Module 11: Problem Complexity and NP-Complete Problems

11.1   Informal and Formal Problem Complexity

 

Characterizing problem complexity:

 

The decision version of a combinatorial optimization problem:

 

11.2   Some Well-Known Problems

 

Before describing what "NP-complete" means, we'll take a look at some well-known problems.
 

Clique:

 

In-Class Exercise 11.2: Draw an 8-vertex graph such that every vertex is in some clique of size 3.
 

Vertex cover

 

In-Class Exercise 11.3: Draw an 8-vertex graph whose smallest vertex cover is of size 2.
 

Satisfiability (SAT)

 

In-Class Exercise 11.4: Is the expression (x1 + x2) (x'1 + x'2+ x'3) (x3 + x2 + x'4) satisfiable?
 

Other problems (optimization versions briefly described):

 

11.3   Transforming an Instance of one Problem into an Instance of Another Problem

 

Consider this unusual approach to solving a problem:

  • Suppose we've written code to find the max of an array of int's: 
    
    Algorithm: max (A) 
    Input: An array of int's A
        // ... find the max value in A and return it ...

  • Now consider this: 
    
    Algorithm: min (A) 
    Input: An array of int's A
        // ... find the min value in A and return it ...
        for i=0 ... 
            B[i] = - A[i]
        endfor
        m = max (B)
        return -m

  • Notice the strange way of "solving" a different problem:
    • We have a solution to problem X.
    • We have a new problem Y.
    • Transform the data, feed it to the solution for X
    • Interpret the result as a solution to Y.
 

Now let's apply this to combinatorial optimization problems:

  • Suppose that:
    1. We can transform a problem instance of clique into an equivalent problem instance of vertex cover.
    2. We can transform a solution to the vertex cover instance into a solution for the clique instance.

  • If a fast algorithm was found for Vertex cover:

    we could use it to solve clique:

 

Example: transforming clique to vertex cover:

  • Suppose we are given an instance of the clique problem, G = (V, E).
    ⇒ Recall: goal is to find largest clique.
  • Example:

  • Define G', the complement of G as follows:
    • G' has the same vertex set V of G.

    • For every two vertices, u, v not connected in G, place an edge in G'.

  • Claim: if G' has a vertex cover of size k then G has a clique of size n - k:
    • Suppose G' has vertex cover of size k (k = 3 in example).

    • Consider the other vertices:

      There are no edges between them (because all other edges are "covered" by the vertex cover).

    • This means, there MUST be an edge between every such pair in G:

    • Thus, there is a clique of size n - k in G.

  • Performing the transformation has two parts:
    1. Transforming G to G':
      • Processes each pair of vertices once
        ⇒ O(n2) time to transform an instance of clique into an instance of vertex cover.
    2. Transform vertex cover solution to clique solution:
      • Build clique graph
        ⇒ O(n2) time (worst-case).
    Thus, both transformations take polynomial-time.
 

Decision version:

  • Decision version of clique: Is there a clique of size at least k?

  • Decision version of vertex cover: Is there a vertex cover of size at most m?

  • What we have shown: size-(n-k) vertex-cover in G' ⇔ size-k clique in G.

  • Thus, we can create a solution for the decision version of clique:

 

Time taken:

  • Suppose the vertex-cover algorithm takes time P1(n).

  • Next, suppose the transformation takes time P2(n).

  • We have a clique algorithm that takes time P1(n) + P2(n), still a polynomial.

 

Some other known transformations:

  • Satisfiability (3-SAT) to clique.
    ⇒ suppose the transformation takes time P3(n).
    ⇒ we can build an algorithm for 3-SAT:

  • Hamiltonian-tour to 3-SAT.
    ⇒ suppose the transformation takes time P4(n).
    ⇒ we can build an algorithm for Hamiltonian-tour

 

In-Class Exercise 11.5: Consider the following problem:

  • There are n items with weights w1, w2, ..., wn respectively.
  • There are K people who are to carry these items on a trip.
  • We would like to distribute the weights as evenly as possible. Let us call this the weight distribution problem.
Answer the following questions:
  • Formulate the problem as a combinatorial optimization problem.
  • State the decision version of the problem.
  • Show that the decision version of Bin-Packing Problem (BPP) can be transformed into the decision version of this problem.
 

11.4   The Theory of NP-Completeness (Abridged)

 

NP problems:

  • NP problems are those decision problems whose candidate solutions can be verified in polynomial-time.

