Cisc320 Homework 3 Solutions. Question 1: Original. Insert 170. Insert 34. TA: Matt Saponaro

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1 Question 1: Original Insert 170 Insert 34

2 Insert 28

3 Insert 15

4 Insert 9

5 Insert 7

6 Question 2: Proof by induction. Base Case: Consider a red-black tree with n = 2 nodes. o The root must be a black node as defined by the red-black tree properties. o The child must be a red node in order to maintain the black-height property. Induction Step: Let k be any integer such that kk 2. Suppose that for all integers ii, such that 1 ii kk. The statement P, P(i) is true In other words, there exists one red node when inserting k nodes We ve assumed P is true for the base case out to some arbitrary integer k, and now we want to show that this gets us that P(k+1) is true. From the definition of P(k), we know that an insert of k nodes will have atleast 1 red node. Thus we need to consider adding 1 more node to k-node red-black tree.

7 We will examine each insertion case into a RB-tree to see if the RB-properties are withheld for the additional node. If all cases are true, then P(k+1) is true. Case1: red uncle The newly inserted node will be red, while the parent and uncle will be black, and the grandparent (via induction hypothesis) will be. Thus, there will be at least one red node Case2: right child, black uncle The newly inserted node and the parent (via induction hypothesis) will be red (then, propagation which will maintain RB-properties; after propagation, at least the new child will be red). Case3: left child, black uncle The newly inserted node and the grandparent (via induction hypothesis) will be red (then, propagation which will maintain RB-properties; after propagation, at least the new child will be red.). Similarly for Case 4, 5, and 6, which are mirrored versions of cases 1, 2, and 3. Since all cases are true for P(k+1), we can conclude by the induction principle, P(n) is true for all n > 1. Question 3: Need to prove that for nn NN. nn = 2 13ii + kk; 0 < kk < 2 13, ii NN. And can be decomposed into 2*i 13 s + k. Ie. That any number can be decomposed into a multiple of an even number of 13 s + k. Thus, we just must show that kk; 0 < kk < 26 can be decomposed into the minimal amount of coins using the maximum possible coin (the greedy algorithm). We need to show that: 1) K can be decomposed using the coin set 2) And k is optimal. Proof of 1): The below program outputs a set for all values of k Proof of 2): The below program checks all decompositions of k and compares it against the greedy solution the greedy solution has a minimal amount of coins for k.

8 Python code: def greedydecomposition(k, coinset): greedydecomposition = [] for coin in coinset: #Note: coinset is sorted in descending order while k - coin >= 0: greedydecomposition += [coin] k -= coin if k == 0: return greedydecomposition else: return -1 #returns all possible decompositions of k using the coinset def alldecompositions(k, coinset, allpossibledecompositions, currentdecomposition): if k == 0: allpossibledecompositions += [currentdecomposition] return allpossibledecompositions else: for coin in coinset: if(k - coin >= 0): currentdecomptemp = currentdecomposition[:] currentdecomptemp += [coin] alldecompositions(k - coin, coinset, allpossibledecompositions, currentdecomptemp) return allpossibledecompositions #returns size of the optimal decompositions of k def optimaldecompositionlength(k, coinset): initialdecomposition = [] initialpossibilities = [] ordereddecompositionsbysize = sorted(alldecompositions(k, coinset, initialpossibilities, initialdecomposition), key=len) return len(ordereddecompositionsbysize[0]) def main(): coinset = [13, 9, 5, 3, 1] sorted(coinset,reverse=true) #enforces descending order for k in range(1,coinset[0]*2): if greedydecomposition(k,coinset) == -1: print("no decomposition exists for", k) elif len(greedydecomposition(k, coinset))!= optimaldecompositionlength(k, coinset): print("not optimal", k) else: print("optimal for", k, greedydecomposition(k,coinset)) print("done") main()

9 Output: ('optimal for', 1, [1]) ('optimal for', 2, [1, 1]) ('optimal for', 3, [3]) ('optimal for', 4, [3, 1]) ('optimal for', 5, [5]) ('optimal for', 6, [5, 1]) ('optimal for', 7, [5, 1, 1]) ('optimal for', 8, [5, 3]) ('optimal for', 9, [9]) ('optimal for', 10, [9, 1]) ('optimal for', 11, [9, 1, 1]) ('optimal for', 12, [9, 3]) ('optimal for', 13, [13]) ('optimal for', 14, [13, 1]) ('optimal for', 15, [13, 1, 1]) ('optimal for', 16, [13, 3]) ('optimal for', 17, [13, 3, 1]) ('optimal for', 18, [13, 5]) ('optimal for', 19, [13, 5, 1]) ('optimal for', 20, [13, 5, 1, 1]) ('optimal for', 21, [13, 5, 3]) ('optimal for', 22, [13, 9]) ('optimal for', 23, [13, 9, 1]) ('optimal for', 24, [13, 9, 1, 1]) ('optimal for', 25, [13, 9, 3]) Question 4: a) O(nlog(n)) algorithm MergeSort(A) Cost: O(nlogn) int sum = 0; for I in range(len(a)): Cost(O(n)) o sum += A[i]; o if(sum > B): return A[0 i-1] return A T(n) = O(nlogn) + O(n) = O(nlogn) b) O(n) algorithm Question4a(A,B,searchSpaceStart, searchspaceend currentsumofa): MedianOfMedianIndex = MedianOfMedians(A[searchSpaceStart searchspaceend]) Cost(O(n))

10 Partition(A, medianofmedianindex) #Use median of medians as your pivot Cost(O(n)) currentpartitionsum = Sum(A[0 medianofmedianindex]) Cost(O(n)) If currentsumofa > B: i. Question4a(A,B, searchspacestart, medianofmedianindex, currentsumofa) Else: i. Question4a(A,B, medianofmedianindex, SearchSpaceEnd, currentsumofa + currentpartitionsum) T(n) = T(7n/10) + O(n) Question 5: In order to show either of these algorithms are optimal, we need to determine the optimal solution: DDDD : CCCC a) Cost at every gas station is dd iiff. Thus the total cost is nn dd iiff CCCC ii as seen in the diagram CCCC above denoted by a). Need to show equal to optimal FF nn dd CCCC ii ii FF DD which matches the optimal solution. CCCC b) The algorithm denoted overshoots B and has waste (which is not optimal) if the distance traveled to the final gas station!= the distance the car can travel given F and C.