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4/16/2015 MATH 224 – Discrete Mathematics Counting Basic counting techniques are important in many aspects of computer science. For example, consider the.

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Presentation on theme: "4/16/2015 MATH 224 – Discrete Mathematics Counting Basic counting techniques are important in many aspects of computer science. For example, consider the."— Presentation transcript:

1 4/16/2015 MATH 224 – Discrete Mathematics Counting Basic counting techniques are important in many aspects of computer science. For example, consider the two code segments below. What is the value of x after executing the code on the right, after executing the code on the left? 1 x = 0; for (i=0; i < m; i++)for(i=0; i < m; i++) { x = x+1; for (j=0; j { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/3366611/12/slides/slide_0.jpg", "name": "4/16/2015 MATH 224 – Discrete Mathematics Counting Basic counting techniques are important in many aspects of computer science.", "description": "For example, consider the two code segments below. What is the value of x after executing the code on the right, after executing the code on the left. 1 x = 0; for (i=0; i < m; i++)for(i=0; i < m; i++) { x = x+1; for (j=0; j

2 4/16/2015 MATH 224 – Discrete Mathematics Counting In the code segments below what happens to x? What is the value of x after executing the code on the right, after executing the code on the left? 2 x = 0; for (i=0; i < m; i++)for(i=0; i < m; i++) { x = x+1; for (j=n; j>1; j=j/2) for (j=n; j > 1; j=j/2) x = x+1; x = x+1; } // rof

3 4/16/2015 MATH 224 – Discrete Mathematics Counting License plate numbers Consider a license plate that has six characters where the first 3 are digits and the last three are upper case letters. If zeros are not allowed for the first two digits and there are five three letter sequences that are not allowed, how many different license plates are possible. 3 With three decimal digits there are 1000 possible numbers. There are 100 numbers with a leading zero and 100 numbers with a zero as the second digit. As a result there are 1000 – 100 – 100 + 10 = 810 possible 3 digit sequences. Why was 10 added? What about 9*9*10 possible values? Then there are 26 3 = 17,576 three character sequence with five not allowed. So we have a total of 810*17,571 = 14,236,560.

4 4/16/2015 MATH 224 – Discrete Mathematics Counting IPv4 Internet Addresses Each address consists of 4 bytes, that is written as four decimal numbers. For example, the internet address of the CS web pages is 146.163.150.3. Each of the four parts is written as a decimal number that may take on values between 0 and 255. What is the total number of addresses that are available? 256 4 = 4,294,967,296 Some of these are not used as address, but have special meaning. So just as an example, the first byte may not be 0 or 255, and the last byte may not be 0 or 255. Let’s determine the number of addresses taking these restrictions into account. (Note that Example 16 on Page 341 of our text gives the complete set of restrictions for IPv4 addresses.) For the last byte not being 0 or 255 we get 2*2 24 =33,554,432, and also for the first byte 33,554,432 for all ones or all zeroes. So we have a total of 4,294,967,296 + 4*2 16 −33,554,432 −33,554,432 = 4,228,120,576. Why was 4*2 16 added to the total? How does this compare with the result in our text? Why do you think that even more addresses are needed? IPv6 uses 8 bytes for addresses giving 256 8 = 18,446,744,073,709,551,616 possibilities before subtracting those that are used for a general address. 4

5 4/16/2015 MATH 224 – Discrete Mathematics Permutations Permutations count the number of ways distinct objects can be arranged. So for example three different blocks can be arranged in 3! or six ways. How many ways can four different blocks be arranged? How many was can n different blocks be arrange. 5

6 4/16/2015 MATH 224 – Discrete Mathematics Permutations If we choose r objects out of n objects that is called the number of r-permutation from a set of n elements, P(n, r). We assume that n ≥ r ≥ 0. How many ways can four blocks where only two are selected be arranged? P(n, r) = n!/(n-r)!, so in this case we have 4!/2! = 24/2 = 12. Note when n = r the denominator is 0!. In order to make this formula work and for other reasons, 0! is defined to be 1. 4/16/2015 6

