#include <stdio.h>
#include <stdlib.h>
int main() {
int i, j, n, maxi = 0;
printf("\n Introduce the number:\n");
scanf("%d", &n);
for (j = 1; j <= n; j++)
{
i = 0;
while (i < j) {
i++;
if (j == i * i) {
if (j > maxi) {
maxi = j;
printf("%d", maxi);
}
}
}
}
return 0;
}
I have to find the greatest perfect square smaller than than a number n, I succeeded in finding all the perfect squares that are smaller than the number n but because each time it finds a perfect square it displays it I couldn't think of any way to compare all the perfect square that were found (or at least that's what I think the problem is) so I would appreciate some help. I already know that you could also solve this problem using a more simpler method ( like the one below ) and if you have any other ideas on how to solve it I'd like to hear them.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main()
{
int n,j;
printf("\n Your number:\n");
scanf("%d",&n);
j=(int)sqrt(n);
printf("%d",j*j);
return 0;
}
You only need a single loop here. Check if i*i <= n. If so, set maxi to i*i and increment i:
int n, i = 1, sq = 1;
printf("\n Introduce the number:\n");
scanf("%d", &n);
while (i*i <= n) {
sq = i*i;
i++;
}
printf("sq=%d\n", sq);
Find the greatest perfect square that is less than or equal to n
For n>=0, this is akin to finding the integer square root of n.
unsigned greatest_perfect_square(unsigned x) {
unsigned root = usqrt(x);
return root * root;
}
if you have any other ideas on how to solve it I'd like to hear them.
The order of complexity to find the square root is O(bit-width-of-type-n). e.g. 16 iterations.
#include <limits.h>
unsigned usqrt(unsigned x) {
unsigned y = 0;
unsigned xShifted = 0;
const unsigned MSBit = UINT_MAX - UINT_MAX/2;
// This constant relies on no padding and bit width even
const unsigned TwoBitCount_N = sizeof(x) * CHAR_BIT / 2;
for (unsigned TwoBitCount = TwoBitCount_N; TwoBitCount > 0; TwoBitCount--) {
// Shift `xShifted` 2 places left while shifting in the 2 MSbits of x
xShifted <<= 1;
if (x & MSBit) {
xShifted |= 1;
}
x <<= 1;
xShifted <<= 1;
if (x & MSBit) {
xShifted |= 1;
}
x <<= 1;
// Shift the answer 1 bit left
y <<= 1;
// Form test value as y*2 + 1
unsigned Test = (y << 1) | 1;
// If xShifted big enough ...
if (xShifted >= Test) {
xShifted -= Test;
// Increment answer
y |= 1;
}
}
return y;
}
OP's method is far far slower. Even the inner loop takes O(sqrt(n)) time.
Note:
OP's code: j == i * i is subject to overflow and leads to the incorrect answer when j is larger.
j/i == i performs a like test without overflow.
#Jonathan Leffler suggested a Newton-Raphson approximation approach. Some lightly tested code below works quite fast, often taking only a few iterations.
I suspect this is O(log(bit-width-of-type-n)) for the main part, yet of course still O(log(bit-width-of-type-n)) for bit_width().
Both of the functions could be improved.
unsigned bit_width(unsigned x) {
unsigned width = 0;
while (x) {
x /= 2;
width++;
}
return width;
}
unsigned usqrt_NR(unsigned x) {
if (x == 0) {
return 0;
}
unsigned y = 1u << bit_width(x)/2;
unsigned y_previous;
unsigned diff;
unsigned diff1count = 0;;
do {
y_previous = y;
y = (y + x/y)/2;
diff = y_previous < y ? y - y_previous : y_previous - y;
if (diff == 1) diff1count++;
} while (diff > 1 || (diff == 1 && diff1count <= 1));
y = (y_previous + y)/2;
return y;
}
This minimizes the number of multiplications: it looks for the first square which is larger than n, meaning that the perfect square immediately before was the solution.
