Mixing
Assuming X = A + B * C, where A & B are integers, C is irrational; Find A & B given X & C
0
$begingroup$
I'm looking for an efficient algorithm which can solve this problem. Can I do better than with the following brute force algoritm in C++?
pair<float, float> factorize(float value) {
float min_err = 1.0f;
float best_a = 0.0f, best_b = 0.0f;
const float c = sqrt(3);
for(int b = 0, max = value / c; b <= max; b++) {
float t = value - c * b;
float err = abs(t - round(t));
if(err < min_err) {
min_err = err;
best_b = b;
best_a = t;
}
}
return {best_a, best_b};
}
It simply looks through every possible combination of A & B and tries to find the best one.
algorithms prime-factorization mixed-integer-programming
asked Jan 1 at 21:07
KrzysiuKrzysiu
11
$endgroup$
|
show 3 more comments
0
$begingroup$
I'm looking for an efficient algorithm which can solve this problem. Can I do better than with the following brute force algoritm in C++?
pair<float, float> factorize(float value) {
float min_err = 1.0f;
float best_a = 0.0f, best_b = 0.0f;
const float c = sqrt(3);
for(int b = 0, max = value / c; b <= max; b++) {
float t = value - c * b;
float err = abs(t - round(t));
if(err < min_err) {
min_err = err;
best_b = b;
best_a = t;
}
}
return {best_a, best_b};
}
It simply looks through every possible combination of A & B and tries to find the best one.
algorithms prime-factorization mixed-integer-programming
asked Jan 1 at 21:07
KrzysiuKrzysiu
11
$endgroup$
$begingroup$
If $A$ and $B$ are integers, they can be negative. Your algorithm doesn't account for that. Also, notice that $X-BC$ is an integer.
$endgroup$
– John Douma
Jan 1 at 21:16
$begingroup$
Not sure if it's good site for this stuff.
$endgroup$
– Jakobian
Jan 1 at 21:16
$begingroup$
John: I ignored negative numbers to simplify the algorithm. As for X - BC being integer: yes, I was thinking about using the fractional part of X to somehow figure out B, but I have no idea how to do that.
$endgroup$
– Krzysiu
Jan 1 at 21:21
1
$begingroup$
If $C$ is irrational then $S(C)={n+mC,|,n,min mathbb Z}$ is dense in $mathbb R$ so (unless you restrict $A,B$ in some way, e.g. making them positive) you can get arbitrarily close to $X$.
$endgroup$
– lulu
Jan 1 at 21:23
$begingroup$
You want to find $B$ such that the fractional part of $BC$ equals the fractional part of $X$.
$endgroup$
– John Douma
Jan 1 at 21:23
|
show 3 more comments
0
0
0
$begingroup$
I'm looking for an efficient algorithm which can solve this problem. Can I do better than with the following brute force algoritm in C++?
pair<float, float> factorize(float value) {
float min_err = 1.0f;
float best_a = 0.0f, best_b = 0.0f;
const float c = sqrt(3);
for(int b = 0, max = value / c; b <= max; b++) {
float t = value - c * b;
float err = abs(t - round(t));
if(err < min_err) {
min_err = err;
best_b = b;
best_a = t;
}
}
return {best_a, best_b};
}
It simply looks through every possible combination of A & B and tries to find the best one.
algorithms prime-factorization mixed-integer-programming
asked Jan 1 at 21:07
KrzysiuKrzysiu
11
$endgroup$
I'm looking for an efficient algorithm which can solve this problem. Can I do better than with the following brute force algoritm in C++?
pair<float, float> factorize(float value) {
float min_err = 1.0f;
float best_a = 0.0f, best_b = 0.0f;
const float c = sqrt(3);
for(int b = 0, max = value / c; b <= max; b++) {
float t = value - c * b;
float err = abs(t - round(t));
if(err < min_err) {
min_err = err;
best_b = b;
best_a = t;
}
}
return {best_a, best_b};
}
It simply looks through every possible combination of A & B and tries to find the best one.
algorithms prime-factorization mixed-integer-programming
algorithms prime-factorization mixed-integer-programming
asked Jan 1 at 21:07
KrzysiuKrzysiu
11
asked Jan 1 at 21:07
KrzysiuKrzysiu
11
asked Jan 1 at 21:07
KrzysiuKrzysiu
11
asked Jan 1 at 21:07
KrzysiuKrzysiu
11
asked Jan 1 at 21:07
KrzysiuKrzysiu
11
11
$begingroup$
If $A$ and $B$ are integers, they can be negative. Your algorithm doesn't account for that. Also, notice that $X-BC$ is an integer.
$endgroup$
– John Douma
Jan 1 at 21:16
$begingroup$
Not sure if it's good site for this stuff.
$endgroup$
– Jakobian
Jan 1 at 21:16
$begingroup$
John: I ignored negative numbers to simplify the algorithm. As for X - BC being integer: yes, I was thinking about using the fractional part of X to somehow figure out B, but I have no idea how to do that.
