Mixing
Swapping registers in an old calculator
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I came up with this problem inspired by the limitations of an old non-scientific calculator I owned years ago (the two registers were the display, and an internal memory for an additional number).
We have a primitive calculator with only two registers $R_1$ and $R_2$, and the following four operations:
$R_1+R_2to R_2$ (add the content of register $R_1$ to register $R_2$.)
$-R_1+R_2to R_2$ (subtract the content of register $R_1$ from register $R_2$.)
$R_1+R_2to R_1$ (add the content of register $R_2$ to register $R_1$.)
$R_1-R_2to R_1$ (subtract the content of register $R_2$ from register $R_1$.)
For instance, if $R_1=x$ (register $R_1$ contains number $x$) and
$R_2=y$ ($R_2$ contains $y$), after applying the operation $R_1+R_2to
R_2$ we end up with $R_1=x$ and $R_2=x+y$.
Assume that initially we have $R_1=x$ and $R_2=y$, where $x$ and $y$ are arbitrary numbers. For each of the following tasks describe a sequence of operations that would allow us to perform it, or prove that the task cannot be performed (task 1 is very easy, the real challenge is about task 2):
Task 1. Swap the contents of registers $R_1$ and $R_2$ changing the sign of $y$ in the process, so we would end up with $R_1=-y$, $R_2=x$.
Task 2. Swap the contents of registers $R_1$ and $R_2$, so that we would end up with $R_1=y$, $R_2=x$.
reachability
asked Jan 3 at 5:43
mlerma54mlerma54
1705
$endgroup$
add a comment |
$begingroup$
I came up with this problem inspired by the limitations of an old non-scientific calculator I owned years ago (the two registers were the display, and an internal memory for an additional number).
We have a primitive calculator with only two registers $R_1$ and $R_2$, and the following four operations:
$R_1+R_2to R_2$ (add the content of register $R_1$ to register $R_2$.)
$-R_1+R_2to R_2$ (subtract the content of register $R_1$ from register $R_2$.)
$R_1+R_2to R_1$ (add the content of register $R_2$ to register $R_1$.)
$R_1-R_2to R_1$ (subtract the content of register $R_2$ from register $R_1$.)
For instance, if $R_1=x$ (register $R_1$ contains number $x$) and
$R_2=y$ ($R_2$ contains $y$), after applying the operation $R_1+R_2to
R_2$ we end up with $R_1=x$ and $R_2=x+y$.
Assume that initially we have $R_1=x$ and $R_2=y$, where $x$ and $y$ are arbitrary numbers. For each of the following tasks describe a sequence of operations that would allow us to perform it, or prove that the task cannot be performed (task 1 is very easy, the real challenge is about task 2):
Task 1. Swap the contents of registers $R_1$ and $R_2$ changing the sign of $y$ in the process, so we would end up with $R_1=-y$, $R_2=x$.
Task 2. Swap the contents of registers $R_1$ and $R_2$, so that we would end up with $R_1=y$, $R_2=x$.
reachability
asked Jan 3 at 5:43
mlerma54mlerma54
1705
$endgroup$
add a comment |
$begingroup$
I came up with this problem inspired by the limitations of an old non-scientific calculator I owned years ago (the two registers were the display, and an internal memory for an additional number).
We have a primitive calculator with only two registers $R_1$ and $R_2$, and the following four operations:
$R_1+R_2to R_2$ (add the content of register $R_1$ to register $R_2$.)
$-R_1+R_2to R_2$ (subtract the content of register $R_1$ from register $R_2$.)
$R_1+R_2to R_1$ (add the content of register $R_2$ to register $R_1$.)
$R_1-R_2to R_1$ (subtract the content of register $R_2$ from register $R_1$.)
For instance, if $R_1=x$ (register $R_1$ contains number $x$) and
$R_2=y$ ($R_2$ contains $y$), after applying the operation $R_1+R_2to
R_2$ we end up with $R_1=x$ and $R_2=x+y$.
Assume that initially we have $R_1=x$ and $R_2=y$, where $x$ and $y$ are arbitrary numbers. For each of the following tasks describe a sequence of operations that would allow us to perform it, or prove that the task cannot be performed (task 1 is very easy, the real challenge is about task 2):
Task 1. Swap the contents of registers $R_1$ and $R_2$ changing the sign of $y$ in the process, so we would end up with $R_1=-y$, $R_2=x$.
