Trying to write Pari code to solve the above question.
I've got no experience in using Pari, but here's some useful advice:
n is Carmichael if and only if it is composite and, for all a with 1 < a < n which are relatively prime to n, the congruence a^(n-1) = 1 (mod n) holds. To use this definition directly, you need:
1) An efficient way to test if a and n are relatively prime
2) An efficient way to compute a^(n-1) (mod n)
For the first -- use the Euclidean algorithm for greatest common divisors. It is most efficiently computed in a loop, but can also be defined via the simple recurrence gcd(a,b) = gcd(b,a%b) with basis gcd(a,0) = a. In C this is just:
unsigned int gcd(unsigned int a, unsigned int b){
return b == 0? a : gcd(b, a%b);
}
For the second point -- almost the worst possible thing you can do when computing a^k (mod n) is to first compute a^k via repeated multiplication and to then mod the result by n. Instead -- use exponentiation by squaring, taking the remainder (mod n) at intermediate stages. It is a divide-and-conquer algorithm based on the observation that e.g. a^10 = (a^5)^2 and a^11 = (a^5)^2 * a. A simple C implementation is:
unsigned int modexp(unsigned int a, unsigned int p, unsigned int n){
unsigned long long b;
switch(p){
case 0:
return 1;
case 1:
return a%n;
default:
b = modexp(a,p/2,n);
b = (b*b) % n;
if(p%2 == 1) b = (b*a) % n;
return b;
}
}
Note the use of unsigned long long to guard against overflow in the calculation of b*b.
To test if n is Carmichael, you might as well first test if n is even and return 0 in that case. Otherwise, step through numbers, a, in the range 2 to n-1. First check if gcd(a,n) == 1 Note that if n is composite then you must have at least one a before you reach the square root of n with gcd(a,n) > 1). Keep a Boolean flag which keeps track of whether or not such an a has been encountered and if you exceed the square root without finding such an a, return 0. For those a with gcd(a,n) == 1, compute the modular exponentiation a^(n-1) (mod n). If this is ever different from 1, return 0. If your loop finishes checking all a below n without returning 0, then the number is Carmichael, so return 1. An implementation is:
int is_carmichael(unsigned int n){
int a,s;
int factor_found = 0;
if (n%2 == 0) return 0;
//else:
s = sqrt(n);
a = 2;
while(a < n){
if(a > s && !factor_found){
return 0;
}
if(gcd(a,n) > 1){
factor_found = 1;
}
else{
if(modexp(a,n-1,n) != 1){
return 0;
}
}
a++;
}
return 1; //anything that survives to here is a carmichael
}
A simple driver program:
int main(void){
unsigned int n;
for(n = 2; n < 100000; n ++){
if(is_carmichael(n)) printf("%u\n",n);
}
return 0;
}
output:
C:\Programs>gcc carmichael.c
C:\Programs>a
561
1105
1729
2465
2821
6601
8911
10585
15841
29341
41041
46657
52633
62745
63973
75361
This only takes about 2 seconds to run and matches the initial part of this list.
This is probably a somewhat practical method for checking if numbers up to a million or so are Carmichael numbers. For larger numbers, you should probably get yourself a good factoring algorithm and use Korseldt's criterion as described in the Wikipedia entry on Carmichael numbers.
Related
I want to calculate ab mod n for use in RSA decryption. My code (below) returns incorrect answers. What is wrong with it?
unsigned long int decrypt2(int a,int b,int n)
{
unsigned long int res = 1;
for (int i = 0; i < (b / 2); i++)
{
res *= ((a * a) % n);
res %= n;
}
if (b % n == 1)
res *=a;
res %=n;
return res;
}
You can try this C++ code. I've used it with 32 and 64-bit integers. I'm sure I got this from SO.
template <typename T>
T modpow(T base, T exp, T modulus) {
base %= modulus;
T result = 1;
while (exp > 0) {
if (exp & 1) result = (result * base) % modulus;
base = (base * base) % modulus;
exp >>= 1;
}
return result;
}
You can find this algorithm and related discussion in the literature on p. 244 of
Schneier, Bruce (1996). Applied Cryptography: Protocols, Algorithms, and Source Code in C, Second Edition (2nd ed.). Wiley. ISBN 978-0-471-11709-4.
Note that the multiplications result * base and base * base are subject to overflow in this simplified version. If the modulus is more than half the width of T (i.e. more than the square root of the maximum T value), then one should use a suitable modular multiplication algorithm instead - see the answers to Ways to do modulo multiplication with primitive types.
