On Thinking Outside the Box (Column 656)

With God’s help

A few weeks ago I saw, in one of the tabloids distributed in synagogues, the following riddle in an ad for the Lev Academic Center.

At the bottom they wrote that the solution indicates that the solver “thinks outside the box.” This prompted some gloomy reflections about what “thinking outside the box” actually means. Is solving a riddle like this really outside-the-box thinking?

Analysis

First, let’s try to solve the riddle. My assumption is that each shape represents some number. There are three shapes here—sphere, pyramid, and cube—each stands for a number. An additional assumption is that each spatial position/relationship between shapes represents a mathematical operation between the numbers they stand for. We have two positions: a shape above a shape, and a shape (or stack of shapes) next to a shape (stack). Each such relation represents some mathematical operation (which may or may not be the same).

At first glance I assume the numbers should be integers, and the operations should be taken from the basic/simple mathematical operations (addition, subtraction, multiplication, division, and perhaps exponentiation). Why? Because that seems reasonable and natural.

With these assumptions, the solution comes rather naturally. Start with the top equation. It has two identical “stacks,” each made of two shapes. It therefore makes sense to begin by decoding the “next to” relation. Should we multiply the two stacks, add them together, or subtract them? Simple elimination shows it can’t be multiplication, since 24 has no integer square root. It also can’t be subtraction, because then the result would have to be 0 and not 24. Nor can it be exponentiation, since there’s no integer that solves XX=24X^X=24. What remains, then, is addition.

This immediately yields two preliminary results: (1) placing stacks next to each other denotes addition; and (2) the value of each stack in the top equation is 12. Now we must ask what “a shape above a shape” means. As noted, it could be any of the five operations listed above (including addition itself). A first proposal is that it’s a different operation, since different spatial configurations encode different mathematical operations. Perhaps it is exponentiation—but only if we assume the cube is 12 and the pyramid is 1. On that assumption, the bottom equation gives the sphere as −5, and then we get 12+(−5+1)=712 + (−5 + 1) = 7. According to this, in the third equation we obtain (−5)12=1,955,078,125(−5)^{12} = 1,955,078,125.

Well, that can be ruled out, because the bottom of the riddle shows that the result must give us a day of the month, i.e., an integer between 1 and 31. This also rules out the infinite set of solutions with non-integers. We must therefore return to our assumptions.

Note that the instructions for solving the riddle speak of three equations (exercises?), where the third must be solved using the first two. Seemingly the riddle is self-contained, and there is no reference to the hint at the bottom of the picture; it’s merely the result of the solution and not a hint. Yet if the result must indeed be an integer day, that does rule out the solution I gave above—so I must use that as a hint as well. It seems the riddle’s wording is imprecise.

All right, so it’s natural to try decoding “shape above shape” as multiplication. In that case, if the numbers are integers, it’s reasonable—checking the top equation—to set the cube to 4 and the pyramid to 3 (or vice versa). Returning to the second equation, in both cases the circle is 1. But then the result of the third equation changes: it will be either 4 or 3 respectively. Both are whole numbers that yield a calendar date. Those two dates this year fall on Wednesday and Thursday, so neither can be ruled out (I assume Lev won’t hold an open day on Shabbat). We’re left with two different options, both of which fit. There’s no way to decide between them. I presume this is a mistake by Lev—perhaps they didn’t think far enough outside the box (or maybe I missed something?).

By the way, this isn’t the only solution. Without the “day of the month” hint at the bottom, other solutions become possible. For example, still taking “above” to be multiplication, assign to the upper stacks the numbers 2 and 6. If the cube is 2 and the pyramid is 6, the circle below comes out non-integral (1.2), and the final result is 2.4—again not an integer. If we flip them—cube 6 and pyramid 2—then the circle is 0.5. That’s not integral either, but the final result becomes 6/0.5=126/0.5 = 12 (which this year falls on a Friday—legitimate for an open day).

