Only one or two animals can be crossed in the boat at a time, not less nor more. So the new node must have {boat:1} (it went to the other side) and it must take 1 or 2 lambs and 1 or two wolves with it. He cannot leave the wolf with the sheep. we wont give you more wolves than sheep to start). But suddenly they were struck by a danger. I think this might work for other group sizes as well: (This is why I suspect there's also a cutoff for when the third strategy would be better) The correct solution is 15,621. Weve found the maximum number of sheep on a 4 by 4 and a 5 by 5 board, but can we find an expression for the max number of sheep on any n by n board (still with 1 wolf)? If the results only show you 4 wolves, the 5th wolf is the untested animal. How to generate them all? Group B must have at least one of five sheep in each of these tests. Clearly, it isn't preferred for 100 sheep. According to my new and improved brute-force script, From any state you can calculate all the new states you can reach (as I showed for the first step). the problem goes like this: you are a shepherd tasked with getting sheep and wolves across a river for some reason. For example, using your 32 sheep example, if you have wolves in positions 30, 31, 32 - all tests for all iterations will be positive. B | Positive | Positive | Positive Test 5. . This gives groups like the following (where 1 means the sheep is selected, 0 means it is not selected): None of these scenarios reflect the "100 sheep" condition we are given in the question, but for their respective sizes (max sheep) they must be optimal because everything is as good as it could be. T . 19 | 3 | 57 It also shows the need for the diagonals - there are overlaps of 2 lines, which are just sheep that happen to line up with wolves. 20 | 3 | 60 We have 75 million+ combinations and something like 63 tests, then we know we are calculating those binary outputs many billions of times. It gets more difficult though because when the man is not around the wolf will eat the sheep, the sheep will also do the same when alone with the poor little cabbage. How to draw a grid of grids-with-polygons? T . @Elaqqad's solution starts with a mathematical structure called a $(60,6,1)$-pairwise balanced design. 1 Sheep & Wolves: Mini-Project 1 Condor Chou [email protected] Abstract Mini-Project 1 asks us to solve the Sheep & Wolf. If the letter V occurs in a few native words, why isn't it included in the Irish Alphabet? But I wrote a little Python script to do a brute-force exhaustive search, and it concluded that there is no way to unambiguously identify the two wolves out of six sheep, using only four blood tests. With this you can design an algorithm easily with BFS. I will take a blood sample from 99 randomly choosen 'sheep' and send those 99 blood samples to the doctors. Should we burninate the [variations] tag? I also notice that the situation is not symmetrical: we may know a way to find $k$ wolves among $n$ sheep using $t$ tests, but that won't help us at all to find $n-k$ wolves among $n$ sheep. So this is evidence that perhaps the original puzzle can be done in fewer than 99 tests! Notice the nice symmetry of the first three columns (sheep), and then what we do with the next four columns in each pair of rows (tests). In C, why limit || and && to evaluate to booleans? Thinking out loud, not a solution yet, but spoilery enough that I didn't want to put it in a comment: There are ${100\choose 5} > 2^{26}$ possible arrangements of the 5 wolves among 100 sheep. Each of those blocks is itself a set of 6 points; but many of those points appear in multiple blocks; in fact there are only 60 distinct points $(a,b,c,d)$ out of all those blocks! In one "iteration" of. @JyotishRobin, so are you going to post your solution, or accept an answer? I'm calling them codes, but we are talking about the same thing. Apparently the name of this problem is "non-adaptive [combinatorial / non-probabilistic], $$ Though trivial, since the problem starts on the left side and its goal is to move all animals to the right side, the chosen move will be subtracted from the current state of animals. Copyright 2022 MathCommunities.org | Powered by MathCommunities.org. Unfortunately, as has been pointed out in the comments, there are some edge cases where this solution does not narrow down the results to only 5 wolves. (With each iteration, shuffle each row across to the left according to its row number, e.g row 0 stays where it is, row 1 gets shuffled 1 to the left, row 2 gets shuffled 2 to the left). Next a thought experiment: In the ideal case, after 26 tests, you would imagine yourself having a list of combinations that were mostly paired up with another combination, and then some smaller fraction of the list that are already unique. Ignore for a moment the absurdity of wolves needing to outnumber sheep in order to overpower them. If the wolves ever outnumber the