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Still looking for other suggestions, though. If I can't find a good fix, I might force myself to dust off my Python books.
ah, the experiences of lists of random numbers. Javascript makes something like this super easy:Code: [Select]var numArr = Array(20);
for(var i = 0; i < numArr.length;i++){
numArr[i] = i;
}
numArr.sort(function(){return Math.random()-.5;});
console.log(numArr);
It is impossible to randomly shuffle three values in a bounded number of coin tosses. Thus, this method either makes some permutations more probable than others or it uses an unbonded number of comparisons, depending on how "sort" works. For example, on my browser,Code: [Select]ile = Array(); for(t=0; t<1000000; t++) { arr = Array(1,2,3); arr.sort(function(){return Math.random()-.5;}); s=arr[0]+","+arr[1]+","+arr[2]; if(ile[s]) ile[s]++; else ile[s] = 1; } ilereturns 1,2,3 and 2,1,3 twice as frequently as other permutations.
EDIT: With 4 elements some permutations turn out to appear 180 times more frequently than others. Even if you look just at the first element: 1 and 2 appear with probability 28%, 3 with 18% and 4 with 25%. The probabilities of choosing each element seem out to be more and more different when the number of elements increased (with 17 elements, 1..17, they range from 3.5% for 6 to 15.8% for 1). If you wrote a game using this, it could be completely different on a different browser.
var ile = {};
for(var t = 0; t < 1000000; t++){
var arr = [1,2,3];
var arr2 = [];
var s;
while(arr.length > 0){
var index = (Math.random() * arr.length) >> 0;
arr2.push(arr[index]);
if(arr2.length == 1){
s = arr2[0];
} else {
s = s + "," + arr2[arr2.length-1];
}
arr.splice(index,1);
}
if(!ile[s]){
ile[s] = 1;
} else {
ile[s]++;
}
}
ile;
1,2,3: 166352
1,3,2: 166225
2,1,3: 166572
2,3,1: 166947
3,1,2: 166894
3,2,1: 167010
var numArr = Array(20);
for(var i = 0; i < numArr.length;i++){
numArr[i] = i;
}
numArr.sort(function(){return Math.random()-.5;});
console.log(numArr);
some outputs from the above code:[15, 18, 10, 1, 7, 6, 9, 8, 4, 19, 11, 5, 12, 13, 14, 2, 3, 16, 17, 0]
[10, 17, 5, 15, 4, 9, 13, 1, 19, 11, 0, 12, 6, 2, 18, 14, 7, 3, 16, 8]
[19, 2, 3, 4, 5, 6, 8, 7, 13, 11, 0, 9, 1, 14, 15, 16, 17, 12, 10, 18]
var arr = [0,1,2,3,4,5,6,7,8,9];
var endArr = [];
while(arr.length > 0){
var loc = (Math.random() * arr.length) >> 0; //bitwise operator >> 0 does a shift 0 and gives an int. Math.random() returns a value between 0 and 1.
endArr.push(arr[loc]); //add the picked element to the destination array.
arr.splice(loc,1); //remove the just added element from the source array; arr.length is now 1 less.
}
? = doesn't matter
# = wall
X = spot of interest (and is a wall)
@ = current spot
###
#X#
?@?
+–––––––––––––––+
+––––––––––––+roguelikes rock+–––––––––+
| +–––––––––––––––+ |
| |
| |
| a short treatise on why roguelikes |
| are the best thing since the games |
| began to exist... |
| |
| |
+––––––––––––––––––––––––––––––––––––––+
Woah, wait a minute. If you can play the same game twice, this means the game does not have permadeath. You can start over with exactly the same game world and if you make the same moves, the same sequence of events will occur. If you die, you can just repeat your game up to a safe number of turns earlier and continue as if nothing happened.
Obviously, this could be automated, but that wouldn't even be necessary in principle.
I don't need a deterministic RNG. I need just a RNG.
Huh. Somehow I missed the "self" option.
I don't like the idea of 1d(lifeMax*.3). That's excessively unreliable for such an important effect. (lifeMax*.15)d2 or a similar formula would at least ensure the player gets something in exchange for their turn and MP.
It's really easy to survive indefinitely. Just leave one zombie alive at the end of each wave and run from it until your health and magic are full. Then spam bayonet attacks and vivid touch (the only spell that matters) until there's one survivor so you can repeat the process.