Yes, I meant 3 groups - though if you go bottom-up rather than top-down it's similar (I was thinking of the 9-coin version, where it's the same thing!). One of the nine identically looking coins is fake. Any other combination will fall into one of these two groups, like [(2)(45) and (13)], etc. Similarly we can analyze the right subtree. Pick one genuine coin from any of weighed groups, and proceed with (ABCD) as explained in Problem 3. Analysis: Given (N + 1) coins, one is genuine and the rest N can be genuine or only one coin is defective. Blue leaves are valid outcomes, and red leaves are impossible cases. In the worst case, how many minimum number of weighing are required to figure out the odd coin? Analysis: In general, if we know that the coin is heavy or light, we can trace the coin in log3(N) trials (rounded to next integer). We need to make use that in the groupings. Write pseudo code for the divide-into-three algorithm for the fake-coin problem. Here we need one more weighing to check a genuine coin against 1 or 2. o BINARY SEARCH o FAKE COIN PROBLEM , dividing on 3 piles, complexity is less, though log 3 n o RUSSIAN PEASANT MULTIPLICATION o JOSEPHUS PROBLEM. We need two weighing in worst case. Given that a coin is heavier, verify that 3 trials are sufficient to find the odd coin among 12 coins, because 32 < 12 < 33. The required decision tree should result in minimum of (2N + 1) leaves. If(i=l-1) // Reducing the problem to two coins. There can't be $2$ coins that go through the same weighing course (because we won't be able to know which one is fake). (1234) > (5678), i.e. The decision tree confirms the fact (try to draw). Specifically read section 5.5 and section 11.2 including exercises. b. Name the coins as 1, 2, 3, 4 and 5. 1, 2, 3 are genuine and 4th coin can be figured out lighter or heavier in one more trial. The genuine coins all have the same weight and the fake coins all have the same weight, but the fake coins are lighter than the real ones. For n = 2k, this requires exactly k = log2(n)weighings. We started with (N + 1) = 5 coins where N = 4, we end up with (2N + 1) = 9 leaves. In this video, the Fake Coin problem is discussed This video has no prerequisites. (1234) = (5678), both groups are equal. Make sure that your algorithm handles properly all values of n, not only those that are multiples of 3. We need a decision tree with atleast (2N + 1) leaves correspond to the outputs. Difficult: Given a two pan fair balance and N identically looking coins out of which only one coin may be defective. Your friend knows that there are either f or d (the number we are trying to disprove) fake coins in the pile. Among them are f fake coins. The solution to this problem for a given set of coin … Infact, we can get 27 leaves of 3 level full 3-ary tree, but only we got 11 leaves including impossible cases. And also, we know a set of coins being lighter and a set of coins being heavier. Rearranging k and N, we can weigh maximum of N <= (3k – 3)/2 coins in k trials. this forum made possible by our volunteer staff, including ... it would depend on the specific fake coin problem you are trying to solve. With just c-3 (i.e. We know that groups (1234) and (5678) are genuine and defective coin may be in (ABCD). Place one coin on one side of the balance scale, one on the other, and put one coin aside. With the second group, weigh (12) and (34). In the figure I took (3, 2) where 3 is confirmed as genuine. B[j++]=j // Marking the coins with a number. How can I get my machine's name/IP by java? If n is even, divide the coins into two equal stacks and put them on each side of the balance. Solve Problem 4 with N = 8 and N = 13, How many minimum trials are required in each case? If the number of coins, c = (n^3 - 3)/2 for some positive integer value of n, then the solution can be found. post an algorithm that will determine the real coin Set up a recurrence relation for the number of weighing in the divide-into three algorithm for the fake-coin problem and solve it for n = 3k. We are given 5 coins, a group of 4 coins out of which one coin is defective (we don’t know whether it is heavier or lighter), and one coin is genuine. Decrease-by-Constant-Factor Example:Factor Example: Fake-Coin Problem Decrease-by-factor-2 algorithm: if n=1 the coin is fake else ddd o o op oivide the coins into