Is this one-liner correct for calculating the middle int between a, b and c?
Note: I don't want to calculate the mean, just the one that's in the middle if these 3 were ordered from smallest to greatest (median).
Example: the middle value of 1, 77 and 34 is 34.
public static int midpoint(int a, int b, int c) {
return a <= b ? (c <= a ? a : c <= b ? c : b ) : c <= b ? b : c <= a ? c : a;
}
I tried three various options and counted time taken by each one of them in the iteration of 1,000,000
All of them give same result but 2 and 3 are executed in almost zero milliseconds
Performance Test
int iterationCount = 1000000; // 1 million iterations
Long t1 = Calendar.getInstance().getTimeInMillis();
for (int i = 0; i < iterationCount; i++) {
middle = a <= b ? (c <= a ? a : c <= b ? c : b) : c <= b ? b : c <= a ? c : a;
}
Long t2 = Calendar.getInstance().getTimeInMillis();
// #2 Using Collection + List + Sort
Long t3 = Calendar.getInstance().getTimeInMillis();
for (int i = 0; i < iterationCount; i++) {
List<Integer> list = Arrays.asList(a, b, c);
Collections.sort(list);
middle = list.get(1);
}
Long t4 = Calendar.getInstance().getTimeInMillis();
// #3Using Math's min and max
Long t5 = Calendar.getInstance().getTimeInMillis();
for (int i = 0; i < iterationCount; i++) {
middle = Math.max(Math.min(a, b), Math.min(Math.max(a, b), c));
}
Long t6 = Calendar.getInstance().getTimeInMillis();
System.out.println("Time Taken #1 -> " + (t2 - t1));
System.out.println("Time Taken #2 -> " + (t4 - t3));
System.out.println("Time Taken #3 -> " + (t6 - t5));
Output
Time Taken #1 -> 16
Time Taken #2 -> 71
Time Taken #3 -> 6
Looking at the output the option #3 is the best in terms of the performance.
Best of three
middle = Math.max(Math.min(a, b), Math.min(Math.max(a, b), c));
Is this one-liner correct?
Let's step through it:
return a <= b ? (c <= a ? a : c <= b ? c : b ) : c <= b ? b : c <= a ? c : a;
suppose a <= b. Then if c <= a, return a. Thus, c <= a <= b and a is the correct thing to return. On the other hand, if c <= b, then c <= b < a (b/c we know that c <= a is not true). So, b is appropriate to return.
Now suppose a <= b is false - then b > a. Now if c <= b, we have c <= b < a, so b is correct. If c <= b is false, then we have c > b and b > a so we will return a. Notice that you have an extra comparison. (we already know that b > a, and we've just learned that c > b, which gives us the ordering)
Notice two things about this one-liner: while it turned out that your logic was correct, it was difficult and time-consuming to determine that correctness, and it doesn't scale (imagine this one-liner for five items...)
The correct approach is to sort the list and take the middle item.
(which is exactly what you asked for: "just the one that's in the middle if these 3 were ordered from smallest to greatest.")
Math.max(Math.min(a, b), Math.min(Math.max(a, b), c));
Works for all cases.