Mixing
Splitting $f * 2^e$ Across Digits
$begingroup$
I have a number $v = f * 2^e$ where $f$ has up to nine digits, and $0 le e le 3$. I'm interested in splitting, for example, $v=123456789 * 2^2$ into three parts: $f * 2^2 to 493, 827, 156$.
The intuitive (to me) operations are:
$2^2 * v / 10^6$ % $10^3 = 493$
$2^2 * v / 10^3$ % $10^3 = 827$
$2^2 * v / 10^0$ % $10^3 = 156$
But I know (from an algorithm paper I'm reading) that you can do:
- $v / (10^3 / 2^2) / 10^3 = 493$
$v / (10^3 / 2^2)$ % $10^3 = 827$
$v$ % $(10^3 / 2^2) * 2^2 = 156$
The equivalence of the operations are not apparent to me. Can someone provide some clarity about how the operations are equivalent?
If down voting, please indicate to me in a comment how to improve my question or a more appropriate place to ask my question.
algorithms
asked Jan 22 at 3:43
okovkookovko
1013
$endgroup$
add a comment |
$begingroup$
I have a number $v = f * 2^e$ where $f$ has up to nine digits, and $0 le e le 3$. I'm interested in splitting, for example, $v=123456789 * 2^2$ into three parts: $f * 2^2 to 493, 827, 156$.
The intuitive (to me) operations are:
$2^2 * v / 10^6$ % $10^3 = 493$
$2^2 * v / 10^3$ % $10^3 = 827$
$2^2 * v / 10^0$ % $10^3 = 156$
But I know (from an algorithm paper I'm reading) that you can do:
- $v / (10^3 / 2^2) / 10^3 = 493$
$v / (10^3 / 2^2)$ % $10^3 = 827$
$v$ % $(10^3 / 2^2) * 2^2 = 156$
The equivalence of the operations are not apparent to me. Can someone provide some clarity about how the operations are equivalent?
If down voting, please indicate to me in a comment how to improve my question or a more appropriate place to ask my question.
algorithms
asked Jan 22 at 3:43
okovkookovko
1013
$endgroup$
add a comment |
$begingroup$
I have a number $v = f * 2^e$ where $f$ has up to nine digits, and $0 le e le 3$. I'm interested in splitting, for example, $v=123456789 * 2^2$ into three parts: $f * 2^2 to 493, 827, 156$.
The intuitive (to me) operations are:
$2^2 * v / 10^6$ % $10^3 = 493$
$2^2 * v / 10^3$ % $10^3 = 827$
$2^2 * v / 10^0$ % $10^3 = 156$
But I know (from an algorithm paper I'm reading) that you can do:
- $v / (10^3 / 2^2) / 10^3 = 493$
$v / (10^3 / 2^2)$ % $10^3 = 827$
$v$ % $(10^3 / 2^2) * 2^2 = 156$
The equivalence of the operations are not apparent to me. Can someone provide some clarity about how the operations are equivalent?
If down voting, please indicate to me in a comment how to improve my question or a more appropriate place to ask my question.
algorithms
asked Jan 22 at 3:43
okovkookovko
1013
$endgroup$
I have a number $v = f * 2^e$ where $f$ has up to nine digits, and $0 le e le 3$. I'm interested in splitting, for example, $v=123456789 * 2^2$ into three parts: $f * 2^2 to 493, 827, 156$.
The intuitive (to me) operations are:
$2^2 * v / 10^6$ % $10^3 = 493$
$2^2 * v / 10^3$ % $10^3 = 827$
$2^2 * v / 10^0$ % $10^3 = 156$
But I know (from an algorithm paper I'm reading) that you can do:
- $v / (10^3 / 2^2) / 10^3 = 493$
$v / (10^3 / 2^2)$ % $10^3 = 827$
$v$ % $(10^3 / 2^2) * 2^2 = 156$
The equivalence of the operations are not apparent to me. Can someone provide some clarity about how the operations are equivalent?