Task 2. Swap the contents of registers $R_1$ and $R_2$, so that we would end up with $R_1=y$, $R_2=x$.
reachability
asked Jan 3 at 5:43
mlerma54mlerma54
1705
$endgroup$
I came up with this problem inspired by the limitations of an old non-scientific calculator I owned years ago (the two registers were the display, and an internal memory for an additional number).
We have a primitive calculator with only two registers $R_1$ and $R_2$, and the following four operations:
$R_1+R_2to R_2$ (add the content of register $R_1$ to register $R_2$.)
$-R_1+R_2to R_2$ (subtract the content of register $R_1$ from register $R_2$.)
$R_1+R_2to R_1$ (add the content of register $R_2$ to register $R_1$.)
$R_1-R_2to R_1$ (subtract the content of register $R_2$ from register $R_1$.)
For instance, if $R_1=x$ (register $R_1$ contains number $x$) and
$R_2=y$ ($R_2$ contains $y$), after applying the operation $R_1+R_2to
R_2$ we end up with $R_1=x$ and $R_2=x+y$.
Assume that initially we have $R_1=x$ and $R_2=y$, where $x$ and $y$ are arbitrary numbers. For each of the following tasks describe a sequence of operations that would allow us to perform it, or prove that the task cannot be performed (task 1 is very easy, the real challenge is about task 2):
Task 1. Swap the contents of registers $R_1$ and $R_2$ changing the sign of $y$ in the process, so we would end up with $R_1=-y$, $R_2=x$.
Task 2. Swap the contents of registers $R_1$ and $R_2$, so that we would end up with $R_1=y$, $R_2=x$.
reachability
reachability
asked Jan 3 at 5:43
mlerma54mlerma54
1705
asked Jan 3 at 5:43
mlerma54mlerma54
1705
asked Jan 3 at 5:43
mlerma54mlerma54
1705
asked Jan 3 at 5:43
mlerma54mlerma54
1705
asked Jan 3 at 5:43
mlerma54mlerma54
1705
1705
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Task 1 was answered. Task 2:
Not possible. We always have $begin{bmatrix}R_1 \ R_2end{bmatrix} = Mbegin{bmatrix}x \ yend{bmatrix}$ for some matrix $M$ that depends only on the sequence of operations. Initially $M = begin{bmatrix}1 & 0 \ 0 & 1end{bmatrix}$, and the given operations cause $M$ to be left-multiplied by $begin{bmatrix}1 & 0 \ 1 & 1end{bmatrix}$, $begin{bmatrix}1 & 0 \ -1 & 1end{bmatrix}$, $begin{bmatrix}1 & 1 \ 0 & 1end{bmatrix}$, and $begin{bmatrix}1 & -1 \ 0 & 1end{bmatrix}$ respectively. All of these matrices have determinant $1$, so we will always have $det M = 1$, which makes it impossible to reach the desired $begin{bmatrix}0 & 1 \ 1 & 0end{bmatrix}$ with determinant $-1$.
answered Jan 3 at 10:25
Anders KaseorgAnders Kaseorg
56136
$endgroup$
$begingroup$
I had a feeling this was true. Very neat proof.
$endgroup$
– Dr Xorile
Jan 3 at 15:23
add a comment |
$begingroup$
Task 1
1. $R_1 - R_2to R_1$ (now $R_1 = x-y$, $R_2 = y$)
2. $R_1 + R_2to R_2$ (now $R_1 = x-y$, $R_2 = x$)
3. $R_1 - R_2to R_1$ (now $R_1 = -y$, $R_2 = x$)
answered Jan 3 at 6:12
jafejafe
18.4k351179
$endgroup$
2
$begingroup$
Task 2 the second operation isn't in the list of allowed operations
$endgroup$
– Dr Xorile
Jan 3 at 6:26
2
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@DrXorile Oops, so it seems...
