P81??Can you solve the prisoner boxes riddle

Your favorite band is great at playing music, but not so great at being organized. They keep misplacing their instruments on tour, and it's driving their manager mad. On the day of the big concert, the band wakes up to find themselves tied up in a windowless, soundproof practice room. Their manager explains what's happening. Outside, there are ten large boxes. Each contains one of your instruments, but don't be fooled by the pictures - they've been randomly placed. I'm going to let you out one at a time. While you're outside, you can look inside any five boxes before security takes you back to the tour bus. You can't touch the instruments or in any way communicate what you find to the others. No marking the boxes, shouting, nothing. If each one of you can find your own instrument, then you can play tonight. Otherwise, the label is dropping you. You have three minutes to think about it before we start. The band is in despair. After all, each musician only has a 50% chance of finding their instrument by picking five random boxes. And the chances that all ten will succeed are even lower - just 1 in 1024. But suddenly, the drummer comes up with a valid strategy that has a better than 35% chance of working. Can you figure out what it was? Pause the video on the next screen if you want to figure it out for yourself! Answer in: 3 Answer in: 2 Answer in: 1 Here's what the drummer said: Everyone first open the box with the picture of your instrument. If your instrument is inside, you're done. Otherwise, look at whatever's in there, and then open the box with that picture on it. Keep going that way until you find your instrument. The bandmates are skeptical, but amazingly enough, they all find what they need. And a few hours later, they're playing to thousands of adoring fans. So why did the drummer's strategy work? Each musician follows a linked sequence that starts with the box whose outside matches their instrument and ends with the box actually containing it. Note that if they kept going, that would lead them back to the start, so this is a loop. For example, if the boxes are arranged like so, the singer would open the first box to find the drums, go to the eighth box to find the bass, and find her microphone in the third box, which would point back to the first. This works much better than random guessing because by starting with the box with the picture of their instrument, each musician restricts their search to the loop that contains their instrument, and there are decent odds, about 35%, that all of the loops will be of length five or less. How do we calculate those odds? For the sake of simplicity, we'll demonstrate with a simplified case, four instruments and no more than two guesses allowed for each musician. Let's start by finding the odds of failure, the chance that someone will need to open three or four boxes before they find their instrument. There are six distinct four-box loops. One fun way to count them is to make a square, put an instrument at each corner, and draw the diagonals. See how many unique loops you can find, and keep in mind that these two are considered the same, they just start at different points. These two, however, are different. We can visualize the eight distinct three-box loops using triangles. You'll find four possible triangles depending on which instrument you leave out, and two distinct paths on each. So of the 24 possible combinations of boxes, there are 14 that lead to faliure, and ten that result in success. That computational strategy works for any even number of musicians, but if you want a shortcut, it generalizes to a handy equation. Plug in ten musicians, and we get odds of about 35%. What if there were 1,000 musicians? 1,000,000? As n increases, the odds approach about 30%. Not a guarantee, but with a bit of musician's luck, it's far from hopeless. Hi everybody, if you liked this riddle, try solving these two.

