Algorithm 1: Piecewise-linear Algorithm

Ø The number of $a_i\ge 0\left(i=1,2,\cdots ,m\right)$ is C, let t = C/m.

Ø If $t\ge 0.5$ , the minimax point is searched from the left $x=-1$ , otherwise, from $x=1$ . We only describe the search process of start from the left, and the right is the same. Solving the intersection coordinates $\left(X_i,Y_i\right)$ of the $x=-1$ and $y_i\left(i=1,\cdots ,m\right)$ , let $Y=\left(Y_1,\cdots ,Y_m\right)$ and $Y_p=\max \left(Y\right)$ , if the corresponding $a_p\ge 0$ , then terminate the search and $x=-1$ is the solution of the subproblem.

Ø If $a_p\le 0$ , then solving the intersection coordinates $\left({\tilde{X}}_j,{\tilde{Y}}_j\right)$ of the straight line with a slope bigger than $a_p$ and the current line $y_p$ .

Ø Here, the number of lines where the slope is larger than $a_p$ is recorded as w, remember $\tilde{Y}={\tilde{Y}}_1,\cdots ,{\tilde{Y}}_W$ , let ${\tilde{Y}}_h=\max \left(\tilde{Y}\right)$ , and judge whether the abscissa ${\tilde{X}}_h$ is out of bounds, if ${\tilde{X}}_h>1$ , then x = 1 is the solution of the subproblem, otherwise judge whether or not $a_h\ge 0$ , if $a_h\ge 0$ , then stop to research, and ${\tilde{X}}_h$ is the solution of the subproblem, otherwise repeat the step 3, until the minimax point is found.