I'm solving Project Euler problem 16, I've ended up with a code that can logically solve it, but is unable to process as I believe its overflowing or something? I tried int64 in place of int but it just prints 0,0. If i change the power to anything below 30 it works, but above 30 it does not work, Can anyone point out my mistake? I believe its not able to calculate 2^1000.
// PE_16 project main.go
package main
import (
"fmt"
)
func power(x, y int) int {
var pow int
var final int
final = 1
for pow = 1; pow <= y; pow++ {
final = final * x
}
return final
}
func main() {
var stp int
var sumfdigits int
var u, t, h, th, tth, l int
stp = power(2,1000)
fmt.Println(stp)
u = stp / 1 % 10
t = stp / 10 % 10
h = stp / 100 % 10
th = stp / 1000 % 10
tth = stp / 10000 % 10
l = stp / 100000 % 10
sumfdigits = u + t + h + th + tth + l
fmt.Println(sumfdigits)
}
Your approach to this problem requires exact integer math up to 1000 bits in size. But you're using int which is 32 or 64 bits. math/big.Int can handle such task. I intentionally do not provide a ready made solution using big.Int as I assume your goal is to learn by doing it by yourself, which I believe is the intent of Project Euler.
As noted by #jnml, ints aren't large enough; if you wish to calculate 2^1000 in Go, big.Ints are a good choice here. Note that math/big provides the Exp() method which will be easier to use than converting your power function to big.Ints.
I worked through some Project Euler problems about a year ago, doing them in Go to get to know the language. I didn't like the ones that required big.Ints, which aren't so easy to work with in Go. For this one, I "cheated" and did it in one line of Ruby:
Removed because I remembered it was considered bad form to show a working solution, even in a different language.
Anyway, my Ruby example shows another thing I learned with Go's big.Ints: sometimes it's easier to convert them to a string and work with that string than to work with the big.Int itself. This problem strikes me as one of those cases.
Converting my Ruby algorithm to Go, I only work with big.Ints on one line, then it's easy to work with the string and get the answer in just a few lines of code.
You don't need to use math/big. Below is a school boy maths way of doubling a decimal number as a hint!
xs holds the decimal digits in least significant first order. Pass in a pointer to the digits (pxs) as the slice might need to get bigger.
func double(pxs *[]int) {
xs := *pxs
carry := 0
for i, x := range xs {
n := x*2 + carry
if n >= 10 {
carry = 1
n -= 10
} else {
carry = 0
}
xs[i] = n
}
if carry != 0 {
*pxs = append(xs, carry)
}
}
Related
Suppose I have two variables, that only use 6 bits:
var a byte = 31 // 00011111
var b byte = 50 // 00110010
The first (a) have more one bits than the b, however the b is greater than a of course, so is not possible use a > b.
To achieve what I need, I do one loop:
func countOneBits(byt byte) int {
var counter int
var divider byte
for divider = 32; divider >= 1; divider >>= 1 {
if byt & divider == divider {
counter++
}
}
return counter
}
This works, I can use countOneBits(a) > countOneBits(b)...
But I don't think is the best solution for this case, I don't think this need a loop and because of it I'm here.
Have a better alternative (in performance aspect) to count how many 1 have in six bits?
Given that the input is a single byte probably a lookup table is the best option... only takes 256 bytes and you get code like
var count = bitcount[input];
Given that this function will be available in the packagemath/bits in the next Go release (1.9 this August) here is the code for a 32-bit integer.
// OnesCount32 returns the number of one bits ("population count") in x.
func OnesCount32(x uint32) int {
return int(pop8tab[x>>24] + pop8tab[x>>16&0xff] + pop8tab[x>>8&0xff] + pop8tab[x&0xff])
}
Where the pop8tab is defined here. And for your question in particular : 8bits
func OnesCount8(x uint8) int {
return int(pop8tab[x])
}
It is also possible to count bits with binary operations. See this bit twiddling hacks.
func bitSetCount(v byte) byte {
v = (v & 0x55) + ((v>>1) & 0x55)
v = (v & 0x33) + ((v>>2) & 0x33)
return (v + (v>>4)) & 0xF
}
You'll have to benchmark to see if this is faster than the lookup table which is the simplest to implement.
