What makes the permanent a lot more difficult than the determinant?

Question

The permanent of an $n$-by-$n$ matrix $A = \left( a_{i,j} \right)$ is defined as follows

$$ \operatorname{perm}(A) = \sum\limits_{\sigma \in S_n} \prod\limits_{i=1}^n a_{i,\sigma(i)} \tag{1} $$

On the other hand, the determinant of $A$ is defined as follows

$$ \det(A) = \sum\limits_{\sigma \in S_n} \operatorname{sgn}(\sigma)\prod\limits_{i=1}^n a_{i,\sigma_i} \tag{2} $$

where $\operatorname{sgn}(\sigma)$ denotes the signature of $\sigma$. I know that the calculation of $(1)$ and $(2)$ are very different since $(1)$ is hard to compute and $(2)$ is easy — in $\mathcal{NP}$ and in $\mathcal{P}$, respectively. Can someone explains to me why there is such a difference? How does $\operatorname{sgn}(\sigma)$ change the complexity of the calculation?

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P.S. Please note that

• I am not professional in this area.

• I do not read related papers. If I have to, maybe I will.

• I know that there is a lot of discussion in the SE website in its different forums but I still do not get the huge difference between $(1)$ and $(2)$.

Answer 1

Expanding on Mark Bennet's comment:

Using the information on this link (by adapting the proof to Permanents)

When you row reduce a determinant, this happens

$$\det(A)=\left|\begin{matrix} a_{11}&\cdots&a_{1n}\\ \vdots&\ddots&\vdots\\ b_{i1}&\cdots&b_{in}\\ \vdots&\ddots&\vdots\\ a_{n1}&\cdots&a_{nn}\\ \end{matrix}\right|=\left|\begin{matrix} a_{11}&\cdots&a_{1n}\\ \vdots&\ddots&\vdots\\ 0&\cdots& b_{in}-\frac{b_{i1}}{a_{11}}a_{1n}\\ \vdots&\ddots&\vdots\\ a_{n1}&\cdots&a_{nn}\\ \end{matrix}\right|+\frac{b_{i1}}{a_{11}}\left|\begin{matrix} a_{11}&\cdots&a_{1n}\\ \vdots&\ddots&\vdots\\ a_{11}&\cdots&a_{1n}\\ \vdots&\ddots&\vdots\\ a_{n1}&\cdots&a_{nn}\\ \end{matrix}\right|$$

However, the 2nd term is zero thanks to the alternating property of the odd permutation terms in the determinant thus killing it off to zero, thus as mentioned the determinant remains unchanged and you have the nice

$$\det(AB)=\det(A)\det(B)$$

Trying to "row reduce" the permanent, however...

$$\text{Perm}(A)=\overset{+}{\left.\begin{matrix} \\ \\ \\ \\ \\ \\ \end{matrix} \right|}\begin{matrix} a_{11}&\cdots&a_{1n}\\ \vdots&\ddots&\vdots\\ b_{i1}&\cdots&b_{in}\\ \vdots&\ddots&\vdots\\ a_{n1}&\cdots&a_{nn}\\ \end{matrix}\overset{+}{\left|\begin{matrix} \\ \\ \\ \\ \\ \\ \end{matrix} \right.}= \overset{+}{\left.\begin{matrix} \\ \\ \\ \\ \\ \\ \end{matrix} \right|}\begin{matrix} a_{11}&\cdots&a_{1n}\\ \vdots&\ddots&\vdots\\ 0&\cdots& b_{in}-\frac{b_{i1}}{a_{11}}a_{1n}\\ \vdots&\ddots&\vdots\\ a_{n1}&\cdots&a_{nn}\\ \end{matrix}\overset{+}{\left|\begin{matrix} \\ \\ \\ \\ \\ \\ \end{matrix} \right.}+\frac{b_{i1}}{a_{11}} \overset{+}{\left.\begin{matrix} \\ \\ \\ \\ \\ \\ \end{matrix} \right|}\begin{matrix} a_{11}&\cdots&a_{1n}\\ \vdots&\ddots&\vdots\\ a_{11}&\cdots&a_{1n}\\ \vdots&\ddots&\vdots\\ a_{n1}&\cdots&a_{nn}\\ \end{matrix}\overset{+}{\left|\begin{matrix} \\ \\ \\ \\ \\ \\ \end{matrix} \right.}$$

and the 2nd terms persists because there is no alternating property to kill it off.

Therefore when you attempt to carry out a modified Gaussian elimination taking account of the above property, then for every row addition you do, you spawn one extra term.

So unlike determinants, instead of having only at most $\frac{(n-1)(n-2)}{2}$ row additions and only one determinant term to do, you now have $\frac{(n-1)(n-2)}{2}$ permanent terms to deal with, and thus the problem is multiplied at least $\frac{(n-1)(n-2)}{2}$ times just for the 1st $\frac{(n-1)(n-2)}{2}$ "Gaussian elimination" you did to get the original term to an upper triangular form, but you still have another $\frac{(n-1)(n-2)}{2}$ terms (the spawned terms) need to be done the same thing, and good luck finding a way to terminate the operation somewhere since these extra terms never vanish.

This 'spawning property' is the reason why the following holds instead:

$$\text{Perm}(AB)\neq\text{Perm}(A)\text{Perm}(B)$$