Problem Statement: The Fibonacci word sequence of bit strings is defined as:
F(0) = 0, F(1) = 1
F(n − 1) + F(n − 2) if n ≥ 2
For example : F(2) = F(1) + F(0) = 10, F(3) = F(2) + F(1) = 101, etc.
Given a bit pattern p and a number n, how often does p occur in F(n)?
Input:
The first line of each test case contains the integer n (0 ≤ n ≤ 100). The second line contains the bit
pattern p. The pattern p is nonempty and has a length of at most 100 000 characters.
Output:
For each test case, display its case number followed by the number of occurrences of the bit pattern p in
F(n). Occurrences may overlap. The number of occurrences will be less than 2^63.
Sample input: 6 10 Sample output: Case 1: 5
I implemented a divide and conquer algorithm to solve this problem, based on the hints that I found on the internet: We can think of the process of going from F(n-1) to F(n) as a string replacement rule: every '1' becomes '10' and '0' becomes '1'. Here is my code:
#include <string>
#include <iostream>
using namespace std;
#define LL long long int
LL count = 0;
string F[40];
void find(LL n, char ch1,char ch2 ){//Find occurences of eiher "11" / "01" / "10" in F[n]
LL n1 = F[n].length();
for (int i = 0;i+1 <n1;++i){
if (F[n].at(i)==ch1&&F[n].at(i+1)==ch2) ++ count;
}
}
void find(char ch, LL n){
LL n1 = F[n].length();
for (int i = 0;i<n1;++i){
if (F[n].at(i)==ch) ++count;
}
}
void solve(string p, LL n){//Recursion
// cout << p << endl;
LL n1 = p.length();
if (n<=1&&n1>=2) return;//return if string pattern p's size is larger than F(n)
//When p's size is reduced to 2 or 1, it's small enough now that we can search for p directly in F(n)
if (n1<=2){
if (n1 == 2){
if (p=="00") return;//Return since there can't be two subsequent '0' in F(n) for any n
else find(n,p.at(0),p.at(1));
return;
}
if (n1 == 1){
if (p=="1") find('1',n);
else find('0',n);
return;
}
}
string p1, p2;//if the last character in p is 1, we can replace it with either '1' or '0'
//p1 stores the substring ending in '1' and p2 stores the substring ending in '0'
for (LL i = 0;i<n1;++i){//We replace every "10" with 1, "1" with 0.
if (p[i]=='1'){
if (p[i+1]=='0'&&(i+1)!= n1){
if (p[i+2]=='0'&&(i+2)!= n1) return;//Return if there are two subsequent '0'
p1.append("1");//Replace "10" with "1"
++i;
}
else {
p1.append("0");//Replace "1" with "0"
}
}
else {
if (p[i+1]=='0'&&(i+1)!= n1){//Return if there are two subsequent '0'
return;
}
p1.append("1");
}
}
solve(p1,n-1);
if (p[n1-1]=='1'){
p2 = p1;
p2.back() = '1';
solve(p2,n-1);
}
}
main(){
F[0] = "0";F[1] = "1";
for (int i = 2;i<38;++i){
F[i].append(F[i-1]);
F[i].append(F[i-2]);
}//precalculate F(0) to F(37)
LL t = 0;//NumofTestcases
int n; string p;
while (cin >> n >> p) {
count = 0;
solve(p,n);
cout << "Case " << ++t << ": " << count << endl;
}
}
The above program works fine, but with small inputs only. When i submitted the above program to codeforces i got an answer wrong because although i shortened the pattern string p and reduces n to n', the size of F[n'] is still very large (n'>=50). How can i modify my code to make it works in this case, or is there another approach (such as dynamic programming?). Many thanks for any advice.
More details about the problem can be found here: https://codeforces.com/group/Ir5CI6f3FD/contest/273369/problem/B
I don't have time now to try to code this up myself, but I have a suggested approach.
First, I should note, that while that hint you used is certainly accurate, I don't see any straightforward way to solve the problem. Perhaps the correct follow-up to that would be simpler than what I'm suggesting.
My approach:
Find the first two ns such that length(F(n)) >= length(pattern). Calculating these is a simple recursion. The important insight is that every subsequent value will start with one of these two values, and will also end with one of them. (This is true for all adjacent values -- for any m > n, F(m) will begin either with F(n) or with F(n - 1). It's not hard to see why.)
Calculate and cache the number of occurrences of the pattern in this these two Fs, but whatever index shifting technique makes sense.