P82??Can you solve the prisoner hat riddle

You and nine other individuals have been captured by super intelligent alien overlords. The aliens think humans look quite tasty, but their civilization forbids eating highly logical and cooperative beings. Unfortunately, they're not sure whether you qualify, so they decide to give you all a test. Through its universal translator, the alien guarding you tells you the following: You will be placed in a single-file line facing forward in size order so that each of you can see everyone lined up ahead of you. You will not be able to look behind you or step out of line. Each of you will have either a black or a white hat on your head assigned randomly, and I won't tell you how many of each color there are. When I say to begin, each of you must guess the color of your hat starting with the person in the back and moving up the line. And don't even try saying words other than black or white or signaling some other way, like intonation or volume; you'll all be eaten immediately. If at least nine of you guess correctly, you'll all be spared. You have five minutes to discuss and come up with a plan, and then I'll line you up, assign your hats, and we'll begin. Can you think of a strategy guaranteed to save everyone? Pause the video now to figure it out for yourself. Answer in: 3 Answer in: 2 Answer in: 1 The key is that the person at the back of the line who can see everyone else's hats can use the words "black" or "white" to communicate some coded information. So what meaning can be assigned to those words that will allow everyone else to deduce their hat colors? It can't be the total number of black or white hats. There are more than two possible values, but what does have two possible values is that number's parity, that is whether it's odd or even. So the solution is to agree that whoever goes first will, for example, say "black" if he sees an odd number of black hats and "white" if he sees an even number of black hats. Let's see how it would play out if the hats were distributed like this. The tallest captive sees three black hats in front of him, so he says "black," telling everyone else he sees an odd number of black hats. He gets his own hat color wrong, but that's okay since you're collectively allowed to have one wrong answer. Prisoner two also sees an odd number of black hats, so she knows hers is white, and answers correctly. Prisoner three sees an even number of black hats, so he knows that his must be one of the black hats the first two prisoners saw. Prisoner four hears that and knows that she should be looking for an even number of black hats since one was behind her. But she only sees one, so she deduces that her hat is also black. Prisoners five through nine are each looking for an odd number of black hats, which they see, so they figure out that their hats are white. Now it all comes down to you at the front of the line. If the ninth prisoner saw an odd number of black hats, that can only mean one thing. You'll find that this strategy works for any possible arrangement of the hats. The first prisoner has a 50% chance of giving a wrong answer about his own hat, but the parity information he conveys allows everyone else to guess theirs with absolute certainty. Each begins by expecting to see an odd or even number of hats of the specified color. If what they count doesn't match, that means their own hat is that color. And everytime this happens, the next person in line will switch the parity they expect to see. So that's it, you're free to go. It looks like these aliens will have to go hungry, or find some less logical organisms to abduct.

P83??Can you solve the Ragnarok riddle

Ragnarok. The fabled end of the world, when giants, monsters, and Norse gods battle for the future. The gods were winning handily until the great serpent J?rmungandr emerged. It swallowed Valhalla, contorted itself across the land, and then merged into one continuous body with no head and no tail. As it begins to digest Valhalla, an exhausted Odin explains that he has just enough power to strike the creature with one final bolt of lightning. If you magnify his blast with your fabled hammer, Mj?lnir, it should pierce the massive serpent. You’ll run with super-speed along the serpent’s body. When you hold your hammer high, Odin will strike it with lightning and split J?rmungandr open at that point. Then, you’ll need to continue running along its body until every part of it is destroyed. You can’t run over the same section twice or you’ll fall into the already blasted part of the snake. But you can make multiple passes through points where the creature intersects its own body. If you leave any portion un-zapped, J?rmungandr will magically regenerate, Odin’s last power will be spent, and Valhalla will fall forever. What path can you take to destroy the serpent? Pause now to figure it out yourself! Answer in 3 2 1 One powerful way to solve problems is to simplify. And in this case, we can focus our attention on the two things that are important for our path: intersections and the stretches of snake between them. Or, as they’re referred to in graph theory, nodes and edges. The edges are important because they’re what we need to travel. And the nodes matter because they connect the edges, and are where we may need to make choices as we run from edge to edge. This simplification into nodes and edges leaves us with a ubiquitous and important mathematical object known as a graph, or network. We just need to figure out how to travel what mathematicians call an Eulerian path, which traces every edge exactly once. Instead of looking at the path as a whole, let’s zoom in on a single node. During some moment in your run, you’ll enter that node, and then exit it. That takes care of two edges. If you enter again, you’ll need to exit again too, which requires another pair of edges. So every point along your path will have edges that come in pairs. One edge in each pair will function as entrance; the other as exit. And that means that the number of edges coming out of every node must be even. There are just two exceptions: the start and end points, where you can exit without entering, or vice versa. If we look at the network formed by the serpent again, and number how many edges emerge from each node, a pattern jumps out that fits what we just saw. Every node has an even number of edges emerging from it, except two. So one of these must be the start of your route, and the other the end. Interestingly enough, any connected network that has exactly 2 nodes with an odd number of edges will also contain an Eulerian path. The same is true if there are no nodes with an odd number of edges— in that case the path starts and ends in the same spot. So knowing that, let’s return to our full graph. We can begin by taking care of this edge here. Now we can zig-zag back and forth across the whole snake until we reach the end. And that's just one solution— it helps to be systematic, but you’re likely to happen upon many others once you know where to begin and end your run. You hold your hammer high at the opportune moment, and Odin sends the world-saving surge of lightning at you. Then you run like you’ve never run before. If you can pull this off, surely nothing could stop the might of the Norse Gods. And if something like that were out there, slouching its way towards you… well, that would be a story for another day.

