Given $x y + x z + * y z * +$ recover the tree, write it in usual notation and simplify
$begingroup$
Given the boolean expression given in reverse Polish notation $$x y + x z + * y z * +$$ recover the tree, write it in usual notation and simplify.
The usual notation is
$$begin{array}{ll}
&x y + x z + * y z * +\
iff&(x+y) x z + * y z * +\
iff&((x+y)+x) z * y z * +\
iff&(((x+y)+x)*z) y z * +\
iff&(((x+y)+x)*z) (y*z) +\
iff&(((x+y)+x)*z)+(y*z)\
end{array}$$
The recovery tree is
Finally, the simplification is
$$begin{array}{ll}
&(((x+y)+x)*z)+(y*z)\
iff&(2*x+y)*z+y*z\
iff&2*x*z+y*z+y*z\
iff&2*z*(x+y)
end{array}$$
Is that correct? Is it possible to write $2*xequiv2x$ and so on?
Thanks!
discrete-mathematics trees polish-notation
asked Nov 3 '18 at 5:48
manoooohmanooooh
6751518
$endgroup$
|
show 2 more comments
$begingroup$
Given the boolean expression given in reverse Polish notation $$x y + x z + * y z * +$$ recover the tree, write it in usual notation and simplify.
The usual notation is
$$begin{array}{ll}
&x y + x z + * y z * +\
iff&(x+y) x z + * y z * +\
iff&((x+y)+x) z * y z * +\
iff&(((x+y)+x)*z) y z * +\
iff&(((x+y)+x)*z) (y*z) +\
iff&(((x+y)+x)*z)+(y*z)\
end{array}$$
The recovery tree is
Finally, the simplification is
$$begin{array}{ll}
&(((x+y)+x)*z)+(y*z)\
iff&(2*x+y)*z+y*z\
iff&2*x*z+y*z+y*z\
iff&2*z*(x+y)
end{array}$$
Is that correct? Is it possible to write $2*xequiv2x$ and so on?
Thanks!
discrete-mathematics trees polish-notation
asked Nov 3 '18 at 5:48
manoooohmanooooh
6751518
$endgroup$
1
$begingroup$
Your third line is incorrect: $x~z~+$ translates to $(x+z)$ before it is multiplied by $(x+y)$.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 6:11
$begingroup$
@FabioSomenzi oh, thanks! So it would be $(x+y)*(x+y)+(y*z)$?
$endgroup$
– manooooh
Nov 3 '18 at 6:18
1
$begingroup$
The second $x+y$ is actually $x+z$, and then you can simplify a bit.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 6:24
1
$begingroup$
It's a Boolean expression, isn't it? So, $+$ is OR and $*$ is AND.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 6:35
1
$begingroup$
Right. You should post the answer, because it's your solution. I just gave a little nudge.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 17:35
|
show 2 more comments
$begingroup$
Given the boolean expression given in reverse Polish notation $$x y + x z + * y z * +$$ recover the tree, write it in usual notation and simplify.
The usual notation is
$$begin{array}{ll}
&x y + x z + * y z * +\
iff&(x+y) x z + * y z * +\
iff&((x+y)+x) z * y z * +\
iff&(((x+y)+x)*z) y z * +\
iff&(((x+y)+x)*z) (y*z) +\
iff&(((x+y)+x)*z)+(y*z)\
end{array}$$
The recovery tree is
Finally, the simplification is
$$begin{array}{ll}
&(((x+y)+x)*z)+(y*z)\
iff&(2*x+y)*z+y*z\
iff&2*x*z+y*z+y*z\
iff&2*z*(x+y)
end{array}$$
Is that correct? Is it possible to write $2*xequiv2x$ and so on?
Thanks!
discrete-mathematics trees polish-notation
asked Nov 3 '18 at 5:48
manoooohmanooooh
6751518
$endgroup$
Given the boolean expression given in reverse Polish notation $$x y + x z + * y z * +$$ recover the tree, write it in usual notation and simplify.
The usual notation is
$$begin{array}{ll}
&x y + x z + * y z * +\
iff&(x+y) x z + * y z * +\
iff&((x+y)+x) z * y z * +\
iff&(((x+y)+x)*z) y z * +\
iff&(((x+y)+x)*z) (y*z) +\
iff&(((x+y)+x)*z)+(y*z)\
end{array}$$
The recovery tree is
Finally, the simplification is
$$begin{array}{ll}
&(((x+y)+x)*z)+(y*z)\
iff&(2*x+y)*z+y*z\
iff&2*x*z+y*z+y*z\
iff&2*z*(x+y)
end{array}$$
Is that correct? Is it possible to write $2*xequiv2x$ and so on?
Thanks!
discrete-mathematics trees polish-notation
discrete-mathematics trees polish-notation
asked Nov 3 '18 at 5:48
manoooohmanooooh
6751518
asked Nov 3 '18 at 5:48
manoooohmanooooh
6751518
asked Nov 3 '18 at 5:48
manoooohmanooooh
6751518
asked Nov 3 '18 at 5:48
manoooohmanooooh
6751518
asked Nov 3 '18 at 5:48
manoooohmanooooh
6751518
6751518
1
$begingroup$
Your third line is incorrect: $x~z~+$ translates to $(x+z)$ before it is multiplied by $(x+y)$.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 6:11
$begingroup$
@FabioSomenzi oh, thanks! So it would be $(x+y)*(x+y)+(y*z)$?
