An Engineer’s Guide to Pictorial Puzzles

Pictorial puzzles are everywhere! They show up in the puzzle of interpreting Heliographics, in classics like Sherlock Holmes’ puzzle in the Adventures of the Dancing Men and as picture puzzles and quizzes on social media.

Somewhere on the internet, on Facebook specifically, I’ve seen this puzzle floating around:

Such posts usually generate various comments, each one with his or her own answer. And, usually, the Original Post person responds with an animated GIF shake of the head:

The problem is not HARD per-se, it is just ill posed. Here is why.

Let’s take the EASY problem:

Notice that I’ve edited the problem to make it an easy one to solve.

The clocks line now has three identical clocks, each showing 9'o’clock. The calculator line and the bulbs line is unchanged from the original.

The Question line has two changes: I’ve made the bulbs and calculator identical to the ones from the lines above.

Ok. Now it is an easy problem: Each 9'o’clock face is 7 (=21/3), each calculator is 10 (30/3) and each bulb is 15 (the +/- cancel out). Ergo,

The answer to the EASY problem is 7+10x45 = 457

Notice though, in the ORIGINAL problem: There are two “types” of clocks: One is 9'o’clock while the other is 3'o’clock. There are two types of calculators: 1234 and 1224, and, there are two types of bulbs: one with 5 sunrays and the other with four sunrays! Ah ha!! Those are the gotchya images that usually trip up the problem solver.

However, this does not make it a HARD problem. It simply makes it an ILL POSED problem. Here is why:

As long as the solution contains logic to explain all the information provided, it becomes a valid solution. There is no such thing as a correct solution to an ILL POSED problem. Here is an example:

The so-called correct solution is as follows:

Assume that 9'o’clock is 3 times 3'o’clock. This solves the clock line to give 3'o’clock equal to 21/7 = 3; Ergo, 9'o’clock is 9.

The calculator line contains 3 of 1234. 1234 sum to 10; three times 10 is 30! The question line however has a calculator showing 1224. 1224 sum to 9; extending the logic that a calculator showing 1224 is 9.

The clock line is essentially one clock with 5 sunrays which sums to 15. By some logic, this makes one sunray equal to 3! Using this logic, the solution line has bulbs of 4 sunrays each making three such bulbs 4x3=12. Three bulbs in total is 12x3=36.

The solution line therefore is 9+ 9 x 36 = 333.

However, this is not the only solution that fits!

Say 9'o’clock is x. Now 3'o’clock is 9'o’clock + 6 hours. This makes the clock line: 3x+6=21, ergo, x=5.

In the question line, the calculator unknown is 1224. Say calc1224 is y. Now this makes calc1234 = y + 10. This makes the calculator line: 3y+30=30, ergo, y=0.

Now that we have y=0, it makes the bulbs irrelevant! We could use absolutely any consistent logic for the bulbs and it won’t change this version of our solution.

The solution line therefore becomes 5 + 0 x SomethingFinite= 5.

The second solution is equally valid. Nothing makes it more or less correct than the so-called correct solution.

Ill posed problems do not have a unique solution.

Sherlock Holmes describes this wonderfully in The Adventures of the Dancing Men,

“What one man can invent another can discover.”

Except, Sherlock needed a lot more information to crack the Dancing Men problem than one note! He converted an ill posed problem to a well posed problem by waiting for more notes and Boom!