return(n <2)?3:((l[0]> l[1])*2| l[0]>= l[1])& f(n -1, l +1);
}
int f_ungolfed(int n,int* l){
int case1 =0, case2 =0, recursion =0;
if(n <2){// Analogous to the ternary conditional I used - n < 2 means we have a single-element/empty list
return3;// Handles the vacuous-truth scenario for single lists
}else{
case1 = l[0]> l[1];// The first case - are the two numbers in the list strictly decreasing? (case1 is 1 if so)
case2 = l[0]>= l[1];// The second case - are the two numbers strictly non-increasing (case2 is 1 if so)
recursion = f_ungolfed(n -1, l +1);// Recursively call ourselves on the "rest" of the list (that is, everything other than the first element). Consider that comparison is transitive, and that we already have a vacuous-truth scenario covered.
case1 *=2;// Shift case1's value over to the left by one bit by multiplying by 2. If case1 was 1 (0b01), it's now 2 (0b10) - otherwise it's still 0 (0b00)
return(case1 | case2)& recursion;
// The bitwise OR operator (|) will combine any 1-bits from case1's value (either 0b10 or 0b00) or case2's value (either 0b01 or 0b00) into either 3, 2, 1, or 0 (0b11, 0b10, 0b01, or 0b00 respectively).
// The bitwise AND operator (&) will combine only matching 1-bits from (case1|case2) and the return value of the recursive call - if recursion = 0b11 and case1|case2 = 0b01, then the return value will be 0b01.