P84??Can you solve the rebel supplies riddle

You’re overseeing the delivery of crucial supplies to a rebel base deep in the heart of enemy territory. To get past Imperial customs, all packages must follow a strict protocol: if a box is marked with an even number on the bottom, it must be sealed with a red top. The boxes are already being loaded onto the transport when you receive an urgent message. One of the four boxes was sealed incorrectly, but they lost track of which one. All the boxes are still on the conveyor belt. Two are facing down: one marked with a four, and one with a seven. The other two are facing up: one with a black top, another with a red one. You know that any violation of the protocol will get the entire shipment confiscated and put your allies in grave danger. But any boxes you pull off for inspection won’t make it onto this delivery run, depriving the rebels of critically needed supplies. The transport leaves in a few moments, with or without its cargo. Which box or boxes should you grab off the conveyor belt? Pause the video now if you want to figure it out for yourself! Answer in: 3 Answer in: 2 Answer in: 1 It may seem like you need to inspect all four boxes to see what’s on the other side of each. But in fact, only two of them matter. Let’s look at the protocol again. All it says is that even-numbered boxes must have a red top. It doesn’t say anything about odd-numbered boxes, so we can just ignore the box marked with a seven. What about the box with a red top? Don’t we need to check that the number on the bottom is even? As it turns out, we don’t. The protocol says that if a box has an even number, then it should have a red top. It doesn’t say that only boxes with even numbers can have red tops, or that a box with a red top must have an even number. The requirement only goes in one direction. So we don’t need to check the box with the red lid. We do, however, need to check the one with the black lid, to make sure it wasn’t incorrectly placed on an even-numbered box. If you initially assumed the rules imply a symmetrical match between the number on the box and the type of lid, you’re not alone. That error is so common, we even have a name for it: affirming the consequent, or the fallacy of the converse. This fallacy wrongly assumes that just because a certain condition is necessary for a given result, it must also be sufficient for it. For instance, having an atmosphere is a necessary condition for being a habitable planet. But this doesn’t mean that it’s a sufficient condition – planets like Venus have atmospheres but lack other criteria for habitability. If that still seems hard to wrap your head around, let’s look at a slightly different problem. Imagine the boxes contain groceries. You see one marked for shipment to a steakhouse and one to a vegetarian restaurant. Then you see two more boxes turned upside down: one labeled as containing meat, and another as containing onions. Which ones do you need to check? Well, it’s easy – make sure the meat isn’t being shipped to the vegetarian restaurant, and that the box going there doesn’t contain meat. The onions can go to either place, and the box bound for the steakhouse can contain either product. Why does this scenario seem easier? Formally, it’s the same problem – two possible conditions for the top of the box, and two for the bottom. But in this case, they’re based on familiar real-world needs, and we easily understand that while vegetarians only eat vegetables, they’re not the only ones who do so. In the original problem, the rules seemed more arbitrary, and when they’re abstracted that way, the logical connections become harder to see. In your case, you’ve managed to get enough supplies through to enable the resistance to fight another day. And you did it by thinking outside the box – both sides of it.

