Google Code Jam 2011 Qualification Round

This Saturday Google Code Jam 2011 Qualification Round took place. I didn't have much time to spend on the problems, but still have solved a couple and proceeded to Round 1.

The ones that I solved are Magicka and Candy Splitting.

Magicka

First I tried to solve it with some cunning string substitution involving regular expressions. But in the end the simple simulation did the trick. I actually was a little bit disappointed after the official solutions were announced, because I thought that there should be some trick in here.

#!/usr/bin/ruby

lines = ARGF.readlines

t = lines.first.to_i
(1..t).each do |test_id|
  arr = lines[test_id].split " "
  c = arr[0].to_i
  combines = arr[1..c].inject({}){|memo, e| memo[e[0..1]]=e[2]; memo[e[1]+e[0]]=e[2]; memo }
  d = arr[c+1].to_i
  opposed = arr[c+2..c+1+d]
  n = arr[c+2+d].to_i
  elems = arr[c+3+d] #.split("").collect {|i| i.to_sym }

  result = []
  elems.split("").each do |elem|
    result << elem
    if result.length > 1
      last2 = result[-2..-1].join
      if combines.key?(last2)
        result.pop(2)
        result = result << combines[last2]
      end
      opposed.each do |o|
        if result.include?(o[0]) && result.include?(o[1])
          result = []
          break
        end
      end
    end
  end

  puts "Case ##{test_id}: [#{result.join(", ")}]"
end

Candy Splitting

This one was a little bit more tricky and the author's solution is much more simpler and clear than mine. I made it using old school brute force, just like Goro (strength is my strength :) )

#!/usr/bin/ruby

def s( candies, pile1, pile2 )
 #p "c:#{candies}, 1:#{pile1}, 2:#{pile2}"
 if candies.empty?
  if !pile1.empty? && !pile2.empty? && pile1.inject(0){|memo, i| memo ^ i} == pile2.inject(0){|memo, i| memo ^ i}
   return pile1
  else
   return false
  end
 end
 x = candies.shift
 p1 = pile1.clone
 return s( candies, p1 << x, pile2) || s( candies, pile1, pile2 << x)
end

lines = ARGF.readlines

t = lines.first.to_i
(1..t).each do |test_id|
 n_candies = lines[test_id*2-1]
 candies = lines[test_id*2].split(" ").collect{|i| i.to_i }

 r = s( candies.sort.reverse, [], [] )
 r = !r ? "NO" : r.inject(:+)

 puts "Case ##{test_id}: #{r}"
end

Determine If Number Is Prime

The SICP book is still being read...

Now I am going to introduce some algorithms that allow us to say whether the number is prime or not. Again we will start from the simplest one.

Take numbers 1 by 1 starting from 2 to √n and check, if it divides n. The complexity of the algorithm is Θ(√n).

(define (prime? n)
  (define (divides? k)
    (= (remainder n k) 0))
  (define (find-divisor n test)
    (cond ((> (sqr test) n) n)
          ((divides? test) test)
          (else (find-divisor n (+ test 1)))))

  (= n (find-divisor n 2)))

Fermat Primality Test

There is a test for primality with Θ(log(n)) complexity. It is based on Fermat little theorem.

an ≡ a (mod n), where 1 ≤ a < n

Which means, that if n is prime, then for any integer a, an − a will be evenly divisible by n.

;this procedure calculates baseexp mod m without using big numbers
(define (expmod base exp m)
  (cond ((= exp 0) 1)
        ((even? exp) (remainder (sqr (expmod base (/ exp 2) m)) m))
        (else (remainder (* base (expmod base (- exp 1) m)) m))))

(define (fermat-test n)
  (define (try-it a)
    (= (expmod a n n) a))
  (try-it (+ 1 (random (- n 1)))))

Unfortunately there are some issues with this test - it is probabilistic. If the test fails, n is certainly composite. But if it passes, it does not guarantee that n is prime, although it shows high probability for n to be. So we can run as many tests as we need to make the mistake probability as low as possible.

