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Design of Algorithms by Induction. Bibliography: [Manber]- Chap 5. Algorithm Analysis <-> Design. Given an algorithm, we can Analyze its complexity Proof its correctness But: Given a problem, how do we design a solution that is both correct and efficient ?

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design of algorithms by induction

Design of Algorithms by Induction

Bibliography: [Manber]- Chap 5

algorithm analysis design
Algorithm Analysis <-> Design
  • Given an algorithm, we can
    • Analyze its complexity
    • Proof its correctness
  • But:
    • Given a problem, how do we design a solution that is both correct and efficient ?
      • Is there a general method for this ?
induction proofs vs design of algorithms by induction 1
Induction proofs vs Design of algorithms by induction (1)
  • Induction used to prove that a statement P(n) holds for all integers n:
    • Base case: Prove P(0)
    • Assumption: assume that P(n-1) is true
    • Induction step: prove that P(n-1) implies P(n) for all n≥0
    • Strong induction:when we assume P(k) is true for all k<=n-1 and use this in proving P(n)
induction proofs vs design of algorithms by induction 2
Induction proofs vs Design of algorithms by induction (2)
  • Induction used in algorithm design:
    • Base case: Solve a small instance of the problem
    • Assumption: assume you can solve smaller instances of the problem
    • Induction step: Show how you can construct the solution of the problem from the solution(s) of the smaller problem(s)
design of algorithms by induction1
Design of algorithms by induction
  • Represents a fundamental design principle that underlies techniques such as divide & conquer, dynamic programming and even greedy
  • Question: how to reduce the problem to a smaller problem or a set of smaller problems ? (n -> n-1, n/2, n/4, …?)
  • Examples:
    • The successful party problem [Manber 5.3]
    • The celebrity problem [Manber 5.5]
    • The skyline problem [Manber 5.6]
    • One knapsack problem [Manber 5.10]
    • The Max Consecutive Subsequence [Manber 5.8]
the successful party problem
The successful party problem
  • Problem: you are arranging a party and have a list of n persons that you could invite. In order to have a successful party, you want to invite as many people as possible, but every invited person must be friends with at least k of the other party guests. For each person, you know his/her friends. Find the set of invited people.
successful party example
Successful party - Example

K=3

Ann

Bob

Finn

Chris

Ellie

Dan

successful party direct approach
Successful party – Direct approach
  • Direct approach for solving: remove persons who have less than k friends
  • But: which is the right order of removing ?
    • Remove all persons with less than k friends, then deal with the persons that are left without enough friends ?
    • Remove first one person, then continue with affected persons ?

Instead of thinking about our algorithm

as a sequence of steps to be executed,

think of proving a theorem that the algorithm exists

the design by induction approach
The design by induction approach
  • Instead of thinking about our algorithm as a sequence of steps to be executed, think of proving a theorem that the algorithm exists
  • We need to prove that this “theorem” holds for a base case, and that if it holds for “n-1” this implies that it holds for “n”
successful party solution
Successful party - Solution
  • Induction hypothesis:We know how to find the solution (maximal list of invited persons that have at least k friends among the other invited persons), provided that the number of given persons is <n
  • Base case:
    • n<= k: no person can be invited
    • n=k+1:if every person knows all of the other persons, everybody gets invited. Otherwise, no one can be invited.
successful party solution1
Successful party - Solution
  • Inductive step:
    • Assume we know how to select the invited persons out of a list of n-1, prove that we know how to select the invited persons out of a list of n, n>k+1
    • If all the n persons have more than k friends among them, all n persons get invited
    • Else, there exists at least one person that has less than k friends. Remove this person and the solution is what results from solving the problem for the remaining n-1 persons
successful party conclusions
Successful party - Conclusions
  • Designing by induction got us to a correct algorithm, because we designed it proving its correctness
    • Every invited person knows at least k other invited
    • We got the maximum possible number of invited persons
  • The best way to reduce a problem is to eliminate some of its elements.
    • In this case, it was clear which persons to eliminate
    • We will see examples where the elimination is not so straightforward
the celebrity problem
The Celebrity problem
  • Problem: A celebrity in a group of people is someone who is known by everybody but does not know anyone. You are allowed to ask anyone from the group a question such as “Do you know that person?” pointing to any other person from the group. Identify the celebrity (if one exists) by asking as few questions as possible
  • Problem:
  • Given a n*n matrix “know” with know[p, q] = true if p knows q and know[p, q] = false otherwise.
  • Determine whether there exists an i such that:
    • Know[j; i] = true (for all j, j≠ i) and Know[i; j] = false (for all j, j≠i )
celebrity solution 1
Celebrity - Solution 1
  • Brute force approach: ask questions arbitrary, for each person ask questions for all others => n*(n-1) questions asked
celebrity solution 2
Celebrity - Solution 2
  • Use induction:
  • Base case: Solution is trivial for n=1, one person is a celebrity by definition.
  • Assumption: Assume we leave out one person and that we can solve the problem for the remaining n-1 persons.
    • If there is a celebrity among the n-1 persons, we can find it
    • If there is no celebrity among the n-1 persons, we find this out
  • Induction step:
  • For the n'th person we have 3 possibilities:
    • The celebrity was among the n-1 persons
    • The n’th person is the celebrity
    • There is no celebrity in this group
celebrity sol 2 induction step
Celebrity - Sol 2- Induction step
  • We must solve the problem for n persons, assuming that we know to solve it for n-1
  • We leave out one person.
  • We choose this person randomly – let it be the n’th person
  • Case 1: There is a celebrity among the n-1 persons, say p. To check if this is also a celebrity for the n'th person
    • check if know[n, p] and not know[p, n]
  • Case 2: There is no celebrity among the n-1 persons. In this case, either person n is the celebrity, or there is no celebrity in the group.
    • To check this we have to find out if know[i, n] and not know[n, i], for all i <>n.
celebrity solution 21
Celebrity – Solution 2