$endgroup$
– Krzysiu
Jan 1 at 21:21
1
$begingroup$
If $C$ is irrational then $S(C)={n+mC,|,n,min mathbb Z}$ is dense in $mathbb R$ so (unless you restrict $A,B$ in some way, e.g. making them positive) you can get arbitrarily close to $X$.
$endgroup$
– lulu
Jan 1 at 21:23
$begingroup$
You want to find $B$ such that the fractional part of $BC$ equals the fractional part of $X$.
$endgroup$
– John Douma
Jan 1 at 21:23
|
show 3 more comments
$begingroup$
If $A$ and $B$ are integers, they can be negative. Your algorithm doesn't account for that. Also, notice that $X-BC$ is an integer.
$endgroup$
– John Douma
Jan 1 at 21:16
$begingroup$
Not sure if it's good site for this stuff.
$endgroup$
– Jakobian
Jan 1 at 21:16
$begingroup$
John: I ignored negative numbers to simplify the algorithm. As for X - BC being integer: yes, I was thinking about using the fractional part of X to somehow figure out B, but I have no idea how to do that.
$endgroup$
– Krzysiu
Jan 1 at 21:21
1
$begingroup$
If $C$ is irrational then $S(C)={n+mC,|,n,min mathbb Z}$ is dense in $mathbb R$ so (unless you restrict $A,B$ in some way, e.g. making them positive) you can get arbitrarily close to $X$.
$endgroup$
– lulu
Jan 1 at 21:23
$begingroup$
You want to find $B$ such that the fractional part of $BC$ equals the fractional part of $X$.
$endgroup$
– John Douma
Jan 1 at 21:23
$begingroup$
If $A$ and $B$ are integers, they can be negative. Your algorithm doesn't account for that. Also, notice that $X-BC$ is an integer.
$endgroup$
– John Douma
Jan 1 at 21:16
$begingroup$
If $A$ and $B$ are integers, they can be negative. Your algorithm doesn't account for that. Also, notice that $X-BC$ is an integer.
$endgroup$
– John Douma
Jan 1 at 21:16
$begingroup$
Not sure if it's good site for this stuff.
$endgroup$
– Jakobian
Jan 1 at 21:16
$begingroup$
Not sure if it's good site for this stuff.
$endgroup$
– Jakobian
Jan 1 at 21:16
$begingroup$
John: I ignored negative numbers to simplify the algorithm. As for X - BC being integer: yes, I was thinking about using the fractional part of X to somehow figure out B, but I have no idea how to do that.
$endgroup$
– Krzysiu
Jan 1 at 21:21
$begingroup$
John: I ignored negative numbers to simplify the algorithm. As for X - BC being integer: yes, I was thinking about using the fractional part of X to somehow figure out B, but I have no idea how to do that.
$endgroup$
– Krzysiu
Jan 1 at 21:21
1
1
$begingroup$
If $C$ is irrational then $S(C)={n+mC,|,n,min mathbb Z}$ is dense in $mathbb R$ so (unless you restrict $A,B$ in some way, e.g. making them positive) you can get arbitrarily close to $X$.
$endgroup$
– lulu
Jan 1 at 21:23
$begingroup$
If $C$ is irrational then $S(C)={n+mC,|,n,min mathbb Z}$ is dense in $mathbb R$ so (unless you restrict $A,B$ in some way, e.g. making them positive) you can get arbitrarily close to $X$.
$endgroup$
– lulu
Jan 1 at 21:23
$begingroup$
You want to find $B$ such that the fractional part of $BC$ equals the fractional part of $X$.
$endgroup$
– John Douma
Jan 1 at 21:23
$begingroup$
You want to find $B$ such that the fractional part of $BC$ equals the fractional part of $X$.
$endgroup$
– John Douma
Jan 1 at 21:23
|
show 3 more comments
0
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$begingroup$
If $A$ and $B$ are integers, they can be negative. Your algorithm doesn't account for that. Also, notice that $X-BC$ is an integer.
$endgroup$
– John Douma
Jan 1 at 21:16
$begingroup$
Not sure if it's good site for this stuff.
$endgroup$
– Jakobian
Jan 1 at 21:16
$begingroup$
John: I ignored negative numbers to simplify the algorithm. As for X - BC being integer: yes, I was thinking about using the fractional part of X to somehow figure out B, but I have no idea how to do that.
$endgroup$
– Krzysiu
Jan 1 at 21:21
1
$begingroup$
If $C$ is irrational then $S(C)={n+mC,|,n,min mathbb Z}$ is dense in $mathbb R$ so (unless you restrict $A,B$ in some way, e.g. making them positive) you can get arbitrarily close to $X$.
$endgroup$
– lulu
Jan 1 at 21:23
$begingroup$
You want to find $B$ such that the fractional part of $BC$ equals the fractional part of $X$.
$endgroup$
– John Douma
Jan 1 at 21:23