$endgroup$
– jafe
Jan 3 at 6:27
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
$begingroup$
Task 1 was answered. Task 2:
Not possible. We always have $begin{bmatrix}R_1 \ R_2end{bmatrix} = Mbegin{bmatrix}x \ yend{bmatrix}$ for some matrix $M$ that depends only on the sequence of operations. Initially $M = begin{bmatrix}1 & 0 \ 0 & 1end{bmatrix}$, and the given operations cause $M$ to be left-multiplied by $begin{bmatrix}1 & 0 \ 1 & 1end{bmatrix}$, $begin{bmatrix}1 & 0 \ -1 & 1end{bmatrix}$, $begin{bmatrix}1 & 1 \ 0 & 1end{bmatrix}$, and $begin{bmatrix}1 & -1 \ 0 & 1end{bmatrix}$ respectively. All of these matrices have determinant $1$, so we will always have $det M = 1$, which makes it impossible to reach the desired $begin{bmatrix}0 & 1 \ 1 & 0end{bmatrix}$ with determinant $-1$.
answered Jan 3 at 10:25
Anders KaseorgAnders Kaseorg
56136
$endgroup$
$begingroup$
I had a feeling this was true. Very neat proof.
$endgroup$
– Dr Xorile
Jan 3 at 15:23
add a comment |
$begingroup$
Task 1 was answered. Task 2:
Not possible. We always have $begin{bmatrix}R_1 \ R_2end{bmatrix} = Mbegin{bmatrix}x \ yend{bmatrix}$ for some matrix $M$ that depends only on the sequence of operations. Initially $M = begin{bmatrix}1 & 0 \ 0 & 1end{bmatrix}$, and the given operations cause $M$ to be left-multiplied by $begin{bmatrix}1 & 0 \ 1 & 1end{bmatrix}$, $begin{bmatrix}1 & 0 \ -1 & 1end{bmatrix}$, $begin{bmatrix}1 & 1 \ 0 & 1end{bmatrix}$, and $begin{bmatrix}1 & -1 \ 0 & 1end{bmatrix}$ respectively. All of these matrices have determinant $1$, so we will always have $det M = 1$, which makes it impossible to reach the desired $begin{bmatrix}0 & 1 \ 1 & 0end{bmatrix}$ with determinant $-1$.
answered Jan 3 at 10:25
Anders KaseorgAnders Kaseorg
56136
$endgroup$
$begingroup$
I had a feeling this was true. Very neat proof.
$endgroup$
– Dr Xorile
Jan 3 at 15:23
add a comment |
$begingroup$
Task 1 was answered. Task 2:
Not possible. We always have $begin{bmatrix}R_1 \ R_2end{bmatrix} = Mbegin{bmatrix}x \ yend{bmatrix}$ for some matrix $M$ that depends only on the sequence of operations. Initially $M = begin{bmatrix}1 & 0 \ 0 & 1end{bmatrix}$, and the given operations cause $M$ to be left-multiplied by $begin{bmatrix}1 & 0 \ 1 & 1end{bmatrix}$, $begin{bmatrix}1 & 0 \ -1 & 1end{bmatrix}$, $begin{bmatrix}1 & 1 \ 0 & 1end{bmatrix}$, and $begin{bmatrix}1 & -1 \ 0 & 1end{bmatrix}$ respectively. All of these matrices have determinant $1$, so we will always have $det M = 1$, which makes it impossible to reach the desired $begin{bmatrix}0 & 1 \ 1 & 0end{bmatrix}$ with determinant $-1$.
answered Jan 3 at 10:25
Anders KaseorgAnders Kaseorg
56136
$endgroup$
Task 1 was answered. Task 2:
Not possible. We always have $begin{bmatrix}R_1 \ R_2end{bmatrix} = Mbegin{bmatrix}x \ yend{bmatrix}$ for some matrix $M$ that depends only on the sequence of operations. Initially $M = begin{bmatrix}1 & 0 \ 0 & 1end{bmatrix}$, and the given operations cause $M$ to be left-multiplied by $begin{bmatrix}1 & 0 \ 1 & 1end{bmatrix}$, $begin{bmatrix}1 & 0 \ -1 & 1end{bmatrix}$, $begin{bmatrix}1 & 1 \ 0 & 1end{bmatrix}$, and $begin{bmatrix}1 & -1 \ 0 & 1end{bmatrix}$ respectively. All of these matrices have determinant $1$, so we will always have $det M = 1$, which makes it impossible to reach the desired $begin{bmatrix}0 & 1 \ 1 & 0end{bmatrix}$ with determinant $-1$.
answered Jan 3 at 10:25
Anders KaseorgAnders Kaseorg
56136
answered Jan 3 at 10:25
Anders KaseorgAnders Kaseorg
56136
answered Jan 3 at 10:25
Anders KaseorgAnders Kaseorg
56136
answered Jan 3 at 10:25
Anders KaseorgAnders Kaseorg
56136
56136
$begingroup$
I had a feeling this was true. Very neat proof.
