim for truthfulness. $\endgroup$ – jpmc26 Feb 14 '18 at 0:57
  • 1
    $\begingroup$ Even "Prove or disprove" still leaves you as an implicit given that the proposition is provable or disprovable. If you can then somehow show that it isn't disprovable, you've proven it, and vice versa. $\endgroup$ – Monty Harder Feb 14 '18 at 21:04
  • 2
    $\begingroup$ @Shufflepants: You missed MontyHarder's point there. He's suggesting that one could write on the answer sheet a proof that the proposition is not disprovable, and then follow it with the sentence "Therefore the proposition must be provable, or else you wouldn't have asked us to either prove or disprove it. Q.E.D." It's just a one-level-more-meta version of Ross's original snark. $\endgroup$ – Quuxplusone Feb 16 '18 at 5:41
12
$\begingroup$

Bertrand's paradox (not to be confused with Russel's paradox) is a paradox in probability based on the following problem:

Given a circle, choose a chord at random. What is the probability that the chord is longer than the side of an equilateral triangle inscribed in the circle?

where, depending on how you choose the chord, you may end up with wildly varying answers (the wikipedia article mentions three examples, all of which may be justified as uniform, giving $\frac12, \frac13$ and $\frac14$ respectively).

One possible resolution, also detailed in that Wikipedia article, uses a technique similar to yours, arguing that this problem is symmetric in the choice of circle, so the distribution of possible chords as lines in the plane should look the same regardless of which circle we end up with. Here it is called the "maximum ignorance principle".

$\endgroup$
7
$\begingroup$

Since it doesn't seem to have been mentioned anywhere yet, if anyone is curious here's how to solve OP's puzzle without the meta assumption (though note how the black-black vs. red-red ambiguity naturally pops up at the end, and that one might naturally be lead to OP's argument as a refinement at that point):

If any participant sees 4 blacks or 4 reds combined on the other two (ie, they each have 2 blacks, or they each have 2 reds), then he knows that he must have two of the other color. So a first round of all "no" answers says that no two participants have two of the same color. The colors on each of the three players head must then combine to form one of the following four rows (potentially after reordering):

$$\begin{array}{ccc} (b,b)&(b,r)&(b,r)\\ (r,r)&(b,r)&(b,r)\\ (b,b)&(b,r)&(r,r)\\ (b,r)&(b,r)&(b,r)\\ \end{array}.$$

So #1, on his second turn, looks at the other two participants heads. By picking any two of the entries in each of the above rows, we see that the possibilities that #1 can see on #2 and #3 form the first two entries of one of the following rows, with the third entry telling us which of the possible total state spaces we could be in: $$\begin{array}{cc} A & B & \text{appears in row...}\\ (b,b) & (b,r)&1,3\\ (b,r) & (b,r)&1,2,4\\ (b,r) & (r,r)&2,3\\ (b,b) & (r,r)&3\\ \end{array}.$$ Since #1 says "no" the second time around, this means he does not see (b,b)+(r,r) on the other two participants.

So when #2 gets his second chance, he knows combined the three of them have at least two (b,r)'s at most one of (b,b) or (r,r). If he saw that #1 and #3 had (b,r) each, he would not know if he had two blacks or two reds, or even (b,r), and would have to answer "no". However, since he answered "yes" then he must have only seen one (b,r) among the other two, and so could deduce that he must have the second mandatory (b,r) pair on his own head.

$\endgroup$
  • $\begingroup$ This approach extends to $n \geq 3$ men under the assumption that there are at most $n+1$ dots of each color. It's always the second in line to get the answer on the second turn. $\endgroup$ – Fabio Somenzi Feb 15 '18 at 8:35
6
$\begingroup$

Since this is a Math forum, let me give a mathematical answer, instead of just another example of a 'riddle'. My example conveys the heart of the issue you are interested in, and stresses an unrelated lesson that math majors often struggle with.

