[IF Competition] The Grand Quest

As in the previous post, I'll simply begin my game with saying some non-spoilery things about this game. That should keep the people on Planet IF happy.

The Grand Quest is a game by Owen Parish, whose Cacophony I recently reviewed on the IFDB. That game was radically non-linear, and gave the player very little guidance; The Grand Quest is almost its exact opposite. Here we simply have a linear series of connected rooms, and you can only progress to the next one once you have solved the puzzle.

Spoilers begin here.

The story and the setting are really just window dressing: the puzzles make or break this game. I was favourably impressed by the first puzzle: it's a kind of riddle, with some wordplay, and although you might not want a whole game based on such thoughts it was nevertheless amusing.

Unfortunately, this quality was not maintained throughout the rest of the game. Some of the puzzles barely qualified for that name, and their inclusion in the work is a mystery to me: the library puzzle, for instance, but the nightmare puzzle was even worse. Was that an attempt at emotion? If so, it utterly failed; if not, what was its role within the game?

The puzzle with the coins wasn't very good either. Nothing in the game has prepared me to explore absolutely everything, and once the sacks and even the ground under the tables didn't yield anything, I concluded that the puzzle must have a cleverer solution. Apparently not. Also, when you write:

Now, let’s see you use your head. Divide all the coins in this room into two piles of equal size, one on each table.

why is the puzzle not to make two piles of equal size, but to make two piles of equal value?

The puzzle with the key was not very original, but servicable.

Luckily, there was still one good puzzle to come: the playing cards. Quite hard, maybe not entirely satisfying, but certainly fun to wrestle with. I think I spent at least an hour getting to know the system and attempting to find a solution. For those who are interested, I'll analyse the system below.

The ending was... well, I don't know. I suppose the story was shallow enough that the game cannot be harmed by such tricks.

Analysis of the playing card puzzle

We have two cards, which we can call X and Y, with a value and a suit; and a machine, which changes the value and the suit. The result of the transformation depends on two things: whether the right box is open or closed, and the order of the cards.

Experimentation will show that the values transform according to the folowing rules:

Open: [+1, -1]
Closed: [+1, +2]

Where the first entry is the change in the value of the first card, and the second entry is the change in the value of the second card. The values are 2, 3, 4, ..., 10, Jack, Queen, King, Ace, and cycling back to 2.

The suits are a little harder, but you will find the following:

[D, *] -> [C, *]
[*, D] -> [*, H]
[H, *] -> [S, *]
[*, H] -> [*, D]
[S, *] -> [D, *]
[*, S] -> [*, C]
[C, *] -> [H, *]
[*, C] -> [*, S]

This is not as arbitrary as it looks. The first card entered simply cycles one step through the row C -> H -> S -> D -> C -> ..., while the second card cycles two steps through this same row.

This means that we have four different transformations for our cards X and Y, which we can write like this:

A: [+1 (+1), -1 (+2)]    (Open, X entered first.)
B: [-1 (+2), +1 (+1)]    (Open, Y entered first.)
C: [+1 (+1), +2 (+2)]    (Closed, X entered first.)
D: [+2 (+2), +1 (+1)]    (Closed, Y entered first.)

Here, the first number designates the value of the card, while the number between brackets designates the suit of the card. Transformation A increases the value of X by 1, decreases the value of Y by 2, cycles the suit of X 1 step, and cycles the suit of Y 2 steps; and so on.

We start with the Four of Diamond and the Six of Spades. Assume that the former must be transformed into the Ace of Clubs and the latter into the Ace of Spades. (I haven't checked whether the other solution is possible as well.)

In that case, the total transformation must be [+10 (+1), +8 (+0)]. (And of course, the value between brackets is modulo 4, the other is modulo 14.)

You can simply check that 4A + 5C + B + D leads to this transformation. Perhaps the easiest way to gain this answer is by doing some linear algebra? It might be tedious, because of the modulos. I didn't try; this is the transformation from the walkthrough.

The order of the transformations does not matter. (Something that, judging from the walkthrough, the author may not have realised.) So, now you can simply type it in in any order you wish. The hardest part is remembering which cards are X and Y, because if you forget that, you are lost. :)