In order to calculate pow(a,b) % n to be used for RSA decryption, the best algorithm I came across is Primality Testing 1) which is as follows:
int modulo(int a, int b, int n){
long long x=1, y=a;
while (b > 0) {
if (b%2 == 1) {
x = (x*y) % n; // multiplying with base
}
y = (y*y) % n; // squaring the base
b /= 2;
}
return x % n;
}
See below reference for more details.
1) Primality Testing : Non-deterministic Algorithms – topcoder
Usually it's something like this:
while (b)
{
if (b % 2) { res = (res * a) % n; }
a = (a * a) % n;
b /= 2;
}
return res;
The only actual logic error that I see is this line:
if (b % n == 1)
which should be this:
if (b % 2 == 1)
But your overall design is problematic: your function performs O(b) multiplications and modulus operations, but your use of b / 2 and a * a implies that you were aiming to perform O(log b) operations (which is usually how modular exponentiation is done).
Doing the raw power operation is very costly, hence you can apply the following logic to simplify the decryption.
From here,
Now say we want to encrypt the message m = 7, c = m^e mod n = 7^3 mod 33
= 343 mod 33 = 13. Hence the ciphertext c = 13.
To check decryption we compute m' = c^d mod n = 13^7 mod 33 = 7. Note
that we don't have to calculate the full value of 13 to the power 7
here. We can make use of the fact that a = bc mod n = (b mod n).(c mod
n) mod n so we can break down a potentially large number into its
components and combine the results of easier, smaller calculations to
calculate the final value.
One way of calculating m' is as follows:- Note that any number can be
expressed as a sum of powers of 2. So first compute values of 13^2,
13^4, 13^8, ... by repeatedly squaring successive values modulo 33. 13^2
= 169 ≡ 4, 13^4 = 4.4 = 16, 13^8 = 16.16 = 256 ≡ 25. Then, since 7 = 4 + 2 + 1, we have m' = 13^7 = 13^(4+2+1) = 13^4.13^2.13^1 ≡ 16 x 4 x 13 = 832
≡ 7 mod 33
Are you trying to calculate (a^b)%n, or a^(b%n) ?
If you want the first one, then your code only works when b is an even number, because of that b/2. The "if b%n==1" is incorrect because you don't care about b%n here, but rather about b%2.
If you want the second one, then the loop is wrong because you're looping b/2 times instead of (b%n)/2 times.
Either way, your function is unnecessarily complex. Why do you loop until b/2 and try to multiply in 2 a's each time? Why not just loop until b and mulitply in one a each time. That would eliminate a lot of unnecessary complexity and thus eliminate potential errors. Are you thinking that you'll make the program faster by cutting the number of times through the loop in half? Frankly, that's a bad programming practice: micro-optimization. It doesn't really help much: You still multiply by a the same number of times, all you do is cut down on the number of times testing the loop. If b is typically small (like one or two digits), it's not worth the trouble. If b is large -- if it can be in the millions -- then this is insufficient, you need a much more radical optimization.
Also, why do the %n each time through the loop? Why not just do it once at the end?
Calculating pow(a,b) mod n
A key problem with OP's code is a * a. This is int overflow (undefined behavior) when a is large enough. The type of res is irrelevant in the multiplication of a * a.
The solution is to ensure either:
the multiplication is done with 2x wide math or
with modulus n, n*n <= type_MAX + 1
There is no reason to return a wider type than the type of the modulus as the result is always represent by that type.
// unsigned long int decrypt2(int a,int b,int n)
int decrypt2(int a,int b,int n)
Using unsigned math is certainly more suitable for OP's RSA goals.
Also see Modular exponentiation without range restriction
// (a^b)%n
// n != 0
// Test if unsigned long long at least 2x values bits as unsigned
#if ULLONG_MAX/UINT_MAX - 1 > UINT_MAX
unsigned decrypt2(unsigned a, unsigned b, unsigned n) {
unsigned long long result = 1u % n; // Insure result < n, even when n==1
while (b > 0) {
if (b & 1) result = (result * a) % n;
a = (1ULL * a * a) %n;
b >>= 1;
}
return (unsigned) result;
}
#else
unsigned decrypt2(unsigned a, unsigned b, unsigned n) {
// Detect if UINT_MAX + 1 < n*n
if (UINT_MAX/n < n-1) {
return TBD_code_with_wider_math(a,b,n);
}
a %= n;
unsigned result = 1u % n;
while (b > 0) {
if (b & 1) result = (result * a) % n;
a = (a * a) % n;
b >>= 1;
}
return result;
}
#endif
int's are generally not enough for RSA (unless you are dealing with small simplified examples)
you need a data type that can store integers up to 2256 (for 256-bit RSA keys) or 2512 for 512-bit keys, etc
Here is another way. Remember that when we find modulo multiplicative inverse of a under mod m.
