I'm coding the spigot algorithm for displaying digits of pi in ada, but my output is wrong and I can't figure out why
I've tried messing with the range of my loops and different ways to output my data but nothings worked properly
with ada.integer_text_io; use ada.integer_text_io;
with Ada.Text_IO; use Ada.Text_IO;
procedure Spigot is
n : constant Integer := 1000;
length : constant Integer := 10*n/3+1;
x,q,nines,predigit :Integer :=0;
a: array (0..length) of Integer;
begin
nines:=0;
predigit:=0;
for j in 0..length loop
a(j):=2;
end loop;
for j in 1..n loop
q:=0;
for i in reverse 1..length loop
x:=10*a(i) + q*i;
a(i):= x mod (2*i-1);
q:= x/(2*i-1);
end loop;
a(1):= q mod 10;
q:=q/10;
if q = 9 then
nines:=nines+1;
elsif q = 10 then
put(predigit+1);
for k in 0..nines loop
put("0");
end loop;
predigit:=0;
nines:=0;
else
put(predigit);
predigit:=q;
if nines/=0 then
for k in 0..nines loop
put("9");
end loop;
nines:=0;
end if;
end if;
end loop;
put(predigit);
end Spigot;
so it should just be displayed at 0 3 1 4 1 5 9 2 6 5 3 5 8 9... but the output i get is 0 3 1 4 1 599 2 6 5 3 5 89... it should only be 1 digit at a time and also the outputted values for pi aren't completely correct
I don't know the algorithm well enough to talk about why the digits are off, but I did notice some issues:
Your array is defined with bounds 0 .. Length, which would give you 1 extra element
In your loop that does the calculation, you loop from 1..length, which is ok, but you don't adjust the variable i consistently. The array indices need to be one less than the i's used in the actual calculations (keep in mind they still have to be correctly in bounds of your array). For example
x:=10*a(i) + q*i;
needs to either be
x:=10*a(i-1) + q*i;
or
x:=10*a(i) + q*(i+1);
depending on what you decide your array bounds to be. This applies to multiple lines in your code. See this Stackoverflow thread
You assign A(1) when your array starts at 0
Your loops to print out "0" and "9" should be either 1..length or 0 .. length-1
When you print the digits using Integer_Text_IO.Put, you need to specify a width of 1 to get rid of the spaces
There might be more, that's all I saw.
I think you are translating this answer.
You need to be more careful of your indices and your loop ranges; for example, you’ve translated
for(int i = len; i > 0; --i) {
int x = 10 * A[i-1] + q*i;
A[i-1] = x % (2*i - 1);
q = x / (2*i - 1);
}
as
for i in reverse 1..length loop
x:=10*a(i) + q*i;
a(i):= x mod (2*i-1);
q:= x/(2*i-1);
end loop;
The loop ranges are the same. But in the seocnd line, the C code uses A[i-1], whereas yours uses a(i); similarly in the third line.
Later, for
for (int k = 0; k < nines; ++k) {
printf("%d", 0);
}
you have
for k in 0..nines loop
put("0");
end loop;
in which the C loop runs from 0 to nines - 1, but yours runs from 0 to nines. So you put out one more 0 than you should (and later on, likewise for 9s).
Also, you should use put (predigit, width=> 0).
I am self taught and thought that I understood recursion, but I can not solve this problem:
What is returned by the call recur(12)?
What is returned by the call recur (25)?
public static int recur (int y)
{
if(y <=3)
return y%4;
return recur(y-2) + recur(y-1) + 1;
}
Would someone please help me with understanding how to solve these problems?
First of all, I assume you mean:
public static int recur(int y)
but the results of this method are discovered by placing a print statement at the beginning of the method:
public static int recur(int y)
{
System.out.println(y);
if(y <=3)
return y % 4;
return recur(y-2) + recur(y-1) + 1;
}
I am not sure what you mean by what is returned because there are several returns, though. Anyway, these are the steps to figure this out:
is 12 <= 3? No
recur(10) Don't proceed to the next recursion statement yet
is 10 <= 3? No
recur(8) Don't proceed to the next recursion statement yet
Continue this pattern until y <= 3 is true. Then you return y % 4 (whatever that number may be).
Now you are ready to go to the second recursive statement in the most recent recur() call. So, recur(y - 1).
