Greedy algorithms that don’t work

\(b)\) Choose the course \(x\) that starts first, discard all classes that conflict with \(x\), and recurse.

S = [1, 2, 3]
F = [9, 3, 4]

• Course 1 gets chosen; other two are discarded.

\(d)\) Choose the course \(x\) with shortest duration, discard all classes that conflict with \(x\), and recurse.

S = [1, 3, 4]
F = [4, 5, 9]

• Course 2 gets chosen; other two are discarded.

Exchange Arguments: The General Pattern

• Assume that there is an optimal solution that is different from the greedy solution.
• Find the “first” difference between the two solutions.
• Argue that we can exchange the optimal choice for the greedy choice without making the solution worse (although the exchange might not make it better).

Notice the “scare quotes” around “first”.

Draw a picture first

\(c)\) Choose the course \(x\) that starts last, discard all classes that conflict with \(x\), and recurse.

• Compare another optimal solution to greedy solution.
• Find the “first” difference.
• Exchange, without making it worse.

Use your own words

Supposing GREEDYSCHEDULE returns \(m\), let \(g_1, g_2,\ldots,g_m\) be the classes the algorithm chose (in order of being chosen, so in reverse chronological order). Suppose also that \(g_1,g_2,\ldots, g_{j-1}, c_j, c_{j+1},\ldots, c_{m'}\) is also a compatible list of classes with \(m' \geq m\).

Since \(g_j\) is part of the first solution, it must end before \(g_{j−1}\) and since it was chosen by the algorithm it must start after all the classes excluding \(g_1, g_2,\ldots, g_{j-1}\), or specifically it must start after \(c_{j+1}\). This means \(g_j\) can replace \(c_j\), giving another solution: \(g_1,g_2,\ldots,g_{j−1},g_j,c_{j+1},\ldots,c_{m'}\).

This can continue, replacing all \(c_i\)’s with \(g_i\)’s so that \(m=m'\), which proves that the greedy solution is optimal.

Less confusing? Events in chronological order

Suppose this greedy algorithm returns \(m\). Let \(g_1, g_2, \ldots, g_m\) be the events chosen by this greedy algorithm.

Suppose that \(c_1,c_2, \ldots, c_{j−1},c_j, g_{j+1}, \ldots, g_{m'}\) is another sequence of compatible events, with \(m'\geq m\).

Since event \(g_j\) is part of the first solution, it must end before \(g_{j+1}\). Given the nature of the greedy algorithm, event \(g_j\) starts at the same time or after all other events ending before \(g_{j+1}\). In particular, it must start after or at the same time as \(c_j\). Therefore, we can replace \(c_j\) with \(g_j\) and obtain another solution \(c_1,c_2, \ldots, c_{j−1},g_j,g_{j+1},\ldots, g_{m'}\). Continuing this way, all of the \(c_i\)’s can be replaced with \(g_j\)’s. Therefore \(m=m'\), and the greedy solution is optimal.

Scheduling Tasks in Sequence

• Problem: schedule tasks on a single processor, without swapping.
  • minimize the average completion time of all the tasks.
• Example: tasks \(a_1\) and \(a_2\) take \(t_1=2\) seconds and \(t_2=3\) seconds to complete.
  • average completion time:
    • Do \(a_1\) before \(a_2\): Task \(a_1\) would finish at time 2, and \(a_2\) would finish at time 5, for an average completion time of \((2+5)/2 = 3.5\).
    • Do \(a_2\) before \(a_1\): The average completion time \(= (3+5)/2 = 4\)

Given tasks \(a_1, a_2, \ldots, a_n\) with running times \(t_1, t_2, \ldots t_n\), describe an algorithm that will determine the ordering of these tasks that minimizes the average completion time.

Table Groups

Greedy Process Scheduling

1. Let T[1..n] be an array such that \(T[i]\) is the time it takes to do task \(a_i\). Describe a greedy algorithm that returns a sequence \(g_1, g_2, \ldots g_n\) such that executing the tasks in the order \(a_{g_1}, a_{g_2}, \ldots a_{g_n}\) has the minimum possible average completion time.

2. Prove your algorithm is correct:

   • Suppose that \(c_1, c_2, \ldots, c_n\) is another sequencing of these events that has minimal average completion time.
   • Let \(c_j\) be the first index in this solution that differs from your greedy solution.
   • Argue that you can replace \(c_j\) with \(g_j\), and the solution will still be optimal.