Get the solution to the bonus riddle here: brilliant.org/TedEdGiantIron/! Also, the first 833 of you who sign up for a PREMIUM subscription will get 20% off the annual fee. Riddle on, riddlers!
does the combination need to be one that is tested? Because I came up with a scenario in which the worst case would mean that the remaining combination are the good ones.
Actually,the "diet" you are thinking of is called slimming, where people eat less to lose weight. Using "calorie-controlled diet" is more appropriate. And yes,I am a huge nerd.
actually, there is: if they were regular size batteries, testing more than 7 pairs would be very easy, it takes very little time to do so. But by being giant, moving them takes a lot of time, and thus you are limited to 7 tries only
This riddle seems quite simple when you first hear it, but turns out to have a much more elegant solution than expected. That’s what makes it a superb riddle.
I think we’re missing the point here. If I don’t iron his shirt, my boss will eat me? The solution is clear: unionise and overthrow the tyrannical employer!
After about 3 minutes, this video started sounding like "if you have 4 good batteries and battery on the giant iron, then your giant iron battery in the good battery pile, would only be 2 possible combinations of the giant battery. And if you remember, the 1 bad battery with the testing all the good batteries would take too long for the good batteries to a giant testing"
I have a little question. Imagine:What if it were different? @starvalentinobeats 1.when... on the first 3 battery pile, you find out that the iron works? Then there must be 2 batteries work, then you test again on 3 possible combinations. The other three is the same.. When it does live, we know we need to test other 3 possibilities to know who is living. Summed up, There we have already 6 tries. And then how is the other two? We have acquired two living batteries or even more in the other piles. Then we could know, what fate shall this energy source have.. How do you find it? 2. When it were like this video instead, when 3 combinations o.n a pile fail, then you know only one or none is. WORKING? This is the real big question : BUT HOW DO YOU TELL, WHICH ONE THEN IS THE WORKING ONE from these fail combinations ? There is some fishhy things on this ridfle..
I think the bonus problem is the same as the other but in reverse. We need all the working batteries in the iron, meaning we're looking to have two dead batteries out of the iron rather than two working ones in it. If you divide them into the same groups, you can put one group of 3 and the group of 2 into the iron leaving you with one space. Test each from the other group of 3 in turn and if it doesn't work that means that that group contains at least two good batteries. Swap them out with the other group of 3 and repeat. If you get the same results then the same must be true and there are at least two good batteries in that set meaning that the group of 2 will both be dead.
Brillant solution. Tho i still don't undertsand why the iron needs 6 batteries if it only need 4 to work, they could just have it be 4, it would end up being the same
@@kentlab3850let’s see if I can help. Consider the original problem, but change to logic to where the iron works if any of the two batteries are good, and we’re trying to find a pair of dead batteries. It’s not hard to see that this doesn’t change the solution, we’re still trying to find a pair of identical batteries. So we do the same process of splitting the batteries into two groups of three and a group of two, and test each pair within the groups of three. Heres the final connection: testing whether a pair has 2 dead batteries is the same as testing whether the remaining 6 batteries have 4 good ones. Every test with the 6 battery iron is functionally the same as the test with the 2 battery iron, just with the charged reversed. That’s why it needs 6 slots. Also here’s another answer I thought of that’s probably simpler: In both the original riddle and the bonus riddle, you need to find a pair of identical batteries. In the original you need to find 2 good batteries, in the bonus you need to find 2 dead batteries. If the iron had 4 slots and needed 4 good batteries, that would require you to find all 4 identical batteries, which is a harder problem than only finding a pair.
@@xavierburval4128 that's an old comment so i'm not sure about the details but i think i was joking about the fact that having extra space for dead batteries in the iron is a bit silly. It wasn't really about the math since the problem with 4 batteries space would be pretty uninteresting if i recall properly
I did it a different way. 1. Test A&B 2. Test C&D 3. Test E&F If none worked, now you know that G&H has either one or two good batteries. If one, then each other pair also has one good one. If both G&H are good, then only one of the other pairs can have two bad ones. 4. Test G&A 5. Test G&B If A&B were both bad, C&D must contain a good one and also H must be good. Or, in the case that each pair had one bad one, G was bad, so again, H would be good and C&D must contain a good one. So finally, 6. Test H&C 7. H&D must work if none of the other tries did.
I came up with same logic. Both of them are correct & in fact, the combinations are similar, just arranged & explained differently. Two groups of 3 & remaining 2: Group 1(A&B, G&A, G&B), Group 2(C&D, H&C, H&D), E&F.
If G&H are both good, that means that the other three pairs must be distributed like so (or else one try would have worked), although not necessarily in this order: good & bad good & bad bad & bad So when you test G with A and then G with B and neither work, A&B must have been the two bad ones. That means that C&D must contain at least one good one. So in the case that G&H are both good, H is good and either C or D must be good if you got that far.
There's another solution: Divide batteries into 4 pairs. Check 3 of the pairs. Check 1st remaining battery with both batteries from one of the pairs. Check 2nd remaining battery with both batteries from a different pair. I'll let you figure it out yourself why it will always work.
edit: i actually realized this was the same solution as the TED-ED one, just rearranged in groups and order. whoops. this works :D it's only the 2nd valid solution I've come across so far, the TED-ED one and this one i came up with the same solution, and i know why it'll always work how long did it take you to come up with this? took me 45min-1hr. took a lot of wrong turns and was actually pretty close to the solution for the while but never put two and two together into the solution
piotrkakol1992 but you will lose a chance of getting it right in 4th try itself by not testing the last pair of batteries as both the remaining batteries can be good ones
For the Bonus Riddle, the answer is exactly the same as the regular riddle, just mirrored in a way. Instead of testing for 2 good batteries to put in the iron, you have to search for 2 bad batteries to leave out. Since good and bad batteries occur equally, you do the exact same grouping and testing pairs strategy as in the first riddle, only you leave the pairs out instead of putting them in.
@@HT-vd4in it’s the same, the information you get is just the opposite, if it works, it means the two you left out is dead If it doesn’t work, atleast one of the batteries you left out is working You’re just testing for dead batteries to leave out instead of working batteries to use
@@HT-vd4in you number the batteries from A to H step 1, insert batteries but leave AB out if it doesn’t work at least of AB is a working battery Step 2, insert batteries but leave BC out if it doesn’t work at least one of BC is a working battery Step 3, insert batteries but leave AC out if it doesn’t work at least one of AC is a working battery ( so among ABC at least 2 are working ) Step 4, insert batteries but leave DE out if it doesn’t work at least one of DE is a working battery Step 5, insert batteries but leave EF out if it doesn’t work at least one of EF is a working battery Step 6, insert batteries but leave DF out if it doesn’t work at least one of DF is a working battery ( so among DEF at least 2 are working ) By now you would have checked 6 times, and you can conclude that GH are both dead batteries if the iron hasn’t worked yet as among ABC two of the batteries are working and among DEF two of the batteries are working So for the 7th step you just insert ABCDEF and leave GH out, this will guarantee that the iron will work if it hasn’t yet after step 6
Alas, I thought this was a straightforward one, but I failed to solve it. I'm embarrassed. I think I got the bonus riddle though. It's actually the same exact problem as the original riddle, but in reverse. In the original riddle, you're choosing which sets of 2 to put in, whereas in the bonus riddle, you're choosing which sets of 2 to leave out. In the original riddle, you're looking for 2 good ones (to ultimately put in), whereas in the bonus riddle, you're looking for 2 dead ones (to ultimately leave out). So the answer is still 7 tries, using a similar procedure.
@@soprecise7993 You still can group them in the same way: ABC, DEF, GH. However, when you are "testing" AB, you actually put CDEFGH into the iron. Therefore, the bonus question is the same as the main question, just that you need to see it from another angle.
I was thinking about making 4 random pairs and trying them all, then if it doesn't work, you know there's one good battery in every pair, so if you grab two of the pairs you can find the right two, but that can take up to 4 more tries, which makes a total of 8.
Rachelle, you’re bang on except you’ve forgotten that, if the “up to 4 more tries”, you already know the result of two of the tries because you trialled them when you tested out the four pairs at random. This is my explanation: 4 good batteries and 4 bad batteries, right? Separate them into four random pairs and test all 4 pairs. If one of the pairs makes the iron work, great! But if not (you’re 4 tries in at this stage), that means that each pair consists of one good and one bad battery. Take any two pairs at random and discard the other 4 batteries. You now have 4 batteries of which 2 are good and 2 are bad. Swap the pairings and try the first pair - either you’ll now have matched up the two good batteries and the iron will work (after attempt 5) or you’ll have matched up the two bad batteries and it won’t work, so you then know the other two batteries are both good and it’ll work on try 6! Edit: actually on those 5th and 6th tries, I’ve realised it’s possible to, again, pair up a good battery with a bad on both occasions, but if this happens, you just pair up the two batteries that haven’t been paired up yet and those will be two good ones 🙂
you can get it with 3 more tries, If you checked (1/2) and (3/4) in part one, you can continue with (1/3) and (2/4) and (1/4). If non of them work you only have (2/3) remaining and it must be the right one.