there is POPCNT golang version:
https://github.com/tmthrgd/go-popcount
I use the following code in my test:
package main
import "fmt"
import "math/big"
func main() {
input := "3333333333333333333.......tested with 100'000x3 , tested with 1'000'0000x3, tested with 10'000'000x3"
bi := big.NewInt(0)
if _, ok := bi.SetString(input, 10); ok {
fmt.Printf("number = %v\n", bi)
testval := new(big.Int)
testval.SetString("3", 10)
resultat, isGanzzahl := myDiv(bi, testval)
fmt.Printf("isGanzzahl = %v, resultat = %v\n", isGanzzahl, resultat)
} else {
fmt.Printf("error parsing line %#v\n", input)
}
}
func myDiv(minuend *big.Int, subtrahend *big.Int) (*big.Int, bool) {
zerotest := big.NewInt(0)
modulus := new(big.Int)
modulus = modulus.Mod(minuend, subtrahend)
if zerotest.Cmp(modulus) == 0 {
res := big.NewInt(0)
res.Quo(minuend, subtrahend)
return res, true
} else {
return big.NewInt(0), false
}
}
100'000 x 3 / 3 == not even a quater second
1'000'000 x 3 / 3 == 9.45 seconds
10'000'000 x 3 / 3 == 16.1 minute
Im looking for a way to make this happens much much faster. If i would like to do this multithreaded in go how do i do this with go-routines? Is there a faster way to do a division with larger numbers?
As this is just for testing i planned to use Numbers in the range of 100'000'000 - 1'000'000'000 digits (which would then be 1GB of ram usage). But 1 billion digits would not work because it would take years to complete.
What would then happen if it is N / M ? Where N=1billion digit, M=10million digit. Is this even possible on a powerful home computer?
How would it look / or what do i have to change to being able to distribute this work to multiple small computer (for example AWS)?
If your number is more than 100000 digits long, you need to use Fast Fourier Transform for multiplication and division: https://en.wikipedia.org/wiki/Multiplication_algorithm#Fourier_transform_methods . The basic idea is to treat numbers as polynomials with x being power of 10 (or power of 2 if you want binary system). Multiply polynomials using Fast Fourier Transform and then propagate carry to get a number from a polynomial. I.e. if we need to multiply 19 by 19 and we use x = 101, we will have (1 * x + 9) * (1 * x + 9) = x2 + 18 * x + 81. Now we propagate carry to convert polynomial back to number: x2 + 18 * x + 81 = x2 + (18 + 8) * x + 1 = x2 + 26 * x + 1 = (1 + 2) * x2 + 6 * x + 1 = 3 * x2 + 6 * x + 1 = 361. The trick is that polynomials can be multiplied efficiently (O(N*log(N) time) using Fast Fourier Transform. The coefficients of the product polynomial are larger than digits, so you need to choose x carefully in order to avoid integer overflow or precision problems.
There unlikely to be a golang library for that so you will need to write it yourself. Here are a few short FFT implementations you can use as a starting point:
http://codeforces.com/contest/632/submission/16449753 http://codeforces.com/contest/632/submission/16445979 http://codeforces.com/contest/632/submission/16449040
http://codeforces.com/contest/632/submission/16448169
If you decide to use FFT modulo prime, see this post for a good choice of the modulo: http://petr-mitrichev.blogspot.com/2015/04/this-week-in-competitive-programming.html
I want to convert a float64 number, let's say it 1.003 to 1003 (integer type). My implementation is simply multiply the float64 with 1000 and cast it to int.
package main
import "fmt"
func main() {
var f float64 = 1.003
fmt.Println(int(f * 1000))
}
But when I run that code, what I got is 1002 not 1003. Because Go automatically stores 1.003 as 1.002999... in the variable. What is the correct approach to do this kind of operation on Golang?
Go spec: Conversions:
Conversions between numeric types
When converting a floating-point number to an integer, the fraction is discarded (truncation towards zero).
So basically when you convert a floating-point number to an integer, only the integer part is kept.
If you just want to avoid errors arising from representing with finite bits, just add 0.5 to the number before converting it to int. No external libraries or function calls (from standard library) required.
Since float -> int conversion is not rounding but keeping the integer part, this will give you the desired result. Taking into consideration both the possible smaller and greater representation:
1002.9999 + 0.5 = 1003.4999; integer part: 1003
1003.0001 + 0.5 = 1003.5001; integer part: 1003
So simply just write:
var f float64 = 1.003
fmt.Println(int(f * 1000 + 0.5))
To wrap this into a function:
func toint(f float64) int {
return int(f + 0.5)
}
// Using it:
fmt.Println(toint(f * 1000))
Try them on the Go Playground.
