If Statement Efficiency Help

Newbie & Debugging Questions
#1

ok, so I have an if statement that is huge :

if( slotNo[0] == 1 && slotNo[1] != 1 && slotNo[2] != 1 && slotNo[3] != 1 && slotNo[4] != 1 && slotNo[5] != 1)
		{
			winnings = 2;
		}

and I would like to know if there is a more efficient way of doing this if statement. I don’t even know how reliable that statement is.

My if statements are gonna get worse too, because I’m going to have a lot of or’s and ands to check.

So if anyone has any advice on how I could do this more reliably and more efficiently would be very appreciated.

#2

If you are doing any unnecessary checking, then you can remove that and it’ll be more efficient. In practice, however, a few things in an if statement won’t change much - you should be thinking more about if this is the right way (fits into your game model) rather than if it’s efficient. If your program is being slow 99% it’s not this.

#3

That’s okay then :slight_smile: I’m doing a slot machine game, and therefore, the single cherry can appear anywhere on the screen and it’ll give 5 credits, so i need to check if it appears in only one place by checking what slotNo[] it appears in and that it doesnt appear in any other one.

#4

What you need is a frequency table (or histogram)


int[] histogram = new int[numSlotIcons];
for(int slotVal: slotNo) {
   histogram[slotVal]++;
}



if(histogram[1] == 1) {
   // only 1 slot had a '1'
}
if(histogram[3] == 1) {
   // only 1 slot had a '3'
}

for(int freq: histogram) {
   if(freq == numSlotIcons) {
       // all slots had the same value
   }
}
#5

Ok, if you had a lot of slots, this could be more managable:


int b=slotNo[0];
for(int i=1;i<slotNo.lenght;i++) b*=slotNo[i];
if(b==1) winnings=2;

In the end, your current if statment is faster.
If that would matter in your context.

#6

riven could you explain a little bit more about your code, i understand that its kind of like a look up table, but i dont understand the part where i enter the values into it and how to look into it.

Also, Damocles, what does your b stand for?

#7

sorry, did not see that it was !=1
forget that code sample

#8

Say you have 5 slots, with the values: 1,3,2,2,0

Now you create a frequency table, with 4 slots (for the values 0,1,2,3)

For every slot value, you increment the entry in the frequency table:
freqTable[1] += 1
freqTable[3] += 1
freqTable[2] += 1
freqTable[2] += 1
freqTable[0] += 1

Your frequency table now holds these values:
1,1,2,1 (1x0, 1x1, 2x2, 1x3)

If you want to know how often ‘1’ was found in the slots, you look it up in the frequency table:
int numberOfOnes = freqTable[1];
int numberOfTwos = freqTable[2];

If numberOfOnes equals 1, then you know that ‘1’ was only found in 1 slot, regardless of which slot it was.

#9

Thanks riven, that’ll help so much when i come to check for more than one of the same slot at the same time :smiley:

#10

I used this approach in a Poker AI (for Java4K)

Checking for something like Full House becomes trivial:


// Ace=12, King=11, Queen=10, Jack=9, 10=8, ..., 2=0
int minDigit = 100;
int maxDigit = 0;
int maxNumberSet = 0;
int maxSymbolSet = 0;
int[] freqCardNumbers = new int[13];
int[] freqCardSymbols = new int[4];
for(Card card: hand) {
   maxNumberSet = Math.max(maxNumberSet, freqCardNumbers[card.digit] += 1);
   maxSymbolSet = Math.max(maxSymbolSet, freqCardSymbols[card.symbol] += 1);
   minDigit = Math.min(minDigit, card.digit);
   maxDigit = Math.max(maxDigit, card.digit);
}



// build histogram of histogram (for pairs check)
int[] freqSetsOfNumbers = new int[4+1];
for(int freq: freqCardNumbers) {
   freqSetsOfNumbers[freq]++;
}

// build histogram of histogram (for flush check)
int[] freqSetsOfSymbols = new int[hand.length+1];
for(int freq: freqCardSymbols) {
   freqSetsOfSymbols[freq]++;
}

if(maxNumberSet == 1 && (maxDigit-minDigit) == 5 && maxSymbolSet == 5) {
   if(maxDigit == 12) {
      // royal flush
   }
   else {
      // straight flush
   }
}
else if(freqSetsOfNumbers[4] == 1) {
   // four of a kind
}
else if(freqSetsOfNumbers[2] == 1 && freqSetsOfNumbers[3] == 1) {
   // full house
}
else if(freqSetsOfSymbols[5] == 1) {
   // flush
}
else if(maxNumberSet == 1 && (maxDigit-minDigit) == 5) {
   // straight
}
else if(freqSetsOfNumbers[3] == 1) {
   // three of a kind
}
else if(freqSetsOfNumbers[2] == 2) {
   // two pairs
}
else if(freqSetsOfNumbers[2] == 1) {
   // one pair
}
else {
   // high card
}

Ofcourse the code in Poker4K was more complex, as it had to compare two hands of equal type.

#11

Well my goal is to create a range of casino games, and my next game idea was a video poker game, so this approach is very helpful thank you, i thought that my coding was going to become complicated with a lot of if statements, containing a lot of different conditions.