## klimenkov Group Title $\lim_{n\rightarrow\infty}\frac{\sqrt[n]{n!}}{n}$ 2 years ago 2 years ago

1. myko

maybe like this: $\sqrt[n]{n!}=\sqrt[n]{n(n-1)(n-2)\cdots1} =\sqrt[n]{n}\sqrt[n]{n-1}\cdots \sqrt[n]{1}=1$ so limit is equal to 0

2. myko

@klimenkov

3. klimenkov

Are you sure that $\lim_{n\rightarrow\infty}\sqrt[n]{n!}=1$

4. myko

look the steps from my comment befor. It looks ok

5. myko

all the roots at the right hand side: $\sqrt[n]{n}=\sqrt[n]{n-1}=\cdots=\sqrt[n]{1}=1$

6. myko

so their product too

7. klimenkov

No, it's not ok. Because you multiply an infinite quantity or 1. As we know $$1^{\infty}={}?$$.

8. myko

$1^{\infty} =1*1*\cdots*1=1$

9. klimenkov

Very nice. What can you say about this pretty limit? $\lim_{n\rightarrow\infty}\left(1+\frac1n\right)^n$It is $$1^{\infty}$$.

10. myko

this happens when you talk about functions. The reason of indetermination of 1^infinity is because of that. But in this case there are no functions involved. That's my point

11. myko

in this case there is just number one multiplyed infinitly many times. And it happens after the limit was taken

12. klimenkov

Ok. What about this? $\lim_{n\rightarrow\infty}\sum_{k=1}^n\frac1n$Is it 0?

13. myko

this is a harmonic series. It is not convergent

14. myko

15. klimenkov

Look at the denominator carefully please. I hopr you will try to get what I'm saying.

16. myko

sry, but i don't

17. klimenkov

Can you find this? $\lim_{n\rightarrow\infty}\sum_{k=1}^n\frac1n$

18. myko

another way to try this: $\lim \sqrt[n]{\frac{n!}{n^n}} = 0$

19. myko

infinity

20. klimenkov

Can you show the way you solve it?

21. myko

I don't remmeber the formal proof of n!/n^n =0, but it's evident, if you try a few first terms of this sequence. There are some posts about it if you google a bit

22. klimenkov

@myko, see this and tell me what is your mistake? http://www.wolframalpha.com/input/?i=Limit+%28n!%29^%281%2Fn%29%2Fn+n-%3Einfinity

23. TuringTest

@mahmit2012 a little help here?

24. mahmit2012

|dw:1351370532404:dw|

25. TuringTest

very nice, my turn...

26. mahmit2012

|dw:1351370740820:dw|

27. mahmit2012

|dw:1351370823511:dw|

28. mahmit2012

|dw:1351370856590:dw|

29. TuringTest

${\sqrt[n]{n!}\over n}=\exp\left(\frac1n\ln(n!)-\ln n\right)=\exp\left({n\ln n-n+O(n)-n\ln n\over n}\right)$$=\exp(-1)=\frac1e$

30. mukushla

*

31. myko

ya I was wrong. Here is another way to solve it. As we know root test is stronger than cuotient test, so the folowing inequality holds: $\lim \inf \frac{a_{n+1}}{a_{n}}\leq\lim \inf \sqrt[n]{a_{n}}\leq \lim \sup \sqrt[n]{a_{n}} \leq \lim \sup\frac{a_{n+1}}{a_{n}}$ let $a_{n} = \frac{n!}{n^{n}}$ then $\frac{a_{n+1}}{a_{n}}=\frac{1}{(1+\frac{1}{n})^{n}}=\frac{1}{e}$ this means that $\lim \sqrt[n]{a_{n}} = \frac{1}{e}$

32. myko

@mahmit2012 $\lim \frac{a_{n+1}}{a_{n}}=\lim \sqrt[n]{a_{n}}$ only if a_n is convergent, what is not implied in this question

33. klimenkov

Nice. But I have one more interesting method to find it. $\lim_{n\rightarrow\infty} \frac{\sqrt[n]{n!}}{n}=\lim_{n\rightarrow\infty}\sqrt[n]{\frac{n!}{n^n}}=\lim_{n\rightarrow\infty}\sqrt[n]{\frac1n\cdot\frac2n\cdots\frac n n}=A$$\ln A=\lim_{n\rightarrow\infty}\frac1n(\ln\frac1n+\ln\frac2n+\ldots+\ln\frac n n)=\int_0^1\ln x dx=-1$$\lim_{n\rightarrow\infty} \frac{\sqrt[n]{n!}}{n}=A=e^{-1}=\frac1e$


