Client Service Portal

Finding Longest Increasing Subsequence in O(nlogn) time. In the last post, longest increasing subsequence, we discussed brute force and dynamic programming based solutions. Example: Input: [10,9,2,5,3,7,101,18] Output: 4 Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4. Making 1 as new sequence will create new sequence which is largest. For example. Proof: Suppose it is not and that there exists some where either or .We will prove neither that case is possible. This article has taken some inspiration from: http://stackoverflow.com/questions/6129682/longest-increasing-subsequenceonlogn and the comments provided by readers under these articles. If longest sequence for more than one indexes, pick any one. This is an implementation of Longest Increasing Subsequence in C. // Returns the length of the longest increasing subsequence. For example, given [10, 9, 2, 5, 3, 7, 101, 18], the longest increasing subsequence is [2, 3, 7, 101]. The length of the LCS is 6. It is easier to come out with a dynamic programming solution whose time complexity is O (N ^ 2). How do we decide when to replace and when to continue with the old element in the list of subsequences? The Longest Increasing Subsequence (LIS) problem is to find the length of the longest subsequence of a given sequence such that all elements of the subsequence are sorted in increasing order. 1. The complexity of this algorithm is O(nlogn) as for each element in the array, it requires O(logn) time to find the ceiling of it and put it at the correct position. This is called the Longest Increasing Subsequence (LIS) problem. We will analyze this problem to explain how to master dynamic programming from the shallower to the deeper. Use Longest Common Subsequence on with and . The link has explanation of approach mentioned in the Wiki. The question is, when will it be safe to add or replace an element in the existing sequence? Find longest monotonically increasingsubsequence (LIS) in the array. Brute-Force (TLE) - O(2^n) time. Let max[i] represent the length of the longest increasing subsequence so far. “end element of smaller list is smaller than end elements of larger lists”. Example 1 . An increasing subsequence contains elements A[i] and A[j] only if i < j and A[i] <  A[j]. Given below is code to find length of LIS (updated to C++11 code, no C-style arrays), edit For the time being, forget about recursive and DP solutions. ), we may end up querying ceil value using binary search (log i) for many A[i]. We can solve the problem recursively and dynamic programming (DP) technique. By gautam94, 6 years ago, , - - -I am having trouble understanding the nlogn algorithm for solving the LIS problem. Instead of two sequences, 3 can replace 5 in the sequence {2, 5}. Design an algorithm to construct all increasing lists of equal longest size. Therefore, T(n) < O( log N! ) 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, Maximum size rectangle binary sub-matrix with all 1s, Maximum size square sub-matrix with all 1s, Longest Increasing Subsequence Size (N log N), Median in a stream of integers (running integers), Median of Stream of Running Integers using STL, Minimum product of k integers in an array of positive Integers, K maximum sum combinations from two arrays, K maximum sums of overlapping contiguous sub-arrays, K maximum sums of non-overlapping contiguous sub-arrays, k smallest elements in same order using O(1) extra space, Find k pairs with smallest sums in two arrays, k-th smallest absolute difference of two elements in an array, Find the smallest and second smallest elements in an array, Maximum and minimum of an array using minimum number of comparisons, Reverse digits of an integer with overflow handled, Write a program to reverse digits of a number, Write a program to reverse an array or string, Rearrange array such that arr[i] >= arr[j] if i is even and arr[i]<=arr[j] if i is odd and j < i, Rearrange positive and negative numbers in O(n) time and O(1) extra space, Rearrange array in alternating positive & negative items with O(1) extra space | Set 1, Rearrange array in alternating positive & negative items with O(1) extra space | Set 2, Design an algorithm to construct the longest increasing list, Longest Monotonically Increasing Subsequence Size (N log N): Simple implementation, Longest Increasing Subsequence using Longest Common Subsequence Algorithm, Construction of Longest Increasing Subsequence (N log N), Longest Bitonic Subsequence in O(n log n), Longest Common Increasing Subsequence (LCS + LIS), Construction of Longest Increasing Subsequence(LIS) and printing LIS sequence, Find the Longest Increasing Subsequence in Circular manner, C/C++ Program for Longest Increasing Subsequence, C++ Program for Longest Increasing Subsequence, Java Program for