This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 5 years ago.
I am quite new to learning python and was playing around with the math functions. I tried to create a function that would allow you to find certain powers e.g. squares, cubes.
Why is it when I run the code below it lists most of the powers required but manages to miss some.
def more_powers():
print "For which power do you wish to find: "
power = int(raw_input("> "))
print "Choose the upperbound: "
n = int(raw_input("> "))
for num in range(2, n):
for base in range(2, num):
if log(num, base) / power == 1:
print "%d is a power of %d." % (num, base)
else:
base += 1
For which power do you wish to find:
> 3
Choose the upperbound:
> 5000
8 is a power of 2.
27 is a power of 3.
64 is a power of 4.
343 is a power of 7.
512 is a power of 8.
729 is a power of 9.
1331 is a power of 11.
1728 is a power of 12.
2197 is a power of 13.
2744 is a power of 14.
3375 is a power of 15.
4096 is a power of 16.
As you can see it misses out the equivalent power for 5, 6, 10 and 17.
Hint:
>>> log(125, 5) / 3 == 1
False
>>> log(125, 5)
3.0000000000000004
>>> log(216, 6) / 3 == 1
False
>>> log(216, 6)
3.0000000000000004
Related
This question already has an answer here:
How to use arrayformula with formulas that do not seem to support arrayformulas?
(1 answer)
Closed 4 months ago.
Using an array formula I want to find the max value of each row of a range and get the resulting range to work with it further.
The problem occurs as soon as I add the MAX() statement since it does seem to behave strangely within an array formula. Even if you ad commands which will give you multiple values within the MAX() statement it does always only return one single value.
E.g. this will give you the ranges which I want to get the max of:
=ARRAYFORMULA(ADDRESS(ROW(E1:E11); COLUMN() + 1; 4) & ":" & ADDRESS(ROW(E1:E11); COLUMN() + 4; 4))
The result looks like the following:
F1:I1
F2:I2
F3:I3
F4:I4
F5:I5
F6:I6
F7:I7
F8:I8
F9:I9
F10:I10
F11:I11
If I now add INDIRECT() to make those to actual ranges and add MAX() it should return the max of each of those ranges since the array formula should go through the ROW(E1:11) as it did bevor. However, the result of this new formula
=ARRAYFORMULA(MAX(INDIRECT(ADDRESS(ROW(E1:E11); COLUMN() + 1; 4) & ":" & ADDRESS(ROW(E1:E11); COLUMN() + 4; 4))))
rather is one single value, the maximum of the first range.
I have even tried to bypass the problem by adding an IF() statement for the array formula to iterate through the rows. Doing so, it did give me a result for all 11 rows, however, the result always was the same (the max of the first row).
The new formula:
=ARRAYFORMULA(IF(ROW(E1:E11) = ROW(E1:E11); MAX(INDIRECT(ADDRESS(ROW(E1:E11); COLUMN() + 1; 4) & ":" & ADDRESS(ROW(E1:E11); COLUMN() + 4; 4))); ""))
The new result (left column are the results of the formula, trying to get the max of each row to its right):
10 7 10 4 1
10 10 8 1 2
10 4 5 9 4
10 10 10 2 2
10 10 10 5 10
10 10 6 9 5
10 4 5 7 3
10 6 9 4 7
10 5 5 7 3
10 9 2 3 10
10 10 3 9 10
=QUERY(TRANSPOSE(QUERY(TRANSPOSE(F1:I),
"select "®EXREPLACE(JOIN( , ARRAYFORMULA(IF(LEN(F1:F&G1:G&H1:H&I1:I),
"max(Col"&ROW(F1:F)-ROW(F1)+1&"),", ""))), ".\z", "")&"")),
"select Col2")
I'm trying to solve programming question, a term called "FiPrima". The "FiPrima" number is the sum of prime numbers before, until the intended prime tribe.
INPUT FORMAT
The first line is an integer number n. Then followed by an integer number x for n times.
OUTPUT FORMAT
Output n number of rows. Each row must contain the xth "FiPrima" number of each line.
INPUT EXAMPLE
5
1 2 3 4 5
OUTPUT EXAMPLE
2
5
10
17
28
EXPLANATION
The first 5 prime numbers in order are 2, 3, 5, 7 and 13.
So:
The 1st FiPrima number is 2 (2)
The 2nd FiPrima number is 5 (2 + 3)
The 3rd FiPrima number is 10 (2 + 3 + 5)
The 4th FiPrima number is 17 (2 + 3 + 5 + 7)
The 5th FiPrima number is 28 (2 + 3 + 5 + 7 + 13)
CONSTRAINTS
1 ≤ n ≤ 100
1 ≤ x ≤ 100
Can anyone create the code ?
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.
This question already has answers here:
What is the result of % in Python?
(20 answers)
Closed 6 years ago.
for x in xrange(12):
if x % 2 == 1:
continue
print x
i know what it does, but the language doesn't make sense to me. In particular the second line is where i am lost.
if x % 2 == 1 means "if x modulo 2 equals 1".
Modulo (or mod) is the remainder after division. So, for example:
3 mod 2 = 1
12 mod 5 = 2
15 mod 6 = 3
For x mod 2, you're there's a remainder if and only iff x is odd. (Because all even numbers are divisible by two with 0 remainder.) Likewise, odd numbers will always have a remainder of 1.
So x % 2 == 1 returns true if x is odd.
This question already has answers here:
Closed 10 years ago.
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