two piles of ⎣n/2⎦cooa,ago ao oddins each, leaving one extra coin if n is odd weigh the two piles if they weigh the same return the one extra coin as the fake coin else continue with the lighter of the two piles e.g. If n is even, divide the coins into two equal stacks and put them on each side of the balance. Creativity is allowing yourself to make mistakes; art is knowing which ones to keep. C) (3) = (4), yielding ambiguity. From the above two examples, we can ensure that the decision tree can be used in optimal way if we can reveal atleaset one genuine coin. Any other group can’t generate full 3-ary tree, try yourself. One stack will be heavier than the other. The left side of tree corresponds to the case (G1) < (23). OK, you have n coins. 4. For instance, if both coins 1 and 2 are counterfeit, either coin 4 or 5 is wrongly picked. Apache Tomcat, SSL: how to add new algorithm. if they weigh the same then discard all of them and continue with the coins of the third pile else continue with the lighter of the first two piles. Consider the first group, pairs (1, 2) and (3, 4). //Output:- The position of fake coin or appropriate message. We don’t know, whether all coins are genuine or any defective one is present. (1234) < (5678), i.e. we are missing valid outputs under some branches that leading to more number of trials. It can’t lead to best tree. Problem Statement. Keep this tiny ad: current ranch time (not your local time) is, https://coderanch.com/t/730886/filler-advertising. Count inversions in an array | Set 3 (Using BIT), Segment Tree | Set 2 (Range Minimum Query), XOR Linked List – A Memory Efficient Doubly Linked List | Set 2, proto van Emde Boas Trees | Set 1 (Background and Introduction), Self-Balancing-Binary-Search-Trees (Comparisons), Remove edges connected to a node such that the three given nodes are in different trees, Proto Van Emde Boas Trees | Set 4 | Deletion, Dynamic Segment Trees : Online Queries for Range Sum with Point Updates, LCA for general or n-ary trees (Sparse Matrix DP approach ), Segment Trees | (Product of given Range Modulo m), Difference between Backtracking and Branch-N-Bound technique, K Dimensional Tree | Set 1 (Search and Insert), Write Interview
Please write to us at contribute@geeksforgeeks.org to report any issue with the above content. University of Pittsburgh, 2013 Although a global solution for the Traveling Salesman Problem does not yet exist, there are algorithms for an existing local solution. Else The coin problem (also referred to as the Frobenius coin problem or Frobenius problem, after the mathematician Ferdinand Frobenius) is a mathematical problem that asks for the largest monetary amount that cannot be obtained using only coins of specified denominations. THE TRAVELING SALESMAN PROBLEM Corinne Brucato, M.S. You have 12 identically looking coins out of which one coin may be lighter or heavier. e.g. Divide and conquer. We can group the coins in two different ways, [(12, 34)] or [(1, 2) and (3, 4)]. Let us shuffle coins as (1235) and (4BCD) as new groups (there are different shuffles possible, they also lead to minimum weighing). Use of balance consists of placing a certain number of coins on each pan, checking the state of the balance and then removing them all. Set j=i and B[l] 2. Here are the detailed conditions: 1) All 12 coins look identical. The leaf nodes on left branch are named to reflect these outcomes. Repeat step 3 while j!=l. The counterfeit coin problem With n coins all the same weight except for one which could weigh more or less, determine the minimum number of weighings x which must be performed in balance scales to identify whether this coin exists and whether it is heavier or lighter than the rest. Given a 3-pan balance (4 outcomes) and N coins, how many minimum trials are needed to figure out odd coin. There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors. Mike Simmons wrote:Even if you don't know the fake coin is lighter, you can do it more efficiently than that. Label the coins as 1, 2, 3, 4 and G (genuine). c. We need two more weighings on right subtree as well. The later case could be (2) < (3) yielding 2 as lighter coin, or (2) > (3) yielding 3 as lighter. Note that we are unable to utilize two outcomes of 3-ary trees. An algorithm for reaching Ivo's answer: You can find the fake out of up to 121 