$n$ cycles in $n×n$ lattice graph average cycle length standard deviation 2 2 2.000000 2.000000 3 14 5.714286 2.249717 4 214 10.710280 2.771717 5 9350 17.462246 3.152186 6 1222364 25.768157 3.724317 7 487150372 35.805114 4.284285 8 603841648932 47.479949 4.756736 9 2318527339461266 60.709770 5.221515 10 27359264067916806102 75.502314 5.707011 11 988808811046283595068100 91.876952 6.201849 12 109331355810135629946698361372 109.838303 6.697420
We can immediately construct the ZDD for the sets of edges containing exactly $n$ out of 4 particular edges. Thus, constructing such a ZDD for each clue and intersecting them with the ZDD of simple cycles yields the solutions.
However, we can solve puzzles faster if we first construct the ZDD $Z$ representing sets that satisfy all the clues, and then follow the algorithm for constructing the simple cycle ZDD, except we respect constraints represented in $Z$. Indeed, computing the simple cycle ZDD for larger puzzles is too expensive, while this approach can sometimes solve them cheaply.