Note:
Be careful when you apply this in case of negative numbers! For example if you have a value of -1.003, then you probably want the result to be -1003. But if you add 0.5 to it:
-1002.9999 + 0.5 = -1002.4999; integer part: -1002
-1003.0001 + 0.5 = -1002.5001; integer part: -1002
So if you have negative numbers, you have to either:
subtract 0.5 instead of adding it
or add 0.5 but subtract 1 from the result
Incorporating this into our helper function:
func toint(f float64) int {
if f < 0 {
return int(f - 0.5)
}
return int(f + 0.5)
}
As Will mentions, this comes down to how floats are represented on various platforms. Essentially you need to round the float rather than let the default truncating behavior to happen. There's no standard library function for this, probably because there's a lot of possible behavior and it's trivial to implement.
If you knew you'd always have errors of the sort described, where you're slightly below (1299.999999) the value desired (1300.00000) you could use the math library's Ceil function:
f := 1.29999
n := math.Ceil(f*1000)
But if you have different kinds of floating error and want a more general sorting behavior? Use the math library's Modf function to separate the your floating point value by the decimal point:
f := 1.29999
f1,f2 := math.Modf(f*1000)
n := int(f1) // n = 1299
if f2 > .5 {
n++
}
fmt.Println(n)
You can run a slightly more generalized version of this code in the playground yourself.
This is probably likely a problem with floating points in general in most programming languages though some have different implementations than others. I wouldn't go into the intricacies here but most languages usually have a "decimal" approach either as a standard library or a third party library to get finer precision.
For instance, I've found the inf.v0 package largely useful. Underlying the library is a Dec struct that holds the exponents and the integer value. Therefore, it's able to hold 1.003 as 1003 * 10^-3. See below for an example:
package main
import (
"fmt"
"gopkg.in/inf.v0"
)
func main() {
// represents 1003 * 10^-3
someDec := inf.NewDec(1003, 3)
// multiply someDec by 1000 * 10^0
// which translates to 1003 * 10^-3 * 1000 * 10^0
someDec.Mul(someDec, inf.NewDec(1000, 0))
// inf.RoundHalfUp rounds half up in the 0th scale, eg. 0.5 rounds to 1
value, ok := someDec.Round(someDec, 0, inf.RoundHalfUp).Unscaled()
fmt.Println(value, ok)
}
Hope this helps!
I have a big.Int variable, and wish to find the cube root of it.
Is this implemented somewhere in the library? The Exp function seems to only take an integer, and big.Rat seems to be lacking Exp entirely.
Sadly but there is no such function in math/big package. This means that you have to roll out something on your own. One of the easiest to understand and implement is the Newton's method.
All you need is to select some starting number x_0 and use the recursive formula
You have to use it in the following way: Let your integer is b. Then your x^3 = b^3 and your f(x) = x^3 - b^3 and f'(x) = 3 * x^2.
So you need to iterate:
x_{n+1}=x_n - \frac{x_{n}^{3} + b^3}{3x_{n}^{2}}
(check this link with the image of the formula, SO sucks in inserting math formulas).
You start from a guess and ends when previous x_n is close to the next one. How close is for you to decide.
P.S.1 you can look for more complicated numerical methods to find roots (you will need less iterations to converge)
P.S.2 if you need, I wrote a method for calculating arbitrary precision square roots in Go.
