Given an undirected graph with positive edge costs, choose a subset of edges such that there are no cycles and the sum of the cost is maximal.
The input consists of several graphs, each defined with number of vertices n, number of edges m, and m triples x,y,c to indicate an edge between x and y of cost c. The vertices are numbered from 0 to n - 1. It is assumed that 1 ≤ n ≤ 104, 0 ≤ m ≤ 5 n, and 1 ≤ c ≤ 105. there may be more than one edge between two vertices, and even edges with x = y.
#include <iostream>
#include <vector>
using namespace std;
using P = pair<int,int>;
using VE = vector<int>;
using VP = vector<P>;
using VVE = vector<VP>;
int n,m;
VVE G;
VE cost;
VE vist;
VE pare;
int maxim(int x){
if(cost[x] != -1) return cost[x];
cost[x] = 0;
for(P y: G[x]){
if(cost[x] <= y.second + maxim(y.first)){
cost[x] = y.second + maxim(y.first);
}
}
return cost[x];
}
int main() {
while(cin >> n >> m){
G = VVE(n);
cost = VE(n,-1);
pare = VE(n,-1);
for(int i = 0; i < m; ++i){
int x,y,c; cin >> x >> y >> c;
G[x].push_back(P(y,c));
G[y].push_back(P(x,c));
}
int mx = -1;
for(int i = 0; i < n; ++i){
if(mx <= maxim(i)){
mx = maxim(i);
}
}
cout << mx << endl;
}
}
This is my code and I don't know how to solve the problem. I would appreciate help. As you can see the graph is read as a vector of vectors. In which each pair indicates that node x goes to node y with cost c.
As a commenter pointed out, this is the maximum spanning tree problem (which is the same as the minimum spanning tree problem, just negate the costs). That problem can be solved with the greedy algorithm. Initially place every node into a heap on its own. Then in a loop consider the edges in decreasing cost order. If the two endpoints of the considered edge are in the same heap, discard the edge. Otherwise select it and merge the heaps. When you have only one heap left, you can stop, and the selected edges form your solution.
Related
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.
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 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.
I am trying to write a program which prints Pythagorean triples (a^2 + b^2 = c^2) for a given range N where a<=b<=c<=N.
#include <stdio.h>
int main()
{
int a = 0, b = 0, c = 0, N, T,c2;
scanf("%d", &T);
while(T--)
{
int counter = 0;
scanf("%d", &N);
{
for (c = 0; c <=N; c++)
{
for (b = 0; b < c; b++)
{
for (a = 0; a < b; a++)
{
c2 = c*c;
if (a*a + b*b == c2 )
//if(sqrt (pow(a,2) + pow(b,2)) == c)
{
++counter;
printf("\n %d , %d, %d \n",a,b,c); }
}
}
}
}
printf("%d\n", counter);
}
return 0;
}
This works well for N<1000. For higher N, say 10000 this takes a lot of time.
Is there any better way to optimize this prog or any better algorithm instead of brute force, so that it takes less time to compute for higher N ?
By number theory, Pythagorean triples are parametrized by (2pq, p^2-q^2, p^2+q^2). You can enumerate over these and just abort whenever c > N. This is of course optimal since you do as many computations as there are triples...
First, you can compute c2 as soon as a new value of c is available:
for (c = 0; c <=N; c++)
{
/* compute c2 here */
This saves the time to compute it over and over for each b and a.
The same goes for b: it is possible to compute the square of b as soon as b is available, instead of for each value of a. Your compiler may automatically apply these optimizations, but not necessarily.
Lastly, there is only one value of a that can make the equation true, and this value is sqrt(c2 - b2). For large values of b, it is much faster to compute this expression and to check whether it makes the equation true than to test all values between 0 and b. If you use double-precision computation for sqrt(c2 - b2), then floating-point approximations will not be an issue until N is about 226.
You can reduce one variable by the following calculations. If we assume that
a = m^2 - n^2, b = 2mn, c = m^2 + n^2
Lets check if it satisfies a^2 + b^2 = c^2:
a^2+b^2 = (m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2 = c^2.
Now, we can iterate through all possible m and n and generate corresponding a, b, c.
It's the fastest method I ever used. However, I don't know if there exist any O(1) or O(log(n)) mathematical solutions.