  • Example: TSP
    • Recall decision version: is there a tour of length at most L?
    • Candidate solution is a tour.
    • Given a particular tour, how long does it take to check whether its length is at most L?
      ⇒ O(n) time.
      ⇒ TSP is an NP-problem.

  • Example: BPP
    • Decision version: can the given items fit into at most K bins?
    • Candidate solution is an assignment.
    • How long does it take to check whether the assignment uses at most K bins?
      ⇒ O(n2) (worst-case, given assignment matrix).
      ⇒ BPP is an NP-problem.

  • Most combinatorial problems are NP problems.

  • Note: NP = "Nondeterministic Polynomial-time" (theoretical reason for this name).

  • Formal way of stating a problem is an NP-problem: the problem is in the language NP.
 

First fundamental result:

  • If you can solve k-SAT in polynomial time, you can solve any problem in NP in polynomial time.
          ⇒ Every NP-problem is no harder than k-SAT

  • Proof is elaborate and uses a construction with Turing machines.

  • Landmark result proved by Steve Cook in 1972.
 

Second fundamental result(s):

  • Many combinatorial optimization problems can be transformed into another combinatorial optimization problem in polynomial-time.
          ⇒ These problems are "equally hard" (to within a polynomial)

  • We have already seen examples of this transformation.

  • First few landmark results proved by Richard Karp in 1973.
    Thousands of such transformations since then.
 

NP-Completeness:

  • A problem is NP-complete if:
    1. It is an NP-problem.
            ⇒ The problem is no harder than k-SAT
    2. Some other existing NP-complete problem can be transformed into it in polynomial-time.
            ⇒ It is at least as hard as other NP-complete problems (including k-SAT).
 

Major implication of the two results:

  • If you can solve ANY ONE NP-complete problem in polynomial-time you can solve EVERY ONE of them in polynomial-time.

  • Thus, they are all of (polynomially) equivalent complexity.

  • The classes P and NP:
    • Let P be the set of problems solvable in polynomial-time
      (e.g., MST is in P).
    • NP is the set of problems described above.
    • If a polynomial-time algorithm is found for any NP-complete problem, then P = NP.

  • The most famous outstanding problem in computer science: is P = NP?

  • Most experts believe P is not equal to NP.

  • Note: just about every interesting combinatorial problem turns out to be NP-complete (thousands of them).
 

What to do?

  • Approach 1: solve problems approximately
    ⇒ fast algorithm, possibly non-optimal solution.

  • Approach 2: solve problems by exhaustive search
    ⇒ optimal solution, but really slow algorithm.
 

11.5   Approximate Solutions

 

Key ideas in approximation:

  • Devise algorithm to make "intelligent guesses" and arrive at a "good" (but not optimal) solution.

  • Evaluating an algorithm: how close to optimal are its solutions?
    • Theoretical analysis: prove a bound (e.g., no worse than twice optimal).
    • Experimental: evaluate algorithm using randomly-generated data.

  • Typical theoretical analysis:
    • Let C* be the optimal cost.
    • Let CA be the cost of solution produced by algorithm A.
      ⇒ CA \(\geq\) C*
    • Question: how far off is CA from C* ?
    • Try to prove: CA < k C* (for small k).
    • TSP example: we will show that CA < 2 C*.
      ⇒ No worse than twice optimal
 

An approximation algorithm for (Euclidean) TSP:

  • Euclidean TSP satisfies the triangle inequality:
    • Consider any three points p1, p2 and p3
    • Then, dist(p1, p2) ≤ dist(p1, p3) + dist(p3, p2).

  • The algorithm:
    1. First find the minimum spanning tree (using any MST algorithm).
    2. Pick any vertex to be the root of the tree.
    3. Traverse the tree in pre-order.
    4. Return the order of vertices visited in pre-order.
 

In-Class Exercise 11.6: Write pseudocode for generating a pre-order. What is the pre-order for this tree (starting at A)?

