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# chapter1 - Chapter 1–Solutions February 2 2007 2 1.1 A...

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Unformatted text preview: Chapter 1–Solutions February 2, 2007 2 1.1. A single digit number is at most b − 1, therefore the sum of any three such numbers is at most 3 b − 3. On the other hand, a two-digit number can be as large as b 2 − 1. It is enough to show that b 2 − 1 ≥ 3 b − 3. Indeed, b 2 − 1 − 3 b − 3 = ( b − 1) · ( b − 2), which is ≥ 0 for b ≥ 2. 1.2. For a number N we need ⌈ log 2 ( N + 1) ⌉ binary digits and ⌈ log 10 ( N + 1) ⌉ decimal digits. By the logarithm conversion formula: ⌈ log 10 ( N + 1) ⌉ = ⌈ log 2 ( N + 1) · log 2 (10) ⌉ ≤ ⌈ log 2 ( N + 1) ⌉·⌈ log 2 (10) ⌉ = 4 · ⌈ log 2 ( N + 1) ⌉ . For very large numbers, ⌈ log 2 ( N + 1) · log 2 (10) ⌉ ≃ log 2 ( N + 1) · log 2 (10) so the required ratio is approximately equal to log 2 (10) ≃ 3 . 8. 1.3. The minimum depth of a d-ary tree is achieved when all nodes have precisely d children. In that case, the depth is log d ( n ). For any d-ary tree, we have depth D ≤ log d ( n ) = log( n ) log( d ) = Ω( log( n ) log( d ) ) 1.4. We can lower bound n ! as parenleftBig n 2 parenrightBig · ··· · parenleftBig n 2 parenrightBig bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright n 2 terms ≤ 1 · 2 · 3 · ··· · parenleftBig n 2 parenrightBig · parenleftBig n 2 + 1 parenrightBig · ··· · n bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright n 2 terms and upper bound it as 1 · 2 · 3 · ··· · n bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright n terms ≤ n · ··· · n bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright n terms Hence, parenleftBig n 2 parenrightBig n 2 ≤ n ! ≤ n n , n 2 log( n 2 ) ≤ log( n !) ≤ n log n, 1 2 ( n log n − n ) ≤ log( n !) ≤ n log n. 1.5. Upper bound: n summationdisplay i =1 1 i =1 + 1 2 + 1 3 + 1 4 + ··· ≤ ≤ 1 + 1 2 + 1 2 + 1 4 + 1 4 + 1 4 + 1 4 + ··· + 1 2 k + ··· + 1 2 k bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright 2 k terms = = 1 + 1 + ··· + 1 + 1 bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright O (log n ) terms = = O (log n ) 3 Lower bound: n summationdisplay i =1 1 i =1 + 1 2 + 1 3 + 1 4 + ··· ≥ ≥ 1 + 1 2 + 1 4 + 1 4 + 1 8 + 1 8 + 1 8 + 1 8 ··· + 1 2 k + ··· + 1 2 k bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright 2 k − 1 terms = =1 + 1 2 + 1 2 + ··· + 1 2 + 1 2 bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Ω(log n ) terms = =Ω(log n ) 1.6. First observe that multiplication of a number N by a one-digit binary number b results in the number 0 if b = 0 and N if b = 1. Also, multiplication of a binary number by 2 k for any power of 2 results in shifting N to the left k times and adding k digits equal to 0 at the end....
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chapter1 - Chapter 1–Solutions February 2 2007 2 1.1 A...

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