coins, but you cannot necessarily distinguish if the fake is heavy or light. If one of the sides of the scale is heavier, the fake coin is on that side. fake coin problem 1. Similar way we can also solve the right subtree (third outcome where (1234) > (5678)) in two more weighing. Similar problem was provided in one of the exercises of the book “Introduction to Algorithms by Levitin”. Let us name the coins as 1, 2, … 8, A, B, C and D. We can combine the coins into 3 groups, namely (1234), (5678) and (ABCD). We can easily rule out groups like [(123) and (45)], as we will get obvious answer. We can best group them as [(G1, 23) and (4)]. The following diagram explains the procedure. Solution. From the above analysis we may think that k = 2 trials are sufficient, since a two level 3-ary tree yields 9 leaves which is greater than N = 4 (read the problem once again). Consider the second outcome where (1234) < (5678). We know that one coin is lighter. Find the minimum number of coins required to form any value between 1 to N,both inclusive.Cumulative value of coins should not exceed N. Coin denominations are 1 Rupee, 2 Rupee and 5 Rupee.Let’s Understand the problem using the following example. we go on to right subtree. A 3-ary tree at k-th level will have 3k leaves and hence we need 3k >= N. In other-words, in k trials we can examine upto 3k coins, if we know whether the defective coin is heavier or lighter. Write pseudocode for the divide-into-three algorithm for the fake-coin problem. You are encouraged to draw decision tree while reading the procedure. we go on to left subtree or (12) > (34) i.e. The output (1) can be solved in two more weighing as special case of two pan balance given in Problem 3. We now have some information on coin purity. 1. Analysis: Given N coins, all may be genuine or only one coin is defective. first group is less in weight than second group. At one point, it was known as the Counterfeit Coin Problem: Find a single counterfeit coin among 12 coins, knowing only that the counterfeit coin has a weight which differs from that of a good coin. One weighing. We need to shuffle the weighed two groups in such a way that we end up with smaller height decision tree. The later instance could be (2) < (3) yielding 3 as heavier or (2) > (3) yielding 2 as heavier. Because there can be N leaves to be lighter, or N leaves to be heavier or one genuine case, on total (2N + 1) leaves. Problem 4 describes this approach of 12 coins. There are at most n-1 fake coins that are either lighter or heaver than the real coin. Make sure that your algorithm handles properly all values of n, not only those that are multiples of 3. b. How many number of weighing are required in worst case to figure out the odd coin, if present? Let us start with relatively simple examples. The problem is, we're only allowed the use of a marker (to make notes on the coins) and three uses of a balance scale. If coins 0 and 13 are deleted from these weighings they give one generic solution to the 12-coin problem. We can get (3) > (2) in which 2 is lighter, or (3) = (2) in which 1 is lighter. Weigh (1234) and (5678). separate 1, 2 and 3. Experience, A) (3) < (4) yielding 4 as defective heavier coin, OR, B) (3) > (4) yielding 3 as defective heavier coin OR. Let us consider (3, 4) as the analogy for (1, 2) is similar. Since any coin among N coins can be defective, we need to get a 3-ary tree having minimum of N leaves. Below are example problems : Binary search; Fake-coin problems; Russian peasant multiplication; Variable-Size-Decrease : In this variation, the size-reduction pattern varies from one iteration of an algorithm to another. Set up a recurrence relation for the number of weighings in the divide-into-three algorithm for the fake-coin problem and solve it for n = 3k. Else If (A[i]>A[l])// Else If (A[i] (4BCD) yielding either 5 as heavier coin or 4 as lighter coin, at the expense of one more weighing. Sort the list in alphabetical order by selection sort. Here's an algorithm that compares as many coins as possible at once. If you're already bored with the thread up to this point, this post will only make matters worse. Observe the above figure that not all the branches are generating leaves, i.e. a. The outcome can be three ways. Problem 3: (Special case of two pan balance). Dennis Deems wrote:Divide and conquer. By using our site, you
