Write a program for solving summation puzzles by enumerating and testing all possible configurations. Using your program, solve the three puzzles given in Section 5.3.3.
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P 5.28: Write a program for solving summation puzzles by enumerating and testing all possible configurations. Using your program, solve the three puzzles given in Section 5.3.3.
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- 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)Please help. How to solve this on Java language by using only Dynamic Programming? Thank you. We are given a set A of integers. Check whether we can partition A into three subsets with equal sums. It is important to note that all elements should be included in the equally divided subsets; no element should be left out, and none of the elements should be repeated in the subsets.For example: A={4,7,6,2,10,7,10,2} has such a partition: {4,2,10}, {7,7,2},{6,10}Input: A set, A, of integersOutput: if we can partition A into three subsets with equal sums, print subsets otherwise present a suitable message.Consider the following recursive definition of list reversal. For a list L, rev(L) is defined: rev([]) = []rev(x : L) = rev(L) + [x] Please prove by induction on L that concatenation is left cancellative: L+M = L+N implies M = N. Please prove by induction on L that rev(rev(L)) = L for all lists L. Please prove by induction on L that rev(L + M ) = rev(M ) + rev(L) for all lists L and M . (Can utilize fact that concatenation = associative) Using 1, 2, and 3 to prove that concatenation is right cancellative as well: L + N = M + N implies L = M.
- Implement the following function that returns a vector of vectors of Item, each of which is a subset of elements chosen from the given vector a[first…last]: /* if a is {1,2,3}, first=0, last=2, the function shall returns a vector of the following vectors: {}, {1}, {2}, {3}, {1,2},{1,3}, {2,3},{1,2,3}, all subsets of a[0…2]. Precondition: last-first+1>=1, i.e., there is at least one element in the a[first…last] Note 1)if the length of a[first…last] is n, then the function should return a vector of 2n vectors 2) The order of these subsets does not need to match what’s listed here… */ vector<vector<int>> subsets (const vector<int> & a, int first, int last) codeGiven the following sets: U= {1,2,3,4,5,6,7,8,}, A={1,4,5,7}, B= {2,5,6,7,}, and C= {3,4,6,7} Complete the following set operations: a. A U(BUC) b. (A N (B N C))' . c. (A N B) U ( A N C) d. (A N B')U (A N C')Give a recursive definition for the set of all strings of a’s and b’s that begins with an a and ends in a b. Say, S = { ab, aab, abb, aaab, aabb, abbb, abab..} Let S be the set of all strings of a’s and b’s that begins with a and ends in a b. The recursive definition is as follows – Base:... Recursion: If u ∈ S, then... Restriction: There are no elements of S other than those obtained from the base and recursion of S.
- Correct answer will be upvoted else downvoted. Computer science. Presently Nezzar has a beatmap of n particular focuses A1,A2,… ,An. Nezzar might want to reorder these n focuses so the subsequent beatmap is great. Officially, you are needed to find a change p1,p2,… ,pn of integers from 1 to n, to such an extent that beatmap Ap1,Ap2,… ,Apn is great. In case it is unthinkable, you ought to decide it. Input The primary line contains a solitary integer n (3≤n≤5000). Then, at that point, n lines follow, I-th of them contains two integers xi, yi (−109≤xi,yi≤109) — directions of point Ai. It is ensured that all focuses are unmistakable. Output In case there is no arrangement, print −1. In any case, print n integers, addressing a legitimate change p. In case there are numerous potential replies, you can print any.9. Is the following statement TRUE or FALSE? If it is TRUE, prove it, and if it is FALSE, provide a counter-example.If f: S→T is a function, where |S| > |T| (and both are finite sets), then there exist elements s1 and s2 in S such that fs1()=fs2()Please help me Josephus Problem is a theoretical problem related to a certain counting-out game. On thiscase, people are standing in a circle waiting to be executed. After a specified number ofpeople are skipped, the next person is executed. The procedure is repeated with theremaining people, starting with the next person, going in the same direction and skippingthe same number of people, until one person remains, and is freed.Arrange the numbers 1 , 2, 3 , ... consecutively (say, clockwise) in a circle. Now removenumber 2 and proceed clockwise by removing every other number, among those thatremain, until one number is left. (a) Let denote the final number which remains. Find formula for .(b) If there are 70 people, what is the safe number (the number that remains)?
- Q#03: Consider a queue that is a Queue-Type object and the size of the array implementation of aqueue is 200. In addition, consider that the value of Queue-Front is 199 and the value of Queue-Rear is 125. a. After adding an element to queue, what will be the values of Queue-Front and Queue-Rear?b. After removing an element from queue, what will be the values of Queue-Front and Queue-Rear?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? '())---> #tComputing 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#?