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Traversing through a branch node with value 0 at index i will cause the number of 0s in D to increase by C_i * (d_i-1) -1. Leaf nodes will cause a decrease of 1 to the same quantity.
In the simplest case, where there are no 0 nodes, it's trivial to see that traversal will complete; the tree gets read as a large series of nested for loops.
In a tree with 0s only as leaves, it is also trivial to see that no expansion of the tree occurs, and execution will terminate.
(the above is why I do replacements with t(x,3) before recursing, so I can guarantee to not get stuck like that and miss out on growth.)
Now for a slightly more complicated case, with one branch 0. It will replace all the leaf nodes after it in traversal order with 0s, but after that the tree becomes one with only 0s as leaves again.
Starting to generalize, a replacement will make future branches larger but will make future replacements smaller and smaller until they become nonexistent. Since it only affects future branches, there's no way to make recursion infinite.
Eventually, the only 0s will be on leaves, which will replace themselves and then execution will terminate.