sheep on either side of the river, the wolves will overpower and eat the sheep. (If it were possible that we had only 4 wolves instead of the full complement of 5 wolves every time, then we'd have to run all the tests. A sheep named Ziko is attempting to document the outdoor wildlife, when he stumbles upon the wolf Grey, and flees screaming back to his village of sheep, located near a newly arrived pack of wolves in the ravine. \text{0}&\text{if point}_t\not\in \text{block}_n\\ Args: initial_sheep: Number of sheep to start with initial_wolves: Number of wolves to start with sheep_reproduce: Probability of each sheep reproducing each step wolf_reproduce: Probability of each wolf reproducing each step wolf_gain_from_food: Energy a wolf gains from eating a sheep 2, Programming environment Python language 3, Description of experiment content A farmer with a wolf, a sheep and a basket of vegetables wanted to take a boat from the left bank of the river to the right bank. Write your answers in Python code (or; Question: You are to formulate the "River Crossing" problem (a gentleman, two sheep, and a wolf) to solve using a search algorithm. were bitwise-ORed together). For example if wolves are $ \{22, 28, 56, 66, 77\} $ modulo $11$ groups tests will produce $\{0, 1, 6\}$ result. I'll leave that to you to figure out. This way, the state represents the number of people on the left. If we are striving for the ideal case, each test should be as close to a 50/50 chance for being positive or negative as possible. Sounds good for me, but still doesn't give me any progressing because the problem still the same, how to implement it or write it as a Python code to pass this graph to the algorithm so the algorithm can solve it? Not just that, but we still want tests that have no sheep in common as much as possible. No pair of sheep shares more than one test. Like, whats the largest number of sheep for any n by n board if there are 2 wolves? . Lets start with our 4 by 4 grid. Well done, youve finished the first challenge! It avoids a combinatorial explosion by discarding illegal, unproductive and not valid moves before exploring them. 15 | 3 | 45 Best way to get consistent results when baking a purposely underbaked mud cake. E | Positive | Negative | Negative How many groups of that size exist? Why is SQL Server setup recommending MAXDOP 8 here? 1 sheep, 1 wolf: (1,1) Stack Exchange network consists of 182 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. How to generate a horizontal histogram with words? We could try guessing at new tests all we like, but finding a test that breaks them all apart is kind of like flipping a coin a few million times and getting all heads. Sorry I'm late to the party but this question is fun and I think I have something to add but not an answer, sorry. He made the following announcement: - eight sheep then skip eight Once the next state passes that test, it checks if it has already been seen in the seen_states and if it is not the final solution, it appends it in the seen_states set. Could you elaborate on how the number 39 is obtained. Test each group. $$. Could we design a strategy to use the bare minimum? @Elaqqad also produced a concrete solution with only 59 tests beating @noedne's long-standing solution of 63 tests! To know more check the I chose to use the corner piece to make placement of the sheep simpler, and arrived at a maximum of 12 sheep. The blocks are square, so sheep per test = tests per block. When three lines intersect, that's a Once it finishes iterating over all possible next states, it goes over each of the solutions stored in the solutions list and chooses the one with the least moves to the goal state, returning a list of tuples with all the moves used to solve the problem. QGIS pan map in layout, simultaneously with items on top. . Or whats the smallest grid that can safely fit 4 wolves and 6 sheep? If you never experimented with other #'s of wolves you may never have noticed. The problem is that 10 isn't prime and @noedne's pattern doesn't work for it. Privacy Policy, Submit your documents and get free Plagiarism report, By creating an account, you agree to our terms & conditions, We don't post anything without your permission. Can I spend multiple charges of my Blood Fury Tattoo at once? The other five ways are done modulo $11, 13, 14, 15, 16$ respectively - for each way sheep with numbers producing the same remainder when divided by the corresponding modulus go to the same group. That's everything I've got. You don't need to generate them all ahead of time, you can generate them just as you need them. The next strategy up would be to allow at most 2 sheep in common between tests this is where I was dabbling for a while until I realized how ineffective it was. