Mixing
Closed Form Addition of BCD numbers
$begingroup$
Binary Coded Decimal (BCD) number representation is a 4-bit encoding which maps numbers 0-9 to their counterpart binary codes. Addition of BCD numbers can be formulated as follows:
- z = a + b (If z < 10)
- z = a + b - 10 (If z >= 10)
For instance, If we add 7 (0111) and 8 (1000), we get 15-10=5, and a carry for a higher order. My question: How can we formulate the aforementioned formula without comparison operator; i.e. what is a closed form formula for addition of 2 BCD numbers (like for example z = a+b-3ab+64a^2-128b^3, this is just an example form of what I need)?
Thanks
logic arithmetic decimal-expansion binary-programming
asked Jan 20 at 20:16
msjmsj
61
$endgroup$
add a comment |
$begingroup$
Binary Coded Decimal (BCD) number representation is a 4-bit encoding which maps numbers 0-9 to their counterpart binary codes. Addition of BCD numbers can be formulated as follows:
- z = a + b (If z < 10)
- z = a + b - 10 (If z >= 10)
For instance, If we add 7 (0111) and 8 (1000), we get 15-10=5, and a carry for a higher order. My question: How can we formulate the aforementioned formula without comparison operator; i.e. what is a closed form formula for addition of 2 BCD numbers (like for example z = a+b-3ab+64a^2-128b^3, this is just an example form of what I need)?
Thanks
logic arithmetic decimal-expansion binary-programming
asked Jan 20 at 20:16
msjmsj
61
$endgroup$
add a comment |
$begingroup$
Binary Coded Decimal (BCD) number representation is a 4-bit encoding which maps numbers 0-9 to their counterpart binary codes. Addition of BCD numbers can be formulated as follows:
- z = a + b (If z < 10)
- z = a + b - 10 (If z >= 10)
For instance, If we add 7 (0111) and 8 (1000), we get 15-10=5, and a carry for a higher order. My question: How can we formulate the aforementioned formula without comparison operator; i.e. what is a closed form formula for addition of 2 BCD numbers (like for example z = a+b-3ab+64a^2-128b^3, this is just an example form of what I need)?
Thanks
logic arithmetic decimal-expansion binary-programming
asked Jan 20 at 20:16
msjmsj
61
$endgroup$
Binary Coded Decimal (BCD) number representation is a 4-bit encoding which maps numbers 0-9 to their counterpart binary codes. Addition of BCD numbers can be formulated as follows:
- z = a + b (If z < 10)
- z = a + b - 10 (If z >= 10)
For instance, If we add 7 (0111) and 8 (1000), we get 15-10=5, and a carry for a higher order. My question: How can we formulate the aforementioned formula without comparison operator; i.e. what is a closed form formula for addition of 2 BCD numbers (like for example z = a+b-3ab+64a^2-128b^3, this is just an example form of what I need)?
Thanks
logic arithmetic decimal-expansion binary-programming
logic arithmetic decimal-expansion binary-programming
asked Jan 20 at 20:16
msjmsj
61
asked Jan 20 at 20:16
msjmsj
61
asked Jan 20 at 20:16
msjmsj
61
asked Jan 20 at 20:16
msjmsj
61
asked Jan 20 at 20:16
msjmsj
61
61
add a comment |
add a comment |
1 Answer
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votes
$begingroup$
It will be an enormous expression but you can do it. If we note that $z=7$ when $a=4$ and $b=3$ we can have a term on the right that is $a(a-1)(a-2)(a-3)(a-5)ldots (a-9)b(b-1)(b-2)(b-4)ldots (b-9)$ divided by a constant. The point is that this term is $0$ unless $a=4,b=3$ and we choose the constant to make the product $7$ when $a=4,b=3$. Another $99$ terms will get the job done. If you write it out, you can feed it to a computer algebra package to expand and collect terms for you.
answered Jan 20 at 20:24
Ross MillikanRoss Millikan
298k24200373
$endgroup$
$begingroup$
Good idea, but this leads to real coefficients. I can only use integer coefficients or powers of 2 and 10 (positive or negative). Do you have any more idea what we can do?
$endgroup$
– msj
Jan 20 at 21:23
$begingroup$
They are rational, so you can store them as numerator and denominator. They will be around $9!^2$ so too large for $32$ bit integers
$endgroup$
– Ross Millikan
Jan 20 at 21:33
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
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active
oldest
votes
$begingroup$
It will be an enormous expression but you can do it. If we note that $z=7$ when $a=4$ and $b=3$ we can have a term on the right that is $a(a-1)(a-2)(a-3)(a-5)ldots (a-9)b(b-1)(b-2)(b-4)ldots (b-9)$ divided by a constant. The point is that this term is $0$ unless $a=4,b=3$ and we choose the constant to make the product $7$ when $a=4,b=3$. Another $99$ terms will get the job done. If you write it out, you can feed it to a computer algebra package to expand and collect terms for you.
answered Jan 20 at 20:24
Ross MillikanRoss Millikan
298k24200373
$endgroup$
$begingroup$
Good idea, but this leads to real coefficients. I can only use integer coefficients or powers of 2 and 10 (positive or negative). Do you have any more idea what we can do?