Is y <= 3? If so, return y % 4. If not, do a process similar to step 1
once you return you add the result of recur(y - 2) + recur(y - 1) + 1. This will be a number of course.
continue this process for many iterations.
Recursion is difficult to follow and understand sometimes even for advanced programmers.
Here is a very common (and similar) problem for you to look into:
Java recursive Fibonacci sequence
I hope this helps!
Good luck!
I have removed the modulus there since any nonegative n less than 4 will just become n so I ended up with:
public static int recur (int y)
{
return y <= 3 ?
y :
recur(y-2) + recur(y-1) + 1;
}
I like to start off by testing the base case so what happens for y when they are 0,1,2,3? well the argument so 0,1,2,3 of course.
What about 4? Well then it's not less or equal to 3 and you get to replace it with recur(4-2) + recur(4-1) + 1 which is recur(2) + recur(3) + 1. Now you can solve each of the recur ones since we earlier established became its argument so you end up with 2 + 3 + 1.
Now doing this for 12 or 25 is exactly the same just with more steps. Here is 5 which has just one more step:
recur(5); //=>
recur(3) + recur(4) + 1; //==>
recur(3) + ( recur(2) + recur(3) + 1 ) + 1; //==>
3 + 2 + 3 + 1 + 1; // ==>
10
So in reality the recursion halts the process in the current iteration until you have an answer that the current iteration adds together so I could have done this in the opposite way, but then I would have paused every time I now have used a previous calculated value.
You should have enough info to do any y.
This is nothing more than an augmented Fibonacci sequence.
The first four terms are defined as 0, 1, 2, 3. Thereafter, each term is the sum of the previous two terms, plus one. This +1 augmentation is where it differs from the classic Fibonacci sequence. Just add up the series by hand:
0
1
2
3
3+2+1 = 6
6+3+1 = 10
10+6+1 = 17
17+10+1 = 28
...
I was asked on an interview to find all numbers only divisible by 3, 5 and 7. I purposed we can make check like
if (num%3==0 || num%5==0 || num%7==0)
return true
else
return false.
But in this case if we have 6 it will pass the test but its also divisible by 2 so this doesn't work. Can you purpose something?
I am using java. Find mean to check if some number is divisible only to this number
I would approach this by removing all of the factors of 3, 5, and 7 from the original number, and seeing what's left.
while(num % 3 == 0)
{
num = num / 3;
}
while(num % 5 == 0)
{
num = num / 5;
}
while(num % 7 == 0)
{
num = num / 7;
}
return (num == 1);
I won't give you a Java algorithm, as it should be fairly easy to implement.
You can just:
1. check if (n%3 == 0)
2. if it is, set n /= 3 and repeat step 1.
3. do the same for the number 5 and 7
4. now if n != 1, return false, else return true
In a Java algorithm:
// n is some random natural number
if (n == 1 || n == 0)
return false
while (!n%3)
{
n /= 3;
}
while (!n%5)
{
n /= 5;
}
while (!n%7)
{
n /= 7;
}
if (n == 1)
{
return true;
}
else
{
return false;
}
It's not the best syntax, I'm just giving an straight-forward implementation of the algorithm presented above.
We first note that 1 is a member of the set. Although it is not divisible by 3, 5 or 7, neither is it divisible by any number other than 3, 5 or 7, so we will say that 1 is in the set. This conforms to the mathematical definition of the set { x = 3i · 5j · 7k | i, j, k ≥ 0 }.
One method is to count from 1, adding 2 at each step, and checking if the number is divisible only by 3, 5 and 7. That's slow because it does a lot of work that immediately gets discarded, since there are many fewer numbers divisible only by 3, 5 and 7 than there are odd numbers.
A better approach is to generate the desired numbers directly, by induction. The number 1 is in the set, and for any x in the set, so are 3 x, 5 x and 7 x. So the algorithm to generate all numbers divisible only by 3, 5 and 7, in order, is:
1. Initialize a priority queue with the number 1.
2. Pop the smallest number in the priority queue, call it x.
3. Add 3x, 5x and 7x to the priority queue.
4. Output x as the next integer in the set.
5. If you want more output, go to Step 2.
6. Halt.
I implemented both algorithms; you can see them at http://ideone.com/YwnAQ8. The brute-force method takes a little over ten seconds to find the 203 members of the 3,5,7 set less than a million; the priority queue does the same calculation in a hundredth of a second, a thousand times faster. The priority queue implementation used there is explained at my blog. You can also see the set of 3,5,7 numbers at OEIS.
