# week-5 - 1 HW1 2 All three statements were incorrect a if f...

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Unformatted text preview: 1 HW1 - 2 All three statements were incorrect. a. if f ( n ) = O(F( n )) and g ( n ) = O(G( n )), then f ( n )/ g ( n ) = O( F( n )/G( n ) ) Counter-Example: f( n ) = n 3 F( n ) = n 3 n 3 = O( n 3 ) g( n ) = n G( n ) = n 2 n = O( n 2 ) f( n )/g( n ) = n 3 / n = n 2 F( n )/G( n ) = n 3 / n 2 = n n 2 ≠ O( n ) HW1 - 2 b. f ( n ) = O( g ( n )) implies that 2 f ( n ) = O(2 g ( n ) ) Counter-Example: f( n ) = 2 n g( n ) = n 2 n = O( n ) 2 2 n ≠ O(2 n ) HW1 - 2 c. For all functions f ( n ) and g ( n ), either f ( n ) = O( g ( n )) or g ( n ) = O( f ( n )) Counter-Example: NOT sin & cos, but partial credit was given for them because they have the right idea. Consider f( n ) = n if n is even; f( n ) = 2 n if n is odd. g( n ) = n 2 2 HW1 - 5 a . Estimate the total amount of money in an armored car completely full of \$20 bill. Standard assumptions: Size and shape of the cargo area Size of a stack of \$20 bills. Results: \$18,000,000 \$34,560,000 \$42,000,000 \$104,000,000 \$145,152,000 \$147,456,000 \$165,888,000 \$275,480,000 \$829,440,000 HW1 - 5 b . Estimate how many cubic miles of water flow out of the mouth of the Mississippi River each day. Standard assumptions: Flow rate of river Size and shape of river mouth Results: 0.012 mi 3 /day 0.15 0.3 0.57 0.864 1.2 48 169.44 1188 Real Numbers? Width = 2,000-3,500 feet Depth ≤ 100 feet Speed = 2 miles per hour Width × Depth × Speed × 24 hours/day ≈ 0.66 mi 3 /day Actual number is closer to 0.5 mi 3 /day, or 611,000 ft 3 /sec (My own approximation without looking anything up was 30 mi 3 /day) 3 HW1 - 5 c . Estimate the number lego blocks in a life-size replica of the Statue of Liberty. Standard assumptions: Size and shape of a brick Height, width and depth of statue (box assumed). Results: 588,144 777,600 938,667 1,170,000 35,280,000 500,000,000 1,042,000,000 11,000,000,000 13,824,000,000 Lego Liberty - “real numbers” Height = 152’ 2” (base to torch) Height = 305’ 1” (from foundation) Height = 111’ 1” (heel to head) Thickness at waist = 35” Average Lego (?) Lego Liberty - “real numbers” Lego Liberty Height = 2’9” tall Contains 2882 pieces Scale = 152’2” / 2’9” = 55.3 Pieces = 2882 x 55.3 3 = 4.87 x 10 8 (487,000,000) 4 HW1 - 6 a . Show T( n ) = 2T( n /2 + 17) + n is O( n log n ) Assume T( n /2 + 17) ≤ c ( n /2 + 17) log ( n /2 + 17) Prove: T( n ) ≤ c n log n T( n ) ≤ 2( c ( n /2 + 17) log ( n /2 + 17)) + n ≤ 2( c ( n /2 + 17) log ( n /2 + 17)) + n (drop floors) ≤ 2( c ( n /2 + 17) log (3 n /4)) + n (for n > 68) = c ( n + 34) log (3 n /4) + n = c ( n + 34) (log n- log(4/3)) + n = c ( n log n- n log(4/3) + 34log n- 34log(4/3)) + n < c ( n log n- n log(4/3) + 34log n ) + n ≤ c n log n- c n log(4/3) + c 34log n + n ≤ c n log n- n + c 34log n ( c > 2/log(4/3)) ≤ c n log n HW1 - 6 a . Show T( n ) = 2T( n /2 + 17) + n is O( n log n ) Why can’t we say: Assume holds for T( n /2 + 17) T( n ) ≤ 2( c (...
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week-5 - 1 HW1 2 All three statements were incorrect a if f...

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