We could also decode “shape above shape” as subtraction (and not multiplication). That opens many possibilities. Each of the upper stacks must yield 12. A subtraction that gives 12 could be, say, 18−618 – 6, i.e., cube 6 and pyramid 18. Using that in the second equation, the circle is 17, and the last equation yields −11. The hint rules this out.

A third option is to decode “shape above shape” as addition as well, just like “next to.” (Exactly as in a pair of equations with two variables XX and YY, there’s nothing that prevents us from getting a solution where X=YX=Y.) In that case there are 12 ways to place the cube and pyramid. If, for example, we take the cube to be 8 and the pyramid 4, the second equation gives the circle as −4, and the third equation yields 4—a perfectly legitimate result. I didn’t check further, but I assume there are additional solutions in this direction.

What if we decode the vertical stack as division? For example, cube 36 and pyramid 3, but then the final result isn’t an integer (we get 9). It seems the other outcomes here are also fractional, so this isn’t a viable option.

That should suffice before we consider what “outside-the-box” thinking might be.

Which solution is “outside the box”?

Even before I noticed the hint at the bottom, my first impression was that the solution was 3 or 4. My assumption was integers, with adjacency meaning addition, and the vertical relation being some other operation. I immediately assumed multiplication. As noted, this produces two different dates, yet it still seemed clear this was the intended answer, and I took the duplicity as their mistake. Suppose I got a reasonable, natural result—why assume it’s the correct one? Why not consider the possibility of identical operations (addition and addition), or addition and exponentiation? Even taking the hint into account, we still saw several other possibilities—for instance, allowing fractional values for the shapes so long as the final result is an integer between 1 and 31; alternatively, treating the vertical relation as addition and allowing negative values for the shapes so that the date comes out sensible. These two solutions may feel less natural and less “obvious,” but why conclude from that that they’re wrong?

I’d go further. The first solution I presented is the result of natural, expected reasoning—inside-the-box thinking. The other two actually give us outside-the-box solutions. In particular, those two yield a single date, whereas the “natural” solution yields two possible dates. So why am I still sure the intended solution is one of the “natural” ones? Simple: because it’s clear to me they expected an inside-the-box solution, not an outside one. Solutions with fractional or negative numbers are “outside the box,” and therefore not the ones they were waiting for.

The conclusion is that the Lev Academic Center is actually testing who thinks inside the box, not who thinks outside it—and perhaps we can even say they themselves think inside the box. They absolutely want systematic, orderly, good thinking; that is, they want the admitted student to be someone capable—but only so long as he thinks inside the box, not outside it. A person who thinks outside the box will arrive on 12.7 and won’t be admitted. The one who picks 4 via the second solution might show up on the right day yet still not get accepted (because in exams they filter out “outside-the-box” thinking), or show up on the wrong day and not get accepted. A person who thinks inside the box will conclude the result is either 3 or 4, and will therefore come on both days and get in. Their error will, at worst, make him work harder, but in the end—no worries—if he thinks sufficiently inside the box, he’ll be admitted.

To clarify better what “inside” and “outside” the box mean, let me turn to a familiar argument by Wittgenstein.

Wittgenstein on Rule-Following

I discussed this before (see, e.g., here, and also this PDF) about the famous “rule-following” argument. Suppose that on a psychometric exam you’re given the sequence:

…2, 4, 6, 8, 10 — and asked for the next term. The obvious answer is 12—i.e., the even numbers (with the recursive rule “add 2,” and then add 2 again, and so on). But Wittgenstein argues that the “next number” could, just as well, be anything—say 16, because the rule could be: add 2, add 2, add 2, then multiply by 2, then add 2 again, etc. Now consider the sequence …3, 5, 7. The natural response is 9, on the assumption we’re dealing with odds (or primes, in which case the next would be 11). But equally well it could be any other number you wish.

He goes further: any given numerical sequence can be continued in any way whatsoever; for every such continuation there exists some suitable rule. For example, take the second sequence 3, 5, 7… Suppose we find a student who wrote −4.7 as the next term. Has he necessarily made a mistake? Not at all. Here’s a rule that justifies his continuation.

