recursion data structures and algorithms (60-254)

RecursionData Structures and Algorithms (60-254)

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Topics

•Recursion•Divide and Conquer•Dynamic Programming•The Greedy Paradigm

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Recursion

Recursion involves:Self-reference

as in a recursive definition, or in a recursive method

The chain of self-reference is terminated by a base case,which we must define separately.

Let us illustrate this with some examples.

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Examples

Example 1:The factorial of a non-negative integer, n, is defined as follows:

n! n (n-1)! n 11 n = 0

Example 2:The GCD of two non-negative integers, not both 0, is defined as follows:

gcd(m, n) gcd(n, m mod n) if n > 0 m if n = 0

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Examples

Example 3:The product of two positive integers m and n is defined as follows:

P(m, n) m + P(m, n-1) if n > 1m if n = 1

Example 4:A classical example of a problem which is nicely solved by recursion is:

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Towers of Hanoi

We have:n disks in order of decreasing size, which are stacked on

a peg, say A.We want to:

Move these disks to another peg, say C, so that they are in the same decreasing order of size.

using a third peg, say B, as workspace, andunder the following rules:

• We can move only one disk at a time.• We cannot place a larger disk on top of a smaller one in any move.

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Recursive Solution

Move (n, A, C, B) how many? from to workspace Move (n, A, C, B)

Move (n-1, A, B, C) +Move (1, A, C, B) +Move (n-1, B, C, A)

Homework: How many moves to solve with n disks?

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Examples

Example 5:Another nice example of recursion:

already discussed in Stacks.Involves printing the digits of a number n, say n = 1234

from right to left, say 4321The digits are input in the reverse order that we want to print them.We want to postpone printing the least significant digits until we have printed the more significant one.Recursion is one way to achieve this.

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Algorithm RecursiveReverseNumberInput: A non-negative number, nOutput: The digits of n in reverse order.

procedure printDecimal(int n)Print n mod 10if (n >= 10)

printDecimal(n div 10)

Compare this 3 line recursive algorithm with the 10 line iterative (stack based) algorithm from chapter 3. What if we want to print digits in the same order?

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Implementation of Recursion

Recursion is implemented by usinga stack of activation records.

Each activation record stores all parametersrelated to the present call, so thatit can be completed when we return to this call.

Why a stack?Because calls are completed in the reverse orderin which they are generated.

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Tail recursion optimization

printDecimal(int n)

Print n mod 10

if (n >= 10)

printDecimal(n div 10)

printDecimal(int n)

loop: Print n mod 10

if (n >=10)

n=n div 10; goto loop

Tail recursion is a special case of recursion where the last operation is a recursive call.

Compilers can easily transform tail recursive code into equivalent iterative code

Can dramatically decrease stack space used and increase efficiency.

Homework: Does C do tail recursion optimization? C++? Java? C#?

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Proving Program Correctness

• Close connection between recursive programs and mathematical induction

• Can use mathematical induction to aid in proving program correctness for recursive programs

• How?• Prove the program produces the correct output for the base case.• Assume it produces the correct output for the case n=k• Prove it produces the correct output for n=k+1• Homework: Prove correctness of the recursive towers of Hanoi

implementation

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An Inefficient Use of Recursion

Recursion is a powerful tool but there are limitations Computing Fibonacci numbersThe nth Fibonacci number is given by:

Fn = Fn-1 + Fn-2 if n > 2

F0 = 0 , F1 = 1 Here are the first few:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …

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Algorithm Fibonacci (Recursive)

Input: A positive integer nOutput: The sequence of Fibonacci numbersprocedure Fibonacci(int n)

if (n == 0 or n == 1)return n;

elsereturn Fibonacci(n-1) + Fibonacci(n-2)

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Inefficient

• Why is it inefficient using recursion for Fibonacci?

F5

F4 F3

F3 F2

F2 F1

F1 F0

F1 F0

F2 F1

F1 F0

Homework: Design an iterative Fibonacci algorithm

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Divide-and-Conquer

An important paradigm …in algorithm design

Divide:Problem into sub-problems

Conquer:By merging solutions of sub-problems …

into solution to the original problem.

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Maximum Contiguous Subsequence Problem

• Find maximum contiguous subsequence of first half.• Find maximum contiguous subsequence of second half.• MCS max. of these two?

a1, a2, … , an/2 | an/2+1, an/2+2, …, an

first half second half

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Straddling sequence

What if max. contiguous subsequence straddles both halves?Example:

-1, 2, 3, | 4, 5, -6MCS in first half is

2, 3MCS in second half is

4,5 However …

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Observation

The MCS of the given sequence is:2, 3, 4, 5

It straddles both halves!Observation:

For a straddling sequence,an/2 is the last element in first half, and

an/2+1 is the first element in the second half.