Longest Increasing Subsequence, Python program for Longest Increasing Subsequence, Longest Increasing consecutive subsequence, Printing longest Increasing consecutive subsequence, Length of the longest increasing subsequence such that no two adjacent elements are coprime, Length of longest increasing index dividing subsequence, Maximize sum of all elements which are not a part of the Longest Increasing Subsequence, Longest Increasing Subsequence having sum value atmost K, Maximum Sum Increasing Subsequence | DP-14, Given an array A[] and a number x, check for pair in A[] with sum as x, Stack Data Structure (Introduction and Program), Write Interview In case of our original array {2, 5, 3}, note that we face same situation when we are adding 3 to increasing sequence {2, 5}. Update – 17 July, 2016: Quite impressive reponses from the readers and few sites referring the post, feeling happy as my hardwork helping others. Analyse to ensure that the upper and lower bounds are also O( N log N ). Given an unsorted array of integers, find the length of longest increasing subsequence. Consider an input array A = {2, 5, 3}. Size of this array in worst case will be n. To append to the list, add another element in the auxiliary array. The request is to help yourself. The solution is not unique for all pair of strings. The complexity of the brute force solution is exponential whereas for the dynamic programming approach it is O(n2). We can optimize on this, observe that we … 1 Longest Common Subsequence Definition: The longest common subsequence or LCS of two strings S1 and S2 is the longest subsequence common between two strings. We will analyze this problem to explain how to master dynamic programming from the shallower to the deeper. 4. Length of Longest Increasing Subsequence is 6 — Venki . When I start writing content to explain the reader, I realized I didn’t understand the cases. I realized I have already covered the algorithm in another post. The maximum length of this array is that of input. Design an algorithm to construct the longest decreasing list.. Alternate implementation using lower_bound() in C++: — Venki. ... We can easily prove that tails is a increasing array. All the thought process for the solution triggered by a note in the book ‘Introduction to Algorithms by Udi Manber’, I strongly recommend to practice the book. If A[i] is largest among all end candidates of active lists, clone the largest active list, and append A[i] to it. Our strategy determined by the following conditions. Let’s take an example and see how it works with an array A = [ 0, 8, 4, 12, 2, 10, 6, 14]. A[5] with value 10. How can we extend the existing sequences with 8? Ex. Querying length of longest is fairly easy. These cookies do not store any personal information. A simple way of finding the longest increasing subsequence is to use the Longest Common Subsequence (Dynamic Programming) algorithm. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. = O(N log N). We also maintain a counter to keep track of auxiliary array length. Run through few examples on paper. Proof: Lets use the method of induction: Base case : Trivially true. We will use an auxiliary array to keep end elements. Involve me and I will understand.” So, pick a suit from deck of cards. Java Solution 1 - Naive . Based on the current number being considered, update these active lists. http://stackoverflow.com/questions/2631726/how-to-determine-the-longest-increasing-subsequence-using-dynamic-programming. → Output: Longest Increasing subsequence: 7 Actual Elements: 1 7 11 31 61 69 70 NOTE: To print the Actual elements – find the index which contains the longest sequence, print that index from main array. I leave it as an exercise to the reader to understand how it works. Let’s revisit the problem statement: Given an array of integers, find the length of the longest increasing subsequence. Please write to us at contribute@geeksforgeeks.org to report any issue with the above content. Last Edit: a day ago. Obviously, it can’t extend either. (⁡ ()) time. Question is – Can we find the longest increasing subsequence in O(nlogn) complexity? If A[i] is the smallest among all end candidates of active lists, start a new active list with A[i] of length 1. This website uses cookies to improve your experience while you navigate through the website. Box stacking problem. Note that at any instance during our construction of active lists, the following condition is maintained. Idea. We can replace 11 with 8, as there is potentially best candidate (9) that can extend the new series {2, 3, 7, 8} or {2, 5, 7, 8}. I just created two increasing sequences to make explanation simple. Start moving backwards and pick all the indexes which are in sequence (descending). It is important to understand what happening to end elements. recursively search: include or not include the next position, and find the maximum of them. The invariant is to maintain lists of increasing sequences and update them based on the next number. Say, the next element is 1. Therefore the length is 4. Lists = [ [0], [0, 2], [0,2,10] ] and [0, 4, 12] is discarded. 