A question I got on my last interview:
Design a function f, such that:
f(f(n)) == -n
Where n is a 32 bit signed integer; you can't use complex numbers arithmetic.
If you can't design such a function for the whole range of numbers, design it for the largest range possible.
Any ideas?
You didn't say what kind of language they expected... Here's a static solution (Haskell). It's basically messing with the 2 most significant bits:
f :: Int -> Int
f x | (testBit x 30 /= testBit x 31) = negate $ complementBit x 30
| otherwise = complementBit x 30
It's much easier in a dynamic language (Python). Just check if the argument is a number X and return a lambda that returns -X:
def f(x):
if isinstance(x,int):
return (lambda: -x)
else:
return x()
How about:
f(n) = sign(n) - (-1)ⁿ * n
In Python:
def f(n):
if n == 0: return 0
if n >= 0:
if n % 2 == 1:
return n + 1
else:
return -1 * (n - 1)
else:
if n % 2 == 1:
return n - 1
else:
return -1 * (n + 1)
Python automatically promotes integers to arbitrary length longs. In other languages the largest positive integer will overflow, so it will work for all integers except that one.
To make it work for real numbers you need to replace the n in (-1)ⁿ with { ceiling(n) if n>0; floor(n) if n<0 }.
In C# (works for any double, except in overflow situations):
static double F(double n)
{
if (n == 0) return 0;
if (n < 0)
return ((long)Math.Ceiling(n) % 2 == 0) ? (n + 1) : (-1 * (n - 1));
else
return ((long)Math.Floor(n) % 2 == 0) ? (n - 1) : (-1 * (n + 1));
}
Here's a proof of why such a function can't exist, for all numbers, if it doesn't use extra information(except 32bits of int):
We must have f(0) = 0. (Proof: Suppose f(0) = x. Then f(x) = f(f(0)) = -0 = 0. Now, -x = f(f(x)) = f(0) = x, which means that x = 0.)
Further, for any x and y, suppose f(x) = y. We want f(y) = -x then. And f(f(y)) = -y => f(-x) = -y. To summarize: if f(x) = y, then f(-x) = -y, and f(y) = -x, and f(-y) = x.
So, we need to divide all integers except 0 into sets of 4, but we have an odd number of such integers; not only that, if we remove the integer that doesn't have a positive counterpart, we still have 2(mod4) numbers.
If we remove the 2 maximal numbers left (by abs value), we can get the function:
int sign(int n)
{
if(n>0)
return 1;
else
return -1;
}
int f(int n)
{
if(n==0) return 0;
switch(abs(n)%2)
{
case 1:
return sign(n)*(abs(n)+1);
case 0:
return -sign(n)*(abs(n)-1);
}
}
Of course another option, is to not comply for 0, and get the 2 numbers we removed as a bonus. (But that's just a silly if.)
Thanks to overloading in C++:
double f(int var)
{
return double(var);
}
int f(double var)
{
return -int(var);
}
int main(){
int n(42);
std::cout<<f(f(n));
}
Or, you could abuse the preprocessor:
#define f(n) (f##n)
#define ff(n) -n
int main()
{
int n = -42;
cout << "f(f(" << n << ")) = " << f(f(n)) << endl;
}
This is true for all negative numbers.
f(n) = abs(n)
Because there is one more negative number than there are positive numbers for twos complement integers, f(n) = abs(n) is valid for one more case than f(n) = n > 0 ? -n : n solution that is the same same as f(n) = -abs(n). Got you by one ... :D
UPDATE
No, it is not valid for one case more as I just recognized by litb's comment ... abs(Int.Min) will just overflow ...
I thought about using mod 2 information, too, but concluded, it does not work ... to early. If done right, it will work for all numbers except Int.Min because this will overflow.
UPDATE
I played with it for a while, looking for a nice bit manipulation trick, but I could not find a nice one-liner, while the mod 2 solution fits in one.
f(n) = 2n(abs(n) % 2) - n + sgn(n)
In C#, this becomes the following:
public static Int32 f(Int32 n)
{
return 2 * n * (Math.Abs(n) % 2) - n + Math.Sign(n);
}
To get it working for all values, you have to replace Math.Abs() with (n > 0) ? +n : -n and include the calculation in an unchecked block. Then you get even Int.Min mapped to itself as unchecked negation does.