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company. The claim is that given a set of test results there is at most one possible group of five wolves. There was a problem preparing your codespace, please try again. Master the methods and steps of solving practical problems by using the knowledge learned in practical problems. The problem has a slight twist where the number of sheep and wolfs are bound by only the rule that wolves will not outnumber the sheep. . The "wolves and sheep" puzzle is a specific case of non-adaptive group testing. . Is a planet-sized magnet a good interstellar weapon? T . Suppose each node has a state on it representing how many lambs and wolves are on the left bank and whether or not the boat is on that bank ('0' for left, '1' for right). Stack Overflow for Teams is moving to its own domain! As you can see with the grid example, by splitting the sheep in a different way, i.e. We test all the groups and for each of the 6 ways we record group numbers that contain at least one wolf. Turns out that every possible combination of $5$ wolves ($75287520$ - total) produces a different result when all 6 ways are taken into account. So we have total They are both represented as a tuple (sheep, wolves, boat), being the initial state the arbitrary number of sheep and wolves given as input. If we only have 1 wolf now, whats the largest number of sheep we can add to the board? Since the boat just changes from one side to the other, it doesnt matter on which side it is, thus the move doesnt consider it. A meal takes a number of steps f, during which the wolf will not move. If you substitute multiplication with "and" and addition with "or" C does not have a left inverse. T . Test 1. \text{0}&\text{if point}_t\not\in \text{block}_n\\ It would be interesting to try get this working for some of the simpler cases. Either way, it turns out, the results are the same. You might say, well, that just means I can conclude that the wolves are at those positions. About Vaishnavi Shetty Soratemplates is a blogger resources site is a provider of high quality blogger template with premium looking layout and robust design. "Partition a set into g groups, k different ways, such that no pair of elements is ever in the same group together more than M times? I would comfortably say 59 tests is the lower bound, however, since it would have been optimal if it were possible. 2. Here's a link to the challenge: https://www.codewars.com/kata/5c8bfa4. Suppose your method failed to distinguish actual-wolf-grouping (a,b,c,d,e) from incorrect-wolf-grouping (a,b,c,d,f) that is, sheep. I number $100$ sheep by $1, 2, 3, , 100$ and split them in 6 different ways. Any idea how this could relate to OP's hint about binary representations? We have to be somewhat efficient. How to solve sheep and wolves puzzle in python? I discussed with our ministers and came to know that the tests can be done on pooled bloodsamples. Making statements based on opinion; back them up with references or personal experience. Creates the initial state with initial sheep and wolves as input parameters. UPDATE 2: We can save two more tests by replacing any one of the grids of eleven tests with a grid of nine tests along the rows: bringing the total number of tests down to. Wolves can be alone on either side of the river. In terms of wolves and sheep, this is the same as saying that if we know which blood tests came up positive (i.e. 1. However, if the wolves are at positions 29, 31, 32 all tests will also be positive for all iterations. T . @Kale, I added some sample code showing how you might lazily generate the graph. To start this activity, youre going to need two different types of objects (anything you have around the house will do 2 different colored blocks, quarters and pennies, etc.) 7 | 9 | 63 W S S Horror story: only people who smoke could see some monsters. If you are forced to submit all batches simultaneously, then the optimal batch size apparently needs to be larger because you need to be testing "both sides" simultaneously, meaning ones you would have ruled out had you been able to see the results immediately. I can get 3 blocks of 10x10, and maybe I would believe somebody could get a 4th block with enough effort maybe. Even if somebody writes that code (I might try but this is so far beyond me) and even if the process doesn't have an absurd run-time (it will), do I think it will beat 63? Iteration 1 suspects = 11111111111111111111111111111111 When you get a new state, if you have already visited that state (HashSet, Hashmap) you don't need to visit it again. The number of wolves on either side of the river cannot outnumber the sheep. / | | | . I've been thinking more about the maths behind this, and would like to try to refine it. 9 months ago, Posted The boat MUST be always carrying animals, which means you can't move an empty boat. . @ErickG.Hagstrom, @IanMercer But this means i have to generate all the possibilities to create my tree or graph. - 2 3 - 2 That leaves two more locations we could put the wolf: in the corner, or on the