Rose: I think it game-in-1 has been posted several months ago. And considering that there are only 32768 different codes, and 41817 games have been played, one would expect 2 - 3 games where the code was solved on the first try.
I thought this variant up myself, but after I thought of it, I also saw a reference to it as a modification of Mastermind in a journal paper somewhere. The idea is that after Player 1 submits their guess, Player 2 submits their guess and also has the option of choosing the number of black and white pegs seen by Player 1 for their guess (so the response could be a lie). Each player would only get one lie per game. Also, obviously if the code was guessed correctly, no lie could be used.
AbigailII: Aargh, I would never be able to play giant Logic with colours, whatever the colours chosen :-) But yes, an alternate piece set would be nice ! Otherwise, something should be done about this grey colour. Orange would be OK if well half-way between yellow and red, green a bit more problematic for colour-blinds. A darker grey would probably do it for everybody. Anyway, a set with letters or numbers would be great. There are probably people who are more colour-blind than me here.
Serendipity: Indeed, the colours are: white, gray, brown, blue, black, red, yellow and pink. My preference would be to replace gray with green (or perhaps orange), as I sometimes find it hard to distinguish between gray and white. But who knows, perhaps one day we'll see 'Giant Logic' with green, light-blue, orange, purple, beige, maroon and violet as additionally colours.
Having said that, perhaps we could have alternative game pieces for Logic; for instance balls with numbers or letters (in contrasting colours, black letters on white balls for instance). We have game piece options for other games as well.
I don't remember whether I have asked that before, but could we please have colours that are easier to distinguish for colour-blind people ? As they are, I can hardly distinguish the pink from the light green, while all others are OK (hint : most colour-blind people have a red vs green deficiency).
Gordon Shumway: The ability to leave an empty spot is identical to adding another colour.
The general formula for the number of codes with N spots and C colours is CN for regular logic; for unique logic, the formula is C!/(C-N)!, assuming there are at least as many colours as there are spots (otherwise, you wouldn't have any possible codes).
coan.net: Currently, Logic has 32768 different codes to choose from. Most games seem to finish in 5 or 6 moves. Your "small" logik only has 4096 possibilities (and that's assuming there isn't a reduction in the number of colours). Most games wouldn't go past move 4. "Large" logik would up the number of codes to 262144. Probably requiring 6 to 7 moves.
As for unique logik, with 8 colours and 5 positions, there would be 6720 codes to choose from. Just over 1 in 5 of the possible codes does not duplicate a colour.
Once a color peg is identified by the game program, and the program issues a black peg is issued to a solution place, the program ignores it? I don't think so. If you move that peg in your line up to an incorrect position, the black peg goes away.
BIG BAD WOLF:The pegs did not go in the right position on my first finished game,if you want the game ID# when you get a correct position the peg is suppose to be placed there,which is not the case,they move from left to right regardless of the peg color.
I hope someone can clarify my mind in some way,or do i just not understand the game.
This may be a dumb question, but can someone explain why it is necessary to wait until both players have made a guess to see the pegs for that guess? Say I am player 1, why can't I hit Move and stay here, and see the results of that guess? How would this offer me an advantage over player 2??
I am sure this is one of those things I will say, "DUH" to myself once it is explained. LOL
Czuch Czuckers: Well I would not want to help on any ongoing games (as to not cheat), but I've played this game on other sites before playing here and have my own way of playing - when I get some more time, I'll write up a short article on how I play - maybe a new addition to brainrook.com site.
BIG BAD WOLF: Thanks, I think that helps a little, i will just have to play some more and figure it all out!
So, if my opponent uses all red and I use two reds in my guess, I will know that I have two colors correct and in the correct position, but I will not know for sure if either one of the red ones are correct or not?
Czuch Czuckers: example: you guess all red - opponent has 1 red in code
You get 1 black peg since one of your guesses match the opponents pieces as the same color & same place.
You can not get more then 1 peg for each of your guesses. Black pegs overrule.
.... wait, let me explain this differently.
First thing computer does is looks for exact matches in your guess and opponenets placement. If the computer finds an exact match (same color & same position), the computer gives you a black peg and from that point on ignores that position.
So from then on, the computer only looks at the remaining 4 positions.
After the computer finds no more exact matches (black pegs), it then starts to look for any color matches in the wrong position (again, only at the 4 remaining positions) If it finds any, it will then give you a white peg and then "ignore" those pieces in the rest of it's findings.
The game is finished when one or both codes are solved. Since the results of moves are determined at the same time, it is possible that both players solve the opponent's code at the same turn. In this case, the game is a draw. Otherwise, if one player solves the code, he wins the game.
Since this takes away the advantage of the player going first, I'm glad this rule is there.
I guess I am confused where, if I guess all red, and my opponent has used one red, then I have one guess that is the correct color and is in the right position, and 4 colors that are the right color but in the wrong position. Why only one black peg then?
Say for example my opponent has a red one in the last position and I guess that color in the first position, I will get a white peg for that, but if I guess a red one in both the first and last positions, The one in the first position doesnt get anything this time?