UPDATE
Inspired by another answer I am going to explain how the function works and how to construct such a function.
Lets start at the very beginning. The function f is repeatedly applied to a given value n yielding a sequence of values.
n => f(n) => f(f(n)) => f(f(f(n))) => f(f(f(f(n)))) => ...
The question demands f(f(n)) = -n, that is two successive applications of f negate the argument. Two further applications of f - four in total - negate the argument again yielding n again.
n => f(n) => -n => f(f(f(n))) => n => f(n) => ...
Now there is a obvious cycle of length four. Substituting x = f(n) and noting that the obtained equation f(f(f(n))) = f(f(x)) = -x holds, yields the following.
n => x => -n => -x => n => ...
So we get a cycle of length four with two numbers and the two numbers negated. If you imagine the cycle as a rectangle, negated values are located at opposite corners.
One of many solution to construct such a cycle is the following starting from n.
n => negate and subtract one
-n - 1 = -(n + 1) => add one
-n => negate and add one
n + 1 => subtract one
n
A concrete example is of such an cycle is +1 => -2 => -1 => +2 => +1. We are almost done. Noting that the constructed cycle contains an odd positive number, its even successor, and both numbers negate, we can easily partition the integers into many such cycles (2^32 is a multiple of four) and have found a function that satisfies the conditions.
But we have a problem with zero. The cycle must contain 0 => x => 0 because zero is negated to itself. And because the cycle states already 0 => x it follows 0 => x => 0 => x. This is only a cycle of length two and x is turned into itself after two applications, not into -x. Luckily there is one case that solves the problem. If X equals zero we obtain a cycle of length one containing only zero and we solved that problem concluding that zero is a fixed point of f.
Done? Almost. We have 2^32 numbers, zero is a fixed point leaving 2^32 - 1 numbers, and we must partition that number into cycles of four numbers. Bad that 2^32 - 1 is not a multiple of four - there will remain three numbers not in any cycle of length four.
I will explain the remaining part of the solution using the smaller set of 3 bit signed itegers ranging from -4 to +3. We are done with zero. We have one complete cycle +1 => -2 => -1 => +2 => +1. Now let us construct the cycle starting at +3.
+3 => -4 => -3 => +4 => +3
The problem that arises is that +4 is not representable as 3 bit integer. We would obtain +4 by negating -3 to +3 - what is still a valid 3 bit integer - but then adding one to +3 (binary 011) yields 100 binary. Interpreted as unsigned integer it is +4 but we have to interpret it as signed integer -4. So actually -4 for this example or Int.MinValue in the general case is a second fixed point of integer arithmetic negation - 0 and Int.MinValue are mapped to themselve. So the cycle is actually as follows.
+3 => -4 => -3 => -4 => -3
It is a cycle of length two and additionally +3 enters the cycle via -4. In consequence -4 is correctly mapped to itself after two function applications, +3 is correctly mapped to -3 after two function applications, but -3 is erroneously mapped to itself after two function applications.
So we constructed a function that works for all integers but one. Can we do better? No, we cannot. Why? We have to construct cycles of length four and are able to cover the whole integer range up to four values. The remaining values are the two fixed points 0 and Int.MinValue that must be mapped to themselves and two arbitrary integers x and -x that must be mapped to each other by two function applications.
To map x to -x and vice versa they must form a four cycle and they must be located at opposite corners of that cycle. In consequence 0 and Int.MinValue have to be at opposite corners, too. This will correctly map x and -x but swap the two fixed points 0 and Int.MinValue after two function applications and leave us with two failing inputs. So it is not possible to construct a function that works for all values, but we have one that works for all values except one and this is the best we can achieve.
Using complex numbers, you can effectively divide the task of negating a number into two steps:
multiply n by i, and you get n*i, which is n rotated 90° counter-clockwise
multiply again by i, and you get -n
The great thing is that you don't need any special handling code. Just multiplying by i does the job.
But you're not allowed to use complex numbers. So you have to somehow create your own imaginary axis, using part of your data range. Since you need exactly as much imaginary (intermediate) values as initial values, you are left with only half the data range.