for (i = 1; i <= n; i++) {
if (i*i > n) {
break;
}
}
i--;
// i*i is your answer
On some platforms it might be useful to exploit the fact that (i+1)*(i+1) = i*i + 2*i + 1, or in other words, if you already have i^2, (i+1)^2 is obtained by adding i to it twice, and incrementing by 1; and at the beginning, 0^2 is 0 to prime the cycle.
for (i = 0, sq = 0; i < n; i++) {
sq += i; // Or on some platforms sq += i<<1 instead of two sums
sq += i; // Some compilers will auto-optimize "sq += 2*i" for the platform
sq++; // Or even sq += ((2*i)|1) as adding 1 to even numbers is OR'ing 1
if (sq > n) {
break;
}
// if sq is declared as signed integer, a possible overflow will
// show it as being negative. This way we can still get a "correct" result
// with i the smallest root that does not overflow.
// In 16-bit arithmetic this is 181, root of 32761; next square would be
// 33124 which cannot be represented in signed 16-bit space.
if (sq < 0) {
break;
}
}
// (i*i) is your answer
Is there any way to find nth root of the number without any external library in C? I'm working on a bare metal code so there is no OS. Also, no complete C is there.
You can write a program like this for nth root. This program is for square root.
int floorSqrt(int x)
{
// Base cases
if (x == 0 || x == 1)
return x;
// Staring from 1, try all numbers until
// i*i is greater than or equal to x.
int i = 1, result = 1;
while (result < x)
{
if (result == x)
return result;
i++;
result = i*i;
}
return i-1;
}
You can use the same approach for nth root.
Here there is a C implementation of the the nth root algorithm you can find in wikipedia. It needs an exponentiation algorithm, so I also include an implementation of a basic method for exponentiation by squaring that you can find also find in wikipedia.
double npower(double const base, int const n)
{
if (n < 0) return npower(1/base, -n)
else if (n == 0) return 1.0;
else if (n == 1) return base;
else if (n % 2) return base*npower(base*base, n/2);
else return npower(base*base, n/2);
}
double nroot(double const base, int const n)
{
if (n == 1) return base;
else if (n <= 0 || base < 0) return NAN;
else {
double delta, x = base/n;
do {
delta = (base/npower(x,n-1)-x)/n;
x += delta;
} while (fabs(delta) >= 1e-8);
return x;
}
}
Some comments on this:
The nth root algorithm in wikipedia leaves freedom for the initial guess. In this example I set it up to be base/n, but this was just a guess.
The macro NAN is usually defined in <math.h>, so you would need to define it to be suitable for your needs.
Both functions are implemented in a very rough and simple way, and their performance can be greatly improved with careful thought.
The tolerance in this example is set to 1e-8 and should be changed to something different. It should probably be proportional to the value of the base.
You can try the nth_root C function :
// return a number that, when multiplied by itself nth times, makes N.
unsigned nth_root(const unsigned n, const unsigned nth) {
unsigned a = n, b, c, r = nth ? n + (n > 1) : n == 1 ;
for (; a < r; b = a + (nth - 1) * r, a = b / nth)
for (r = a, a = n, c = nth - 1; c && (a /= r); --c);
return r;
}
Source
Suppose I have an array of bytes from a secure PRNG, and I need to generate a number between 1 and 10 using that data, how would I do that correctly?
Think of the array as one big unsigned integer. Then the answer is simple:
(Big_Number % 10) + 1
So all that is needed is a method to find the modulus 10 of big integers. Using modular exponentiation:
#include <limits.h>
#include <stdlib.h>
int ArrayMod10(const unsigned char *a, size_t n) {
int mod10 = 0;
int base = (UCHAR_MAX + 1) % 10;
for (size_t i = n; i-- > 0; ) {
mod10 = (base*mod10 + a[i]) % 10;
base = (base * base) % 10;
}
return mod10;
}
void test10(size_t n) {
unsigned char a[n];
// fill array with your secure PRNG
for (size_t i = 0; i<n; i++) a[i] = rand();
return ArrayMod10(a, n) + 1;
}
There will be a slight bias as 256^n is not a power of 10. With large n, this will rapidly decrease in significance.