Understanding Time Complexity Of Algorithms | Bits N Tricks. We could also say it is linear on the number of entries in the table but that is less commonly used. Imagine the time it will take to search a Letter of Daniel, Eric, Jayesh or any Employee. Therefore, by using hash table, we can achieve linear time complexity for finding the duplicate. Total comparisons in Bubble sort is: n ( n – 1) / 2 ≈ n 2 – n Best case 2: O (n ) Average case : O (n2) Worst case : O (n2) 3. The time required is flat, an O(1) constant time complexity. The main points in these lecture slides are:Time Complexity, Complexity of Algorithms, Execution Time, Space Complexity, Worst Case Analysis, Division of Integers, Number of Comparisons, Binary Search, Average Case Complexity, Complexity of Bubble Sort. In the later case, the search terminates in failure with n comparisons. If we assume we needed to search the array n times the total worst case run time of the linear searches would be O (n^2)). Why so important? You do it all the time in real life!. Amortized time per operation using a bounded priority queue[1] logarithmic time DLOGTIME O(log n) log n, log(n 2) Binary search polylogarithmic time poly(log n) (log n)2 fractional power O(nc) where 0 < c < 1 n1/2, n2/3 Searching in a kd-tree linear time O(n) n Finding the smallest item in an unsorted array "n log star n" time O(n log* n). The idea behind linear search is to compare the search item with the elements in the list one by one (using a loop) and stop as soon as we get the first copy of the search element in the list. By this logic, we can say that painting pictures is slower than baking cookies. This is said to run at O(n); it’s run time increases at an order of magnitude proportional to n. Solving linear equations can be reduced to a matrix-inversion problem, implying that the time complexity of the former problem is not greater than the time complexity of the latter. Totally it takes '4n+4' units of time to complete its execution and it is Linear Time Complexity. The search time increases proportionately to the number of new items introduced. If we're running a statement. Lookups on arrays and objects are going to be constant time if you access them directly. Firstly, we analyze the time complexity of the iterative algorithm and then recursive. The notation Ο(n) is the formal way to express the upper bound of an algorithm's running time. The time complexity of a heuristic search algorithm depends on the accuracy of the heuristic function. Solving a system of linear equations has a complexity of at most O (n 3). This video explains the time complexity analysis for binary search. One example is the binary search algorithm. The array to be searched is reduced by half in every iteration. com: Time complexity of an algorithm: In computer science , the time complexity of an algorithm quantifies the amount of time taken by an algorithm to run as a function. > But what is: > > == the time complexity of string indexing? Is it constant? Yes. when first breaks, you know X(last but one fall - success) and Y(last fall - failure). Linear Differential Equations and Oscillators is the first book within Ordinary Differential Equations with Applications to Trajectories and Vibrations, Six-volume Set. Here that is linear time. The improvement of the proposed linear-time algorithm compared with ECL2 (Yu et al. Given an arbitrary network of interconnected nodes, each with an initial value, we study the number of time-steps required for some (or all) of the nodes to gather all of the initial values via a linear iterative strategy. Lets say I have the list 10,20,30,40,50,60,30,40,50. O(N)—Linear Time: Linear Time Complexity describes an algorithm or program who’s complexity will grow in direct proportion to the size of the input data. It costs us space. characterises a function based on growth of function C. A linear search runs in at worst linear time and makes at most n comparisons, where n is the length of the list. But a balanced binary search tree is always OlgN. Space Complexity. Data Structures for Beginners: Arrays, HashMaps, and Lists. For example, a "linear" running time can also. $\begingroup$ @Olologin can you share any references to understand how to calculate time complexities for complex equations? I want to understand the priority of matrix, inverse, transpose etc of different orders. Search for "Journey Into Complexity" Books in the Search Form now, Download or Read Books for FREE, just by Creating an Account to enter our library. So, we can write this as Ω(n). Examples: binary search. For example, if the heuristic evaluation function is an exact estimator, then A* runs in linear time, expanding only those nodes on an optimal solution path. now do a linear search starting from X(conservative but accurate second step - slow). For example -. During each iteration, the first remaining element of the input is only compared with the right-most element of the sorted subsection of the array. So, an algorithm taking X second or 2X + 3 seconds have the same complexity. However, in my previous experiments, it appears to be O (N), namely linear complexity!. Time Complexity of Bisection Search is O(log n). The complexity of an algorithm is usually taken to be its worst-case complexity, unless specified otherwise. Here is an. Dual first-order methods are essential techniques for large-scale constrained convex optimization. As a set, they are the fourth volume in the series Mathematics and Physics Applied to Science and Technology. algorithm solving a Boolean satis ability problem on n variables is improved i it takes time O(2cn) for some constant c < 1, i. The best case time in linear search is for the first element i. For databases, this means that the time execution would be directly proportional to the table size: as the number of rows in the table grows, the time for the query grows. See full list on towardsdatascience. In the later case, the search terminates in failure with n comparisons. The complexity of an algorithm is usually taken to be its worst-case complexity, unless specified otherwise. • Matlab implements sparse linear algebra based on i,j,s format. Dual first-order methods are essential techniques for large-scale constrained convex optimization. [143, 144, 145, 99]) that Depth-First Search (DFS) and Breadth-First Search (BFS) run in linear time in graphs, and that using these techniques one can obtain linear time algorithms (on a RAM) for many interesting graph. The new distance measures can be computed in linear time complexity in the histogram size. veri es in linear time whether a given spanning tree T of a graph G = (V;E) is a minimum spanning tree. Stochastic Diffusion Search is an alternative solution for invariant pattern recognition and focus of attention. O(n log n) - sorting a list. DTIME[2polylogn]. However complexity for above written implementations is O(). Download Binary search program. , 2017) is in the overhead on enumerating bin pairs. Space Complexity. …humans are incredibly good at linking cause and effect – sometime too good. Now considering the worst case in which the search element does not exist in the list of size N then the Simple Linear Search will take a total of 2N+1. Since all letters are placed in one bucket, Put and Get operation will no longer have time complexity of O(1) because put and get operation has to scan each letter inside the bucket for matching key. Multiply to get n*log(n). Write a linear-time filter IntegerSort. Another simple yet important function. Binary Search source. Yields of experiment are expected to provide an information related to the complexity of the algorithm in LibSVM and knowing the running-time indicator of training and testing both for C++ and Java. How many elements of the input sequence need to be checked on the average, assuming that the element being searched for is equally likely to be any element in the array? How about in the worst case? What are the average-case and worst-case running times of linear search in $\theta$-notation? Justify your answers. Examples: binary search. All have polynomial time complexity while some allow very long steps in favorable circumstances. Time complexity of a related-key attack: “Thus, the total time complexity of Step 2(b) is about 2256 ·2167. With an average time complexity of O(log log n), interpolation search beats binary search's O(log n) easily. Time Complexity : θ ( n ) Space Complexity : O(1) Linear Search Example. On the other hand, if you search for a word in a dictionary, the search will be faster because the words are in sorted order, you know the order and can quickly decide if you need to turn to earlier pages or later pages. At each time-step in this strategy, each node in the network transmits a weighted linear combination of its previous transmission and the most recent transmissions of its. hyperparameter Search: Grid search and random search Train & Run time space & time complexity. It went through the entire list so it took linear time. On an unsorted array Binary Search is almost twice as slow as Linear Search with worst Time Complexity of O(n²) and that is not even considering unbalanced trees. Time Complexity of Binary Search Algorithm is O(log 2 n). Motivation: A crucial phenomenon of our times is the diminishing marginal returns of investments in pharmaceutical research and development. Polynomial time means n O(1), or n c for some constant c. Huan Li, Zhouchen Lin; 21(33):1−45, 2020. Total comparisons in Bubble sort is: n ( n – 1) / 2 ≈ n 2 – n Best case 2: O (n ) Average case : O (n2) Worst case : O (n2) 3. Totally it takes '4n+4' units of time to complete its execution and it is Linear Time Complexity. Linear time is when an algorithm depends on the input size. Because of this, time complexity increases. Reducing the number of generations, i. Big-O Notation • We specify the largest term using big-O notation. \ReaderPrograms\ReaderFiles\Chap02\OrderedArray\orderedArray. It measures the worst case time complexity or the longest amount of time an algorithm can possibly take to complete. Linear Complexity: O(n) A linear task’s run time will vary depending on it’s input value. The asymptotic complexity is defined by the most efficient (in terms of whatever computational resource one is considering) algorithm for solving the game; the most common complexity measure (computation time) is always lower-bounded by the logarithm of the asymptotic state-space complexity, since a solution algorithm must work for every. Time Complexity of. This linear search has a time complexity of O(n). Self-balanced Binary Search Trees. However, in my previous experiments, it appears to be O (N), namely linear complexity!. The time required to search an element using a linear search algorithm depends on the size of the list. algorithm solving a Boolean satis ability problem on n variables is improved i it takes time O(2cn) for some constant c < 1, i. The time complexity of Linear Search is O (n). Finding the median in a list seems like a trivial problem, but doing so in linear time turns out to be tricky. The search time increases proportionately to the number of new items introduced. It makes an exponential workspace and solves the problems with exponential complexity in a polynomial (even linear) time. linear search time complexity. Gorky University Publishers, Gorky (1985) (in Russian) Google Scholar. The time complexity has to do with the critical opeations being performed. Multiply to get n*log(n). Serial Search - Analysis. Thus: I use three elements as the threshold when I will switch to Dictionary lookups from List loops. Linear Search is an example for Linear Time Complexity. Linear Time: O(n) An algorithm is said to run in linear time if its time execution is directly proportional to the input size, i. O(n) Linear: Time to complete the work grows in a 1 to 1 relation to input size. In this set of Solved MCQ on Searching and Sorting Algorithms in Data Structure, you can find MCQs of the binary search algorithm, linear search algorithm, sorting algorithm, Complexity of linear search, merge sort and bubble sort and partition and exchange sort. Finding the median in a list seems like a trivial problem, but doing so in linear time turns out to be tricky. Linear search has linear-time complexity; binary search has log-time complexity. Conversely, giv. Data Structures and Algorithms Objective type Questions and Answers. com: Time complexity of an algorithm: In computer science , the time complexity of an algorithm quantifies the amount of time taken by an algorithm to run as a function. Browse other questions tagged time-complexity linear-algebra matrices or ask your own question. • DEMO • Conclusion: Maybe I can SCALE well … Solve O(10^12) problems in O(10^12). See full list on yourbasic. The time complexity is the sum of time spent in all calls plus some extra preprocessing time. BIG O Notation – Time Complexity and Space Complexity Binary search is a technique used to search sorted data sets. This video explains the time complexity analysis for binary search. See full list on freecodecamp. We therefore take the complexity of inverted index search to be (as discussed in Section 2. Linear search performs equality comparisons and Binary search performs ordering comparisons; Let us look at an example to compare the two: Linear Search to find the element “J” in a given sorted list from A-X. for(i=0; i N; i++) { for(j=0; j. Hence the complexity is O(n). Time Complexity of Linear Search Algorithm is O (n). It depends on the condition given in the for loop. Linear Search vs Binary Search. linear regression and the correlation coefficient. Time complexity. The time complexity of suﬃx tree construction has been shown to be equiv-alent to that of sorting [7]. i it is exponentially better than a brute force search. This calculation will be independent of implementation details and programming language. Strictly, we should say the average complexity is $$\mathcal{O}(n)$$. Linear Time Complexity: O(n) When time complexity grows in direct proportion to the size of the input, you are facing Linear Time Complexity, or O(n). As a set, they are the fourth volume in the series Mathematics and Physics Applied to Science and Technology. algorithm solving a Boolean satis ability problem on n variables is improved i it takes time O(2cn) for some constant c < 1, i. So during the execution of an algorithm, the total time required that will be decided in the time complexity. We could also say it is linear on the number of entries in the table but that is less commonly used. So there is no advantage of binary search over linear search if every search is on a fresh array. Search time is proportional to the list size. For N = 1024 it is 80% faster, and I guess the performance ratio should converge to two at infinity. > How are strings stored in Python? As arrays? As linked lists?. Complexity Classes. Linear time complexity might sound inefficient when you image input sizes in the billions, but linear time isn't actually too bad. More than 1 Million Books in Pdf, ePub, Mobi, Tuebl and Audiobook formats. Thus, the time complexity of this recursive function is the product O(n). For the analysis to correspond usefully to the actual execution time, the time required to perform a fundamental step must be guaranteed to be bounded above by a constant. The best case for a linear search algorithm is to have the value x for which one is searching located at the first position of the ADT. Total comparisons in Bubble sort is: n ( n – 1) / 2 ≈ n 2 – n Best case 2: O (n ) Average case : O (n2) Worst case : O (n2) 3. In case of a sorted array, Binary Search is faster but the caveat here is also on how arrays are treated by the Language Translator. Time Complexity of Linear Search Algorithm is O (n). It is conjectured that the indistinguishability of photons is responsible for the computational complexity of linear optics. Consider a sorted array of 16 elements. the time required to complete the above operation increases linearlywith respect to 'n' (input). Time complexity of algorithms An algorithm is a collection of steps that process a given input to produce an output. See full list on yourbasic. Time Complexity : θ ( n ) Space Complexity : O(1) Linear Search Example. The search stops when the item is found or when the search has examined each item without success. A sorted array is required New insert() Searching a sorted array by repeatedly dividing the search interval in half. For example. Don't overanalyze O(N). Thus, we have-. For example, for a function f(n) Ο(f(n)) = { g(n) : there exists c > 0 and n 0 such that f(n) ≤ c. Hence, this is another difference between linear search and binary search. Time Complexity. The search time increases proportionately to the number of new items introduced. More than 1 Million Books in Pdf, ePub, Mobi, Tuebl and Audiobook formats. DTIME[2polylogn]. The linear search with break becomes faster than counting linear search shortly after N = 128. Linear time: O(n). The cases are as follows − Best Case − Here the lower bound of running time is calculated. Answer: d Explanation: It is practical to implement linear search in the situations mentioned in When the list has only a few elements and When performing a single search in an unordered list, but for larger elements the complexity becomes larger and it makes sense to sort the list and employ binary search or hashing. The space complexity is also. 2012: J Paul Gibson T&MSP: Mathematical Foundations MAT7003/ L9-Complexity&AA. Informally, this means that the running time increases at most linearly with the size of the input. The order of growth (e. If there are NO nested loops we can probably guess the complexity of the code we looking at would be in the O(n). A(n) = $\frac{n + 1}{2}$ However, I am having trouble coming up with the Average Case complexity in the case where half of the elements in the size n array are duplicates. In this case, the insertion sort algorithm has a linear running time (i. Linear search is a very basic and simple search algorithm. The List has an O(N) linear time complexity. The time complexity of suffix tree construction has been shown to be equivalent to that of sorting: O(n) for a constant-size alphabet or an integer alphabet and O(n log n) for a general alphabet. ) Combinatorial- Algebraic Methods in Applied Mathematics, pp. With a faster sorter like merge-sort, which is O(N*log(N. In linear search we simply iterate over elements and check whether it is the desired element or not. Imagine the time it will take to search a Letter of Daniel, Eric, Jayesh or any Employee. doubling n, time increases only by a factor of c. This is a more mathematical way of expressing running time, and looks more like a function. The time taken to search a given element will increase if the number of elements in the array increases. Here that is linear time. Therefore, much research has been invested into discovering algorithms exhibiting linear time or, at least, nearly linear time. Hourly Update. If you had to search for a name in a directory by reading. Introduction The time complexity of a given algorithm can be obtained from theoretical analysis and computational analysis according to the algorithm’s running. • for selection sort, C(n)=n2/2-n/2 n2/2 • In addition, we’ll typically ignore the coefficient of the largest term (e. Time complexity of Bubble sort in Best Case is O(N). As we will see in the next chapter, kNN's effectiveness is close to that of the most accurate learning methods in text classification (Table 15. In [15] a chosen plaint-text linear attack was suggested and in [5] time complexity of the attack rst stage was reduced by using Fast Fourier Transform. The time complexity of suﬃx tree construction has been shown to be equiv-alent to that of sorting [7]. This search requires only one unit of space to store the element to be searched. I know the answer is O(n), but is this correct: The first element has probability $1/n$ and requires 1 comparison; the second probability $1/(n-1)$ and requires 2 comparisons. In our previous tutorial we discussed about Linear search algorithm which is the most basic algorithm of searching which has some disadvantages in terms of time complexity, so to overcome them to a level an algorithm based on dichotomic (i. Time Complexity For Linked Lists; Time Complexity; Time Complexity In While Loop; My Future Plan Because Of My Teacher? Python Mini-challenge: "Lucky" Numbers; Advance Code Not Hardware? Efficiency Of Linear Search Vs Binary Search In Unsorted List; Algorithm Not Efficient Enough; Cryptography And Data Structure; Filling List(s) With Random Numbers. These problems will introduce things (like the variable i above) just to waste your time. This requires to scan the array completely and check each element for the array that we need to search. Linear Search is an example for Linear Time Complexity. Linear search runs in at worst linear time and makes at most n comparisons, where n is the length of the list. In linear search, we have to check each node/element. O(N)—Linear Time: Linear Time Complexity describes an algorithm or program who’s complexity will grow in direct proportion to the size of the input data. Linear time or O(n). 