If down voting, please indicate to me in a comment how to improve my question or a more appropriate place to ask my question.
algorithms
algorithms
asked Jan 22 at 3:43
okovkookovko
1013
asked Jan 22 at 3:43
okovkookovko
1013
edited Jan 22 at 6:45
okovko
asked Jan 22 at 3:43
okovkookovko
1013
asked Jan 22 at 3:43
okovkookovko
1013
asked Jan 22 at 3:43
okovkookovko
1013
1013
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Which is $v$? Is it $123456789$ or $493827156$? If you are using integer divide, splitting it into three pieces is as easy as:
$v/10^6$ to get the top three,
$v%10^3$ to get the bottom three and
$(v/10^3)%10^3$ to get the middle three.
The paper is using $v=123456789$ because it divides the modulus by $2^2$. They will work because $2^2$ divides evenly into $10^3$. If it didn't, it would be better to multiply $v$ by $2^2$ before doing the divides.
Your operations will be close for $v=123456789$ but not correct. $v/10^6=123$, so when you multiply it by $2^2$ you get $492$, which carries through the rest unchanged. You lose the carry because you divided before multiplying by $4$.
answered Jan 22 at 4:16
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
$v = f * 2^e$. For $f=123456789$ and $e=2$, $v = 493827156$. I hope that makes it clearer. So, you need need to factor in the exponent part like I did. But I see your point that multiplying by $2^2$ before doing the divides / modulus eliminates the need for a second modulus.
$endgroup$
– okovko
Jan 22 at 6:41
$begingroup$
As well as preventing errors from integer rounding.
$endgroup$
– okovko
Jan 22 at 6:50
$begingroup$
Thanks, after seeing your answer, I was able to completely reason through the way the paper does it.
$endgroup$
– okovko
Jan 22 at 7:16
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Which is $v$? Is it $123456789$ or $493827156$? If you are using integer divide, splitting it into three pieces is as easy as:
$v/10^6$ to get the top three,
$v%10^3$ to get the bottom three and
$(v/10^3)%10^3$ to get the middle three.
The paper is using $v=123456789$ because it divides the modulus by $2^2$. They will work because $2^2$ divides evenly into $10^3$. If it didn't, it would be better to multiply $v$ by $2^2$ before doing the divides.
Your operations will be close for $v=123456789$ but not correct. $v/10^6=123$, so when you multiply it by $2^2$ you get $492$, which carries through the rest unchanged. You lose the carry because you divided before multiplying by $4$.
answered Jan 22 at 4:16
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
$v = f * 2^e$. For $f=123456789$ and $e=2$, $v = 493827156$. I hope that makes it clearer. So, you need need to factor in the exponent part like I did. But I see your point that multiplying by $2^2$ before doing the divides / modulus eliminates the need for a second modulus.
$endgroup$
– okovko
Jan 22 at 6:41
$begingroup$
As well as preventing errors from integer rounding.
$endgroup$
– okovko
Jan 22 at 6:50
$begingroup$
Thanks, after seeing your answer, I was able to completely reason through the way the paper does it.
$endgroup$
– okovko
Jan 22 at 7:16
add a comment |
$begingroup$
Which is $v$? Is it $123456789$ or $493827156$? If you are using integer divide, splitting it into three pieces is as easy as:
$v/10^6$ to get the top three,
$v%10^3$ to get the bottom three and
$(v/10^3)%10^3$ to get the middle three.
The paper is using $v=123456789$ because it divides the modulus by $2^2$. They will work because $2^2$ divides evenly into $10^3$. If it didn't, it would be better to multiply $v$ by $2^2$ before doing the divides.
Your operations will be close for $v=123456789$ but not correct. $v/10^6=123$, so when you multiply it by $2^2$ you get $492$, which carries through the rest unchanged. You lose the carry because you divided before multiplying by $4$.
answered Jan 22 at 4:16
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
$v = f * 2^e$. For $f=123456789$ and $e=2$, $v = 493827156$. I hope that makes it clearer. So, you need need to factor in the exponent part like I did. But I see your point that multiplying by $2^2$ before doing the divides / modulus eliminates the need for a second modulus.
$endgroup$
– okovko
Jan 22 at 6:41
$begingroup$
As well as preventing errors from integer rounding.
$endgroup$
– okovko
Jan 22 at 6:50
$begingroup$
Thanks, after seeing your answer, I was able to completely reason through the way the paper does it.
$endgroup$
– okovko
Jan 22 at 7:16
add a comment |
$begingroup$
Which is $v$? Is it $123456789$ or $493827156$? If you are using integer divide, splitting it into three pieces is as easy as:
$v/10^6$ to get the top three,
$v%10^3$ to get the bottom three and
$(v/10^3)%10^3$ to get the middle three.