$endgroup$
– msj
Jan 20 at 21:23
$begingroup$
They are rational, so you can store them as numerator and denominator. They will be around $9!^2$ so too large for $32$ bit integers
$endgroup$
– Ross Millikan
Jan 20 at 21:33
add a comment |
$begingroup$
It will be an enormous expression but you can do it. If we note that $z=7$ when $a=4$ and $b=3$ we can have a term on the right that is $a(a-1)(a-2)(a-3)(a-5)ldots (a-9)b(b-1)(b-2)(b-4)ldots (b-9)$ divided by a constant. The point is that this term is $0$ unless $a=4,b=3$ and we choose the constant to make the product $7$ when $a=4,b=3$. Another $99$ terms will get the job done. If you write it out, you can feed it to a computer algebra package to expand and collect terms for you.
answered Jan 20 at 20:24
Ross MillikanRoss Millikan
298k24200373
$endgroup$
$begingroup$
Good idea, but this leads to real coefficients. I can only use integer coefficients or powers of 2 and 10 (positive or negative). Do you have any more idea what we can do?
$endgroup$
– msj
Jan 20 at 21:23
$begingroup$
They are rational, so you can store them as numerator and denominator. They will be around $9!^2$ so too large for $32$ bit integers
$endgroup$
– Ross Millikan
Jan 20 at 21:33
add a comment |
$begingroup$
It will be an enormous expression but you can do it. If we note that $z=7$ when $a=4$ and $b=3$ we can have a term on the right that is $a(a-1)(a-2)(a-3)(a-5)ldots (a-9)b(b-1)(b-2)(b-4)ldots (b-9)$ divided by a constant. The point is that this term is $0$ unless $a=4,b=3$ and we choose the constant to make the product $7$ when $a=4,b=3$. Another $99$ terms will get the job done. If you write it out, you can feed it to a computer algebra package to expand and collect terms for you.
answered Jan 20 at 20:24
Ross MillikanRoss Millikan
298k24200373
$endgroup$
It will be an enormous expression but you can do it. If we note that $z=7$ when $a=4$ and $b=3$ we can have a term on the right that is $a(a-1)(a-2)(a-3)(a-5)ldots (a-9)b(b-1)(b-2)(b-4)ldots (b-9)$ divided by a constant. The point is that this term is $0$ unless $a=4,b=3$ and we choose the constant to make the product $7$ when $a=4,b=3$. Another $99$ terms will get the job done. If you write it out, you can feed it to a computer algebra package to expand and collect terms for you.
answered Jan 20 at 20:24
Ross MillikanRoss Millikan
298k24200373
answered Jan 20 at 20:24
Ross MillikanRoss Millikan
298k24200373
answered Jan 20 at 20:24
Ross MillikanRoss Millikan
298k24200373
answered Jan 20 at 20:24
Ross MillikanRoss Millikan
298k24200373
298k24200373
$begingroup$
Good idea, but this leads to real coefficients. I can only use integer coefficients or powers of 2 and 10 (positive or negative). Do you have any more idea what we can do?
$endgroup$
– msj
Jan 20 at 21:23
$begingroup$
They are rational, so you can store them as numerator and denominator. They will be around $9!^2$ so too large for $32$ bit integers
$endgroup$
– Ross Millikan
Jan 20 at 21:33
add a comment |
$begingroup$
Good idea, but this leads to real coefficients. I can only use integer coefficients or powers of 2 and 10 (positive or negative). Do you have any more idea what we can do?
$endgroup$
– msj
Jan 20 at 21:23
$begingroup$
They are rational, so you can store them as numerator and denominator. They will be around $9!^2$ so too large for $32$ bit integers
$endgroup$
– Ross Millikan
Jan 20 at 21:33
$begingroup$
Good idea, but this leads to real coefficients. I can only use integer coefficients or powers of 2 and 10 (positive or negative). Do you have any more idea what we can do?
$endgroup$
– msj
Jan 20 at 21:23
$begingroup$
Good idea, but this leads to real coefficients. I can only use integer coefficients or powers of 2 and 10 (positive or negative). Do you have any more idea what we can do?
$endgroup$
– msj
Jan 20 at 21:23
$begingroup$
They are rational, so you can store them as numerator and denominator. They will be around $9!^2$ so too large for $32$ bit integers
$endgroup$
– Ross Millikan
Jan 20 at 21:33
$begingroup$
They are rational, so you can store them as numerator and denominator. They will be around $9!^2$ so too large for $32$ bit integers
$endgroup$
– Ross Millikan
Jan 20 at 21:33
add a comment |
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