무슨말인지모르겠다면 아래글을 쭉. The number of steps and time required to solve a problem is based on input size. Best Case. It makes an exponential workspace and solves the problems with exponential complexity in a polynomial (even linear) time. Also, each algorithm's time complexity is explained in separate video lectures. best-case: this is the complexity of solving the problem for the best input. Solving a system of linear equations has a complexity of at most O (n 3). This means the bigger the number of wine bottles in our system, the more time it will take. Definition of time complexity in the Definitions. In the linear search problem, the best case occurs when x is present at the first location. This research includes both software and hardware methods. A few common algorithmic complexities: O(log n) - binary search. If you were to find the name by looping through the list entry after entry, the time complexity would be O (n). Since binary search algorithm splits array in half every time, at most log 2 N steps are performed. So, we can write this as Ω(n). Complexity theory argues that systems are complex interactions of many parts which cannot be predicted by accepted linear equations. The measure for the working storage an algorithm needs is called space complexity. , the work is O (1) comparison. Time complexity of neural network. A single iteration (loop) over all the elements in the array gives us a complexity of O(n). The time complexity function expresses that dependence. For typical values of n = 30, m = 30, and q = 5, the time complexity would be 4500, which is much higher than 110. The asymptotic complexity is defined by the most efficient (in terms of whatever computational resource one is considering) algorithm for solving the game; the most common complexity measure (computation time) is always lower-bounded by the logarithm of the asymptotic state-space complexity, since a solution algorithm must work for every. , c ~ 2d)! Contrast with exponential: For any constant c, there is a d such that n → n+d increases time. The Idea of time complexity is not to calculate how much time an algorithm will take to complete, but to compute the order of magnitude of time for the completion of computation by an algorithm. Finding the median in a list seems like a trivial problem, but doing so in linear time turns out to be tricky. Linear search is linear O(N) Binary search depends on if the tree is balanced or not. Informally, this means that the running time increases at most linearly with the size of the input. ” Most cryptanalytic papers discuss certiﬁcational attacks: Data complexity — just slightly less than the entire code book. The average to the worst case of this kind of search is a linear complexity or O(n). Informática Educativa [email protected] If there are NO nested loops we can probably guess the complexity of the code we looking at would be in the O(n). Operation count: In this technique, we consider the operations in the given algorithm or program that contribute to the execution time and count how many times those operations will be performed. These have yielded near-linear time algorithms for many diverse problems. I know the answer is O(n), but is this correct: The first element has probability $1/n$ and requires 1 comparison; the second probability $1/(n-1)$ and requires 2 comparisons. It was experimentally found in [6, 7] that time complexity of Matsui’s attack on DES may be decreased with a better ranking of the values of relevant sub-key bits, though data complexity and. The List has an O(N) linear time complexity. I know that for an array of size n distinct elements, the Average Case complexity for linear search is as follows:. This obviously requires a constant number of comparison operations, i. We then verify if these times look like the time complexity we're expecting (constant, linear, or polynomial (quadratic or greater)). This study proposes linear time complexity sorting algorithms for nearest level control-based BE and TR MMC models to further accelerate the EMT simulation of the equivalent MMC-HVdc models. Which one of the following is the tightest upper bound that represents the time complexity of inserting an object into a binary search tree of n nodes? A) O(1) (B) O(log n ). The time complexity of suﬃx tree construction has been shown to be equiv-alent to that of sorting [7]. complexity addressing precisely the kind of problem raised in the last two paragraphs: Given a computational prob-lem, can it be solved by an e cient algorithm? For many common computational tasks (such as nding a solution of a set of linear equations) there is a polynomial-time algo-rithm that solves them|this class of problems is called P. Eight time complexities that every programmer should know. While that isn’t bad, O (log. This video explains the time complexity analysis for binary search. BIG O Notation – Time Complexity and Space Complexity Binary search is a technique used to search sorted data sets. Generate an hypothesis: The running time is about 1 x 10-10 x N 3 seconds 4. The time complexity of a heuristic search algorithm depends on the accuracy of the heuristic function. 14 Code sample for Linear Regression. O(n log n) Linearithmic: This is a nested loop, where the inner loop runs in log n time. 39+ o (1)) en ln n to en ln n + O ( n ) in expectation and with high probability, which is tight up to. Motivation: A crucial phenomenon of our times is the diminishing marginal returns of investments in pharmaceutical research and development. If the element is found then its position is displayed. University of Toronto - Fall 2000 Department of Computer Science Week 12 - Complexity & Searching Complexity The complexity of an algorithm is the amount of a resource, such as time, that the algorithm requires. It is generally one of the first algorithms taught in computer science courses because it is a good algorithm to learn to build intuition about sorting. During the study of discrete mathematics, I found this course very informative and applicable. In most of the cases, you are going to see these kind of Big-O running time in your code. for(i=0; i N; i++) { for(j=0; j. Linear search performs equality comparisons and Binary search performs ordering comparisons; Let us look at an example to compare the two: Linear Search to find the element “J” in a given sorted list from A-X. Firstly, we analyze the time complexity of the iterative algorithm and then recursive. Let's take an array int arr [] = { 2,1,7,5,9} Suppose we have to search an element 5. This is an example of logarithmic complexity. Time Complexity of Sorting Algorithms Let's check the time complexity of mostly used sorting algorithms. In my knowledge, the time complexity should be at least O (N^2) or O (NlogN) (the N is number of links), considering it is a graph problem. More than 1 Million Books in Pdf, ePub, Mobi, Tuebl and Audiobook formats. I Linear: RHS is a sum of multiples of previous terms of the sequence (linear combination of previous terms). The number of steps and time required to solve a problem is based on input size. In Binary search half of the given array will be ignored after just one comparison. Complexity Classes. What we are left with is the fact that the time in sequential search grows linearly with the input, while in binary search it grows logarithmically -. In my knowledge, the time complexity should be at least O (N^2) or O (NlogN) (the N is number of links), considering it is a graph problem. See full list on towardsdatascience. \ReaderPrograms\ReaderFiles\Chap02\OrderedArray\orderedArray. all of the mentioned. The search time increases proportionately to the number of new items introduced. While that isn’t bad, O (log. Aaronson and Arkhipov argued in section 1. Complexity International-- journal for scientific papers dealing with any area of complex systems research. A single iteration (loop) over all the elements in the array gives us a complexity of O(n). If you give a condition in the inner loop that will always terminate the inner loop and/or outer loop without executing n times for all elements, then it will have less than O(n) time. Here, n is the number of elements in the linear array. It means we generate a vector that has 5 elements, and these elements are bounded in [-11,11]. Time complexity is a function dependent from the value of n. (And if the number. At each time-step in this strategy, each node in the network transmits a weighted linear combination of its previous transmission and the most recent transmissions of its. Lookups on arrays and objects are going to be constant time if you access them directly. So, the time complexity of binary search is O(log2n). It will be easier to understand after learning O(n), linear time complexity, and O(n^2), quadratic time complexity. compares each element with the value being searched for, and stops when either the value is found or the end of the array is encountered. Linear search is iterative whereas Binary search is Divide and conquer. Binary Search Algorithm and its Implementation. When the input is a random permutation, the rank of the pivot is uniform random from 0 to n − 1. At each time-step in this strategy, each node in the network transmits a weighted linear combination of its previous transmission and the most recent transmissions of its. In the best-case scenario, the element is present at the beginning of the list and in the worst-case, it is present at the end. all of the mentioned. Linear search is a perfect example. It concisely captures the important differences in the asymptotic growth rates of functions. By this logic, we can say that painting pictures is slower than baking cookies. java logarithms Complexity of algorithm Time complexity Space complexity Time complexity: in big O notation. This time complexity is a marked improvement on the O(N) time complexity of Linear Search. Linear-Time Sorting. Also, each algorithm's time complexity is explained in separate video lectures. The second one runs in time sublinear in d, assuming the edit distance is not too small. Linear time: O(n). The best algorithms for sorting a random array have a run time of O(n * log n). Always takes the same time. Linear search is a perfect example. i it is exponentially better than a brute force search. The Idea of time complexity is not to calculate how much time an algorithm will take to complete, but to compute the order of magnitude of time for the completion of computation by an algorithm. • VERY difficult to develop. This is an example of logarithmic complexity. Time complexity of a related-key attack: “Thus, the total time complexity of Step 2(b) is about 2256 ·2167. Hence Bisection Search is way better than Linear Search. Search time is proportional to the list size. In this post I’m going to walk through one of my favorite algorithms, the median-of-medians approach to find the median of a list in deterministic linear time. 2 , page 15. fractal image compression time complexity image compression critical issue video sequence alternative method large number computer animation multi-dimensional nearest neighbor search data storage logarithmic time decoding phase transmission time encoding step linear time image portion multi-dimensional search sequential search data compression. So, the time complexity of binary search is O(log2n). Using the hypothesis, make a prediction: When N =. Although the limiting factor for linear cryptanalysis attacks is usually the data complexity, such an improvement is relevant and can be motivated both by practical and theoretical reasons, as the following scenarios underline. The number of steps and time required to solve a problem is based on input size. Time complexity Posted 28 December 2015 - 04:35 PM Hi guys,lets say I have algorithm ,which finds ,when the number in a list is bigger than the next one. It concisely captures the important differences in the asymptotic growth rates of functions. Here, n is the number of elements in the linear array. Linear time is the best possible time complexity in situations where the algorithm has to sequentially read its entire input. Program for Recursive and Non-Recursive Binary Search in C++ - Analysis Of Algorithms / Data Structures. Visualize high dimensional data. Let us consider an algorithm of sequential searching in an array. Like an array, a linear list stores a collection of objects of a certain type, usually denoted as Time complexity, space complexity, and the O-notation : 2. It went through the entire list so it took linear time. com Linear Time Complexity. We could also say it is linear on the number of entries in the table but that is less commonly used. Huan Li, Zhouchen Lin; 21(33):1−45, 2020. Time Complexity : θ ( n ) Space Complexity : O(1) Linear Search Example. Time complexity of linear search -O(n) , Binary search has time complexity O(log n). time complexity of a linear cryptanalysis attack using algorithm 2. This is an example of logarithmic complexity. Note that an algorithm might take diﬀerent amounts of time on inputs of the. If you were to find the name by looping through the list entry after entry, the time complexity would be O (n). Lets say I have the list 10,20,30,40,50,60,30,40,50. The second one runs in time sublinear in d, assuming the edit distance is not too small. That is, I'm looking for references that looks like the following. complexity = in between logN and N. Here, although your array is of a fixed size, the time needed to complete the operation is still a linear function of the number of elements in the array. The main points in these lecture slides are:Time Complexity, Complexity of Algorithms, Execution Time, Space Complexity, Worst Case Analysis, Division of Integers, Number of Comparisons, Binary Search, Average Case Complexity, Complexity of Bubble Sort. BIG O Notation – Time Complexity and Space Complexity Binary search is a technique used to search sorted data sets. On the other hand, searching is currently one of the most used methods for finding solution for problems in real life, that the blind search algorithms are accurate, but their time complexity is exponential such as breadth. At each time-step in this strategy, each node in the network transmits a weighted linear combination of its previous transmission and the most recent transmissions of its. // Time complexity: O(1) // Space complexity: O(1) int x = 15; x += 6; System. This requires to scan the array completely and check each element for the array that we need to search. This first book consists of chapters 1 and 2 of the fourth volume. starts in the middle, then see if the value being searched for is greater or less than the middle value. As a rule of thumb, it is best to try. 2) and, assuming average document length does not change over time,. O(1) indicates that the algorithm used takes "constant" time, ie. Sequential search write a sequential search function and then find the best, worst, and average case time complexity. Time complexity is commonly estimated by counting the number of elementary operations performed by the algorithm, supposing that each elementary operation takes a fixed amount of time to perform. Thus, the amount of time taken and the number of elementary operations performed by the algorithm are taken to differ by at most a constant factor. Time Complexity of Binary Search Algorithm is O(log 2 n). linear: sorting twice the number of elements takes quite a bit more than just twice as much time; searching (using binary search) through a sorted list twice as long, takes a lot less than twice as much time. This paper reports some new results on the average time complexity of EAs. In: Markov, A. Serial Search - Analysis. Linear search is rarely used practically because other search algorithms such as the binary search algorithm and hash tables allow significantly faster searching comparison to Linear search. what we do is we simply loop over array and check whether it is. For a linear-time algorithm, if the problem size doubles, the number of operations also doubles. Algorithm analysis is an important part of a broader computational complexity theory, which provides theoretical estimates for the resources needed by any. The measure for the working storage an algorithm needs is called space complexity. Let n represent the size of the array arr. We could also say it is linear on the number of entries in the table but that is less commonly used. Always takes the same time. algorithm runs in near-linear time, namely d1+ε for any ﬁxed ε > 0. Reducing the number of generations, i. 1 of [] that the exchange symmetry of identical bosons creates an effective entanglement (a kind of artificial entanglement), which would be the origin of the computational complexity in linear optics. The running time of the loop is directly proportional to N. Therefore, by using hash table, we can achieve linear time complexity for finding the duplicate. characterises a function based on growth of function C. In order to be able to classify algorithms we have to define limiting behaviors for functions describing the given algorithm. The time complexity of above algorithm is O(n). Here, complexity refers to the time complexity of performing computations on a multitape Turing machine. These Multiple Choice Questions (mcq) should be practiced to improve the Data Structure skills required for various interviews (campus interview, walk-in interview, company interview), placement, entrance exam and other competitive examinations. Diagram above is from Objective-C Collections by NSScreencast. Trees Data Structures for Beginners. Space complexity : O (1) O(1) O (1) or (O (n) O(n) O (n)) We sorted nums in place here - if that is not allowed, then we must spend linear additional space on a copy of nums and sort the copy instead. The time complexity of ECL2 is O (n + M ϵ 1 w 2) , where O(n) is the time complexity of scoring and binning, and O (M ϵ 1 w 2) is the time complexity of enumerating bin pairs. Sort an array of 0's, 1's and 2's in linear time complexity; Checking Anagrams (check whether two string is anagrams or not) Relative sorting algorithm; Finding subarray with given sum; Find the level in a binary tree with given sum K; Check whether a Binary Tree is BST (Binary Search Tree) or not; 1[0]1 Pattern Count. near-linear time. As a rule of thumb, it is best to try. However, in my previous experiments, it appears to be O (N), namely linear complexity!. The search stops when the item is found or when the search has examined each item without success. For the analysis to correspond usefully to the actual execution time, the time required to perform a fundamental step must be guaranteed to be bounded above by a constant. A few common algorithmic complexities: O(log n) - binary search. Dual first-order methods are essential techniques for large-scale constrained convex optimization. The best case time in linear search is for the first element i. This linear search has a time complexity of O(n). Aaronson and Arkhipov argued in section 1. given two natural numbers $$n$$ and $$m$$, are they relatively prime?. O(N)—Linear Time: Linear Time Complexity describes an algorithm or program who’s complexity will grow