The following problems appeared in The Riddler. It’s a problem about knots! Or rather, not-knots.
Imagine a framed picture suspended by a cord that’s hanging on two nails. If the picture were hung “normally,” you’d expect the removal of one nail to leave the picture hanging from the other (albeit a bit askew). But how can you hang a picture on two nails so that if you remove either nail the picture will crash to the floor?
What about three nails? What about four nails? What about on two red nails and two blue nails such that the picture falls if both red nails are removed and if both blue nails are removed but remains hanging if one of each color is removed?
Here is my solution for the case of two nails:
Here is a way to turn this problem into a mathematical statement. If we call the nails $A$ and $B$, then hanging the wire “normally” corresponds to $A \wedge B.$ In other words, the frame only falls if we pull $A$ and $B$. What we’d like to find is $A \vee B$ ($A$ or $B$). I found the solution for this two-nail case by trial-and-error. Here is a diagram illustrating $A \wedge B$ and $A \vee B$.
Note that the over/under of the wire is important! If we don’t pay attention to this, then we could end up with an overhand knot and then the picture frame would not fall!
Here is my solution for the cases of three and four nails:
While we could solve the two-nail case by trial-and-error, the case of three or more nails is substantially more complicated. So we need to think more mathematically (and in this case, recursively) about the problem. What we’re trying to find, using the notation of the previous part, is $A \vee B \vee C$, which is the same as $A \vee (B \vee C)$ by associativity. If we start with the solution to the two-nail problem and simply add a third nail right next to $B$, then we obtain $A \vee (B \wedge C)$. Then, we can recursively apply the first solution twice to $(B \wedge C)$ to turn it into $(B \vee C)$. We must apply it twice because there are two wires that go above $B$ and $C$. This is best explained by a diagram, so without further ado:
We now see how the case of four nails can be solved. By applying the technique we used for three nails twice. Namely:
A\wedge B \wedge C \wedge D \implies
(A\wedge B) \vee (C\wedge D) \implies
(A \vee B) \vee (C \vee D)
\]Note that we also could have solve it via $A \vee (B \vee (C \vee D))$, but the knot would end up being far more complicated! Here is a diagram of the solution for four nails:
Two red and two blue
The final puzzle asks us to imagine two blue nails and two red nails, where the picture frame falls if both red or both blue are removed, but not if one red and one blue is removed. If $A$ and $B$ are blue and $B$ and $C$ are red, then the corresponding mathematical statement is simply:
(A \wedge B) \vee (C \wedge D)
\]But this is a case we’ve already solved! Indeed, it was an intermediate step on the way to our solution of the four-nail problem. Here is the solution with the nails colored: