
EssaysThe Riddle of the Potionsby Prefect Marcus Danger lies before you, while safety lies behind, The Riddle of the Potions seldom gets the respect it merits. It is the last challenge faced by Harry and Hermione before entering the chamber where the Philosopher's Stone is kept safely hidden in the Mirror of Erised. Usually when reading the book, we speed past the riddle, admire Hermione for solving it, enjoy Harry and Hermione's parting scene, and then race on the next chapter for the book's climax. The riddle is left in the dust. I have often wondered if the riddle was for real. Can one solve the puzzle given the clues available? To put it another way, did Rowling really put forth the effort to devise a real puzzle, or did she just put together some rhyming verses that sounded good, without any real thought of working them out? After all, there are some 420 possible solutions to a puzzle involving 3 poisons, 2 wines, 1 forward potion, and 1 backward potion. I am happy to say that it can be logically proven that she did put forth the time and effort to create a real puzzle. The riddle leads inexorably to two and only two possible solutions. If we knew the location of the smallest bottle, we would be able to narrow it down to one.
First some definitions:
The bottles will be number one through seven,
beginning on the left. They look like this:
So let us begin.
CLUE 1: "One among us seven will let you move ahead."
CLUE 2: "Another will transport the drinker back instead."
CLUE 3: "Two among our number hold only nettle wine."
CLUE 4: "Three of us are killers, waiting hidden in line."
CLUE 5: "First, however slyly the poison tries to hide 
You will always find some on nettle wine's left side;"
CLUE 6: "Second, different are those who stand at either end,
But if you would move onward, neither is your friend;"
CLUE 7: "Third, as you see clearly, all are different size,
Neither dwarf nor giant holds death in their insides;"
CLUE 8: "Fourth, the second left and the second on the right 
Are twins once you taste them, though different at first sight."
Putting It Together:
If 3 is W, then R1 is satisfied by bottles #2, #3, #6, and #7.
Bottles #4 and #5 can be F P or P F. Thus two possible solutions are:
Going back to B P 3 4 5 P W, if 4
is W, then 3 must be P (R1) leaving 5 to be F. If 5 is W, then 4 must be
P (R1) leaving 3 to be F. Thus two more possible solutions are:
That exhausts all the possible solutions if Bottle #1 were B.
Now let's look at the solutions if Bottle #1 were P (C1,R3).
We know that Bottle #2 must be P or W (C5, C7). If it is P, then so must
Bottle #6 (R7). This forces bottles #3 and #7 to be W (R1). The two possible
solutions here are:
Now if Bottle #2 was a W, then #6 would be also (R7),
forcing Bottle #5 to be P. Bottle #7 cannot be a W
(there are only two and they are at #2 and #6.), F (R4), or P (R2). It
must be B. This leaves the final possible solutions with Bottle #1 being
P as:
In summary, the eight possible solutions are:
Not bad for starting out with 420 possible solutions! This is the best we can do with the clues given. R5 and R6, which tell us that the largest and the smallest bottles are not poison, cannot be used directly since Rowling does not tell us the relative sizes of the bottles. However, if we work backwards from the solution provided by Hermione, we can narrow the field to the most likely two solutions. Consider R5 and R6. Rowling does not tell us the different sizes of the bottles, other than inform us later that F is the smallest bottle. We also learn from Hermione that Bottle #7 is B. What can we infer from that?
Note that there are two and only two solutions that have
Bottle #7 being B. These are also the only two
solutions that do not have P for Bottles #2 and #6. Therefore, if either
Bottle #2 or #6 were the largest bottle, we are left with only two possible
solutions:
Notice the only difference between the two solutions is the swapping of Bottles #3 and #4. One is F, the other is a P. If either one was the smallest, we know it can't be poison (R6). That forces the smallest bottle to be F.
So our solution looks like this:
Conclusions:
© 2004 by Prefect Marcus 