in direct proportion to the size of the input data. what we do is we simply loop over array and check whether it is. In case of a sorted array, Binary Search is faster but the caveat here is also on how arrays are treated by the Language Translator. Let us assume that given an array whose elements order is not known. Consider that we have an algorithm, and we are calculating the time. Time Complexity of Bisection Search is O(log n). Tests are robust , non-parametric statistical tests, since timing is noisy (so need to be robust), and noise can take various forms (so non-parametric, since no particular model of noise). One example is the binary search algorithm. The number of operations in the best case is constant (not dependent on n). The time complexity of suﬃx tree construction has been shown to be equiv-alent to that of sorting [7]. Download Binary search program. Alright, so we have linear-over-n many logarithmic-over-n loops. java graph-algorithms competitive-programming dfs binary-search-tree common-algorithms time-complexity implementation bfs longest-common-subsequence binary-search segment-tree binary-indexted-tree two-pointers space-complexity all-pairs-shortest-path matching-algorithm maximal-bipartite-matching lower-bound lowest-common-ancestor. Linear Search is sequential search which scans one item at a time. Space Complexity. In order to speed up the static analyses formulated using the Dyck-CFL reachability problems, we propose an efficient algorithm of O(n) time for the Dyck-CFL reachability problem when the graph considered is a bidirected tree with specific constraints, while a naïve algorithm runs in O(n2) time. This requires to scan the array completely and check each element for the array that we need to search. This web page gives an introduction to how recurrence relations can be used to help determine the big-Oh running time of recursive functions. See full list on yourbasic. As investigated in [ ],theHPPcanbesolvedusing. Time complexity of Bubble sort in Worst Case is O(N^2), which makes it quite inefficient for sorting large data volumes. All have polynomial time complexity while some allow very long steps in favorable circumstances. We show an improved algorithm for the satis ability problem for circuits of constant depth and linear size. We will see more about Time Complexity in future. In a serial search, we step through an array (or list) one item at a time looking for a desired item. for(i=0; i N; i++) { for(j=0; j. Linear Time: O(n) An algorithm is said to run in linear time if its time execution is directly proportional to the input size, i. Which one of the following is the tightest upper bound that represents the time complexity of inserting an object into a binary search tree of n nodes? A) O(1) (B) O(log n ). It measures the worst case time complexity or the longest amount of time an algorithm can possibly take to complete. O(1) is the best possible time complexity! Data structures like hash tables make clever use of algorithms to pull off constant time operations and speed things up dramatically. So, the time complexity of binary search is O(log2n). Eight time complexities that every programmer should know. O(n²) – Quadratic Time. Serial Search - Analysis. The time complexity of an algorithm is commonly expressed using big O notation , which excludes coefficients and lower order terms. The first is the way used in lecture - "logarithmic", "linear", etc. The improvement of the proposed linear-time algorithm compared with ECL2 (Yu et al. Here you will learn about python binary search with program and algorithm. This search requires only one unit of space to store the element to be searched. Time complexity. These problems will introduce things (like the variable i above) just to waste your time. The time complexity of a heuristic search algorithm depends on the accuracy of the heuristic function. For instance, it is known since the 1960s and 70s (e. Linear Time: O(n) An algorithm is said to run in linear time if its time execution is directly proportional to the input size, i. Hence time complexity of the Binary search is O(LogN). The best case time in linear search is for the first element i. Most of the time we can speak on sorting integers in linear time, but as we can see later this is not the only case. Hourly Update. Time complexity Posted 28 December 2015 - 04:35 PM Hi guys,lets say I have algorithm ,which finds ,when the number in a list is bigger than the next one. It is easy to see that $$\widetilde{\mathcal {S}}$$ can be obtained in one pass through $$\widetilde{\mathcal {A}}$$ and $$\widetilde{\mathcal {B}}$$, therefore in linear time. Time complexity (linear search vs binary search) 1. I n linear search, we need to write more code whereas in binary search we need to write less code. algorithm solving a Boolean satis ability problem on n variables is improved i it takes time O(2cn) for some constant c < 1, i. now do a linear search starting from X(conservative but accurate second step - slow). So the option is 'B'. 무슨말인지모르겠다면 아래글을 쭉. In this book, Keith Morrison introduces complexity theory to the world of education, drawing out its implications for school leadership. For better understanding,lets take an example: given an array arr[]={12,11,4,0,3,5} and we want to search whether 5 present in the given array or not. In Binary search half of the given array will be ignored after just one comparison. Nested for loops are the perfect example of this category. What we are left with is the fact that the time in sequential search grows linearly with the input, while in binary search it grows logarithmically -. The complexity of linear search algorithm is. Time complexity is commonly estimated by counting the number of elementary operations performed by the algorithm, supposing that each elementary operation takes a fixed amount of time to perform. Run time is O (log N) Sample code for ordered array. To measure Time complexity of an algorithm Big O notation is used which: A. In order to speed up the static analyses formulated using the Dyck-CFL reachability problems, we propose an efficient algorithm of O(n) time for the Dyck-CFL reachability problem when the graph considered is a bidirected tree with specific constraints, while a naïve algorithm runs in O(n2) time. Indeed, 100 cookies don’t take much longer than 12 cookies — provided you have a big enough bowl. • DEMO • Conclusion: Maybe I can SCALE well … Solve O(10^12) problems in O(10^12). Time complexity (linear search vs binary search) 1. linear regression and the correlation coefficient. Tests are robust , non-parametric statistical tests, since timing is noisy (so need to be robust), and noise can take various forms (so non-parametric, since no particular model of noise). Binary search. So the option is 'B'. This study proposes linear time complexity sorting algorithms for nearest level control-based BE and TR MMC models to further accelerate the EMT simulation of the equivalent MMC-HVdc models. From this we can see these equations are similar and our equation matches linear equation. Counting sort and radix sort assume that the input consists of integers in a small. Since binary search has a best case efficiency of O(1) and worst case (average case) efficiency of O(log n), we will look at an example of the worst case. As the number increases so does the time difference. For a general alphabet, suﬃx tree construction has time bound of Θ(nlogn). Here is the official definition of time complexity. Hence number of times while loop will execute will determine the complexity of the algorithm. If you had to search for a name in a directory by reading. If we assume we needed to search the array n times the total worst case run time of the linear searches would be O (n^2)). All complex systems can be seen as a number of nodes joined together – lines and junctions – or in the case of the human brain, long spindly nerve cells and synapses. Linear Search Simple search from the first element to the last till we find the required element. So, the time complexity of binary search is O(log2n). It has a complexity of n 2. Hourly Update. Let's take an array int arr [] = { 2,1,7,5,9} Suppose we have to search an element 5. the complexity. The time complexity function expresses that dependence. As a set, they are the fourth volume in the series Mathematics and Physics Applied to Science and Technology. Finally, together with the analysis, it is concluded that the linear time complexity is validated based on the experiments. This is a more mathematical way of expressing running time, and looks more like a function. On the other hand, searching is currently one of the most used methods for finding solution for problems in real life, that the blind search algorithms are accurate, but their time complexity is exponential such as breadth. at 11:59pm • Asymptotic analysis Asymptotic Analysis CSE 373 Data Structures & Algorithms Ruth Anderson Spring 2007 04/04/08 2 Linear Search vs Binary Search Linear Search Binary Search Best Case Asymptotic Analysis Worst Case So … which algorithm is better?. Time complexity (linear search vs binary search) 1. Time Complexity and the divide and conquer strategy Or : how to measure algorithm run-time And : design efficient algorithms Oct. Worst-case running time - the algorithm finds the number at the end of the list or determines that the number isn't in the list. Complexity and running time Factors: algorithmic complexity, startup costs, additional space requirements, use of recursion (function calls are expensive and eat stack space), worst-case behavior, assumptions about input data, caching, and behavior on already-sorted or nearly-sorted data; Worst-case behavior is important for real-time systems. For a general alphabet, suﬃx tree construction has time bound of Θ(nlogn). That is, I'm looking for references that looks like the following. Solving linear equations can be reduced to a matrix-inversion problem, implying that the time complexity of the former problem is not greater than the time complexity of the latter. Motivation: A crucial phenomenon of our times is the diminishing marginal returns of investments in pharmaceutical research and development. This first book consists of chapters 1 and 2 of the fourth volume. It's not easy trying to determine the asymptotic complexity (using big-Oh) of recursive functions without an easy-to-use but underutilized tool. This technique is probably the easiest to implement and is applicable to many situations. Definition of time complexity in the Definitions. Algorithmic Complexity Notes on Notation: Algorithmic complexity is usually expressed in 1 of 2 ways. Data Structure MCQ - Complexity. In linear search we simply iterate over elements and check whether it is the desired element or not. for(i=0; i N; i++) { for(j=0; j. If the values match it will return success. For a general alphabet, suﬃx tree construction has time bound of Θ(nlogn). In our previous tutorial we discussed about Linear search algorithm which is the most basic algorithm of searching which has some disadvantages in terms of time complexity, so to overcome them to a level an algorithm based on dichotomic (i. In a serial search, we step through an array (or list) one item at a time looking for a desired item. See full list on freecodecamp. Now considering the worst case in which the search element does not exist in the list of size N then the Simple Linear Search will take a total of 2N+1. Algorithm analysis is an important part of a broader computational complexity theory, which provides theoretical estimates for the resources needed by any. Linear search is rarely used practically because other search algorithms such as the binary search algorithm and hash tables allow significantly faster searching comparison to Linear search. In computer science, the time complexity of an algorithm quantifies the amount of time taken by an algorithm to run as a function of the length of the string representing the input:226. The running time of the two loops is proportional to. Results Here, then, as a concrete example, is a plot of the run-times of the most interesting algorithms on an Intel Core i7 running at 2. O(1): Constant Time Complexity. O(n) - finding the largest item in an unordered list. This first book consists of chapters 1 and 2 of the fourth volume. Since binary search algorithm splits array in half every time, at most log 2 N steps are performed. Best Case. larger search space of constituent trees (compared to the space of dependency trees) would make it unlikely that accurate parse trees could be built deterministically, we show that the precision and recall of constituents produced by our parser are close to those produced by statistical parsers with higher run-time complexity. Time Complexity : This section explains the importance of time complexity analysis, the asymptotic notations to denote the time complexity of algorithms. The best case for a linear search algorithm is to have the value x for which one is searching located at the first position of the ADT. 2) and, assuming average document length does not change over time,. Most algorithms, however, are built from many combinations of these. The "Binary Search Time Complexity" Lesson is part of the full, Tree and Graph Data Structures course featured in this preview video. Here complexity is said to be linear. This time complexity is a marked improvement on the O(N) time complexity of Linear Search. So time complexity in the best case would be Θ(1) Most of the times, we do worst case analysis to analyze algorithms. Suppose varMin=-11, and varMax=11, and varSize=5. So there must be some type of behavior that algorithm is showing to be given a complexity of log n. Constant time compelxity, or O(1), is just that: constant. This time complexity of binary search remains unchanged irrespective of the element position even if it is not present in the array. It will be easier to understand after learning O(n), linear time complexity, and O(n^2), quadratic time complexity. first do a binary search (agressive first step - fast) with 1 bulb. The time factor when determining the efficiency of algorithm is measured by. all of the mentioned. Totally it takes '4n+4' units of time to complete its execution and it is Linear Time Complexity. worst case, the time for insertion is proportional to the number of elements in the array, and we say that the worst-case time for the insertion operation is linear in the number of elements in the array. As we learned in the previous tutorial that the time complexity of Linear search algorithm is O(n), we will analyse the same and see. When the input is a random permutation, the rank of the pivot is uniform random from 0 to n − 1. If the element is found then its position is displayed. So for any value of n it will give us linear time. For the analysis to correspond usefully to the actual execution time, the time required to perform a fundamental step must be guaranteed to be bounded above by a constant. compares each element with the value being searched for, and stops when either the value is found or the end of the array is encountered. Gorky University Publishers, Gorky (1985) (in Russian) Google Scholar. If you had to search for a name in a directory by reading. Most algorithms, however, are built from many combinations of these. Time and space complexity depends on lots of things like. However, in my previous experiments, it appears to be O (N), namely linear complexity!. Neglecting the constant value 5 the complexity would be N as loop will run N times so it does not fit the definition of linear time. We observe how space complexity evolves when the algorithm’s input size grows, just as we do for time complexity. We then verify if these times look like the time complexity we're expecting (constant, linear, or polynomial (quadratic or greater)). O(n) is for linear complexity, O(n 2) is for quadratic. Worst Case time complexity is O(n) which means that value was not found in the array (or found at the very last index) which means that we had to iterate n times to reach to that conclusion. In case of the monks, the number of turns taken to transfer 64 disks, by following the above rules, will be 18,446,744,073,709,551,615; which will surely take a lot of time!!. > How are strings stored in Python? As arrays? As linked lists?. O(N^2) because it sorts only one item in each iteration and in each iteration it has to compare n-i elements. This video is meant for educational. Introduction The time complexity of a given algorithm can be obtained from theoretical analysis and computational analysis according to the algorithm’s running. So, an algorithm taking X second or 2X + 3 seconds have the same complexity. This means that as the input grows, the algorithm takes proportionally longer to complete. The idea behind linear search is to compare the search item with the elements in the list one by one (using a loop) and stop as soon as we get the first copy of the search element in the list. Learn more about time complexity of neural network. The "Binary Search Time Complexity" Lesson is part of the full, Tree and Graph Data Structures course featured in this preview video. Unbalanced binary search tree can turn into a linked list in the worst case if the elements added are in descending order so O(N) time complexity. For typical values of n = 30, m = 30, and q = 5, the time complexity would be 4500, which is much higher than 110. What we are left with is the fact that the time in sequential search grows linearly with the input, while in binary search it grows logarithmically -. If we plot the graph of an+b for different values of n we will see that it is a straight line. These approximation and runtime guarantees are signiﬁcantly better then the bounds known for worst-case inputs, e. What’s the maximum number of loop iterations? log2n That is, we can’t cut the search region in half more than that many times. Time complexity (linear search vs binary search) 1. java that reads from standard input a sequence of integers that are between 0 and 99 and prints to standard output the same integers in sorted order. The time complexity of linear search is O(n), meaning that the time taken to execute increases with the number of items in our input list lys. For example. Morzhakov, N. Linear Homogeneous Recurrences De nition A linear homogeneous recurrence relation of degree k with constant coe cients is a recurrence relation of the form an = c1an 1 + c2an 2 + + ck an k with c1;:::;ck 2 R , ck 6= 0. the complexity. In my knowledge, the time complexity should be at least O (N^2) or O (NlogN) (the N is number of links), considering it is a graph problem. • DEMO • Conclusion: Maybe I can SCALE well … Solve O(10^12) problems in O(10^12). Linear time or O(n). Algorithm Complexity When N doubles Examples Constant 1 increases fixed times No loop Logarithmic log N increases constant Binary search Linear N doubles Traverse an array Linearithmic NlogN more than doubles Quick/Merge Sort, FFT Quadratic N^2 increases fourfold B Cubic N^3 increases eightfold NxN matrix multiplication Exponential 2^N running time squares!. Bubble sort is a simple, inefficient sorting algorithm used to sort lists. At each time-step in this strategy, each node in the network transmits a weighted linear combination of its previous transmission and the most recent transmissions of its. Total comparisons in Bubble sort is: n ( n – 1) / 2 ≈ n 2 – n Best case 2: O (n ) Average case : O (n2) Worst case : O (n2) 3. See full list on yourbasic. Know Thy Complexities! Hi there! This webpage covers the space and time Big-O complexities of common algorithms used in Computer Science. The time complexity of linear search is O(N) while binary search has O(log 2 N). Run time is O (log N) Sample code for ordered array. i it is exponentially better than a brute force search. Given an arbitrary network of interconnected nodes, each with an initial value, we study the number of time-steps required for some (or all) of the nodes to gather all of the initial values via a linear iterative strategy. If connections are sparse, then sparse math can be used for the gradient computations, etc leading to reduced complexity.


