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M. T. Goodrich and R. Tamassia
John Wiley & Sons
Solution of Exercise R-1.
The numbers in the first row are quite large. The table below calculates it approxi- mately in powers of 10. People might also choose to use powers of 2. Being close to the answer is enough for the big numbers (within a few factors of 10 from the answers shown).
1 Second 1 Hour 1 Month 1 Century
log n 210
6 ≈ 10300000 23.^6 ×^10
9 ≈ 1010
9 22.^6 ×^10
12 ≈ 100.^8 ×^10
12 23.^1 ×^10
15 ≈ 1010
15 √ n ≈ 1012 ≈ 1. 3 × 1019 ≈ 6. 8 × 1024 ≈ 9. 7 × 1030 n 106 3. 6 × 109 ≈ 2. 6 × 1012 ≈ 3. 12 × 1015 n log n ≈ 105 ≈ 109 ≈ 1011 ≈ 1014 n^2 1000 6 × 104 ≈ 1. 6 × 106 ≈ 5. 6 × 107 n^3 100 ≈ 1500 ≈ 14000 ≈ 1500000 2 n^ 19 31 41 51 n! 9 12 15 17
M. T. Goodrich and R. Tamassia
John Wiley & Sons
Solution of Exercise R-1.
The Loop1 method runs in O ( n ) time.
M. T. Goodrich and R. Tamassia
John Wiley & Sons
Solution of Exercise R-1.
The Loop3 method runs in O ( n^2 ) time.
M. T. Goodrich and R. Tamassia
John Wiley & Sons
Solution of Exercise R-1.
The Loop4 method runs in O ( n^2 ) time.
M. T. Goodrich and R. Tamassia
John Wiley & Sons
Solution of Exercise R-1.
By the definition of big-Omega, we need to find a real constant c > 0 and an integer constant n 0 ≥ 1 such that n^3 log n ≥ cn^3 for n ≥ n 0. Choosing c = 1 and n 0 = 2, shows n^3 log n ≥ cn^3 for n ≥ n 0 , since log n >≥ 1 in this range.
M. T. Goodrich and R. Tamassia
John Wiley & Sons
Solution of Exercise C-1.
To say that Al’s algorithm is “big-oh” of Bill’s algorithm implies that Al’s algo- rithm will run faster than Bill’s for all input greater than somenon-zero positive integer n 0. In this case, n 0 = 100.
M. T. Goodrich and R. Tamassia
John Wiley & Sons
Solution of Exercise C-1.
One possible solution is f ( n ) = n^2 + ( 1 + sin ( n )).
M. T. Goodrich and R. Tamassia
John Wiley & Sons
Solution of Exercise C-1.
The induction assumes that the set of n − 1 sheep without a and the set of n − 1 sheep without b have sheep in common. Clearly this is not true with the case of 2 sheep. If a base case of 2 sheep could be shown, then the induction would be valid.
M. T. Goodrich and R. Tamassia
John Wiley & Sons
Solution of Exercise C-1.
For convenience assume that n is even. Then
n
i = 1
log 2 i / geq
n
i = n 2 + 1
log 2
n 2
n 2
log 2
n 2
which is Ω( n log n ).
M. T. Goodrich and R. Tamassia
John Wiley & Sons
Solution of Exercise C-1.
Let Bn − 1 , Bn − 2 ,..., B 1 , B 0 be the n bottles of wine. Consider any binary string b = bn − 1 bn − 2 ... b 1 b 0 of length n. A string b corresponds to a test as follows. If b has k 1’s, we give a taste tester to drink a sample that consists of exactly k drops: a drop of bottle Bi is included to the sample only if bi = 1. The idea is to have that sufficient collection of tests b 1 , b 2 ,..., bT running in parallel, so that, when (after a month) their outcome is known, we can deterministically identify the poisoned bottle. There are several ways for someone to construct the tests. A naive solution uses n tests, each of them having a drop from only one distinct bottle. We can do much better. An efficient construction of the tests will be in such a way so that a binary search is performed in the sequence of the bottles. For simplicity assume that n is a power of 2, i.e., n = 2 k^ for some integer k. We define B ( t ) to be the binary string of length t that consists of t 2 consecutive 0’s and followed by 2 t 1’s. Let the first test be b 1 = B ( n ). Given the output of this test, our search space is reduced by half: if b 1 is positive then we know that the poisoned bottle is one of Bn 2 − 1 ,..., B 1 , B 0 , otherwise one of Bn − 1 , Bn − 2 ,..., Bn 2.
