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The Handshake Problem

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n people are in a room Each person shakes hands with each other person exactly once. How many handshakes will take place?

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Example: n = 5

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Person 1 shakes hands with the other four and leaves. Running Total: 4

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Person 1 shakes hands with the other four and leaves. Running Total: 4

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Person 2 shakes hands with the other three and leaves. Running Total: 4 + 3

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Person 2 shakes hands with the other three and leaves. Running Total: 4 + 3

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Person 3 shakes hands with the other two and leaves. Running Total: 4 + 3 + 2

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Person 3 shakes hands with the other two and leaves. Running Total: 4 + 3 + 2

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Person 4 shakes hands with the other one and leaves. Running Total: 4 + 3 + 2 + 1

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Person 5 has no one left to shake with. Running Total: 4 + 3 + 2 + 1

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Person 5 has no one left to shake with. Running Total: 4 + 3 + 2 + 1 = 10

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So when n = 5, it takes 1+2+3+4 handshakes In general, it takes 1+2+3+ … + (n-1) handshakes. Ex: If there were 10 people, there would be 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 handshakes.

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Let’s count the handshakes another way. Count each hand in a handshake as a “half- handshake”. Two half-handshakes make a whole handshake.

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When n = 5 Each person has to shake hands with the other four. So each person contributes 4 half- handshakes. Since there are 5 people, this is a total of 5 ⋅ 4 = 20 half handshakes, or 10 whole handshakes.

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So for a general n, the total number of handshakes is n(n-1)∕2. Ex: if n = 10, the number of handshakes is

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If n = 50, then the number of handshakes is This is much easier than adding up 1+ 2 + ⋅⋅⋅ + 49

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By counting in two different ways, we determined that the number of handshakes is both 1 + 2 + ⋅⋅⋅ + (n-1) and n(n-1)/2 Since these formulas count the same things, we have established the identity 1 + 2 + ⋅⋅⋅ + (n-1) = n(n-1)/2

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Let f(n) = 1 + 2 + ⋅⋅⋅ + (n-1). We’ve seen that

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Let f(n) = 1 + 2 + ⋅⋅⋅ + (n-1). We’ve seen that To get a function for 1 + 2 + ⋅⋅⋅ + n, we replace each n with n+1

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The formula 1 + 2 + ⋅⋅⋅ + (n-1) + n = n(n+1)/2 Was discovered by Carl Friedrich Gauss when he was a student in primary school.

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Gauss’s teacher wanted to keep Gauss busy, so he gave him the assignment of adding all the numbers from 1 to 100. Gauss produced the correct answer in a matter of seconds. His teacher was impressed. And annoyed.

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Gauss realized that it is easy to add up all the numbers twice. 1 + 2 + 3 + ⋅⋅⋅ + 99 + 100 100 + 99 + 98 + ⋅⋅⋅ + 2 + 1 101 + 101 + 101 + ⋅⋅⋅ + 101 + 101 =101(100). Dividing this by 2 gives a sum of 101 ⋅ 50=5050

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We can do the same trick for any n: 1 + 2 + ⋅⋅⋅ + (n-1) + n n + (n-1) + ⋅⋅⋅ + 2 + 1 n+1 + n+1 + ⋅⋅⋅ + n+1 + n+1 = (n+1)n So 1 + ⋅⋅⋅ + n = (n+1)n/2

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