2. Recursive with Memoization (MLE) We are adding an element A[i] to these lists. Discard all other lists of the same length as that of this modified list. It seems like a lot of things need to be done just for maintaining the lists and there is significant space complexity required to store all of these lists. The longest increasing subsequence in this example is not unique: for instance,     {0, 4, 6, 9, 11, 15} or We will verify the end elements of all the lists to find a list whose end element is smaller than A[i] (floor value). Even though it may look complex at first time, once if we understood the logic, coding is simple. Induction hypothesis: Suppose we have processed i-1 elements and the length of the set is LIS[i-1], i.e the length of the LIS possible with first i-1 elements. 2. To find the smallest number which is greater than the current number, we can use binary search algorithm. Yet, there is a potential that the new smallest element can be start of an LIS. The decision to take for each element being considered is whether we create new active subsequences with length 3 with element 9 in them or continue with 11. The following algorithm shows how to add/replace the new elements in the existing lists or to create a new list with it. We claim that D [ i + 1] is the length of longest increasing subsequence ending at A [ i + 1]. Contribute to mission-peace/interview development by creating an account on GitHub. Your Task: Complete the function longestSubsequence() which takes the input array and its size as input parameters and returns the length of the longest increasing subsequence. Next, we go to A[1] which is 8. Requesting to run through some examples after reading the article, and please do your work on paper (don’t use editor/compiler). Instead of getting the longest increasing subarray, how to return the length of longest increasing subsequence? Given an array of random numbers. For example, the length of LIS for {10, 22, 9, 33, 21, 50, 41, 60, 80} is 6 … Level: MediumAsked In: Amazon, Facebook, Microsoft Understanding the Problem. Initial content preparation took roughly 6 hours to me. Whatever the content you are seeing in the gray colored example is from these pages. Same as A[5] We will clone the list which has end smaller than A[6], extend it, and discard all other lists which have the same length. First of all, can 8 be part of LIS? For example, the length of LIS for {10, 22, 9, 33, 21, 50, 41, 60, 80} is … If the input is [1, 3, 2, 3, 4, 8, 7, 9], the output should be 5 because the longest increasing subsequence is [2, 3, 4, 8, 9]. You also have the option to opt-out of these cookies. These elements will extend the existing sequences. I suspect, many readers might not get the logic behind CeilIndex (binary search). The longest subsequence is not necessarily contiguous, or unique. Also, model your solution using DAGs. Our output will be 4, as {2,3,5,8} is the longest increasing subsequence. Please share if you find something wrong or missing. By gautam94, 6 years ago, , - - -I am having trouble understanding the nlogn algorithm for solving the LIS problem. In the worst case the array divided into N lists of size one (note that it does’t lead to worst case complexity). It is required to understand above strategy to devise an algorithm. Given below was my personal experience. 2. Note that we are dealing with end elements only. Application. Necessary cookies are absolutely essential for the website to function properly. It is mandatory to procure user consent prior to running these cookies on your website. For A[2] with value 4, A[i] is less than the end of one of the list and greater than the end of other. Let us consider another sample A = {2, 5, 3}. Last Edit: October 24, 2018 3:27 AM. and replace an number with A[i], if there exists a number A[j] such that if E > A[i] < A[j], it means, the new number falls somewhere between A[j] and E. What if A[i] is smaller than all elements in the present list of subsequences? Lists = [ [0], [0, 2], [0,2,6] ] and [0, 2, 10] is discarded. Attention reader! Link to CPP implementation. In this case, we have to create a new list and add A[i] into it. Recursion leads to exponential algorithm as we solve overlapped subproblems again and again, and DP is quadratic algorithm. Longest Increasing Subsequence Problem with O(nlogn) complexity. Also, if you want to contribute to the website, please refer to Publishing and contact us. Given an unsorted array of integers, find the