Applied Algorithms

1
Applied Algorithms

• Lecture
• Backtracking

2
Announcements

• ACM Student Chapter Meeting
• Where    FAB-150. Fourth Avenue Building, across
  the hallway from the Linux Lab
•  When     Wednesday, June 1. 1130AM
•  Topic    Become an Undergrad Researcher at PSU!
• Final Exam Date
• Monday, June 6, 2005
• 1230 1420 PM

3
Checking for primality

• Bool prime(int n) // test for primality of n
• int i int m
• if (n2) return TRUE
• if (nlt2) return FALSE
• m sqrt(n)1
• for (i2iltmi)
• if (ni 0) return FALSE
• return TRUE

4
Sum of 4 primes

• Why wont this work?
• for (i2, iltn, i)
• if (prime(i))
• for (j2, jltn, j)
• if (prime(j))
• for (k2, kltn, k)
• if (prime(k))
• for (l2, lltn, l)
• if (prime(l))
• if (ijkltarget)
• printf(ld ld ld ld\n,I,j,k,l)

5
What if we looped over just primes

• Precompute all the primes from 1 to target
• Int primessize
• Void initPrimes(int largest)
• How many primes are there between 1 and
  10,000,000 ?

6
Lets explore?

• int countPrimes(int n)
• int i int count
• count 0
• for (i2iltni)
• if (prime(i)) count
• if (i100000 0)
• printf("i ld, num primes
  ld\n",i,count)
• printf("num primes ld density
  ld\n",count,n/count)

7
Analyze

• Are there a lot of primes?
• Whats the average distance between primes?
• Is this number big?
• Does this suggest a solution?

8
Largest distance between primes

• int enumerate(int n)
• int i int count int last int maxdif
• count 0
• last 2
• maxdif 0
• for (i3iltni)
• if (prime(i))
• count
• maxdif max(i-last,maxdif)
• last i
• if (i100000 0)
• printf("i ld, num primes ld, maxdif
  ld\n
• ,i,count,maxdif)

9
Can we find 3 primes?

• Why is this more feasible?
• Should be easy to find 3 primes that add up to a
  small integer since we need only loop through
  primes up to that integer.
• If max154, how many small primes will we need?

10

• int firstThirty()
• int i int count
• count 0
• for (i2countlt30i)
• if (prime(i))
• count
• printf("ld,",i)

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Strategy

• / An array of the first 30 small primes /
• int sp30
• 2,3,5,7,11,13,17,19,23,29,31,37,41
• ,43,47,53,59,61,67,71,73,79,83,89,97
• ,101,103,107,109,113
• int ps2003
• / ps6 2,2,2 psi stores 3 primes that
  add up to i
• ps7 2,2,3
• ps8 2,3,3
• . . .
• ps199 3,83,113
• /

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Computing ps

• int init_ps_sub(n)
• int j int k int l
• for (j0jlt30j) // Find 3 small primes
• for (k0klt30k) // such that p1p2p3 n
• for (l0llt30l) // store them in the "ps"
  array
• if (spjspkspln)
• psn0 spj
• psn1 spk
• psn2 spl
• return 0
• printf("Can't find 3 small primes for
  ld\n",n)

13
Computing 4 primes

• int test(int n)
• int i int diff
• if (nlt8) printf("Imposible\n") return 0
• for (in-6 igt2 i--) // find largest prime
  at least
• if (prime(i)) // 6 less than n, call
  it "i"
• diff n-i // then lookup 3 small
  primes
• printf("ld ld ld ld\n
• ,i,psdiff0,psdiff1
• ,psdiff2)
• return 0

14
In Class Problems

• Queue 8.6.3
• Page 179 of the text
• We will discuss this in class

15
Queues of N people of different height
How can we count how many can see to the left? To
the right?
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Problem

• Given a queue of fixed size N.
• A fixed number of people who can see to the left.
• A fixed number of people who can see to the
  right.
• How many way can we arrange the N people to get
  the matching left and right counts?

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Solution?

• Compute all possible ways to arrange N people.
• Throw away those that dont have the correct left
  and right count.
• Arranging things in different order is called a
  permutation.

18
Computing permutations

• What are the permutations of 1,2,3 ?
• Chose an element.
• Choices are 1, 2, or 3
• For each choice prepend it to the permutations of
  whats left after removing the choice.
• Candidates 1,2,3
• (1,2,3) (2,1,3) (3,1,2)
• What are the permutations of 2,3?
• 2,3 and 3,2

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• Permutations 1,2,3
• candidates (1,2,3) (2,1,3)(3,1,2)
• permutations 2,3 2,3, 3,2
• 1,3 1,3,3,1
• 1,2 1,2,2,1
• 1,2,3 1,3,2
• 2,1,3,2,3,1
• 3,1,2,3,2,1

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• candidates xs f xs
• where f before (choiceafter)
• (choice,reverse beforeafter)
• (f (choicebefore) after)
• f before
• gen2 Int -gt Int -gt Int -gt Int
• gen2 chosen all (reverse chosen)all
• gen2 chosen toAdd all add pairs all
• where pairs candidates toAdd
• add all all
• add ((choice,remains)pairs) all
• add pairs
• (gen2 (choicechosen)
• remains all)

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How many permutations are there?

• 1,2,3,4
• 4 choices
• For each choice we prepend it to the permutations
  of list of size three.
• Perms( 0) 1 (the empty sequence)
• Perms( x1) (x1) Perms( x)
• Perms( n) n!

22
We need some trick

• Find the tallest person, and then break it into 2
  sub problems. Where can the tallest person be
  given left and right numbers?

23
Bounding the tallest

• If left n, then tallest must be at position at
  least n,
• If right m, then tallest must be at position at
  most Max-m1
• So tallest must be between
• Left ? tallest ? N-Right1
• Left3
• Right3

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Solving the sub problem

• k exact number of views to the left required
• n size of the permutation required
• chosen is the prefix of the permutation so far
• remains the set left to choose the next element
  from
• p2 k n chosen remains all
• length chosen n (reverse
  chosen,remains)all
• True add pairs all
• where pairs candidates remains
• add all all
• add ((x,rs)pairs) all
• let next (xchosen)
• in if (possible k n next)
• then (p2 k n next rs (add pairs
  all))
• else (add pairs all)

25
Whats possible?

• Sometimes we know a chosen sequence cant lead to
  a good solution.
• possible k n xs viewl lt k
• viewl n - len gt k
• where len length xs
• viewl countRight xs

26
All possible winners

• try2 n left right
• n1 lt leftright
• True concat (map each positionsN)
• where positionsN left .. n - right 1
• each i concat(map f leftChoices)
• where leftChoices p2 (left-1) (i-1)
• 1..n-1
• f (lefts,remains)
• concat(map add (p2 (right-1)
  
• (n-i)
• remains
  ))
• where add (rights,_)
• lefts n
  reverse rights

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Todays Assignments

• Read for next time
• Chapter 9 of the text. pp 188-218
• Be prepared to answer questions in class next
  Friday from the reading.
• Programming assignment
• 8.6.8 Bigger Square Please
• Page 186
• Write a solution
• Submit your solution (until you get it right)
• Hand in both your program, and the judge output.
• Those who volunteer to discuss their program get
  class participation points. Email me solutions
  before noon on Friday, May 6.