Which set of preferences for three candidates is impossible?

Question

Hi recently i appeared in an aptitude,there was a problem that i realy cant understand please provide some idea, how to solve it.( sorry to for poor English.)

(Question)-> Three candidates, Amar, Birendra and Chanchal stand for the local election. Opinion polls are conducted and show that fraction a of the voters prefer Amar to Birendra, fraction b prefer Birendra to Chanchal and fraction c prefer Chanchal to Amar. Which of the following is impossible?

(a) (a, b, c) = (0.51, 0.51, 0.51);
(b) (a, b, c) = (0.61, 0.71, 0.67);
(c) (a, b, c) = (0.68, 0.68, 0.68);
(d) (a, b, c) = (0.49, 0.49, 0.49);
(e) None of the above.

Answer 1

I assume that people are asked whom they like most and whom they like least of A, B and C and the most liked cannot be the same as the least liked. We have to solve a system of linear equations an inequalities.

We define the three letter variable as $abc$ the relative frequency that A is liked most and C is liked least by a person. This should mean A is preferred to B and C and B to C. We now get the folowing four equations:

$$\begin{eqnarray} abc + acb + cab & = & a \\ bac + bca + abc & = & b \\ cba + cab +bca & = & c \\ abc + acb + bca + bac + cab + cba & = & 1 \end{eqnarray} $$ The last one means that all frequencies sum up to $1$. solving this system of linear equations we can eliminate 4 of the 6 variables and get

$$ \begin{eqnarray} abc &=& -x-y+b \\ bca &=& x \\ cab &=& y+a+c-1 \\ cba &=& -x-y-a+1 \\ bac &=& y \\ acb &=& x-b-c+1 \end{eqnarray} $$

(here we introduced the new variable $x$ and $y$ instead of $bca$ and $bac$ to make the notation simpler) now we know that $$0 \le abc \le1, \; 0 \le acb \le 1, \ldots $$

and therefore

$$ \begin{eqnarray} 0 &\le& -x-y+b &\le& 1 \\ 0 &\le& x &\le& 1 \\ 0 &\le& y+a+c-1 &\le& 1 \\ 0 &\le& -x-y-a+1 &\le& 1 \\ 0 &\le& y &\le& 1 \\ 0 &\le& x-b-c+1 &\le& 1 \end{eqnarray} $$

from the first and the fourth inequation we get $$ b-1 \le x+y \le b$$ and $$ -a \le x+y \le 1-a$$ and therefore $$\min{\{-a,b-1\}} \le x+y \le \min{\{b,1-a\}}$$

we know that $ 0 \le a$, $b\le 1$, $x \ge 0$ and $y \ge 0$. Therefore $$0 \le x+y \le \min{\{b,1-a\}} \tag{1}$$ From the for remaining inequalities we get $$ \max{\{0,c+b-1\}} \le x \le \min{\{1,c+b\}} \tag{2}$$ $$ \max{\{0,1-a-c\}} \le y \le \min{\{1,2-a-c\}} \tag{3}$$

Geometrically each of the inequalities $(1)$, $(2)$ and $(3)$ is a stripe in the plane bounded by two parallel lines. In $(1)$ this lines intersect the $x$-axis in $135$ degrees. The left line (lower bound) is through the origin. In $(2)$ the lines are parallel to the $y$-axis and in $(3)$ they are parallel to the $y$-axis. The solution (intersection) of $(2)$ and $(3)$ is a rectangle with sides that are parallel to the axis. This rectangle always lays left from the lower left bound of the $(1)$-stripe. The stripe and the rectangle have a non-empty intersection if the lower left vertex of the rectangle is in the stripe. This is only the case if it is left to the right (upper) bound of the stripe.

These inequalities have a solution is the sum of the LHS of $(2)$ and $(3)$ is smaller than the RHS of $(1)$ and the sum of the RHS of $(2)$ and $(3)$ is larger than the LHS of $(1)$. The latter is always the case so the remaining inequality is

$$\max{\{0,c+b-1\}} + \max{\{0,1-a-c\}} \le \min{\{b,1-a\}} \tag{4}$$

From this it follows that answer $c$ is not possible.

If there is a solution, than there is a solution with $$x=\max{\{0,c+b−1\}}$$ $$y=\max{\{0,1−a−c\}}$$ and therefore

$$ \begin{eqnarray} abc & = & b-\max{\{0 , c+b-1\}}-\max{\{0 , -c-a+1\}} \\ bca & = & \max{\{0 , c+b-1\}} \\ cab & = & c+a+\max{\{0 , -c-a+1\}}-1 \\ cba & = & -a- \max{\{0 , c+b-1\}}-\max{\{0 , -c-a+1\}}+1 \\ bac & = & \max{\{0 , -c-a+1\}} \\ acb & = & -c-b+ \max{\{0 , c+b-1\}}+1 \end{eqnarray} $$