For F(n+1) (and all subsequent values) calculate by adding together
The count for F(n)
The count for F(n - 1)
The count for those spanning both F(n) and F(n - 1). We can achieve that by testing every breakdown of pattern into (nonempty) prefix and suffix values (i.e., splitting at every internal index) and counting those where F(n) ends in prefix and F(n - 1) starts with suffix. But we don't have to have all of F(n) and F(n - 1) to do this. We just need the tail of F(n) and the head of F(n - 1) of the length of the pattern. So we don't need to calculate all of F(n). We just need to know which of those two initial values our current one ends with. But the start is always the predecessor, and the end oscillates between the previous two. It should be easy to keep track.
The time complexity then should be proportional to the product of n and the length of the pattern.
If I find time tomorrow, I'll see if I can code this up. But it won't be in C -- those years were short and long gone.
Collecting the list of prefix/suffix pairs can be done once ahead of time
I'm starting to practice Dynamic Programming and I just can't wrap my head around this question:
Question:
A child is running up a staircase with n steps and can hop either 1 step, 2 steps, or 3 steps at a time. Implement a method to count how many possible ways the child can run up the stairs.
The solution from the cracking the coding interview book is like this:
"If we thought about all the paths to the nth step, we could just build them off the paths to the three previous steps. We can get up to the nth stop by any of the following:
Going to the (n-1) step and hopping 1 step
Going to the (n-2) step and hopping 2 steps
Going to the (n-3) step and hopping 3 steps"
Therefor to find the solution you just add the number of these path together !
That's what loses me ! Why isn't the answer like this: add number of those paths then add 3 ? Since if you are on step n-1 or n-2 or n-3, there are 3 ways to get the nth step? I understand that if you write down the answers for the first 4 bases cases (assuming that n=0 returns 1) You can see the fibonacci-like pattern. But you may not also see it so it's difficult.
And then they came up with this code:
public static int countWaysDP(int n, int[] map) {
if (n < 0)
return 0;
else if (n == 0)
return 1;
else if (map[n] > -1)
return map[n];
else {
map[n] = countWaysDP(n - 1, map) + countWaysDP(n - 2, map) + countWaysDP(n - 3, map);
return map[n]; }
}
So my second question. How does it return 1 when n == 0. Even if I accept that fact, I still can't figure out a way to solve it if I return 0 when n == 1.
Hope this makes sense.
Thank you
Here is how I wrapped my head around this-
From the book -
On the very last hop, up to the nth step, the child could have
done either a single, double, or triple step hop. That is, the last
move might have been a single step hop from step n-1, a double
step hop from step n-2, or a triple step hop from n-3. The
total number of ways of reaching the last step is therefore the sum of
the number of ways of reaching each of the last three steps
You are correctly contemplating -
Why isn't the answer like this: add number of those paths then add 3 ?
Since if you are on step n-1 or n-2 or n-3, there are 3 ways to get
the nth step?
The problem with such a base case is that it will be applicable only if n >= 3. You clearly will not add 3 if there are only 2 steps.
Let's break down the individual cases and understand what exactly is the base case here.
n=0
There are no stairs to climb.
Total number of ways = 0
n=1
Total number of ways = 1StepHop from (n-1)
Number of ways to do 1StepHop from Step 0(n-1) = 1
Total number of ways = 1
n=2
Total number of ways = 2StepHop from (n-2) + 1StepHop from (n-1)
Number of ways to do 2StepHop to reach Step 2 from Step 0(n-2) = 1
Number of ways to do 1StepHop to reach Step 2 from Step 1(n-1) = 1 (Previous answer for n=1)
Total number of ways = 1 + 1 = 2
n=3
Total number of ways = 3StepHop from (n-3) + 2StepHop from (n-2) + 1StepHop from (n-1)
Number of ways to do 3StepHop to reach Step 3 from Step 0(n-3) = 1
Number of ways to do 2StepHop to reach Step 3 from Step 1(n-2) = 2 (From previous answer for n = 2)
Number of ways to do 1StepHop to reach Step 3 from Step 2 = 1 (From previous answer for n=1)
Total number of ways = 1 + 2 + 1 = 4
Observation -
As you can see from above, we are correctly accounting for the last step in each case. Adding one for each of -> 1StepHop from n-1, 2StepHop from n-2 and 3StepHop from n-3.