The number $555,555$ can decompose, as the product of two factors of three digits, in how many ways?

The number $555,555$ can decompose, as the product of two factors of three digits, in how many ways?

I've seen the answer to the question, and there is only one way: Since $555, 555 = 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 37$, the only way to combine the factors to achieve expressing it as a product of two three-digit numbers is $(3 \cdot 7 \cdot 37) (5 \cdot 11 \cdot 13)$. Regardless of this, I struggle to understand how the answer was formulated. Can someone show me the procedure?

Sorry if the question is poorly phrased, it is a rough translation of the original problem in Spanish.

• list all the divisors in order. $\sqrt{555555} \approx 745.36.$ One of your divisors must be larger than that but still smaller than 1000. The other divisor in your pair will be between 555.56 and 745.36. I guess you can find them by hand, look at all the products of two primes, then three primes. You do not need to do four primes because you already did two – Will Jagy Apr 16 '18 at 2:34
• Yeah, two primes is too small, the biggest is $13 \cdot 37 = 481.$ So, three primes each – Will Jagy Apr 16 '18 at 2:38

Nothing wrong with the approach Will gives in the comments. Here's another way. Obviously $555,555=555\times1001$, but $1001=7\times11\times13$ is a little too large. The way to make it a little smaller is to swap the factor 7 with the factor 5 in 555, which gives you your solution, $(3\times7\times37)(5\times11\times13)$.

The factors $7,11,13$ can't be used together, since $(7)(11)(13)=1001$.

So one of the groups, group $1$ say, must have exactly two of the factors $7,11,13$.

Hence the factors $3$ and $5$ can't both be in group $1$, else the product of the factors in group $1$ would be at least $(3)(5)(7)(11) > (7)(11)(13)$.