@@xiaoshen194 REASONING: It’s surprisingly simple. Let’s get rid of “good” and “bad” for a second and just have battery types “A” and “B”, four of each. Now we have our two irons, one from the main riddle, and the other from the bonus riddle. Let’s say Iron A has 2 slots and requires 2 A batteries, while Iron B has 6 slots and requires 4 B batteries. Now imagine putting two A batteries into Iron A, making it power on, and putting each of the 6 other batteries in Iron B. Because Iron A has no B batteries, all 4 B batteries go to Iron B, making it turn on. What if we put at least one B battery into Iron A? In that case, Iron A would fail to start, since it doesn’t have 2 A batteries. And because there is one B battery in Iron A, there is 3 or less B batteries remaining to put in Iron B; not enough to power it on. So, in conclusion, whenever Iron A is on, Iron B is also on, and whenever Iron is off, Iron B is also off. That makes the main riddle and bonus riddle essentially the same question. Now for the question: What is the least amount of tries do you need so that (iron) has 4 Type B batteries in its 6 slots? This is the same question as “What is the least amount of tries do you need so that (iron) has 2 Type A(Good) batteries in its 2 slots?” The answer to this question is seven. Need proof? Go to 0:55. (It’s probably not possible to do it in 6 tries or under). And since the answer to that question is seven, the answer to the Iron B question, or the bonus riddle, must also be seven. And so, there’s our answer. It’s like the same question to “How many tries is needed so that two batteries excluded from Iron B both don’t work?” They’re the same thing; all that is needed to solve the problem is to see the connection.
The answer to the bonus riddle is also "7". we'll use the same method, but somehow oppositely since we now use 6 batteries and leave 2 every try. So we exclude two batteries (say A&B) from the tries. (So basically pin them in every try). and divide the rest into two groups (CDE - FGH). Then in every try in each group we will insert 1 battery from the Group and exclude 2. Group (CDE) Tries: 1. ABCFGH (if failed then D&E have at least 1 working) 2. ABDFGH (if failed then C&E have at least 1 working) 2. ABEFGH (if failed then C&D have at least 1 working) and if the three tries all field > then C,D&E have at least 2 working) ==>1 same goes for Group (FGH): 1. ABCDEF (if failed then G&H have at least 1 working) 2. ABCDEG (if failed then F&H have at least 1 working) 2. ABCDEH (if failed then F&G have at least 1 working) and if the three tries all field > then C,D&E have at least 2 working) ==>2 So, if if both groups have at least 2 working batteries, and there are total 4 working batteries. then each group have 2 working batteries (2+2=4) then A&B are surely not-working batteries. So the 7th try would be CDEFGH and it will surely succeed.
There is an alternate solution that also takes 7 tries. Divide the batteries into 4 pairs. 1. Test AB 2. Test CD 3. Test EF 4. Test AG 5. Test BG 6. Test CH 7. Test DH It will work every time.
This is actually the same solution phrased differently. Three groups in your solution are ABG, CDH, EF. You also have a different order for testing than shown here but that doesn't actually matter.
I had a different solution. That seems more efficient First split into 4 piles of 2. Try each pair. Assuming it fails on all, we know that for each set of 2, exactly one was a bad battery. Then we pick 2 of these pairs, and try every other combination with them that wasn't tried. So if AB-CD was tried, we try AC-BD, and AD-BC. One of those is guaranteed to be working. That gives a solution in at most 6 guesses. We don't even need 7th.
Actually, that's 8 guesses. AB, CD, EF, GH, AC, BD, AD, BC. You are correct that one of those is guaranteed to be working, but if it's BC, then you got it on the 8th try and have failed the puzzle in the worse case.
I don't remember what led to this mistake, but I think you are right, was likely wrong at that time. I think I viewed AC-BD as one test, and likewise with AD-BC
Bonus riddle is 7 tries, because it’s the same riddle in a different way. Original riddle says: pick 2 you put in, and if they are both good, you win. Bonus riddle basically says: pick 2 you don’t put in, and if they’re both bad you win(since that means there’s 2 bad batteries and 4 good ones in the iron, thus making it work). You just put in the following sets: CDEFGH, BDEFGH, ADEFGH: if all fail, there can’t have been a combination of 2 bad batteries between A, B and C, thus meaning at least 2 are good. Next, we test out ABCFGH, ABCEGH, ABCDGH, and if all fail again, we know that D, E and F have no 2 bad batteries in them, so at least 2 are good. That means the combination ABCDEF will always work, since ABC has 2 good ones, DEF has 2 good ones, 2+2=4, and the iron requires exactly that many good batteries. Now here’s hoping none of those batteries died during that testing phase...
I think the answer to the bonus riddle is 3. Say we have batteries a-h, first we try a-f, if it doesn't work, we try d-h, if that doesn't work, we try a,b,c,f,g,h. It must definitely work in one of these three tries.
You could do it like this (Bonus answer mentioned too) - First try to put the first two batteries, 1 and 2, in the iron. If it doesn't work, one or both of them are faulty. Try with 3 and 4. Also put 5 and 6 inside. If the light didnt work for all 3 of these pairs, there's only two possibilities: A. Three batteries were faulty, one among 1 and 2, one among 3 and 4, and one among 5 and 6. This means one among 7 and 8 is faulty. B. We've already gotten past all four faulty batteries. Two of the three pairs have one bad battery, and both of the last are bad. Maybe there was one bad battery among 1 and 2, one among 3 and 4, and BOTH 5 and 6 were faulty. This means both 7 and 8 are good. Take the first of the last two batteries, number 7. Test it against 5 and 6. If it doesnt work, it could be because 7 was the bad battery in possibility A, meaning 8 is the good one. It could also be because 5 and 6 were the bad ones in possibility B, meaning 8 is still good (because 7 and 8 were both good). We've now used up 5 tries - 1 and 2, 3 and 4, 5 and 6, 7 and 5, 7 and 6. We use the last two on 8. We know 8 is good, and one among 1 and 2 is good. We test 8 with 1, and finally 8 with 2. The iron turns on, leaving you with just enough time to slip out of a window and look back as the un-offed iron burns the giants' house down. The bonus answer is still 7. If you use 6 of the batteries, you have 2 left in hand. Both cant be dead, because then the 4 live ones are in the iron. Either they're both alive, or one's alive and one's dead. Say they're both alive. There are 4 dead and 2 alive batteries in the torch. Say it's like this - A D D A D D We can group them in 2s (A D) (D A) (D D) Since there are 4 Ds and 2 As, one group is guaranteed to have two Ds. We just replace each group with what we have in hand. Step 1: (A A) (D A) (D D) - Replaced A D with the A A in hand. We now have A D. Step 2: (A A) (A D) (D D) - Replaced D A with the A D in hand. We now have D A. Step 3: (A A) (A D) (D A) - Victory! Say it's a dead-alive pair. There are 3 dead and 3 alive batteries in the torch. Say it's like this - (A D) (D D) (D A) As long as there's a pair of Ds, the same tactic as before will work. But what if it's like this? (A D) (A D) (A D) We'll do the same thing. Step 1: (D A) (A D) (A D) - Replaced A D with the D A in hand. We now have A D. Step 2: (D A) (A D) (A D) - Replaced A D with the A D in hand. We now have A D. Step 3: (D A) (A D) (A D) - Replaced A D with the A D in hand. We now have A D. If we haven't won after step 3, there's no (D D) pair. Now, we just have to test both the batteries in our hand against both of the last pair. The possibilities are D D, D A, A D and A A, so it takes 4 more tries, bringing the answer to 7.
For the bonus riddle, last case: Won't the total number of trials be 8? 1(initial)+3(replacing each group inside iron)+4(getting a working pair from 2 DA pairs)
@@JMCJollyRoger You begin with nothing inside the iron. Suppose the first 6 batteries you take already had 4live ones. This is the initial step. You assumed that 6 batteries were already inside the iron and iron was not working, which is incorrect assumption.
The answer to the bonus riddle is the exact same but in inverse: Now you have to find two dead batteries instead of full ones. At least one of the groups of 3, 3 and 2 batteries has to have 2 dead batteries, so you test them the same way by making sure the 2 batteries left out are always of the same group, and you should get the iron working with 7 tries or less.
You would better drop the battery because dead battery bounces more than the good battery. In this way you would easily find it out. It also takes only 6 tries or less.
Except two things. One, how are you going to drop them? And two, the batteries are way larger than normal double A batteries so they probably don’t bounce the same. Try bouncing a car battery, it doesn’t work
I think I got the answer! Divide the batteries 3,3,2 then try the first 3 (all possibilities are 3) and then the next 3 (same, there's 3 possibilities) and if that didn't worked then try the remaining 2 (should be both good considering the groups of 3 not working!!)
No it really isn't, aside from the fact that this puzzle has four dead batteries and the temple one only has two possessed explorers in the temple riddle, you don't need to figure out exactly who's lying, just the correct way out with the two possessed ones making it more difficult.It came down to figuring out who to trust and either going with majority or process of elimination. The battery one, on the other hand, requires you to find exactly two working batteries out of the eight and you need to make sure both are working. It comes down to you having to make sure you manage to, with certainty, be able to pick out two batteries, both working. Also in the battery one, you can test different pairs while, in the temple one, you only have enough time for one trip out, one trip back and one more in what you think to be the right direction
There's a bit of a mix-up here. The narrator says at 3:08 that we are required to find _the number of tries to get the iron working._ The answer to this question will be 7. However, the solution to this bonus riddle at brilliant.org/weekly-problems/2018-11-26/intermediate/?p=2, is actually the solution for a slightly different question- "In the worst-case scenario, _how many tests do you need so you know of a particular set of batteries which will get your device working?"_ We need 6 tries to find the set of batteries that will get the iron working and 7 tries to get the iron working. So, the answer to this particular riddle, given in this video is 7 (and not 6).
I came up with a solution different from TED. It involves splitting the 8 batteries into 4 pairs like some others' solutions below, but where they test all 4 pairs first, their solution at worst case requires 8 tests. I figured out how I could do it in 7 guaranteed by testing 3 pairs first, then *not* testing the fourth pair. So label the pairs A, B, C, and D, with D being the one not tested. If none of the first 3 pairs work, then that means they all have at least 1 dead battery. With 4 dead batteries, that means D has at most 1 dead battery, therefore at least 1 working battery. If D has 1 working and 1 dead, then that means it is true for all pairs. If D has 2 working batteries, that means all 4 dead batteries are in the first 3 pairs, which via pigeonhole means 1 of those pairs has 2 dead batteries (the converse is similarly true). For tests 4 and 5, test battery D1 against both of pair A. If the iron still does not work, then either D1 doesn't work (making all 4 pairs 1 working/1 dead), or D1 is working but both of A are dead (therefore both of D are working). Either way, it proves D2 is a working battery, and both pairs B and C have 1 working/1 dead battery. Tests 6 and 7 will be testing D2 against both from pair B (or C as it doesn't matter which). Since D2 is proven working and one of B has to be working, it is guaranteed by test 7 at latest the iron will be working.
