Solution: Roller Coasters
Answer: MOVING PLATFORM
Written by Jonathan
The first steps is to solve the clues, which resolve to 10 ten-letter roller coaster names at the associated theme parks.
|Clue (theme park)||Word|
|Playland Park (2 wds.)||CRAZY MOUSE|
|California's Great America||RAILBLAZER|
Next, we notice that the flavor text tells us that every roller coaster does not stay level. This clues that we should fill in the grid from left to right such that every word moves up or down one space with each succeeding column. This is a powerful piece of information, because it implies that all the roller coasters in alternate rows are linked, that is, their movement must be the same.
Once this is identified, there are various solve paths that can be taken to find the following completed grid. One such solve path is detailed in the appendix below. (See the appendix also for a definition of odd and even.)
Once solved, we obtain the answer, MOVING PLATFORM.
To start, it helps to figure out which words belong in the odd positions and which belong in the even positions. To do this, first isolate all the letters given in odd starting positions and those starting in even starting positions, and note down which words can provide these letters. In order to deduce this, it will help to realize that the requirement of moving up and down in every column partitions the grid into two sets of letters, in a checkerboard-like fashion. Furthermore, each set of cells is made up of 6 words.
For convenience, we will call the black cells odd and the white cells even.
There are a couple of ways to solve this logic puzzle, but they all need the main piece of logic that the paths taken by the odd and even roller coasters must be the same.
- First we start by trying to classify the words into odd and even based on unique letters in each column/position of the words. The K in column 5 has to belong to GATEKEEPER (odd), which implies the T in column 3 belongs to AFTERSHOCK (even). The C in the column 5 has to belong to SWITCHBACK (even).
- We can also start by finding out the paths of the words for each parity. There is no word that has DS in the 8th and 9th column, nor SK in the 9th and 10th column, hence the even path has to end ↑↑.
- The Ks in the last column have to be SWITCHBACK and AFTERSHOCK. Because we already know the C in column 5 belongs to SWITCHBACK, the only way to join that C with either the ending ACK/OCK is to make the path of evens end in ↓↓↓↑↑, which fixes the positions of the two words.
- Since GATEKEEPER was determined to be odd, the T in column 3 must belong to AFTERSHOCK, which means that the even path must go ↓↑↑ in columns 3 to 5. (The I in column 2 blocks AFTERSHOCK from going ↑ in column 3.)
- Continuing the chain for SWITCHBACK we find that the A in column 1 forces the even path to start ↑, completing the even path.
- We are now guaranteed that the remaining given A in column 1 belongs to APOCALYPSE (checked by the S in column 9) and given S in column 1 belongs to SIDEWINDER (checked by the D in column 8), both of which we can fill completely.
- We also know that the last even word must have a D in position 2 and E in position 4. The only unused word satisfying this constraint is DIVERTICAL, which completes the evens.
- We are done with evens, so now let us move on to odds. Given this, we know the remaining words belong to odd: CRAZYMOUSE, GATEKEEPER, MINDBENDER, RAILBLAZER and REVOLUTION.
- We can start in the top left corner of the grid. We know none of these words starts with RI, therefore the first move has to be ↓, and the word starting in the top left cell is MINDBINDER. The word below it must either be RAILBLAZER or REVOLUTION.
- Since none of these words have an E in column 4, we know the path cannot be ↓↓ (otherwise, MINDBENDER will intersect the E) or ↑↓ or ↓↑ (otherwise, the word starting R will intersect the E). Therefore the next moves must be ↑↑
- However, we do know that GATEKEEPER is the only word with an E in column 4. Working backwards, we can identify where GATEKEEPER must have begun from. Even further, the K in column 5 must belong to it, since no other word has a K in column 5. Along with the fact that the next L in column 6 blocks the path of GATEKEEPER, we deduce the next move for odd is ↓↓.
- The N in column 7 must belong to MINDBENDER, thus the next move for odd must be ↑.
- The P in column 8 must belong to GATEKEEPER, thus the next move for odd must be ↓.
- The word in between MINDBENDER and GATEKEEPER now reads an L in column 6, fixing that word as RAILBLAZER.
- Now we look at the end of MINDBENDER and RAILBLAZE, as well as the E in the last column. The last two moves must be ↓↓, because it cannot be ↑↑ (RAILBLAZER will intersect the E) or ↑↓ or ↓↑ (MINDBENDER will intersect the E). This fixes the entire path of odd.
- Finally, the only unused clue is the V in column 3, which must belong to REVOLUTION. We can work backwards to figure out where it started, and this places CRAZYMOUSE, which completes the grid.
This was my first puzzle written for this hunt. The puzzle originally used any 10 letter word, and had a step to extract using braille at the end. External testsolvers remarked that the braille step felt out of place. With way more freedom in my choice of words, the obvious choice was to use roller coaster names.
As seen from the appendix, quite a number of clues are there mostly for checking, so probably at least one clue could be removed with this puzzle still being unique. But as an opening easy logic puzzle, we gave an extra check to make sure every word was checked at least once, and should open up a fair amount of different logical paths to arrive at the final answer. Hopefully this served as an easy enough puzzle for the start of the hunt, given the many harder puzzles to come.