When possible, we should group the coins in such a way that every branch is going to yield valid output (in simple terms generate full 3-ary tree). The fake coin problem can be solved recursively using the decrease-by-constant factor strategy. You might have seen 3-pan balance in science labs during school days. Finding the Fake Coin We have been studying decrease-and-conquer, so it's not too surprising that a decrease-and-conquer algorithm works here. One of the coins is a counterfeit coin. In this video, the Fake Coin problem is discussed This video has no prerequisites. Decrease by a constant factor algorithms are very efficient especially when the factor is greater than 2 as in the fake-coin problem. Without a reference coin Considering best out come of balance, we can group the coins in two different ways, [(1, 2), (3, 4) and (5)], or [(12), (34) and (5)]. As explained earlier ternary tree at level k, can have utmost 3k leaves and we need a tree with leaves of 3k > (2N + 1). https://www.geeksforgeeks.org/decision-trees-fake-coin-puzzle This can be solved in one weighing (read Problem 1). If the scale balances, the coin that we put aside is fake. Since the total possible outcomes are (2(N + 1) + 1), number of weighing (trials) are given by the height of ternary tree, k >= log3[2(N + 1) + 1]. Remember to group coins such that the first weighing reveals atleast one genuine coin. We arrived at impossible cases due to the assumptions made earlier on the path. In the above problem, under any possibility we need only two weighing. You are given 101 coins, of which 51 are genuine and 50 are counterfeit. And perhaps "groups of 3" should be "3 groups"? Outcomes (2) and (3) are special. 1. Replace coin 1 from Sn-1 with coins 1, 2, and 3 in... 2. Please use ide.geeksforgeeks.org, generate link and share the link here. Objective: Given a set of coins and amount, Write an algorithm to find out how many ways we can make the change of the amount using the coins given. In each recursive call, divide the total coins as follows: • If n =1, the coin is the fake coin and return it as a fake coin • If n =2, compare them and call the algorithm recursively on the lighter coin. Prove that no algorithm can be faster that yours. How can we trace which coin, if any, is odd one and also determine whether it is lighter or heavier in minimum number of trials in the worst case? An algorithm is a procedure or formula for solving a problem, based on conducting a sequence of specified actions. Note: each coin can be heavier or less than the real coin and they are not the same weight. Given 5 coins out of which one coin is lighter. – – – by Venki. Defective coin may be in (ABCD) group. Add a third weighting that separates the coins in each group of three. of ECE, Arasu Engineering College, Kumbakonam, Tamilnadu ,India #5Assistant professor,Dept. Note the equality sign. If we proceed as in Problem 1, we will not generate best decision tree. Devise a brute-force algorithm to identify the stack with the fake coins and determine its worst-case e ﬃ ciency class. You are only allowed 3 weighings on a two-pan balance and must also determine if the counterfeit coin … In other words, we need atleast k > log3(2N + 1) weighing to find the defective one. On each weighing, number of coins on each side has to be the same, #L=#R (because there is no meaning to fake/real coins if I don't weigh them against the same number of coins … Note that it is impossible to solve above 4 coins problem in two weighing. We are given 4 coins, out of which only one coin may be defective. You are given two pan fair balance. Bi-set or tri-set? Note that it impossible to get (3) < (2), it contradicts our assumption leaned to left side. We still have to worry about dealing with an odd number during a weighing, but This is possible in two ways, either 1 should be lighter or either of (2, 3) should be heavier. a. Answer: a. Algorithm for the fake-coin problem: If n mod 3 =1 then divide the coins into the piles of sizes k, k and K=1, K+1 ALGORITHM fakeCoin(n) if n=1 then the coin is fake else divide the coins into three piles. Just put the first... 3. What is the minimum number of weighings needed to identify the stack with the fake coins? ...
To find the fake coin out of 3 coins in 2 weightings...
1 <--> 2
1 <--> 3
...
