Mixing
Is there a way to avoid BigInteger/BigDecimal?
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I need to compute something like this (pseudocode):
// a, b, x, y are long, x,y <= 10^12
long i = (a - n)/(x*y)
and
long j = (b - n)/(x*y) - ceiling
Sometimes x*y doesn't fit in long. I would like to avoid BigDecimal/BigInteger usage as it is too costly and not needed anywhere else. Is there a smart math solution e.g. with two longs or smth like that?
Thank you!
UPDATE: Sorry, guys, one more constraint: I also have a variable computed as following (maybe it can be rewritten as well):
sum += x*y
I need to recalculate it to compare with another variable to stop the cycle.
java algorithm optimization bigdecimal
asked Jan 3 at 8:29
Dmitry SenkovichDmitry Senkovich
2,14111541
add a comment |
I need to compute something like this (pseudocode):
// a, b, x, y are long, x,y <= 10^12
long i = (a - n)/(x*y)
and
long j = (b - n)/(x*y) - ceiling
Sometimes x*y doesn't fit in long. I would like to avoid BigDecimal/BigInteger usage as it is too costly and not needed anywhere else. Is there a smart math solution e.g. with two longs or smth like that?
Thank you!
UPDATE: Sorry, guys, one more constraint: I also have a variable computed as following (maybe it can be rewritten as well):
sum += x*y
I need to recalculate it to compare with another variable to stop the cycle.
java algorithm optimization bigdecimal
asked Jan 3 at 8:29
Dmitry SenkovichDmitry Senkovich
2,14111541
Comments are not for extended discussion; this conversation has been moved to chat.
– Samuel Liew♦
Jan 3 at 23:13
add a comment |
I need to compute something like this (pseudocode):
// a, b, x, y are long, x,y <= 10^12
long i = (a - n)/(x*y)
and
long j = (b - n)/(x*y) - ceiling
Sometimes x*y doesn't fit in long. I would like to avoid BigDecimal/BigInteger usage as it is too costly and not needed anywhere else. Is there a smart math solution e.g. with two longs or smth like that?
Thank you!
UPDATE: Sorry, guys, one more constraint: I also have a variable computed as following (maybe it can be rewritten as well):
sum += x*y
I need to recalculate it to compare with another variable to stop the cycle.
java algorithm optimization bigdecimal
asked Jan 3 at 8:29
Dmitry SenkovichDmitry Senkovich
2,14111541
I need to compute something like this (pseudocode):
// a, b, x, y are long, x,y <= 10^12
long i = (a - n)/(x*y)
and
long j = (b - n)/(x*y) - ceiling
Sometimes x*y doesn't fit in long. I would like to avoid BigDecimal/BigInteger usage as it is too costly and not needed anywhere else. Is there a smart math solution e.g. with two longs or smth like that?
Thank you!
UPDATE: Sorry, guys, one more constraint: I also have a variable computed as following (maybe it can be rewritten as well):
sum += x*y
I need to recalculate it to compare with another variable to stop the cycle.
java algorithm optimization bigdecimal
java algorithm optimization bigdecimal
asked Jan 3 at 8:29
Dmitry SenkovichDmitry Senkovich
2,14111541
asked Jan 3 at 8:29
Dmitry SenkovichDmitry Senkovich
2,14111541
edited Jan 3 at 8:57
Dmitry Senkovich
asked Jan 3 at 8:29
Dmitry SenkovichDmitry Senkovich
2,14111541
asked Jan 3 at 8:29
Dmitry SenkovichDmitry Senkovich
2,14111541
asked Jan 3 at 8:29
Dmitry SenkovichDmitry Senkovich
2,14111541
2,14111541
Comments are not for extended discussion; this conversation has been moved to chat.
– Samuel Liew♦
Jan 3 at 23:13
add a comment |
Comments are not for extended discussion; this conversation has been moved to chat.
– Samuel Liew♦
Jan 3 at 23:13
Comments are not for extended discussion; this conversation has been moved to chat.
– Samuel Liew♦
Jan 3 at 23:13
Comments are not for extended discussion; this conversation has been moved to chat.
– Samuel Liew♦
Jan 3 at 23:13
add a comment |
2 Answers
2
active
oldest
votes
You could divide two times
y = a / (b * c)
y1 = a / b
y = y1 / c
It's hard to understand how you use this sum exactly and how the compared value is calculated but would it be possible to store the result of the comparison instead to avoid large values such as if result > 0 then sum is larger, if result < 0 then other value is larger,...
answered Jan 3 at 8:34
Joakim DanielsonJoakim Danielson
10.6k3725
seems to be reasonable but I also need to store the current sum of suchm*w, sorry, I've updated the question
– Dmitry Senkovich
Jan 3 at 8:42
add a comment |
I've noticed that the value I'm comparing the sum with is long. So I've solved the problem in the following way:
- I'm calculating
ilike this: (long) (((a - n) / (double) x) / y)) - Here goes
j: (long) Math.ceil((b - n) / (double) x*y)
The case above overflows of course. But I prevent overflowing doing the following try-catch before:
try {
xy = Math.multiplyExact(x, y);
...
} catch (ArithmeticException ex) {
// some handling
break;
}
This trick made by code work faster enough.