Then
a and m must be coprime with each other.
We can use gcd extended for calculating modulo multiplicative inverse.
For computing ab mod m when a and b can have more than 105 digits then its tricky to compute the result.
Below code will do the computing part :
#include <iostream>
#include <string>
using namespace std;
/*
* May this code live long.
*/
long pow(string,string,long long);
long pow(long long ,long long ,long long);
int main() {
string _num,_pow;
long long _mod;
cin>>_num>>_pow>>_mod;
//cout<<_num<<" "<<_pow<<" "<<_mod<<endl;
cout<<pow(_num,_pow,_mod)<<endl;
return 0;
}
long pow(string n,string p,long long mod){
long long num=0,_pow=0;
for(char c: n){
num=(num*10+c-48)%mod;
}
for(char c: p){
_pow=(_pow*10+c-48)%(mod-1);
}
return pow(num,_pow,mod);
}
long pow(long long a,long long p,long long mod){
long res=1;
if(a==0)return 0;
while(p>0){
if((p&1)==0){
p/=2;
a=(a*a)%mod;
}
else{
p--;
res=(res*a)%mod;
}
}
return res;
}
This code works because ab mod m can be written as (a mod m)b mod m-1 mod m.
Hope it helped { :)
use fast exponentiation maybe..... gives same o(log n) as that template above
int power(int base, int exp,int mod)
{
if(exp == 0)
return 1;
int p=power(base, exp/2,mod);
p=(p*p)% mod;
return (exp%2 == 0)?p:(base * p)%mod;
}
This(encryption) is more of an algorithm design problem than a programming one. The important missing part is familiarity with modern algebra. I suggest that you look for a huge optimizatin in group theory and number theory.
If n is a prime number, pow(a,n-1)%n==1 (assuming infinite digit integers).So, basically you need to calculate pow(a,b%(n-1))%n; According to group theory, you can find e such that every other number is equivalent to a power of e modulo n. Therefore the range [1..n-1] can be represented as a permutation on powers of e. Given the algorithm to find e for n and logarithm of a base e, calculations can be significantly simplified. Cryptography needs a tone of math background; I'd rather be off that ground without enough background.
For my code a^k mod n in php:
function pmod(a, k, n)
{
if (n==1) return 0;
power = 1;
for(i=1; i<=k; $i++)
{
power = (power*a) % n;
}
return power;
}
#include <cmath>
...
static_cast<int>(std::pow(a,b))%n
but my best bet is you are overflowing int (IE: the number is two large for the int) on the power I had the same problem creating the exact same function.
I'm using this function:
int CalculateMod(int base, int exp ,int mod){
int result;
result = (int) pow(base,exp);
result = result % mod;
return result;
}
I parse the variable result because pow give you back a double, and for using mod you need two variables of type int, anyway, in a RSA decryption, you should just use integer numbers.
I have a C code off finding large perfect numbers below,
#include <stdio.h>
int main ()
{
unsigned long long num,i,sum;
while (scanf ("%llu",&num) != EOF && num)
{
sum = 1;
for (i=2; i*i<=num; i++)
{
if (num % i == 0)
{
if (i*i == num)
sum += i;
else
sum += (i + num/i);
}
}
if (sum == num)
printf ("Perfect\n");
else if (sum > num)
printf ("Abundant\n");
else
printf ("Deficient\n");
}
return 0;
}
I tried to find whether a number is perfect, abundant or deficient. I run a loop upto the square root of numto minimize the runtime. It works fine <= 10^15 but for the larger values it takes too long time to execute.
For example,for the following input sets,
8
6
18
1000000
1000000000000000
0
this code shows the following outputs,
Deficient
Perfect
Abundant
Abundant
Abundant
But, for 10^16 it doesn't respond quickly.
So, is there any better way to find a perfect number for too long values? Or is there any better algorithm to implement here??? :)
Yes, there is a better algorithm.
Your algorithm is basically the simple one--adding up the divisors of a number to find... the sum of the divisors of a number (excluding itself). But you can use the number-theoretic formula for finding the sum of the divisors of a number (including itself). If the prime numbers dividing n are p1, p2, ..., pk and the powers of those primes in the canonical decomposition of n are a1, a2, ..., ak, then the sum of the divisors of n is
(p1**(a1+1) - 1) / (p1 - 1) * (p2**(a2+1) - 1) / (p2 - 1) * ...
* (pk**(ak+1) - 1) / (pk - 1)
You can find the prime divisors and their exponents more quickly than finding all the divisors of n. Subtract n from that expression above and you get the sum you want.