Now looking at the code, the case where we return 1 if n==0 is a bit counter-intuitive since we already saw that the answer should be 0 if n==0. -
public static int countWaysDP(int n, int[] map) {
if (n < 0)
return 0;
else if (n == 0)
return 1; <------------- this case is counter-intuitive
else if (map[n] > -1)
return map[n];
else {
map[n] = countWaysDP(n - 1, map) + countWaysDP(n - 2, map) + countWaysDP(n - 3, map);
return map[n];
}
From the observation, you can see that this counter intuitive case of n==0 is actually the one which is accounting for the final step - 1StepHop from n-1, 2StepHop from n-2 and 3StepHop from n-3.
So hitting n==0 case makes sense only during recursion - which will happen only when the initial value of n is greater than 0.
A more complete solution to this problem may have a driver method which handles that case outside of the core recursive algorithm -
int countWays(int n) {
if (n <= 0 ) return 0;
int[] map = new int[n+1];
for(int i = 0; i<n+1; i++){
map[i] = -1;
}
return countWaysDP(n, map);
}
Hope this is helpful.
You can find the solution on
https://github.com/CrispenGari/Triple-Step-Algorithim/blob/master/main.cpp .
int count_Ways(int n){
if(n<0){
return 0;
}else if(n==0){
return 1;
}else{
return count_Ways(n-1) +count_Ways(n-2) + count_Ways(n-3);
}
}
int main(){
cout<<"Enter number of stairs: ";
int n;
cin>>n;
cout<<"There are "<< count_Ways(n)<<" possible ways the child can run up
thestairs."<<endl;
return 0;
}
I have spent a lot of time to learn about implementing/visualizing dynamic programming problems using iteration but I find it very hard to understand, I can implement the same using recursion with memoization but it is slow when compared to iteration.
Can someone explain the same by a example of a hard problem or by using some basic concepts. Like the matrix chain multiplication, longest palindromic sub sequence and others. I can understand the recursion process and then memoize the overlapping sub problems for efficiency but I can't understand how to do the same using iteration.
Thanks!
Dynamic programming is all about solving the sub-problems in order to solve the bigger one. The difference between the recursive approach and the iterative approach is that the former is top-down, and the latter is bottom-up. In other words, using recursion, you start from the big problem you are trying to solve and chop it down to a bit smaller sub-problems, on which you repeat the process until you reach the sub-problem so small you can solve. This has an advantage that you only have to solve the sub-problems that are absolutely needed and using memoization to remember the results as you go. The bottom-up approach first solves all the sub-problems, using tabulation to remember the results. If we are not doing extra work of solving the sub-problems that are not needed, this is a better approach.
For a simpler example, let's look at the Fibonacci sequence. Say we'd like to compute F(101). When doing it recursively, we will start with our big problem - F(101). For that, we notice that we need to compute F(99) and F(100). Then, for F(99) we need F(97) and F(98). We continue until we reach the smallest solvable sub-problem, which is F(1), and memoize the results. When doing it iteratively, we start from the smallest sub-problem, F(1) and continue all the way up, keeping the results in a table (so essentially it's just a simple for loop from 1 to 101 in this case).
Let's take a look at the matrix chain multiplication problem, which you requested. We'll start with a naive recursive implementation, then recursive DP, and finally iterative DP. It's going to be implemented in a C/C++ soup, but you should be able to follow along even if you are not very familiar with them.
/* Solve the problem recursively (naive)
p - matrix dimensions
n - size of p
i..j - state (sub-problem): range of parenthesis */
int solve_rn(int p[], int n, int i, int j) {
// A matrix multiplied by itself needs no operations
if (i == j) return 0;
// A minimal solution for this sub-problem, we
// initialize it with the maximal possible value
int min = std::numeric_limits<int>::max();
// Recursively solve all the sub-problems
for (int k = i; k < j; ++k) {
int tmp = solve_rn(p, n, i, k) + solve_rn(p, n, k + 1, j) + p[i - 1] * p[k] * p[j];
if (tmp < min) min = tmp;
}
// Return solution for this sub-problem
return min;
}
To compute the result, we starts with the big problem:
solve_rn(p, n, 1, n - 1)
The key of DP is to remember all the solutions to the sub-problems instead of forgetting them, so we don't need to recompute them. It's trivial to make a few adjustments to the above code in order to achieve that:
/* Solve the problem recursively (DP)
p - matrix dimensions
n - size of p
i..j - state (sub-problem): range of parenthesis */
int solve_r(int p[], int n, int i, int j) {
/* We need to remember the results for state i..j.
