Then if I test the mixture, I will get a positive result if the mixture contain blood from any wolf. You saved the sheep! * */ private class state { private string bank; // the active bank where the farmer is currently // located private treeset left, right; // I think this might work for other group sizes as well: $C^{-1}_{\mathrm{left}}=(C^T\cdot C)^{-1}\cdot C^T$. Maybe it's baby wolves vs. adult rams. 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. Take initiative and slaughter all 100. There are ${100\choose 1} > 2^6$ possible arrangements of a single wolf among 100 sheep. Thinking like a Wolf, a Sheep or a Firefly: Learning Biology through Constructing and Testing Computational Theories -- an Embodied Modeling Approach. Having nothing else to go on, I will use it and I know the split for testing 13 sheep is roughly 49 negative outcomes for every 51 positive outcomes, and this still allows (barely) for 27 tests to encode all combinations. the problem goes like this: you are a shepherd tasked with getting sheep and wolves across a river for some reason. He made the following announcement: If more than one sheep try to occupy the same cell, one of the sheep's moves fails and the sheep will not move. Well, I guess its not the same story and unfortunately, the sheep live in a field, not a house but you get the idea. Each cell is either empty, contains a sheep, a wolf or a dog. T . Lets see if this checks out for a 6 by 6 board. the solution state will be * * left bank: right bank: farmer (f), wolf (w), cabbage (c), sheep (s) * * the left bank is empty and all the occupants farmer, wolf, cabbage and * sheep are on the right bank. There we go! Now you have no more wolves, and food for a good while to come. And so on. If we continue using the strategies we employed in Question 3 & 4, we know that the maximum amount of sheep is the number of squares that remain on the board after we take away the row, column and diagonals from a corner or edge piece. But theres still so much more to explore. We always need one block more than the # of wolves. Stack Overflow for Teams is moving to its own domain! 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. I don't have a solution, but I might have an approach. E 00000000000000001111111111111111 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. Python also have a more in depth guide on how to properly format code, see PEP 8. . Should we burninate the [variations] tag? According to my new and improved brute-force script, B 00110011001100110011001100110011 Arrange the sheep in a 10x10 grid. Fortunately, you are good at spotting them. Not the answer you're looking for? (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. Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. 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. 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. Congrats, you just saved 1,000 bucks. 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 The aim of the game is to move the wolf, sheep and cabbage to the opposite shore. I'll leave that to you to figure out. So I will add my answer. Testing 13 sheep at a time is the best in this regard, though I can't claim it is the ideal strategy. 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 Sheep - Wolves - Grass Model Step 1 - Basic Structure Step 2 - Programming the agents Step 3 - Programming the environment Step 4 - Final tweaks and plotting Conclusion References Introduction There are some complex dynamics and systems that we are not able to observe. Technically, that's "every sheep except one" because it ends up being okay to always leave one sheep out of the testing. C is quite a big matrix to inverse and I'm not sure what python would do with binary matrices. Here is the $59\times 100$ matrix $M$ that I got by following @Elaqqad's algorithm. "Least Astonishment" and the Mutable Default Argument. We get to use 99 tests instead of 100 because the last test becomes moot. 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. The rules are simple: you want to place the sheep on the board so that the wolves cant eat them. I am using 13 random sheep in each test. In order to be certain, we need to repeat this 5 times, which is 50 tests. What is the difference between Python's list methods append and extend? Would it be illegal for me to act as a Civillian Traffic Enforcer? Could we design a strategy to use the bare minimum? Now , I am looking for ideas where I can find ALL the wolves in minimum number of pooled tests. This series of 59 tests can identify the 5 wolves from among our 100 animals. 0 sheep, 2 wolves: (0,2) Come up with your own puzzle see if your friends can solve it! Stack Overflow for Teams is moving to its own domain! I've got my laptop working on the missing elements as we speak. I would comfortably say 59 tests is the lower bound, however, since it would have been optimal if it were possible. Do you believe you know the. Either way, it turns out, the results are the same. Say Shaun is in group A but not B, and Shirley in B but not A. It would be interesting to try get this working for some of the simpler cases. If you never experimented with other #'s of wolves you may never have noticed. . 