Let now the second test be b 2 = B ( n 2 )|| B ( n 2 ), where || denotes string concate- nation. Clearly tests b 1 and b 2 reduce the search space to one forth of the ini- tial. We proceed in the same way: b 3 = B ( n 4 )|| B ( n 4 )|| B ( n 4 )|| B ( n 4 ) and, generally, bi = B ( (^) i + n 1 )|| B ( (^) i + n 1 )|| B ( (^) i + n 1 )|| B ( (^) i + n 1 ), for 1 ≤ i ≤ k , where k = log n. Combining all k tests b 1 , b 2 , ..., bk the search space is finally reduced to only one bottle. A similar reasoning can be followed in the case that n is not an exact power of 2; in that case, the number of tests are ⌈ n ⌉. We give an example for n = 13. The tests could be:
test 1: b 1 =11111100 0 0000
test 2: b 2 =11100011 1 0000
test 3: b 3 =10010010 0 1100
test 4: b 4 =11011011 0 1010,
M. T. Goodrich and R. Tamassia
John Wiley & Sons
Solution of Exercise C-1.
Start at the upper left of the matrix. Walk across the matrix until a 0 is found. Then walk down the matrix until a 1 is found. This is repeated until the last row or column is encountered. The row with the most 1’s is the last row which was walked across. Clearly this is an O ( n )-time algorithm since at most 2 · n comparisons are made.
M. T. Goodrich and R. Tamassia
John Wiley & Sons
Solution of Exercise C-1.
Using the two properties of the array, the method is described as follows.
- Starting from element A [ n − 1 , 0 ], we scan A moving only to the right and upwards.
- If the number at A [ i , j ] is 1, then we add the number of 1s in that column ( i + 1) to the current total of 1s
- Otherwise we move up one position until we reach another 1.
An example of the traversal of the array is shown in the figure 0.1.
array A
Figure 0.1: The traversal of the array.
The pseudo-code of the algorithm is shown below. The running time is O ( n ). In the worst case, you will visit at most 2 n − 1 places in the array. In the case that the diagram has all 0s in rows 2 through n and 1s in the first row, then there will be n − 1 iterations of the for loop at constant time (since it will never enter the while loop) and 1 iteration of the while loop which has n iterations of constant time.
M. T. Goodrich and R. Tamassia
John Wiley & Sons
Solution of Exercise C-2.
Name the two stacks as E and D , for we will enqueue into E and dequeue from D. To implement enqueue( e ), simply call E .push( e ). To implement dequeue(), simply call D .pop(), provided that D is not empty. If D is empty, iteratively pop every element from E and push it onto D , until E is empty, and then call D .pop(). For the amortized analysis, charge $2 to each enqueue, using $1 to do the push into E. Imagine that we store the extra cyber-dollar with the element just pushed. We use the cyber-dollar associated with an element when we move it from E to D. This is sufficient, since we never move elements back from D to E. Finally, we charge $1 for each dequeue to pay for the push from D (to remove the element returned). The total charges for n operations is O ( n ); hence, each operation runs in O ( 1 ) amortized time.
M. T. Goodrich and R. Tamassia
John Wiley & Sons
Solution of Exercise C-2.
The number of permutations of n numbers is n! = n ( n − 1 )( n − 2 )... 3 · 2 · 1. The idea is to compute this product only with increments of a counter. We use a stack for that purpose. Starting with the call Enumerate( 0 , S ), where S is a stack of n elements. the pseudo code should be like that:
Algorithm Enumerate( t , S ){ k = S .size(); while (! S .isEmpty() ) { S.pop(); t ++; S ′: a new stack of size k − 1 Enumerate( t , S ′); } }