Is the same solution but you arrive to it after a different though process. In both cases you divide in the groups A1, A2 and D1; B1, B2 and D2; C1 and C2. Still, great job getting to the solution.
There is another ways. First, you make a four pairs from the batteries. Then, you try each one of the four pairs, if non of them work, you know that every pair include one working battery and one unworking battery. You take two pairs and do all the possible combinations with them. Thats it.
That will take 10 tries... Unless you only take the batteries from different pair and test against each other after you know theres 1 good each, but that will still take 8 tries total. Giant is hence angry and you are dinner 🍽
Came up with a different solution. Started off similarly to the video. Pick 3 batteries and test the possible combinations. 1.) A&B 2.) A&C 3.) B&C If none worked, then there were at least two bad batteries and, at most, one good battery. The remaining 5 batteries have either 3 good batteries and 2 bad batteries or 4 good batteries and 1 bad battery. Test pairs of the remaining batteries (there should be one unpaired battery). 4.) D&E 5.) F&G If neither worked, then both pairs each have a single bad battery, and the last, unpaired battery (H) is a good battery (since there were at most 2 bad batteries remaining). Test the last battery with each of the batteries in one of the pairs. 6.) H&D 7.) H&E One of these two must work.
6 tries are sufficient here. The only was that "AB CD EF GH" doesnt work is if you have managed to create pairs of 1 working and 1 dead batteries. If "AC" doesnt work, it means you picked dead batteries from each pair. Therefore, BD must work. 6 tries needed only.
Doesn't always work. Good batteries: B C E G Bad batteries: A D F H AB - 1 good, 1 bad CD - 1 good, 1 bad EF - 1 good, 1 bad GH - 1 good, 1 bad AC - 1 good, 1 bad BD - 1 good, 1 bad AD - 2 bad Sorry.
I love these riddles and the feeling you get when you solve them... What a rush. I believe people give up on themselves too quickly when attempting to solve these; don't get me wrong, they're fun to watch, but there's an extra layer of excitement when you're able to solve them. I've discovered that I am only able to solve these sorts of problems in my head. It might sound whacky, but I encourage you to try it. On paper it's easy to get hung up on the battery combinations themselves, because that's the normal approach. In your head, the problem is less stable and you can actually picture the scenario. There's one other advantage; when solving on paper, you are inclined to attempt to dive into every detail; in your head, you are more able to see the bigger picture and think conceptually. I can't be the only one this is true for. Try it out and see if you that works for you. It sounds counterintuitive, but give it a try. Find a nice relaxing spot, close your eyes, and use the power of your mind to figure out the riddle. 🤔
@@GabriTell I understand feeling that way, but I don't think it's true. You solved it! I've been solving riddles and puzzles for a long time, so I've learned some tips and tricks, both from the riddles I've been able to solve and the riddles I haven't been able to solve. And I've gotten a lot faster over time. The more riddles you solve the faster you'll get.
There is one more way.. First we divide the batteries in four groups of 2 batteries in each Then we test all the four groups and if non of it works then it means Each group has one working battery then we select any two group an test with one battery from both the groups If that doesn't work then the remaining two are the working once In this method there is a 70% chance that we can find the working pair in less than 6 tyres
That does not work. Suppose that you know that AB and CD are pairs with one good and one bad battery. If you test the pairs AC, AD and BC last, then I can make you wrong by having B and D be the good batteries and A and C be the bad batteries. Note that if a pair fails, either both of them are bad or one is good and one is bad.
Lol the bonus riddle is exactly the same. Each time you test 6 batteries it will work only if the 2 remaining ones are empty! So... ORIGINAL: Finding 2 out of the 4 charged batteries, in a pool of 8 BONUS: Finding 2 out of the 4 empty batteries, in a pool of 8
*BONUS RIDDLE:* Testing 6 batteries to see if there are 4 good ones in the group is equivalent to test if the two batteries you left out are bad ones. You need BOTH of these batteries to be bad in order for the iron to work. The method of testing is indirect: in order to test two batteries, you have to put the remaining 6 in the iron. If it does not light up, it means that at least one of the batteries you left out is a good one. With this in mind, we can deduce there is a big similarity with the first riddle: you can test two batteries each time, however, the iron will work when the tested ones are both bad (opposite to the first riddle, where they have to be good). Finally, we can come to the conclusion that given that there are 4 good batteries and 4 bad ones, the number of tries it takes in both riddles is the same: 7. (In order to imagine this better we can think of a ghost iron that only works with 2 bad batteries. Following the indirect method of testing described above, the riddle becomes identical to the original one) *Edit: Reading through the comments I found @Swaroop Pandit explained more or less the same thing before I even wrote this, so full credits to you as well man ;)*
I believe the solution to the bonus riddle to be 4(worst case scenario) and 2 on average (calculating all possible outcomes). Here's my reasoning: Make 2 piles of batteries with 4 batteries in each pile. We know there are 4 Good Batteries (G) and 4 Bad Batteries (B) so our piles would look like one of the following possibilities: A) (Pile 1) GGGG (Pile 2) BBBB; B) (Pile 1) GGGB (Pile 2) BBBG; C) (Pile 1) GGBB (Pile 2) GGBB; We would insert all batteries from one pile into the Iron and divide the other pile into two subsets of two (e.g. case A Pile 2 subset 1 BB, case A Pile 2 subset 2 BB), and try both subsets of Pile 2 with Pile 1 (2 tries in total). If the Iron doesn't turn on after the first attempt, take out all batteries, put in all batteries from Pile 2 and make subsets of 2 from Pile 1 and insert them into the iron(2 more tries). It is guaranteed that by the 4th try the Iron will turn on. Here are all possible scenarios: For case A): a)We initially pick Pile 1(GGGG) and the first half of Pile 2(BB) and the Iron will turn on. b)We initially pick Pile 2 and put the first half of Pile 1 in the iron (BBBBGG - 1st try). Take out first half of Pile 1 and put in the other (BBBBGG - 2nd try). The iron didn't turn on so remove Pile 2 from the iron and insert Pile 1 into it together with the first half of Pile 2(GGGGBB - 3rd try). The iron will turn on. For case B): a) We initially pick Pile 1(GGGB) and insert the first half of Pile 2(BB) (GGGBBB - 1st try). The iron doesn't turn on so remove the first half of Pile 2 and insert the last half of Pile 2(BG) (GGGBBG - 2nd try). We have 4 batteries so the iron will turn on. b) We initially pick Pile 2(BBBG) and insert the second half of Pile 1(GB) (BBBGGB - 1st try). The iron doesn't turn on so we remove the second half of Pile 1 and insert the first half (GG) (BBBGGG - 2nd try). The iron doesn't turn on so we remove Pile 1 and insert Pile 1 together with the first half of Pile 2(BB) (GGGBBB - 3rd try). The iron doesn't turn on so remove the first half of Pile 2 and insert the last half of Pile 2(BG) (GGGBBG - 4th try). We have 4 batteries so the iron will turn on. For case C) (both piles are identical so there will only be one subcase): We pick Pile 1 (GGBB) and insert it into the Iron together with the second half of Pile 2(BB) (GGBBBB - 1st try). The iron doesn't turn on so we remove the second half of Pile 2 and insert the first half of Pile 2 (GG) (GGBBGG - 2nd try). The iron turns on as we have 4 batteries. We can notice that subcase b) always contains subcase a). Since testing any subcase takes 2 tries, and if we always end up with the worst case scenario, it can take a maximum of 4 tries. If my solution isn't good or isn't optimal, I'm sorry but this is all I could come up with. Would love to see some better solutions!
The only problem with your solution is that in case C, it's not guaranteed that the batteries will be in the order GGBB/BBGG. They could be GBGB or BGBG. In that case, it will not turn on. It's a really good idea, though!
The bonus riddle is same as the first one... Just think of it this way: You need 4/6 working meaning you need to sort out the two which aren't working. Now, divide them up in the same format as the solution to the first one... 3,3,2... And whenever he chooses two batteries for the first one, you choose these to be left out instead of plugged in.. So in short, you are looking at finding 2 incorrect batteries that need to be removed from the mix.. makes the number of tries same as that in the video... 7
“you can’t rely on luck, you need to assume the worst possible outcome and plan accordingly” this is exactly what i’ve been saying about how governments should’ve handled the pandemic. actually that applies to any situation where people could be hurt or die.
6 tries (at most) solution: 1. Divide them into 4 pairs and try it one pair at a time. 2. If all 4 pairs failed, there is only one good battery in each pair. 3. Swap any 2 batteries in 2 of the pairs. 4. Since there are at least 2 good batteries in one of the pair. Done! Example: Try 1. AB 2. CD 3. EF 4. GH If all failed, swap B & C. 5. AC 6. BD
The bonus riddle is simpler than it seems. If you need four good batteries in the iron, then what you are really doing is choosing two batteries NOT to put in, and you need those two to be DEAD. It's an identical problem in reverse. Follow the same procedure, except leaving out the two batteries you've chosen.