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Computing a max. straddling sequenceIn the example:

Last element = 3First element = 4

Computing a max. straddling sequence:

a1, a2, a3, a4, |a5, a6, a7, a8

Find maximum (max1) of:

a4

a4 + a3

a4 + a3 + a2

a4 + a3+ a2+ a1

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Computing a max. straddling sequenceFind the maximum of:

a5

a5 + a6

a5 + a6 + a7

a5 + a6+ a7+ a8

Let the maximum be max2Required maximum:

max1 + max2Required subsequence:

Glue subsequence for max1and subsequence for max2

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Complexity

Theorem:For an n-element sequence,

the complexity of finding a maximumstraddling sequence is O(n)

Proof:

We look at each element exactly once!

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Putting things together …

S = a1, a2, … , an/2, an/2+1, an/2+2, …, an

S1 = a1, a2, … , an/2

S2 = an/2+1, an/2+2, …, an

Max ( S ) = Max ( Max (S1),

Max (S2), Max (straddling sequence) )

Max (S1) and Max (S2) are found recursively.

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Complexity of Recursive MCS

How?T(n/2) = time to solve a problem of size n/2

O(n) = complexity of finding a

Maximum straddling subsequenceWhat is a solution to (1)?

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Approach 1

# of levels in the treeof recursive calls?

1 + log2 n

O(n) work needed to go fromone level to the next higher level.

Therefore, we have O (n log n)

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Graphically

4444

nnnn

n

n / 2 n / 2

n / 4 n / 4 n / 4 n / 4

n/8 n/8 n/8 n/8 n/8 n/8 n/8 n/8

n

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nn

…

log n +1

A total of… O (n log n)

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Approach 2

20 T(n) 2 T(n/2) + c n,21 T(n/2) 2 T(n/4) + c n/2,22 T(n/4) 2 T(n/8) + c n/4,23 T(n/8) 2 T(n/16) + c n/8,………

Adding up, we have:20 T(n) + 21 T(n/2) + 22 T(n/4) + 23 T(n/8) + … 2 T(n/2) + 22 T(n/4) + 23 T(n/8) + …

+ c n + 2 c n/2 + 22 c n/4 + …

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Moral of the story

T(n) c n + c n + c n + …

log2 n terms

T(n) c n log2 n

Recursive MCS takes… T(n) = O (n log n)whereas non-recursive MCS takes…

T(n) = O (n) Divide-and-conquer Not always best solution !!

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A General Recurrence Relation

For T(n) = A T(n/B) + O(nk) where A 1, B > 1

Such recurrences do occur in practice…

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Matrix Multiplication

or

A > B2

where T(n) is the time-complexity of

multiplying two nxn matrices.

In (1), A = 8, B = 2, k = 2(2), A = 7, B = 2, k = 2

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Tree of Recursive Calls (n = 8)

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n = 8

n = 4 n = 4

n = 2 n = 2 n = 2 n = 2

n =1 n =1 n =1 n =1 n =1 n =1 n =1 n =1

1

2

3

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Divide and Conquer

Another example:Find both the maximum and minimum of a set of n elements, S

Naive solution:Find maximum: n – 1 comparisonsFind minimum: n – 2 comparisons

Total: 2n - 3 comparisonsCan we do better?? ... Yes …

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Algorithm Divide-and-Conquer

procedure maxMin(S)if |S| = 2 // S = {a,b}

return (max(a,b), min(a,b))else

divide S into two subsets,

say S1 and S2, each with half of elements.

(max1, min1) maxMin(S1)

(max2, min2) maxMin(S2)return( max(max1, max2), min(min1, min2) )

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Number of comparisons?

T(n) = # of comparisons on n elementsT(n) = 1 n = 2

= 2T(n/2) + 2, n > 2Solution:

n=22 T(4) = 2T(2) + 2 = 4n=23 T(8) = 2T(4) + 2 = 10n=24 T(16) = 2T(8) + 2 = 22

…n=2k T(2k) = 3 2k-1 – 2 (Guess pattern and prove by induction)

= 3/2 n - 2 vs. 2 n - 3

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Solving by Repeated Substitution

T(n) = 1, n = 2= 2T(n/2) + 2, n > 2

T(n/2) = 2T(n/2/2) + 2 = 2T(n/22) + 2

T(n) = 2[2T(n/22) + 2] + 2 {since T(n/2) = 2T(n/2/2) + 2 = 2T(n/22) + 2}= 22T(n/22 ) + 22 + 21

= 22[2T(n/23) + 2] + 22 + 21 {since T(n/22) = 2T(n/22/2) + 2 = 2T(n/23) + 2}= 23T(n/23 ) + 23 + 22 + 21

…= 2kT(n/2k ) + 2k + … + 23 + 22 + 21 {Continue until n/2k = 2 (base case). So 2k=n/2}= n/2 T(2) + 2(2k-1 + … + 23 + 22 + 21 + 20) = n/2 + 2(2k - 1)= n/2 + 2(n/2 - 1)= n/2 + 2n/2 – 2= 3n/2 - 2

Note: 2k-1 + … + 23 + 22 + 21 + 20 is a geometric series Sum = a(1-rk)/(1-r) where a=1, r=2 = (1 – 2k)/(1-2)

= 2k – 1

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Note

However…T(n) = O(n) !!!

for both the naive andthe divide-and-conquer algorithm

For the curious!!Approximately 3n/2 – 2 comparisons are both

necessary and sufficientto find the maximum and minimum

of a set of n elements.

