Homework For LuLz

[John]

Simplify the following Boolean expression. The final expression should (be logically equivalent) have a weight of four (meaning F = a + b where a and b are some literal in the expression in its complimented or uncomplimented form).
' indicates the literals complement
a + b indicates a OR b
ab indicates a AND b


F = y’z’ + x’yz + x’y + xyz + xz’

[morefood2001]

@methodz: sorry to steal a question

@John, I believe that the answer is F=y+z'

[Datab0x]

@methodz - if you also do logic help I think I may hit this thread up when I get my next assignment. I dont want someone to program for me, I just find that I take way too long coming up with an attack plan for the projects. So i need help with that.

[MeTh0Dz|Reb0rn]

I'll code my solution and you can just use it as a guide.

[WingedPanther]

Originally Posted by John

Simplify the following Boolean expression. The final expression should (be logically equivalent) have a weight of four (meaning F = a + b where a and b are some literal in the expression in its complimented or uncomplimented form).
' indicates the literals complement
a + b indicates a OR b
ab indicates a AND b


F = y’z’ + x’yz + x’y + xyz + xz’

Originally Posted by morefood2001

@methodz: sorry to steal a question

@John, I believe that the answer is F=y+z'

Odd, I'm getting F=y'z' + yz + x'y + xz', assuming AND binds more strongly than OR.

[John]

There are several methods to find the solution mine is far from the simplest but it is the most explicit.

(x + x') => 1
(1 + x) => 1

(1)
F = y’z’ + x’yz + x’y + xyz + xz’
F = y’z’ +yz(x' + x) + x’y + xz’
F = y’z’ +yz + x’y + xz’

(2)
F = y’z’ +yz + x’y + xz’(y + y')
F = y’z’ +yz + x’y + xyz’ + xy'z’

(3)
F = y’z’ +yz + x’y(z + z') + xyz’ + xy'z’
F = y’z’ +yz + x’yz + x’yz' + xyz’ + xy'z’

(4)
F = y’z’ + yz + x’yz + x’yz' + xyz’ + xy'z’
F = y’z’ + (1 + x')yz + x’yz' + xyz’ + xy'z’
F = y’z’ + yz + x’yz' + xyz’ + xy'z’

(5)
F = y’z’ + yz + x’yz' + xyz’ + xy'z’
F = y’z’ + yz + (x’ + x)yz' + xy'z’
F = y’z’ + yz + yz' + xy'z’

(6)
F = y’z’ + yz + yz' + xy'z’
F = yz + yz' + (1 + x)y'z’
F = yz + yz' + y'z’

(7)
F = yz + yz' + y'z’ + yz'

(8)
F = yz + yz' + y'z’ + yz'
F = y(z + z') + yz' + y'z’
F = y + yz' + y'z’

(9)
F = y + yz' + y'z’
F = y + z'(y + y')
F = y + z'

Q.E.D.

[chili5]

Wow, I have no idea what the above math is.

If your bored:

Have the user guess at random number. Have the computer tell the user either “Higher” or “Lower” until correct. The program will show the total number of guesses. Also, if the user is a moron, tell them so. For example, assume the correct number is 12. If the user inputs 7 and your program informs the user “higher”, if at any later time the user inputs a number of 7 or lower, the program is permitted to call the user a “moron”. Call the user an idiot if they repeat a guessed number.

I haven't tried it yet, but what you come up with would be interesting.

[Aereshaa]

Since you didn't specify a language, I tried that kind of game in Ruby, the most recent language I've added to my repertoire. This one not only keeps track of maximum and minimum, but leaves the user to deduce them!

class GuessGame
    def initialize
        @correct = rand(1000) + 1 # Correct guess.
        @max = 1000 # Maximum and minimum for
        @min = 1    # "homing in" on the answer.
        @last = nil # The last guess.
        @guesses = 0
    end
    def guess (x)
        @guesses += 1
        curr = (x - @correct).abs
        if x = @correct
            print "You got it!\n"
        end
        if @last
            lastd = (@last - @correct).abs 
            if lastd < curr
                print "Colder\n"
            elsif lastd > curr
                print "Warmer\n"
                if x > @last && @last > @min 
                    @min = @last # Increase the minimum.
                elsif x < @last && @last < @max
                    @max = @last # Decrease the maximum.
                end  
            end
            if x > @max || x < @min
                print "Idiot!\n" # Problem exists between keyboard and chair.
            end 
        end
        @last = x; # This guess becomes the last guess
    end
    attr_reader :max, :min
end

game = GuessGame.new
continue = true # Continue until the user exits.
while(continue)
    print "Guess: " # Prompt.
    cmd = gets 
    if cmd =~ /exit|quit|bye/i # Exit commands; case insensitive.
        print "See ya!\n"
        exit
#   elsif cmd =~ /max|min/i # For testing.
#       print game.max, "\n", game.min, "\n"
    elsif cmd =~ /^[0-9]+/
        game.guess cmd.to_i
    end 
end

What do you think?

[WingedPanther]

@chili5: it's formal logic.

@John: I found your solution... non-intuitive so used a Venn-diagram approach: I did a Venn diagram for each of the original 5 pieces, then the union (+) and it's trivial to observe that the result is y+z'

[John]

Originally Posted by WingedPanther

@John: I found your solution... non-intuitive so used a Venn-diagram approach: I did a Venn diagram for each of the original 5 pieces, then the union (+) and it's trivial to observe that the result is y+z'

That wasn't actually my solution, but a solution provided my professor. I too find it very non-intuitive. He merely used it as an example to show it is indeed logical, but more importantly, there is a need for a simpler method. And I suppose that method is a Karnaugh map, although I have not studied it yet. When I do dive deeper into this logic, I'm sure your venn-diagrams will make more sense, and you will probably receive a question or two from me along the way.