P85??Can you solve the river crossing riddle

As a wildfire rages through the grasslands, three lions and three wildebeest flee for their lives. To escape the inferno, they must cross over to the left bank of a crocodile-infested river. Fortunately, there happens to be a raft nearby. It can carry up to two animals at a time, and needs as least one lion or wildebeest on board to row it across the river. There's just one problem. If the lions ever outnumber the wildebeest on either side of the river, even for a moment, their instincts will kick in, and the results won't be pretty. That includes the animals in the boat when it's on a given side of the river. What's the fastest way for all six animals to get across without the lions stopping for dinner? Pause here if you want to figure it out for yourself. Answer in: 3 Answer in: 2 Answer in: 1 If you feel stuck on a problem like this, try listing all the decisions you can make at each point, and the consequences each choice leads to. For instance, there are five options for who goes across first: one wildebeest, one lion, two wildebeest, two lions, or one of each. If one animal goes alone, it'll just have to come straight back. And if two wildebeest cross first, the remaining one will immediately get eaten. So those options are all out. Sending two lions, or one of each animal, can actually both lead to solutions in the same number of moves. For the sake of time, we'll focus on the second one. One of each animal crosses. Now, if the wildebeest stays and the lion returns, there will be three lions on the right bank. Bad news for the two remaining wildebeest. So we need to have the lion stay on the left bank and the wildebeest go back to the right. Now we have the same five options, but with one lion already on the left bank. If two wildebeest go, the one that stays will get eaten, and if one of each animal goes, the wildebeest on the raft will be outnumbered as soon as it reaches the other side. So that's a dead end, which means that at the third crossing, only the two lions can go. One gets dropped off, leaving two lions on the left bank. The third lion takes the raft back to the right bank where the wildebeest are waiting. What now? Well, since we've got two lions waiting on the left bank, the only option is for two wildebeest to cross. Next, there's no sense in two wildebeest going back, since that just reverses the last step. And if two lions go back, they'll outnumber the wildebeest on the right bank. So one lion and one wildebeest take the raft back leaving us with one of each animal on the left bank and two of each on the right. Again, there's no point in sending the lion-wildebeest pair back, so the next trip should be either a pair of lions or a pair of wildebeest. If the lions go, they'd eat the wildebeest on the left, so they stay, and the two wildebeest cross instead. Now we're quite close because the wildebeest are all where they need to be with safety in numbers. All that's left is for that one lion to raft back and bring his fellow lions over one by one. That makes eleven trips total, the smallest number needed to get everyone across safely. The solution that involves sending both lions on the first step works similarly, and also takes eleven crossings. The six animals escape unharmed from the fire just in time and begin their new lives across the river. Of course, now that the danger's passed, it remains to be seen how long their unlikely alliance will last.

P86??Can you solve the rogue AI riddle

A hostile artificial intelligence called NIM has taken over the world’s computers. You’re the only person skilled enough to shut it down, and you’ll only have one chance. You’ve broken into NIM’s secret lab, and now you’re floating in a raft on top of 25 stories of electrified water. You’ve rigged up a remote that can lower the water level by ejecting it from grates in the sides of the room. If you can lower the water level to 0, you can hit the manual override, shut NIM off, and save the day. However, the AI knows that you’re here, and it can lower the water level, too, by sucking it through a trapdoor at the bottom of the lab. If NIM is the one to lower the water level to 0, you’ll be sucked out of the lab, resulting in a failed mission. Control over water drainage alternates between you and NIM, and neither can skip a turn. Each of you can lower the water level by exactly 1, 3, or 4 stories at a time. Whoever gets the level exactly to 0 on their turn will win this deadly duel. Note that neither of you can lower the water below 0; if the water level is at 2, then the only move is to lower the water level 1 story. You know that NIM has already computed all possible outcomes of the contest, and will play in a way that maximizes its chance of success. You go first. How can you survive and shut off the artificial intelligence? Pause here if you want to figure it out for yourself. Answer in: 3 Answer in: 2 Answer in: 1 You can’t leave anything up to chance - NIM will take any advantage it can get. And you’ll need to have a response to any possible move it makes. The trick here is to start from where you want to end and work backwards. You want to be the one to lower the water level to 0, which means you need the water level to be at 1, 3, or 4 when control switches to you. If the water level were at 2, your only option would be to lower it 1 story, which would lead to NIM making the winning move. If we color code the water levels, we can see a simple principle at play: there are “l(fā)osing” levels like 2, where no matter what whoever starts their turn there does, they’ll lose. And there are winning levels, where whoever starts their turn there can either win or leave their opponent with a losing level. So not only are 1, 3, and 4 winning levels, but so are 5 and 6, since you can send your opponent to 2 from there. What about 7? From 7, all possible moves would send your opponent to a winning level, making this another losing level. And we can continue up the lab in this way. If you start your turn 1, 3, or 4 levels above a losing level, then you’re at a winning level. Otherwise, you’re destined to lose. You could continue like this all the way to level 25. But as a shortcut, you might notice that levels 8 through 11 are colored identically to 1 through 4. Since a level’s color is determined by the levels 1, 3, and 4 stories below it, this means that level 12 will be the same color as level 5, 13 will match 6, 14 will match 7, and so on, In particular, the losing levels will always be multiple of 7, and two greater than multiples of 7. Now, from your original starting level of 25, you have to make sure your opponent starts on a losing level every single turn— if NIM starts on a winning level even once, it’s game over for you. So your only choice on turn 1 is to lower the water level by 4 stories. No matter what the AI does, you can continue giving it losing levels until you reach 0 and trigger the manual override. And with that, the crisis is averted. Now, back to a less stressful kind of surfing.

