method of long division based on 10s complement
$begingroup$
There are two ways of performing subtraction using ten's complement, using an algorithm similar to two's complement on most (practically all?) CPUs. Is there a good way to structure a division problem so it's convenient to use this algorithm?
Here's a sample of the first way:
145
- 67
-------- complement of minuend
854
+ 67
-------- addition
921
-------- complement of result
078
Here's a sample of the second way, as a bonus it demonstrates what happens when the minuend has fewer digits than the subtrahend and how carries are resolved.
145
- 67
--------- complement of subtrahend
145
+ 932
--------- addition
1077
--------- "complement" of result (subtract 1000 and add 1).
78
Despite the vertical space they consume here, these methods are fairly convenient for pencil and paper use, with one glaring exception.
Used as a "subroutine" in long division, the constant need to swap digits with difference-from-9 counterparts is very impractical, as is the fact that the "implicit zeroes" to the right of the current partial quotient need to be complemented and changed into 9s.
28r8
_______
31 ) 876 0 #1 start
62 20 #2 partial quotient
876 #3 dup #1
379 #4 complement #2
1255 #5 add #4, #5
256 #6 complement #5
248 8 #7 partial quotient
256 #8 dup #6
751 #9 complement of #7
1007 #10 add #8, #9
8 #11 complement of #10
Is there a way of doing long division with this method of subtraction that doesn't involve repeating digits so many times?
arithmetic
asked Dec 16 '18 at 2:33
Gregory NisbetGregory Nisbet
734612
$endgroup$
add a comment |
$begingroup$
There are two ways of performing subtraction using ten's complement, using an algorithm similar to two's complement on most (practically all?) CPUs. Is there a good way to structure a division problem so it's convenient to use this algorithm?
Here's a sample of the first way:
145
- 67
-------- complement of minuend
854
+ 67
-------- addition
921
-------- complement of result
078
Here's a sample of the second way, as a bonus it demonstrates what happens when the minuend has fewer digits than the subtrahend and how carries are resolved.
145
- 67
--------- complement of subtrahend
145
+ 932
--------- addition
1077
--------- "complement" of result (subtract 1000 and add 1).
78
Despite the vertical space they consume here, these methods are fairly convenient for pencil and paper use, with one glaring exception.
Used as a "subroutine" in long division, the constant need to swap digits with difference-from-9 counterparts is very impractical, as is the fact that the "implicit zeroes" to the right of the current partial quotient need to be complemented and changed into 9s.
28r8
_______
31 ) 876 0 #1 start
62 20 #2 partial quotient
876 #3 dup #1
379 #4 complement #2
1255 #5 add #4, #5
256 #6 complement #5
248 8 #7 partial quotient
256 #8 dup #6
751 #9 complement of #7
1007 #10 add #8, #9
8 #11 complement of #10
Is there a way of doing long division with this method of subtraction that doesn't involve repeating digits so many times?
arithmetic
asked Dec 16 '18 at 2:33
Gregory NisbetGregory Nisbet
734612
$endgroup$
add a comment |
$begingroup$
There are two ways of performing subtraction using ten's complement, using an algorithm similar to two's complement on most (practically all?) CPUs. Is there a good way to structure a division problem so it's convenient to use this algorithm?
Here's a sample of the first way:
145
- 67
-------- complement of minuend
854
+ 67
-------- addition
921
-------- complement of result
078
Here's a sample of the second way, as a bonus it demonstrates what happens when the minuend has fewer digits than the subtrahend and how carries are resolved.
145
- 67
--------- complement of subtrahend
145
+ 932
--------- addition
1077
--------- "complement" of result (subtract 1000 and add 1).
78
Despite the vertical space they consume here, these methods are fairly convenient for pencil and paper use, with one glaring exception.
Used as a "subroutine" in long division, the constant need to swap digits with difference-from-9 counterparts is very impractical, as is the fact that the "implicit zeroes" to the right of the current partial quotient need to be complemented and changed into 9s.
28r8
_______
31 ) 876 0 #1 start
62 20 #2 partial quotient
876 #3 dup #1
379 #4 complement #2
1255 #5 add #4, #5
256 #6 complement #5
248 8 #7 partial quotient
256 #8 dup #6
751 #9 complement of #7
1007 #10 add #8, #9
8 #11 complement of #10
Is there a way of doing long division with this method of subtraction that doesn't involve repeating digits so many times?
arithmetic
asked Dec 16 '18 at 2:33
Gregory NisbetGregory Nisbet
734612
$endgroup$
There are two ways of performing subtraction using ten's complement, using an algorithm similar to two's complement on most (practically all?) CPUs. Is there a good way to structure a division problem so it's convenient to use this algorithm?
Here's a sample of the first way:
145
- 67
-------- complement of minuend
854
+ 67
-------- addition
921
-------- complement of result
078
Here's a sample of the second way, as a bonus it demonstrates what happens when the minuend has fewer digits than the subtrahend and how carries are resolved.
145
- 67
--------- complement of subtrahend
145
+ 932
--------- addition
1077
--------- "complement" of result (subtract 1000 and add 1).
78
Despite the vertical space they consume here, these methods are fairly convenient for pencil and paper use, with one glaring exception.
Used as a "subroutine" in long division, the constant need to swap digits with difference-from-9 counterparts is very impractical, as is the fact that the "implicit zeroes" to the right of the current partial quotient need to be complemented and changed into 9s.
28r8
_______
31 ) 876 0 #1 start
62 20 #2 partial quotient
876 #3 dup #1
379 #4 complement #2
1255 #5 add #4, #5
256 #6 complement #5
248 8 #7 partial quotient
256 #8 dup #6
751 #9 complement of #7
1007 #10 add #8, #9
8 #11 complement of #10
Is there a way of doing long division with this method of subtraction that doesn't involve repeating digits so many times?
arithmetic
arithmetic
asked Dec 16 '18 at 2:33
Gregory NisbetGregory Nisbet
734612
asked Dec 16 '18 at 2:33
Gregory NisbetGregory Nisbet
734612
asked Dec 16 '18 at 2:33
Gregory NisbetGregory Nisbet
734612
asked Dec 16 '18 at 2:33
Gregory NisbetGregory Nisbet
734612
asked Dec 16 '18 at 2:33
Gregory NisbetGregory Nisbet
734612
734612
add a comment |
add a comment |
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