(define (fermat-prime? n times)
  (cond ((= times 0) true)
        ((fermat-test n) (fermat-prime? n (- times 1)))
        (else false)))

Unfortunately that is not all. There are some numbers that pass the test (for every a!), without being prime. These are called Carmichael numbers. So to determine, if n is prime, we need to check if it passes the Fermat test and is not a Carmichael number, or use a similar Miller–Rabin primality test, that cannot be tricked.

Miller–Rabin Primality Test

It starts from an alternate form of Fermat's Little Theorem:

an-1 ≡ 1 (mod n), where 1 ≤ a < n

The difference is that whenever we perform the squaring step in expmod, we check to see if we have discovered a "nontrivial square root of 1 modulo p", that is, a number not equal to 1 or n - 1 whose square is equal to 1 mod n. It is possible to prove that if such a nontrivial square root of 1 exists, then n is not prime. It is also possible to prove that if n is an odd number that is not prime, then, for at least half the numbers a < n, computing an-1 in this way will reveal a nontrivial square root of 1 mod n.

(define (miller-rabin-prime? n times)
  (cond ((= times 0) true)
        ((miller-rabin-test n) (miller-rabin-prime? n (- times 1)))
        (else false)))

(define (miller-rabin-test n)
  (define (try-it a)
    (= (expmod a (- n 1) n) 1))
  (try-it (+ 1 (random (- n 1)))))

(define (expmod base exp m)
  (define (!= x y) (not (= x y)))

  (cond ((= exp 0) 1)
        ((and (!= base 1) 
              (!= base (- m 1)) 
              (= (remainder( sqr base) m) 1)) 0)
        ((even? exp) (remainder (sqr (expmod base (/ exp 2) m)) m))
        (else (remainder (* base (expmod base (- exp 1) m)) m))))

Fibonacci Numbers

Continuing reading the SICP book...

In every book about algorithms I have read there is always a chapter about the Fibonacci numbers. Mainly on the different algorithms of calculating the n-th Fibonacci number and comparing their order of growth. First comes the algorithm, based on definition:

(define (fib n)
  (cond ((= n 0) 0)
        ((= n 1) 1)
        (else (+ (fib (- n 1))
                 (fib (- n 2))))))

This function is good at explaining tree recursion, but is awful for calculating Fibonacci numbers, as it makes too many unnecessary calculations. The order of growth is Θ(Fib(n)), which means exponential. The next comes iterative funcion, which solves the task much better, using less resources and time. Its order of growth is Θ(n) - linear.

(define (fib n)
  (fib-iter 1 0 n))

(define (fib-iter a b count)
  (if (= count 0)
     b
     (fib-iter (+ a b) a (- count 1))))

But there is also an algorithm that calculates Fibonacci number with Θ(log n) complexity. There, in the book they give you the function with gaps and some pointers ( no C here :) ), how to fill them.

(define (fib n)
  (fib-iter 1 0 0 1 n))

(define (fib-iter a b p q count)
  (cond ((= count 0) b)
        ((even? count)
         (fib-iter a
                   b
                   [??] calculate p'
                   [??] calculate q'
                   (/ count 2)))
        (else (fib-iter (+ (* b q) (* a q) (* a p))
                        (+ (* b p) (* a q))
                        p
                        q
                        (- count 1)))))

The authors remind us of the transformation of the state variables a and b in the previous (iterative) fib-iter procedure: a ← a + b and b ← a. Provided these state changes are labeled transformation T, applying T repeatedly for n iterations starting with a = 1 and b = 0 will produce the pair a = Fib(n + 1) and b = Fib(n). So the Fibonacci numbers are produced by the n^th power of the transformation T (Tⁿ), starting with the pair (1, 0).

Now consider the family of transformations T_pq which transforms the pair (a, b) according to the following rules:

a ← bq + aq + ap
b ← bp + aq

Where transformation T is just a special case of T_pq, where p = 0 and q = 1. Now it's up to you to find the transformation T_p'q' such so, if we apply T_pq twice, the effect is the same as using a single transformation T_p'q' of the same form. Compute p' and q' in terms of p and q and put into the function.

I will not provide the solution here, in case you want to handle it yourself. But if you are eager to see the result, visit some other blog that has the solution. :)