P87??Can you solve the sea monster riddle

According to legend, once every thousand years a host of sea monsters emerges from the depths to demand tribute from the floating city of Atlantartica. As the ruler of the city, you’d always dismissed the stories… until today, when 7 Leviathan Lords rose out of the roiling waters and surrounded your city. Each commands 10 giant kraken, and each kraken is accompanied by 12 mermites. Your city’s puny army is hopelessly outmatched. You think back to the legends. In the stories, the ruler of the city saved his people by feeding the creatures a ransom of pearls. The pearls would be split equally between the leviathans lords. Each leviathan would then divide its share into 11 equal piles, keeping one, and giving the other 10 to their kraken commanders. Each kraken would then divide its share into 13 equal piles, keeping one, and distributing the other twelve to their mermite minions. If any one of these divisions left an unequal pile or leftover pearl, the monsters would pull everyone to the bottom of the sea. Such was the fate of your fabled sister city. You rush to the ancient treasure room and find five chests, each containing a precisely counted number of pearls prepared by your ancestors for exactly this purpose. Each of the chests bears a number telling how many pearls it contains. Unfortunately, the symbols they used to write digits 1,000 years ago have changed with time, and you don’t know how to read the ancient numbers. With hundreds of thousands of pearls in each chest, there’s no time to recount. One of these chests will save your city and the rest will lead to its certain doom. Which do you choose? Pause the video to figure it out yourself. Answer in 3 Answer in 2 Answer in 1 There isn’t enough information to decode the ancient Atlantartican numeral system. But all hope is not lost, because there’s another piece of information those symbols contain: patterns. If we can find a matching pattern in arabic numerals, we can still pick the right chest. Let’s take stock of what we know. A quantity of pearls that can appease the sea monsters must be divisible by 7, 11, and 13. Rather than trying out numbers at random, let’s examine ones that have this property and see if there are any patterns that unite them. Being divisible by 7, 11, and 13 means that our number must be a multiple of 7, 11, and 13. Those three numbers are all prime, so multiplying them together will give us their least common multiple: 1001. That’s a useful starting place because we now know that any viable offering to the sea monsters must be a multiple of 1001. Let’s try multiplying it by a three digit number, just to get a feel for what we might get. If we try 861 times 1001, we get 861,861, and we see something similar with other examples. It’s a peculiar pattern. Why would multiplying a three-digit number by 1001 end up giving you two copies of that number, written one after the other? Breaking down the multiplication problem can give us the answer. 1001 times any number x is equal to 1000x + x. For example, 725 times 1000 is 725,000, and 725 x 1 is 725. So 725 x 1001 will be the sum of those two numbers: 725,725. And there’s nothing special about 725. Pick any three-digit number, and your final product will have that many thousands, plus one more. Even though you don’t know how to read the numbers on the chests, you can read which pattern of digits represents a number divisible by 1001. As with many problems, trying concrete examples can give you an intuition for behavior that may at first look abstract and mysterious. The monsters accept your ransom and swim back down to the depths for another thousand years. With the proper planning, that should give you plenty of time to prepare for their inevitable return.

