jmc

By QM-AM, $$\sqrt{\frac{a^2+(1-b{)}^2}{2}}\ge \frac{a+(1-b)}{2},$$so $\sqrt{a^2+(1-b{)}^2}\ge \frac{1}{\sqrt{2}}(a+(1-b)).$ Similarly, $$\begin{array}{rl}\sqrt{b^2+(1-c{)}^2}& \ge \frac{1}{\sqrt{2}}(b+(1-c)),\\ \sqrt{c^2+(1-d{)}^2}& \ge \frac{1}{\sqrt{2}}(c+(1-d)),\\ \sqrt{d^2+(1-a{)}^2}& \ge \frac{1}{\sqrt{2}}(d+(1-a)).\end{array}$$Adding these inequalities, we get $$\sqrt{a^2+(1-b{)}^2}+\sqrt{b^2+(1-c{)}^2}+\sqrt{c^2+(1-d{)}^2}+\sqrt{d^2+(1-a{)}^2}\ge 2\sqrt{2}.$$Equality occurs when $a=b=c=d=\frac{1}{2}.$

Since $a$ and $1-b$ are nonnegative, $$\sqrt{a^2+(1-b{)}^2}\le \sqrt{a^2+2a(1-b)+(1-b{)}^2}=\sqrt{(a+(1-b){)}^2}=a+1-b.$$Similarly, $$\begin{array}{rl}\sqrt{b^2+(1-c{)}^2}& \le b+1-c,\\ \sqrt{c^2+(1-d{)}^2}& \le c+1-d,\\ \sqrt{d^2+(1-a{)}^2}& \le d+1-a.\end{array}$$Adding all these inequalities, we get $$\sqrt{a^2+(1-b{)}^2}+\sqrt{b^2+(1-c{)}^2}+\sqrt{c^2+(1-d{)}^2}+\sqrt{d^2+(1-a{)}^2}\le 4.$$Equality occurs when $a=b=c=d=0,$ and $a=b=c=d=1.$

If we set $a=b=c=d=t,$ then $$\sqrt{a^2+(1-b{)}^2}+\sqrt{b^2+(1-c{)}^2}+\sqrt{c^2+(1-d{)}^2}+\sqrt{d^2+(1-a{)}^2}=4\sqrt{t^2+(1-t{)}^2}.$$In the range $0\le t\le 1,$ $4\sqrt{t^2+(1-t{)}^2}$ takes on all the values from $2\sqrt{2}$ to 4, so the possible values of the expression is the interval $\boxed{[2\sqrt{2},4]}.$