Why CLRS example on residual networks does not follows its formula?
$begingroup$
I am learning algorithms to solve Maximum Flow problem by reading the CRLS book and confused by the following figure:
That is:
A flow in a residual network provides a roadmap for adding flow to the
original flow network. If $f$ is a flow in $G$ and $f'$ is a flow in
the corresponding residual network $G_f$, we define $f uparrow f'$,
the augmentation of flow $f$ by $f'$, to be a function from $V times V$ to
$R$, defined by
$$(f uparrow f')(u, v) = begin{cases} f(u,v) + f'(u, v) - f'(v, u) &
> text{if (u,v) $in$ E} \ 0 & text{otherwise} end{cases}$$
How the flow network in (c), for example $(s, v_2)$ got the flow 12 ?
If we follow the formula, it must have a flow 5:
$8 + 5 - 8 = 5$
algorithms network-flow
asked Apr 7 at 15:52
maksadbekmaksadbek
1185
$endgroup$
add a comment |
$begingroup$
I am learning algorithms to solve Maximum Flow problem by reading the CRLS book and confused by the following figure:
That is:
A flow in a residual network provides a roadmap for adding flow to the
original flow network. If $f$ is a flow in $G$ and $f'$ is a flow in
the corresponding residual network $G_f$, we define $f uparrow f'$,
the augmentation of flow $f$ by $f'$, to be a function from $V times V$ to
$R$, defined by
$$(f uparrow f')(u, v) = begin{cases} f(u,v) + f'(u, v) - f'(v, u) &
> text{if (u,v) $in$ E} \ 0 & text{otherwise} end{cases}$$
How the flow network in (c), for example $(s, v_2)$ got the flow 12 ?
If we follow the formula, it must have a flow 5:
$8 + 5 - 8 = 5$
algorithms network-flow
asked Apr 7 at 15:52
maksadbekmaksadbek
1185
$endgroup$
add a comment |
$begingroup$
I am learning algorithms to solve Maximum Flow problem by reading the CRLS book and confused by the following figure:
That is:
A flow in a residual network provides a roadmap for adding flow to the
original flow network. If $f$ is a flow in $G$ and $f'$ is a flow in
the corresponding residual network $G_f$, we define $f uparrow f'$,
the augmentation of flow $f$ by $f'$, to be a function from $V times V$ to
$R$, defined by
$$(f uparrow f')(u, v) = begin{cases} f(u,v) + f'(u, v) - f'(v, u) &
> text{if (u,v) $in$ E} \ 0 & text{otherwise} end{cases}$$
How the flow network in (c), for example $(s, v_2)$ got the flow 12 ?
If we follow the formula, it must have a flow 5:
$8 + 5 - 8 = 5$
algorithms network-flow
asked Apr 7 at 15:52
maksadbekmaksadbek
1185
$endgroup$
I am learning algorithms to solve Maximum Flow problem by reading the CRLS book and confused by the following figure:
That is:
A flow in a residual network provides a roadmap for adding flow to the
original flow network. If $f$ is a flow in $G$ and $f'$ is a flow in
the corresponding residual network $G_f$, we define $f uparrow f'$,
the augmentation of flow $f$ by $f'$, to be a function from $V times V$ to
$R$, defined by
$$(f uparrow f')(u, v) = begin{cases} f(u,v) + f'(u, v) - f'(v, u) &
> text{if (u,v) $in$ E} \ 0 & text{otherwise} end{cases}$$
How the flow network in (c), for example $(s, v_2)$ got the flow 12 ?
If we follow the formula, it must have a flow 5:
$8 + 5 - 8 = 5$
algorithms network-flow
algorithms network-flow
asked Apr 7 at 15:52
maksadbekmaksadbek
1185
asked Apr 7 at 15:52
maksadbekmaksadbek
1185
asked Apr 7 at 15:52
maksadbekmaksadbek
1185
asked Apr 7 at 15:52
maksadbekmaksadbek
1185
asked Apr 7 at 15:52
maksadbekmaksadbek
1185
1185
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
That's not what the formula gives you. As the caption says, the capacity of the augmenting path in the residual network in (b) is $4$. Therefore we send 4 units of flow along the augmenting path from $s$ to $t$, namely, the path $s to v_2 to v_3 to t$. In particular, $f(s,v_2)=8$, $f'(s,v_2)=4$, and $f'(v_2,s)=0$, so the updated flow is $8+4-0=12$.
answered Apr 7 at 17:50
D.W.♦D.W.
103k12129294
$endgroup$
add a comment |
$begingroup$
It is explained in part (b) of the caption of Figure 26.4.
The residual network $G_f$ with augmenting path $p$ shaded; its residual capacity is $c_f(p)=c_f(v_2,v_3)=4$.
Since the capacity of path $p$ is 4 (not 5), we find a flow $f'$ in the residual network $G_f$ that is defined by $f'(s,v_2)=f'(v_2,v_3)=f'(v_3,t)=4$. So for the network flow $fuparrow f'$ in (c), we have
$$ (fuparrow f')(v_2, v_3)=f(v_2,v_3)+f'(v_2,v_3) = 8+4=12.$$
answered Apr 7 at 17:51
Apass.JackApass.Jack
14.2k1940
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "419"
};
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: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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
},
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%2fcs.stackexchange.com%2fquestions%2f106608%2fwhy-clrs-example-on-residual-networks-does-not-follows-its-formula%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
That's not what the formula gives you. As the caption says, the capacity of the augmenting path in the residual network in (b) is $4$. Therefore we send 4 units of flow along the augmenting path from $s$ to $t$, namely, the path $s to v_2 to v_3 to t$. In particular, $f(s,v_2)=8$, $f'(s,v_2)=4$, and $f'(v_2,s)=0$, so the updated flow is $8+4-0=12$.
answered Apr 7 at 17:50
D.W.♦D.W.
