Can you solve the vampire hunter riddle? – Dan Finkel
Articles Blog

Can you solve the vampire hunter riddle? – Dan Finkel

November 20, 2019


The greatest challenge a vampire hunter
can take on is to bring sunlight
into a vampire’s lair. You’ve stealthily descended into the
darkness of a vampire cave, setting a sequence of mirrors as you go. When the sun reaches
the right angle in the sky, a focused beam of light will ricochet
along the mirrors, strike your diffuser, and illuminate the great chamber
where the vampires sleep. You set the final mirror and sneak through an opening
in the corner of the great chamber. The diffuser must be wall-mounted, but the walls are crowded with coffins, which you don’t dare disturb. The only open spots are in the other
three corners of the room. The light will enter through the southwest
corner at a 45 degree angle and bounce off the perfectly smooth
metallic walls until it hits one of the
other three corners. But which corner will it hit? You know the room is a rectangle 49 meters
wide and 78 meters long. You could probably find the answer by drawing a scale model of the room
and tracing the path of the light, but the sun will be in its
place in just minutes, and you’ve got no time to spare. Fortunately, there’s a different way
to solve this puzzle that’s both simple and elegant. So in which corner should
you place the diffuser to flood the vampire lair with sunlight? Pause the video if you want to figure
it out for yourself. Answer in 2 Answer in 1 You could tackle this problem
by examining smaller rooms, and you’d find a lot
of interesting patterns. But there’s one insight that can unravel
this riddle in almost no time at all. Let’s draw the chamber
on a coordinate grid, with the Southwest corner
at the point (0,0). The light passes through grid points with coordinates that are either
both even or both odd. This is true even after it bounces off
one or more walls. Another way of thinking about it is this: since the light travels at a
45 degree angle, it always crosses the diagonal
of a unit square. Traveling 1 meter horizontally
changes the x coordinate from even to odd or vice versa. Traveling 1 meter vertically
changes the y coordinate from even to odd or vice versa. Traveling diagonally – as the light
does here – does both at once, so the x and y coordinates of any points
the light passes through must be both even, or both odd. This observation is more
powerful than it seems. In particular, it means that we have a
way to identify the kinds of points the light won’t ever go through If one of the coordinates is even
and the other is odd, the light will miss them. That means it’ll miss the top
two corners of the room, since those points have one even
and one odd coordinate. The Southeast corner is the
only option for the diffuser. And indeed, when that precious
beam of sunlight enters the hall, it bounces between the walls and strikes the Southeast corner, spot on. The vampires, sensing the intrusion, burst from their coffins
and turn to dust in the light. It was a “high stakes” test, and you passed with flying colors.

Only registered users can comment.

  1. On what grounds the top left corner is eliminated and bottom right is taken knowing the fact that both have zeros in it…

  2. 49,29,20,69,9,40,38,11,60,18,31,
    47,2,51,27,22,71,7,42,36,13,62,
    16,33,45,4,53,25,74,4,45,33,16,
    65,13,36,42,7,56,22,27,76,2,47,
    31,18,67,11,38,40,9,58,20,29,49

  3. I thought of it with symmetry. After you've drawn the whole pattern, you shouldn't be able to tell (from just the drawing of the pattern) if the light came from the south east corner, or the one it hit. To hit the north west or south east corners, the pattern has to have bilateral symmetry. To hit the north west corner and have bilateral symmetry, it has to hit the middle of the west or east walls, but it can't since that's an odd length wall. To hit the north east corner, it has to have 180 degree rotational symmetry, and I guess it can't fit some reason. The video's proof is simpler.

  4. I got it by trigonometry. I think this was a fluke on my part, but I'll post the solution.

    The light comes in at a 45° angle.
    It must create a right angled triangle.
    The light hits the wall, which means it has a side length of 49 metres.
    As it is a 45° angled triangle, this means that the other side is the same, and the other angle.

    The light ray bounces off the north wall 49 metres into the room, and bounces back to the south wall, at the same angle, and hits the south wall at the 78 metre mark.

    As the room is 78 metres long, it hits one of the corners. The southeast corner.

  5. Just put it into one corner if it works fine. If not try the others.
    Or antoher mathematical aproach:
    Being in the corner means that the light traveled exactly 49 meters or exactly 0 meters north (assuming that every time you travel south you travel negative distance) and it has to travel exactly 78 or 0 meters to the east (assuming that traveling west counts as negetively traveling east)
    We want to find the first point where both of that is true at the same time.
    I will also show a method that will work on every given rectangle with lenght of natural numbers.