length of longest increasing subsequence. Your algorithm should run in O(n^2) complexity. Further, we add one more element, say 8 to the array i.e. Inspired by http://www.geeksforgeeks.org/longest-monotonically-increasing-subsequence-size-n-log-n/ int lengthOfLIS ( vector < int >& nums) { vector < int > res; for ( int i= 0 ; i i) such that E < A[i] < A[j] or (E > A[i] < A[j] – for replace). Note that S is not the LIS itself. I have implemented the algorithm given here on page number 6. ), we are not sure whether adding 8 will extend the series or not. The longest increasing subsequence in the given array is [ 0,2,6,14] with a length of 4. Since we are searching for the longest increasing subsequence then we take the maximum value for all possible k < i + 1 and so D [ i + 1] is the maximum value plus 1 as defined above. The loop runs for N elements. We can optimize on this, observe that we use only ends of the list and their sizes. - It is an increasing subsequence; - There exists an increasing subsequence (in the input read so far) with the same lenght of the sequence stored in S, and terminating in the same way of the sequence stored in S. - Such increasing subsequence is as long as possible. Let us add two more elements, say 7, 11 to the array. Assume there is 9 in the input array, say {2, 5, 3, 7, 11, 8, 7, 9 …}. Since the approach is offline (what we mean by offline? We extend a list by adding element to auxiliary array. 14 VIEWS. In computer science, the longest increasing subsequence problem is to find a subsequence of a given sequence in which the subsequence's elements are in sorted order, lowest to highest, and in which the subsequence is as long as possible. We will find the list which has end less than A[i], in this case, the first list containing [0]. Clone it and append A[2] to it and discard all other lists of the same length. brightness_4 The above O(nlogn) code will go wrong in {2, 6, 7, 4, 1, 2, 9, 5, 8} case. If we want to add 8, it should come after 7 (by replacing 11). // Note that this is looking for the longest strictly increasing subsequence. algorithm Longest Increasing Sequence는 “순서” 또는 “아이템들을 연속해서 조합할 때의 최대” 등을 구할 때 매우 많이 사용되는 방법이다. This subsequence has length 6; the input sequence has no 7-member increasing subsequences. 0. flyseeksky 99. You will never forget the approach. To replace just overwrite the smallest number which is greater than the current number. By observation we know that the LIS is either {2, 3} or {2, 5}. We would love to publish your article and at the same time, will pay you too. But opting out of some of these cookies may have an effect on your browsing experience. If the next element is 10 we know that adding 9 to subsequence leads us to longer subsequences rather than keeping 11. Given an array of random numbers. To understand this process, let’s work out an example. From the observations, we need to maintain lists of increasing sequences. Clone and append A[i] to this list. An Introduction to the Longest Increasing Subsequence Problem. Make a sorted copy of the sequence , denoted as . Input: N = 6 A[] = {5,8,3,7,9,1} Output: 3 Explanation:Longest increasing subsequence 5 7 9, with length 3. Longest Increasing Subsequence in O(nlogn), http://stackoverflow.com/questions/6129682/longest-increasing-subsequenceonlogn. Russian doll envelopes. 4. For example, the length of LIS for {10, 22, 9, 33, 21, 50, 41, 60, 80} is … Given a sequence of elements c 1, c 2, …, c n from a totally-ordered universe, find the longest increasing subsequence. So, can we store the ends of all the lists of an auxiliary array and do operations on them? But, it was a good lesson. S1 : A--AT-- G G C C-- A T A n=10 S2: A T A T A A T T C T A T --m=12The LCS is AATCAT. Note that the latest element 8 is greater than smallest element of any active sequence (will discuss shortly about active sequences). Longest increasing subsequence Longest increasing subsequence. The longest increasing subsequence problem is to find a subsequence of a given sequence in which the subsequence’s elements are in sorted order, lowest to highest, and in which the subsequence is as long as possible. What if we add another element, 11 in this? It looks like readers are not doing any homework prior to posting comments. The task is to find the length of the longest subsequence in a given array of integers such that all elements of the subsequence are sorted in strictly ascending order. If we add it t0 subsequences, the length of the longest subsequence remains 3. Interview questions. The following link worth referring after you do your work. 