   

  • Example:
    • Consider these 7 points:

    • A minimum-spanning tree:

    • Pick vertex A as the root:

    • Traverse in pre-order:

    • Tour:

  • Claim: the tour's length is no worse than twice the optimal tour's length.
    • Let L = the length of the tour produced by the algorithm.
    • Let L* = the length of an optimal tour.
    • Let M = weight of the MST (total length).
    • Observe: if we remove any one edge from a tour, we will get a spanning tree.
      ⇒ L* > M.
    • Now consider a pre-order tree walk from the root, back to the root:

    • Let W = length of this walk.
    • Then, W = 2M (each edge is traversed twice).
    • Thus, W < 2L*.
    • Finally, we will show that L ≤ W and therefore, L < 2L*.
    • To see why, consider the tree walk from B to D:


      ⇒ L takes a shorter route than W (triangle inequality).

    • Thus, L ≤ W.
 

An approximation algorithm for BPP: First-Fit

  • The algorithm:
    1. Consider items in the order given (i.e., arbitrary order).
    2. For each item in order:
      1. Scan current bins in order to see if item fits anywhere.
      2. If not, place item in new bin.

  • Claim: number of bins is no more than twice the optimal number of bins.
    • Let B = bin size.
    • Let S = sum of item sizes.
    • Let b = the number bins used by the algorithm.
    • Let b* = optimal number of bins.
    • Algorithm cannot leave more than two bins only half-full:

    • Thus, the most bins are used when all are just over half-full.
      ⇒ b ≤ S / (B / 2) = 2S / B.
    • Optimal solution cannot be better than fractional optimal solution S / B.
    • Thus, b ≤ 2 b*.

  • It is possible (lengthy proof) to show:
    • b < 1.7 b* + 2 (no more than 70% worse).
    • There are cases when b > 1.7 b* - 1 (at least 70% worse).
      ⇒ b < 1.7b* is the best FF can do.

  • Decreasing First-Fit algorithm: sort in decreasing order first and then apply FF.
    • Best known result: b < 1.22 b* (difficult proof).
 

The world of approximation:

  • For many NP-complete problems, there are results on approximation algorithms.

  • Theoretical results:
    • Proofs that an algorithm is "close" to optimal.
    • Proofs that an algorithm is definitely worse than optimal.
    • Sharper bounds for algorithms.

  • Experimental results:
    • Comparisons with large "real" data sets (e.g, TSP on state-capitals in the U.S.).
    • Comparisons using randomly-generated data.

  • Approximation algorithms for polynomially-solvable problems:
    • Sometimes the polynomial-time solution takes too long for large n (e.g, n3)
      ⇒ a faster approximation-algorithm may be desirable.
 

Known approximation results for TSP:

  • 1992: Unless P=NP, there exists ε>0 such that no polynomial-time TSP heuristic can guarantee CH/C* ≤ 1+ε for all instances satisfying the triangle inequality.

  • 1998: For Euclidean TSP, there is an algorithm that is polyomial for fixed ε>0 such that CH/C* ≤ 1+ε

  • 2006:
          → Largest problem solved optimally: 85,900-city problem.

  • 2009: Best practical heuristic
          → 1,904,711-city problem solved within 0.056% of optimal
 

In-Class Exercise 11.7: Download this template as well as UniformRandom, and implement the First-fit and Decreasing-First-Fit algorithms for Bin Packing.
 

11.6   Exhaustive Search

 

Key ideas:

  • To find the optimal solution of an NP-complete problem, exhaustive search can be used
    ⇒ try every possible "state" in the state-space.

  • Exhaustive search is feasible for small problem sizes.
    (e.g., n < 15)

  • However, there are ways of cleverly avoiding some solutions by pruning the search
    ⇒ can solve larger problems (typically n < 30).

  • Two types of search:
    1. Enumeration: enumerate all possibilities one by one and track the "best".
      • Advantages: Easy to implement, does not require understanding the problem in detail.
      • Disadvantages: Takes no advantage of problem structure.

    2. Constructive state-space search:
      • Builds up solutions in pieces.
      • Conceptually, full-solutions are leaves of a tree.
      • TSP Example: build tours from partial-tours.
      • Two flavors: Backtracking and Branch-and-bound.
      • Backtracking: roughly, a depth-first search of tree.
      • Branch-and-bound: roughly, a breadth-first search of tree.
      • Both can be speeded up with pruning (avoiding unnecessary exploration).

