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## T(n) = 4 T(n/3) + (n)

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**E[Xi] = n/(n-i)**X=X0 + … + Xn/2-1 E[X]= n/(n-i) = (n) 1 n/(n-i)2 for in/2 n/2-1 i=1**compute the median m of A**• check if m occurs in A more than n/2 times n/2 m n/2 -1 n/2 n/2 n/2 If element occurs in A more then n/2 times then it must be a median.**compute the n/3-rd and 2n/3-rd smallest**• elements x and y • 2. check if x or y occurs in A more than n/3 times y x n/3 n/3 n/3**sort array A**• for each b B, using binary search • check whether X-b is in A Step 1 – O(n log n) Step 2 – n * O(log n) = O(n log n)**Reduces to merging k sorted lists problem**(homework 2.1).**1. Find MIN and MAX of A**2. divide the interval [MIN,MAX] into n+1 buckets 3. for each non-empty bucket b, compute minb and maxb in the bucket 4. for each pair of neighboring non-empty buckets b,c compare the interval [maxb,minc] to the longest interval so far.**Optimization problems**Input: Output: Objective:**Activity selection problem**Input:list of time-intervals L Output:a non-overlapping subset S of the intervals Objective:maximize |S| 3,7 2,4 5,8 6,9 1,11 10,12 0,3**Activity selection problem**Input:list of time-intervals L Output:a non-overlapping subset S of the intervals Objective:maximize |S| 3,7 2,4 5,8 6,9 1,11 10,12 0,3 Answer = 3**Activity selection problem**Algorithm 1: 1. sort the activities by the starting time 2. pick the first activity a 3. remove all activities conflicting with a 4. repeat**Activity selection problem**Algorithm 1: 1. sort the activities by the starting time 2. pick the first activity a 3. remove all activities conflicting with a 4. repeat**Activity selection problem**Algorithm 1: 1. sort the activities by the starting time 2. pick the first activity a 3. remove all activities conflicting with a 4. repeat**Activity selection problem**Algorithm 2: 1. sort the activities by length 2. pick the shortest activity a 3. remove all activities conflicting with a 4. repeat**Activity selection problem**Algorithm 2: 1. sort the activities by length 2. pick the shortest activity a 3. remove all activities conflicting with a 4. repeat**Activity selection problem**Algorithm 2: 1. sort the activities by length 2. pick the shortest activity a 3. remove all activities conflicting with a 4. repeat**Activity selection problem**Algorithm 3: 1. sort the activities by ending time 2. pick the activity which ends first 3. remove all activities conflicting with a 4. repeat**Activity selection problem**Algorithm 3: 1. sort the activities by ending time 2. pick the activity which ends first 3. remove all activities conflicting with a 4. repeat**Activity selection problem**Algorithm 3: 1. sort the activities by ending time 2. pick the activity which ends first 3. remove all activities conflicting with a 4. repeat**Activity selection problem**Algorithm 3: 1. sort the activities by ending time 2. pick the activity which ends first 3. remove all activities conflicting with a 4. repeat**Activity selection problem**Algorithm 3: 1. sort the activities by ending time 2. pick the activity a which ends first 3. remove all activities conflicting with a 4. repeat Theorem: Algorithm 3 gives an optimal solution to the activity selection problem.**Activity selection problem**Theorem: Algorithm 3 gives an optimal solution to the activity selection problem. Proof idea: Any solution can be “morphed” to the solution given by the Algorithm 3.**Activity selection problem**Proof idea: Any solution can be “morphed” to the solution given by the Algorithm 3. Proof: W.l.o.g., all solutions below are sorted by the ending time. Let A=[a_1,b_1],…,[a_k,b_k] be the solution output by the Algorithm 3. Let O=[c_1,d_1],…,[c_l,d_l] be an optimal solution, which maximizes the length of the prefix common with A. A O**Activity selection problem**Proof: W.l.o.g., all solutions below are sorted by the ending time. Let A=[a_1,b_1],…,[a_k,b_k] be the solution output by the Algorithm 3. Let O=[c_1,d_1],…,[c_l,d_l] be an optimal solution, which maximizes the length of the prefix common with A. A O Let m-1 be the length of the common prefix of A and O. Assume m k. Then [am,bm] does not intersect [ci,di] for i<m (since it does not intersect [ai,bi] for i<m). Moreover [am,bm] does not intersect [ci,di] for i>m (since bm dm and dm<ci for i>m). Hence we can replace [cm,dm] by [am,.bm], and obtain another optimal solution which contradicts our choice of