그런데 이때 보통 i번째 아이템에 대해 0부터 i-1까지의 아이템을 비교해서 최대값을 갱신하는 O (n*n) 알고리즘 이 흔히 사용된다. 3. Java/Python Binary search O(nlogn) time with explanation. We'll assume you're ok with this, but you can opt-out if you wish. Statement: For each i, length of current set is equal to the length of the largest increasing subsequence. In general, we have set of active lists of varying length. We need not to maintain all the lists. There may be more than one LIS combination, it is only necessary for you to return the length. Show me and I will remember. It seems like a lot of things need to be done just for maintaining the lists and there is significant space complexity required to store all of these lists. Is the above algorithm an online algorithm? It is easier to come out with a dynamic programming solution whose time complexity is O (N ^ 2). Part of MUMmer system for aligning entire genomes. In the last post, longest increasing subsequence, we discussed brute force and dynamic programming based solutions.The complexity of the brute force solution is exponential whereas for the dynamic programming approach it is O(n 2).Question is – Can we find the longest increasing subsequence in O(nlogn) complexity?. The longest increasing subsequence in the given array is [ 0,2,6,14] with a length of 4. We do not care what was prior to them in list. Output: Length of the Longest contiguous subsequence is 4. → input array becomes {2, 5, 3, 7, 11, 8}. Try with few other examples, before reading further. Following the same approach, we will go through all the numbers in the given array. By using our site, you Bridges across the river. This category only includes cookies that ensures basic functionalities and security features of the website. Out of these cookies, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. Note: There may be more than one LIS combination, it is only necessary for you to return the length. 3. Output will be 5 but the original output is 4. In the last post, longest increasing subsequence, we discussed brute force and dynamic programming based solutions.The complexity of the brute force solution is exponential whereas for the dynamic programming approach it is O(n 2).Question is – Can we find the longest increasing subsequence in O(nlogn) complexity?. The longest increasing subsequence problem is to find a subsequence of a given sequence in which the subsequence’s elements are in sorted order, lowest to highest, and in which the subsequence is as long … The number bellow each missile is its height. 738. dietpepsi 10742. What if new element 9 is added to array? I have implemented the algorithm given here on page number 6. To find the length of the longest subsequence, keep track of the length of the auxiliary array because this will be the length of LIS. Also, ensure we have maintained the condition, “end element of smaller list is smaller than end elements of larger lists“. 7 2 8 1 3 4 10 6 9 5. Discarding operation can be simulated with replacement, and extending a list is analogous to adding more elements to array. This website uses cookies to improve your experience. Writing code in comment? 1. Find the longest increasing sub-sequence of cards from the shuffled suit. I will extend the array during explanation. What are the problems you can solve with the longest increasing subsequence? We also use third-party cookies that help us analyze and understand how you use this website. First, suppose that then this means that we have two strictly increasing subsequences that end in .Let the first subsequence be of length and let the second subsequence be of length and so .Since this is a strictly increasing subsequence, we must have . It will be clear with an example, let us take example from wiki {0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15}. Note that I am considering only strictly increasing sequences. The basic idea behind the solution is to keep track of all active subsequences at a given point in time. This subsequence is not necessarily contiguous, or unique. O(n 2) dynamic programming solution. Longest Increasing Subsequence (short for LIS) is a classic problem. Find longest increasing subsequence (LIS) in the array. So, before starting this problem, have a quick overview of Fenwick Tree or Binary Indexed Tree. I finished initial code in an hour. We can store the end elements in an array. The longest increasing subsequence in this example is not unique. We can write it down as an array: enemyMissileHeights = [2, 5, 1, 3, 4, 8, 3, 6, 7] What we want is the Longest Increasing Subsequence of … Prerequisite to understand this problem is knowledge about Fenwick Tree. The observation is, when we encounter new smallest element in the array, it can be a potential candidate to start new sequence. The Longest Increasing Subsequence (LIS) problem is to find the length of the longest subsequence of a given sequence such that all elements of the subsequence are sorted in increasing order. To discard an element, we will trace ceil value of A[i] in auxiliary