Untested code: Detect if a biased result occurred. Calling code could repeatedly call this function with new a array values to get an unbiased result on the rare occasions when bias occurs.
int ArrayMod10BiasDetect(const unsigned char *a, size_t n, bool *biasptr) {
bool bias = true;
int mod10 = 0;
int base = (UCHAR_MAX + 1) % 10; // Note base is usually 6: 256%10, 65536%10, etc.
for (size_t i = n; i-- > 0; ) {
mod10 = (base*mod10 + a[i]) % 10;
if (n > 0) {
if (a[i] < UCHAR_MAX) bias = false;
} else {
if (a[i] < UCHAR_MAX + 1 - base) bias = false;
}
base = (base * base) % 10;
}
*biaseptr = bias;
return mod10;
}
As per the comments follow-up, it seems what you need is modulus operator [%].
You may also need to check the related wiki.
Note: Every time we use the modulo operator on a random number, there is a probability that we'll be running into modulo bias, which ends up in disbalancing the fair distribution of random numbers. You've to take care of that.
For a detailed discussion on this, please see this question and related answers.
It depends on a bunch of things. Secure PRNG sometimes makes long byte arrays instead of integers, let's say it is 16 bytes long array, then extract 32 bit integer like so: buf[0]*0x1000000+buf[1]*0x10000+buf[2]*0x100+buf[3] or use shift operator. This is random so big-endian/little-endian doesn't matter.
char randbytes[16];
//...
const char *p = randbytes;
//assumes size of int is 4
unsigned int rand1 = p[0] << 24 + p[1] << 16 + p[2] << 8 + p[3]; p += 4;
unsigned int rand2 = p[0] << 24 + p[1] << 16 + p[2] << 8 + p[3]; p += 4;
unsigned int rand3 = p[0] << 24 + p[1] << 16 + p[2] << 8 + p[3]; p += 4;
unsigned int rand4 = p[0] << 24 + p[1] << 16 + p[2] << 8 + p[3];
Then use % on the integer
ps, I think that's a long answer. If you want number between 1 and 10 then just use % on first byte.
OK, so this answer is in Java until I get to my Eclipse C/C++ IDE:
public final static int simpleBound(Random rbg, int n) {
final int BYTE_VALUES = 256;
// sanity check, only return positive numbers
if (n <= 0) {
throw new IllegalArgumentException("Oops");
}
// sanity check: choice of value 0 or 0...
if (n == 1) {
return 0;
}
// sanity check: does not fit in byte
if (n > BYTE_VALUES) {
throw new IllegalArgumentException("Oops");
}
// optimization for n = 2^y
if (Integer.bitCount(n) == 1) {
final int mask = n - 1;
return retrieveRandomByte(rbg) & mask;
}
// you can skip to this if you are sure n = 10
// z is upper bound, and contains floor(z / n) blocks of n values
final int z = (BYTE_VALUES / n) * n;
int x;
do {
x = retrieveRandomByte(rbg);
} while (x >= z);
return x % n;
}
So n is the maximum value in a range [0..n), i.e. n is exclusive. For a range [1..10] simply increase the result with 1.
What is the most efficient way given to raise an integer to the power of another integer in C?
// 2^3
pow(2,3) == 8
// 5^5
pow(5,5) == 3125
Exponentiation by squaring.
int ipow(int base, int exp)
{
int result = 1;
for (;;)
{
if (exp & 1)
result *= base;
exp >>= 1;
if (!exp)
break;
base *= base;
}
return result;
}
This is the standard method for doing modular exponentiation for huge numbers in asymmetric cryptography.
Note that exponentiation by squaring is not the most optimal method. It is probably the best you can do as a general method that works for all exponent values, but for a specific exponent value there might be a better sequence that needs fewer multiplications.
For instance, if you want to compute x^15, the method of exponentiation by squaring will give you:
x^15 = (x^7)*(x^7)*x
x^7 = (x^3)*(x^3)*x
x^3 = x*x*x
This is a total of 6 multiplications.
