I have a question about prime numbers algorithm.
why in the following pseudo code i increases by 6 and not by 2 every iteration?
function is_prime(n)
if n ≤ 1
return false
else if n ≤ 3
return true
else if n mod 2 = 0 or n mod 3 = 0
return false
let i ← 5
while i * i ≤ n
if n mod i = 0 or n mod (i + 2) = 0
return false
i ← i + 6
return true
Thanks!
If it increased by 2 it would be testing almost everything twice, that wouldn't make any sense. So I assume you mean: how can it get away with not testing every odd number?
This is because every prime p greater than 3 is of the form 6n±1. Proof:
Consider the remainder r = p mod 6. Obviously r must be odd. Notice also that r cannot be 3, because then p would be divisible by 3, making it not a prime. This leaves only the possibilities 1 and 5, which correspond p being of the form 6n+1 or the form 6n-1 respectively.
The effect is that it avoid testing multiples of 3. Dividing by a multiple of 3 is redundant, because we already know that n is not a multiple of 3, so it cannot be the multiple of a multiple of 3 either.
The assignment in the loop body is i <- i + 6, not i <- i + 2. In the if statement the expression i + 2 just becomes a new value. There is no assignment operator in that expression.
The algorithm is based on the fact that prime numbers can be predicted using the formula 6k ± 1 and this does not apply on 2 and 3.
For instance
(6 * 1) - 1 = 5
(6 * 2) - 1 = 11
(6 * 3) - 1 = 17
The list goes on and on.
Related
I realise that there are several topics already covering this. But my question is not regarding how to build such an algorithm, rather in finding what mistake I have made in my implementation that's causing a single test out of dozens to fail.
The challenge: supplied with a std::list<int> of random numbers, determine the last digit of x1 ^ (x2 ^ (x3 ^ (... ^ xn))). These numbers are large enough, or the lists long enough, that the result is astronomical and cannot be handled by traditional means.
My solution: I chose to use a modular arithmetic approach. In short, the last digit of these huge powers will be the same as that of a reduced power consisting of the first digit of the base (mod 10), raised to the last two digits of the exponent (mod 100). The units in a sequence of powers repeat in patterns of 4 at most, so we can use mod 4 to reduce the exponent, offset by 4 to avoid remainders of 0. At least, this is my understanding of it so far based on the following resources: brilliant / quora.
#include <list>
#include <cmath>
int last_digit(std::list<int> arr)
{
// Break conditions, extract last digit
if (arr.size() == 1) return arr.back() % 10;
if (arr.size() == 0) return 1;
// Extract the last 2 items in the list
std::list<int> power(std::prev(arr.end(), 2), arr.end());
arr.pop_back(); arr.pop_back();
// Treat them as the base and exponent for this recursion
long base = power.front(), exponent = power.back(), next;
// Calculate the current power
switch (exponent)
{
case 0: next = 1; break;
case 1: next = base % 100; break;
default: next = (long)std::pow(base % 10, exponent % 4 + 4) % 100;
}
if (base != 0 && next == 0) next = 100;
// Add it as the last item in the list
arr.push_back(next);
// Recursively deal with the next 2 items in the list
return last_digit(arr);
}
Random example: 123,232 694,027 140,249 ≡ 8
First recrusion: { 123'232, 694'027, 140'249 }
base: 694,027 mod 10 = 7
exponent: 140,249 mod 4 + 4 = 5
next: 75 = 16,807 mod 100 = 7
Second recursion: { 123'232, 7 }
base: 123,232 mod 10 = 2
exponent: 7 mod 4 + 4 = 7
next: 27 = 128 mod 100 = 28
Third recursion: { 28 }
return: 28 mod 10 = 8
The problem: this works for dozens of test cases (like the one above), but fails for 2 2 101 2 ≡ 6.
By hand:
1012 = 10,201
210,201 mod 4 = 0, + 4 = 4
24 = 16 // 6 -correct
Following the algorithm, however:
First recursion: { 2, 2, 101, 2 }
base: 101 mod 10 = 1
exponent: 2 mod 4 + 4 = 6
next: 16 = 1 mod 100 = 1
Second recursion: { 2, 2, 1 } (we can already see that the result is going to be 4)
exponent = 1, next = 2 mod 100 = 2
Third recursion: { 2, 2 }
base: 2 mod 10 = 2
exponent: 2 mod 4 + 4 = 6
next: 26 = 64 mod 100 = 64
Fourth recursion: { 64 }
return 64 mod 10 = 4 // -wrong
In a way, I see what's going on, but I'm not entirely sure why it's happening for this one specific case, and not for dozens of others. I admit I'm rather pushing the limits of my maths knowledge here, but I get the impression I'm just missing a tiny part of the puzzle.