For the last riddle, easy! I take my newly sorted batteries and put them in. Only took one try. (I had time to sort them while the party was going on and it didnt say the baby messed it up or this was while the party was going on)
So, I made a bit of experimentation for the last riddle. If I'm not wrong, if an electrical component is well fabricated, it only is current that makes it work. If you want to turn on an LED with 220 volts, but it only runs with 2 amps, you must have the apropriate resistor for the LED to work correctly. The only way you can make the iron work by placing only 4 batteries, even though it has 6 available slots, i that you connect all batteries in a parallel circuit. The total voltage across the entire circuit, assuming that all batteries produce the same voltage, is the voltage of one battery. If all batteries produce n volts, the net voltage is also of n volts. Furthermore, if only one battery has a different voltage, the circuit will overheat and one of the main batteries, first or last/ top or bottom, will explode do to the amount of current. Actually, for no battery to explode, all batteries should have the same voltage. Also, the first and last batteries are the main ones because of the arrangement: the current through any other battery is considerably smaller than the main ones, therefore, being almost a null contribution for the functioning of the circuit. This makes the iron only needing to have one battery to work, instead of 4.
The first riddle. 1-2 3-4 Now we know that in first four batteries there are at least 2 dead and at most 2 good. 5-6 5-7 6-7 Now we know that there is at most one good battery in (5, 6, 7). So 8 is Good. And in group (1, 2) there is one good battery as well as in group (3, 4). 8-1 8-2
Technically speaking, it's just rearranging the video's solution as you're still ultimately looking at three groups and comparing the combinations within all of them, but moving the group with two batteries to the front of your tries (1,2,8) (5,6,7) (3,4). Still though, the logic makes sense.
Rio C actually i looked at this closer and I deemed it to be a fundamentally different solution as both the “connectivity of the groups” you are testing is different and the deductive logic used is different.
@@madisonking8057 The deductive logic to get this set of pairs to be tested is different, but in the end, you are still trying the same 7 pairs, knowing that at least one of them is a pair of good batteries. Different logic, same result :)
You can also divide them in 2 groups of 4. You try all the possible combinations (6) of one of the two groups, and if they all fail you pick two random batteries from the other group
Divide the batteries into groups of 2. In the worst scenario (4 tries) no one works, it means that every group has 1 good and 1 dead. You take the batteries from 2 groups and test them until you find the 2 that works good (worst scenario, 2 more tries). Thats how you can get 2 good batteries in 6 tries or less
Take 4 sets in beginning. It will work with 8 steps.Took time figuring it out had to share it. Step 1 : Check first set . Step 2 : Check second set. Now you know that in first two sets there can be 2,3or 4 dead battery if didn't work . Let's say there were three or four dead ones then checking other two sets guarantee to give one working set.( Step 3 and 4 ) Step 5 : Take one battery out and try it with a battery from other set . Step 6 : Try with another battery of that set . Step 7 : The one you kept in Step 5 is definitely dead toss it out replace with the one you toke out in Step 5. Step 8 : The one you introduced in Step 6 is definitely a good one now. Don't know you include Step 8 as a step . You figured out after 7 tries which were good ones but by then you were food because you might not had time to change batteries.
@@Araqius you are right, my way will give me 8 or less but probability i would need more than 4 turns is 1*4/7*1*3/5*1*2/3*1*1 = 22.8% more than 6 is 11.4% more than 7 would be 5.7% when actual solution from video let you solve more than 6 turns is: (C1/4 * C2/4)/(C3/8) * (C1/3*C2/2)/(C3/5) = 12.8 % i will be ok with my impure 5.7% =)
I did it in 6; Test two, If they don't work, throw them away. The worst is that you threw 1 good and 1 bad, so 3 good and 3 bad left Do this again and you will end up with 2 good and 2 bad the worst Now try all the possible outcomes, In worst, you'll end up in 6 tries.
My solution was this: I take 4 batteries (6 possible pairs). If none of them works, there's only working batteries left and it doesn't matter which two of them I take.
Get the solution to the bonus riddle here: brilliant.org/TedEdGiantIron/! Also, the first 833 of you who sign up for a PREMIUM subscription will get 20% off the annual fee. Riddle on, riddlers!
I watch all of your videos. So please reply.
Cool puzzle and cool solution!
Krishna Kumar You forgot the possibility of GB-GB-GB-GB. It would take a few more tries to see which are good and which are bad.
does the combination need to be one that is tested? Because I came up with a scenario in which the worst case would mean that the remaining combination are the good ones.
the link does not seem to be working, it just redirects me to the homepage
wow if I’m ever in a situation where I have to iron a giant shirt with a giant iron and a bunch of batteries I’ll know exactly what to do
tHaNkS tEd Ed
th-cam.com/video/PkKAHroBTtA/w-d-xo.html
and if there are exactly 8 batteries of which 4 are dead and 4 are good
AND you only have 7 tries for some arbitrary reason
@@georgplaz because the batteries are apparently dungbells.
Royal Lobster 78 don’t self advertise
In the last riddle, you only need to try once because the giant baby didn’t messed it up this time.
lol
Man,I think,""""YOU DESERVE MORE RESPECT THAN TRUMP FOR THIS ANSWER!!!!!!!!!!!!!!""""""
yeah!
Well, I think of that too. LOL 2000 IQ
😂🤣😁😆😄😃😀😀
"Or else you'll end up being the main course tonight"
*Looks at the size of the manager*
If that's the main course, I think the giants are on diet
@Otaku potato
LOL
Actually,the "diet" you are thinking of is called slimming, where people eat less to lose weight. Using "calorie-controlled diet" is more appropriate. And yes,I am a huge nerd.
Ikr
They got to look fly for that party after all.
LOL
Nobody:
Ted-Ed Riddle Master: You die if you can't solve this riddle.
Fair
Not funny
Exactly what would happen in real life
(This is a joke, don't take this seriously)
It would take way less effort to roll the batteries than carry them on your back, you know.
hey, you're not s'pposed to use logic and common sense on these bogus riddles!
bla bla they are good riddles, because they are hard. But to create a good riddle, you need to create a difficult situation, hence the restrictions.
@@chrisding1976 maybe...
ZHBG 9224 bfdi profile picture
I DIDNT KNOW
I love how there's absolutely NO good reason this riddle has to be about a giant family's servant or an iron.
Paradoxic comment
actually, there is: if they were regular size batteries, testing more than 7 pairs would be very easy, it takes very little time to do so. But by being giant, moving them takes a lot of time, and thus you are limited to 7 tries only
@@silvanobianchini6924 i will still put all my power for the 8th try cuz i dun wanna be someone's food
But also, just to make the pun Giant Iron.
I saw a video from someone else with the same riddle about an ordinary torch
me watching knowing full well i can never solve any of these damn riddles
I agree
Me too sis
I manage intuitively or not at all
A couple of them have been solvable. this aint one of them though
I just watch these videos so I can give the riddles to my family.
This riddle seems quite simple when you first hear it, but turns out to have a much more elegant solution than expected. That’s what makes it a superb riddle.
Just do the old "lick the battery to see if it zaps you" trick
its giant battery the shock might just kill you on the lick 👅
@@solitarystride That's fair
@@PyroGothNerd yes
@@solitarystride well, it's die either way
@@emiliatan5792 True everyone dies sooner or later 😉
Gotta give props to Ted Ed for coming up with very well justified back stories for all of their riddles
Just plug the iron to the electric socket.
kirby march Barcena nah, the plug is too heavy
I’m going to do bouncing life hack but dang, I can’t make these big o’ battery bounce
wahhaj musthafa Loosen your Aspect Corset to make ant physics stronger
But it has no power cord...
Doesn't have a cord :(
Girl, who would serve that tiny of a human "as a main course" to those giants? smh.
Clearly he would be the garnish ON whatever their main course was.
Stratelier Do you know what you’re saying
You dont know. You've never eaten a human.
I hope
I’m pretty sure that the human *is* a garnish.
DJ Tile Turnip That’s where you’re wrong.
Why don’t u bounce the batteries to see if they good or not
Because that won't work!
That's what I was thinking too
shoaib kayani why??
How do you propose to bounce these batteries that are as tall as you are in a way that allows you to still get the shirt ironed on time?
I am not so strong to bounce batteries
I think we’re missing the point here.
If I don’t iron his shirt, my boss will eat me?
The solution is clear: unionise and overthrow the tyrannical employer!
but its just you
but how will you do that, if even his baby is 3 times bigger than u?
George Quilitz that only makes unionising easier 😉😂
Paul Dominique Schal I admit, I may be oversimplifying the complexities of the political-economics of fantasy universes.
I had read unionise as un-ion-ise and kept wondering where they displayed the ionising gun in the puzzle😅
Step 1: Ask the batteries if you have green eyes
Step 2: Ask the giant baby if you can leave
Ratin correct
You forgot to ask them “if you’re powered, would you say ozo?”
@@bonyfax2345 lol
And all boxes with an even number must have a red lid ontop
Ratin just stop everyone is doing this like bruh we don’t care
After about 3 minutes, this video started sounding like "if you have 4 good batteries and battery on the giant iron, then your giant iron battery in the good battery pile, would only be 2 possible combinations of the giant battery. And if you remember, the 1 bad battery with the testing all the good batteries would take too long for the good batteries to a giant testing"
starvalentinobeatz 8 possible tries, so no
I have a little question. Imagine:What if it were different?
@starvalentinobeats
1.when...
on the first 3 battery pile, you find out that the iron works? Then there must be 2 batteries work, then you test again on 3 possible combinations.
The other three is the same.. When it does live, we know we need to test other 3 possibilities to know who is living.
Summed up, There we have already 6 tries. And then how is the other two? We have acquired two living batteries or even more in the other piles. Then we could know, what fate shall this energy source have.. How do you find it?
2. When it were like this video instead, when 3 combinations o.n a pile fail, then you know only one or none is. WORKING? This is the real big question :
BUT HOW DO YOU TELL, WHICH ONE THEN IS THE WORKING ONE from these fail combinations ?
There is some fishhy things on this ridfle..
@@queenofunderland I think you just solved the riddle!!!!
@@queenofunderland i dont quite get what u mean. Can u elaborate?
