Hi

It's seems to be not too hard and I'm hoping it's the right place to ask:

The rules of a new game with two players:
The first player choose four integers (x y z u) and the second player should guess the four numbers in order to win.
In each turn the second player can choose four integers (a b c d) and the first player tells him the sum:
ax + by + cz + du

What is the minimum number of turns the second player will need to win? and how he will solve it?

Four turns, first turn (a=1,b=0,c=0,d=0) will find x, second turn (b=1,a=c=d=0) will find y, and in the same way he will find c and, total four turns.

What do you think? This is the right answer?
Any faster way to solve it?
Any comment will be appreciated.

Thanks

2. Originally Posted by ron3
Hi

It's seems to be not too hard and I'm hoping it's the right place to ask:

The rules of a new game with two players:
The first player choose four integers (x y z u) and the second player should guess the four numbers in order to win.
In each turn the second player can choose four integers (a b c d) and the first player tells him the sum:
ax + by + cz + du

What is the minimum number of turns the second player will need to win? and how he will solve it?

Four turns, first turn (a=1,b=0,c=0,d=0) will find x, second turn (b=1,a=c=d=0) will find y, and in the same way he will find c and, total four turns.

What do you think? This is the right answer?
Any faster way to solve it?
Any comment will be appreciated.

Thanks
Hi,

about the place to ask your question, I can't see the connection between your problem and probability or statistics. It would rather have fit in the "number theory" or "algebra" section. Anyway I have an answer.

Indeed, suppose we know the value of only three linear combinations (arbitrarily chosen), so that we have three equations satisfied by $x,y,z,u$. Instead of looking for integer solutions, let us first look at real solutions. Geometrically, the set of the quadruplets $(x,y,z,u)\in\mathbb{R}^4$ that satisfy the three given equations is a line in $\mathbb{R}^4$ with rational slope. (remark: it may also be a larger subspace if the linear combinations are dependent, but this only makes things worse)
So all you know given three linear combinations is that $(x,y,z,u)$ is an integer point on a given line in $\mathbb{R}^4$ with rational direction. However, there are infinitely many integer points on such a line (there is at least one by assumption, and we can add multiples of an integer-valued direction vector of the line), so that the data we have does not allow to find which is the solution.
On the other hand, the problem becomes easy if for instance you know that $x,y,z,u\in\{0,\ldots,M\}$, and only one linear combination would suffice. For instance, $x + 10^k y + 10^{2k} z + 10^{3k} u$, where $k$ is chosen so that $10^k>10 M$, is a natural integer that looks like (in base 10): $u0\cdots0z0\cdots0y0\cdots0 x$ and it is thus possible to "read" $x,y,z,u$ from it. If $x,y,z,u$ can be arbitrarily large, then 4 linear combinations are necessary. I must say I was really disappointed when I realized that there was no clever strategy...