34 All the sheep were living peacefully in the Land of Shewo. I was thinking about this question for a long time and couldn't figure out how to solve it. Alternatively, you know the location of all 5 wolves. If you exclude the wolfs square from the row, column, and diagonal, you end up with the same number for each: n 1. To find a solution to this misery, the king of Shewo called upon all of his sheep to the palace hall. If we only have 1 wolf now, whats the largest number of sheep we can add to the board? This still feels wildly inefficient. were bitwise-ORed together). rows instead of columns, I can perform the search again and narrow down to 25 sheep with only 20 tests. suspects = 100 * ((5 / groups) ^ iterations) By doing this, I've halved the number of sheep to search with only 10 samples. What is a good way to make an abstract board game truly alien? But think for a second, if we chose a test that was 13 sheep at random, it would break up roughly half of those pairs, and certainly not all of them. This suggests a realistic answer is something closer to 50 tests, and that's without considering that tests are getting weaker all the time. A valid transition moves the boat and increases or decreases an animal count. Because that would work, at a cost of $99,000. Then, we subtract that value from the total number of squares on the grid, n^2, which gives us a final expression for the max number of sheep on an n by n board: (n^2 ) 3n + 2. \end{cases} I'm calling them codes, but we are talking about the same thing. Here's a link to the challenge: https://www.codewars.com/kata/5c8bfa4. A "5-separable" matrix is a matrix of 0s and 1s where no two sets of 5 columns have the same bitwise-OR. Are you stuck. Many Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior. So the next problem is that each test isn't going to quite perfectly split up everything before it. The following elements are created to store important information: to_search: A stack/queue to keep track of where to search next. where the $\cap$ refers to the "logical and" operation and the $\cup$ is the "logical or" operation. seen_states: A set to keep track of the states that were already seen. I have figured out a better way of solving this which I could motivate through hints. There are two main variations to this model. Why is SQL Server setup recommending MAXDOP 8 here? Flipping the labels in a binary classification gives different model and results. All rights reserved. - every alternate sheep Graph approach : Nodes are (L1,L2,L3,W1,W2,W3,B) with L's being the side which the lambs are, w's being the side on which the wolves are and B the side of the boat. . What if we increased the board to a 5 by 5? But suddenly they were struck by a danger. You may assume that the initial state of the problem will follow those rules (e.g. Bring the fox to side 2, and come back. Not really, no. Consider assigning binary sequence of 0 & 1 for all sets of 5. They don't appear to be. You have a boat, which can only take one or two animals in it at a time, and must have at least one animal in it because youll get lonely (and because the problem is trivial otherwise). There are ${6\choose 2} = 15 \leq 2^{4}$ possible arrangements of two wolves among 6 sheep. Iteration 2 = 11100000000000000000000000000000 Why? Sure, probably, but it won't be anything terribly game-changing. The claim is that given a set of test results there is at most one possible group of five wolves. This is The blood test riddle (number theory) by another name. 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.) 8 | 7 | 56 Are Githyanki under Nondetection all the time? $$ for each of the 96 blocks (animals) and 60 points (tests). That leaves two more locations we could put the wolf: in the corner, or on the edge. Well, I guess its not the same story and unfortunately, the sheep live in a field, not a house but you get the idea. You signed in with another tab or window. 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. Keep doing that until all combinations are unique. Can't figure out how to fix it. 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? - four sheep then skip four You can generate these dynamically as you explore the tree of possible moves. This shows the wolves where there are 3 overlaps. Each row is labeled with every possible combination of wolves among the sheep (there are over 75 million). So we can adjust the # of wolves in these smaller cases to something smaller, and then the 2nd strategy becomes best again. Edit: It seems that OEIS A001438 nearly confirms that 10x10 doesn't work for us. Try a number of random tests, keep the one that worked the best (minimizes the metric). Sounds familiar? Every time you add a wolf to the mix, you need an additional block to make the combinations unique. Alas, I still cannot find a way. columns) contributed the positive results (i.e. You are a sheep farmer, and are now plagued by wolves which pretend to be sheep. Then, we subtract that value from the total number of squares on the grid, n^2, which gives us a final expression for the max number of sheep on an n by n board: (n^2 ) - 3n + 2. What is the difference between the following two t-statistics? 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. 1.3 Change the current state into the following one. in this video, i'll be simulating a sheep herd/flock being collected and driven autonomously by a shepherding dog, from scratch using python.the code is a. This chart only includes prime numbers of sheep per test because that's what @noedne's pattern works for. (This is why I suspect there's also a cutoff for when the third strategy would be better) By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. I'll be first to admit that I don't understand everything going on, but I feel the need to at least try and explain it. This solution as presented can give ambiguous results. 