I tried to visualize this on the following figure, assuming signed 8-bit data. You would have to scale this for 32-bit integers. The allowed range for initial n is -64 to +63.
Here's what the function does for positive n:
If n is in 0..63 (initial range), the function call adds 64, mapping n to the range 64..127 (intermediate range)
If n is in 64..127 (intermediate range), the function subtracts n from 64, mapping n to the range 0..-63
For negative n, the function uses the intermediate range -65..-128.
Works except int.MaxValue and int.MinValue
public static int f(int x)
{
if (x == 0) return 0;
if ((x % 2) != 0)
return x * -1 + (-1 *x) / (Math.Abs(x));
else
return x - x / (Math.Abs(x));
}
The question doesn't say anything about what the input type and return value of the function f have to be (at least not the way you've presented it)...
...just that when n is a 32-bit integer then f(f(n)) = -n
So, how about something like
Int64 f(Int64 n)
{
return(n > Int32.MaxValue ?
-(n - 4L * Int32.MaxValue):
n + 4L * Int32.MaxValue);
}
If n is a 32-bit integer then the statement f(f(n)) == -n will be true.
Obviously, this approach could be extended to work for an even wider range of numbers...
for javascript (or other dynamically typed languages) you can have the function accept either an int or an object and return the other. i.e.
function f(n) {
if (n.passed) {
return -n.val;
} else {
return {val:n, passed:1};
}
}
giving
js> f(f(10))
-10
js> f(f(-10))
10
alternatively you could use overloading in a strongly typed language although that may break the rules ie
int f(long n) {
return n;
}
long f(int n) {
return -n;
}
Depending on your platform, some languages allow you to keep state in the function. VB.Net, for example:
Function f(ByVal n As Integer) As Integer
Static flag As Integer = -1
flag *= -1
Return n * flag
End Function
IIRC, C++ allowed this as well. I suspect they're looking for a different solution though.
Another idea is that since they didn't define the result of the first call to the function you could use odd/evenness to control whether to invert the sign:
int f(int n)
{
int sign = n>=0?1:-1;
if (abs(n)%2 == 0)
return ((abs(n)+1)*sign * -1;
else
return (abs(n)-1)*sign;
}
Add one to the magnitude of all even numbers, subtract one from the magnitude of all odd numbers. The result of two calls has the same magnitude, but the one call where it's even we swap the sign. There are some cases where this won't work (-1, max or min int), but it works a lot better than anything else suggested so far.
Exploiting JavaScript exceptions.
function f(n) {
try {
return n();
}
catch(e) {
return function() { return -n; };
}
}
f(f(0)) => 0
f(f(1)) => -1
For all 32-bit values (with the caveat that -0 is -2147483648)
int rotate(int x)
{
static const int split = INT_MAX / 2 + 1;
static const int negativeSplit = INT_MIN / 2 + 1;
if (x == INT_MAX)
return INT_MIN;
if (x == INT_MIN)
return x + 1;
if (x >= split)
return x + 1 - INT_MIN;
if (x >= 0)
return INT_MAX - x;
if (x >= negativeSplit)
return INT_MIN - x + 1;
return split -(negativeSplit - x);
}
You basically need to pair each -x => x => -x loop with a y => -y => y loop. So I paired up opposite sides of the split.
e.g. For 4 bit integers:
0 => 7 => -8 => -7 => 0
1 => 6 => -1 => -6 => 1
2 => 5 => -2 => -5 => 2
3 => 4 => -3 => -4 => 3
A C++ version, probably bending the rules somewhat but works for all numeric types (floats, ints, doubles) and even class types that overload the unary minus:
template <class T>
struct f_result
{
T value;
};
template <class T>
f_result <T> f (T n)
{
f_result <T> result = {n};
return result;
}
template <class T>
T f (f_result <T> n)
{
return -n.value;
}
void main (void)
{
int n = 45;
cout << "f(f(" << n << ")) = " << f(f(n)) << endl;
float p = 3.14f;
cout << "f(f(" << p << ")) = " << f(f(p)) << endl;
}
x86 asm (AT&T style):
; input %edi
; output %eax
; clobbered regs: %ecx, %edx
f:
testl %edi, %edi
je .zero
movl %edi, %eax
movl $1, %ecx
movl %edi, %edx
andl $1, %eax
addl %eax, %eax
subl %eax, %ecx
xorl %eax, %eax
testl %edi, %edi
setg %al
shrl $31, %edx
subl %edx, %eax
imull %ecx, %eax
subl %eax, %edi
movl %edi, %eax
imull %ecx, %eax
.zero:
xorl %eax, %eax
ret
Code checked, all possible 32bit integers passed, error with -2147483647 (underflow).