I reckon this post is long and arduous enough as it is. If anyone has any insights into where I'm going wrong, I'd appreciate some pointers.
There's a lot of problems regarding the modulo of a really big number and a lot of the sol'ns back there was basically based on basic number theory. Fermat's Little Theorem, Chinese Remainder Theorem, and the Euler's Phi Function can all help you solve such problems. I recommend you to read "A Computational Introduction to Number Theory and Algebra" by Victor Shoup. It'll help you a lot to better simplify and approach number theory-related questions better. Here's the link for the book in case you'll browse this answer.
Is there efficient way to downscale number of elements in array by decimal factor?
I want to downsize elements from one array by certain factor.
Example:
If I have 10 elements and need to scale down by factor 2.
1 2 3 4 5 6 7 8 9 10
scaled to
1.5 3.5 5.5 7.5 9.5
Grouping 2 by 2 and use arithmetic mean.
My problem is what if I need to downsize array with 10 elements to 6 elements? In theory I should group 1.6 elements and find their arithmetic mean, but how to do that?
Before suggesting a solution, let's define "downsize" in a more formal way. I would suggest this definition:
Downsizing starts with an array a[N] and produces an array b[M] such that the following is true:
M <= N - otherwise it would be upsizing, not downsizing
SUM(b) = (M/N) * SUM(a) - The sum is reduced proportionally to the number of elements
Elements of a participate in computation of b in the order of their occurrence in a
Let's consider your example of downsizing 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 to six elements. The total for your array is 55, so the total for the new array would be (6/10)*55 = 33. We can achieve this total in two steps:
Walk the array a totaling its elements until we've reached the integer part of N/M fraction (it must be an improper fraction by rule 1 above)
Let's say that a[i] was the last element of a that we could take as a whole in the current iteration. Take the fraction of a[i+1] equal to the fractional part of N/M
Continue to the next number starting with the remaining fraction of a[i+1]
Once you are done, your array b would contain M numbers totaling to SUM(a). Walk the array once more, and scale the result by N/M.
Here is how it works with your example:
b[0] = a[0] + (2/3)*a[1] = 2.33333
b[1] = (1/3)*a[1] + a[2] + (1/3)*a[3] = 5
b[2] = (2/3)*a[3] + a[4] = 7.66666
b[3] = a[5] + (2/3)*a[6] = 10.6666
b[4] = (1/3)*a[6] + a[7] + (1/3)*a[8] = 13.3333
b[5] = (2/3)*a[8] + a[9] = 16
--------
Total = 55
Scaling down by 6/10 produces the final result:
1.4 3 4.6 6.4 8 9.6 (Total = 33)
Here is a simple implementation in C++:
double need = ((double)a.size()) / b.size();
double have = 0;
size_t pos = 0;
for (size_t i = 0 ; i != a.size() ; i++) {
if (need >= have+1) {
b[pos] += a[i];
have++;
} else {
double frac = (need-have); // frac is less than 1 because of the "if" condition
b[pos++] += frac * a[i]; // frac of a[i] goes to current element of b
have = 1 - frac;
b[pos] += have * a[i]; // (1-frac) of a[i] goes to the next position of b
}
}
for (size_t i = 0 ; i != b.size() ; i++) {
b[i] /= need;
}
Demo.
You will need to resort to some form of interpolation, as the number of elements to average isn't integer.
You can consider computing the prefix sum of the array, i.e.
0 1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9 10
yields by summation
0 1 2 3 4 5 6 7 8 9
1 3 6 10 15 21 28 36 45 55
Then perform linear interpolation to get the intermediate values that you are lacking, like at 0*, 10/6, 20/6, 30/5*, 40/6, 50/6, 60/6*. (Those with an asterisk are readily available).
0 1 10/6 2 3 20/6 4 5 6 40/6 7 8 50/6 9
1 3 15/3 6 10 35/3 15 21 28 100/3 36 45 145/3 55
Now you get fractional sums by subtracting values in pairs. The first average is
(15/3-1)/(10/6) = 12/5
I can't think of anything in the C++ library that will crank out something like this, all fully cooked and ready to go.
So you'll have to, pretty much, roll up your sleeves and go to work. At this point, the question of what's the "efficient" way of doing it boils down to its very basics. Which means:
1) Calculate how big the output array should be. Based on the description of the issue, you should be able to make that calculation even before looking at the values in the input array. You know the input array's size(), you can calculate the size() of the destination array.
2) So, you resize() the destination array up front. Now, you no longer need to worry about the time wasted in growing the size of the dynamic output array, incrementally, as you go through the input array, making your calculations.