Induction: In order to prove that the sum of the first $n$ integers is $n(n+1)/2$, you assume the formula is correct for a given $n$ and use this information to prove the same formula for $n+1$. In essence you say "IF there is a formula, it has to be this one..."

Extremal problems: Which 2D figure of perimeter 1 covers the most area? Which line between 2 points is the shortest? You usually get a proposed solution and prove that it does not improve upon infinitesimal deformations. In essence, you prove that "IF there is an optimal shape, it has to be the one you propose..."

There are many situations like those above. A complete proof requires proving that there IS a solution (induction base, or a compactness argument). This is the lesson that math majors should internalize.

But most often, a proof assumes implicitly that there is a solution. In fact, one uses that assumption without even noticing it.

$\endgroup$
  • 5
    $\begingroup$ I realise this is mostly a matter of phrasing, but I just want to make something absolutely clear: a correct proof never actually assumes there is a solution. At some point, you may assume that a solution exists and deduce what it must then be; however, you then must check that this is actually a solution. This is equivalent to showing that there is only one solution, and then showing a solution, thus finding all the solutions. (I'm not trying to contradict your answer, just I feel the wording of the final paragraph may lead someone to think that proofs assume a solution) $\endgroup$ – Sam T Feb 12 '18 at 22:38
  • $\begingroup$ This induction example seems not to fit - you have to somehow come up with the idea for what formula you're going to prove before you prove it. The direct non-inductive argument (doubling a right triangle to make an $n$ by $n + 1$ rectangle) actually produces the formula. Probably many examples of induction are like this. $\endgroup$ – Ibrahim Tencer Feb 13 '18 at 6:10
  • $\begingroup$ But induction still involves proving that your formula is correct for a specific case, before extrapolating to the general by showing that if it's true for n it's also true for n+1. I don't really think it that counts as a meta assumption $\endgroup$ – Mick O'Hea Feb 13 '18 at 17:28
  • $\begingroup$ Insightful answer. And outside the realm of mathematics, for instance in the humanities, the basic assumption most often used is that there is no true solution. (Examples: war, murder, insanity.) Of course, when it's assumed there is no solution, it's hardly surprising that no solution is found. $\endgroup$ – Wildcard Feb 14 '18 at 4:28
  • $\begingroup$ This is certainly not how I think of induction. Induction is "prove a property for everything by means of first proving the property for whatever is the most primitive kind of thing, then proving that all possible ways of constructing a less primitive thing from a more primitive thing must preserve the property". No assumption; just construction. $\endgroup$ – Patrick Stevens Feb 14 '18 at 22:18
6
$\begingroup$

While the other very entertaining answers have been providing more examples of problems/riddles where similar techniques work, I think that it's just as important to recognize that each one works (or not) on a case-by-case basis, and there is no universal magic that works in all cases.

Consider for example the simple problem of "determine the maximum area of a rectangle with integer sides and a perimeter of $20$". A solution must obviously exist, since there are only a finite number of such rectangles. Then, applying the same line of reasoning as here...

Since the game is symmetric w.r.t. the colors red and black, the game would have the same answer-pattern if we switch the colors.

...it follows that neither side is distinguished in any way, so the problem is entirely symmetric in the two sides, and therefore the solution must be the one where the sides are equal so the rectangle is a square with side length $\,5\,$ and area $\,25\,$.

It turns out that that is indeed the correct answer, right? Yes, but had the problem asked for the rectangle with "the minimum area", instead, the same line of reasoning would have lead to the same square as the answer, which is wrong in that case - correct answer is the rectangle $\,1$x$9\,$.

Point of the above is that symmetry in the conditions does not always imply a corresponding symmetry in the solutions. For a more involved math example borrowed from another MSE question, both extrema of $\,(a^2-ab+b^2)(b^2-bc+c^2)(c^2-ca+a^2)\,$ for non-negative $\,a,b,c\,$ that satisfy $\,a+b+c=3\,$ do in fact exist, but neither is attained at $\,a=b=c=1\,$.