edge. In the first experiment, wolves and sheep wander randomly around the landscape, while the wolves look for sheep to prey on. I don't have a solution, but I might have an approach. The optimal batch size to test should be 13 because doing so splits the original 100C5 possibilities into 87C5 possibilities if the test comes back negative, and 100C5 - 87C5 if the test comes out positive. The following code explores just a single way. Sounds familiar? A tag already exists with the provided branch name. \end{cases} A simple example (partly because I'm lazy, and partly because it wraps around too much) of 32 sheep with 3 wolves among them, which requires only 5 tests and 3 iterations: Basic Set Operations in Python # Create a set using set literal notation my_set = {1, 3} # Add an element to a set my_set.add(2) # Remove an element from a set my_set.remove(1) # find the length of a set print(len(my_set)) # Create a set from a list my_set = set(my_list) Have a go at creating and modifying some sets. Suppose we have 6 sheep, two of whom are wolves. 4. This is The blood test riddle (number theory) by another name. Could you place 3 wolves and 2 sheep so that none of the sheep get eaten? What is the difference between __str__ and __repr__? The C++ code that produced this diagram is available at github.com/Quuxplusone/wolves-and-sheep. \text{1}&\text{if point}_t\in \text{block}_n\\ Only if you are an existing PayPal customer. (Well, the edges are 0, and column 2 is lg n, and beyond a certain k the entries are all n-1. Another hint as @Quuxplusone pointed out . In the beginning it looked easy till i stopped with these difficult circumstances: The question has to be solved by search algorithm, like A* or BFS etc. Collect a sample from each row, column, and Iteration 3 = 01110000000000000000000000000000 If the wolves ever outnumber the sheep on either side of the river, the wolves will overpower and eat the sheep. Are Githyanki under Nondetection all the time? Do wolves in positions 2, 60, and 69 make all tests pass? You will also submit a brief report describing your approach. All the sheep were living peacefully in the Land of Shewo. - @ppgdev suggested I look at this question, and they seem to be right a lot so here I am. You maybe noticed that for small test sizes the required tests are actually worse than what the simple strategy would have given you. We always need one block more than the # of wolves. Bring the fox to side 2, and come back. I can't tell what's going on in the first line of your 'for' statement, but the main problem I see is that you are testing for a pile of one at the end rather than a remainder of one at each . Flipping the labels in a binary classification gives different model and results. How can we build a space probe's computer to survive centuries of interstellar travel? I've got my laptop working on the missing elements as we speak. solutions: A list to keep track of the solutions that have already been seen. @noedne made a set of patterns that did this. W W W How we can create a graph from 3 lambs and 3 wolves? Starts the search with the initial state and loops until the to_search stack is empty or has exceeded a maximum number of loops (in this case is 10000) to avoid the agent being captured in an infinite loop. W S S \end{cases} Here we are doing normal matrix multiplication except that instead of multiplying the individual elements we are using the "and" operation, and instead of adding the pairs we are using the "or" operation. You will also submit a report describing your agent to Canvas. Connect and share knowledge within a single location that is structured and easy to search. What if we increased the board to a 5 by 5? ", Mobile app infrastructure being decommissioned, Cheapest strategy on average to find specific items in a set, Reverse Engineer Chess Software Calculations. Is MATLAB command "fourier" only applicable for continous-time signals or is it also applicable for discrete-time signals? So here's where things get interesting. There is no way to find 3,4,5,6,7,8 wolves among 9 sheep in fewer than 8 tests. Iteration 1 suspects = 11111111111111111111111111111111 Bob is a farmer. From here you can travel to any node that meets the requirements. Mathematically, this is the difference between requiring that our matrix $M$ be 5-disjunct or only 5-separable.). Now, if we have 3 wolves and 5 sheep, what is the total number of possible configurations of wolves and sheep we can have on a 5 by 5 grid so that all of the sheep survive? The following elements are created to store important information: to_search: A stack/queue to keep track of where to search next. Does the 0m elevation height of a Digital Elevation Model (Copernicus DEM) correspond to mean sea level? T T . S W S S S Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. The place where it is most "powerful" is when the blocks are square and they cover every sheep