$endgroup$
– Dr Xorile
Jan 3 at 15:23
add a comment |
$begingroup$
I had a feeling this was true. Very neat proof.
$endgroup$
– Dr Xorile
Jan 3 at 15:23
$begingroup$
I had a feeling this was true. Very neat proof.
$endgroup$
– Dr Xorile
Jan 3 at 15:23
$begingroup$
I had a feeling this was true. Very neat proof.
$endgroup$
– Dr Xorile
Jan 3 at 15:23
add a comment |
$begingroup$
Task 1
1. $R_1 - R_2to R_1$ (now $R_1 = x-y$, $R_2 = y$)
2. $R_1 + R_2to R_2$ (now $R_1 = x-y$, $R_2 = x$)
3. $R_1 - R_2to R_1$ (now $R_1 = -y$, $R_2 = x$)
answered Jan 3 at 6:12
jafejafe
18.4k351179
$endgroup$
2
$begingroup$
Task 2 the second operation isn't in the list of allowed operations
$endgroup$
– Dr Xorile
Jan 3 at 6:26
2
$begingroup$
@DrXorile Oops, so it seems...
$endgroup$
– jafe
Jan 3 at 6:27
add a comment |
$begingroup$
Task 1
1. $R_1 - R_2to R_1$ (now $R_1 = x-y$, $R_2 = y$)
2. $R_1 + R_2to R_2$ (now $R_1 = x-y$, $R_2 = x$)
3. $R_1 - R_2to R_1$ (now $R_1 = -y$, $R_2 = x$)
answered Jan 3 at 6:12
jafejafe
18.4k351179
$endgroup$
2
$begingroup$
Task 2 the second operation isn't in the list of allowed operations
$endgroup$
– Dr Xorile
Jan 3 at 6:26
2
$begingroup$
@DrXorile Oops, so it seems...
$endgroup$
– jafe
Jan 3 at 6:27
add a comment |
$begingroup$
Task 1
1. $R_1 - R_2to R_1$ (now $R_1 = x-y$, $R_2 = y$)
2. $R_1 + R_2to R_2$ (now $R_1 = x-y$, $R_2 = x$)
3. $R_1 - R_2to R_1$ (now $R_1 = -y$, $R_2 = x$)
answered Jan 3 at 6:12
jafejafe
18.4k351179
$endgroup$
Task 1
1. $R_1 - R_2to R_1$ (now $R_1 = x-y$, $R_2 = y$)
2. $R_1 + R_2to R_2$ (now $R_1 = x-y$, $R_2 = x$)
3. $R_1 - R_2to R_1$ (now $R_1 = -y$, $R_2 = x$)
answered Jan 3 at 6:12
jafejafe
18.4k351179
edited Jan 3 at 6:27
answered Jan 3 at 6:12
jafejafe
18.4k351179
answered Jan 3 at 6:12
jafejafe
18.4k351179
answered Jan 3 at 6:12
jafejafe
18.4k351179
18.4k351179
2
$begingroup$
Task 2 the second operation isn't in the list of allowed operations
$endgroup$
– Dr Xorile
Jan 3 at 6:26
2
$begingroup$
@DrXorile Oops, so it seems...
$endgroup$
– jafe
Jan 3 at 6:27
add a comment |
2
$begingroup$
Task 2 the second operation isn't in the list of allowed operations
$endgroup$
– Dr Xorile
Jan 3 at 6:26
2
$begingroup$
@DrXorile Oops, so it seems...
$endgroup$
– jafe
Jan 3 at 6:27
2
2
$begingroup$
Task 2 the second operation isn't in the list of allowed operations
$endgroup$
– Dr Xorile
Jan 3 at 6:26
$begingroup$
Task 2 the second operation isn't in the list of allowed operations
$endgroup$
– Dr Xorile
Jan 3 at 6:26
2
2
$begingroup$
@DrXorile Oops, so it seems...
$endgroup$
– jafe
Jan 3 at 6:27
$begingroup$
@DrXorile Oops, so it seems...
$endgroup$
– jafe
Jan 3 at 6:27
add a comment |
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