$\endgroup$
  • 12
    $\begingroup$ The faulty assumption here is that there is only one solution. The question's reasoning would not work if the correct riddle solution was "either two red or two black", but by the form of the riddle it can be inferred that is not a valid solution. The minimum area rectangle is "1x9 or 9x1" $\endgroup$ – immibis Feb 13 '18 at 1:10
  • 2
    $\begingroup$ @immibis That's a good math point, but it's not always obvious in more casual speak. Even in my (very simplistic) rectangle example, the question supposedly asks for the "minimum area", not a specific rectangle, and even if it were to ask for a specific rectangle, 1x9 and 9x1 could easily be taken to be the "same" rectangle depending on how you tilt your head. My intention was not to indiscriminately put down symmetry-based arguments at large, but to show why certain care is needed when using them. $\endgroup$ – dxiv Feb 13 '18 at 1:24
  • 1
    $\begingroup$ Is this supposed to be a perimeter of 20? $\endgroup$ – Peter Kagey Feb 13 '18 at 4:27
  • $\begingroup$ @PeterKagey Or a semiperimeter of 10 ;-) Thanks for catching that, corrected now. $\endgroup$ – dxiv Feb 13 '18 at 4:29
  • $\begingroup$ The symmetric assumption only works for the dots because there is no geometric structure and the two types of dots are distinguishable. In the case of rectangles, there are geometric considerations which invalidate the assumption. It's not that the symmetric assumption doesn't hold in the dots case, it's just that you'd need to go through a little bit more work to rigorously prove that assumption. $\endgroup$ – Shufflepants Feb 13 '18 at 16:53
4
$\begingroup$

Maybe it is not exactly what you're after, but here is one meta-argument that I like:

Q: Are there two irrational numbers $x$ and $y$ such that $x^y$ is a rational?

A: Either $x=\sqrt{2}^\sqrt{2}\in \mathbb{Q}$, in which case the answer is yes, or $x=\sqrt{2}^\sqrt{2}\not\in \mathbb{Q}$, in which case $x^\sqrt{2}=2$ implies that the answer is also yes.

$\endgroup$
  • 3
    $\begingroup$ @ToddSewell It is transcendental by Gelfond-Schneider theorem :) en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem $\endgroup$ – Arnaud Mortier Feb 12 '18 at 16:33
  • 13
    $\begingroup$ How is that a meta-argument? $\endgroup$ – Wojowu Feb 12 '18 at 18:26
  • 8
    $\begingroup$ But this is just the law of excluded middle, probably THE classical argument of propositional logic. $\endgroup$ – Hyperplane Feb 12 '18 at 20:08
  • 5
    $\begingroup$ This doesn't quite seem appropriate. What is the meta-assumption here? As far as I can tell, this is a complete and correct non-constructive proof of the above theorem. You do not need to know anything about the existence of such a proof, or the solvability of the problem (as being asked this question might indicate) $\endgroup$ – DreamConspiracy Feb 13 '18 at 13:30
  • 1
    $\begingroup$ @Hyperplane And the one which spectacularly gives you the wrong solution to this 'puzzle' previously commented under the question. $\endgroup$ – immibis Feb 14 '18 at 4:55
0
$\begingroup$

This is why I think all puzzle/riddle compilations should include ones impossible to solve, with no solution or simply "not interesting" at all.

One example is chess puzzles. I am yet to find a chess puzzle library that both includes extremely normal game conditions, where e.g. indeed taking that pawn is the right thing to do, and other variants (where there's actually a winning queen sacrifice). This I feel would make me a better chess player abecause it reflects what you experience across the game.

I also feel the same with my job. I'm a researcher and also a riddle lover. The real-life "riddles" I come across everyday at work might not have a solution at all! I might be a better researcher if I train my critical thinking with a collection of riddles that may have an ingenious solution, a dumb solution, or no solution at all

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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