the handshake problem

The Handshake Problem

The Handshake Problem

• n people are in a room• Each person shakes hands with each other

person exactly once.• How many handshakes will take place?

Example: n = 5

Example: n = 5

Person 1 shakes hands with the other four and leaves.

Running Total: 4

Person 1 shakes hands with the other four and leaves.

Running Total: 4

Person 2 shakes hands with the other three and leaves.

Running Total: 4 + 3

Person 2 shakes hands with the other three and leaves.

Running Total: 4 + 3

Person 3 shakes hands with the other two and leaves.

Running Total: 4 + 3 + 2

Person 3 shakes hands with the other two and leaves.

Running Total: 4 + 3 + 2

Person 4 shakes hands with the other one and leaves.

Running Total: 4 + 3 + 2 + 1

Person 5 has no one left to shake with.

Running Total: 4 + 3 + 2 + 1

Person 5 has no one left to shake with.

Running Total: 4 + 3 + 2 + 1 = 10

• 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 = 45handshakes.

Let’s count the handshakes another way.

• Count each hand in a handshake as a “half-handshake”.

• Two half-handshakes make a whole handshake.

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.

So for a general n, the total number of handshakes is n(n-1)∕2.

Ex: if n = 10, the number of handshakes is

If n = 50, then the number of handshakes is

This is much easier than adding up

1+ 2 + + 49⋅⋅⋅

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⋅⋅⋅

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

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

The formula 1 + 2 + + (n-1) + n = n(n+1)/2⋅⋅⋅

Was discovered byCarl Friedrich Gauss when he was a student in primary school.

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.

Gauss realized that it is easy to add up all the numbers twice.

1 + 2 + 3 + ⋅⋅⋅ + 99 + 100100 + 99 + 98 + ⋅⋅⋅ + 2 + 1 101 + 101 + 101 + ⋅⋅⋅ + 101 + 101

=101(100).

Dividing this by 2 gives a sum of 101 50=5050⋅

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