P88? ?Can you solve the secret sauce riddle

One of the top chefs from Pasta Palace has been kidnapped by operatives from Burger Bazaar hoping to learn the location of their secret sauce recipe. Little do they know that a third party— Sausage Saloon— has sent you to take advantage of the situation. As their top spy, your skills range from infiltration and subterfuge, to safecracking and reading faces for signs of deception. You’ve tracked the captors to where they’re holding the chef prisoner. From your hiding spot, you can see him on the other side of the glass, while in front of you an interrogator wearing headphones speaks into a microphone. “We already know the recipe is on the 13th floor of the bank vault, in a safe deposit box numbered between 13 and 1300. Now tell us… Is the number less than 500?” You can’t hear the chef’s answer, but you can see that he’s lying. The interrogator, however, falls for it. He follows up by asking, “Is it a perfect square?” Again you can’t hear the answer but can tell the chef is lying, while the interrogator takes him at his word. He then asks, “Is it a perfect cube?” This time the chef answers truthfully. The interrogator thinks for a minute and says, “Good. Now if you just tell me whether or not the number’s second digit is a one, we’ll be done here.” But as the chef starts to answer, the interrogator stands up, blocking your view. Within moments he rushes out of the room, announcing that he’s got the answer and is sending agents to retrieve the recipe. You know that the Burger Bazaar people have the wrong box number. But can you figure out the right one and retrieve the recipe yourself? Pause the video to figure it out for yourself. Answer in 3 Answer in 2 Answer in 1 The key here is to work backwards. We don’t know what the chef answers to the final question or whether he answers truthfully. But we do know that by the time the interrogator asks it, he’s narrowed the options down to two numbers– one where the second digit is 1, and one where it isn’t. Our goal, then, is to find answers to the previous questions that lead to just two possibilities. Of the three constraints offered, the one that narrows our options the most is if the number is a perfect cube. That leaves us with only eight answers between 13 and 1300. So let’s assume the answer to the third question was a truthful YES. Now, let’s look at the second question. If the chef answered YES to the number being a perfect square, it would narrow the interrogator’s options to just 64 and 729– the only numbers in our range that are both a square and a cube. But neither of these has a 1 as the second digit. So the given answer to the second question must’ve been NO. And that also means we can eliminate these two squares from the interrogator's list, leaving only six numbers. Now for the first question, which allows us to divide this list. If the chef answered YES to the number being less than 500, we’d have four options, which is too many. But a NO leaves us with two numbers greater than 500, one of which does have a 1 as its second digit. We don’t know which of these numbers the interrogator thinks is correct. But that doesn’t matter– remember, his conclusion was based on lies. You, on the other hand, are now in a position to reconstruct the truth. First, the chef said the number was greater than 500 but lied, meaning it’s actually less than 500. Second, the chef said it wasn’t a perfect square but lied again, meaning the number is indeed a square. And finally, he truthfully confirmed that it was also a cube. And as we’ve already seen, the only number under 500 that’s both a square and a cube is 64. You find the secret recipe and are gone before anyone’s the wiser. Corporate espionage is not an easy game— but sometimes, that’s just how the sausage is made.