$endgroup$
– manooooh
Nov 3 '18 at 6:18
1
$begingroup$
The second $x+y$ is actually $x+z$, and then you can simplify a bit.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 6:24
1
$begingroup$
It's a Boolean expression, isn't it? So, $+$ is OR and $*$ is AND.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 6:35
1
$begingroup$
Right. You should post the answer, because it's your solution. I just gave a little nudge.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 17:35
|
show 2 more comments
1
$begingroup$
Your third line is incorrect: $x~z~+$ translates to $(x+z)$ before it is multiplied by $(x+y)$.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 6:11
$begingroup$
@FabioSomenzi oh, thanks! So it would be $(x+y)*(x+y)+(y*z)$?
$endgroup$
– manooooh
Nov 3 '18 at 6:18
1
$begingroup$
The second $x+y$ is actually $x+z$, and then you can simplify a bit.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 6:24
1
$begingroup$
It's a Boolean expression, isn't it? So, $+$ is OR and $*$ is AND.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 6:35
1
$begingroup$
Right. You should post the answer, because it's your solution. I just gave a little nudge.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 17:35
1
1
$begingroup$
Your third line is incorrect: $x~z~+$ translates to $(x+z)$ before it is multiplied by $(x+y)$.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 6:11
$begingroup$
Your third line is incorrect: $x~z~+$ translates to $(x+z)$ before it is multiplied by $(x+y)$.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 6:11
$begingroup$
@FabioSomenzi oh, thanks! So it would be $(x+y)*(x+y)+(y*z)$?
$endgroup$
– manooooh
Nov 3 '18 at 6:18
$begingroup$
@FabioSomenzi oh, thanks! So it would be $(x+y)*(x+y)+(y*z)$?
$endgroup$
– manooooh
Nov 3 '18 at 6:18
1
1
$begingroup$
The second $x+y$ is actually $x+z$, and then you can simplify a bit.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 6:24
$begingroup$
The second $x+y$ is actually $x+z$, and then you can simplify a bit.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 6:24
1
1
$begingroup$
It's a Boolean expression, isn't it? So, $+$ is OR and $*$ is AND.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 6:35
$begingroup$
It's a Boolean expression, isn't it? So, $+$ is OR and $*$ is AND.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 6:35
1
1
$begingroup$
Right. You should post the answer, because it's your solution. I just gave a little nudge.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 17:35
$begingroup$
Right. You should post the answer, because it's your solution. I just gave a little nudge.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 17:35
|
show 2 more comments
1 Answer
1
active
oldest
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$begingroup$
No, it is not correct. As @FabioSomenzi said in comments, the expression must be $(x+y)*(x+z)+(y*z)$, which has as a tree
and after applying some properties ends up with $xvee(ywedge z)$.
answered Dec 24 '18 at 19:50
manoooohmanooooh
6751518
$endgroup$
add a comment |
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$begingroup$
No, it is not correct. As @FabioSomenzi said in comments, the expression must be $(x+y)*(x+z)+(y*z)$, which has as a tree
and after applying some properties ends up with $xvee(ywedge z)$.
answered Dec 24 '18 at 19:50
manoooohmanooooh
6751518
$endgroup$
add a comment |
$begingroup$
No, it is not correct. As @FabioSomenzi said in comments, the expression must be $(x+y)*(x+z)+(y*z)$, which has as a tree
and after applying some properties ends up with $xvee(ywedge z)$.
answered Dec 24 '18 at 19:50
manoooohmanooooh
6751518
$endgroup$
add a comment |
$begingroup$
No, it is not correct. As @FabioSomenzi said in comments, the expression must be $(x+y)*(x+z)+(y*z)$, which has as a tree
and after applying some properties ends up with $xvee(ywedge z)$.
answered Dec 24 '18 at 19:50
manoooohmanooooh
6751518
$endgroup$
No, it is not correct. As @FabioSomenzi said in comments, the expression must be $(x+y)*(x+z)+(y*z)$, which has as a tree
and after applying some properties ends up with $xvee(ywedge z)$.
answered Dec 24 '18 at 19:50
manoooohmanooooh
6751518
answered Dec 24 '18 at 19:50
manoooohmanooooh
6751518
answered Dec 24 '18 at 19:50
manoooohmanooooh
6751518
answered Dec 24 '18 at 19:50
manoooohmanooooh
6751518
6751518
add a comment |
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1
$begingroup$
Your third line is incorrect: $x~z~+$ translates to $(x+z)$ before it is multiplied by $(x+y)$.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 6:11
$begingroup$
@FabioSomenzi oh, thanks! So it would be $(x+y)*(x+y)+(y*z)$?
$endgroup$
– manooooh
Nov 3 '18 at 6:18
1
$begingroup$
The second $x+y$ is actually $x+z$, and then you can simplify a bit.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 6:24
1
$begingroup$
It's a Boolean expression, isn't it? So, $+$ is OR and $*$ is AND.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 6:35
1
$begingroup$
Right. You should post the answer, because it's your solution. I just gave a little nudge.
$endgroup$
– Fabio Somenzi
Nov 3 '18 at 17:35