Function Celebrity_Sol2(n:integer) returns integer

if n = 1 then return 1

else

p = Celebrity_Sol2(n-1);

if p != 0 then

// p is the celebrity among n-1

if( knows[n,p] and not knows[p,n] ) then

return p

end if

end if

forall i = 1..n-1

if (knows[n, i] or not knows[i, n]) then

return 0 // there is no celebrity

end if

end for

return n // n is the celebrity

celebrity solution 2 analysis
Celebrity – Solution 2 Analysis

T(n)

Function Celebrity_Sol2(n:integer) returns integer

if n = 1 then return 1

else

p = Celebrity_Sol2(n-1);

if p != 0 then

// p is the celebrity among n-1

if( knows[n,p] and not knows[p,n] ) then

return p

end if

end if

forall i = 1..n-1

if (knows[n, i] or not knows[i, n]) then

return 0 // there is no celebrity

end if

end for

return n // n is the celebrity

T(n-1)

O(n)

celebrity solution 2 analysis1
Celebrity – Solution 2 Analysis
  • T(n)=T(n-1) + n
  • T(1)=1
  • T(n)=1+2+3+ +n=n*(n+1)/2
  • T(n) is O(n2)
  • We have reduced a problem of size n to a problem of size n-1. We then still have to relate the n-th element to the n-1 other elements, and here this is done by a sequence which is O(n), so we get an algorithm of complexity O (n2), which is the same as the brute force.
  • If we want to reduce the complexity of the algorithm to O(n), we should have T(n)=T(n-1)+c
celebrity solution 3
Celebrity – Solution 3
  • The key idea here is to reduce the size of the problem from n persons to n-1, but in a clever way – by eliminating someone who is a non-celebrity.
  • After each question, we can eliminate a person
    • if knows[i,j] then i cannot be a celebrity => elim i
    • if not knows[i,j] then j cannot be a celebrity => elim j
celebrity solution 31
Celebrity - Solution 3
  • Use induction:
  • Base case: Solution is trivial for n=1, one person is a celebrity by definition.
  • Assumption: Assume we leave out one person who is not a celebrity and that we can solve the problem for the remaining n-1 persons.
    • If there is a celebrity among the n-1 persons, we can find it
    • If there is no celebrity among the n-1 persons, we find this out
  • Induction step:
  • For the n'th person we have only 2 possibilities left:
    • The celebrity was among the n-1 persons
    • There is no celebrity in this group
celebrity sol 3 induction step
Celebrity - Sol 3- Induction step
  • We must solve the problem for n persons, assuming that we know to solve it for n-1
  • We leave out one person. In order to decide which person to elim, we ask a question (to a random person i about a random person j).
  • The eliminated person is e=i or e=j, and we know that e is not a celebrity
  • Case 1: There is a celebrity among the n-1 persons that remain after eliminating e, say p. To check if this is also a celebrity for the person e
    • check if know[e, p] and not know[p, e]
  • Case 2: There is no celebrity among the n-1 persons. In this case, there is no celebrity in the group. It is no need any more to check if e is a celebrity !
celebrity solution 32
Celebrity – Solution 3