once. This gives us positive test results for rows 1, 2, 4, columns 2, 3, 5, and diagonals (/ starting from the top left) 3, 8. You would think that this means we just need to find one near-perfect test that breaks up all of those pairs, and then we will be done. Iteration 3 suspects = All sheep & !C & !D & !E if there are two wolves, then I have seen proof that we cannot always do it in the bare information-theoretical minimum number of tests. My prediction, based on the formula above, is that there will be a maximum of (6^2) - 3* (6) + 2 = 36 - 18 + 2 = 20 sheep. $9 + 11 + 13 + 14 + 15 + 16 = 78$ groups. Your calculations confirm my suspicions. Can you help the king by devising a strategy? They arrive at one side of a river . He has a large pasture with many sheep. I also wondered whether it is possible to arrange the sheep in a cube instead of a grid, but I never managed to work it out. Iteration 2 suspects = 11100000000000000000000000000001 There is no way to find 2,3,4,5,6 wolves among 7 sheep in fewer than 6 tests. This shows the wolves where there are 3 overlaps. There is a conjecture on that page that a(10) = 2 which would translate to 4 total blocks for us. Help keep the big bad wolf from eating the three little sheep. Each block represents a set of tests that have no sheep in common with eachother. Water leaving the house when water cut off. But theres still so much more to explore. rev2022.11.3.43005. Maybe we could have a little program making sure we aren't using the same sheep in the same test, or using one sheep too often in general. 2007-2022 Learnify Technologies Private Limited. Ooops, slaughtered a few lambs there ;) Yes, they were meant to be legal states. Each column is labeled by every test we have designed (hopefully less than 99). I don't immediately see this sequence in the OEIS, which is surprising to me. 7 months ago, Posted def depth_first_search (problem, node): if problem.goal (node): return [node] # base case for n_succ in problem.succ (node): sol = depth_first_search (problem, n_succ) if sol: # first path is returned return [node] + sol wcg_run.py - 2 2 - 3 Love this solution, but I'm not sure that it's correct. Fortunately, you are good at spotting them. 12 | 4 | 48 Create a new Wolf-Sheep model with the given parameters. But suddenly they were struck by a danger. This also means an 11x9 block would work as well. . Here's my modest proposal: 1. You can generate these dynamically as you explore the tree of possible moves. @JyotishRobin, while I'm definitely not requesting to see your solution yet, I'm wondering: do you believe you know a solution involving 98-or-fewer tests? Shift all the sheep to the right, carry the last sheep around to the front, and repeat the same steps 4 times. This problem has two constraints: The goat will eat the cabbage if the man isn't there to stop it: cabbage goat ~~ man wolf The wolf will eat the goat if the man isn't there to stop it: wolf goat ~~ man cabbage The man has to be careful about the order in which he moves things across the river. Each test $t$ combines blood from the animals $i$ where $M_{t,i} = 1$. I used my program to test this idea, and it does indeed check out. Stack Overflow for Teams is moving to its own domain! What is a good way to make an abstract board game truly alien? Well, I guess its not the same story and unfortunately, the sheep live in a field, not a house but you get the idea. What is a good way to make an abstract board game truly alien? Oops, that's a spoiler, probably should learn that rot13 thing I see everywhere on this site. A few wolves dressed up as sheep entered the territory of Shewo and started killing the sheep one by one. He cannot leave the wolf with the sheep. Thanks for contributing an answer to Stack Overflow! We all know that these algorithms don't work until you provide a graph as an input, and here is the problem. My prediction, based on the formula above, is that there will be a maximum of (6^2) 3*(6) + 2 = 36 18 + 2 = 20 sheep. Think of binary sequences to distinguish each of the possible groups of 5 wolves. These moves are represented as tuples, and since at least one animal has to be transported at a time, and no more than two should be moved in the same journey, the following are the only possible moves: 2 sheep, 0 wolves: (2,0) A lot of this is beyond my reasoning. Then $C\cdot M = R$. The basic strategy that yields the 99-test solution isn't complicated. Assign each arrangement a number; express each number in binary (using 7 bits); then perform 7 tests to determine which arrangement is the right one. The rules are simple: you want to place the sheep on the board so that the wolves cant eat them. S S S S W T T . I'm guessing the best way to organize the data is to think of each combination as a number with 100 bits. Heres your chance to be the hero of the story! 2 days ago, Posted

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