103k12129294
$endgroup$
add a comment |
$begingroup$
That's not what the formula gives you. As the caption says, the capacity of the augmenting path in the residual network in (b) is $4$. Therefore we send 4 units of flow along the augmenting path from $s$ to $t$, namely, the path $s to v_2 to v_3 to t$. In particular, $f(s,v_2)=8$, $f'(s,v_2)=4$, and $f'(v_2,s)=0$, so the updated flow is $8+4-0=12$.
answered Apr 7 at 17:50
D.W.♦D.W.
103k12129294
$endgroup$
add a comment |
$begingroup$
That's not what the formula gives you. As the caption says, the capacity of the augmenting path in the residual network in (b) is $4$. Therefore we send 4 units of flow along the augmenting path from $s$ to $t$, namely, the path $s to v_2 to v_3 to t$. In particular, $f(s,v_2)=8$, $f'(s,v_2)=4$, and $f'(v_2,s)=0$, so the updated flow is $8+4-0=12$.
answered Apr 7 at 17:50
D.W.♦D.W.
103k12129294
$endgroup$
That's not what the formula gives you. As the caption says, the capacity of the augmenting path in the residual network in (b) is $4$. Therefore we send 4 units of flow along the augmenting path from $s$ to $t$, namely, the path $s to v_2 to v_3 to t$. In particular, $f(s,v_2)=8$, $f'(s,v_2)=4$, and $f'(v_2,s)=0$, so the updated flow is $8+4-0=12$.
answered Apr 7 at 17:50
D.W.♦D.W.
103k12129294
answered Apr 7 at 17:50
D.W.♦D.W.
103k12129294
answered Apr 7 at 17:50
D.W.♦D.W.
103k12129294
answered Apr 7 at 17:50
D.W.♦D.W.
103k12129294
103k12129294
add a comment |
add a comment |
$begingroup$
It is explained in part (b) of the caption of Figure 26.4.
The residual network $G_f$ with augmenting path $p$ shaded; its residual capacity is $c_f(p)=c_f(v_2,v_3)=4$.
Since the capacity of path $p$ is 4 (not 5), we find a flow $f'$ in the residual network $G_f$ that is defined by $f'(s,v_2)=f'(v_2,v_3)=f'(v_3,t)=4$. So for the network flow $fuparrow f'$ in (c), we have
$$ (fuparrow f')(v_2, v_3)=f(v_2,v_3)+f'(v_2,v_3) = 8+4=12.$$
answered Apr 7 at 17:51
Apass.JackApass.Jack
14.2k1940
$endgroup$
add a comment |
$begingroup$
It is explained in part (b) of the caption of Figure 26.4.
The residual network $G_f$ with augmenting path $p$ shaded; its residual capacity is $c_f(p)=c_f(v_2,v_3)=4$.
Since the capacity of path $p$ is 4 (not 5), we find a flow $f'$ in the residual network $G_f$ that is defined by $f'(s,v_2)=f'(v_2,v_3)=f'(v_3,t)=4$. So for the network flow $fuparrow f'$ in (c), we have
$$ (fuparrow f')(v_2, v_3)=f(v_2,v_3)+f'(v_2,v_3) = 8+4=12.$$
answered Apr 7 at 17:51
Apass.JackApass.Jack
14.2k1940
$endgroup$
add a comment |
$begingroup$
It is explained in part (b) of the caption of Figure 26.4.
The residual network $G_f$ with augmenting path $p$ shaded; its residual capacity is $c_f(p)=c_f(v_2,v_3)=4$.
Since the capacity of path $p$ is 4 (not 5), we find a flow $f'$ in the residual network $G_f$ that is defined by $f'(s,v_2)=f'(v_2,v_3)=f'(v_3,t)=4$. So for the network flow $fuparrow f'$ in (c), we have
$$ (fuparrow f')(v_2, v_3)=f(v_2,v_3)+f'(v_2,v_3) = 8+4=12.$$
answered Apr 7 at 17:51
Apass.JackApass.Jack
14.2k1940
$endgroup$
It is explained in part (b) of the caption of Figure 26.4.
The residual network $G_f$ with augmenting path $p$ shaded; its residual capacity is $c_f(p)=c_f(v_2,v_3)=4$.
Since the capacity of path $p$ is 4 (not 5), we find a flow $f'$ in the residual network $G_f$ that is defined by $f'(s,v_2)=f'(v_2,v_3)=f'(v_3,t)=4$. So for the network flow $fuparrow f'$ in (c), we have
$$ (fuparrow f')(v_2, v_3)=f(v_2,v_3)+f'(v_2,v_3) = 8+4=12.$$
answered Apr 7 at 17:51
Apass.JackApass.Jack
14.2k1940
edited Apr 7 at 17:52
answered Apr 7 at 17:51
Apass.JackApass.Jack
14.2k1940
answered Apr 7 at 17:51
Apass.JackApass.Jack
14.2k1940
answered Apr 7 at 17:51
Apass.JackApass.Jack
14.2k1940
14.2k1940
add a comment |
add a comment |
Thanks for contributing an answer to Computer Science 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%2fcs.stackexchange.com%2fquestions%2f106608%2fwhy-clrs-example-on-residual-networks-does-not-follows-its-formula%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