Function Celebrity_Sol3(S:Set of persons) return person

if card(S) = 1 return S(1)

pick i, j any persons in S

if knows[i, j] then // i no celebrity

elim=i

else // if not knows[i, j] then j no celebrity

elim=j

p = Celebrity_Sol3(S-elim)

if p != 0 and knows[elim,p] and not knows[p,elim]

return p

else

return 0 // no celebrity

end if

end if

end function Celebrity_Sol3

celebrity solution 3 analysis
Celebrity – Solution 3 Analysis

Function Celebrity_Sol3(S:Set of persons) return person

if card(S) = 1 return S(1)

pick i, j any persons in S

if knows[i, j] then // i no celebrity

elim=i

else // if not knows[i, j] then j no celebrity

elim=j

p = Celebrity_Sol3(S-elim)

if p != 0 and knows[elim,p] and not knows[p,elim]

return p

else

return 0 // no celebrity

end if

end if

end function Celebrity_Sol3

T(n)=T(n-1)+c

Algorithm is O(n)

celebrity conclusions
Celebrity - Conclusions
  • The size of the problem should be reduced (from n to n-1) in a clever way
  • This example shows that it sometimes pays off to expend some effort ( in this case – one question) to perform the reduction more effectively
the skyline problem
The Skyline Problem
  • Problem: Given the exact locations and heights of several rectangular buildings, having the bottoms on a fixed line, draw the skyline of these buildings, eliminating hidden lines.
the skyline problem1
The Skyline Problem
  • A building Bi is represented as a triplet (Li, Hi, Ri)
  • A skyline of a set of n buildings is a list of x coordinates and the heights connecting them
  • Input (1,11,5), (2,6,7), (3,13,9), (12,7,16), (14,3,25), (19,18,22), (23,13,29), (24,4,28)
  • Output (1, 11), (3, 13), (9, 0), (12, 7), (16, 3), (19, 18), (22, 3), (23, 13), (29, 0)
skyline solution 1
Skyline – Solution 1
  • Base case: If number of buildings n=1, the skyline is the building itself
  • Inductive step: We assume that we know to solve the skyline for n-1 buildings, and then we add the n’th building to the skyline
adding one building to the skyline
Adding One Building to the Skyline
  • We scan the skyline, looking at one horizontal line after the other, and adjusting when the height of the building is higher than the skyline height

Worst case:

O(n)

Bn

skyline solution 1 analysis
Skyline – Solution 1 Analysis

T(n)

Algorithm Skyline_Sol1(n:integer) returns Skyline

if n = 1 then return Building[1]

else

sky1 = Skyline_Sol1(n-1);

sky2 = MergeBuilding(Building[n], sky1);

return sky2;

end algorithm

T(n-1)

O(n)

  • T(n)=T(n-1)+O(n)
  • O(n2)
skyline solution 2
Skyline – Solution 2
  • Base case: If number of buildings n=1, the skyline is the building itself
  • Inductive step: We assume that we know to solve the skylines for n/2 buildings, and then we merge the two skylines
merging two skylines
Merging two skylines
  • We scan the two skylines together, from left to right, match x coordinates and adjust height where needed

n1

n2

Worst case:

O(n1+n2)

skyline solution 2 analysis
Skyline – Solution 2 Analysis

T(n)

Algorithm Skyline_Sol2(left, right:integer) returns Skyline

if left=right then return Building[left]

else

middle=left+(left+right)/2

sky1 = Skyline_Sol2(left, middle);

sky2 = Skyline_Sol2(middle+1, right);

sky3 = MergeSkylines(sky1, sky2);

return sky3;

end algorithm

T(n/2)

T(n/2)

O(n)

