最終更新: 2011-08-20T02:11+0900
ナンバープレイス。30秒くらいかかります。数独ソルバは前にも書いたことがある(20100511)。その時とくらべてしたことといえば優先度を付けたくらい。場合によってそれが効果的なのは Problem 83のとき(20110325p01.03)に実感してる。
def main
tl3digits = []
numarr = ''
DATA.each_line{|line| line.chomp!
if /^\d{9}$/ =~ line
numarr << line
tl3digits << solve(numarr)[0,3].to_i if numarr.size == 9*9
else
numarr = ''
end
}
puts "#{tl3digits.size} puzzles were solved."
puts "The sum of 3-digit numbers is #{tl3digits.inject(&:+)}."
end
def solve(nums)
pns = possible_numbers(nums) # [[index,X,Y,...],...]
pns_shortest = 10
pns_shortest_index = -1
pns.each_with_index{|pn,i|
if pn.size-1 < pns_shortest
pns_shortest = pn.size-1
pns_shortest_index = i
end
}
if 10 <= pns_shortest
return nums # solved!
elsif 0 == pns_shortest
return nil # false branch. no solution.
else
pn = pns[pns_shortest_index]
index = pn.shift
return pn.map{|n|
nums_ = nums.dup
nums_[index] = n
solve(nums_)
}.compact[0] # nil or solution
end
end
def possible_numbers(nums)
pns = []
(0...9).each{|y|
(0...9).each{|x|
i = index(x,y)
next unless nums[i] == ?0
pns << [i] + ((?1..?9).to_a - determined_numbers(nums, h_indices(x,y)) - determined_numbers(nums, v_indices(x,y)) - determined_numbers(nums, b_indices(x,y)))
}
}
return pns
end
def index(x,y)
return x + y*9
end
def h_indices(x,y)
return y*9 ... (y+1)*9
end
def v_indices(x,y)
return (0...9).map{|y| x + y*9 }
end
def b_indices(x,y)
i = index(x-x%3, y-y%3)
return [i, i+1, i+2, i+9, i+10, i+11, i+18, i+19, i+20]
end
def determined_numbers(nums, indices)
return indices.map{|i| nums[i] }.reject{|n| n == ?0 }
end
main;
__END__
content of sudoku.txt here.
Bignumがある Rubyでは関係ないけど、64ビット整数を前提に小細工(ループ展開)。
ten = 56866
978807.times{
ten = (ten * 256) % 10000000000
}
p (ten + 1) % 10000000000
場違いに簡単な問題。
max_lineno, max_number = 0, 0
lineno = 0
DATA.each_line{|line| line.chomp!; next if line.empty?; lineno += 1
base, exp = *line.split(",", 2).map(&:to_i)
number = exp * Math.log(base)
max_lineno, max_number = lineno, number if max_number < number
}
p max_lineno
__END__
content of base_exp.txt here.
スクリプトは不完全。4つの数{1,2,5,8}と四則演算とカッコを使って 36を作る方法がわからない。
Ops = [:+, :-, :*, :/].freeze
Ops3 = Ops.product(Ops, Ops).freeze
max_fnxn = 0 # fnxn: first non-expressive number
max_fnxn_c4 = nil
(0..9).to_a.combination(4){|c4|
xns = [true]
c4.permutation.to_a.product(Ops3).each{|_| nums, ops = *_
num = nums.last.to_f
3.times{|i|
num = num.send(ops[i], nums[i])
}
xns[num.to_i] = true if num.finite? and 0 < num.to_i and num.ceil == num.floor
} # 24 * 64
fnxn = xns.index(nil) || xns.length
if max_fnxn < fnxn
max_fnxn = fnxn
max_fnxn_c4 = c4
end
} # 210
puts "{#{max_fnxn_c4.sort.join(',')}} forms 1 to #{max_fnxn-1} numbers."
例えば、(5 - (1/2)) * 8 = 36。数字の順列 ABCDと演算子の順列___をそのまま連結して A_(B_(C_D)) を作るだけでは全ての式を網羅できてなかった。左右対称になるのを除いて、次の 3通りに順列と順列を組み合わせた。((A_B)_C)_D, (A_(B_C))_D, (A_B)_(C_D)
Ops = [:+, :-, :*, :/].freeze
Ops3 = Ops.product(Ops, Ops).freeze
Evaluate = lambda{|exp|
s = []
until exp.empty?
h = exp.shift
if h.kind_of?(Symbol)
s[-2] = s[-2].send(h, s[-1])
s.pop
else
s.push(h)
end
end
return s[-1]
}
max_fnxn = 0 # fnxn: first non-expressive number
max_fnxn_c4 = nil
(0..9).to_a.combination(4){|c4|
xns = [true]
c4.map(&:to_f).permutation.to_a.product(Ops3).each{|_| nums, ops = *_
[
[nums[0], nums[1], ops[0], nums[2], ops[1], nums[3], ops[2]], # ((A_B)_C)_D
[nums[0], nums[1], nums[2], ops[0], ops[1], nums[3], ops[2]], # (A_(B_C))_D
[nums[0], nums[1], ops[0], nums[2], nums[3], ops[1], ops[2]] # (A_B)_(C_D)
].each{|exp|
num = Evaluate.call(exp)
xns[num.to_i] = true if num.finite? and 0 < num.to_i and num.ceil == num.floor
} # 3
} # 24 * 64
fnxn = xns.index(nil) || xns.length
if max_fnxn < fnxn
max_fnxn = fnxn
max_fnxn_c4 = c4
end
} # 210
puts "{#{max_fnxn_c4.sort.join(',')}} forms 1 to #{max_fnxn-1} numbers."