Similarly, the factor $37$ can't be in group $1$, else the product of the factors in group $1$ would be at least $(37)(7)(11) > (7)(11)(13)$.

Label the other group as group $2$.

Thus, group $2$ contains

• $37$
• Exactly one of $7,11,13$.
• At least one of $3,5$.

But since $(37)(27) = 999$, the factors in group $2$ other than $37$ must have a product which is at most $27$.

It follows that neither of the factors $11$ or $13$ is in group $2$, since $(11)(3) > 27$, and $(13)(3) > 27$.

So the factor $7$ must be in group $2$, and the factors $11,13$ must be in group $1$.

Since the factor $7$ is in group $2$, the factor $5$ can't be in group $2$, since $(7)(5) > 27$.

Hence, the factor $5$ must be in group $1$, and the factor $3$ must be group $2$.

Thus, group $2$ has the factors $37,7,3$, and group $1$ has the factors $11,13,5$.

It is intelligent brute force. The largest a three digit number can be is $999$ so you need to find a factor of $555,555$ that is between $556$ and $999$. The other will also be in that range so you are done. Next note that $3 \cdot 5 \cdot 7=105$ which is too small by itself and too large multiplied by any of the other factors, so two of $3,5,7$ have to be in one factor and one in the other. $11\cdot 13 \cdot 37 \gt 999$ so again two of those need to be in one factor and one in the other. We are down to $18$ combinations to try, three singletons from $3,5,7$ times all the one or two combinations from $11,13,37$. I missed Will Jagy's point that you need three factors in each set, so that decreases the number to try to $9$.

• I also thought that trying to solve for the same question just that for 2 factors of 2 digits and then I would split one of the digits into 2. Would this help me solve this type of question faster? Also, the question doesn't require me to state the numbers, but it does require me to state the amount of ways I can decompose it. – Brian Blumberg Apr 16 '18 at 2:44
• I don't understand the first sentence. What does it mean to split one of the digits into $2$? Having gotten down to $9$ possibilities it won't take long to try them all. You could note that $3\cdot 11\cdot 37$ is too large, so $37$ must be by itself and $11,13$ must be in the other set. Now just try each of $3,5,7$ times $11\cdot 13=143$ to see which are in range and find that only $5$ works. – Ross Millikan Apr 16 '18 at 2:53
• Yeah, never mind my question. I forgot to delete it after I realized that. Thanks. – Brian Blumberg Apr 16 '18 at 3:01

The prime factor are $3 · 5 · 7 · 11 · 13 · 37$ so there are $2^6=64$ factors and $32$ complement pairs. Just list them all but don't bother with those that are less than $555555/999=556$

Toss out, $1,3,5,7,11,13,37,3*5,3*7,3*11,3*13,3*37,3*5*7,3*5*11,3*5*13, 3*5*37,3*7*11,3*7*13$ (that's 18 that are too small).

$3*7*37=777$ and its compliment is $5*11*13=715$. (That's 1 that is acceptible)

We can continuing tossing out $3*11*13$ and $3*11*37$and $3*13*37$ are too high so we toss them. (That's 21 that are unaccptible)

$3*5*7*11$ is too high so there wonvt be any more factors that are multiples of $3$ in range. Hence no other complements which aren't multiples of $3$ will be in range either.

Of the 9 we haven'haven't considered: $3*5*7*13,3*5*7*37,3*5*11*13,3*5*11*37,3*5*13*37,3*7*11*13,3*7*11*37,3*7*13*37,3*11*13*37$ are all too big.

So that was exhaustive.

We can write $$555,555=5\times111,111$$ and notice that $111=37\times3$, so we have $$555,555=5\times37\times3003$$ and since $1001=7\times11\times13$, the prime factorization is $$555,555=3\times5\times7\times11\times13\times37.$$

If we multiply each of the three combinations $11\times13$, $11\times37$ and $13\times37$ by $3$, we see that only $3\times(13\times37)$ exceeds $1000$ which is a four-digit number.

Hence $37$ must pair with two other one-digit numbers, and $11\times13$ must pair with either $3$ or $5$ since $7\times11\times13>1000$.

If $11\times13$ pairs with $3$, then the other product must be $$37\times(5\times7)>1000$$ which is not a three-digit number.

Therefore, the only possible combination for $555,555=P_1\times P_2$ is $$P_1=5\times11\times13,\quad P_2=3\times7\times37.$$

I think the procedure might be based on Fermat method.

We can calculate that the number $555555$ can be expressed as a difference of two squares:

$746^2 - 31^2=556516-961$

Number $31$ is the distance from $746$ both ways which gives $715$ and $777$ accordingly. From here the smallest prime factor of:

$777$ is number $3$ which equals to $3\cdot259$ and for

$715$ is number $5$ which equals to $5\cdot143$

Further factorisation results in final solution where $143=11\cdot13$ and $259=7\cdot37$ therefore:

$(5\cdot11\cdot13)(3\cdot7\cdot37)$ can satisfy equation.

If $555{,}555=ab$ with $a,b\lt1000$, then we must also have $a,b\gt555$ (e.g., if $b\lt1000$, then $a=555{,}555/b\gt555{,}555/1000\gt555$). Now since $555{,}555=3\cdot5\cdot7\cdot11\cdot13\cdot37$, we may assume, without loss of generality, that $a=37k$. From $a\gt555=15\cdot37$, we see that $k\gt15$, and from $a\lt1000$, we see that $k\le\lfloor1000/37\rfloor=27$. The only product $k$ of the primes $3$, $5$, $7$, $11$, and $13$ that falls in the interval $15\lt k\le27$ is $k=3\cdot7=21$. Thus $a=37\cdot21=777$, $b=5\cdot11\cdot13=715$ is the only factorization of $555{,}555$ into two three-digit numbers.


# Contrapositive in change from “no self-defeating object” to “every object can be defeated”?

Here, Terence Tao presents a collection of similar mathematical arguments that he calls "no self-defeating object" (examples are Euclids proof of the infinitude of the primes and Cantor's theorem). In the second post, he remarks that one can reformulate these "no-self defeating object" arguments to get a "every object can be defeated"-version.

The simplest example "no self-defeating object" goes as follows:

Proposition 1 (No largest natural number). There does not exist a natural number N that is larger than all the other natural numbers.

Proof: Suppose for contradiction that there was such a largest natural number N. Then N+1 is also a natural number which is strictly larger than N, contradicting the hypothesis that N is the largest natural number.

The corresponding "every object can be defeated"-version is:

Proposition 1′. Given any natural number N, one can find another natural number N' which is larger than N.

Proof. Take N' := N+1.

Terence Tao also remarks:

"This is done by converting the “no self-defeating object” argument into a logically equivalent “any object can be defeated” argument, with the former then being viewed as an immediate corollary of the latter. This change is almost trivial to enact (it is often little more than just taking the contrapositive of the original statement), but it does offer a slightly different “non-counterfactual” (or more precisely, “not necessarily counterfactual”) perspective on these arguments which may assist in understanding how they work."

My question: What has the contrapositive to do with the change from "no self-defeating object" to "every object can be defeated"?

As I understand it, the "no self-defeating object"-version is of the form "$\neg\exists x: P(x)$" and the "every object can be defeated"-version is "$\forall x:\neg P(x)$". That these are equivalent is de Morgan for quantifiers, what does it have to do with contrapositives?

• – Henning Makholm Oct 9 '16 at 12:14
• @HenningMakholm: Ah, thanks. So do you think that Terence Tao just compared the change from $\neg\exists x : P(x)$ to $\forall x : \neg P(x)$ to taking the contrapositive, but is not saying that this is the same as taking contrapositives? – user376483 Oct 9 '16 at 12:45
• I'm not privy to Tao's thoughts, so the most I'm saying is that I too see a similarity between this kind of rewriting and contraposition, and once mused that it might be useful to call it contraposition. (Not many here agreed with that, though -- but it nice to see perhaps Tao might). – Henning Makholm Oct 9 '16 at 12:48
• One way to make it look more like a contrapositive is to view it as a change from "if $m$ is larger than all natural numbers then $m$ is not a natural number" to "if $m$ is a natural number then $m$ is not larger than all natural numbers". Of course, Tao only says that the process he describes is "often" "little more" than just taking the contrapositive, he does not claim that what he is doing is literally a contrapositive. I do find that mathematicians and logicians often use "contrapositive" informally in a broader sense than its formal meaning. – Carl Mummert Oct 9 '16 at 16:42

If, as is standard in presentations of intuitionistic logic, you treat $\lnot \phi$ as $\phi \Rightarrow \mathsf{false}$ then the role of the contrapositive here becomes clear: the contrapositive of:

$$(\exists N \in \Bbb{N}\cdot\forall m\in \Bbb{N}\cdot N > m) \Rightarrow \mathsf{false}$$

is:

$$\lnot \mathsf{false} \Rightarrow \lnot(\exists N \in \Bbb{N}\cdot\forall m\in \Bbb{N}\cdot N > m)$$

which, using De Morgan's laws and a tiny bit of arithmetic reasoning (to make things agree with Tao's presentation) is equivalent to: $$\forall N \in \Bbb{N}\cdot\exists m\in \Bbb{N}\cdot m > N.$$

This transformation giving Tao's Proposition $1'$ makes clear the innate constructive nature of the reasoning that is presented in disguise in the proof by contradiction in Tao's Proposition $1$.

• After all, this also relies on De Morgan's rule, and I don't see why you first interpret $\phi$ as $\phi\implies\bot$ and form the contrapositive of $(\exists N \in \Bbb{N}\cdot\forall m\in \Bbb{N}\cdot N > m) \Rightarrow \mathsf{false}$. One could just form the contrapositive of $\exists N \in \Bbb{N}\cdot\forall m\in \Bbb{N}\cdot N \geq m$ which is $\exists N \in \Bbb{N}\cdot\forall m\in \Bbb{N}\cdot N > m$. – user377104 Oct 25 '16 at 17:24
• Of course it relies on De Morgan's laws: the point is that the end result is a constructive truth that captures exactly what the proof actually proves. The reason for interpreting $\lnot \phi$ as $\phi \Rightarrow \mathsf{false}$ is because the notion of contrapositive is usually associated with implications: which is the point of the question: "what does this have to do with contrapositives?" The two existentially quantified statements that you claim are contrapositives are not contrapositives in any sense of the term that I am aware of. – Rob Arthan Oct 25 '16 at 19:16


# Tag Info

13

The contrapositive of the statement If $\overbrace{\text{$ab$and$a+b$have the same parity}}^{\large P}$, then $\overbrace{\text{$a$is even and$b$is even}}^{\large Q}$. is If $\overbrace{\text{$a$is odd or$b$is odd}}^{\large\lnot Q}$, then $\overbrace{\text{$ab$and$a+b$have different parities}}^{\large\lnot P}$. Note that $Q$ is the ...

