I made a solver for the Totenreich Claw Machine Puzzle.

There are no guarantees this will work, and it could be wildly inaccurate due to my limited testing sample size. This is the best formula I could determine for how the game calculates what placements win.

I looked deeper into this because I don't believe Treyarch would make the puzzle completely random, allowing for softlocks. Also, the game has to determine by how much to move the needle, so there have to be some formula.

Here's a write-up on what I did and my findings:

I measured the pixel position of the needle indicator via screenshots when compared to the left wall (right where the red ends). Since I measured pixels of a screenshot, there's a margin of error of ~1-2px.

Baseline (empty board, no uranium) = 27px, green zone = 513-595px raw.

I found a base formula for placements with a single group which fits all tested configs with 2 or fewer U (uranium) in the group:

f(n) = 6.4n² + 10.6n

where n = total nodes in a connected group (uranium + rods connected to it). Groups with no uranium don't score.

Final score = 27 (baseline) + sum of f(n) across all groups containing U.

It seems shape, and edge count doesn't matter. Only node count per group is used.

3U bonus: if all 3 uraniums end up touching the same rod group (one big group), that group's score gets multiplied by 1.25 instead of just being f(n).

Unknown: what the bonus is when the 3 uraniums split into separate groups (like 2 in one group + 1 in another). A combination of two groups (6 and 3) scores about 104px higher than the base formula predicts, so there's clearly some bonus happening, but I only have 2 data points for split 3U setups and can't figure out the exact formula yet.