Pairing unit fractions
$begingroup$
(I was suggested to post here from stack overflow so hey guys!)
I have been given a problem where fractions between 1/2 - 1/1000 have to be added to create the longest sequence of all unique unit fractions.
The rules on constructing these fractions:
Let create a set: D , D is only to hold unique unit fractions , sub-fractions can add up to the same fraction for example:
1/10
/
1/110 + 1/11 1/35 + 1/14
All sub-fractions can be held within the set as long as they themselves are not the same fractions once we are adding them together it is ok for them to total up to the same root.
The goal:
The fractions have to be added in a way to sum up to exactly 1. Any sub-fractions are not allowed to be over 1000 it is explicitly between 2 and 1000 hence the fractions which make up 1/1000 would not be applicable ( 1/1004 + 1/251000 ).
What currently I found to be the most effective:
Find the two lowest multiples which make-up the current fraction that I am looking at so for e.g 1/6 = A = 3 , B = 2. And now we complete the following equation: C = (A+B)*A , D = (A+B)*B. Now C & D are the sub-fractions which add up to my initial fraction
1/6
/
1/15 1/10
In code:
public static int provideFactorsSmallest(int n) {
int k = new int[2];
for(int i = 2; i <= n - 1; i++) {
if(n % i == 0) {
k[0] = i;
break;
}
}
for(int i = k[0] + 1; i <= n - 1 && k[0] != 0; i++) {
//System.out.println("I HAVE BEEN ENTERED");
if(k[0] * i == n) {
k[1] = i;
int firstTerm = k[0];
int secondTerm = k[1];
k[0] = (firstTerm + secondTerm) * firstTerm;
k[1] = (firstTerm + secondTerm) * secondTerm;
return k;
}
}
return null;
}
My question:
What would be the most effective way to pair and group the numbers to achieve possible longest fraction sequence?
java algorithm
edited 7 hours ago
mdfst13
17.9k62157
asked 8 hours ago
MKUltraMKUltra
111
New contributor
MKUltra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
(I was suggested to post here from stack overflow so hey guys!)
I have been given a problem where fractions between 1/2 - 1/1000 have to be added to create the longest sequence of all unique unit fractions.
The rules on constructing these fractions:
Let create a set: D , D is only to hold unique unit fractions , sub-fractions can add up to the same fraction for example:
1/10
/
1/110 + 1/11 1/35 + 1/14
All sub-fractions can be held within the set as long as they themselves are not the same fractions once we are adding them together it is ok for them to total up to the same root.
The goal:
The fractions have to be added in a way to sum up to exactly 1. Any sub-fractions are not allowed to be over 1000 it is explicitly between 2 and 1000 hence the fractions which make up 1/1000 would not be applicable ( 1/1004 + 1/251000 ).
What currently I found to be the most effective:
Find the two lowest multiples which make-up the current fraction that I am looking at so for e.g 1/6 = A = 3 , B = 2. And now we complete the following equation: C = (A+B)*A , D = (A+B)*B. Now C & D are the sub-fractions which add up to my initial fraction
1/6
/
1/15 1/10
In code:
public static int provideFactorsSmallest(int n) {
int k = new int[2];
for(int i = 2; i <= n - 1; i++) {
if(n % i == 0) {
k[0] = i;
break;
}
}
for(int i = k[0] + 1; i <= n - 1 && k[0] != 0; i++) {
//System.out.println("I HAVE BEEN ENTERED");
if(k[0] * i == n) {
k[1] = i;
int firstTerm = k[0];
int secondTerm = k[1];
k[0] = (firstTerm + secondTerm) * firstTerm;
k[1] = (firstTerm + secondTerm) * secondTerm;
return k;
}
}
return null;
}
My question:
What would be the most effective way to pair and group the numbers to achieve possible longest fraction sequence?
java algorithm
edited 7 hours ago
mdfst13
17.9k62157
asked 8 hours ago
MKUltraMKUltra
111
New contributor
MKUltra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
(I was suggested to post here from stack overflow so hey guys!)
I have been given a problem where fractions between 1/2 - 1/1000 have to be added to create the longest sequence of all unique unit fractions.
The rules on constructing these fractions:
Let create a set: D , D is only to hold unique unit fractions , sub-fractions can add up to the same fraction for example:
1/10
/
1/110 + 1/11 1/35 + 1/14
All sub-fractions can be held within the set as long as they themselves are not the same fractions once we are adding them together it is ok for them to total up to the same root.
