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Discrete Mathematics

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Phase 5 Individual Project
03/23/2014
Math 203
PPCC

Part I: Look up a roulette wheel diagram. The following sets are defined:
A = the set of red numbers
B = the set of black numbers
C = the set of green numbers
D = the set of even numbers
E = the set of odd numbers
F = {1,2,3,4,5,6,7,8,9,10,11,12}
Answers:
AUB- {All BLACK and RED numbers}
A∩D- {All numbers that are both RED and EVEN}
B∩C- {NO numbers intersect between these two sets}
CUE- {All ODD numbers and 00, 0}
B∩F- {2,4,6,10,11}
E∩F- {1,3,5,7,9,11}
Part II: The implementation of the program that runs the game involves testing. One of the necessary tests is to see if the simulated spins are random. Create an n-ary relation, in …show more content…

Demonstrate how the search would go if you used:
A sequential search starts with the first number and goes in order to the last number. It will take until the element number 12 find match
A binary search is always used on the sorted array.
State the runtime for each of the searches, in this example, and for general data sets of size n. Address the issue of the order of the data in binary searching.
* The runtime for the sequential search is N.
* The runtime for the binary search isn't found
Suppose an algorithm that processes a data set of size 8 has a runtime of 72. The same algorithm has a runtime of 110 when applied to a data set of size 10; and when applied to a data set of size 20, it has a runtime of 420. Using big-O notation, state the runtime for this algorithm for the general case of a data set of size n. - O(n2 + n)
Suppose you develop an algorithm that processes the first element of an array (length of n), then processes the first 2 elements, then the first 3 elements, and so on, until the last iteration of a loop, when it processes all elements. Thus, if n = 4, the runtime would be 1 + 2 + 3 + 4 = 10.
Create a table that depicts the runtime for arrays of length 1 to 10. Would you expect the general runtime to be O(n), O(n2), O(n3), or some other function of n? Explain.
1= 1= 1
2 = 1 + 2= 3
3 = 1

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