Uses globals...but so?
bool done = false
f(int n)
{
int out = n;
if(!done)
{
out = n * -1;
done = true;
}
return out;
}
This Perl solution works for integers, floats, and strings.
sub f {
my $n = shift;
return ref($n) ? -$$n : \$n;
}
Try some test data.
print $_, ' ', f(f($_)), "\n" for -2, 0, 1, 1.1, -3.3, 'foo' '-bar';
Output:
-2 2
0 0
1 -1
1.1 -1.1
-3.3 3.3
foo -foo
-bar +bar
Nobody ever said f(x) had to be the same type.
def f(x):
if type(x) == list:
return -x[0]
return [x]
f(2) => [2]
f(f(2)) => -2
I'm not actually trying to give a solution to the problem itself, but do have a couple of comments, as the question states this problem was posed was part of a (job?) interview:
I would first ask "Why would such a function be needed? What is the bigger problem this is part of?" instead of trying to solve the actual posed problem on the spot. This shows how I think and how I tackle problems like this. Who know? That might even be the actual reason the question is asked in an interview in the first place. If the answer is "Never you mind, assume it's needed, and show me how you would design this function." I would then continue to do so.
Then, I would write the C# test case code I would use (the obvious: loop from int.MinValue to int.MaxValue, and for each n in that range call f(f(n)) and checking the result is -n), telling I would then use Test Driven Development to get to such a function.
Only if the interviewer continues asking for me to solve the posed problem would I actually start to try and scribble pseudocode during the interview itself to try and get to some sort of an answer. However, I don't really think I would be jumping to take the job if the interviewer would be any indication of what the company is like...
Oh, this answer assumes the interview was for a C# programming related position. Would of course be a silly answer if the interview was for a math related position. ;-)
I would you change the 2 most significant bits.
00.... => 01.... => 10.....
01.... => 10.... => 11.....
10.... => 11.... => 00.....
11.... => 00.... => 01.....
As you can see, it's just an addition, leaving out the carried bit.
How did I got to the answer? My first thought was just a need for symmetry. 4 turns to get back where I started. At first I thought, that's 2bits Gray code. Then I thought actually standard binary is enough.
Here is a solution that is inspired by the requirement or claim that complex numbers can not be used to solve this problem.
Multiplying by the square root of -1 is an idea, that only seems to fail because -1 does not have a square root over the integers. But playing around with a program like mathematica gives for example the equation
(18494364652+1) mod (232-3) = 0.
and this is almost as good as having a square root of -1. The result of the function needs to be a signed integer. Hence I'm going to use a modified modulo operation mods(x,n) that returns the integer y congruent to x modulo n that is closest to 0. Only very few programming languages have suc a modulo operation, but it can easily be defined. E.g. in python it is:
def mods(x, n):
y = x % n
if y > n/2: y-= n
return y
Using the equation above, the problem can now be solved as
def f(x):
return mods(x*1849436465, 2**32-3)
This satisfies f(f(x)) = -x for all integers in the range [-231-2, 231-2]. The results of f(x) are also in this range, but of course the computation would need 64-bit integers.
C# for a range of 2^32 - 1 numbers, all int32 numbers except (Int32.MinValue)
Func<int, int> f = n =>
n < 0
? (n & (1 << 30)) == (1 << 30) ? (n ^ (1 << 30)) : - (n | (1 << 30))
: (n & (1 << 30)) == (1 << 30) ? -(n ^ (1 << 30)) : (n | (1 << 30));
Console.WriteLine(f(f(Int32.MinValue + 1))); // -2147483648 + 1
for (int i = -3; i <= 3 ; i++)
Console.WriteLine(f(f(i)));
Console.WriteLine(f(f(Int32.MaxValue))); // 2147483647
prints:
2147483647
3
2
1
0
-1
-2
-3
-2147483647
Essentially the function has to divide the available range into cycles of size 4, with -n at the opposite end of n's cycle. However, 0 must be part of a cycle of size 1, because otherwise 0->x->0->x != -x. Because of 0 being alone, there must be 3 other values in our range (whose size is a multiple of 4) not in a proper cycle with 4 elements.