@@queenofunderland lol u only need 2 batteries.. so if first group is work than udont need to fnd another batteries.. lol
I think the bonus problem is the same as the other but in reverse. We need all the working batteries in the iron, meaning we're looking to have two dead batteries out of the iron rather than two working ones in it. If you divide them into the same groups, you can put one group of 3 and the group of 2 into the iron leaving you with one space. Test each from the other group of 3 in turn and if it doesn't work that means that that group contains at least two good batteries. Swap them out with the other group of 3 and repeat. If you get the same results then the same must be true and there are at least two good batteries in that set meaning that the group of 2 will both be dead.
They haven't been messed up this time
Brillant solution. Tho i still don't undertsand why the iron needs 6 batteries if it only need 4 to work, they could just have it be 4, it would end up being the same
@@kentlab3850let’s see if I can help.
Consider the original problem, but change to logic to where the iron works if any of the two batteries are good, and we’re trying to find a pair of dead batteries. It’s not hard to see that this doesn’t change the solution, we’re still trying to find a pair of identical batteries. So we do the same process of splitting the batteries into two groups of three and a group of two, and test each pair within the groups of three. Heres the final connection: testing whether a pair has 2 dead batteries is the same as testing whether the remaining 6 batteries have 4 good ones. Every test with the 6 battery iron is functionally the same as the test with the 2 battery iron, just with the charged reversed. That’s why it needs 6 slots.
Also here’s another answer I thought of that’s probably simpler:
In both the original riddle and the bonus riddle, you need to find a pair of identical batteries. In the original you need to find 2 good batteries, in the bonus you need to find 2 dead batteries. If the iron had 4 slots and needed 4 good batteries, that would require you to find all 4 identical batteries, which is a harder problem than only finding a pair.
@@xavierburval4128 that's an old comment so i'm not sure about the details but i think i was joking about the fact that having extra space for dead batteries in the iron is a bit silly.
It wasn't really about the math since the problem with 4 batteries space would be pretty uninteresting if i recall properly
@@kentlab3850 ooooh I see what you mean now.
I did it a different way.
1. Test A&B
2. Test C&D
3. Test E&F
If none worked, now you know that G&H has either one or two good batteries. If one, then each other pair also has one good one. If both G&H are good, then only one of the other pairs can have two bad ones.
4. Test G&A
5. Test G&B
If A&B were both bad, C&D must contain a good one and also H must be good. Or, in the case that each pair had one bad one, G was bad, so again, H would be good and C&D must contain a good one. So finally,
6. Test H&C
7. H&D must work if none of the other tries did.
I came up with same logic. Both of them are correct & in fact, the combinations are similar, just arranged & explained differently. Two groups of 3 & remaining 2: Group 1(A&B, G&A, G&B), Group 2(C&D, H&C, H&D), E&F.
What happens if G&H are both good? You need another go to test it
If G&H are both good, that means that the other three pairs must be distributed like so (or else one try would have worked), although not necessarily in this order:
good & bad
good & bad
bad & bad
So when you test G with A and then G with B and neither work, A&B must have been the two bad ones. That means that C&D must contain at least one good one.
So in the case that G&H are both good, H is good and either C or D must be good if you got that far.
i tried this way, but you need a maximum of 8 tries in the worst case scenario
What distribution needs 8 tries?
TedEd: *”Can you solve the **_________** riddle?”*
Me every time: Nuh.
"giant iron"
And yet I watch every video
"or you'll end up the main course"
Boi we are like smaller then their fingers,
ya lol XD
There's another solution:
Divide batteries into 4 pairs. Check 3 of the pairs. Check 1st remaining battery with both batteries from one of the pairs. Check 2nd remaining battery with both batteries from a different pair.
I'll let you figure it out yourself why it will always work.
it took me 5 minutes to verify this solution, but it's perfectly valid for me, thank you
@@GordonHugenay Glad to have helped. :-)
edit: i actually realized this was the same solution as the TED-ED one, just rearranged in groups and order. whoops.
this works :D it's only the 2nd valid solution I've come across so far, the TED-ED one and this one
i came up with the same solution, and i know why it'll always work
how long did it take you to come up with this? took me 45min-1hr. took a lot of wrong turns and was actually pretty close to the solution for the while but never put two and two together into the solution
piotrkakol1992 but you will lose a chance of getting it right in 4th try itself by not testing the last pair of batteries as both the remaining batteries can be good ones
In the end you are doing the exact same thing only trying the same combinations in different order.
I prefer the iron giant riddle
Ok
Hi
@@buddyclem7328 I love the doctor.
Doctor Who?
Exactly.
Quit watching just2good I see
Why
Solution:
Throw away the batteries right when they die.
Simple.
Or buy rechargeables and recharge them immediately when they die
They might be saving them to take to the recycling depot. It'd be way too much trouble to make a trip there every time a battery dies.
NoriMori, but put them in a cup? And put that cup right next to the cup with the working batteries?
Yeah, and let’s not forget that they were in a baby’s reach
@@colonelcorneliuscornwall3547 that's the reason this mess happened anyway
Thanks! Now I’m prepared if I ever get stuck in the COMPLETELY LIKELY situation of being the butler for a giant.
Why Giants do not hire a Giant Butler??🤔
maybe giant butlers are too expensive
Because giant butlers are too lazy
I dunno just quit the job..
But then you'll lose your spectacular dental and retirement plans!
Get eaten
But they pay a lot
What is it’s really just slavery and the giants call it a job to make it sound better
rio ranger!
For the Bonus Riddle, the answer is exactly the same as the regular riddle, just mirrored in a way. Instead of testing for 2 good batteries to put in the iron, you have to search for 2 bad batteries to leave out. Since good and bad batteries occur equally, you do the exact same grouping and testing pairs strategy as in the first riddle, only you leave the pairs out instead of putting them in.
@@HT-vd4in it’s the same, the information you get is just the opposite, if it works, it means the two you left out is dead
If it doesn’t work, atleast one of the batteries you left out is working
You’re just testing for dead batteries to leave out instead of working batteries to use
How exactly are you doing this?
Tell me step by step?
@@HT-vd4in you number the batteries from A to H
step 1, insert batteries but leave AB out if it doesn’t work at least of AB is a working battery
Step 2, insert batteries but leave BC out if it doesn’t work at least one of BC is a working battery
Step 3, insert batteries but leave AC out if it doesn’t work at least one of AC is a working battery ( so among ABC at least 2 are working )
Step 4, insert batteries but leave DE out if it doesn’t work at least one of DE is a working battery
Step 5, insert batteries but leave EF out if it doesn’t work at least one of EF is a working battery
Step 6, insert batteries but leave DF out if it doesn’t work at least one of DF is a working battery ( so among DEF at least 2 are working )
By now you would have checked 6 times, and you can conclude that GH are both dead batteries if the iron hasn’t worked yet as among ABC two of the batteries are working and among DEF two of the batteries are working
So for the 7th step you just insert ABCDEF and leave GH out, this will guarantee that the iron will work if it hasn’t yet after step 6
@@derekhansj thanks. Wow 🤩
Alas, I thought this was a straightforward one, but I failed to solve it. I'm embarrassed.
I think I got the bonus riddle though. It's actually the same exact problem as the original riddle, but in reverse. In the original riddle, you're choosing which sets of 2 to put in, whereas in the bonus riddle, you're choosing which sets of 2 to leave out. In the original riddle, you're looking for 2 good ones (to ultimately put in), whereas in the bonus riddle, you're looking for 2 dead ones (to ultimately leave out). So the answer is still 7 tries, using a similar procedure.
I just came to this conclusion as well. And I was working so much extra math and logic to this bonus problem at first.
You guys didn’t have to be embarrassed because everyone who clicks on these riddle videos knows damn well they might not be able to solve them.
@@gayaraaukwaththa9170 I'm not even trying to find the answer myself at this point lol
You can’t test a singular group of 3 batteries when it requires 4 so how would u do it
@@soprecise7993 You still can group them in the same way: ABC, DEF, GH. However, when you are "testing" AB, you actually put CDEFGH into the iron. Therefore, the bonus question is the same as the main question, just that you need to see it from another angle.
“Has six battery slots”
*shows 2*
Batteries can stack like flashlights
I was thinking about making 4 random pairs and trying them all, then if it doesn't work, you know there's one good battery in every pair, so if you grab two of the pairs you can find the right two, but that can take up to 4 more tries, which makes a total of 8.
Rachelle, you’re bang on except you’ve forgotten that, if the “up to 4 more tries”, you already know the result of two of the tries because you trialled them when you tested out the four pairs at random. This is my explanation:
4 good batteries and 4 bad batteries, right?
Separate them into four random pairs and test all 4 pairs. If one of the pairs makes the iron work, great!
But if not (you’re 4 tries in at this stage), that means that each pair consists of one good and one bad battery.
Take any two pairs at random and discard the other 4 batteries. You now have 4 batteries of which 2 are good and 2 are bad.
Swap the pairings and try the first pair - either you’ll now have matched up the two good batteries and the iron will work (after attempt 5) or you’ll have matched up the two bad batteries and it won’t work, so you then know the other two batteries are both good and it’ll work on try 6!
Edit: actually on those 5th and 6th tries, I’ve realised it’s possible to, again, pair up a good battery with a bad on both occasions, but if this happens, you just pair up the two batteries that haven’t been paired up yet and those will be two good ones 🙂
you can get it with 3 more tries, If you checked (1/2) and (3/4) in part one, you can continue with (1/3) and (2/4) and (1/4). If non of them work you only have (2/3) remaining and it must be the right one.
@@raptagames I thought so too, unfortunately it must power up in 7 tries, not just know the answer in seven.
How many of you are here just to find answerof the bonus riddle??
At least 3
the answer is seven
6 is the answer
@@jackwuchannel 6 bud
@@xiaoshen194
REASONING:
It’s surprisingly simple.