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Dynamic Programming

Another Paradigm…

Given coins worth c1, c2, …, cn cents,Make up change for k cents,

using the minimum number of coinsof the above denominations.

In order that this be always possible,

We assume that c1 = 1

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Coins problem

The essence of this method is to record (in a table) the solutions to all smaller subproblems.Let min(k)

denote the minimum number of coinsneeded to make k cents of change.

Then,min(k) = min { min(r) + min(k-r) }

0 r k/2 if we know min(1), min(2), …, min(k-1)

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Example

c1 = 1 cent

c2 = 5 cents

c3 = 10 cents

c4 = 25 cents

min(1) = 1, and min(2) = min(1) + min(1) = 2Already know the value of min(2), then

min(3) = min(1) + min(2) = 1 + 2 = 3Again, already know values of min(2) and min(3), so:

min(4) = min{ min(0) + min(4), min(1) + min(3), min(2) + min(2) } = 4

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Example

# of cents Min How solution is found?1 1 min{min(1)+min(0)} 2 2 min{min(0)+min(2), min(1)+min(1)}3 3 min{min(0)+min(3), min(1)+min(2)}4 4 min{min(0)+min(4), min(1)+min(3), min(2)+min(2)}5 1 min{min(0)+min(5), min(1)+min(4), min(2)+min(3)}6 2 min{min(0)+min(6), min(1)+min(5),

min(2)+min(4), min(3)+min(3)}7 3 min{min(0)+min(7), min(1)+min(6),

min(2)+min(5), min(3)+min(4)}

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Example

# of cents Min How solution is found?

8 4 min{min(0)+min(8), min(1)+min(7),

min(2)+min(6), min(3)+min(5), min(4)+min(4)}





min(11) = min{0+11, 1+10, 2+9, 3+8, 4+7, 5+6}

= min{ , 2, 7, 7, 7, 3}

= 2

1+1=2 2+5=7 3+4=7 1+2=3

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Dynamic Programming Solution

Why don’t we use recursion? Quite inefficient !!What do we use instead? Dynamic Programming…The algorithm uses two arrays:coinsUsed: Stores the minimum number of coins needed to make change of k cents, k = 1,…, maxChange.lastCoin: Stores the last coin that was used by the minimum (stored in coinsUsed)It is used to go backwards and list the coin denominations that compose the solution (after the minimum has been found).

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Coins

coinsUsed[0] 0; lastCoin[0] 1for cents 1 to maxChange do // maxChange = k

minCoins centsnewCoin 1for j 1 to diffCoins do // diffCoins = n

if (coins[j] > cents)continue // Cannot use coin j

if (coinsUsed[cents - coins[j]] + 1 < minCoins)minCoins coinsUsed[cents – coins[j]] + 1newCoin coins[j]

coinsUsed[cents] minCoinslastCoin[cents] newCoin

Print “minimum number of coins:”, coinsUsed[maxChange]

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Time Complexity

O ( n k ) where

n = number of coins of different denominationsk = amount of change we want to make

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Greedy Paradigm

An interesting question:Can we reach a global optimum by

a process of incremental optimization?We can, but not always guaranteed to succeed.

Coin change:Greedy does not work !!!

Take n1 of largest coin

n2 of second largest coin…

nk of smallest coin.

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Counter-examples

Coin denominations: 1, 4, 5, 6 cents.Make change for 9 cents.

Greedy: 6 + 1 + 1 + 1 (four coins)Optimal: 4 + 5 (just two coins)

Coin denominations: 1, 10 and 25 centsMake change for 31 cents.

Greedy: 25 + 1 + 1 + 1 + 1 + 1 + 1 ,i.e., 7 coins required

Optimal solution:10 + 10 + 10 + 1

i.e., 4 coins required !!!

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Generating Permutations

Problem:Give an algorithm that

computes the different waysof making k cents in change.

?

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Generating Permutations

Write a recursive programthat generates all permutations of

1, 2, 3, …, nFor example,

if n = 3, the output should be:1 2 32 1 31 3 23 1 23 2 12 3 1

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O(n!)

• Someone asked about O(n!)• List all the combinations of a set of n elements is O(n!)• Traveling Saleman problem has a naïve solution that is O(n!), but there is a dynamic programming solution that is O(n2 2n)• The following is a list of common types of orders and their names:

Notation NameO(1) ConstantO(log(n)) LogarithmicO(log(log(n)) Double logarithmic (iterative logarithmic)O(n) LinearO(n log(n)) Loglinear, Linearithmic, Quasilinear or SupralinearO(n2) QuadraticO(n3) CubicO(nc) Polynomial (different class for each c > 1)O(cn) Exponential (different class for each c > 1)O(n!) FactorialO(nn) Yuck!

recursion data structures and algorithms (60-254)

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