7 MATH 224 – Discrete Mathematics Permutations If we have a pair of dice, how many different patterns, counting order, are possible? This is just the number of 2-permutations of 6 distinct items plus the case were both numbers are the same. Thus we have P(6,2) + 6 = 6!/(6-2)! + 6 = 720/24 + 6 = 36. Do you know a simpler way to determine this number? This may more easily be determined as 6 2. The traveling salesperson problem asks a person to visit a number of different places and then return to the starting point. How many different tours are possible for 5 locations? The answer is 5! if we allow any starting point. What will the number be if we fix the starting point? What if we want to find the shortest tour and the number of cities is large? For example, what are the number of tours if there are 20 cities? If we fix the starting point the number of tours would be 121,645,100,408,832,000. Can we check all of them to find the shortest? 0 1 2 3 4 4/16/2015 7

8 MATH 224 – Discrete Mathematics Combinations Sometimes all we care about is the items that are selected and not the order in which they are selected, e.g., a hand of cards. So for example if we have four different block colors and want to know the number of different sets of two blocks, we have to divided the number of permutations by the number of different orderings. In this case, each of the larger boxes represents a single combination since only the colors matter not the order. 4/16/2015 8

9 MATH 224 – Discrete Mathematics Combinations Combinations count the number of ways of choosing r objects from a set of n objects r-choose n, where order doesn’t count. So C(n, r) = P(n, r)/r!. Here we divide by the number of ways of ordering r objects. In this case we have C(4, 2) = 4!/2!2! = 24/4 = 6. Note that the number of combinations is also the same as the binomial coefficient that we have already seen. 4/16/2015 9

10 MATH 224 – Discrete Mathematics Combinations Combinations are written as C(n, r) often read as n choose r. The formula for combinations is C(n, r) = P(n, r)/P(r, r) = P(n, r)/r! = n!/(n-r)!r! The combinations of n choose r are the same as the binomial coefficients that we have already seen. Note that C(n, r) = C(n, n-r) since C(n, r) = n!/(n-r)!r! and C(n, n-r) = n!/r!(n-r)!. We can use combinations to determine how many different hands of cards can be drawn since order does not matter. If we are playing 5-card draw the number of hands is 52!/47!∙5! = 52∙51∙50∙49∙48/120 = 2,598,960. We can use this to determine the probability of getting a royal flush. To get a royal flush you must have the ace, king, queen, jack, and 10 all from the same suit. Since there are 4 suits and order does not matter, the chances of getting a royal flush is 4/2,598,9760 ≈ 0.00015%. That is pretty low odds. 4/16/2015 10

11 4/16/2015 MATH 224 – Discrete Mathematics Combinations, Binomial Coefficients & Pascals Triangle 1 12 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 19 36 84 126 126 84 36 9 1 1 10 45 120 210 252 210 120 45 10 1 4/16/2015 11 Sum of Row 1 = 2 0 2 = 2 1 4 = 2 2 8 = 2 3 16 = 2 4 32 = 2 5 64 = 2 6 128 = 2 7 256 = 2 8 512 = 2 9 1024 = 2 10 B(j, k) = 1 if k = 0 or j = k B(j –1,k –1)+B(j – 1,k,) all other cases j ≥ k ≥ 0

12 4/16/2015 MATH 224 – Discrete Mathematics Combinations Combinations may also be used to determine the number of ways of getting a particular result when tossing coins. For example, if we want to determine the likely hood of getting 4 heads when tossing a coin 8 times we need to consider the number of ways of getting 4 heads and 4 tails. That is just C(8,4) = 8!/4!4! = 70. Then there are 2 8 = 256 possible outcomes so the probability of getting 4 heads is 70/256 ≈ 0.273. What would C(8,0)+C(8,1) + … + C(8,8) be? Well that would be 1 + 8 + 28 +56 +70 +56 + 28 + 8 + 1 = 256. Do you know what result you always get when adding all the disjoint probabilities? 4/16/2015 12


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