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Check if $x \in \mathbb{R}^n$ satisfies $\|x\|_1\leq 1$ using linear methods only, that is, find $A \in \mathbb{R}^{m\times n}$, $b \in \mathbb{R}^m$ such that

$$Ax \leq b \Leftrightarrow x \in \overline{\mathbb{B}}_{(\mathbb{R}^n, \|\cdot\|_1)}$$

where $m$ is not exponentially dependent on $n$.

My straightforward solution, checking the sums $x_1+\dots+x_n$, $x=(x_1,\dots,x_n)$, with plus replaced by minus in all possible ways, against $b=(1,\dots,1)$, leads to $m=2^n$. Is there a way to do this in non-exponential time? Thanks in advance.

Rodrigo de Azevedo
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Ramen
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  • Are you sure you're not allowed to introduce additional variables? It's quite straightforward if you can. – Michael Grant May 20 '17 at 15:24
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    I do not believe this to be possible as written. The number of faces represented by an inequality $Ax\leq b$ is no more than the number of rows. The number of faces of this norm ball is exponential in $n$. – Michael Grant May 20 '17 at 15:32
  • Thanks, I thought so. I am asking as the above situation was one of many of an exercise that asked to use half-spaces to represent some subsets and asked to try to do so in non-exponential time. I can introduce additional variables, how does this change the situation? – Ramen May 21 '17 at 16:45
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    With $n$ additional variables it can be done with $m=2n+1$ half-spaces. – Michael Grant May 21 '17 at 16:47
  • Actually it can be done with $n-1$ additional variables and $2n$ half-spaces but it's a bit more messy. – Michael Grant May 21 '17 at 16:48
  • Can you give me a hint? – Ramen May 21 '17 at 18:16
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    ;-) Sure, when I get to my desk I can even answer! Basically in the easier case the new variables effectively represent the absolute values of the $x$ values. – Michael Grant May 21 '17 at 18:18

1 Answers1

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As we discussed in the comments: it is not possible to express the $\ell_1$ norm ball without an exponential number of rows, unless you allow for additional variables.

The simplest approach is to introduce new variables $y_i$, $i=1,2,\dots, n$. Then you have: \begin{array}{ll} -x_i \leq y_i & i = 1,2,\dots,n \\ +x_i\leq y_i & i=1,2,\dots, n \\ \sum_i y_i \leq 1 \end{array} So with $n$ additional variables you can express this with $2n+1$ inequalities. However, try and convince yourself that this still produces the same result: \begin{array}{ll} -x_i \leq y_i & i = 1,2,\dots,n \\ +x_i\leq y_i & i=1,2,\dots, n \\ \sum_i y_i = 1 \end{array} All I have done is change the last inequality to an equation. If $\|x\|_1=1$, then this obviously holds, because the only valid values of $y$ are $y_i=|x_i|$. But it will work even if $\|x\|_1<1$; for instance, you can choose $y_i=|x_i|+\frac{1 - \|x\|_1}{n}$. Once you accept the validity of this system, then, you can eliminate one of the $y_i$ values: \begin{array}{ll} -x_i \leq y_i & i = 1,2,\dots,n-1 \\ +x_i\leq y_i & i=1,2,\dots, n-1 \\ -x_n\leq 1 - y_1 - y_2 - \dots - y_{n-1} \\ +x_n\leq 1 - y_1 -y_2 - \dots - y_{n-1} \end{array} So in fact, with $n-1$ additional variables we can express the norm ball with $2n$ inequalities.

Laurențiu Diaconu
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Michael Grant
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