9

Your result is an immediate consequence of the following proposition. Proposition. Suppose $X\subseteq Y$. Then $\mathscr P(X)\subseteq\mathscr P(Y)$. Proof. Let $E\in\mathscr P(X)$. Then $E\subseteq X\subseteq Y$ so that $E\subseteq Y$. Hence $E\in\mathscr P(Y)$. This proves $\mathscr P(X)\subseteq\mathscr P(Y)$. $\Box$ Do you see how your problem is now ...

8

\begin{align} |S| & = |S-T| + |S\cap T| \\[8pt] |T| & = |T-S| + |S\cap T| \end{align} If $|S-T|=|T-S|$, then the two right sides are the same, so the two left sides are the same. We can also write a proof explicitly dealing with bijections. You ask why one would "assume" a bijection exists. The bijection $g$ that you write about is not simply ...

7

You can proceed directly as follows: $2x = (x+y) + (x-y)$ which must be irrational as it is the sum of a rational and an irrational. So $x$ is irrational. Similarly $2y = (x+y) - (x-y)$ is irrational.

6

not induction, but maybe useful to note: firstly, since 3 is a prime, we have $n^3 \equiv_3 n$ (Fermat's little theorem) secondly $2n \equiv_3 -n$ (since $3n = 2n + n \equiv_3 0$) adding these two results: $$n^3 + 2n \equiv_3 n-n =0$$

6

By Spectral Theorem, $A$ is orthogonally similar to a diagonal matrix, i.e $$P^{T}AP=\pmatrix{\mu_1 \\ & \ddots \\ && \mu_n}$$ where $\mu_i>0$ is eigenvalue of $A$, and $\space P^{T}P=P^{-1}P=I$. For any $v$, let $v=Pu$. Then $$v^TAv=u^TP^TAPu=\sum_{k=1}^n\mu_ku_k^2>0$$

5

You need to show us some effort in the future. First, to show two sets are equal, we normally pick an element of the first set, show it is contained in the second, then pick an element in the second, and show it is contained in the first. If we suppose $x \in A$, then $x=2k$ for some integer $k$. Since $x = 2k$, $x = 2(k-1)+2$, and since $k-1$ is an ...

5

Take $N\in\Bbb N$ such that $|a_n-L|<1$ for $n\ge N$ and let $$M=\max\{|a_1|,\ldots,|a_N|,L+1\}.$$

5

Let $f:\{1,\ldots,n\}\to X$ be a surjection. Suppose, for the sake of contradiction, that $X$ has at least $n+1$ distinct elements $\{x_1,\ldots,x_{n+1}\}\subseteq X$. Since $f$ is a surjection, there exists, for each $i\in\{1,\ldots,n+1\}$, some $k_i\in\{1,\ldots,n\}$ such that $f(k_i)=x_i$. Since $f$ is a function and the $(x_i)_{i=1}^{n+1}$ are distinct, ...

4

We have that $F_n>F_{n-1}$ then $$F_{n+1}=F_n+F_{n-1}>2F_{n-1}>2\cdot2^{(n-1)/2}=2^{(n+1)/2}$$

4

HINT: For each $x\in X$, let $A_x=\{k\mid f(k)=x\}$. Then each $A_x$ is non-empty. Use that to construct an injection from $X$ into $\{1,\ldots,n\}$.

4

HINT: No, you can’t assume that $A=C$ and show that the inclusions hold: that’s the converse of what you’re supposed to prove, and an implication and its converse are not logically equivalent. Use the fact that $A\subseteq B$ and $B\subseteq C$ to show that $A\subseteq C$. You’re given that $C\subseteq A$, so the rest is straightforward.

4

The given inequality is equivalent to $a^3-a=a(a^2-1)>0$. By multiplying both sides by $a^2+1$, which is always positive, we get $a(a^2-1)(a^2+1)>a^2+1>0$, or $a^5-a>0$.

4

I would write it as: Let $P(n)$ stand for the expression: $$\forall x\leq n(x\not\in A)$$ Then use the assumption that $A$ has no least element to prove that $P(1)$ and $P(n)\implies P(n+1)$. Thus, we've shown that $\forall x:x\not\in A$, which means $A$ is empty. That's essentially the same as your proof, but uses less set notation.

4

Usually, a proof by contradiction of the statement $p \implies q$ is when you assume that the opposite of the desired conclusion is true (i.e., assume the negation of $q$ is true), and follow a few logical implications until you reach a statement that somehow explicitly or implicitly contradicts an initial assumption from the statement $p$. Meanwhile, a ...

4

The Lambert W-function is a function $W(z)$ which solves $z=W(z)e^{W(z)}$. It is a multi-valued function. In this case, you are trying to solve: $$e^{x\pi i/2} = x$$ of: $$\frac{-\pi i}{2}=\frac{-x\pi i}{2}e^{-x\pi i/2}$$ So $$x\frac{-\pi i}{2} = W(-\pi i/2)$$ or $$x =\frac{2i}{\pi} W\left(\frac{-\pi i}{2}\right)$$ I don't think you can do better ...

3

You won't be surprised to learn that in the last seventy-plus years since Tarski's book was first published in English, many other books have been appeared which will perhaps serve better as introductions to modern logic. And if you have downloaded my Teach Yourself Logic, you will have seen my "entry-level" suggestions on formal logic at the beginning of ...

3

You cannot go like this from $k=0$ to $k=1$ (i.e. $k=1$ cannot be expressed in the form $m+l$ as you wrote).

3

$N(t)$ is not equal to $N(t-s)+N(s)$, but $$N(t) = \Big(N(t) - N(s)\Big) + \Big(N(s)\Big)$$ and the two expressions inside the $\Big(\text{big parentheses}\Big)$ are independent of each other (whereas $N(t-s)$ and $N(s)$ are not independent of each other). So \begin{align} \Pr(N(s) = k \mid N(t) = n) & = \frac{\Pr(N(s)=k\ \&\ N(t) =n) ...

3

Yes, your work is all correct. Except a minor issue: the opposite of $|a_n| < \epsilon$ is $|a_n| \ge \epsilon$, not $|a_n| > \epsilon$. Instead of writing $P_n(X)$ as $|a_n| < \epsilon \; \forall n > X$, you may have found it clearer to write it as $\forall n > X \; |a_n| < \epsilon$. Then your entire statement would have been $$... 3 In general, if A\subseteq B, then \mathscr P (A)\subseteq \mathscr P (B) because every subset of A is a subset of B. More formally, if a\in \mathscr P (A), we need to show that a\in \mathscr P (B). But this is trivial, since if x\in a, then x\in B which implies that a\subseteq B which is the same as a\in \mathscr P (B). Now take ... 3 X\subset Y implies every element of X is an element of Y, so subsets of X are subsets of Y, so \mathcal{P}(X)\subset\mathcal{P}(Y). Finally, for Y=\mathcal{P}(X) you have \mathcal{P}(X)\subset\mathcal{P}(\mathcal{P}(X)). 3 Hint:$$\frac{a_{n+1}}{n}=\frac{a_{n+1}}{n+1}\frac{n+1}{n}.$$Or perhaps more to the point,$$\frac{a_{n+1}}{n+1}=\frac{a_{n+1}}{n}\frac{n}{n+1}.$$We've shown a_{n+1}/n\to l, we know n/(n+1)\to1, hence a_{n+1}/(n+1)\to l. And now this implies that a_n/n\to l. Given \epsilon>0 there exists N so |a_{n+1}/(n+1)-l|<\epsilon for all ... 3 If f(a)=c and f(b)=d, then$$\begin{align} \int_a^b f(x) \,\,dx+\int_c^d f^{-1}(y) \,\,dy &=\int_a^b f(x) \,\,dx+\int_a^b f^{-1}(f(x)) f'(x) \,\,dx\\\\ &=\int_a^b f(x) \,\,dx+\int_a^b x f'(x) \,\,dx\\\\ &=\int_a^b \left(f(x)+x f'(x)\right) \,\,dx\\\\ &=\int_a^b (xf(x))' \,\,dx\\\\ &=bf(b)-af(a)\\\\ &=bd-ac \end{align}$$Now, let ... 3 What would the proper negation look like? It turns out that, in this case, there are a number of ways you can go in how you want to prove this claim, not just via direct proof or contrapositive but also how you frame the question logically as well. I'll outline what I think is the clearest and easiest way of going about it. Claim: Let ... 3 You have the contrapositive right. You must negate P and Q separately and prove that the negation of Q implies the negation of P. To expand on this, for "a and b are even" to be false, you only need one of a and b to be odd, so the negation is "a is not even or b is not even". And for the statement "a+b and ab have the same parity" ... 3 Here is a simple proof that K(n) is not only exponential but 'super' exponential in the sense that for all constants C, there is some n_0 such that |K(n)|\geq C^n for all n\gt n_0. Let's rewrite your series as \sum_n\frac{a_n}{a_{n+1}} so that we don't run out of indices; in other words, K(n)=a_n. (For convenience's sake I'm going to take ... 3 y = \frac{3x^2+2y}{x^2+2}, multiplying both sides of the equation by x^2+2 results in an equivalent equation because that term is never 0 (in the reals at least). You end up with yx^2 + 2y = 3x^2+2y subtract 2y from both sides (always legitimate). yx^2=3x^2 Since x\neq 0 we can divide both sides by x^2 and get y=3 3 Proof: We first must note that \pi_j is the unique solution to \pi_j=\sum \limits_{i=0} \pi_i P_{ij} and \sum \limits_{i=0}\pi_i=1. Let's use \pi_i=1. From the double stochastic nature of the matrix, we have$$\pi_j=\sum_{i=0}^M \pi_iP_{ij}=\sum_{i=0}^M P_{ij}=1 Hence, $\pi_i=1$ is a valid solution to the first set of equations, and to make it a ...

3

You need to show independence of increments, i.e. if $0\le a<b<c<d$ then $(N_1(d)+N_2(d)) - (N_1(c)+N_2(c))$ is independent of $(N_1(b)+N_2(b)) - (N_1(a)+N_2(a))$, and similarly for more than two intervals. You can prove that by using independence of increments of each of the two processes separately plus independence of $N_1$ and $N_2$. You also ...

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# Creating many big sets of small numbers

There are $$n$$ numbers $$a_1,\ldots,a_n\in [0,1]$$.

Their sum is $$\sum_{i=1}^n a_i = s$$, where $$s$$ is some integer.

We want to group them into sets so that the sum of each set is at least $$t$$, where $$t$$ is some integer.

Let $$F(n,s,t)$$ be the largest number of sets that we can always create (for any $$a_i$$).

What is $$F(n,s,t)$$?

Example. $$F(n=8,s=7,t=1)=4$$:

• Proof that $$F(8,7,1)\geq 4$$: We can always create 4 sets by dividing the $$8$$ numbers arbitrarily into $$4$$ pairs. The sum of each pair is at most $$2$$, and the sum of all pairs is $$7$$, so the sum of each pair is at least $$1$$.
• Proof that $$F(8,7,1)\leq 4$$: We cannot always create 5 sets. Suppose for all $$i$$, $$a_i=7/8$$. In any $$5$$ sets, at least one set is a singleton so its sum is less than $$1$$.

