Dot Product Riddle

Von Neumann

I was recently posed a riddle that went like the following:

There are two people. Person A picks three numbers from 0-99. Person B guesses which three numbers that person A has picked. Then, person A gives the dot product of his picked numbers with person B's guessed numbers. The question is how could person B figure out person A's selected numbers in three guesses. Even more challenging is to provide a solution that allows person B to guess the numbers in one guess.

I have a solution to the first part:

Think of person A and person B as having their guess put into vectors [itex]\vec{a}[/itex]=(a[itex]_{1}[/itex], a[itex]_{2}[/itex], a[itex]_{3}[/itex]) and [itex]\vec{b}[/itex]=(b[itex]_{1}[/itex], b[itex]_{2}[/itex], b[itex]_{3}[/itex]) respectively. To get the corresponding component a[itex]_{1}[/itex], person B should select the components (1,0,0) so the dot product will yield a[itex]_{1}[/itex]. Same for a[itex]_{2}[/itex] and a[itex]_{3}[/itex]. Simple enough.

The next part I am stumped. The only clue I was given is that person B's three guesses are not restricted between 0-99. Anyone have any insight?

Ferramentarius

The key is that the numbers have a finite length, and can be separated far enough from each other by multiplication for further examination.

Von Neumann

Hey guys!

I had no idea it was so simple! I was looking into it too much. Thanks for the help.

Adarsh Kumar

This also depends upon the value incorporated --In case one is choosing (7 ,77 ,93) as the three value -it may lead to come with 3 guesses .
Finite length numbers can be taken as simple guesses.Comes handy only with vectors

autodidude

I'm probably way off here but don't you get a single equation with 3 unknowns? (and a restricted domain)

D H

That's one additional guess, not one guess. The conversation would go like this:

Person A: I've picked three numbers from 0-99. Can you guess what they are, in the order in which I picked them? As a hint, I'll tell you the inner product of my numbers and your guess if your guess is wrong.
Person B: OK. Here's my first guess: b₁, b₂, and b₃.
Person A: Hey! That's cheating! It's also wrong. But since I didn't make my rules clear enough, I guess I'll have to tell you that the inner product is c.
Person B: OK! Here's my second guess: a₁, a₂, and a₃.
Person A: Correct.

Two guesses, not one. The puzzle is how to frame the first guess so that the second guess will inevitably be correct.


No. There is a way (there are an infinite number of ways) to formulate the initial guess so that the second guess will always be correct.

autodidude

How?

Mute

Read Ferramentarius' clue again, and note that the guesses for B are not restricted to the range 0-99.