Let’s get rid of “good” and “bad” for a second and just have battery types “A” and “B”, four of each.
Now we have our two irons, one from the main riddle, and the other from the bonus riddle. Let’s say Iron A has 2 slots and requires 2 A batteries, while Iron B has 6 slots and requires 4 B batteries. Now imagine putting two A batteries into Iron A, making it power on, and putting each of the 6 other batteries in Iron B. Because Iron A has no B batteries, all 4 B batteries go to Iron B, making it turn on. What if we put at least one B battery into Iron A? In that case, Iron A would fail to start, since it doesn’t have 2 A batteries. And because there is one B battery in Iron A, there is 3 or less B batteries remaining to put in Iron B; not enough to power it on. So, in conclusion, whenever Iron A is on, Iron B is also on, and whenever Iron is off, Iron B is also off. That makes the main riddle and bonus riddle essentially the same question. Now for the question: What is the least amount of tries do you need so that (iron) has 4 Type B batteries in its 6 slots? This is the same question as “What is the least amount of tries do you need so that (iron) has 2 Type A(Good) batteries in its 2 slots?” The answer to this question is seven. Need proof? Go to 0:55. (It’s probably not possible to do it in 6 tries or under). And since the answer to that question is seven, the answer to the Iron B question, or the bonus riddle, must also be seven. And so, there’s our answer. It’s like the same question to “How many tries is needed so that two batteries excluded from Iron B both don’t work?” They’re the same thing; all that is needed to solve the problem is to see the connection.
Step 1: Confirm you have green eyes
Step 2: Ask to leave because you’re too overwhelmed to solve the riddle
madlad
@@isaaclewis8694 ya
The answer to the bonus riddle is also "7".
we'll use the same method, but somehow oppositely since we now use 6 batteries and leave 2 every try.
So we exclude two batteries (say A&B) from the tries. (So basically pin them in every try).
and divide the rest into two groups (CDE - FGH).
Then in every try in each group we will insert 1 battery from the Group and exclude 2.
Group (CDE) Tries:
1. ABCFGH (if failed then D&E have at least 1 working)
2. ABDFGH (if failed then C&E have at least 1 working)
2. ABEFGH (if failed then C&D have at least 1 working)
and if the three tries all field > then C,D&E have at least 2 working) ==>1
same goes for Group (FGH):
1. ABCDEF (if failed then G&H have at least 1 working)
2. ABCDEG (if failed then F&H have at least 1 working)
2. ABCDEH (if failed then F&G have at least 1 working)
and if the three tries all field > then C,D&E have at least 2 working) ==>2
So, if if both groups have at least 2 working batteries, and there are total 4 working batteries.
then each group have 2 working batteries (2+2=4)
then A&B are surely not-working batteries.
So the 7th try would be CDEFGH and it will surely succeed.
There is an alternate solution that also takes 7 tries.
Divide the batteries into 4 pairs.
1. Test AB
2. Test CD
3. Test EF
4. Test AG
5. Test BG
6. Test CH
7. Test DH
It will work every time.
This is actually the same solution phrased differently. Three groups in your solution are ABG, CDH, EF. You also have a different order for testing than shown here but that doesn't actually matter.
one of the rare riddles I’ve solved here
I had a different solution. That seems more efficient
First split into 4 piles of 2. Try each pair.
Assuming it fails on all, we know that for each set of 2, exactly one was a bad battery.
Then we pick 2 of these pairs, and try every other combination with them that wasn't tried. So if AB-CD was tried, we try AC-BD, and AD-BC. One of those is guaranteed to be working.
That gives a solution in at most 6 guesses. We don't even need 7th.
Actually, that's 8 guesses. AB, CD, EF, GH, AC, BD, AD, BC. You are correct that one of those is guaranteed to be working, but if it's BC, then you got it on the 8th try and have failed the puzzle in the worse case.
I don't remember what led to this mistake, but I think you are right, was likely wrong at that time. I think I viewed AC-BD as one test, and likewise with AD-BC
"Helps you develop your joy of learning. " 3:17 I started laughing so hard!
same
I love how no one takes this logical riddle seriously.
because plenty of them are also dumbasses
3:12 You only need to try once since no one mess it up this time.
I could never solve these riddles and yet they are just so fun to try!
Bonus riddle is 7 tries, because it’s the same riddle in a different way. Original riddle says: pick 2 you put in, and if they are both good, you win. Bonus riddle basically says: pick 2 you don’t put in, and if they’re both bad you win(since that means there’s 2 bad batteries and 4 good ones in the iron, thus making it work).
You just put in the following sets: CDEFGH, BDEFGH, ADEFGH: if all fail, there can’t have been a combination of 2 bad batteries between A, B and C, thus meaning at least 2 are good. Next, we test out ABCFGH, ABCEGH, ABCDGH, and if all fail again, we know that D, E and F have no 2 bad batteries in them, so at least 2 are good. That means the combination ABCDEF will always work, since ABC has 2 good ones, DEF has 2 good ones, 2+2=4, and the iron requires exactly that many good batteries. Now here’s hoping none of those batteries died during that testing phase...
I think the answer to the bonus riddle is 3. Say we have batteries a-h, first we try a-f, if it doesn't work, we try d-h, if that doesn't work, we try a,b,c,f,g,h. It must definitely work in one of these three tries.
@@gowrinarayani247 Your way doesn't work if the good ones are a,c,e,g or b,d,f,h.
You could do it like this (Bonus answer mentioned too) -
First try to put the first two batteries, 1 and 2, in the iron. If it doesn't work, one or both of them are faulty.
Try with 3 and 4. Also put 5 and 6 inside.
If the light didnt work for all 3 of these pairs, there's only two possibilities:
A. Three batteries were faulty, one among 1 and 2, one among 3 and 4, and one among 5 and 6. This means one among 7 and 8 is faulty.
B. We've already gotten past all four faulty batteries. Two of the three pairs have one bad battery, and both of the last are bad. Maybe there was one bad battery among 1 and 2, one among 3 and 4, and BOTH 5 and 6 were faulty. This means both 7 and 8 are good.
Take the first of the last two batteries, number 7. Test it against 5 and 6. If it doesnt work, it could be because 7 was the bad battery in possibility A, meaning 8 is the good one. It could also be because 5 and 6 were the bad ones in possibility B, meaning 8 is still good (because 7 and 8 were both good).
We've now used up 5 tries - 1 and 2, 3 and 4, 5 and 6, 7 and 5, 7 and 6.
We use the last two on 8. We know 8 is good, and one among 1 and 2 is good. We test 8 with 1, and finally 8 with 2. The iron turns on, leaving you with just enough time to slip out of a window and look back as the un-offed iron burns the giants' house down.
The bonus answer is still 7.
If you use 6 of the batteries, you have 2 left in hand. Both cant be dead, because then the 4 live ones are in the iron.
Either they're both alive, or one's alive and one's dead.
Say they're both alive.
There are 4 dead and 2 alive batteries in the torch. Say it's like this -
A D D A D D
We can group them in 2s
(A D) (D A) (D D)
Since there are 4 Ds and 2 As, one group is guaranteed to have two Ds. We just replace each group with what we have in hand.
Step 1:
(A A) (D A) (D D) - Replaced A D with the A A in hand. We now have A D.
Step 2:
(A A) (A D) (D D) - Replaced D A with the A D in hand. We now have D A.
Step 3:
(A A) (A D) (D A) - Victory!
Say it's a dead-alive pair.
There are 3 dead and 3 alive batteries in the torch. Say it's like this -
(A D) (D D) (D A)
As long as there's a pair of Ds, the same tactic as before will work.
But what if it's like this?
(A D) (A D) (A D)
We'll do the same thing.
Step 1:
(D A) (A D) (A D) - Replaced A D with the D A in hand. We now have A D.
Step 2:
(D A) (A D) (A D) - Replaced A D with the A D in hand. We now have A D.
Step 3:
(D A) (A D) (A D) - Replaced A D with the A D in hand. We now have A D.
If we haven't won after step 3, there's no (D D) pair. Now, we just have to test both the batteries in our hand against both of the last pair. The possibilities are D D, D A, A D and A A, so it takes 4 more tries, bringing the answer to 7.
For the bonus riddle, last case: Won't the total number of trials be 8? 1(initial)+3(replacing each group inside iron)+4(getting a working pair from 2 DA pairs)
@@armeetluthra8913 What's the initial step?
@@JMCJollyRoger You begin with nothing inside the iron. Suppose the first 6 batteries you take already had 4live ones. This is the initial step. You assumed that 6 batteries were already inside the iron and iron was not working, which is incorrect assumption.
@@armeetluthra8913 Ah yeah
The answer to the bonus riddle is the exact same but in inverse: Now you have to find two dead batteries instead of full ones.
At least one of the groups of 3, 3 and 2 batteries has to have 2 dead batteries, so you test them the same way by making sure the 2 batteries left out are always of the same group, and you should get the iron working with 7 tries or less.
Good observation.
I puzzled this out all the night and the answer struck me as i was laying in bed!!!!! thanks ted ed for these fantastic riddles :)
Prepare to be a snack for the first time in your life
And last.
Becaus of your profile pic that i thought there is a hair on my tablet
@@fghfhg9599 dark mode ftw
Actually you are the main course
You didnt have to be that savage lmao
*The **_Irony_** in this video is hysterical!*
It fell _flat._
The irnoy is that you tried to make an iron pun in a video with pretty much no irony.
nvm that isn't irony either
That pun tho
Okay... but why hang on to the dead batteries? Just throw them out.
You would better drop the battery because dead battery bounces more than the good battery. In this way you would easily find it out. It also takes only 6 tries or less.
Except two things. One, how are you going to drop them? And two, the batteries are way larger than normal double A batteries so they probably don’t bounce the same. Try bouncing a car battery, it doesn’t work
I think I got the answer! Divide the batteries 3,3,2 then try the first 3 (all possibilities are 3) and then the next 3 (same, there's 3 possibilities) and if that didn't worked then try the remaining 2 (should be both good considering the groups of 3 not working!!)