This can be done in a matrix, which we call dp,
such that dp[i][j] is the best solution for the
state i..j. We initialize everything to 0 first.
static keyword here is just a C/C++ thing for keeping
the matrix between function calls, you can also either
make it global or pass it as a parameter each time.
MAXN is here too because the array size when doing it like
this has to be a constant in C/C++. I set it to 100 here.
But you can do it some other way if you don't like it. */
static int dp[MAXN][MAXN] = {{0}};
/* A matrix multiplied by itself has 0 operations, so we
can just return 0. Also, if we already computed the result
for this state, just return that. */
if (i == j) return 0;
else if (dp[i][j] != 0) return dp[i][j];
// A minimal solution for this sub-problem, we
// initialize it with the maximal possible value
dp[i][j] = std::numeric_limits<int>::max();
// Recursively solve all the sub-problems
for (int k = i; k < j; ++k) {
int tmp = solve_r(p, n, i, k) + solve_r(p, n, k + 1, j) + p[i - 1] * p[k] * p[j];
if (tmp < dp[i][j]) dp[i][j] = tmp;
}
// Return solution for this sub-problem
return dp[i][j];;
}
We start with the big problem as well:
solve_r(p, n, 1, n - 1)
Iterative solution is only to, well, iterate all the states, instead of starting from the top:
/* Solve the problem iteratively
p - matrix dimensions
n - size of p
We don't need to pass state, because we iterate the states. */
int solve_i(int p[], int n) {
// But we do need our table, just like before
static int dp[MAXN][MAXN];
// Multiplying a matrix by itself needs no operations
for (int i = 1; i < n; ++i)
dp[i][i] = 0;
// L represents the length of the chain. We go from smallest, to
// biggest. Made L capital to distinguish letter l from number 1
for (int L = 2; L < n; ++L) {
// This double loop goes through all the states in the current
// chain length.
for (int i = 1; i <= n - L + 1; ++i) {
int j = i + L - 1;
dp[i][j] = std::numeric_limits<int>::max();
for (int k = i; k <= j - 1; ++k) {
int tmp = dp[i][k] + dp[k+1][j] + p[i-1] * p[k] * p[j];
if (tmp < dp[i][j])
dp[i][j] = tmp;
}
}
}
// Return the result of the biggest problem
return dp[1][n-1];
}
To compute the result, just call it:
solve_i(p, n)
Explanation of the loop counters in the last example:
Let's say we need to optimize the multiplication of 4 matrices: A B C D. We are doing an iterative approach, so we will first compute the chains with the length of two: (A B) C D, A (B C) D, and A B (C D). And then chains of three: (A B C) D, and A (B C D). That is what L, i and j are for.
L represents the chain length, it goes from 2 to n - 1 (n is 4 in this case, so that is 3).
i and j represent the starting and ending position of the chain. In case L = 2, i goes from 1 to 3, and j goes from 2 to 4:
(A B) C D A (B C) D A B (C D)
^ ^ ^ ^ ^ ^
i j i j i j
In case L = 3, i goes from 1 to 2, and j goes from 3 to 4:
(A B C) D A (B C D)
^ ^ ^ ^
i j i j
So generally, i goes from 1 to n - L + 1, and j is i + L - 1.
Now, let's continue with the algorithm assuming that we are at the step where we have (A B C) D. We now need to take into account the sub-problems (which are already calculated): ((A B) C) D and (A (B C)) D. That is what k is for. It goes through all the positions between i and j and computes the sub problems.
I hope I helped.
The problem with recursion is the high number of stack frames that need to be pushed/popped. This can quickly become the bottle-neck.
The Fibonacci Series can be calculated with iterative DP or recursion with memoization. If we calculate F(100) in DP all we need is an array of length 100 e.g. int[100] and that's the guts of our used memory. We calculate all entries of the array pre-filling f[0] and f[1] as they are defined to be 1. and each value just depends on the previous two.
If we use a recursive solution we start at fib(100) and work down. Every method call from 100 down to 0 is pushed onto the stack, AND checked if it's memoized. These operations add up and iteration doesn't suffer from either of these. In iteration (bottom-up) we already know all of the previous answers are valid. The bigger impact is probably the stack frames; and given a larger input you may get a StackOverflowException for what was otherwise trivial with an iterative DP approach.
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.