Tom and Sangu played a game. First, the Tom gives Sangu a sequence of different numbers N a1, a2,., an (not really set) and the number K. Sangu should build all the sequences of the series in length K. In each row, he should write down the product of the numbers K - 2: all the features of the following except the smallest and finest element. Sangu wins the game if he can write down all these numbers and tell Tom their product (module 10° +7, as it can be very large). However, Sangu is a lazy child and wants to help him win the game. Count the number Sangu has to tell Tom in python language.

Tom and Sangu played a game. First, the Tom gives Sangu a sequence of different numbers N a1, a2,., an (not really set) and the number K. Sangu should build all the sequences of the series in length K. In each row, he should write down the product of the numbers K - 2: all the features of the following except the smallest and finest element. Sangu wins the game if he can write down all these numbers and tell Tom their product (module 10° +7, as it can be very large). However, Sangu is a lazy child and wants to help him win the game. Count the number Sangu has to tell Tom in python language.

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

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Directions: Use the set- roaster notation to write the following sets, and determine the number of elements that are in a set. 1. Let A={ 1,2,3 }, B={ 3,1,2 },and C={ 1,1,2,3,3,3 }. What are the elements of A? B? and C? How are A, B and C related? 2. Let A={ w,x,y,z } and B= { a,b }. Find AxB? And BxB?
Implement the following Racket functions: You must use recursion, and not iteration. You may not use side-effects (e.g. set!). Transitive? Input: a list of pairs, L. Interpreting L as a binary relation, Transitive? returns #t if L is a transitive relation and #f otherwise. Examples: (display "Transitive? \n") (Transitive? '((a b) (b c) (a c))) ---> #t (Transitive? '((a a) (b b) (c c))) ---> #t (Transitive? '((a b) (b a))) ---> #f (Transitive? '((a b) (b a) (a a))) ---> #f (Transitive? '((a b) (b a) (a a) (b b))) ---> #t (Transitive? '())---> #t
Consider the set A of all strings of letters a-f of length 6. A few elements of this set are dcebac, ebaafe, and abeafa. What is the cardinality of each of the following sets? Please indicate which of the principles of multiplication, addition, or inclusion-exclusion you have used in each part. Part A The set A itself Part B The set B a subset of A of strings of length 6 in which no elements are repeated Part C The set C a subset of A of strings which contain the substring fab Part D The set D a subset of A of strings which contain the substring fab or the substring feebb (or both) Part E The set E of strings which contain either the substring fab or the substring baff (or both)
In class we showed how to calculate the longest increasing subsequence. This requirement turned out toostrict and we are interested to finding the longest close-to-increasing subsequence.We say that a sequence of numbers Z is close-to-increasing if the size of any decreasing substring is atmost 2. So given two consecutive elements in the sequence Zi and Zi+1 it must either be that• Zi < Zi+1• or if Zi > Zi+1 it must be that Zi+1 < Zi+2.Below are some examples of of close-to-increasing sequences:A = 11 10 18 17 25 33 30 34 27 29B = 8 99 33 34 40 11 13 20 23 22As we can see A2 = 10 < 11 = A1 but A3 = 12 > 10 = A2 so the constraint is satisfied. Same is true inother places where monotonicity fails at A7, A11, A14. Similar for sequence B.The following are examples of sequences that are not close-to-increasing:C = 10 11 13 12 18 30 27 22 33 40 44D = 17 16 1 11 13 12 15 19 30 25 33We have C6 = 30 > 27 = C7 and C6 = 27 > 22 = C9 so the constraint is violated. Similarly D1…
It is straightforward and efficient to compute the union of two sets using Boolean values. We may create a new union set by Oring the matching items of the two BitArrays since the union of two sets is a combination of the members of both sets. At other words, if the value in the corresponding place of either BitArray is True, a member is added to the new set.Computing the intersection of two sets is analogous to computing the union, except that the And operator is used instead of the Or operator Using the same technique we used to detect the difference, we can determine if one set is a subset of another. For example, if:setA(index) && !(setB(index))evaluates to False then setA is not a subset of setB.The BitArray Set ImplementationWrite The code for a CSet class based on a BitArray.
Implement the following Racket functions: Reflexive? Input: a list of pairs, L and a list S. Interpreting L as a binary relation over the set S, Reflexive? returns #t if L is a reflexive relation over the set S and #f otherwise. Examples: (display "Reflexive?\n") (Reflexive? '((a a) (b b) (c c)) '(a b c)) ---> #t (Reflexive? '((a a) (b b)) '(a b c)) ---> #f (Reflexive? '((a a) (a s) (b b) (c c)) '(a b c)) ---> #f (Reflexive? '() '()) ---> #t You must use recursion, and not iteration. You may not use side-effects (e.g. set!).
Given the following function, what happens if a[] contains just one element that doesn't match val? int binarySearch(int a[], int first, int last, int val){ if (first > last) return -1; int middle = (first + last) / 2; if (a[middle] == val) return middle; if (a[middle] < val) return binarySearch(a, middle+1, last, val); else return binarySearch(a, first, middle-1, val);} Group of answer choices binarySearch never calls itself again and terminates (recursion never happens) binarySearch calls itself once then terminates (recursion happens once) binarySearch calls itself twice then terminates (recursion happens twice) binarySearch calls itself 3 times then terminates (recursion happens 3 times)
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
a)Write a recursive definition for the set of odd positive integers. b)Use master theorem to find the solution to the recurrence relation f(n) = 4f(n/2) + 2? ! ,when n = 2" , where k is a positive integer and f(1) = 1.
Computing the intersection of two sets is similar to computing the union;only for this operation we use the And operator instead of the Or operator.Similarly, the difference of two sets is found by executing the And operatorwith a member from the first set and the negation of the corresponding member of the second set. We can determine if one set is a subset of another set byusing the same formula we used for finding the difference. For example, if:setA(index) && !(setB(index))evaluates to False then setA is not a subset of setB.Implement The code for a CSet class based on a BitArray in c#?
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.
Solve this and make sure the code compiles orderdp.cpp: #include <fcntl.h> #include <stdlib.h> #include <unistd.h> #include <inttypes.h> #include "orders.h" item::item(uint16_t s, uint16_t q) : sku(s), quantity(q) { } item_list::item_list() : items(NULL), num_products(0) { } void item_list::add(const item& i) { items = (item**)realloc(items, (num_products+1)*sizeof(item*)); items[num_products] = new item(i); num_products++; } order::order(uint16_t o, std::string n) : order_id(o), customer_name(n) { } void order::addItem(const item& i) { product_items.add(i); } order** read_orders(const char* filename, uint16_t* num_orders) { int fd; fd = open(filename, O_RDONLY); int err = fd; if (err == -1) { return NULL; } read(fd, num_orders, sizeof(*num_orders)); order** orders = new order*[*num_orders]; for (int i=0; i<*num_orders; i++) { uint16_t order_id; uint16_t num_products;…