O. Case m>k cannot happen, since Algorithm 3 would output more activities**Coin change problem**Input:a number P (price) Output:a way of paying P using bills, coins Objective:minimize the number of bills, coins used 8.37 = 5 + 1 + 1 + 1 + 0.25 + 0.10 + 0.01 + 0.01 8 bills/coins used**Coin change problem**For simplicity assume that the coin values are: 10,5,1 Pay(N) if N 10 then output(10), Pay(N-10) elif N 5 then output(5), Pay(N-5) elif N 1 then output(1), Pay(N-1)**Coin change problem**coin values: 10,5,1 Pay(N) if N 10 then output(10), Pay(N-10) elif N 5 then output(5), Pay(N-5) elif N 1 then output(1), Pay(N-1) Theorem: For coin values 10,5,1 the Pay algorithm uses the minimal number of coins.**Coin change problem**coin values: 10,5,1 Theorem: For coin values 10,5,1 the Pay algorithm uses the minimal number of coins. Proof: Let N be the amount to be paid. Let the optimal solution be P=A*10 + B*5 + C. Clearly B1 (otherwise we can decrease B by 2 and increase A by 1, improving the solution). Similarly C4. Let the solution given by Pay be P=a*10 + b*5 + c. Clearly b1 (otherwise the algorithm would output 10 instead of 5). Similarly c4.**Coin change problem**coin values: 10,5,1 Proof: Let N be the amount to be paid. Let the optimal solution be P=A*10 + B*5 + C. Clearly B1 (otherwise we can decrease B by 2 and increase A by 1, improving the solution). Similarly C4. Let the solution given by Pay be P=a*10 + b*5 + c. Clearly b1 (otherwise the algorithm would output 10 instead of 5). Similarly c4. From 0 C4 and P=(2A+B)*5+C we have C=P mod 5. Similarly c=P mod 5, and hence c=C. Let Q=(P-C)/5. From 0 B 1 and Q = 2A + B we have B=Q mod 2. Similarly b=Q mod 2, and hence b=B. Thus a=A, b=B, c=C, i.e., the solution given by Pay is optimal.**Coin change problem**For simplicity assume that the coin values are: 4,3,1 Pay(N) if N 4 then output(4), Pay(N-4) elif N 3 then output(3), Pay(N-3) elif N 1 then output(1), Pay(N-1)**Coin change problem**coin values: 4,3,1 6 = 4 + 1 + 1 6 = 3 + 3 Theorem: For coin values 4,3,1 the Pay algorithm uses the minimal number of coins.**Huffman codes**binary character code = assignment of binary strings to characters e.g. ASCII code A = 01000001 B = 01000010 C = 01000011 …. fixed-length code 01000001010000100100001101000001**Huffman codes**binary character code = assignment of binary strings to characters e.g. ASCII code A = 01000001 B = 01000010 C = 01000011 …. fixed-length code 01000001 01000010 01000011 01000001**Huffman codes**binary character code = assignment of binary strings to characters E.g. A = 0 B = 01 C = 11 …. variable-length code 011111111110…**Huffman codes**binary character code = assignment of binary strings to characters DEFINITION: a binary character code is prefix-free, if no code-word is a prefix of another code-word A = 01 B = 101 C = 1011 …. NOT prefix-free (B is a prefix of C)**Optimal prefix-code problem**Input:alphabet, with frequencies Output:prefix code Objective:expected number of bits per character**Optimal prefix-code problem**A 60% B 20% C 10% D 10%**Optimal prefix-code problem**A 60% - 00 B 20% - 01 C 10% - 10 D 10% - 11 2bits per character**Optimal prefix-code problem**A 60% - 0 B 20% - 10 C 10% - 110 D 10% - 111 1.6 bits per character**Optimal prefix-code problem**Huffman ( [a1,f1],[a2,f2],…,[an,fn]) if n=1 then code[a1] else let fi,fj be the 2 smallest f’s Huffman ( [ai,fi+fj],[a1,f1],…,[an,fn] ) code[aj] code[ai] 0 code[ai] code[ai] 1**Optimal prefix-code problem**Lemma 1: Let x,y be the symbols with frequencies fx < fy. Then in an optimal prefix code length(Cx) length(Cy).**Optimal prefix-code problem**Lemma 2: Let C = R 0 be a longest codeword in an optimal code. Then R 1 is also a codeword. Lemma 1: Let x,y be the symbols with frequencies fx < fy. Then in an optimal prefix code length(Cx) length(Cy).**Optimal prefix-code problem**Lemma 3: Let x,y be the symbols with the smallest frequencies. Then there exists an optimal prefix code such that the codewords for x and y differ only in the last bit. Lemma 2: Let C = R 0 be a longest codeword in an optimal code. Then R 1 is also a codeword. Lemma 1: Let x,y be the symbols with frequencies fx < fy. Then in an optimal prefix code length(Cx) length(Cy).**Optimal prefix-code problem**Theorem: the prefix code output by the Huffman algorithm is optimal. Lemma 3: Let x,y be the symbols with the smallest frequencies. Then there exists an optimal prefix code such that the codewords for x and y differ only in the last bit.