  • T(n)=2T(n/2)+O(n)
  • O(n log n)
skyline conclusion
Skyline - Conclusion
  • When the effort of combining the subproblems cannot be reduced, it is more efficient to split into several subproblems of the same type which are of equal sizes
    • T(n)=T(n-1)+O(n) … O(n2)
    • T(n)=2T(n/2)+O(n) … O(n log n)
  • This technique is Divide and conquer
efficiently dividing and combining
Efficiently dividing and combining
  • Solving a problem takes following 3 actions:
    • Divide the problem into a number of subproblems that are smaller instances of the same problem.
    • Solve the subproblems
    • Combine the solutions to the subproblems into the solution for the original problem.
  • Execution time is defined by recurrences like
    • T(size)=Sum (T(subproblem size)) + CombineTime
    • Choosing the right subproblems sizes and the right way of combining them decides the performance !
recurrences
Recurrences
  • T(n)=O(1), if n=1
  • For n>1, we may have different recurrence relations, according to the way of splitting into subproblems and of combining them. Most common are:
    • T(n)=T(n-1)+O(n) … O(n2)
    • T(n)=T(n-1)+O(1) … O(n)
    • T(n)=2T(n/2)+O(n) … O(n log n)
    • T(n)=2T(n/2)+O(1) … O(n)
    • T(n)=a*T(n/b) + f (n) ;
the knapsack problem
The Knapsack Problem
  • This is one of the multiple variants of the knapsack problem, that applies to packaging goods in standard containers of a shipping company that have to be exactly filled up
  • The problem: Given an integer K and n items of different sizes such that the i’th item has an integer size ki, find a subset of the items whose sizes sum to exactly K, or determine that no such subset exist
  • P(n,K)
  • P(i,k) – the first i items and a knapsack of size k
knapsack try 1
Knapsack - Try 1
  • Induction hypothesis: We know how to solve P(n-1, K)
  • Base case: n=1: there is a solution if the single element is of size K
  • Inductive step:
  • Case 1: P(n-1,K) has a solution: we simply do not use the n’th item, we already have the solutiom
  • Case 2: P(n-1,K) has no solution: this means that we must use the n’th item of size kn. This implies that the rest of the items must fit into a smaller knapsack of size K- kn.We have reduced the problem to two smaller subproblems: P(n-1, K) and P(n-1, K-kn).
  • In order to solve this, we need to strenghten the hypothesis
knapsack try 2 solution 1
Knapsack - Try 2 – Solution 1
  • Induction hypothesis: We know how to solve P(n-1, k) for all 0<=k<=K
  • Base case: n=1: there is a solution if the single element is of size k
  • Inductive step:
  • P(n-1, k) and P(n-1, k-kn).
  • We have reduced the problem of size n to 2 problems of size n-1
  • T(n)=2 T(n-1) + O(1)
  • T(n) is O(2n)
knapsack solution 2
Knapsack – Solution 2
  • Solution 1 solved 2n problems P(i,k)
  • Actually, the total number of distinct problems P(i,k) is n*K
  • 2n – n*K is redundant work !
  • Solution: we store all the known results of all problems P(i,k) in an n*K matrix
knapsack conclusions
Knapsack - Conclusions
  • Divide-and-conquer algorithms partition the problem into disjoint subproblems, solve the subproblems, and then combine their solutions to solve the original problem.
  • When the subproblems overlap, a divide-and-conquer algorithm does more work than necessary, repeatedly solving the common subsubproblems.
  • In order to avoid repeated work, the solutions of subproblems are saved and when the same subproblem appears again the saved solution is used
  • This technique is Dynamic programming
finding the maximum consecutive subsequence
Finding the Maximum Consecutive Subsequence
  • Problem: Given a sequence X = (x1, x2, …, xn) of (not necessarily positive) real numbers, find a subsequence xi; xi+1; … ; xj of consecutive elements such that the sum of the numbers in it is maximum over all subsequences of consecutive elements
  • Example: The profit history (in billion $) of the company ProdIncCorp for the last 10 years is given below. Find the maximum amount that ProdIncCorp earned in any contiguous span of years.
max consec subsequence try 1
Max consec subsequence – Try 1
  • Base case: n=1: we have a single element and it is the max subsequence
  • Induction hypothesis:We know how to find the maximum subsequence in a sequence of length < n.
  • We know(we assume) that the maximum consecutive subsequence of x1;…; xn-1 is xi… xj , j<=n-1
  • We have to find the maximum consecutive subsequence of x1;…; xn-1; xn
    • We distinguish 2 cases:
      • Case 1: j=n-1 (the max subseq is a suffix of the given one)
      • Case 2: j<n-1 (the max subseq is no suffix of the given one)
max consec subsequence try 11
Max consec subsequence – Try 1
  • Case 1: j=n-1 (the max subseq is a suffix of the given one)

n-1

n

If x[n] >0 then the max seq will be xi;… xj; xn

n-1

n

If x[n] <0 then the max seq remains xi;… xj

max consec subsequence try 12
Max consec subsequence – Try 1
  • Case 2: j<n-1 (the max subseq is not a suffix)

n-1

n

n-1

n

We need to know also

the maximum subsequence that is a suffix !