Let the rule be a 3rd-degree polynomial with four coefficients:

f(n) = a1 + a2·n + a3·n^2 + a4·n^3

If we want the sequence to be 3, 5, 7, … we choose the coefficients accordingly, by setting up four equations in four unknowns:

f(1) = a1 + a2 + a3 + a4         = 3
f(2) = a1 + 2a2 + 4a3 + 8a4      = 5
f(3) = a1 + 3a2 + 9a3 + 27a4     = 7
f(4) = a1 + 4a2 + 16a3 + 64a4    = −4.7

Solving these is straightforward. The result is:

a1=14.7a_1 = 14.7, a2=−23.11666a_2 = -23.11666, a3=13.7a_3 = 13.7, a4=−2.28333a_4 = -2.28333.

Plugging back in gives:

f(n) = 14.7 − 23.11666·n + 13.7·n^2 − 2.28333·n^3

Substituting n=1,2,3,4n=1,2,3,4 yields exactly the desired values. In other words, the answer written by the student—−4.7—is just as “correct” as any other. This is, of course, only an example; one can do the same for any sequence and any desired continuation in countless ways (choose higher-degree polynomials, other functional forms, etc., tuning the parameters to the target).

Back to outside-the-box thinking

So why won’t the student who wrote −4.7 be admitted to university? Because he thinks outside the box. The psychometric exam expects an inside-the-box answer, and sorts students accordingly—admitting those who think inside the box and filtering out those who think outside it.

You probably suspect I’m criticizing the academic filtering process, but I’m not. This is how it should work. Imagine the process allowed everyone who thinks like that to be admitted: the fellow who wrote −4.7, together with one who wrote 1, another who wrote 1549, a third who wrote 17.3, a fourth who wrote 2e, and so on. If every answer is acceptable, what exactly is the exam testing? How is any filtering being done? In such an exam, only a student who justifies his answer (provides an appropriate f(n)f(n)) would be admitted. Random answers without justification are out (this is not a multiple-choice test but a psychometric exam requiring full solutions).

Now the lecturer walks into class to teach some mathematics. He has no chance. You simply cannot teach such a class. Wittgenstein explains there that every rule we want to teach our students relies on giving examples and generalizing. Suppose we teach them to count the natural numbers in base 10—first from 1 to 10, then to 100, then 1000. The teacher asks a student to continue the count, and he answers without hesitation: e−. The astonished teacher asks how he got that, and the student produces an f(n)f(n) that yields the first thousand naturals followed by e−. Fine, the teacher presses on and teaches up to 10,000. The next student answers 15, and of course provides a function that “justifies” it. In fact, these students don’t need to supply the justifying functions—their brains are wired such that after 1,000, e− just feels natural to them, and to the other one after 10,000, 15 just feels right. This will happen to that teacher in any mathematical subject he tries to teach, elementary or advanced.

The conclusion is that when students’ minds are built in a peculiar way (outside the box), there’s no way to teach them. A teacher can teach only students whose minds are built like his own—i.e., those who have the same box he does. No wonder an institution constructed in a certain way wants to admit only students whose minds fit the institution’s standard. Students with a different mental makeup simply won’t be able to learn there. No wonder Einstein and other geniuses were known as “bad students”: their minds were built differently from their teachers’ and everyone else’s, so teaching them was very hard.

The upshot is that the Lev Academic Center is right in its policy of admitting those who think inside the box. The only question that bothers me is why they claim they’re looking specifically for those who think outside the box. Well, perhaps you need outside-the-box thinking to understand that.

You can find a charming story about “outside-the-box” thinking here.

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References mentioned: mikyab.net post, and this TAU PDF: https://law.tau.ac.il/sites/law.tau.ac.il/files/media_server/law_heb/Law_Society_Culture/books/procedurot/ProceduresRiesenfeld.pdf.