Hope it helps someone!
answered Jan 4 at 9:30
Dmitry SenkovichDmitry Senkovich
2,14111541
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You could divide two times
y = a / (b * c)
y1 = a / b
y = y1 / c
It's hard to understand how you use this sum exactly and how the compared value is calculated but would it be possible to store the result of the comparison instead to avoid large values such as if result > 0 then sum is larger, if result < 0 then other value is larger,...
answered Jan 3 at 8:34
Joakim DanielsonJoakim Danielson
10.6k3725
seems to be reasonable but I also need to store the current sum of suchm*w, sorry, I've updated the question
– Dmitry Senkovich
Jan 3 at 8:42
add a comment |
You could divide two times
y = a / (b * c)
y1 = a / b
y = y1 / c
It's hard to understand how you use this sum exactly and how the compared value is calculated but would it be possible to store the result of the comparison instead to avoid large values such as if result > 0 then sum is larger, if result < 0 then other value is larger,...
answered Jan 3 at 8:34
Joakim DanielsonJoakim Danielson
10.6k3725
seems to be reasonable but I also need to store the current sum of suchm*w, sorry, I've updated the question
– Dmitry Senkovich
Jan 3 at 8:42
add a comment |
You could divide two times
y = a / (b * c)
y1 = a / b
y = y1 / c
It's hard to understand how you use this sum exactly and how the compared value is calculated but would it be possible to store the result of the comparison instead to avoid large values such as if result > 0 then sum is larger, if result < 0 then other value is larger,...
answered Jan 3 at 8:34
Joakim DanielsonJoakim Danielson
10.6k3725
You could divide two times
y = a / (b * c)
y1 = a / b
y = y1 / c
It's hard to understand how you use this sum exactly and how the compared value is calculated but would it be possible to store the result of the comparison instead to avoid large values such as if result > 0 then sum is larger, if result < 0 then other value is larger,...
answered Jan 3 at 8:34
Joakim DanielsonJoakim Danielson
10.6k3725
edited Jan 3 at 12:42
answered Jan 3 at 8:34
Joakim DanielsonJoakim Danielson
10.6k3725
answered Jan 3 at 8:34
Joakim DanielsonJoakim Danielson
10.6k3725
answered Jan 3 at 8:34
Joakim DanielsonJoakim Danielson
10.6k3725
10.6k3725
seems to be reasonable but I also need to store the current sum of suchm*w, sorry, I've updated the question
– Dmitry Senkovich
Jan 3 at 8:42
add a comment |
seems to be reasonable but I also need to store the current sum of suchm*w, sorry, I've updated the question
– Dmitry Senkovich
Jan 3 at 8:42
seems to be reasonable but I also need to store the current sum of such
m*w, sorry, I've updated the question– Dmitry Senkovich
Jan 3 at 8:42
seems to be reasonable but I also need to store the current sum of such
m*w, sorry, I've updated the question– Dmitry Senkovich
Jan 3 at 8:42
add a comment |
I've noticed that the value I'm comparing the sum with is long. So I've solved the problem in the following way:
- I'm calculating
ilike this: (long) (((a - n) / (double) x) / y)) - Here goes
j: (long) Math.ceil((b - n) / (double) x*y)
The case above overflows of course. But I prevent overflowing doing the following try-catch before:
try {
xy = Math.multiplyExact(x, y);
...
} catch (ArithmeticException ex) {
// some handling
break;
}
This trick made by code work faster enough.
Hope it helps someone!
answered Jan 4 at 9:30
Dmitry SenkovichDmitry Senkovich
2,14111541
add a comment |
I've noticed that the value I'm comparing the sum with is long. So I've solved the problem in the following way:
- I'm calculating
ilike this: (long) (((a - n) / (double) x) / y)) - Here goes
j: (long) Math.ceil((b - n) / (double) x*y)
The case above overflows of course. But I prevent overflowing doing the following try-catch before:
try {
xy = Math.multiplyExact(x, y);
...
} catch (ArithmeticException ex) {
// some handling
break;
}
This trick made by code work faster enough.
Hope it helps someone!
answered Jan 4 at 9:30
Dmitry SenkovichDmitry Senkovich
2,14111541
add a comment |
I've noticed that the value I'm comparing the sum with is long. So I've solved the problem in the following way:
- I'm calculating
ilike this: (long) (((a - n) / (double) x) / y)) - Here goes
j: (long) Math.ceil((b - n) / (double) x*y)
The case above overflows of course. But I prevent overflowing doing the following try-catch before:
try {
xy = Math.multiplyExact(x, y);
...
} catch (ArithmeticException ex) {
// some handling
break;
}
This trick made by code work faster enough.
Hope it helps someone!
answered Jan 4 at 9:30
Dmitry SenkovichDmitry Senkovich
2,14111541
I've noticed that the value I'm comparing the sum with is long. So I've solved the problem in the following way:
- I'm calculating
ilike this: (long) (((a - n) / (double) x) / y)) - Here goes
j: (long) Math.ceil((b - n) / (double) x*y)
The case above overflows of course. But I prevent overflowing doing the following try-catch before:
try {
xy = Math.multiplyExact(x, y);
...
} catch (ArithmeticException ex) {
// some handling
break;
}
This trick made by code work faster enough.
Hope it helps someone!
answered Jan 4 at 9:30
Dmitry SenkovichDmitry Senkovich
2,14111541
answered Jan 4 at 9:30
Dmitry SenkovichDmitry Senkovich
2,14111541
answered Jan 4 at 9:30
Dmitry SenkovichDmitry Senkovich
2,14111541
answered Jan 4 at 9:30
Dmitry SenkovichDmitry Senkovich
2,14111541
2,14111541
add a comment |
add a comment |
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Comments are not for extended discussion; this conversation has been moved to chat.
– Samuel Liew♦
Jan 3 at 23:13