From there, you can develop the algorithm for 12 coins in 3 weightings using the method on the site mentioned above. Let us solve the classic “fake coin” puzzle using decision trees. One stack will be heavier than the other. For each coin, x, in Sn-1, replace it with 3x-2, 3x-1 and 3x. If we observe the figure, after the first weighing the problem reduced to “we know three coins, either one can be lighter (heavier) or one among other two can be heavier (lighter)”. Also, the tree is not full tree, middle branch terminated after first weighing. Overall we need 3 weighings to trace the odd coin. We also need to tell whether it is heavier or lighter. The same analogy can be applied when the coin in heavier. The algorithm lets the user specify if the coin is a heavy one or a lighter one or is of an unknown nature. Starting with n coins: If there is one coin left, then it is the fake coin. The 12 coin problem: There are twelve coins which are equally likely to be fake and the fake coin is either more or less heavy. Given N coins with a fake among them which has a weight slightly more than the real one, what is the minimum number of times you need to use the balance to correctly identify the fake one in the worst case? You may well have realized that you can divide the pile in half, weigh the halves, and narrow your focus to the pile that is lighter. 1.1. We are able to solve the 12 coin puzzle in 3 weighing in the worst case. Put the remaining n -2⌊ n /2⌋ coins into a separate pile, C. After reading every problem try to solve on your own. fake coin, and in exactly 10balance weighings, we determine the coin. How to design a tiny URL or URL shortener? The outcome of second trail can be three ways. The middle case (G1) = (23) is self explanatory, i.e. The outcome can be (12) < (34) i.e. S #1, Abinash #2, Surya Sabeson #3, Arun Balaji.A #4, Kavitha.D #5 #1,2,3,4 UG Students, Dept. Otherwise, divide the coins into 2 equal piles, A and B, of size ⌊ n /2⌋ coins. The former instance is obvious when the next weighing (2, 3) is balanced, yielding 1 as heavier. The former instance is obvious when next weighing (2, 3) is balanced, yielding 1 as lighter. To figure out the odd coin, how many minimum number of weighing are required in the worst case? In addition, you know the location of each of the fake coins. Murprhey's law strikes again. The right side of tree corresponds to the case (G1) > (23). This is another problem in which i will show you the advantage of Dynamic programming over recursion. This is possible in two ways, either 1 is heavier or either of (2, 3) should be lighter. We revealed lighter or heavier possibility after first weighing. 3. Infact we should have 11 outcomes since we stared with 5 coins, where are other 2 outcomes? There are the two different variants of the puzzle given below. How do you want to group them? In both the cases, we know that (ABCD) is genuine. acknowledge that you have read and understood our, GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, Segment Tree | Set 1 (Sum of given range), XOR Linked List - A Memory Efficient Doubly Linked List | Set 1, Largest Rectangular Area in a Histogram | Set 1, Design a data structure that supports insert, delete, search and getRandom in constant time. Easy: Given a two pan fair balance and N identically looking coins, out of which only one coin is lighter (or heavier). How can you find odd coin, if any, in minimum trials, also determine whether defective coin is lighter or heavier, in the worst case? If they balance (5) is defective one, otherwise pick the lighter pair, and we need one more weighing to find odd one. We are able to use all outcomes of two level full 3-ary tree. Greg Charles wrote:I knew that math looked wrong. The left subtree is possible in two ways, Further on the left subtree, as second trial, we weigh (1, 2) or (3, 4). You are given n gold coins, and one of them is fake.Assume that all the coins are identical, except that the fake coin is lighter.Given a balance scale, where you can put a bunch of coins on the left and the right and determine which is heavier, design the fastest algorithm for determining the fake coin.. 12 coin puzzle in 3 weighing in case of two pan fair balance N. Below, try to solve the 12 coin puzzle in 3 weighing in groupings... Into 2 equal fake coin problem algorithm, a and b, of which 51 are genuine and defective may! More number of weighing are required in the groupings instance, if present ( read problem 1 leaves! 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