  • Backtracking example with TSP on five vertices:

 

11.7   Exhaustive Enumeration

 

We will consider methods for:

  • Generating all possible subsets of a set:
    • Example: if S = {1, 2, 3} there are eight subsets:
      1. Empty set.
      2. { 1 }
      3. { 2 }
      4. { 3 }
      5. { 1, 2 }
      6. { 1, 3 }
      7. { 2, 3 }
      8. { 1, 2, 3 }
    • Example application: Vertex cover problem
      • Recall: we want the minimal subset of vertices that "covers" the edges.
      • Solution: generate all possible subsets of vertices and try each one.

    In-Class Exercise 11.8: A set of n elements has 2n possible subsets. Why?

  • Generating all possible size-k subsets of a set:
    • Example: if S = {1, 2, 3, 4} and k = 2, there are six combinations:
      1. { 1, 2 }
      2. { 1, 3 }
      3. { 1, 4 }
      4. { 2, 3 }
      5. { 2, 4 }
      6. { 3, 4 }
    • Example application: decision version of clique
      • Recall: Given a graph, is there a clique of size k?
      • Solution: generate all possible vertex subsets of size k, and check whether a clique exists.

  • Generating all possible permutations of a set:
    • Example: if S = {1, 2, 3} there are six permutations:
      1. 1 2 3
      2. 1 3 2
      3. 2 1 3
      4. 2 3 1
      5. 3 1 2
      6. 3 2 1
    • Example application: TSP

Note: we want methods with small memory requirements
⇒ cannot store all combinations or permutations.
⇒ must be able to "generate" next combination/permutation from most recent one.
 

11.8   Generating All Possible Subsets

 

Key ideas:

  • Suppose n = number of elements in set.

  • Bit-pattern equivalence:
    • Example: consider the subset { 1, 3, 4 } of { 1, 2, 3, 4, 5 }
    • Use the bitstring "1 0 1 1 0" to represent this subset.
    • The i-th bit is "1" if the i-th element is in the particular subset.

  • Given a bit-pattern, it's easy to produce the equivalent subset
    ⇒ extract those elements whose bit is "1".

  • Thus, "generating all subsets" is equivalent to "generating all bit patterns".

  • Key observations:
    • There are 2n bit patterns.
    • If the bit-patterns are treated as binary numbers
      ⇒ all we have to do is enumerate the binary numbers 0, 1, ..., 2n-1
    • For each such bit-pattern produced, create the corresponding subset.

  • Example: if S = {1, 2, 3}
    1. "0 0 0" = Empty set.
    2. "0 0 1" = { 3 }
    3. "0 1 0" = { 2 }
    4. "0 1 1" = { 2, 3 }
    5. "1 0 0" = { 1 }
    6. "1 0 1" = { 1, 3 }
    7. "1 1 0" = { 1, 2 }
    8. "1 1 1" = { 1, 2, 3 }
 

In-Class Exercise 11.9: What is the next subset for the following subset (of a 5-element set): 10111?
 

Implementation

  • We will use a boolean array called member
    ⇒ member[i]=true if element i is in the current set.

  • We will write code to generate one element from the previous one:
    • We will use the natural order of binary numbers.
    • Example: if current subset is "1 0 1", the next subset is "1 1 0".
      ⇒ add "1" to current number to get the next.

  • We will write:
    • A method getFirstSubset to initialize and create the first subset.
    • A method getNextSubset to be called for each successive subset.

  • Pseudocode: 
    
Algorithm: getFirstSubset (N)
Input: the size of the set

1.    member = create boolean array of size N;
      // First subset corresponds to "0" in binary. 
2.    for each i, member[i] = false
3.    return member

Output: first subset - the empty set.
    
    
Algorithm: getNextSubset (member)
Input: current subset, represented by boolean array member
      // Start at the rightmost bit. 
1.    i = N-1

      // Binary addition: as long as there are 1's, flip them. 
2.    while i >= 0 and member[i] = true
3.      member[i] = false
4.      i = i -1
5.    endwhile

      // Did we run past the beginning of the array? 
6.    if i < 0
        // Yes, so all were 1's and we're done with generation. 
7.      return null
8.    else
        // No, so we found a zero where the carry-over from binary 
        // addition finally stops. 
9.      member[i] = true
9.      return member
10.   endif

Output: return new subset in which member[i]=true if i is in the subset.
 

In-Class Exercise 11.10: This exercise is optional. Download this template and try implementing the above pseudocode.
 

11.9   Generating All Possible Combinations

 

Key ideas

  • Goal: generate all k-size subsets from the integers 0, ..., n-1

  • We will use the bitstring idea from above
    ⇒ every bitstring will have exactly k one's.
 