TED-Ed: Can-
Me: *interrupts*
Me: no I cannot solve that riddle.
As same as the Temple riddle, only it about battries
O ye
Oh
No it really isn't, aside from the fact that this puzzle has four dead batteries and the temple one only has two possessed explorers
in the temple riddle, you don't need to figure out exactly who's lying, just the correct way out with the two possessed ones making it more difficult.It came down to figuring out who to trust and either going with majority or process of elimination.
The battery one, on the other hand, requires you to find exactly two working batteries out of the eight and you need to make sure both are working. It comes down to you having to make sure you manage to, with certainty, be able to pick out two batteries, both working.
Also in the battery one, you can test different pairs while, in the temple one, you only have enough time for one trip out, one trip back and one more in what you think to be the right direction
Forcea Comedy:“Ooo I’m using big brain words bc I’m big brain”
In this bonus riddle the baby giant didn't knock the batteries over, so it would just take one try😂
There's a bit of a mix-up here. The narrator says at 3:08 that we are required to find _the number of tries to get the iron working._ The answer to this question will be 7.
However, the solution to this bonus riddle at brilliant.org/weekly-problems/2018-11-26/intermediate/?p=2, is actually the solution for a slightly different question-
"In the worst-case scenario, _how many tests do you need so you know of a particular set of batteries which will get your device working?"_
We need 6 tries to find the set of batteries that will get the iron working and 7 tries to get the iron working. So, the answer to this particular riddle, given in this video is 7 (and not 6).
This one was easy, but here’s a harder riddle. How does he flip that giant shirt?! 😂
I appreciate these videos. Thanks TedEd
that's a brilliant riddle. So simple, but took me 15 minutes to solve.
I see , you are the guy that solves 4+ years old puzzles in less than 4 years
I came up with a solution different from TED. It involves splitting the 8 batteries into 4 pairs like some others' solutions below, but where they test all 4 pairs first, their solution at worst case requires 8 tests. I figured out how I could do it in 7 guaranteed by testing 3 pairs first, then *not* testing the fourth pair. So label the pairs A, B, C, and D, with D being the one not tested. If none of the first 3 pairs work, then that means they all have at least 1 dead battery. With 4 dead batteries, that means D has at most 1 dead battery, therefore at least 1 working battery. If D has 1 working and 1 dead, then that means it is true for all pairs. If D has 2 working batteries, that means all 4 dead batteries are in the first 3 pairs, which via pigeonhole means 1 of those pairs has 2 dead batteries (the converse is similarly true). For tests 4 and 5, test battery D1 against both of pair A. If the iron still does not work, then either D1 doesn't work (making all 4 pairs 1 working/1 dead), or D1 is working but both of A are dead (therefore both of D are working). Either way, it proves D2 is a working battery, and both pairs B and C have 1 working/1 dead battery. Tests 6 and 7 will be testing D2 against both from pair B (or C as it doesn't matter which). Since D2 is proven working and one of B has to be working, it is guaranteed by test 7 at latest the iron will be working.
Good job!
Is the same solution but you arrive to it after a different though process.
In both cases you divide in the groups A1, A2 and D1; B1, B2 and D2; C1 and C2.
Still, great job getting to the solution.
Amazing riddles! Though i cant solve a single one😂😂😂😅
There is another ways.
First, you make a four pairs from the batteries.
Then, you try each one of the four pairs, if non of them work, you know that every pair include one working battery and one unworking battery.
You take two pairs and do all the possible combinations with them.
Thats it.
That will take 10 tries...
Unless you only take the batteries from different pair and test against each other after you know theres 1 good each, but that will still take 8 tries total.
Giant is hence angry and you are dinner 🍽
I just realized its 8 and not 7.... my bad.
Came up with a different solution. Started off similarly to the video. Pick 3 batteries and test the possible combinations.
1.) A&B
2.) A&C
3.) B&C
If none worked, then there were at least two bad batteries and, at most, one good battery. The remaining 5 batteries have either 3 good batteries and 2 bad batteries or 4 good batteries and 1 bad battery. Test pairs of the remaining batteries (there should be one unpaired battery).
4.) D&E
5.) F&G
If neither worked, then both pairs each have a single bad battery, and the last, unpaired battery (H) is a good battery (since there were at most 2 bad batteries remaining). Test the last battery with each of the batteries in one of the pairs.
6.) H&D
7.) H&E
One of these two must work.
Intuitively, the bonus riddle looks the same: instead of "find 2 good" (to use), it becomes "find 2 bad" (not to use).
I use these videos as background noise for homework instead of solving them
Goodbye now.
AB CD EF GH AC BD AD. In these 7 tries, you'll find at least one combination of 2 good batteries
no, you'll need to try BC as well, in case both A and D are bad
6 tries are sufficient here. The only was that "AB CD EF GH" doesnt work is if you have managed to create pairs of 1 working and 1 dead batteries. If "AC" doesnt work, it means you picked dead batteries from each pair. Therefore, BD must work. 6 tries needed only.
Doesn't always work.
Good batteries: B C E G
Bad batteries: A D F H
AB - 1 good, 1 bad
CD - 1 good, 1 bad
EF - 1 good, 1 bad
GH - 1 good, 1 bad
AC - 1 good, 1 bad
BD - 1 good, 1 bad
AD - 2 bad
Sorry.
@@adriank7799 Can you explain your solution in detail? I don't think it's possible to solve this with only 6 attempts.
I love these riddles and the feeling you get when you solve them... What a rush.
I believe people give up on themselves too quickly when attempting to solve these; don't get me wrong, they're fun to watch, but there's an extra layer of excitement when you're able to solve them.
I've discovered that I am only able to solve these sorts of problems in my head. It might sound whacky, but I encourage you to try it. On paper it's easy to get hung up on the battery combinations themselves, because that's the normal approach. In your head, the problem is less stable and you can actually picture the scenario. There's one other advantage; when solving on paper, you are inclined to attempt to dive into every detail; in your head, you are more able to see the bigger picture and think conceptually.
I can't be the only one this is true for. Try it out and see if you that works for you. It sounds counterintuitive, but give it a try. Find a nice relaxing spot, close your eyes, and use the power of your mind to figure out the riddle.
🤔
this is 100% the case with me as well. And I suppose, this would be the case for most people if they tried
Sht... It took me like 3 days to solve it... I feel like a frcking retərd 😔
@@GabriTell I understand feeling that way, but I don't think it's true. You solved it!
I've been solving riddles and puzzles for a long time, so I've learned some tips and tricks, both from the riddles I've been able to solve and the riddles I haven't been able to solve. And I've gotten a lot faster over time. The more riddles you solve the faster you'll get.
@@AW-xl6hu Thank you :')
I currently took a Brilliant's course because I'm so interested in this! 👍✨
I’M INSANELY PROUD OF MYSELF FOR ACTUALLY SOLVING THIS ONE!!!
Thank you, this information will surely help me
Actually, you can take two peace’s of foil, attach them to both sides of batteries (one by one), and see if it gets on fire.
yay giant fire make me perish
Lick em- that's all you need to do, lick the batteries
wallah t un ouf
awesome djman uea bit you will get a electric shock to help possibly get hurt alot
They are giant batteries so they might end up killing you... right?
@@ameymorshed4744 *_Say what?_*
That only works with D batteries, but I like your style
i love how i saw the batteries and the mixup and immediately thought *lick them*
There is one more way..
First we divide the batteries in four groups of 2 batteries in each
Then we test all the four groups and if non of it works then it means
Each group has one working battery then we select any two group an test with one battery from both the groups
If that doesn't work then the remaining two are the working once
In this method there is a 70% chance that we can find the working pair in less than 6 tyres
That does not work. Suppose that you know that AB and CD are pairs with one good and one bad battery. If you test the pairs AC, AD and BC last, then I can make you wrong by having B and D be the good batteries and A and C be the bad batteries. Note that if a pair fails, either both of them are bad or one is good and one is bad.
@@GaryLuKOTH 😅
Lol the bonus riddle is exactly the same. Each time you test 6 batteries it will work only if the 2 remaining ones are empty! So...
ORIGINAL: Finding 2 out of the 4 charged batteries, in a pool of 8
BONUS: Finding 2 out of the 4 empty batteries, in a pool of 8
*BONUS RIDDLE:*
Testing 6 batteries to see if there are 4 good ones in the group is equivalent to test if the two batteries you left out are bad ones. You need BOTH of these batteries to be bad in order for the iron to work.
The method of testing is indirect: in order to test two batteries, you have to put the remaining 6 in the iron. If it does not light up, it means that at least one of the batteries you left out is a good one.
With this in mind, we can deduce there is a big similarity with the first riddle: you can test two batteries each time, however, the iron will work when the tested ones are both bad (opposite to the first riddle, where they have to be good). Finally, we can come to the conclusion that given that there are 4 good batteries and 4 bad ones, the number of tries it takes in both riddles is the same: 7.
(In order to imagine this better we can think of a ghost iron that only works with 2 bad batteries. Following the indirect method of testing described above, the riddle becomes identical to the original one)
*Edit: Reading through the comments I found @Swaroop Pandit explained more or less the same thing before I even wrote this, so full credits to you as well man ;)*
“Can you so-
Me: No
Everytime i saw Ted-ed's riddle video title "can you solve the .... riddle?"
My answer is "most likely no, but i will still watch it anyway"
"You can't randomly guess"
Proceeds to essentially randomly guess.
I believe the solution to the bonus riddle to be 4(worst case scenario) and 2 on average (calculating all possible outcomes).