The goal:
The fractions have to be added in a way to sum up to exactly 1. Any sub-fractions are not allowed to be over 1000 it is explicitly between 2 and 1000 hence the fractions which make up 1/1000 would not be applicable ( 1/1004 + 1/251000 ).
What currently I found to be the most effective:
Find the two lowest multiples which make-up the current fraction that I am looking at so for e.g 1/6 = A = 3 , B = 2. And now we complete the following equation: C = (A+B)*A , D = (A+B)*B. Now C & D are the sub-fractions which add up to my initial fraction
1/6
/
1/15 1/10
In code:
public static int provideFactorsSmallest(int n) {
int k = new int[2];
for(int i = 2; i <= n - 1; i++) {
if(n % i == 0) {
k[0] = i;
break;
}
}
for(int i = k[0] + 1; i <= n - 1 && k[0] != 0; i++) {
//System.out.println("I HAVE BEEN ENTERED");
if(k[0] * i == n) {
k[1] = i;
int firstTerm = k[0];
int secondTerm = k[1];
k[0] = (firstTerm + secondTerm) * firstTerm;
k[1] = (firstTerm + secondTerm) * secondTerm;
return k;
}
}
return null;
}
My question:
What would be the most effective way to pair and group the numbers to achieve possible longest fraction sequence?
java algorithm
edited 7 hours ago
mdfst13
17.9k62157
asked 8 hours ago
MKUltraMKUltra
111
New contributor
MKUltra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
(I was suggested to post here from stack overflow so hey guys!)
I have been given a problem where fractions between 1/2 - 1/1000 have to be added to create the longest sequence of all unique unit fractions.
The rules on constructing these fractions:
Let create a set: D , D is only to hold unique unit fractions , sub-fractions can add up to the same fraction for example:
1/10
/
1/110 + 1/11 1/35 + 1/14
All sub-fractions can be held within the set as long as they themselves are not the same fractions once we are adding them together it is ok for them to total up to the same root.
The goal:
The fractions have to be added in a way to sum up to exactly 1. Any sub-fractions are not allowed to be over 1000 it is explicitly between 2 and 1000 hence the fractions which make up 1/1000 would not be applicable ( 1/1004 + 1/251000 ).
What currently I found to be the most effective:
Find the two lowest multiples which make-up the current fraction that I am looking at so for e.g 1/6 = A = 3 , B = 2. And now we complete the following equation: C = (A+B)*A , D = (A+B)*B. Now C & D are the sub-fractions which add up to my initial fraction
1/6
/
1/15 1/10
In code:
public static int provideFactorsSmallest(int n) {
int k = new int[2];
for(int i = 2; i <= n - 1; i++) {
if(n % i == 0) {
k[0] = i;
break;
}
}
for(int i = k[0] + 1; i <= n - 1 && k[0] != 0; i++) {
//System.out.println("I HAVE BEEN ENTERED");
if(k[0] * i == n) {
k[1] = i;
int firstTerm = k[0];
int secondTerm = k[1];
k[0] = (firstTerm + secondTerm) * firstTerm;
k[1] = (firstTerm + secondTerm) * secondTerm;
return k;
}
}
return null;
}
My question:
What would be the most effective way to pair and group the numbers to achieve possible longest fraction sequence?
java algorithm
java algorithm
edited 7 hours ago
mdfst13
17.9k62157
asked 8 hours ago
MKUltraMKUltra
111
New contributor
MKUltra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 7 hours ago
mdfst13
17.9k62157
asked 8 hours ago
MKUltraMKUltra
111
New contributor
MKUltra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 7 hours ago
mdfst13
17.9k62157
edited 7 hours ago
mdfst13
17.9k62157
edited 7 hours ago
mdfst13
17.9k62157
17.9k62157
asked 8 hours ago
MKUltraMKUltra
111
New contributor
MKUltra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
MKUltraMKUltra
111
asked 8 hours ago
MKUltraMKUltra
111
111
New contributor
MKUltra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
MKUltra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
MKUltra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
- In first
forloop we can stop wheni*i>=n. As we need two different divisors, smallest must be less thanMath.sqrt(n)
- Second
forloop looks like simple integer division.