3) So what's left is the actual work: iterating over the input array, and calculating the downsized values.
auto b=input_array.begin();
auto e=input_array.end();
auto p=output_array.begin();
Don't see many other options here, besides brute force iteration and calculations. Iterate from b to e, getting your samples, calculating each downsized value, and saving the resulting value into *p++.
consider that
0 -- is the first
1 -- is the second
2 -- is the third
.....
9 -- is the 10th
11 -- is the 11th
what is an efficient algorithm to find the nth palindromic number?
I'm assuming that 0110 is not a palindrome, as it is 110.
I could spend a lot of words on describing, but this table should be enough:
#Digits #Pal. Notes
0 1 "0" only
1 9 x with x = 1..9
2 9 xx with x = 1..9
3 90 xyx with xy = 10..99 (in other words: x = 1..9, y = 0..9)
4 90 xyyx with xy = 10..99
5 900 xyzyx with xyz = 100..999
6 900 and so on...
The (nonzero) palindromes with even number of digits start at p(11) = 11, p(110) = 1001, p(1100) = 100'001,.... They are constructed by taking the index n - 10^L, where L=floor(log10(n)), and append the reversal of this number: p(1101) = 101|101, p(1102) = 102|201, ..., p(1999) = 999|999, etc. This case must be considered for indices n >= 1.1*10^L but n < 2*10^L.
When n >= 2*10^L, we get the palindromes with odd number of digits, which start with p(2) = 1, p(20) = 101, p(200) = 10001 etc., and can be constructed the same way, using again n - 10^L with L=floor(log10(n)), and appending the reversal of that number, now without its last digit: p(21) = 11|1, p(22) = 12|1, ..., p(99) = 89|8, ....
When n < 1.1*10^L, subtract 1 from L to be in the correct setting with n >= 2*10^L for the case of an odd number of digits.
This yields the simple algorithm:
p(n) = { L = logint(n,10);
P = 10^(L - [1 < n < 1.1*10^L]); /* avoid exponent -1 for n=1 */
n -= P;
RETURN( n * 10^L + reverse( n \ 10^[n >= P] ))
}
where [...] is 1 if ... is true, 0 else, and \ is integer division.
(The expression n \ 10^[...] is equivalent to: if ... then n\10 else n.)
(I added the condition n > 1 in the exponent to avoid P = 10^(-1) for n=0. If you use integer types, you don't need this. Another choice it to put max(...,0) as exponent in P, or use if n=1 then return(0) right at the start. Also notice that you don't need L after assigning P, so you could use the same variable for both.)
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Possible Duplicate:
Generating a sequence using prime numbers 2, 3, and 5 only, and then displaying an nth term (C++)
I've been brainstorming over this forever, and I just can't figure this out. I need to solve the following problem:
Generate the following sequence and display the nth term in the
sequence
2,3,4,5,6,8,9,10,12,15, etc..... Sequence only has Prime numbers
2,3,5
I need to use basic C++, such as while, for, if, etc. Nothing fancy. I can't use arrays simply because I don't know much about them yet, and I want to understand the code for the solution.
I'm not asking for a complete solution, but I am asking for guidance to get through this... please.
My problem is that I can't figure out how to check if the number if the number in the sequence is divisible by any other prime numbers other than 2, 3, and 5.
Also let's say I'm checking the number like this:
for(int i=2; i<n; i++){
if(i%2==0){
cout<<i<<", ";
}else if(i%3==0){
cout<<i<<", ";
}else if(i%5==0){
cout<<i<<", ";
}
It doesn't work simply due to the fact that it'll produce numbers such as 14, which can be divided by prime number 7. So I need to figure out how to ensure that that sequence is only divisible by 2, 3, and 5..... I've found lots of material online with solutions for the problem, but the solutions they have are far too advance, and I can't use them (also most of them are in other languages... not C++). I'm sure there's a simpler way.
The problem with your code is that you just check one of the prime factors, not all of them.
Take your example of 14. Your code only checks if 2,3 or 5 is a factor of 14, which is not exactly what you need. Indeed, you find that 2 is a factor of 14, but the other factor is 7, as you said. What you are missing is to further check if 7 has as only factors 2,3 and 5 (which is not the case). What you need to do is to eliminate all the factors 2,3 and 5 and see what is remaining.
Let's take two examples: 60 and 42
For 60
Start with factors 2
60 % 2 = 0, so now check 60 / 2 = 30.
30 % 2 = 0, so now check 30 / 2 = 15.
15 % 2 = 1, so no more factors of 2.
Go on with factors 3
15 % 3 = 0, so now check 15 / 3 = 5.