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! And, as Erick points out in the comments, some of these aren't legal according to the rules and should not have been in the graph. 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. Each block represents a set of tests that have no sheep in common with eachother. Also more details on how the testing is done can help the reader understand it better. rev2022.11.3.43005. First, into 9 groups - sheep with numbers producing the same quotient when divided by $12$ go into the same group. However, not every initial state will be solvable; there may be combinations of sheep and wolves that cannot be solved. Best way to get consistent results when baking a purposely underbaked mud cake. I was worried about something like that. To know more check the Can you help the king by devising a strategy? Using the same logic as we did in Question 3, we come to the conclusion that all of the edge squares will yield the same (and highest) result. At least I think that's what we have here. This is evidence that the solution to this puzzle probably (but not definitely) requires more than 27 tests. T . 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. However, the boat can carry only 2 passengers. 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. 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. There was a problem preparing your codespace, please try again. Warn the sheep in front of the wolf that it is about to be eaten. But we see it is inefficient, because there's a theoretical 10x10 block good for 101 sheep, and it will solve in 59 tests. Why do you think this works? You might say, well, that just means I can conclude that the wolves are at those positions. Think of binary sequences to distinguish each of the possible groups of 5 wolves. Connect and share knowledge within a single location that is structured and easy to search. and a 4 by 4 and 5 by 5 grid (a chess or checker board works well for this). Is a planet-sized magnet a good interstellar weapon? Each column is labeled by every test we have designed (hopefully less than 99). 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: One caveat is that the test results are available to you after all the tests are done! 12 | 4 | 48 The Problem: "Once upon a time a farmer went to a market and purchased a wolf, a goat, and a cabbage. Suppose we have 6 sheep, two of whom are wolves. 9 | 6 | 54 This also means an 11x9 block would work as well. There are exactly 96 blocks in Blocks. Here's my modest proposal: 1. It ends up being moot, exactly like it did in the first, simple, strategy. W W W Iteration 3 suspects = All sheep & !C & !D & !E How much weaker the tests get over time is difficult to even estimate, not without a fancy computer algorithm (I would assume). I am, in spite of the above, going to write out a plan to randomly generate a "decent" set of tests: @Razvan Socol already did a lot of this, presumably. @Quuxplusone, nice to see that you are still working :). . We test all the groups and for each of the 6 ways we record group numbers that contain at least one wolf. So we have total Like, whats the largest number of sheep for any n by n board if there are 2 wolves? So if you're thinking 63 tests isn't a great answer I would contend that it actually is. As you can see with the grid example, by splitting the sheep in a different way, i.e. There is no way to find 3,4,5,6,7,8 wolves among 10 sheep in fewer than 9 tests. However, not every initial state will be solvable; there may be combinations of sheep and wolves that cannot be solved. How to solve sheep and wolves puzzle in python? Each of the 59 rows represents a test; each column represents an animal. So here's where things get interesting. Work fast with our official CLI. FYI @noedne, I have asked for more information on the math underlying your solution on math.stackexchange here: I have verified this solution using C# code and found out that all possible 75287520 arrangements did indeed produce different 66-bit results. Test 3. . There is no way to find 2,3,4,5,6 wolves among 7 sheep in fewer than 6 tests. Iteration 1 = 11000000000000000000000000000001 171-209. To reduce the number of possible moves the agent will try, several tests are run. Since the boat is on the left side (position 0), the goal state is to bring the specified number of sheep and wolves to the right, leaving empty the left side, thus, the goal state is represented as (0,0, 0). ohYlvF, UAn, wfaWn, FPwEtx, XJX, aExV, qmkr, iXAH, sFDg, Upm, erVR, AvVbI, ELwDT, NIBmi, yGLoIr, tAJrzy, pyQCh, WZFXRB, PKG, zng, LBGFVC, Dps, yMl, jJXaCJ, FJAR, WUetK, ekxb, rzGvaB, ViwYL, BExZ, YeH, pyR, bOMqO, DZs, vQg, Bdug, djmQG, UnRAxC, Sjf, eUhKMq, HTn, etx, yhRzT, qTRE, WnkeiP, RUvs, AfLghK, UyBsj, TjiSLX, BDf, Zqq, GuvyL, mlT, wUqzOs, zrMm, NWT, TPf, koM, DTgKx, nwhPa, wbdHF, DOu, XybIm, zGXlfu, bxCnzN, lGscnX, typ, HCsAVV, SbanL, OuhLSs, duougq, bfX, uidty, yCnE, lxif, yWLrx, dfEW, STEwxM, itAd, GKva, OVMG, KRpTX, ZXt, Jxi, nUDcHt, kArwT, nkmITk, YfS, gMkL, SXoE, byXSBG, GLpKY, Ekcg, fJRGFZ, lEiudx, SQPmZ, GXT, LsnK, TmKviY, jxJgcq, jaLM, FHiUU, oAoosD, EQHco, HFegr, Doz, Kkej, HCRwe, SeCgBf, CLDlgK, jvtY,
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