Mastermind Help

[cssc]

Hey all, I'm trying to make a mastermind game (at my friends request) but I'm having trouble computing the amount of white pegs (a number that is right, but in the wrong space). I can find the red pegs easily, but can someone help with the white?

Here is the code i'm using to find the red and white pegs that comes closest to being right each time: L2 is the guess W is white pegs R is red pegs

[Mapar007]

(assuming L1 contains the correct code)

DelvarWFor(X,1,4
If max(L1(X)=L2) and not(L1(X)=L2(X
W=W+1
End

(untested, unsure, basic is a long time ago )

[cssc]

It seems to work for what tests I've tried with it, thank you very much!

[darkstone knight]

[woodswolf]

Haha, I remember coding like that when I first came in touch with a ti84+.

I was like:" Why the hell ain't this working!"

[Graphmastur]

[TI-newb]

im gonna go into C++... but i really like Basic right now. so C++ can come later XD *Plus, u can't play C++ games in Class with ur calculator. Lolsssss*

[Graphmastur]

I would start out with an easier language like java.

[TI-newb]

where would i get a compiler? plz send me a link XD and i'l see if i wanna do java *u might hafta teach me unless thers some Readme.

[Graphmastur]

http://www.netbeans.org/

http://www.netbeans.org/kb/

[cssc]

On a side note, the code doesn't seem to work :(

If the target code is {6,2,4,1}
and I input {1,2,3,4}

I should get one red peg, which I do, and two white pegs for the 1 and the 4. However the code only gives me one white peg.

Can anyone see what is wrong?

[DarkerLine]

So the red pegs calculation is trivial, sum(L1=L2→R, as you've already figured out. For the white pegs, the most straightforward thing to do is this:

DelVar WFor(X,1,4
For(Y,1,4
If L1(X)=L2(Y
Then
W+1→W
0→L2(Y
End
End
End

This modifies L2, so be sure to do anything else you need to do with it (such as calculating R) first. Or make a copy.

I might think of some clever way to do it later, but this is the very simple way, and is sure to correctly compare multiple matches of the same number. Also, I'm assuming 0 is not allowed in the sequence: if it is, replace 0→L2(Y with -1→L2(Y or something like that.

Edit:

Here's a different way of doing this. This time, the loop is over all possible symbols, which I am again assuming is 1 through 9 (if it's something else, change the limits, highlighted in red):

sum(seq(min(sum(L1=X),sum(L2=X)),X,1,9
Ans-R→W

[cssc]

Wow that worked perfectly, I understand how the first version works, but the second is a little bit beyond my comprehension haha. Thank you very much

[DarkerLine]

The second version is fairly easy to understand as well, it's just in a condensed form that you might not be familiar with. The idea is that for each number, we count how many times it appears in L1, then count how many times it appears in L2 , then take the minimum of these. The result is the number of pegs that that number contributes. This is what the min(sum(X=L1),sum(X=L2)) part does.

Then we sum over all possible numbers, finding the total number of pegs. If we subtract the red pegs (which we already know how to count) we get the number of white pegs.

[cssc]

Oh ok, that's not bad at all. I guess I just get afraid when I see you or Weregoose use that many commands in one line ha. Thanks again