I a result contains a negativ component, then no solution exists. For the frequencies $a$,$b$,$c$ in the OP we get

$$\left[ a=0.51 , b=0.51 , c=0.51 , {\it abc}=0.49 , {\it bca}=0.02 , {\it cab}=0.02 , {\it cba}=0.47 , {\it bac}=0 , {\it acb}=0.0 \right] $$

$$\left[ a=0.61 , b=0.71 , c=0.67 , {\it abc}=0.33 , {\it bca}=0.38 , {\it cab}=0.28 , {\it cba}=0.01 , {\it bac}=0 , {\it acb}=0.0 \right] $$

$$\left[ a=0.68 , b=0.68 , c=0.68 , {\it abc}=0.32 , {\it bca}=0.36 , {\it cab}=0.36 , {\it cba}=-0.04 , {\it bac}=0 , {\it acb}=0.0 \right] $$

$$\left[ a=0.49 , b=0.49 , c=0.49 , {\it abc}=0.47 , {\it bca}=0 , {\it cab}=0.0 , {\it cba}=0.49 , {\it bac}=0.02 , {\it acb}=0.02 \right] $$

you can check this using

$$a=abc+acb+cab$$ $$b=bca+bac+abc$$ $$c=cab+cba+bca$$

The inequality $(4)$ can be investigated further. We can remove the $\min$ and the $\max$ function by distinguishing 8 different cases. I will not write down this lengthy proof but finally the following holds:

• if two of the sums $a+b$, $a+c$, $b+c$ are $\ge 1$ and one is $\le 1$ than there is a solution for $(4)$
• if two of the sums $a+b$, $a+c$, $b+c$ are $\le 1$ and one is $\ge 1$ than there is a solution for $(4)$
• if all of the three sums $a+b$, $a+c$, $b+c$ are $\le 1$ but $a+b+c \ge 1$ than there is a solution for $(4)$
• if all of the three sums $a+b$, $a+c$, $b+c$ are $\ge 1$ but $a+b+c \le 2$ than there is a solution for $(4)$

This proves the following:

A necessary and sufficient condition for $(4)$ (and therefore for th OP) to have a solution is that for $a$,$b$ and $c$ holds $$1 \le a+b+c \le 2$$

Answer 2

You should consider all six possible orderings (by preference) of the three candidates: $ABC,ACB,BAC,BCA, CAB,CBA$. (Assuming no "ties" occur in the people's opinion).

The question is: Can you assign nonnegative numbers to these six orderings ins such a way that $a=ABC+ACB+CAB$, $b=ABC+BAC+BCA$, and $c=BCA+CAB+CBA$? For example, we see that $a+b+c=2ABC+ACB+BAC+2BCA+ 2CAB+CBA$, which implies that $1\le a+b+c\le 2$ is a necessary condition.

Answer 3

Base on the reasoning of my other answer I was leaded to the following answer that is much simpler. It show that the condition stated by @HagenvonEitzen is sufficient bei constructiong explicit solutions.

Lemma: $(a,b,c)$ are possible number (as defined in the OP) if and only if $0 \le a+b+c \le 1$.

I assume that people are asked whom they like most and whom they like least of A, B and C and the most liked cannot be the same as the least liked. We have to solve a system of linear equations an inequalities.

We define the three letter variable as $abc$ the relative frequency that A is liked most and C is liked least by a person. This should mean A is preferred to B and C and B to C.

We now get the folowing four equations:

$$\begin{eqnarray} \tag{1}\\ abc + acb + cab & = & a \\ bac + bca + abc & = & b \\ cba + cab +bca & = & c \\ abc + acb + bca + bac + cab + cba & = & 1 \end{eqnarray} $$

From this we get $$a+b+c \\=(abc + acb + cab)+(bac + bca + abc)+(cba + cab +bca)\\=(abc + acb + bca + bac + cab + cba)+(abc+bca+cab)\\=1+(abc+bca+cab)$$ We can conlude that $$1 \le a+b+c \le 2 \tag{2}$$ because $$0 \le abc+bca+cab \le 1$$

So this is a necessary condition for the problem to have a solution.

Now lets assume that $(2)$ is valid.

Now let's check the three sums $a+b$, $a+c$, $b+c$. Without loss of generality we assume that $a \le b \le c$ and therefore $a+b \le a+c \le b+c$.

If all of these sums are $ \le 1$ we set

$$acb=1-b,cba=b+c-1,bac=b+a-1,bca=-b-c-a+2,cab=0,abc=0$$

if all of these sums are $ \ge 1$ we set

$$acb=b+c+a-1,cba=0,bac=0,bca=b,cab=-b-a+1,abc=-b-c+1$$

the remaining case is that at least one sum is $\ge 1$ and at least one sum is $\le 1$. So we have $a+b \le 1$ and $b+c \ge 1$. We set

$$acb=a,cba=b+c-1,bac=0,bca=1-c,cab=-b-a+1,abc=0$$

It is easy to check that all these defined values lie between $0$ and $1$. So they are valid relative frequencies. And they satisfy $(1)$.

So the condition (2) is also sufficient.

In your sample only the answer c does not fullfill the condition.