Randomized Algorithms

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The Hiring Problem

Hiring cost: \(C_h\)

• Permute candidates

  \(E[X] = E[\sum X_i] = \sum E[X_i] = \sum 1/i = ln(N)\)

  \(O(ln(N)C_h)\)

  • How to permute?

    Assign each element a random priority and sort.

• Hire Only Once

  For first k candidates, we determine the threshold, no hire.

  For remaining candidates, we hire the first that is above the threshold.

  \(P(\)no one at k+1 - i-1 are hired\() = k/(i-1)\) (The maximum of first i is at first j)

  Best \(k = N/e\)

Quick Sort

The pivot always divides the set so that each side is at least \(n/4\).

If the pivot doesn't satisfies, we choose again.

The expected number of choose is \(E = 1/2*1 + 1/2*(E+1)\) ,\(E = 2\)

• Type j problem

  The problem divided by \(3/4\) for \(j\) times.

  \(N(3/4)^{j+1} \leq |S| \leq N(3/4)^j\). \(|S|\) is the problem size.

  There are at most \((4/3)^{j+1}\) problems at type \(j\).

  Number of different types: \(log_{4/3}N = O(logN)\)

Some proofs:

• \(\int_k^N 1/x dx \leq \sum_{i=k}^{N-1} 1/i \leq \int_{k-1}^{N-1} 1/x dx\).

  It is obvious that : \(\int_k^{k+1} 1/x dx \leq \int _k^{k+1} 1/k dx = 1/k\).

  So we have: \(\int_k^N 1/x dx = \int_k^{k+1} 1/x dx + \int_{k+1}^{k+2} 1/x dx +\dots + \int_{N-1}^{N} 1/x dx\leq \sum_{i=k}^{N-1} 1/i \).

  We also have: \(\int_k^{k+1} 1/x dx \geq \int _k^{k+1} 1/(k+1) dx = 1/(k+1)\).

  With similar strategy, we can get: \(\sum_{i=k}^{N-1} 1/i \leq \int_{k-1}^{N-1} 1/x dx\).

  Warp it up, we have \(\int_k^N 1/x dx \leq \sum_{i=k}^{N-1} 1/i \leq \int_{k-1}^{N-1} 1/x dx\).

• Optimum k?

  \(f'(k) = 1/Nln(N/k) + -1/N = 1/N*(ln(N/k) - 1)\)

  \(N/k = e\)

  \(k = N/e\)