It turns out this can be done using "just" 5 multiplications via addition-chain exponentiation.
n*n = n^2
n^2*n = n^3
n^3*n^3 = n^6
n^6*n^6 = n^12
n^12*n^3 = n^15
There are no efficient algorithms to find this optimal sequence of multiplications. From Wikipedia:
The problem of finding the shortest addition chain cannot be solved by dynamic programming, because it does not satisfy the assumption of optimal substructure. That is, it is not sufficient to decompose the power into smaller powers, each of which is computed minimally, since the addition chains for the smaller powers may be related (to share computations). For example, in the shortest addition chain for a¹⁵ above, the subproblem for a⁶ must be computed as (a³)² since a³ is re-used (as opposed to, say, a⁶ = a²(a²)², which also requires three multiplies).
If you need to raise 2 to a power. The fastest way to do so is to bit shift by the power.
2 ** 3 == 1 << 3 == 8
2 ** 30 == 1 << 30 == 1073741824 (A Gigabyte)
Here is the method in Java
private int ipow(int base, int exp)
{
int result = 1;
while (exp != 0)
{
if ((exp & 1) == 1)
result *= base;
exp >>= 1;
base *= base;
}
return result;
}
An extremely specialized case is, when you need say 2^(-x to the y), where x, is of course is negative and y is too large to do shifting on an int. You can still do 2^x in constant time by screwing with a float.
struct IeeeFloat
{
unsigned int base : 23;
unsigned int exponent : 8;
unsigned int signBit : 1;
};
union IeeeFloatUnion
{
IeeeFloat brokenOut;
float f;
};
inline float twoToThe(char exponent)
{
// notice how the range checking is already done on the exponent var
static IeeeFloatUnion u;
u.f = 2.0;
// Change the exponent part of the float
u.brokenOut.exponent += (exponent - 1);
return (u.f);
}
You can get more powers of 2 by using a double as the base type.
(Thanks a lot to commenters for helping to square this post away).
There's also the possibility that learning more about IEEE floats, other special cases of exponentiation might present themselves.
power() function to work for Integers Only
int power(int base, unsigned int exp){
if (exp == 0)
return 1;
int temp = power(base, exp/2);
if (exp%2 == 0)
return temp*temp;
else
return base*temp*temp;
}
Complexity = O(log(exp))
power() function to work for negative exp and float base.
float power(float base, int exp) {
if( exp == 0)
return 1;
float temp = power(base, exp/2);
if (exp%2 == 0)
return temp*temp;
else {
if(exp > 0)
return base*temp*temp;
else
return (temp*temp)/base; //negative exponent computation
}
}
Complexity = O(log(exp))
If you want to get the value of an integer for 2 raised to the power of something it is always better to use the shift option:
pow(2,5) can be replaced by 1<<5
This is much more efficient.
int pow( int base, int exponent)
{ // Does not work for negative exponents. (But that would be leaving the range of int)
if (exponent == 0) return 1; // base case;
int temp = pow(base, exponent/2);
if (exponent % 2 == 0)
return temp * temp;
else
return (base * temp * temp);
}
Just as a follow up to comments on the efficiency of exponentiation by squaring.
The advantage of that approach is that it runs in log(n) time. For example, if you were going to calculate something huge, such as x^1048575 (2^20 - 1), you only have to go thru the loop 20 times, not 1 million+ using the naive approach.
Also, in terms of code complexity, it is simpler than trying to find the most optimal sequence of multiplications, a la Pramod's suggestion.
Edit:
I guess I should clarify before someone tags me for the potential for overflow. This approach assumes that you have some sort of hugeint library.
Late to the party:
Below is a solution that also deals with y < 0 as best as it can.