The paper is using $v=123456789$ because it divides the modulus by $2^2$. They will work because $2^2$ divides evenly into $10^3$. If it didn't, it would be better to multiply $v$ by $2^2$ before doing the divides.
Your operations will be close for $v=123456789$ but not correct. $v/10^6=123$, so when you multiply it by $2^2$ you get $492$, which carries through the rest unchanged. You lose the carry because you divided before multiplying by $4$.
answered Jan 22 at 4:16
Ross MillikanRoss Millikan
299k24200374
$endgroup$
Which is $v$? Is it $123456789$ or $493827156$? If you are using integer divide, splitting it into three pieces is as easy as:
$v/10^6$ to get the top three,
$v%10^3$ to get the bottom three and
$(v/10^3)%10^3$ to get the middle three.
The paper is using $v=123456789$ because it divides the modulus by $2^2$. They will work because $2^2$ divides evenly into $10^3$. If it didn't, it would be better to multiply $v$ by $2^2$ before doing the divides.
Your operations will be close for $v=123456789$ but not correct. $v/10^6=123$, so when you multiply it by $2^2$ you get $492$, which carries through the rest unchanged. You lose the carry because you divided before multiplying by $4$.
answered Jan 22 at 4:16
Ross MillikanRoss Millikan
299k24200374
edited Jan 22 at 4:21
answered Jan 22 at 4:16
Ross MillikanRoss Millikan
299k24200374
answered Jan 22 at 4:16
Ross MillikanRoss Millikan
299k24200374
answered Jan 22 at 4:16
Ross MillikanRoss Millikan
299k24200374
299k24200374
$begingroup$
$v = f * 2^e$. For $f=123456789$ and $e=2$, $v = 493827156$. I hope that makes it clearer. So, you need need to factor in the exponent part like I did. But I see your point that multiplying by $2^2$ before doing the divides / modulus eliminates the need for a second modulus.
$endgroup$
– okovko
Jan 22 at 6:41
$begingroup$
As well as preventing errors from integer rounding.
$endgroup$
– okovko
Jan 22 at 6:50
$begingroup$
Thanks, after seeing your answer, I was able to completely reason through the way the paper does it.
$endgroup$
– okovko
Jan 22 at 7:16
add a comment |
$begingroup$
$v = f * 2^e$. For $f=123456789$ and $e=2$, $v = 493827156$. I hope that makes it clearer. So, you need need to factor in the exponent part like I did. But I see your point that multiplying by $2^2$ before doing the divides / modulus eliminates the need for a second modulus.
$endgroup$
– okovko
Jan 22 at 6:41
$begingroup$
As well as preventing errors from integer rounding.
$endgroup$
– okovko
Jan 22 at 6:50
$begingroup$
Thanks, after seeing your answer, I was able to completely reason through the way the paper does it.
$endgroup$
– okovko
Jan 22 at 7:16
$begingroup$
$v = f * 2^e$. For $f=123456789$ and $e=2$, $v = 493827156$. I hope that makes it clearer. So, you need need to factor in the exponent part like I did. But I see your point that multiplying by $2^2$ before doing the divides / modulus eliminates the need for a second modulus.
$endgroup$
– okovko
Jan 22 at 6:41
$begingroup$
$v = f * 2^e$. For $f=123456789$ and $e=2$, $v = 493827156$. I hope that makes it clearer. So, you need need to factor in the exponent part like I did. But I see your point that multiplying by $2^2$ before doing the divides / modulus eliminates the need for a second modulus.
$endgroup$
– okovko
Jan 22 at 6:41
$begingroup$
As well as preventing errors from integer rounding.
$endgroup$
– okovko
Jan 22 at 6:50
$begingroup$
As well as preventing errors from integer rounding.
$endgroup$
– okovko
Jan 22 at 6:50
$begingroup$
Thanks, after seeing your answer, I was able to completely reason through the way the paper does it.
$endgroup$
– okovko
Jan 22 at 7:16
$begingroup$
Thanks, after seeing your answer, I was able to completely reason through the way the paper does it.
$endgroup$
– okovko
Jan 22 at 7:16
add a comment |
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