1. Minimum Euclidean Distance Pairs Given a list of N coordinates on the X-Y Cartesian plane, find the set of pair/s of coordinates with the minimum Euclidean distance between them. Euclidean distance can be calculated using the formula below: dist((x1, y1), (x2, y2)) = v(x2 - x1)² + (y2 - y1)²

1. Minimum Euclidean Distance Pairs Given a list of N coordinates on the X-Y Cartesian plane, find the set of pair/s of coordinates with the minimum Euclidean distance between them. Euclidean distance can be calculated using the formula below: dist((x1, y1), (x2, y2)) = v(x2 - x1)² + (y2 - y1)²

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

See similar textbooks

Similar questions

def longest_unique_substring(s: str) -> str:"""Given a string <s>, return the longest unique substring that occurs within<s>.A unique substring is a substring within <s> which DOES NOT have anyrepeating characters. As an example, "xd" is unique but "xxd" is not.If there are two equal length unique substrings within <s>, return the onethat starts first (i.e., begins at a smaller index).tips:In order to get your function to run fast, consider using a dictionary tostore the indexes of previously seen characters, from there, you canfollow a set of rules based on each new character you see to determinethe length of the longest unique substring seen so far.>>> longest_unique_substring('aab')'ab' >>> longest_unique_substring('abcabcbb')'abc'"""
def longest_unique_substring(s: str) -> str:"""Given a string <s>, return the longest unique substring that occurs within<s>.A unique substring is a substring within <s> which DOES NOT have anyrepeating characters. As an example, "xd" is unique but "xxd" is not.If there are two equal length unique substrings within <s>, return the onethat starts first (i.e., begins at a smaller index).tips:In order to get your function to run fast, consider using a dictionary tostore the indexes of previously seen characters, from there, you canfollow a set of rules based on each new character you see to determinethe length of the longest unique substring seen so far.>>> longest_unique_substring('aab')'ab' >>> longest_unique_substring('abcabcbb')'abc'""" RESTRICTIONS: - Do not add any imports and do it on python .Do not use recursion. Do not use break/continue.Do not use try-except statements.
def longest_unique_substring(s: str) -> str:"""Given a string <s>, return the longest unique substring that occurs within<s>.A unique substring is a substring within <s> which DOES NOT have anyrepeating characters. As an example, "xd" is unique but "xxd" is not.If there are two equal length unique substrings within <s>, return the onethat starts first (i.e., begins at a smaller index).tips:In order to get your function to run fast, consider using a dictionary tostore the indexes of previously seen characters, from there, you canfollow a set of rules based on each new character you see to determinethe length of the longest unique substring seen so far.>>> longest_unique_substring('aab')'ab'""" RESTRICTIONS: - Do not add any imports and do it on python .Do not use recursion. Do not use break/continue.Do not use try-except statements.
Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses. For example, given n = 3, a solution set is: [ "((()))", "(()())", "(())()", "()(())", "()()()" ] """ def generate_parenthesis_v1(n): defadd_pair(res, s, left, right): ifleft==0andright==0: res.append(s) return ifright>0: add_pair(res, s+")", left, right-1) ifleft>0: add_pair(res, s+"(", left-1, right+1) res= [] add_pair(res, "", n, 0) returnres def generate_parenthesis_v2(n): defadd_pair(res, s, left, right): ifleft==0andright==0: res.append(s) ifleft>0: add_pair(res, s+"(", left-1, right) ifright>0andleft<right: add_pair(res, s+")", left, right-1).
Given: P={a, b, c, d, e, f, g, h, i} and Q={a, e, o, i, u, } List the elements of the sets: A. PnQ 2.PUQ
Define function p:cat represents set concatenation ◦, which concatenates every pair of strings from both sets using racket. We can specify set concatenation as p1 ◦ p2 = {u · v | u ∈ p1 ∧ v ∈ p2}. Alternatively, we give an inductive specification: ∅ ◦ p2 = ∅ p1 ◦ p2 = prefix(u, p2) ∪ (p'1◦ p2) if p1={u} ∪ p'1.