    If we had a rectangle with 10m from south to north and 15m from west to east it is pretty easy. It hits the north west corner. This can be drawn on a paper easily but you can also calculate why. the light will (as shown in the video) always travel 1 meter north or west when traveling 1 meter west or east and that goes the other way around. To travel exactly 0 or 10 meters the total distance (counting north and south as positive) has to be any amount of meters (we now call it "N1") that can be devided by 10. 10/10=1 it will now exactly be somewhere in the north. 20/10=2 it will now be in the south. If we continune this we will get the following results. On every odd number we will be in the north and on every even number in the south. This can be proven by pure logic. Everytime the lights moves exactly 10m from either exactly south or exactly north it reaches the other side, as they are at a 10m distance. So everytime that "N1" increases by 10 the light swaps sides. Also increasing "N1" by 20 means it will return to the side it was at during "N1". Adding an even number to an even number gives an even number, adding an even number to an odd number gives an odd number. So we only need to prove, that it hits the upper side at one specific time where "N1" devided by 10 is odd. For example at "N1" = 10. It now reached north as it started in the south. We do not even need to prove that it hits the bottom side at "N1"/10="an even number", because we have already proven that "N1"/10 increasing by 1 always exactly swaps the sides. An odd number plus an 1 is always an even number. Which means that if "N1"/10=1 has a solution for "N1" at any given point for "N1"/10=an even number it will have traveled exaclty into the south.
    This was pretty long but pretty easy to understand. In my mind while solving the riddle al that was already clear but now it is proven.
    So for 15m and "N2" (the distance traveled from east to west plus the distance traveled from west to east) it is the same. At "N2"/15=1 we have moved exactly 1 lenght. Meaning we are in the east. Having "N2"/15="any even number" we now got back into the west.
    Now the solution starts. (For this easy rectangle) We need to find the lowest number that can be divided by both lenghts of the rectangles sides (we will give that term a variable later on). The smallest common multiple. For 10 and 15 it is easy. 30. So we need to set "N1"="N2"=30 and now devide "N1" by 10 and "N2" by 15 get the results… "N1"/10=3 "N2"/15=2. It is in the north and in the west.
    Now we can do a solution that can be used on any rectangle, we enter from the south west.
    We also add "x", "y", "n1" and "n2".
    x="the distance between north and south" (as "N1" had no unit this will also have non)
    y="the distance between west and east" (as "N2" had no unit this will also have non)
    n1=N1/x
    n2=N2/y
    "x" and "y" will be given by the rectangle. We now need to find the smallest common multiple of "x" and "y". Then we calculate "n1" and "n2". "n1" being even means we need to put our deluminator into the south it being odd means we need to put it into the north. "n2" being even means we need to put it into the west it being odd means we need to put it into the east.
    Walking in from another corner mixes up the meaning of "n1" and "n2".
    Now back to this specific rectangle.
    We can now just use our new formula. We only need to find the smallest common multiple of "x" and "y".
    Now prime numbers come in handy.
    Just take a look at any prime number such as 11 and any other number that is not an multiple of 11 by its own such as 4. There is now common devisor, so we need to take 11 times 4. 44. Our value for "N1"="N2" and "n1" and "n2" are natural numbers.
    This helps as 49 is the square of a prime number. Meaning that it can only be divided by 1, itself and the prime number it is the square of (7). As 7 is no devisor of 78 there is no chance we find a common multiple below 49 times 78. That can be calculated without a calculator. But we do not even need to. We now know that for this square "N1"="N2"=x*y. So we use our formula for "n1" and "n2". n1=N1/x=x*y/x=y=78 n2=N2/y=x*y/y=x=49
    "n1" being even gets us to the south corner "n2" being odd gets us to the east corner.

    What makes this method pretty sweat is that at values for "x" and "y" that are pretty easy for humand mind to cover within secconds (such as a rectangle that is 6m*4m) the method is a bit complecated in comparison. But at values that are harder to cover this method gets even easier.

  6. or maybe the vampires are trying to live in peace why do you have to kill them have recent movies and shows taught us nothing 🙂

  7. Why not just place the diffuser in the corner you came in through? Surely the light will return to the same corner eventually.

  8. In all this time, you could’ve gotten on with staking them through the heart and shoving garlic in their mouths. They’re asleep, they ain’t gonna notice.

  9. If the walls are perfectly metalic, just let the light bounce on it several times, it will flood the room with light in less than a second.

  10. Has anyone made a Castlvania/Belmont jome yet? Fine…

    Here we can see Simon Belmont’s Plan B for when his whip just doesn’t wanna work

  11. This was such a cool riddle, nothing too crazy but still able to blow my mind with seemingly simple stuff I can't believe I didn't get afterwards

  12. Ted Ed these riddles are not riddles. They are complex math problems. The animations are great but they are not riddles. I'm Not saying stop making these because they are fun but posing them as riddles is kinda click bait-y. Like im not even mad because I can't solve them. If I can't solve a riddle I love learning it, but the right answer to these are always convoluted and never satisfying.

  13. I went YOLO and did it, "it has to be south because it's the only possible edge in the south part of the room", :v, also, you said I have bare minutes to put the lights on and you expect me to draw a graf and milimetrically adjust the perfecr angle in which the light will work. Yeah, sounds legit why not…

  14. It is actually false. Light cannot reflect in a dark room it will be instantly absorbed. Rather we can hold the reflecror in our hand and as the sun will come make some stamping noises and reflect the light on them thus leaving no difficulties

  15. 50% comments: DVD Screensaver
    25% comments: LeT's DrAw A cOoRdInAtE gRaPh!
    12.5% comments: I legit guessed
    12.5% comments: Wacky alternate solutions

Leave a Reply

Your email address will not be published. Required fields are marked *