P89? ?Can you solve the secret werewolf riddle

You’re on the trail of a werewolf that’s been terrorizing your town. After months of detective work, you’ve narrowed your suspects to one of five people: the mayor, the tailor, the baker, the grocer, or the carpenter. You’ve invited them to dinner with a simple plan: you’ll slip a square of a rare werewolf antidote into each of their dinners. Unfortunately, your pet goat just ate four of the squares, and you only have one left. Luckily, the remaining square is 50 grams, and the minimum effective dose is 10 grams. If you can precisely divide the square into fifths you’ll have just enough antidote for everyone. You’ll have to use a laser-cutting tool to cut up the square; every other means available to you isn’t precise enough. There are 8 points that can act as starting or ending points for each cut. To use the device, you’ll have to input pairs of points that tell the laser where to begin and end each cut, and then the laser executes all the cuts simultaneously. It’s okay to cut the square into as many pieces as you want, as long as you can group them into 10 gram portions. But you can’t fold the square or alter it otherwise, and you only get one shot at using the laser cutter. The full moon is rising, and in a moment someone will transform and tear you all apart unless you can cure them first. How can you divide the antidote into perfect fifths, cure the secret werewolf, and save everyone? Pause the video now if you want to figure it out for yourself. Answer in 3 Answer in 2 Answer in 1 When it comes to puzzles that involve cutting and rearranging, it’s often helpful to actually take a piece of paper and try cutting it up to see what you can get. If we cut BF and DH we’d get fourths, but we need fifths. Maybe there’s a way to shave a bit off of a quarter to get exactly one fifth. Cutting BE looks good at first, but that last cut takes a off a quarter of a quarter, leaving us with a portion of 3/16: just smaller than a fifth, and not enough to cure a werewolf. What if we started with BE instead? That would also give us a quarter. And is there a way to shave just a bit more off? Both DG and CH look promising. If we make one more cut, from A to F, we may start to notice something. With these four cuts—from B to E, D to G, F to A, and H to C—we’ve got four triangles and a square in the middle. But the pieces that make each triangle can also be rearranged to make a square identical to the middle one. This means that we’ve split the antidote into perfect fifths! What’s interesting about this sort of problem is that while it’s possible to solve it by starting from the geometry, it’s actually easier to start experimenting and see where that gets you. That wouldn’t be as viable if the square had, say, 24 cut points, but with just 8 there are only so many reasonable options. You secretly dose each of the townspeople as the full moon emerges in the sky. And just as you do, a terrible transformation begins. Then, just as suddenly, it reverses. Your measurements were perfect, and the people and animals of the town can rest a little easier.

P90? ?Can you solve the seven planets riddle

Your interstellar police squad has tracked a group of dangerous rebels to a cluster of of seven small planets. Now you must apprehend them quickly before their reinforcements arrive. Of course, the rebels won’t just stay put. They’ll try to dodge you by moving from planet to planet. But you have one major advantage. Every hour, your state-of-the-art cruiser can warp between any two planets, while their beat-up smuggling ship can only jump to an adjacent planet in that same time. These rebels don’t like to stay put. Every time they can relocate, they will. Your scouts tell you that the approaching rebel fleet is 10 hours away. You can’t risk letting the rebels escape. Can you devise a sequence for searching the planets that’s guaranteed to catch them in 10 warps or less, no matter what moves they make? Rounding up the rebels won’t be easy. For one, you have no way of knowing which planet they’re on to begin with. And without that information, it’s hard to determine where they’ll move next. So where do you begin? When tackling problems of this kind it often helps to simplify things, so you can better understand their dynamics. Let’s imagine that this cluster has the same arrangement but no outermost planets, leaving only the four in the center. We still don’t know which planet the rebels start on. But there’s one key feature: the third planet is adjacent to all others, which means the rebels either start there and move somewhere else, or start on one of the other planets and have no choice but to move to planet three. Simply checking planet three twice in a row would do the trick. Adding the three outer planet adds a bit more complexity, but the same strategy remains. We want to check the planets in an order that will eventually corner the rebels. And there’s another insight that can help us: each hour, the rebels move from an even-numbered planet to an odd-numbered planet, or vice versa. This gives us a way to simplify the problem by dividing the planets into two subsets, and tackling each one separately. For starters, let’s assume the rebels begin on an even-numbered planet: either two, four, or six. So we’ll search planet two first. If they’re not there, they must have started on either four or six. which means they can move to three, five, or seven. Planet three at the center gives them the most options for their next move, so we’ll want to check there next. If we don’t find them, they must have been at planet five or seven, meaning they’ll next move to four or six. Let’s now search planet four. If they’re not there, they must have gone to the sixth planet and can only flee to three or seven. If we next scour planet three and don’t find them, we know they went to planet seven and are now cornered. They can only move to planet six, where we’ll apprehend them on our fifth search. Of course, this plan only works assuming that the rebels were on an even-numbered planet in the first hour. But what if that assumption was wrong? In that case, they must’ve started on an odd-numbered planet. And because they move to an adjacent planet every hour, their location must alternate between odd and even-numbered planets. This means that if they were on an odd-numbered planet to start, after five moves, they'd be on an even-numbered planet. So if our first five searches missed them because our assumption that they started on an even-numbered planet was wrong, all we have to do now is repeat the sequence! Searching the planets in order two, three, four, three, six, two, three, four, three, six, leaves the rebels nowhere to run. Thanks to your deductive reasoning, order is restored to the galaxy.