Whole code is equivalent to this:
private static int provideFactorsSmallest_v(int n) {
for (int firstTerm = 2; firstTerm*firstTerm < n; firstTerm++) {
if (n % firstTerm == 0) {
int secondTerm = n / firstTerm;
return new int{
(firstTerm + secondTerm) * firstTerm,
(firstTerm + secondTerm) * secondTerm
};
}
}
return null;
}
answered 6 hours ago
Mikhail EfimovMikhail Efimov
1111
New contributor
Mikhail Efimov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
wow that return new int that got me spiced up!
$endgroup$
– MKUltra
6 hours ago
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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votes
$begingroup$
- In first
forloop we can stop wheni*i>=n. As we need two different divisors, smallest must be less thanMath.sqrt(n)
- Second
forloop looks like simple integer division.
Whole code is equivalent to this:
private static int provideFactorsSmallest_v(int n) {
for (int firstTerm = 2; firstTerm*firstTerm < n; firstTerm++) {
if (n % firstTerm == 0) {
int secondTerm = n / firstTerm;
return new int{
(firstTerm + secondTerm) * firstTerm,
(firstTerm + secondTerm) * secondTerm
};
}
}
return null;
}
answered 6 hours ago
Mikhail EfimovMikhail Efimov
1111
New contributor
Mikhail Efimov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
wow that return new int that got me spiced up!
$endgroup$
– MKUltra
6 hours ago
add a comment |
$begingroup$
- In first
forloop we can stop wheni*i>=n. As we need two different divisors, smallest must be less thanMath.sqrt(n)
- Second
forloop looks like simple integer division.
Whole code is equivalent to this:
private static int provideFactorsSmallest_v(int n) {
for (int firstTerm = 2; firstTerm*firstTerm < n; firstTerm++) {
if (n % firstTerm == 0) {
int secondTerm = n / firstTerm;
return new int{
(firstTerm + secondTerm) * firstTerm,
(firstTerm + secondTerm) * secondTerm
};
}
}
return null;
}
answered 6 hours ago
Mikhail EfimovMikhail Efimov
1111
New contributor
Mikhail Efimov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
wow that return new int that got me spiced up!
$endgroup$
– MKUltra
6 hours ago
add a comment |
$begingroup$
- In first
forloop we can stop wheni*i>=n. As we need two different divisors, smallest must be less thanMath.sqrt(n)
- Second
forloop looks like simple integer division.
Whole code is equivalent to this:
private static int provideFactorsSmallest_v(int n) {
for (int firstTerm = 2; firstTerm*firstTerm < n; firstTerm++) {
if (n % firstTerm == 0) {
int secondTerm = n / firstTerm;
return new int{
(firstTerm + secondTerm) * firstTerm,
(firstTerm + secondTerm) * secondTerm
};
}
}
return null;
}
answered 6 hours ago
Mikhail EfimovMikhail Efimov
1111
New contributor
Mikhail Efimov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
- In first
forloop we can stop wheni*i>=n. As we need two different divisors, smallest must be less thanMath.sqrt(n)
- Second
forloop looks like simple integer division.
Whole code is equivalent to this:
private static int provideFactorsSmallest_v(int n) {
for (int firstTerm = 2; firstTerm*firstTerm < n; firstTerm++) {
if (n % firstTerm == 0) {
int secondTerm = n / firstTerm;
return new int{
(firstTerm + secondTerm) * firstTerm,
(firstTerm + secondTerm) * secondTerm
};
}
}
return null;
}
answered 6 hours ago
Mikhail EfimovMikhail Efimov
1111
New contributor
Mikhail Efimov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 6 hours ago
Mikhail EfimovMikhail Efimov
1111
New contributor
Mikhail Efimov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 6 hours ago
Mikhail EfimovMikhail Efimov
1111
answered 6 hours ago
Mikhail EfimovMikhail Efimov
1111
1111
New contributor
Mikhail Efimov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mikhail Efimov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Mikhail Efimov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
wow that return new int that got me spiced up!
$endgroup$
– MKUltra
6 hours ago
add a comment |
$begingroup$
wow that return new int that got me spiced up!
$endgroup$
– MKUltra
6 hours ago
$begingroup$
wow that return new int that got me spiced up!
$endgroup$
– MKUltra
6 hours ago
$begingroup$
wow that return new int that got me spiced up!
$endgroup$
– MKUltra
6 hours ago
add a comment |
MKUltra is a new contributor. Be nice, and check out our Code of Conduct.
MKUltra is a new contributor. Be nice, and check out our Code of Conduct.
MKUltra is a new contributor. Be nice, and check out our Code of Conduct.
MKUltra is a new contributor. Be nice, and check out our Code of Conduct.
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