I chose these extra weird values to be MIN_INT, MAX_INT, and MIN_INT+1. Furthermore, MIN_INT+1 will map to MAX_INT correctly, but get stuck there and not map back. I think this is the best compromise, because it has the nice property of only the extreme values not working correctly. Also, it means it would work for all BigInts.
int f(int n):
if n == 0 or n == MIN_INT or n == MAX_INT: return n
return ((Math.abs(n) mod 2) * 2 - 1) * n + Math.sign(n)
Nobody said it had to be stateless.
int32 f(int32 x) {
static bool idempotent = false;
if (!idempotent) {
idempotent = true;
return -x;
} else {
return x;
}
}
Cheating, but not as much as a lot of the examples. Even more evil would be to peek up the stack to see if your caller's address is &f, but this is going to be more portable (although not thread safe... the thread-safe version would use TLS). Even more evil:
int32 f (int32 x) {
static int32 answer = -x;
return answer;
}
Of course, neither of these works too well for the case of MIN_INT32, but there is precious little you can do about that unless you are allowed to return a wider type.
I could imagine using the 31st bit as an imaginary (i) bit would be an approach that would support half the total range.
works for n= [0 .. 2^31-1]
int f(int n) {
if (n & (1 << 31)) // highest bit set?
return -(n & ~(1 << 31)); // return negative of original n
else
return n | (1 << 31); // return n with highest bit set
}
The problem states "32-bit signed integers" but doesn't specify whether they are twos-complement or ones-complement.
If you use ones-complement then all 2^32 values occur in cycles of length four - you don't need a special case for zero, and you also don't need conditionals.
In C:
int32_t f(int32_t x)
{
return (((x & 0xFFFFU) << 16) | ((x & 0xFFFF0000U) >> 16)) ^ 0xFFFFU;
}
This works by
Exchanging the high and low 16-bit blocks
Inverting one of the blocks
After two passes we have the bitwise inverse of the original value. Which in ones-complement representation is equivalent to negation.
Examples:
Pass | x
-----+-------------------
0 | 00000001 (+1)
1 | 0001FFFF (+131071)
2 | FFFFFFFE (-1)
3 | FFFE0000 (-131071)
4 | 00000001 (+1)
Pass | x
-----+-------------------
0 | 00000000 (+0)
1 | 0000FFFF (+65535)
2 | FFFFFFFF (-0)
3 | FFFF0000 (-65535)
4 | 00000000 (+0)
:D
boolean inner = true;
int f(int input) {
if(inner) {
inner = false;
return input;
} else {
inner = true;
return -input;
}
}
return x ^ ((x%2) ? 1 : -INT_MAX);
I'd like to share my point of view on this interesting problem as a mathematician. I think I have the most efficient solution.
If I remember correctly, you negate a signed 32-bit integer by just flipping the first bit. For example, if n = 1001 1101 1110 1011 1110 0000 1110 1010, then -n = 0001 1101 1110 1011 1110 0000 1110 1010.
So how do we define a function f that takes a signed 32-bit integer and returns another signed 32-bit integer with the property that taking f twice is the same as flipping the first bit?
Let me rephrase the question without mentioning arithmetic concepts like integers.
How do we define a function f that takes a sequence of zeros and ones of length 32 and returns a sequence of zeros and ones of the same length, with the property that taking f twice is the same as flipping the first bit?
Observation: If you can answer the above question for 32 bit case, then you can also answer for 64 bit case, 100 bit case, etc. You just apply f to the first 32 bit.
Now if you can answer the question for 2 bit case, Voila!
And yes it turns out that changing the first 2 bits is enough.
Here's the pseudo-code
1. take n, which is a signed 32-bit integer.
2. swap the first bit and the second bit.
3. flip the first bit.
4. return the result.
Remark: The step 2 and the step 3 together can be summerised as (a,b) --> (-b, a). Looks familiar? That should remind you of the 90 degree rotation of the plane and the multiplication by the squar root of -1.
If I just presented the pseudo-code alone without the long prelude, it would seem like a rabbit out of the hat, I wanted to explain how I got the solution.