In-Class Exercise 11.11: How many size-3 subsets does a 5-element set have? Write down the corresponding bit patterns.
 

  • Consider a particular subset:


    ⇒ we need a systematic way to re-arrange the 1's.

  • Key ideas:
    • Start with all the 1's bunched up on the left:

    • End with all the 1's bunched up on the right:

    • First, we will make the rightmost 1 try all positions to the right: 
      
    1 1 1 1 1 0 0 0       

    1 1 1 1 0 1 0 0       

    1 1 1 1 0 0 1 0       

    1 1 1 1 0 0 0 1
    • Then, we will move the next-rightmost up by one: 
      
    1 1 1 0 1 1 0 0
    • Now, again we will try all positions for the rightmost 1 (to the right of the recently-moved 1). 
      
    1 1 1 0 1 1 0 0       

    1 1 1 0 1 0 1 0       

    1 1 1 0 1 0 0 1
    • Then, we will move the next-rightmost up by one more and try more combinations: 
      
    1 1 1 0 0 1 1 0       

    1 1 1 0 0 1 0 1
    • Finally, the rightmost two 1's get bunched up on the right: 
      
    1 1 1 0 0 0 1 1
    • Now we move the third 1 over rightwards, and try all combinations of two 1's to its right: 
      
    1 1 0 1 1 1 0 0
    • ... and so on until all 1's are bunched up to the right end.

    • Key observation: at each step we
      • (A) move the rightmost 1 that has a zero to its right.
      • (B) collect next to it all the 1's to its right.

 

Implementation

  • Data structures:
    • member - a boolean array where member[i] = true if element i is in the current set.
    • position - an integer array where positions[i] = 1 if there is a 1 in the i-th position.

  • The code will be spread over three methods:
    • translatePositionsToSet - given a particular arrangement of 1's, create the actual subset.
    • getFirstCombination - initialize and return first combination.
    • getNextCombination - generate next combination by shifting 1's.

  • High-level pseudocode: 
    
Algorithm: getNextCombination ()
Input: current combination. 
1.    r = position of rightmost 1 with a 0 to it's right;
2.    swap (r, r+1)
3.    bunch up all 1's in positions r+2,...,n-1 to the right of r+1
4.    return new combination

  • Detailed pseudocode: 
    
Algorithm: translatePositionsToSet (positions)
Input: an array with k 1's.
1.    Set member array so member[i]=true when position[i]=1
2.    return member
Output: member array
   
    
Algorithm: getFirstCombination (N, k)
Input: integers N and k

      // Create space for arrays. 
1.    Create new arrays member and position;

      // Bunch up the k 1's to the left. 
2.    for i=0 to k-1: positions[i] = 1

      // Return the corresponding subset. 
3.    member = translatePositionsToSet (positions)
4.    return member

Output: initial set elements via the membership boolean array.
   
    
Algorithm: getNextCombination (member)
Input: current combination. Also, current positions array is accessible.

     // The trivial case: 
1.   if N=1
2.     return null
3.   endif

     // Find the rightmost 1 that is followed by a 0. 
4.   r = position of rightmost 1 with a 0 to the right of it;

     // If we couldn't find one, all are bunched up to the right, so we're done. 
5.   if r < 0
6.     return null
7.   endif

     // Otherwise, we've found a 1. Next, count the 1's to the right of it. 
8.   count = find all 1's to the right of position r;

     // Switch the 1 with the zero to its right. 
9.   swap (r, r+1)

     // Now bunch up the 1's to the right of our special 1. 
10.  for j=1 to count
11.    positions[r+1+j] = 1
12.  endfor

     // The remaining are all 0's. 
13.  Set positions[j] = 0, for j > r+1+count

     // This creates the new subset to return. 
14.  member = translatePositionsToSet (positions)

15.  return member

Output: member array for next combination

  • A complete example with N=5 and k=3 (two steps shown per combination): 
    
