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
5681517
$endgroup$
|
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
5681517
$endgroup$
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
5681517
$endgroup$
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
5681517
asked Nov 3 '18 at 5:48
manoooohmanooooh
5681517
asked Nov 3 '18 at 5:48
manoooohmanooooh
5681517
asked Nov 3 '18 at 5:48
manoooohmanooooh
5681517
asked Nov 3 '18 at 5:48
manoooohmanooooh
5681517
5681517
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
votes
$begingroup$
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
5681517
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2982593%2fgiven-x-y-x-z-y-z-recover-the-tree-write-it-in-usual-nota%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
5681517
$endgroup$
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
5681517
$endgroup$
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
5681517
$endgroup$
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
5681517
answered Dec 24 '18 at 19:50
manoooohmanooooh
5681517
answered Dec 24 '18 at 19:50
manoooohmanooooh
5681517
answered Dec 24 '18 at 19:50
manoooohmanooooh
5681517
5681517
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2982593%2fgiven-x-y-x-z-y-z-recover-the-tree-write-it-in-usual-nota%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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