It uses a result of intmax_t for maximum range. There is no provision for answers that do not fit in intmax_t.
powjii(0, 0) --> 1 which is a common result for this case.
pow(0,negative), another undefined result, returns INTMAX_MAX
intmax_t powjii(int x, int y) {
if (y < 0) {
switch (x) {
case 0:
return INTMAX_MAX;
case 1:
return 1;
case -1:
return y % 2 ? -1 : 1;
}
return 0;
}
intmax_t z = 1;
intmax_t base = x;
for (;;) {
if (y % 2) {
z *= base;
}
y /= 2;
if (y == 0) {
break;
}
base *= base;
}
return z;
}
This code uses a forever loop for(;;) to avoid the final base *= base common in other looped solutions. That multiplication is 1) not needed and 2) could be int*int overflow which is UB.
more generic solution considering negative exponenet
private static int pow(int base, int exponent) {
int result = 1;
if (exponent == 0)
return result; // base case;
if (exponent < 0)
return 1 / pow(base, -exponent);
int temp = pow(base, exponent / 2);
if (exponent % 2 == 0)
return temp * temp;
else
return (base * temp * temp);
}
The O(log N) solution in Swift...
// Time complexity is O(log N)
func power(_ base: Int, _ exp: Int) -> Int {
// 1. If the exponent is 1 then return the number (e.g a^1 == a)
//Time complexity O(1)
if exp == 1 {
return base
}
// 2. Calculate the value of the number raised to half of the exponent. This will be used to calculate the final answer by squaring the result (e.g a^2n == (a^n)^2 == a^n * a^n). The idea is that we can do half the amount of work by obtaining a^n and multiplying the result by itself to get a^2n
//Time complexity O(log N)
let tempVal = power(base, exp/2)
// 3. If the exponent was odd then decompose the result in such a way that it allows you to divide the exponent in two (e.g. a^(2n+1) == a^1 * a^2n == a^1 * a^n * a^n). If the eponent is even then the result must be the base raised to half the exponent squared (e.g. a^2n == a^n * a^n = (a^n)^2).
//Time complexity O(1)
return (exp % 2 == 1 ? base : 1) * tempVal * tempVal
}
int pow(int const x, unsigned const e) noexcept
{
return !e ? 1 : 1 == e ? x : (e % 2 ? x : 1) * pow(x * x, e / 2);
//return !e ? 1 : 1 == e ? x : (((x ^ 1) & -(e % 2)) ^ 1) * pow(x * x, e / 2);
}
Yes, it's recursive, but a good optimizing compiler will optimize recursion away.
One more implementation (in Java). May not be most efficient solution but # of iterations is same as that of Exponential solution.
public static long pow(long base, long exp){
if(exp ==0){
return 1;
}
if(exp ==1){
return base;
}
if(exp % 2 == 0){
long half = pow(base, exp/2);
return half * half;
}else{
long half = pow(base, (exp -1)/2);
return base * half * half;
}
}
I use recursive, if the exp is even,5^10 =25^5.
int pow(float base,float exp){
if (exp==0)return 1;
else if(exp>0&&exp%2==0){
return pow(base*base,exp/2);
}else if (exp>0&&exp%2!=0){
return base*pow(base,exp-1);
}
}
In addition to the answer by Elias, which causes Undefined Behaviour when implemented with signed integers, and incorrect values for high input when implemented with unsigned integers,
here is a modified version of the Exponentiation by Squaring that also works with signed integer types, and doesn't give incorrect values:
#include <stdint.h>
#define SQRT_INT64_MAX (INT64_C(0xB504F333))
int64_t alx_pow_s64 (int64_t base, uint8_t exp)
{
int_fast64_t base_;
int_fast64_t result;
base_ = base;
if (base_ == 1)
return 1;
if (!exp)
return 1;
if (!base_)
return 0;
result = 1;
if (exp & 1)
result *= base_;
exp >>= 1;
while (exp) {
if (base_ > SQRT_INT64_MAX)
return 0;
base_ *= base_;
if (exp & 1)
result *= base_;
exp >>= 1;
}
return result;
}
Considerations for this function:
(1 ** N) == 1
(N ** 0) == 1
(0 ** 0) == 1
(0 ** N) == 0
If any overflow or wrapping is going to take place, return 0;
I used int64_t, but any width (signed or unsigned) can be used with little modification. However, if you need to use a non-fixed-width integer type, you will need to change SQRT_INT64_MAX by (int)sqrt(INT_MAX) (in the case of using int) or something similar, which should be optimized, but it is uglier, and not a C constant expression. Also casting the result of sqrt() to an int is not very good because of floating point precission in case of a perfect square, but as I don't know of any implementation where INT_MAX -or the maximum of any type- is a perfect square, you can live with that.