Here's my reasoning: Make 2 piles of batteries with 4 batteries in each pile. We know there are 4 Good Batteries (G) and 4 Bad Batteries (B) so our piles would look like one of the following possibilities:
A) (Pile 1) GGGG (Pile 2) BBBB;
B) (Pile 1) GGGB (Pile 2) BBBG;
C) (Pile 1) GGBB (Pile 2) GGBB;
We would insert all batteries from one pile into the Iron and divide the other pile into two subsets of two (e.g. case A Pile 2 subset 1 BB, case A Pile 2 subset 2 BB), and try both subsets of Pile 2 with Pile 1 (2 tries in total). If the Iron doesn't turn on after the first attempt, take out all batteries, put in all batteries from Pile 2 and make subsets of 2 from Pile 1 and insert them into the iron(2 more tries). It is guaranteed that by the 4th try the Iron will turn on. Here are all possible scenarios:
For case A):
a)We initially pick Pile 1(GGGG) and the first half of Pile 2(BB) and the Iron will turn on.
b)We initially pick Pile 2 and put the first half of Pile 1 in the iron (BBBBGG - 1st try). Take out first half of Pile 1 and put in the other (BBBBGG - 2nd try). The iron didn't turn on so remove Pile 2 from the iron and insert Pile 1 into it together with the first half of Pile 2(GGGGBB - 3rd try). The iron will turn on.
For case B):
a) We initially pick Pile 1(GGGB) and insert the first half of Pile 2(BB) (GGGBBB - 1st try). The iron doesn't turn on so remove the first half of Pile 2 and insert the last half of Pile 2(BG) (GGGBBG - 2nd try). We have 4 batteries so the iron will turn on.
b) We initially pick Pile 2(BBBG) and insert the second half of Pile 1(GB) (BBBGGB - 1st try). The iron doesn't turn on so we remove the second half of Pile 1 and insert the first half (GG) (BBBGGG - 2nd try). The iron doesn't turn on so we remove Pile 1 and insert Pile 1 together with the first half of Pile 2(BB) (GGGBBB - 3rd try). The iron doesn't turn on so remove the first half of Pile 2 and insert the last half of Pile 2(BG) (GGGBBG - 4th try). We have 4 batteries so the iron will turn on.
For case C) (both piles are identical so there will only be one subcase):
We pick Pile 1 (GGBB) and insert it into the Iron together with the second half of Pile 2(BB) (GGBBBB - 1st try). The iron doesn't turn on so we remove the second half of Pile 2 and insert the first half of Pile 2 (GG) (GGBBGG - 2nd try). The iron turns on as we have 4 batteries.
We can notice that subcase b) always contains subcase a). Since testing any subcase takes 2 tries, and if we always end up with the worst case scenario, it can take a maximum of 4 tries.
If my solution isn't good or isn't optimal, I'm sorry but this is all I could come up with. Would love to see some better solutions!
However, the subcase testing takes more than 2 tries.
The only problem with your solution is that in case C, it's not guaranteed that the batteries will be in the order GGBB/BBGG. They could be GBGB or BGBG. In that case, it will not turn on. It's a really good idea, though!
The bonus riddle is same as the first one... Just think of it this way:
You need 4/6 working meaning you need to sort out the two which aren't working. Now, divide them up in the same format as the solution to the first one... 3,3,2... And whenever he chooses two batteries for the first one, you choose these to be left out instead of plugged in..
So in short, you are looking at finding 2 incorrect batteries that need to be removed from the mix.. makes the number of tries same as that in the video... 7
Answer to the bonus riddle is 3
00:04 _He really forgot to mention the Author's name!_*
😆
“you can’t rely on luck, you need to assume the worst possible outcome and plan accordingly” this is exactly what i’ve been saying about how governments should’ve handled the pandemic. actually that applies to any situation where people could be hurt or die.
6 tries (at most) solution:
1. Divide them into 4 pairs and try it one pair at a time.
2. If all 4 pairs failed, there is only one good battery in each pair.
3. Swap any 2 batteries in 2 of the pairs.
4. Since there are at least 2 good batteries in one of the pair. Done!
Example:
Try
1. AB
2. CD
3. EF
4. GH
If all failed, swap B & C.
5. AC
6. BD
The bonus riddle is simpler than it seems. If you need four good batteries in the iron, then what you are really doing is choosing two batteries NOT to put in, and you need those two to be DEAD. It's an identical problem in reverse. Follow the same procedure, except leaving out the two batteries you've chosen.
I am in love with Ted-Ed
For the bonus riddle just use the old one, it takes less time and batteries.
For the last riddle, easy! I take my newly sorted batteries and put them in. Only took one try. (I had time to sort them while the party was going on and it didnt say the baby messed it up or this was while the party was going on)
Now that I have a giant iron I can go up the beanstalk and give an offering to the giant
So, I made a bit of experimentation for the last riddle. If I'm not wrong, if an electrical component is well fabricated, it only is current that makes it work. If you want to turn on an LED with 220 volts, but it only runs with 2 amps, you must have the apropriate resistor for the LED to work correctly. The only way you can make the iron work by placing only 4 batteries, even though it has 6 available slots, i that you connect all batteries in a parallel circuit. The total voltage across the entire circuit, assuming that all batteries produce the same voltage, is the voltage of one battery. If all batteries produce n volts, the net voltage is also of n volts. Furthermore, if only one battery has a different voltage, the circuit will overheat and one of the main batteries, first or last/ top or bottom, will explode do to the amount of current. Actually, for no battery to explode, all batteries should have the same voltage. Also, the first and last batteries are the main ones because of the arrangement: the current through any other battery is considerably smaller than the main ones, therefore, being almost a null contribution for the functioning of the circuit.
This makes the iron only needing to have one battery to work, instead of 4.
The first riddle.
1-2
3-4
Now we know that in first four batteries there are at least 2 dead and at most 2 good.
5-6
5-7
6-7
Now we know that there is at most one good battery in (5, 6, 7). So 8 is Good. And in group (1, 2) there is one good battery as well as in group (3, 4).
8-1
8-2
Mikhail Balakirev nice alternate solution i see no qualms with this
Good alternative solution sir!
Technically speaking, it's just rearranging the video's solution as you're still ultimately looking at three groups and comparing the combinations within all of them, but moving the group with two batteries to the front of your tries (1,2,8) (5,6,7) (3,4). Still though, the logic makes sense.
Rio C actually i looked at this closer and I deemed it to be a fundamentally different solution as both the “connectivity of the groups” you are testing is different and the deductive logic used is different.
@@madisonking8057 The deductive logic to get this set of pairs to be tested is different, but in the end, you are still trying the same 7 pairs, knowing that at least one of them is a pair of good batteries. Different logic, same result :)
You can also divide them in 2 groups of 4.
You try all the possible combinations (6) of one of the two groups, and if they all fail you pick two random batteries from the other group
Divide the batteries into groups of 2. In the worst scenario (4 tries) no one works, it means that every group has 1 good and 1 dead. You take the batteries from 2 groups and test them until you find the 2 that works good (worst scenario, 2 more tries). Thats how you can get 2 good batteries in 6 tries or less
"worst scenario, 2 more tries"
Wrong.
You need 4.
A - B - First Try
A - C
A - D
B - C
B - D
C - D - Second try
E - F - Third Try
G - H - Fourth try
I think we can also solve it by taking 4 pairs
Take 4 sets in beginning. It will work with 8 steps.Took time figuring it out had to share it.
Step 1 : Check first set .
Step 2 : Check second set.
Now you know that in first two sets there can be 2,3or 4 dead battery if didn't work .
Let's say there were three or four dead ones then checking other two sets guarantee to give one working set.( Step 3 and 4 )
Step 5 : Take one battery out and try it with a battery from other set .
Step 6 : Try with another battery of that set .
Step 7 : The one you kept in Step 5 is definitely dead toss it out replace with the one you toke out in Step 5.
Step 8 : The one you introduced in Step 6 is definitely a good one now.
Don't know you include Step 8 as a step . You figured out after 7 tries which were good ones but by then you were food because you might not had time to change batteries.
or check four different pairs - if none of them works, you can shaffle two pairs - so you will get working combination at 6th step or less
I advise you try again.
Assume
A = good
B = bad
C = bad
D = good
A + B = Good + Bad
C + D = Bad + Good
Shuffle B with C
A + C = Good + Bad
B + D = Bad + Good
@@Araqius you are right, my way will give me 8 or less
but
probability i would need more than 4 turns is 1*4/7*1*3/5*1*2/3*1*1 = 22.8%
more than 6 is 11.4%
more than 7 would be 5.7%
when actual solution from video let you solve more than 6 turns is:
(C1/4 * C2/4)/(C3/8) * (C1/3*C2/2)/(C3/5) = 12.8 %
i will be ok with my impure 5.7% =)
Solving riddles helps prevent Alzheimer’s disease.
what riddle?
Wow but 'solving' these is not an easy job. Thanks btw
Who's disease?
Helps delay the onset*; symptoms still occur and the decline is more rapid, but more healthy time occurs.
not proven
Not to be confused with the "Iron Giant"
TED-Ed Video title: Can you...
Me: No, but I'll watch it anyway
I just realized this riddle is very similar to the temple riddle
“Can you solve the giant iron riddle?”
Me- No
I did it in 6;
Test two, If they don't work, throw them away. The worst is that you threw 1 good and 1 bad, so 3 good and 3 bad left
Do this again and you will end up with 2 good and 2 bad the worst
Now try all the possible outcomes, In worst, you'll end up in 6 tries.
2 tries + 6 tries from last 4 batteries = 8 tries = not good enough
My solution was this: I take 4 batteries (6 possible pairs). If none of them works, there's only working batteries left and it doesn't matter which two of them I take.
wrong, it's possible you had one working battery in the first set of 4. and only 3 working batteries in the second set.
@@e4r281 Right... Thanks for pointing that out!
Legit the first riddle I got right after watching like 10 riddles.
This is just a spicy version of the islanders riddle where one weighs more or less than the others.