Here is the fifth maze in the Wormeus series. This time Wormeus is faced with multiple Stickotaurs and multiple Universes. Different universes are connected via … wait for it … drum roll dlrdlrdlrdlrdlrdlrdlr … WORMHOLES, ba-dum-tish!!!!!!!!!!! Wormholes can teleport Wormeus to a different Universe but have no effect on Stickotaurs (therefore every universe always contains exactly one Stickotaur).

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Normal Wormeus rules apply. In addition:

(1) After Wormeus has taken an action, the Stickotaur in the same universe as Wormeus takes two actions and all other Stickotaurs stay still.

(2) If Wormeus occupies the same square as a wormhole icon he can teleport to the corresponding wormhole in a different universe. Teleporting is not mandatory, but it does cost an action.

Your task is to eat all the apples in at most 100 moves. Solutions with clear-cut explanations are encouraged and cool wormhole animations wouldn’t hurt either 😊


1 Answer 1


The second universe seems to be just a simple detour, so let's solve that first.

Clean up the prison in the middle: UeReD eReUe (10 actions)
Bring the 'taur to the wormhole: DLULU DRDDL e (11 actions)
Clean bottom part: ReReR eU (7 actions)
Juke the 'taur into the prison: UDU (3 actions)
Eat the rest of the apples: eUeLe LeLe (9 actions)
Return to the wormhole: DD (2 actions)

There are very few choices to make in this universe, so let's assume this solution (42 actions) is optimal for universe 2, so we have 56 actions left for universe 1. (Or 59, if we don't have to return.)

In universe 1, we have only one shot at getting the 'taur behind the wall on the left, so let's grab it, and clean the middle.

Initial juke: LLRU (4 moves)
Clean up middle: eUeUe ReDeD eReRe U (16 actions)

Now we are standing on the wormhole, and both of the following have to happen with the 'taur still behind the wall on the left. The order in which we do these doesn't matter:

Eat the apple in the nook above: UeD (3 actions)
Universe 2: T[Universe 2]T (44 actions)

Continuing, we can eat the apple on the left with the Stickotaur already on the way to its next prison, so we first liberate the 'taur.

Here, sticky sticky: DDUUL (5 moves)
Into the prison with you: eUeUR e (6 actions)

And all that's left is the

Victory lap: LLeLe LeDeD eDeDe ReRRe Re (22 actions)

Which brings the Stickotaur right up to our final-apple-eating face, and the final action count up to 4+16+3+44+5+6+22, which equals exactly 100. Yay!

  • $\begingroup$ Well done. Not intended to be difficult, but I find that with new rulesets it's usually better to err on the side of "too easy". $\endgroup$
    – happystar
    Jul 15, 2021 at 21:05

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