max consec subsequence try 2
Max consec subsequence – Try 2
  • Stronger Induction hypothesis: We know how to find the maximum subsequence in a sequence of length < n and the maximum subsequence that is a suffix
  • Base case: when n=0, both subsequences are empty (their sums are 0)
  • Inductive step: We add xn to the max suffix. If this sum is bigger than the global max subsequence, then we update it (as well as the new suffix). Otherwise we retain the previous max subsequence. We also need to find the new max suffix, and add xn to its old value. If this sum results negative, we take the empty set as max suffix.
max consec subseq solution
Max consec subseq - Solution

Algorithm Max_Subsequence(IN: X[1..n], OUT: GlobalMax, SuffixMax)

begin

if (n=0) then

GlobalMax:=0;

SuffixMax:=0;

else

Max_Subsequence(X[1..n-1], GlobalMax1, SuffixMax1);

GlobalMax:=GlobalMax1;

if x[n] + SuffixMax1 > GlobalMax1 then

SuffixMax := SuffixMax1 + x[n];

GlobalMax := SuffixMax;

else if x[n] + SuffixMax1 > 0 then

SuffixMax := SuffixMax1 + x[n];

else

SuffixMax := 0;

end

T(n)=T(n-1)+c

Algorithm is O(n)

max consec subseq solution1
Max consec subseq - Solution

Algorithm Max_Subsequence(X,n)

Input: X (array of length n)

Output: Global_Max (The sum of the maximum subsequence)

begin

Global_Max:= 0;

Suffix_Max := 0;

for i=1 to n do

if x[i] + Suffix_Max > Global_Max then

Suffix_Max := SuffixMax + x[i];

Global_Max := Suffix_Max;

else if x[i] + Suffix_Max > 0 then

Suffix_Max := Suffix_Max + x[i];

else

Suffix_Max := 0;

end

It was straightforward to rewrite the solution in a non-recursive way !

max consec subseq conclusion
Max consec subseq - Conclusion
  • Sometimes we cannot prove that P(<n) => P(n)
  • We can add an additional statement Q and prove easier
  • (P(< n) and Q(< n) ) => P(n)
  • Q is a property of the solution, property that we must discover
  • Attention: Q(n) must become part of the induction hypothesis, we must in fact prove:
  • (P(< n) and Q(< n) ) => ( P(n) and Q(n))

This is Strengthen the Induction Hypothesis

conclusions 1
Conclusions (1)
  • What is Design by induction ?
  • An algorithm design method that uses the idea behind induction to solve problems
    • Instead of thinking about our algorithm as a sequence of steps to be executed, think of proving a theorem that the algorithm exists
      • We need to prove that this “theorem” holds for a base case, and that if it holds for “n-1” this implies that it holds for “n”
conclusions 2
Conclusions (2)
  • Why/when should we use Design by induction ?
    • It is a well defined method for approaching a large variety of problems (“where do I start ?”)
      • Just take the statement of the problem and consider it is a theorem to be proven by induction
    • This method is always safe: designing by induction gets us to a correct algorithm, because we designed it proving its correctness
    • Encourages abstract thinking vs coding => you can handle the reasoning in case of complex algorithms
      • A[1..n]; A[n/4+1 .. 3n/4]
      • L, R; L=(R+3L+1)/4, R=(L+3R+1)/4
    • Wecan make it also efficient (see next slide)
conclusions 3
Conclusions (3)
  • The inductive step is always based on a reduction from problem size n to problems of size <n. How to efficiently make the reduction to smaller problems:
    • Sometimes one has to spend some effort to find the suitable element to remove. (see Celebrity Problem)
    • If the amount of work needed to combine the subproblems is not trivial, reduce by dividing in subproblems of equal size –divide and conquer(see Skyline Problem)
    • A reduction to several subproblems might eventually compute the same subproblems again and again. In this case, when the subproblems are not disjoint, try to store these intermediate results – dynamic programming. (see Knapsack Problem)
conclusions 4
Conclusions (4)
  • What if induction does not succeed for the given problem statement?:
    • A straightforward reduction might not give sufficient information to extend the result, in this case one can try to strengthen the induction hypothesis. (see Maximum Subsequence Problem)
      • Try to find relevant properties of the solution and add these to the induction hypothesis