# How to find out if two solutions are equivalent or different?

Given 5 different numbers ($\in \mathbb N$) in a specific brackets pattern like:

$$\left(\left(\left(x_1 + x_2 \right) - x_3\right) \times x_4 \right) / x_5 = \text{result}$$

Only the brackets are fixed, the numbers and the operators can be permutated at will.
And there must be all the four operators.

Does someone knows how could I find out if a permutation of the numbers and of the operators that gives the same result

• is an equivalent solution under the distributive, associative and/or commutative law
• or is a different solution that "just happens" to give the same result

For example:

$\left(\left(\left(6 - 3 \right) \times 2\right) / 1 \right) + 5 = 11$

is equivalent to:

$\left(\left(\left(6 - 3 \right) / 1\right) \times 2\right) + 5 = 11$

but it's different from:

$\left(\left(\left(3 - 2 \right) \times 6\right) / 1 \right) + 5 = 11$

that gives the same result only with this particular choice of numbers: 2,3 and 6.

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@JasperLoy: I consider the first two equations equivalent because they will always return the same result, no matter what numbers I choose since (((a−b)×c)/d)+e=(((a−b)/d)×c)+e. While the fact that (6−3)×2=(3−2)×6 relies on the particular choice of numbers. So at the end even if the same set of numbers is involved I consider the two solutions different. I'm not very good at explaining myself, but I hope to have been more clear, what do you think? –  Rik Poggi Feb 13 '12 at 17:55
@RikPoggi, it seems that we just need to solve the equation(s) f(a,b,c,d,e)=g(a,b,c,d,e). Maybe there are some interesting things in your question, but that may be beyond me. –  puresky Feb 14 '12 at 12:40

For the heck of it I defined the 24 functions $f_1$, $\ldots$, $f_{24}$ resulting from permuting the operation symbols in your expression. Then I listed the $120$ ways of assigning variables $x_1$, $\ldots$, $x_5$ to the placeholders $u$, $v$, $x$, $y$, $z\$ I had in the $f_j$. With this setup I did all $${24\choose 2}\cdot 120 + 24\cdot 119 = 35 976$$ comparisons of two resulting expressions and found 46 matching pairs among them. I spare you the list.
The number $35 976$ is obtained as follows: We have to check all pairs $f_i$ vs. $f_j$, $\ i<j$, where $f_i$ is assigned the standard $5$-tuple $(x_1,x_2,x_3,x_4,x_5)$ of variables and $f_j$ one of the $120$ permutations of this $5$-tuple. In addition we have to check for each $i$ the function $f_i$ with the standard assignment $(x_1,\ldots,x_5)$ vs. the same $f_i$ with each one of the $119$ other assignements of the variables $x_1$, $\ldots$, $x_5$.
It's interesting, I will try it too, how did you made the comparisons? But is not clear to me why there are $35976$ comparisons instead of $2880 \cdot 2879 / 2$ –  Rik Poggi Feb 14 '12 at 14:47