making sure that every new position is different from previous two positions - ios

Question

Below is a method I wrote that takes a random number and makes sure that a sprite does repeat consecutively at the same position. I want to change it so that every new sprite takes a different position of the other two. I am not really getting it right. Please help.
- (float)randomlyChooseXValue {
CGSize s = [[CCDirector sharedDirector] winSize];
int randX = arc4random() % 3;
if (oldRandX != randX) {
if (randX == 0) {
xPos = xPos1*(s.width/480.0);
} else if (randX == 1) {
xPos = xPos2*(s.width/480.0);
} else {
xPos = xPos3*(s.width/480.0);
}
oldRandX = randX;
} else {
[self randomlyChooseXValue];
}
return xPos;
}

Answer 1

If I understand this correctly you need to find 3 random values and the 3rd one should be different then the 1st 2. If this is true you need to store last 2 random values and compare them in the 1st if statement:
- (float)randomlyChooseXValue {
CGSize s = [[CCDirector sharedDirector] winSize];
int randX = arc4random() % 3;
if ((oldRandX1 != randX) && (oldRandX2 != randX)) { //check for 2 values
if (randX == 0) {
xPos = xPos1*(s.width/480.0);
} else if (randX == 1) {
xPos = xPos2*(s.width/480.0);
} else {
xPos = xPos3*(s.width/480.0);
}
oldRandX2 = oldRandX1; //store 1st value to 2nd place
oldRandX1 = randX; //store new value to 1st place
} else {
[self randomlyChooseXValue];
}
return xPos;
}

Answer 2

Since you explained that you are okay with the sequence repeating, then you only need to really make two random choices: the first position, and the direction.
// somewhere, initialize global oldRandX = -1, dir = -1
- (float)randomlyChooseXValue {
CGSize s = [[CCDirector sharedDirector] winSize];
if (oldRandX < 0) {
oldRandX = arc4random() % 3;
dir = arc4random() % 2;
} else if (dir) {
oldRandX = (oldRandX + 1) % 3;
} else {
oldRandX = (oldRandX + 2) % 3;
}
if (randX == 0) {
xPos = xPos1*(s.width/480.0);
} else if (randX == 1) {
xPos = xPos2*(s.width/480.0);
} else {
xPos = xPos3*(s.width/480.0);
}
return xPos;
}
This can generate every possible sequence of the three positions:
0, 1, 2
0, 2, 1
1, 2, 0
1, 0, 2
2, 0, 1
2, 1, 0
and repeat them.
The other answer will achieve the same results, but like your original method, it might take several tries to get it. Your random function can keep choosing the wrong value an unbounded number of times; it becomes a worse problem when there's only one right number to pick.