I've implemented a cube root function for big.Int, both using a simple binary search and Newton's method as per Salvador Dali's formula. Although I am pretty sure there is room for improvement, here is the code that I've got:
var (
n0 = big.NewInt(0)
n1 = big.NewInt(1)
n2 = big.NewInt(2)
n3 = big.NewInt(3)
n10 = big.NewInt(10)
)
func cbrtBinary(i *big.Int) (cbrt *big.Int, rem *big.Int) {
var (
guess = new(big.Int).Div(i, n2)
dx = new(big.Int)
absDx = new(big.Int)
minDx = new(big.Int).Abs(i)
step = new(big.Int).Abs(new(big.Int).Div(guess, n2))
cube = new(big.Int)
)
for {
cube.Exp(guess, n3, nil)
dx.Sub(i, cube)
cmp := dx.Cmp(n0)
if cmp == 0 {
return guess, n0
}
absDx.Abs(dx)
switch absDx.Cmp(minDx) {
case -1:
minDx.Set(absDx)
case 0:
return guess, dx
}
switch cmp {
case -1:
guess.Sub(guess, step)
case +1:
guess.Add(guess, step)
}
step.Div(step, n2)
if step.Cmp(n0) == 0 {
step.Set(n1)
}
}
}
func cbrtNewton(i *big.Int) (cbrt *big.Int, rem *big.Int) {
var (
guess = new(big.Int).Div(i, n2)
guessSq = new(big.Int)
dx = new(big.Int)
absDx = new(big.Int)
minDx = new(big.Int).Abs(i)
cube = new(big.Int)
fx = new(big.Int)
fxp = new(big.Int)
step = new(big.Int)
)
for {
cube.Exp(guess, n3, nil)
dx.Sub(i, cube)
cmp := dx.Cmp(n0)
if cmp == 0 {
return guess, n0
}
fx.Sub(cube, i)
guessSq.Exp(guess, n2, nil)
fxp.Mul(n3, guessSq)
step.Div(fx, fxp)
if step.Cmp(n0) == 0 {
step.Set(n1)
}
absDx.Abs(dx)
switch absDx.Cmp(minDx) {
case -1:
minDx.Set(absDx)
case 0:
return guess, dx
}
guess.Sub(guess, step)
}
}
Surprisingly enough, the simple binary algorithm is also the fastest:
BenchmarkCbrt/binary/10e6-4 100000 19195 ns/op
BenchmarkCbrt/binary/10e12-4 30000 43634 ns/op
BenchmarkCbrt/binary/10e18-4 20000 73334 ns/op
BenchmarkCbrt/newton/10e6-4 30000 47027 ns/op
BenchmarkCbrt/newton/10e12-4 10000 123612 ns/op
BenchmarkCbrt/newton/10e18-4 10000 214884 ns/op
Here is the full code, including tests and benchmarks: https://play.golang.org/p/uoEmxRK5jgs.
There are two random functions f1(),f2().
f1() returns 1 with probability p1, and 0 with probability 1-p1.
f2() returns 1 with probability p2, and 0 with probability 1-p2.
I want to implement a new function f3() which returns 1 with probability p3(a given probability), and returns 0 with probability 1-p3. In the implemetion of function f3(), we can use function f1() and f2(), but you can't use any other random function.
If p3=0.5, an example of implemention:
int f3()
{
do
{
int a = f1();
int b = f1();
if (a==b) continue;
// when reachs here
// a==1 with probability p1(1-p1)
// b==1 with probability (1-p1)p1
if (a==1) return 1;//now returns 1 with probability 0.5
if (b==1) return 0;
}while(1)
}
This implemention of f3() will give a random function returns 1 with probability 0.5, and 0 with probability 0.5. But how to implement the f3() with p3=0.4? I have no idea.
I wonder, is that task possible? And how to implement f3()?
Thanks in advance.
p1 = 0.77 -- arbitrary value between 0 and 1
function f1()
if math.random() < p1 then
return 1
else
return 0
end
end
-- f1() is enough. We don't need f2()
p3 = 0.4 -- arbitrary value between 0 and 1
--------------------------
function f3()
left = 0
rigth = 1
repeat
middle = left + (right - left) * p1
if f1() == 1 then
right = middle
else
left = middle
end
if right < p3 then -- completely below
return 1
elseif left >= p3 then -- completely above
return 0
end
until false -- loop forever
end
This can be solved if p3 is a rational number.
We should use conditional probabilities for this.
For example, if you want to make this for p3=0.4, the method is the following:
Calculate the fractional form of p3. In our case it is p3=0.4=2/5.
Now generate as many random variables from the same distribution (let's say, from f1, we won't use f2 anyway) as the denominator, call them X1, X2, X3, X4, X5.
We should regenerate all these random X variables until their sum equals the numerator in the fractional form of p3.
Once this is achieved then we just return X1 (or any other Xn, where n was chosen independently of the values of the X variables). Since there are 2 1s among the 5 X variables (because their sum equals the numerator), the probability of X1 being 1 is exactly p3.
For irrational p3, the problem cannot be solved by using only f1. I'm not sure now, but I think, it can be solved for p3 of the form p1*q+p2*(1-q), where q is rational with a similar method, generating the appropriate amount of Xs with distribution f1 and Ys with distribution f2, until they have a specific predefined sum, and returning one of them. This still needs to be detailed.