array (again observe the end elements in your rough work), and replace ceil value with A[i]. The complexity is THETA (N log N). For A[0], there are no active lists of subsequences, we will create a new one. There are few requests for O(N log N) algo in the forum posts. Therefore it is possible to do a binary search in tails array to find the one needs update. In the above example, E = 11, A[i] = 8 and A[j] = 9. Bonus: You have learnt Patience Sorting technique partially , Here is a proverb, “Tell me and I will forget. For subsequence, numbers are not necessarily contiguous. I know many of you might have read recursive and dynamic programming (DP) solutions. close, link We scan the lists (for end elements) in decreasing order of their length. Algorithm - Longest Increasing Subsequence. Note that S is not the LIS itself. How can it extend the current sequences {2, 3} or {2, 5}. We add a new number A[i] to the sequence if A[i] > E, E is the last element in subsequence Let us take small samples and extend the solution to large instances. In the worst case (what is worst case input? This is a brilliant example of how longest increasing subsequence (LIS) could be applied. Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. The Longest increasing subsequence is {0, 2, 6, 9, 11, 15} This subsequence has length 6; the input sequence has no 7-member increasing subsequences. If we take a closer look, we can notice that it is O(n) under the assumption that hash insert and search take O(1) time. 3. Could you improve it to O(nlogn) time complexity? Now the increasing sequences are {2, 3, 7, 11} and {2, 5, 7, 11} for the input array {2, 5, 3, 7, 11}. A[6] is 6. These cookies will be stored in your browser only with your consent. Took my note book (I have habit of maintaining binded note book to keep track of my rough work), and after few hours I filled nearly 15 pages of rough work. Each time a new element is to be added, scan all the lists of subsequences in decreasing order of their length. The number of piles is the length of the longest subsequence. . Clone, extend, and discard all the same length subsequences. - It is an increasing subsequence; - There exists an increasing subsequence (in the input read so far) with the same lenght of the sequence stored in S, and terminating in the same way of the sequence stored in S. - Such increasing subsequence is as long as possible. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. A[i] is greater than the ends of all the current lists, we will take the longest one and append A[1] to it. For A[3] with value 12, it is the same case as A[1] since it is greater than all the ends of the current lists, we will clone the longest available list and append it to that. I got to know the link via my recently created Disqus profile. Longest Increasing Subsequence O(n^2) -> O(nlogn), clean code, easy to understand. I know it will be confusing, I will clear it shortly! Level: MediumAsked In: Amazon, Facebook, Microsoft Understanding the Problem. 입력이 10000이하이면 구현이 용이한 O(n*n) 알고리즘을, 그 이상이면 O(NlogN) 알고리즘을 사용하는 것을 추천한다. Same as A[4]. The Longest Increasing Subsequence (LIS) problem is to find the length of the longest subsequence of a given sequence such that all elements of the subsequence are sorted in increasing order. If A[i] is in between, find the list with the largest end number that is smaller than A[i]. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. Profess to ‘know’ is different from real understanding (no disrespect). A[4] with value 2, it has the same case as A[2], Clone the one with largest end which is less than A[4], append A[4] to it and discard all same length lists. What happens now? code. Link to CPP implementation. 2016-02-09 ... 그러므로 O(NlogN)알고리즘을 사용할 수 있어야 한다. Experience. Finding Longest Increasing Subsequence in O(nlogn) time. Design an algorithm to construct the longest increasing list. 1. If yes, how? The Longest Increasing Subsequence (LIS) problem is to find the length of the longest subsequence of a given sequence such that all elements of the subsequence are sorted in increasing order. Please use ide.geeksforgeeks.org, generate link and share the link here. To make it clear, consider the array is {2, 5, 3, 1, 2, 3, 4, 5, 6}. We use cookies to ensure you have the best browsing experience on our website. For example, longest increasing subsequence of [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15] is [0, 2, 6, 9, 11, 15]. Time Complexity: At first look, time complexity looks more than O(n). Don’t stop learning now.

Acacia Confusa Uses, Grid Systems In Graphic Design Review, Fiio M50x Bluetooth Adapter, Fenugreek Hair Spray, Tvn Live Stream Youtube, Miami Real Estate Market Predictions 2020, Calculate With Confidence, 7th Edition 9780323396837, No Longer Slaves Chords Bb,