5 % 3 = 2, so no more factors of 3.
Finish with factors 5
5 % 5 = 0, so now check 5 / 5 = 1
1 % 5 = 1, so no more factors of 5.
We end up with 1, so this number is part of the sequence.
For 42
Again, start with factors 2
42 % 2 = 0, so now check 42 / 2 = 21.
21 % 2 = 1, so no more factors of 2.
Go on with factors 3
21 % 3 = 0, so now check 21 / 3 = 7.
7 % 3 = 1, so no more factors of 3.
Finish with factors 5
7 % 5 = 2, so no more factors of 5.
We end up with 7 (something different from 1), so this number is NOT part of the sequence.
So in your implementation, you should probably nest 3 while loops in your for loop to reflect this reasoning.
Store the next i value in temporary variable and then divide it by 2 as long as you can (eg. as long as i%2 == 0). Then divide by 3 as long as you can. Then by 5. And then check, what is left.
What about this?
bool try_hamming(int n)
{
while(n%2 == 0)
{
n = n/2;
}
while(n%3 == 0)
{
n = n/3;
}
while(n%5 == 0)
{
n = n/5;
}
return n==1;
}
This should return true when n is a hamming nummer and false other wise. So the main function could look something like this
#include<iostream>
using namespace std;
main()
{
for(int i=2;i<100;++i)
{
if(try_hamming(i) )
cout<< i <<",";
}
cout<<end;
}
this schould print out all Hamming numbers less then 100
How is this code working for multiple of 5
bool isMultipleof5(int n)
{
/* If n is a multiple of 5 then we make sure that last
digit of n is 0 */
if ( (n&1) == 1 )
n <<= 1;
float x = n;
x = ( (int)(x*0.1) )*10;
/* If last digit of n is 0 then n will be equal to (int)x */
if ( (int)x == n )
return true;
return false;
}
It first makes n divisable by 2.
Next, it checks if it is divisable by 10 by multiplying with 0.1 and again with 10. The idea that if it is divisable by 10, you will get back to the original, and only then.
So, if the modifies n is divisable by 10 - it is certainly divisable by 5 as well, and since modified n is always divisable by 2, if it is divisable by 5 it will be divisable by 10, and the algorithm works.
NOTE: This is very unsuggested and especially might break with large values due to floating point precision issues. using the % operator should be prefered: return (n % 5) == 0
This is how the code works with some examples.
if ( (n&1) == 1 ) //Checks if the number is odd
n <<= 1; //Multiplies the number by 2 if odd
x = ( (int)(x * 0.1) //Divides the number 10 then truncates any decimal places
* 10 ) //Multiplies it back by 10
if ( (int)x == n ) //If the floating point value equals the (semi) original value its divisible by 5
return true;
return false; //Other wise false
Example:
15 & 1 == 1 //15 is odd
15 <<= 1; //n is now 30
30 / 10 = 3;
3 * 10 = 30; //x is now 30
30 == 30 //15 is a multiple of 5
17 & 1 == 1 //17 is odd
17 <<= 1; //n is now 34
34 / 10 = 3.4;
((int)3.4 = 3) * 10 = 30; //x is now 30
30 != 34 //17 is not a multiple of 5.
As others said though just simply use the mod operator %.
This is how it works:
Double the number. Now anything ending in 5 will be divisible 10 (and also divisible by 5). n <<= 1; (the check for oddness is unnecessary (n&1) == 1)
Divide it by 10, and cast away the fractional part. (int)(x*0.1)
Multiply it by 10, so now we have the same number as in step 1 only if the number in step 1 was already divisible by 10.
The use of floating point to divide by 10 makes this algorithm dangerous and probably incorrect for large values.
Try this
bool isMultipleof5(int n)
{
return (n%5) == 0;
}
A simpler way would be
bool isMultipleof5(int n)
{
return 0 == ( n % 5 ) ;
}
#define IS_MULTIPLE_OF_5(n) (((n)%5) ? 0 : 1)
I'd agree that (n % 5) == 0 would be an ideal solution, but that wasn't really the question.
This code works because it first checks if the input is odd. If it is, it multiplies by two. Since all odd multiples of 5 end with a 5, multiplying by 2 gives a number that ends with 0.
Then it checks if the last digit is 0. This can only happen if it started as a 0 (i.e. was even, we didn't change it) or if it was odd and ended in a 5 (we multiplied by 2). So, if it ends in 0 then the input must have been divisible by 5.
I'd add that this is also an awkward way to check the value of the last digit. I'd suggest n % 10 == 0 instead, but like others mentioned... you could have just used n % 5 == 0 in the first place ;).