First to say, that's a nice problem to tweak one's brain. I managed to solve the problem for p3 = 0.4, for what you just asked for! And I think, generalisation of such problem, is not so trivial. :D
Here is how, you can solve it for p3 = 0.4:
The intuition comes from your example. If we generate a number from f1() five times in an iteration, (see the code bellow), we can have 32 types of results like bellow:
1: 00000
2: 00001
3: 00010
4: 00011
.....
.....
32: 11111
Among these, there are 10 such results with exactly two 1's in it! After identifying this, the problem becomes simple. Just return 1 for any of the 4 combinations and return 0 for 6 others! (as probability 0.4 means getting 1, 4 times out of 10). You can do that like bellow:
int f3()
{
do{
int a[5];
int numberOfOneInA = 0;
for(int i = 0; i < 5; i++){
a[i] = f1();
if(a[i] == 1){
numberOfOneInA++;
}
}
if (numberOfOneInA != 2) continue;
else return a[0]; //out of 10 times, 4 times a[0] is 1!
}while(1)
}
Waiting to see a generalised solution.
Cheers!
Here is an idea that will work when p3 is of a form a/2^n (a rational number with a denominator that is a power of 2).
Generate n random numbers with probability distribution of 0.5:
x1, x2, ..., xn
Interpret this as a binary number in the range 0...2^n-1; each number in this range has equal probability. If this number is less than a, return 1, else return 0.
Now, since this question is in a context of computer science, it seems reasonable to assume that p3 is in a form of a/2^n (this a common representation of numbers in computers).
I implement the idea of anatolyg and Egor:
inline double random(void)
{
return static_cast<double>(rand()) / static_cast<double>(RAND_MAX);
}
const double p1 = 0.8;
int rand_P1(void)
{
return random() < p1;
}
int rand_P2(void)//return 0 with 0.5
{
int x, y; while (1)
{
mystep++;
x = rand_P1(); y = rand_P1();
if (x ^ y) return x;
}
}
double p3 = random();
int rand_P3(void)//anatolyg's idea
{
double tp = p3; int bit, x;
while (1)
{
if (tp * 2 >= 1) {bit = 1; tp = tp * 2 - 1;}
else {bit = 0; tp = tp * 2;}
x = rand_P2();
if (bit ^ x) return bit;
}
}
int rand2_P3(void)//Egor's idea
{
double left = 0, right = 1, mid;
while (1)
{
dashenstep++;
mid = left + (right - left) * p1;
int x = rand_P1();
if (x) right = mid; else left = mid;
if (right < p3) return 1;
if (left > p3) return 0;
}
}
With massive math computings, I get, assuming P3 is uniformly distributed in [0,1), then the expectation of Egor is (1-p1^2-(1-p1)^2)^(-1). And anatolyg is 2(1-p1^2-(1-p1)^2)^(-1).
Speaking Algorithmically , Yes It is possible to do that task done .
Even Programmatically , It is possible , but a complex problem .
Lets take an example .
Let
F1(1) = .5 which means F1(0) =.5
F2(2) = .8 which means F1(0) =.2
Let Suppose You need a F3, such that F3(1)= .128
Lets try Decomposing it .
.128
= (2^7)*(10^-3) // decompose this into know values
= (8/10)*(8/10)*(2/10)
= F2(1)&F2(1)*(20/100) // as no Fi(1)==2/10
= F2(1)&F2(1)*(5/10)*(4/10)
= F2(1)&F2(1)&F1(1)*(40/100)
= F2(1)&F2(1)&F1(1)*(8/10)*(5/10)
= F2(1)&F2(1)&F1(1)&F2(1)&F1(1)
So F3(1)=.128 if we define F3()=F2()&F2()&F2()&F1()&F1()
Similarly if you want F4(1)=.9 ,
You give it as F4(0)=F1(0) | F2(0) =F1(0)F2(0)=.5.2 =.1 ,which mean F4(1)=1-0.1=0.9
Which means F4 is zero only when both are zero which happens .
So making use this ( & , | and , not(!) , xor(^) if you want ) operations with a combinational use of f1,f2 will surely give you the F3 which is made purely out of f1,f2,
Which may be NP hard problem to find the combination which gives you the exact probability.
So, Finally the answer to your question , whether it is possible or not ? is YES and this is one way of doing it, may be many hacks can be made into it this to optimize this, which gives you any optimal way .