I have implemented algorithm that memorizes all computed powers and then uses them when need. So for example x^13 is equal to (x^2)^2^2 * x^2^2 * x where x^2^2 it taken from the table instead of computing it once again. This is basically implementation of #Pramod answer (but in C#).
The number of multiplication needed is Ceil(Log n)
public static int Power(int base, int exp)
{
int tab[] = new int[exp + 1];
tab[0] = 1;
tab[1] = base;
return Power(base, exp, tab);
}
public static int Power(int base, int exp, int tab[])
{
if(exp == 0) return 1;
if(exp == 1) return base;
int i = 1;
while(i < exp/2)
{
if(tab[2 * i] <= 0)
tab[2 * i] = tab[i] * tab[i];
i = i << 1;
}
if(exp <= i)
return tab[i];
else return tab[i] * Power(base, exp - i, tab);
}
Here is a O(1) algorithm for calculating x ** y, inspired by this comment. It works for 32-bit signed int.
For small values of y, it uses exponentiation by squaring. For large values of y, there are only a few values of x where the result doesn't overflow. This implementation uses a lookup table to read the result without calculating.
On overflow, the C standard permits any behavior, including crash. However, I decided to do bound-checking on LUT indices to prevent memory access violation, which could be surprising and undesirable.
Pseudo-code:
If `x` is between -2 and 2, use special-case formulas.
Otherwise, if `y` is between 0 and 8, use special-case formulas.
Otherwise:
Set x = abs(x); remember if x was negative
If x <= 10 and y <= 19:
Load precomputed result from a lookup table
Otherwise:
Set result to 0 (overflow)
If x was negative and y is odd, negate the result
C code:
#define POW9(x) x * x * x * x * x * x * x * x * x
#define POW10(x) POW9(x) * x
#define POW11(x) POW10(x) * x
#define POW12(x) POW11(x) * x
#define POW13(x) POW12(x) * x
#define POW14(x) POW13(x) * x
#define POW15(x) POW14(x) * x
#define POW16(x) POW15(x) * x
#define POW17(x) POW16(x) * x
#define POW18(x) POW17(x) * x
#define POW19(x) POW18(x) * x
int mypow(int x, unsigned y)
{
static int table[8][11] = {
{POW9(3), POW10(3), POW11(3), POW12(3), POW13(3), POW14(3), POW15(3), POW16(3), POW17(3), POW18(3), POW19(3)},
{POW9(4), POW10(4), POW11(4), POW12(4), POW13(4), POW14(4), POW15(4), 0, 0, 0, 0},
{POW9(5), POW10(5), POW11(5), POW12(5), POW13(5), 0, 0, 0, 0, 0, 0},
{POW9(6), POW10(6), POW11(6), 0, 0, 0, 0, 0, 0, 0, 0},
{POW9(7), POW10(7), POW11(7), 0, 0, 0, 0, 0, 0, 0, 0},
{POW9(8), POW10(8), 0, 0, 0, 0, 0, 0, 0, 0, 0},
{POW9(9), 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
{POW9(10), 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
};
int is_neg;
int r;
switch (x)
{
case 0:
return y == 0 ? 1 : 0;
case 1:
return 1;
case -1:
return y % 2 == 0 ? 1 : -1;
case 2:
return 1 << y;
case -2:
return (y % 2 == 0 ? 1 : -1) << y;
default:
switch (y)
{
case 0:
return 1;
case 1:
return x;
case 2:
return x * x;
case 3:
return x * x * x;
case 4:
r = x * x;
return r * r;
case 5:
r = x * x;
return r * r * x;
case 6:
r = x * x;
return r * r * r;
case 7:
r = x * x;
return r * r * r * x;
case 8:
r = x * x;
r = r * r;
return r * r;
default:
is_neg = x < 0;
if (is_neg)
x = -x;
if (x <= 10 && y <= 19)
r = table[x - 3][y - 9];
else
r = 0;
if (is_neg && y % 2 == 1)
r = -r;
return r;
}
}
}
My case is a little different, I'm trying to create a mask from a power, but I thought I'd share the solution I found anyway.