There are some tricks, of course, to find the pis and ais more efficiently: I'll leave that to you.
By the way, if your purpose is just to find the perfect numbers, as in your title, you would do better to use Euclid's formula for even prime numbers. Find the Mersenne prime numbers by examining all 2**p-1 for prime p to see if they are prime--there are shortcuts to doing this as well--then constructing a perfect number from that Mersenne prime. This would leave out any odd perfect numbers, though. If you find any, let the mathematical community know--that would make you world famous.
Of course, the fastest way of all to find perfect numbers is to use the lists already made of some of them.
It is a matter of factorization of numbers. You can read more here: https://en.wikipedia.org/wiki/Integer_factorization
Unfortunately no good news for you - the bigger the number gets, the longer it takes.
To start with your code, try not to multiply i*i each iteration.
Instead of:
for (i=2; i*i<=num; i++)
calculate square root of num first, and then compare
i <= square_root_of_num in the loop.
// Program to determine whether perfect or not
# include <bits/stdc++.h>
using namespace std;
map<long long int, int> mp; // to store prime factors and there frequency
void primeFactors(long long int n)
{
// counting the number of 2s that divide n
while (n%2 == 0)
{
mp[2] = mp[2]+1;
n = n/2;
}
long long int root = sqrt(n);
// n must be odd at this point. So we can skip every even numbers next
for (long long int i = 3; i <= root; i = i+2)
{
// While i divides n, count frequency of i prime factor and divide n
while (n%i == 0)
{
mp[i] = mp[i]+1;
n = n/i;
}
}
// This condition is to handle the case whien n is a prime number
// greater than 2
if (n > 2)
{
mp[n] = mp[n]+1;
}
}
long long int pow(long long int base, long long int exp)
{
long long int result = 1;
base = base;
while (exp>0)
{
if (exp & 1)
result = (result*base);
exp >>= 1;
base = (base*base);
}
return result;
}
int main ()
{
long long num, p, a, sum;
while (scanf ("%lld",&num) != EOF && num)
{
primeFactors(num);
sum = 1;
map<long long int, int> :: iterator i;
for(i=mp.begin(); i!=mp.end(); i++)
{
p = i->first;
a = i->second;
sum = sum*((pow(p,a+1)-1)/(p-1));
}
if (sum == 2*num)
printf ("Perfect\n");
else if (sum > num)
printf ("Abundant\n");
else
printf ("Deficient\n");
mp.clear();
}
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.
This is a contest problem (ACM ICPC South America 2015), it was the hardest in the problem set.
Summary: Given integers N and K, count the number of sequences a of length N consisting of integers 1 ≤ ai ≤ K, subject to the condition that for any x in that sequence there has to be a pair i, j satisfying i < j and ai = x − 1 and aj = x, i.e. the last x is preceded by x − 1 at some point.
Example: for N = 1000 and K = 100 the solution should be congruent to 265428620 modulo (109 + 7). Other examples and details can be found in the problem description.
I tried everything in my knowledge, but I need pointers to know how to do it. I even printed some lists with brute force to find the pattern, but I didn't succeed.
I'm looking for an algorithm, or formula that allows me to get to the right solution for this problem. It can be any language.
EDIT:
I solved the problem using a formula I found on the internet (someone who explained this problem). However, just because I programmed it, doesn't mean I understand it, so the question remains open. My code is here (the online judge returns Accepted):
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
ll mod = 1e9+7;
ll memo[5001][5001];
ll dp(int n, int k){
// K can't be greater than N
k = min(n, k);
// if N or K is 1, it means there's only one possible list
if(n <= 1 || k <= 1) return 1;
if(memo[n][k] != -1) return memo[n][k];
ll ans1 = (n-k) * dp(n-1, k-1);
ll ans2 = k * dp(n-1, k);
memo[n][k] = ((ans1 % mod) + (ans2 % mod)) % mod;
return memo[n][k];
}
int main(){
int n, q;
for(int i=0; i<5001; i++)
fill(memo[i], memo[i]+5001, -1);
while(scanf("%d %d", &n, &q) == 2){
for(int i=0; i<q; i++){
int k;
scanf("%d", &k);
printf("%s%lld", i==0? "" : " ", dp(n, k));
}
printf("\n");
}
return 0;
}
The most important lines are the recursive call, particularly, these lines
ll ans1 = (n-k) * dp(n-1, k-1);
ll ans2 = k * dp(n-1, k);
memo[n][k] = ((ans1 % mod) + (ans2 % mod)) % mod;
Here I show the brute force algorithm for the problem in python. It works for small numbers, but for very big numbers it takes too much time. For N=1000 and K=5 it is already infeasible (Needs more than 100 years time to calculate)(In C it should also be infeasible as C is only 100 times faster than Python). So the problem actually forces you to find a shortcut.
import itertools
def checkArr(a,K):
for i in range(2,min(K+1,max(a)+1)):
if i-1 not in a:
return False
if i not in a:
return False
if a.index(i-1)>len(a)-1-a[::-1].index(i):
return False
return True
def num_sorted(N,K):
result=0
for a in itertools.product(range(1,K+1), repeat=N):
if checkArr(a,K):
result+=1
return result
num_sorted(3,10)
It returns 6 as expected.