Implement the Tic-Tac-Toe game in C++ using functions and 2D array of 3X3 size. Rules of the Game a) The game is to be played between computer and user.b) One of the players chooses ‘O’ and the other ‘X’ to mark their respective cells.c) The game starts with one of the players and the game ends when one of the players has onewhole row/ column/ diagonal filled with his/her respective character (‘O’ or ‘X’).d) If no one wins, then the game is said to be draw.
A fibonacci series is defined as a series where the number at the current index, is the value of the summation of the index preceding it (index -1) and (index-2). Essentially, for a list fibonacci_numbers which is the fibonacci numbers series, fibonacci_numbers[i] = fibonacci_numbers[i-1] + fibonacci_numbers[i-2] The fibonacci series always begins with 0, and then a 1 follows. So an example for fibonacci series up to the first 7 values would be - 0, 1, 1, 2, 3, 5, 8, 13 Complete the fibonacci(n) function, which takes in an index, n, and returns the nth value in the sequence. Any negative index values should return -1. Ex: If the input is: 7 the output is: fibonacci(7) is 13 Important Note Use recursion and DO NOT use any loops. Review the Week 9 class recording to see a variation of this solution. def fibonacci(n): if (n < 0 ): return -1 else: return n fibonacci_numbers = (fibonacci[n - 1] + fibonacci[n - 2]) return fibonacci_numbers(n) # TODO: Write…
Implement the following function, without using any data structure. /* Given two vectors of chars, check if the two vectors are permutations of each other, i.e., they contains same values, in same or different order.e.g., V1=[‘a’,’b’,’a’] and V2=[‘b’,’a’,’a’] stores same multi-set of data points: i.e., both contains two ‘a’, and one ‘b’. e.g., V3=[‘a’,’c’,’t’,’a’] and V4=[‘a’,’c’,’t’] are not same multi-set. V3 contains two ‘a’s, while V4 has only one ‘a’. Note: when considering multiset, the number of occurrences matters. @param list1, list2: two vectors of chars @pre: list1, list2 have been initialized @post: return true if list1 and list2 stores same values (in same or different order); return false, if not. */  bool SameMultiSet (vector<char> list1, vector<char> list2)
Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0. This time, your problem has additional details: Constraints: The solution set must not contain duplicate triplets. The order of the output and the order of the triplets does not matter. 3 <= nums.length <= 3000 -105 <= nums[i] <= 105 Function definition for Java: public List<List<Integer>> threeSum(int[] nums) { // Your code here } Function definition for Python: def threeSum(self, nums: List[int]) -> List[List[int]]: #Your code here Announced Test Cases: Input: nums = [0,1,1] Output: [] Explanation: The only possible triplet does not sum up to 0.Input: nums = [-5,0,5,10,-10,0] Output: [[-10,0,10],[-5,0,5]] Explanation: There are two possible combinations of triplets that satisfy: (-5,0,5) and (-10,0,10). Hint: There are 3 well-known ways to solve this problem!
Implement the Tic-Tac-Toe game in C++ using functions and 2D array of 3X3 size. Rules of the Gamea) The game is to be played between computer and user.b) One of the players chooses ‘O’ and the other ‘X’ to mark their respective cells.c) The game starts with one of the players and the game ends when one of the players has onewhole row/ column/ diagonal filled with his/her respective character (‘O’ or ‘X’).d) If no one wins, then the game is said to be draw. Note:No pointers should be used.The program should be simple and easy.
1. A set of integers are relatively prime to each other if there is no integer greater than 1 that divides all the elements. Furthermore, in Number Theory, it is known that the Euler function,ϕ (n), expresses the number of positive integers less than n that are relatively prime with n. Choose the alternative that has the correct value of ϕ(n) for every n below. A) ϕ(5) = 4 B) ϕ(6) = 2 C) ϕ(10) = 3 D) ϕ(14) = 6 E) ϕ(17) = 16