Obviously, it only works for powers of 2.
Mask1 = 1 << (Exponent - 1);
Mask2 = Mask1 - 1;
return Mask1 + Mask2;
In case you know the exponent (and it is an integer) at compile-time, you can use templates to unroll the loop. This can be made more efficient, but I wanted to demonstrate the basic principle here:
#include <iostream>
template<unsigned long N>
unsigned long inline exp_unroll(unsigned base) {
return base * exp_unroll<N-1>(base);
}
We terminate the recursion using a template specialization:
template<>
unsigned long inline exp_unroll<1>(unsigned base) {
return base;
}
The exponent needs to be known at runtime,
int main(int argc, char * argv[]) {
std::cout << argv[1] <<"**5= " << exp_unroll<5>(atoi(argv[1])) << ;std::endl;
}
I've noticed something strange about the standard exponential squaring algorithm with gnu-GMP :
I implemented 2 nearly-identical functions - a power-modulo function using the most vanilla binary exponential squaring algorithm,
labeled ______2()
then another one basically the same concept, but re-mapped to dividing by 10 at each round instead of dividing by 2,
labeled ______10()
.
( time ( jot - 1456 9999999999 6671 | pvE0 |
gawk -Mbe '
function ______10(_, __, ___, ____, _____, _______) {
__ = +__
____ = (____+=_____=____^= \
(_ %=___=+___)<_)+____++^____—
while (__) {
if (_______= __%____) {
if (__==_______) {
return (_^__ *_____) %___
}
__-=_______
_____ = (_^_______*_____) %___
}
__/=____
_ = _^____%___
}
}
function ______2(_, __, ___, ____, _____) {
__=+__
____+=____=_____^=(_%=___=+___)<_
while (__) {
if (__ %____) {
if (__<____) {
return (_*_____) %___
}
_____ = (_____*_) %___
--__
}
__/=____
_= (_*_) %___
}
}
BEGIN {
OFMT = CONVFMT = "%.250g"
__ = (___=_^= FS=OFS= "=")(_<_)
_____ = __^(_=3)^--_ * ++_-(_+_)^_
______ = _^(_+_)-_ + _^!_
_______ = int(______*_____)
________ = 10 ^ 5 + 1
_________ = 8 ^ 4 * 2 - 1
}
GNU Awk 5.1.1, API: 3.1 (GNU MPFR 4.1.0, GNU MP 6.2.1)
.
($++NF = ______10(_=$___, NR %________ +_________,_______*(_-11))) ^!___'
out9: 48.4MiB 0:00:08 [6.02MiB/s] [6.02MiB/s] [ <=> ]
in0: 15.6MiB 0:00:08 [1.95MiB/s] [1.95MiB/s] [ <=> ]
( jot - 1456 9999999999 6671 | pvE 0.1 in0 | gawk -Mbe ; )
8.31s user 0.06s system 103% cpu 8.058 total
ffa16aa937b7beca66a173ccbf8e1e12 stdin
($++NF = ______2(_=$___, NR %________ +_________,_______*(_-11))) ^!___'
out9: 48.4MiB 0:00:12 [3.78MiB/s] [3.78MiB/s] [<=> ]
in0: 15.6MiB 0:00:12 [1.22MiB/s] [1.22MiB/s] [ <=> ]
( jot - 1456 9999999999 6671 | pvE 0.1 in0 | gawk -Mbe ; )
13.05s user 0.07s system 102% cpu 12.821 total
ffa16aa937b7beca66a173ccbf8e1e12 stdin
For reasons extremely counter-intuitive and unknown to me, for a wide variety of inputs i threw at it, the div-10 variant is nearly always faster. It's the matching of hashes between the 2 that made it truly baffling, despite computers obviously not being built in and for a base-10 paradigm.
Am I missing something critical or obvious in the code/approach that might be skewing the results in a confounding manner ? Thanks.