    1 1 1 0 0       Rightmost 1-followed-by-0 found at position 2

    1 1 0 1 0       Moved the 1 found rightwards and bunched up the 1's to the right


    1 1 0 1 0       Rightmost 1-followed-by-0 found at position 3

    1 1 0 0 1       Moved the 1 found rightwards and bunched up the 1's to the right


    1 1 0 0 1       Rightmost 1-followed-by-0 found at position 1

    1 0 1 1 0       Moved the 1 found rightwards and bunched up the 1's to the right


    1 0 1 1 0       Rightmost 1-followed-by-0 found at position 3

    1 0 1 0 1       Moved the 1 found rightwards and bunched up the 1's to the right


    1 0 1 0 1       Rightmost 1-followed-by-0 found at position 2

    1 0 0 1 1       Moved the 1 found rightwards and bunched up the 1's to the right


    1 0 0 1 1       Rightmost 1-followed-by-0 found at position 0

    0 1 1 1 0       Moved the 1 found rightwards and bunched up the 1's to the right


    0 1 1 1 0       Rightmost 1-followed-by-0 found at position 3

    0 1 1 0 1       Moved the 1 found rightwards and bunched up the 1's to the right


    0 1 1 0 1       Rightmost 1-followed-by-0 found at position 2

    0 1 0 1 1       Moved the 1 found rightwards and bunched up the 1's to the right


    0 1 0 1 1       Rightmost 1-followed-by-0 found at position 1

    0 0 1 1 1       Moved the 1 found rightwards and bunched up the 1's to the right
 

In-Class Exercise 11.12: This exercise is optional. Download this template and try implementing the above pseudocode for generating combinations.
 

11.10   Generating All Possible Permutations

 

Key ideas

  • Note: there are n! permutations of n items.

  • Generating permuations is surprisingly more complicated than generating combinations.

  • There are many approaches - we will use the Trotter-Johnson algorithm.

  • Assume: the n items are the integers 0, ..., n-1.

  • First, consider all 2-permutations:

    We can view this differently:

    • There are 2 permutations.
    • Generate the second from the first by "moving" the largest element backwards.

  • Next, consider all 3-permutations:

  • Now look at the first few 4-permutations:

  • How do we capture the logic?
    • Suppose each item is given a direction (left or right).
    • An element can "move"
      1. in its prescribed direction;
      2. if there's a smaller element next to it.
    • A mobile element is one that can move.
    • Let's look at the 4-permutations again:

    • The key idea: at each step, move the largest mobile element, and change the directions of those that are larger (if any).
 

Implementation

  • High-level pseudocode: 
    
Algorithm: getNextPermutation ()
Input: current permutation. 
1.    Find highest-numbered mobile element.
2.    Move this one step in its current direction.
3.    Reverse its direction and the direction of all higher-numbered elements.
4.    return new permutation.

  • Data structures:
    • permutation - an integer array where permutation[j] = i if element i is in the j-th position of a permutation.
    • pointsLeft - a boolean array where pointsLeft[i] = true if element i points left (it points right, otherwise).

  • The code will be spread over two methods:
    • getFirstPermutation - initialize and return first permutation.
    • getNextPermutation - generate next permutation from the current one.

  • Pseudocode: 
    
Algorithm: getFirstPermutation (N)
Input: the number of items N (Thus, the integers 0, ..., n-1).

     // Create space and initialize. 
1.   Create arrays permutation and pointsLeft.
2.   for each i, permutation[i] = i
3.   for each i, pointsLeft[i] = true
4.   return permutation

Output: the first permutation 0 1 ... (n-1)
   
    
Algorithm: getNextPermutation (permutation)
Input: the current permutation, where permutation[j] = i means
       number i is the j-th position in the permutation.

      // Locate the largest mobile element in the current permutation. 
      // Recall: an element can move only in its current direction and 
      //         if there's a smaller element next to it in that direction. 
1.    m = position of largest mobile element in permutation array;

      // Only the permutation (n-1) ... 2 1 has no mobile elements. 
2.    if no mobile elements
3.      return null

      // Move the largest mobile element in its current direction. 
      // Mustn't forget to swap in both arrays. 
4.    if pointsLeft[m]
5.      swap (permutation, m, m-1)
6.      swap (pointsLeft, m, m-1)
7.    else
8.      swap (permutation, m, m+1)
9.      swap (pointsLeft, m, m+1)
10.   endif

      // Change the direction of all elements larger than the one just moved. 
11.   largestMobileValue = permutation[m]
12.   Switch directions of all elements larger than largestMobileValue;

      // Return newly generated permutation. 
13.   return permutation
 

In-Class Exercise 11.13: This exercise is optional. Download this template and try implementing the above pseudocode for generating permutations.

© 2001, Rahul Simha