CSC317 1 Hiring problem-review - PowerPoint Presentation

Slide1

CSC317

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Hiring

problem-review

Cost to interview (low Ci)Cost to fire/hire … (expensive Ch)n number of candidatesm hired

O (cin + chm)

Independent of order of candidates

depends on order of candidates

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Randomized hiring problem(n)

1. randomly permute the list of candidates

2. Hire-Assistant(n)

Hire

-Assistant

(n)

1. best = 0 //least qualified candidate

2.

for

i = 1

to

n

3.

interview

candidate

i

4.

if

candidate

i

better

than

best

5

.

best = i

6.

hire

candidate

i

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Randomized hiring problem(n)

Instead of worst case

We want to know the cost on average of hiring new applicantsX random variable equal to number times new person hired

i

f candidate

i

is hired

vice versa

For notation, we can drop the I symbol:

i

f candidate

i

is hired

v

ice versa

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Randomized hiring problem(n)

X

random variable equal to number times new person hired

Probability that candidate

i

is hired

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Randomized hiring problem(n)

X

random variable equal to number times new person hired

Probability that candidate

i

is hired

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Randomized hiring problem(n)

X

random variable equal to number times new person hired

Harmonic series

?how come?

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Randomized hiring problem(n)

Cost of hiring a candidate

C

h

On average:

C

h

ln

(n)

Compare to worst case

C

h

n

.

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Quicksort

Input:

n numbersOutput: sorted numbers, e.g., in increasing order

O(n log(n))

on average

Sorts in place, unlike

mergesort

Elegant

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Quicksort

Quicksort(A, p, r)

1.

If p<r2. q = Partition(A,p,r)3. Quicksort(A,p,q-1)4. Quicksort(A,q+1,r)

p

r

Divide-and-conquer; but what is special about it?

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Partition

1. Pick pivot element in array

A (for now, we’ll choose last element)2. Rearrange A so that elements before pivot are smaller, and elements after pivot are largerExample:A = [8 6 1 3 7 5 2 4] PivotAfter partition:A = [1 3 2 4 5 7 6 8] < pivot > pivotPivot in its right positionhttp://www.cs.miami.edu/~burt/learning/Csc517.101/workbook/partition.html

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Partition (more algorithmically)

Partition(A, p, r)

p

r

1.

x

=A[ r ]2. i=p-13. for j = p to r-14. if A[ j ] <= x5. i = i + 1;6. swap A[ i ] with A[ j ]7. swap A[ i+1 ] with A[ r ]8. return i+1

O(?)

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Partition (more algorithmically)

Partition(A, p, r)

p

r

1.

x

=A[ r ]2. i=p-13. for j = p to r-14. if A[ j ] <= x5. i = i + 1;6. swap A[ i ] with A[ j ]7. swap A[ i+1 ] with A[ r ]8. return i+1

O(n)Why? For each j at least one swap.

pivot

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Partition

Loop

invariants?

p

r

i

j

i

keeps track of pivot location

j keeps track of next element to check

Before the for loop at given j

1. All elements in A[p..i] ≤ pivot2. All elements in A[i+1..j-1] > pivot3. A[r] = pivot

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Quicksort

Quicksort(A, p, r)

1.

If p<r2. q = Partition(A,p,r)3. Quicksort(A,p,q-1)4. Quicksort(A,q+1,r)

p

r

When will partition do a bad job?

When the partition is for zero elements versus

n-1

p

r

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Quicksort

Quicksort(A, p, r)

1.

If p<r2. q = Partition(A,p,r)3. Quicksort(A,p,q-1)4. Quicksort(A,q+1,r)

p

r

When will partition do a good job?

When

the partition is roughly equal in terms of numbers smaller than and greater than pivot

Mergesort

anyone?

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Quicksort

Quicksort(A, p, r)

1.

If p<r2. q = Partition(A,p,r)3. Quicksort(A,p,q-1)4. Quicksort(A,q+1,r)

p

r

How do we choose a pivot point that we are good on average?

We choose pivot randomly!

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Quicksort

Quicksort(A, p, r)

1.

If p<r2. q = Partition(A,p,r)3. Quicksort(A,p,q-1)4. Quicksort(A,q+1,r)

p

r

What about a

9 to 1 split?

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Why might a random choice be good enough?

Even a 9 to 1, or 3 to 1 split, is

O(n log n)

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Quicksort on average run time

We’ll prove that

average run time with random pivots for any input array is O(n log n)Randomness is in choosing pivotAverage as good as best case!

Most work of Quicksort in

comparisons

Each

call to partition is constant

plus number

of

comparisons in for

loop

Let

X = total number of comparisons in

all

calls

to Partition