
#1
Feb1013, 09:51 AM

P: 97

I was recently posed a riddle that went like the following:
There are two people. Person A picks three numbers from 099. 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 099. Anyone have any insight? 



#2
Feb1013, 11:13 AM

P: 22

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




#3
Feb1013, 11:30 AM

P: 97

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



#4
Feb1513, 12:06 AM

P: 1

Dot Product Riddle
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 



#5
Feb1513, 07:31 AM

P: 333

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




#6
Feb1513, 07:48 AM

Mentor
P: 14,473

Person A: I've picked three numbers from 099. 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: b_{1}, b_{2}, and b_{3}. 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: a_{1}, a_{2}, and a_{3}. 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. 



#7
Feb1513, 09:03 PM

P: 333





#8
Feb1613, 11:24 AM

HW Helper
P: 1,391




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