I have this ASP code:

$\%$ Input:

road$($b$,$a$).$

road$($c$,$b$).$

road$($g$,$b$).$

road$($b$,$g$).$

road$($a$,$f$).$

road$($f$,$b$).$

road$($b$,$e$).$

road$($e$,$c$).$

road$($c$,$d$).$

blocked$($c$,$d$).$

start$($a$).$

$\%$ Define cleanable roads: Roads that are not blocked

cleanable$($A$,$ B$)$ :$-$ road$($A$,$ B$),$ not blocked$($A$,$ B$).$

$\%$ The sweeper can traverse cleanable roads

reachable$($A$,$ B$)$ :$-$ cleanable$($A$,$ B$).$

$\%$ Define the route: in$\_$route$($A$,$ B$)$ means the sweeper uses the road from A to B

$1\{$ in$\_$route$($A$,$ B$)$ : reachable$($A$,$ B$)\}1$ :$-$ reachable$($A$,$ B$).$

$\%$ Ensure all cleanable roads are either included or excluded from the route

:$-$ cleanable$($A$,$ B$),$ not in$\_$route$($A$,$ B$).$

$\%$ Define visited locations

visited$($X$)$ :$-$ start$($X$).$

visited$($Y$)$ :$-$ in$\_$route$($X$,$ Y$),$ visited$($X$).$

$\%$ Ensure all endpoints of cleanable roads are visited

:$-$ cleanable$($X$,\_),$ not visited$($X$).$

:$-$ cleanable$(\_,$ X$),$ not visited$($X$).$

$\%$ Ensure the route is a closed loop: sweeper must return to its starting point

:$-$ start$($S$),$ not visited$($S$).$

$\%$ Minimize the number of streets cleaned more than once

#minimize $\{1,$ in$\_$route$($A$,$ B$)$ : reachable$($A$,$ B$)\}.$

#show in$\_$route$/2.$

#show visited$/1.$

Which is the solution for my problem described in the image attached

The code outputs:

visited$($a$)$ visited$($f$)$ visited$($b$)$ visited$($g$)$ visited$($e$)$ visited$($c$)$ in$\_$route$($a$,$f$)$ in$\_$route$($f$,$b$)$ in$\_$route$($b$,$g$)$ in$\_$route$($b$,$e$)$ in$\_$route$($e$,$c$)$ in$\_$route$($g$,$b$)$ in$\_$route$($c$,$b$)$ in$\_$route$($b$,$a$)$

it should output:

visited$($a$)$ visited$($f$)$ visited$($b$)$ visited$($g$)$ visited$($b$)$ visited$($e$)$ visited$($c$)$ in$\_$route$($a$,$f$)$ in$\_$route$($f$,$b$)$ in$\_$route$($b$,$g$)$ in$\_$route$($g$,$b$)$ in$\_$route$($b$,$e$)$ in$\_$route$($e$,$c$)$ in$\_$route$($c$,$b$)$ in$\_$route$($b$,$a$)$

In the attached hand$-$drawn image is the problem we are trying to solve $($we want the general solution, but this graph is the perfect test for verification$)$

The graph in the image can be described by the input:

road$($b$,$a$).$

road$($c$,$b$).$

road$($g$,$b$).$

road$($b$,$g$).$

road$($a$,$f$).$

road$($f$,$b$).$

road$($b$,$e$).$

road$($e$,$c$).$

road$($c$,$d$).$

blocked$($c$,$d$).$

start$($a$).$

The problem with the current solution is that: we dont have a $$Single$-$Path$($Postman$/$Euler$)$ Encoding

This problem is essentially the Route Inspection Problem $($aka Chinese Postman Problem$)$ or an Euler$-$like path problem on a directed graph:

$1.$ You must cover $($clean$)$ each edge at least once.

$2.$ You want to minimize how many edges get traversed more than once.

$3.$ You want a single cycle $($start $=$ end$).$

If the directed graph were strictly Eulerian $($every node has equal in$-$degree and out$-$degree, and the graph is strongly connected in the relevant subgraph$),$ then you could do an Euler tour with zero repeats. But whenever a graph is not balanced, you must duplicate edges to $$patch up$$ in$-$degree vs$.$ out$-$degree, or to remain strongly connected.

Crucially, you also need to encode the idea of a single contiguous route. Merely saying $$include all edges, and we must visit them all at least once$$ does not guarantee that in the final solution, after traveling b$$g$,$ your next move is forced to be g$$b$.$

Please fix my solution to include this funcitonality.

Test it with the input:

road$($b$,$a$).$

road$($c$,$b$).$

road$($g$,$b$).$

road$($b$,$g$).$

road$($a$,$f$).$

road$($f$,$b$).$

road$($b$,$e$).$

road$($e$,$c$).$

road$($c$,$d$).$

blocked$($c$,$d$).$

start$($a$).$

The output should be:

visited$($a$)$ visited$($f$)$ visited$($b$)$ visited$($g$)$ visited$($b$)$ visited$($e$)$ visited$($c$)$ in$\_$route$($a$,$f$)$ in$\_$route$($f$,$b$)$ in$\_$route$($b$,$g$)$ in$\_$route$($g$,$b$)$ in$\_$route$($b$,$e$)$ in$\_$route$($e$,$c$)$ in$\_$route$($c$,$b$)$ in$\_$route$($b$,$a$)$