Similarly, whenever $$n$$ is even, $$F(n,n-1,1)=n/2$$.

What else is known on the function $$F$$?

Currently I am particularly interested in the case $$t=2$$, but I will be happy for any more general references.

UPPER BOUND: $$F(n,s,t)\leq \lfloor {s+1\over t+1}\rfloor$$. Proof. Suppose that $$s+1$$ numbers equal $$s/(s+1)$$ and the other $$n-s-1$$ numbers equal $$0$$. To create a set with sum at least $$t$$, we need $$t+1$$ nonzeros. So we can create at most $$\lfloor {s+1\over t+1}\rfloor$$ such sets.

• @bof I added the upper bound that I had in mind. It is similar but not identical to yours. I am not sure about the lower bound. – Erel Segal-Halevi Jun 24 at 19:51

For $$n\in\mathbb N$$ and $$s,t\in\mathbb R$$ with $$0\lt t\le s\le n$$, let $$F(n,s,t)$$ be the greatest integer $$m$$ such that any family of $$n$$ numbers $$a_1,\dots,a_n\in[0,1]$$ with $$a_1+\cdots+a_n=s$$ can be partitioned into $$m$$ subfamilies, each with sum $$\ge t$$.

Lemma 1. If $$k\in\mathbb N$$ and $$s\le k\le n$$, then $$F(n,s,t)\le\left\lfloor\frac k{\lceil kt/s\rceil}\right\rfloor$$.

Lemma 2. If $$n\gt s$$ then $$F(n,s,t)\le\left\lfloor\frac{\lfloor s+1\rfloor}{\lfloor t+1\rfloor}\right\rfloor$$.

Proof. Put $$k=\lfloor s+1\rfloor$$ in Lemma 1.

Lemma 3. $$F(n,s,t)\ge\left\lfloor\frac{s+1}{t+1}\right\rfloor$$.

Proof. Let $$m=\left\lfloor\frac{s+1}{t+1}\right\rfloor\lt s+1$$, so that $$t\le\frac{s+1}m-1=\frac sm-\frac{m-1}m$$. We may assume that $$m\ge2$$.

Lat $$a_1,\dots,a_n\in[0,1]$$ be given, $$a_1+\cdots+a_n=s$$. For notational convenience we assume that $$a_1,\dots,a_p\gt0$$ while $$a_{p+1}=\cdots=a_n=0$$.

Partition the interval $$[0,s]$$ into $$m$$ equal subintervals $$J_1,\dots,J_m$$, indexed from left to right; that is, $$J_i=[c_{i-1},c_i]$$ where $$c_i=\frac{is}m$$. Then $$|J_i|=\frac sm\gt1-\frac1m$$.

Also partition $$[0,s]$$ into subintervals $$A_1,\dots,A_p$$ of respective lengths $$|A_i|=a_i$$. Let $$\mathcal A=\{A_1,\dots,A_p\}$$.

Each interval $$A\in\mathcal A$$ will be assigned to at most one of the intervals $$J_1,\dots,J_m$$, and (some of) the numbers $$a_1,\dots,a_p$$ will be assigned correspondingly to $$m$$ groups. Namely, an interval $$A\in\mathcal A$$ is assigned to the interval $$J_i=[c_{i-1},c_i]$$ if it satisfies one of the following three conditions: $$A\subseteq J_i;$$ $$i\gt1,\ \ c_{i-1}\in A,\ \ \frac{|A\cap J_i|}{|A|}\gt\frac{i-1}m;$$ $$i\lt m,\ \ c_i\in A,\ \ \frac{|A\cap J_i|}{|A|}\gt\frac{m-i}m.$$ It is important to note that no interval $$A\in\mathcal A$$ is assigned to more than one $$J_i$$.

Now the set of intervals assigned to $$J_i$$ covers $$J_i$$, except possibly for an interval at the left of length $$\le\frac{i-1}m|A|\le\frac{i-1}m$$, and an interval at the right of length $$\le\frac{m-i}m|A|\le\frac{m-i}m$$. Therefore, the sum of the lengths of intervals assigned to $$J_i$$ is $$\ge\frac sm-\frac{i-1}m-\frac{m-i}m=\frac sm-\frac{m-1}m\ge t$$.

Theorem. If $$t\in\mathbb N$$ and $$n\gt s$$, then $$F(n,s,t)=\left\lfloor\frac{s+1}{t+1}\right\rfloor$$.

Proof. Lemmas 2 and 3.

• Yes, for some applications it may be interesting to evaluate $F(n,s,t)$ for $s,t$ rational numbers. Alternatively we can scale the numbers up to integers. We need to add just one more parameter: each number $a_i$ is in $[0,q]$, for some integer $q\geq 1$. I think that in this case your technique leads to a lower bound of $\lfloor{s+q\over t+q}\rfloor$, but I have to verify – Erel Segal-Halevi Jun 27 at 18:10
• Instead of separating lemmas 1 and 2, can you have just one lemma in which the number of nonzero terms is $\lfloor s+1 \rfloor$? It seems to cover both cases: when $s$ is an integer, $\lfloor s+1 \rfloor = \lceil s+1 \rceil = s+1$, and when $s$ is not an integer, $\lfloor s+1 \rfloor =\lceil s\rceil$. – Erel Segal-Halevi Jun 30 at 10:09
• $F \leq \lfloor s+1 \rfloor / \lfloor t+1 \rfloor$. Since you have $\lfloor s+1 \rfloor$ nonzero terms, and each term equals $s / \lfloor s+1 \rfloor < 1$. So in each subfamily with sum $t$, there must be at least $\lfloor t+1 \rfloor$ such elements. – Erel Segal-Halevi Jun 30 at 13:50
• Looks good, thanks! Now the only gap that remains is when $t$ is not an integer - the upper bound has $\lfloor t+1\rfloor$ in the denominator and the lower bound has $t+1$. – Erel Segal-Halevi Jul 3 at 15:57
• The upper bound in Lemma 2 is the result of setting $k=\lfloor s+1\rfloor$ in Lemma 1, but this is not necessarily the optimal value of $k$. For instance, $F(n,1,0.4)\le2$ by Lemma 2, but (assuming $n\ge3$) by setting $k=3$ in Lemma 1 we get $F(n,1,0.4)\le1$. – bof Jul 3 at 18:19


# Relation between matrices

Maybe this is not a usual question for this forum... I have the following two matrices, mat1 and mat2, respectively:

mat1={{0.0178885,-0.0178885,-0.00894427,0.00894427,0.,0.,-0.00894427,0.00894427,0.,0.},{-0.0178885,0.0178885,0.00894427,-0.00894427,0.,0.,0.00894427,-0.00894427,0.,0.},{-0.00894427,0.00894427,0.0178885,-0.0178885,-0.00894427,0.00894427,0.,0.,0.,0.},{0.00894427,-0.00894427,-0.0178885,0.0178885,0.00894427,-0.00894427,0.,0.,0.,0.},{0.,0.,-0.00894427,0.00894427,0.00894427,-0.00894427,0.,0.,0.,0.},{0.,0.,0.00894427,-0.00894427,-0.00894427,0.00894427,0.,0.,0.,0.},{-0.00894427,0.00894427,0.,0.,0.,0.,0.0178885,-0.0178885,-0.00894427,0.00894427},{0.00894427,-0.00894427,0.,0.,0.,0.,-0.0178885,0.0178885,0.00894427,-0.00894427},{0.,0.,0.,0.,0.,0.,-0.00894427,0.00894427,0.00894427,-0.00894427},{0.,0.,0.,0.,0.,0.,0.00894427,-0.00894427,-0.00894427,0.00894427}};
mat2={{0.0198382,-0.0198382,-0.00991908,0.00991908,0.,0.,-0.00991908,0.00991908,0.,0.},{-0.0198382,0.0198382,0.00991908,-0.00991908,0.,0.,0.00991908,-0.00991908,0.,0.},{-0.00991908,0.00991908,0.0203862,-0.0203862,-0.0104672,0.0104672,0.,0.,0.,0.},{0.00991908,-0.00991908,-0.0203862,0.0203862,0.0104672,-0.0104672,0.,0.,0.,0.},{0.,0.,-0.0104672,0.0104672,0.0104672,-0.0104672,0.,0.,0.,0.},{0.,0.,0.0104672,-0.0104672,-0.0104672,0.0104672,0.,0.,0.,0.},{-0.00991908,0.00991908,0.,0.,0.,0.,0.0203862,-0.0203862,-0.0104672,0.0104672},{0.00991908,-0.00991908,0.,0.,0.,0.,-0.0203862,0.0203862,0.0104672,-0.0104672},{0.,0.,0.,0.,0.,0.,-0.0104672,0.0104672,0.0104672,-0.0104672},{0.,0.,0.,0.,0.,0.,0.0104672,-0.0104672,-0.0104672,0.0104672}};

If I plot the matrices, as well as their ratio, I obtain a visual representation of these:

Quiet@List[MatrixPlot[mat1], MatrixPlot[mat2], MatrixPlot[mat1/mat2]]

I think that there exists a numerical relation between mat1 and mat2, but I can't find it. Can anyone help me?

• Should your Dssmt and Dss be the same thing? – Roman May 9 '19 at 15:16
• @Roman there is a numerical relation among them, but it isn't a simple scaling operation by means of a scalar – Gae P May 9 '19 at 15:25
• Please make your code self-contained so that people can help without needing to guess. – Roman May 9 '19 at 15:46

Numerically all 100 differences between 100 corresponding elements of matrices correspond fall into just 5 symmetric values:

Round[Union[Flatten[mat1-mat2]],.0001]

{-0.0025,-0.0019,-0.0015,-0.001,0.,0.001,0.0015,0.0019,0.0025}

Which looks a lot like some small noise imposed on:

{-25,-20,-15,-10,0,10,15,20,25}/10000

It is actually quite easy to understand if you visualize your data. First of all you can see that the values of both matrices follow each other closely

ListPlot3D[{mat1,mat2},InterpolationOrder->0,PlotStyle->{Red,Blue},
BoxRatios->1,Mesh->None,SphericalRegion->True,PlotLegends->{"mat1","mat1"}]

You can see the also see the differences and realize, - for some elements mat1 is greater (more red), sometimes mat (more blue), and sometimes the same (white):

ListPlot3D[Rescale[Ds-Dss],InterpolationOrder->0,ColorFunction->
"TemperatureMap",BoxRatios->1,Mesh->None,SphericalRegion->True]

You also can see that all total

In[]:= Length[Flatten[mat1-mat2]]
Out[]= 100

differences between corresponding elements fall into just 9 different values and if you take in account symmetry - just 5 values:

BarChart[Union[Flatten[mat1 - mat2]], PlotTheme -> "Detailed"]

-- lets call them levels. And now you can find the statistics of how many differences correspond to a specific level -- and see it is symmetric:

ListLinePlot[Sort[Tally[Flatten[mat1 - mat2]]], PlotRange -> All, PlotTheme -> "Business"]