How to find LCM of {1, 2, ..., n} where 0 < n < 10001 in fastest possible way. The one way is to calculate n! / gcd (1,2,.....,n) but this can be slow as number of testcases are t < 501 and the output should be LCM ( n! ) % 1000000007
Code for the same is:
#include<bits/stdc++.h>
using namespace std;
#define p 1000000007;
int fact[10001] = {1};
int gcd[10001] = {1};
int main()
{
int i, j;
for( i = 2;i < 10001; i++){
fact[i] = ( i * fact[i-1]) % p;
}
for(i=2 ;i < 10001; i++){
gcd[i] =__gcd( gcd[i-1], i );
}
int t;
cin >> t;
while( t-- ) {
int n;
cin >> n;
int res = ( fact[n] / gcd[n] );
cout << res << endl;
}
return 0;
}
But this code is not performing as well. Why?
Your current solution is not correct, as has been mentioned in the comments.
One way to solve this is to realize that the LCM of those numbers is just the product of all the "largest" powers of distinct primes less than or equal to n. That is, find each prime p less than or equal to n, then find the largest power of each of those primes such that it's still less than or equal to n, and multiply those together. For a given p, said power can be expressed in pseudocode as:
p ** math.Floor(math.Log(n) / math.Log(p))
Here's an implementation in Golang that returns immediately:
http://play.golang.org/p/8s4eE_CQ9Y
$ time go run lcm.go
5793339670287642968692270879166240098634860297998518825393138351148979300145773182308832598
<several lines later>
800000
real 0m0.225s
user 0m0.173s
sys 0m0.044s
For completeness, the full code from that playground link is pasted here:
package main
import (
"fmt"
"math"
"math/big"
)
func main() {
n := 10001
primes := make([]int, 1, n)
primes[0] = 2
SIEVE:
for i := 3; i <= n; i++ {
for _, p := range primes {
if i%p == 0 {
continue SIEVE
}
}
primes = append(primes, i)
}
logN := math.Log(float64(n))
lcm := big.NewInt(1)
for _, p := range primes {
floatP := float64(p)
e := math.Floor(logN / math.Log(floatP))
lcm.Mul(lcm, big.NewInt(int64(math.Pow(floatP, e))))
}
fmt.Println(lcm)
}
I would calculate this in completely different way: the LCM of {1,...,n} is a product of all primes p[i]<=n, each in power floor(log(n)/log(p[i])). This product is divisible by all numbers up to n, and this is the least such number. Your main trouble is to calculate table of primes then.
I'm going to suggest something less dynamic, but it will increase your speed dramatically. Set up a factorial table (perhaps an array) and store pre-calculated factorial integer representations there. That way, it's a simple O(1) operation, versus O(n). Here's a reference table, but you may also precalculate those yourself: http://www.tsm-resources.com/alists/fact.html It's okay to do so, because those values will never change. If we're talking optimization for speed, then why not store the values we know, rather than calculate them each time?
If, however, you're opposed to storing these calculations beforehand, I suggest looking at optimized algorithms and work your way from there:
Here are two excellent resources for faster factorial calculation algorithms:
http://www.luschny.de/math/factorial/conclusions.html
http://www.luschny.de/math/factorial/scala/FactorialScalaCsharp.htm
This is very simple but it seems to run fast enough. Probably Amit Kumar Gupta's idea is faster. Stack overflow around n = 9500 on my machine but that could be fixed by memoizing the function and building up the memo from small numbers to larger numbers. I didn't take the modulus but that fix is easy, particularly if the modulus is prime. Is it?
import java.math.BigInteger;
public class LCM {
// compute f(n) = lcm(1,...n)
public static BigInteger f(int n) {
if (n == 1) return BigInteger.ONE;
BigInteger prev = f(n-1);
return prev.divide(prev.gcd(BigInteger.valueOf(n)))
.multiply(BigInteger.valueOf(n));
}
public static void main(String[] args) {
int n = Integer.parseInt(args[0]);
System.out.println("f(" + n + ") = " + f(n));
}
}