How to get the end in algorithm2e The 2019 Stack Overflow Developer Survey Results Are InEnd word is hidden in algorithm blockalgorithm2e problemRemove “end” in algorithm2eHow to resolve the error: “Missing { inserted.” inside an algorithm2eMissing end statement in algorithm2e environmentIndent the block with the algorithm2e usepackageWrite the else if block with algorithm2eHow to get a “tight” boxed algorithm with the algorithm2e packagealgorithm2e - Undefined control sequence. endHow to show up “end” in algorithm2e?Remove the semicolon at the end of the line (algorithm2e)

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How to get the end in algorithm2e



The 2019 Stack Overflow Developer Survey Results Are InEnd word is hidden in algorithm blockalgorithm2e problemRemove “end” in algorithm2eHow to resolve the error: “Missing { inserted.” inside an algorithm2eMissing end statement in algorithm2e environmentIndent the block with the algorithm2e usepackageWrite the else if block with algorithm2eHow to get a “tight” boxed algorithm with the algorithm2e packagealgorithm2e - Undefined control sequence. endHow to show up “end” in algorithm2e?Remove the semicolon at the end of the line (algorithm2e)










3















I am using algorithm2e to typeset an algorithm. I like the appearance of the following, but I am not able to get the end to close the `




begin




`. Any way to do this?



documentclassarticle
usepackage[longend,ruled,vlined]algorithm2e

begindocument

beginalgorithm
captionPHORIZONTAL Algorithmlabelalg:phorizontal
KwIn$R$: relation; $Pr$: set of simple predicates
KwOut$F_R$: set of horizontal fragments of $R$
Begin
$Pr' leftarrow $COM_MIN($R, Pr$) ;
determine the set $M$ of minterm predicates ;
determine the set $I$ of implications among $p_i in Pr'$ ;
ForEach$m_i in M$
If $m_i$ is contradictory according to $I$$M leftarrow M - m_i$

$F_R = sigma_m_i R, m_i in M, forall m_i in M$ ;

endalgorithm
enddocument


enter image description here










share|improve this question






















  • Place SetAlgoLined at the inception point of your algorithm.

    – Siba Mishra
    Mar 30 at 12:53











  • THanks. Will try that as well.

    – ozsu
    Mar 30 at 16:29















3















I am using algorithm2e to typeset an algorithm. I like the appearance of the following, but I am not able to get the end to close the `




begin




`. Any way to do this?



documentclassarticle
usepackage[longend,ruled,vlined]algorithm2e

begindocument

beginalgorithm
captionPHORIZONTAL Algorithmlabelalg:phorizontal
KwIn$R$: relation; $Pr$: set of simple predicates
KwOut$F_R$: set of horizontal fragments of $R$
Begin
$Pr' leftarrow $COM_MIN($R, Pr$) ;
determine the set $M$ of minterm predicates ;
determine the set $I$ of implications among $p_i in Pr'$ ;
ForEach$m_i in M$
If $m_i$ is contradictory according to $I$$M leftarrow M - m_i$

$F_R = sigma_m_i R, m_i in M, forall m_i in M$ ;

endalgorithm
enddocument


enter image description here










share|improve this question






















  • Place SetAlgoLined at the inception point of your algorithm.

    – Siba Mishra
    Mar 30 at 12:53











  • THanks. Will try that as well.

    – ozsu
    Mar 30 at 16:29













3












3








3








I am using algorithm2e to typeset an algorithm. I like the appearance of the following, but I am not able to get the end to close the `




begin




`. Any way to do this?



documentclassarticle
usepackage[longend,ruled,vlined]algorithm2e

begindocument

beginalgorithm
captionPHORIZONTAL Algorithmlabelalg:phorizontal
KwIn$R$: relation; $Pr$: set of simple predicates
KwOut$F_R$: set of horizontal fragments of $R$
Begin
$Pr' leftarrow $COM_MIN($R, Pr$) ;
determine the set $M$ of minterm predicates ;
determine the set $I$ of implications among $p_i in Pr'$ ;
ForEach$m_i in M$
If $m_i$ is contradictory according to $I$$M leftarrow M - m_i$

$F_R = sigma_m_i R, m_i in M, forall m_i in M$ ;

endalgorithm
enddocument


enter image description here










share|improve this question














I am using algorithm2e to typeset an algorithm. I like the appearance of the following, but I am not able to get the end to close the `




begin




`. Any way to do this?



documentclassarticle
usepackage[longend,ruled,vlined]algorithm2e

begindocument

beginalgorithm
captionPHORIZONTAL Algorithmlabelalg:phorizontal
KwIn$R$: relation; $Pr$: set of simple predicates
KwOut$F_R$: set of horizontal fragments of $R$
Begin
$Pr' leftarrow $COM_MIN($R, Pr$) ;
determine the set $M$ of minterm predicates ;
determine the set $I$ of implications among $p_i in Pr'$ ;
ForEach$m_i in M$
If $m_i$ is contradictory according to $I$$M leftarrow M - m_i$

$F_R = sigma_m_i R, m_i in M, forall m_i in M$ ;

endalgorithm
enddocument


enter image description here







algorithm2e






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Mar 30 at 12:18









ozsuozsu

1,13611126




1,13611126












  • Place SetAlgoLined at the inception point of your algorithm.

    – Siba Mishra
    Mar 30 at 12:53











  • THanks. Will try that as well.

    – ozsu
    Mar 30 at 16:29

















  • Place SetAlgoLined at the inception point of your algorithm.

    – Siba Mishra
    Mar 30 at 12:53











  • THanks. Will try that as well.

    – ozsu
    Mar 30 at 16:29
















Place SetAlgoLined at the inception point of your algorithm.

– Siba Mishra
Mar 30 at 12:53





Place SetAlgoLined at the inception point of your algorithm.

– Siba Mishra
Mar 30 at 12:53













THanks. Will try that as well.

– ozsu
Mar 30 at 16:29





THanks. Will try that as well.

– ozsu
Mar 30 at 16:29










1 Answer
1






active

oldest

votes


















5














Referring to the answert by @egreg in End word is hidden in algorithm block , the vlined option suppresses the printing of “end”. Just remove the option:



documentclassarticle
usepackage[longend,ruled]algorithm2e

begindocument

beginalgorithm
captionPHORIZONTAL Algorithmlabelalg:phorizontal
KwIn$R$: relation; $Pr$: set of simple predicates
KwOut$F_R$: set of horizontal fragments of $R$
Begin
$Pr' leftarrow $COM_MIN($R, Pr$) ;
determine the set $M$ of minterm predicates ;
determine the set $I$ of implications among $p_i in Pr'$ ;
ForEach$m_i in M$
If $m_i$ is contradictory according to $I$$M leftarrow M - m_i$

$F_R = sigma_m_i R, m_i in M, forall m_i in M$ ;

endalgorithm
enddocument


enter image description here






share|improve this answer























  • Thank you. I was hoping I could keep both, but apparently that is not possible.

    – ozsu
    Mar 30 at 16:02











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5














Referring to the answert by @egreg in End word is hidden in algorithm block , the vlined option suppresses the printing of “end”. Just remove the option:



documentclassarticle
usepackage[longend,ruled]algorithm2e

begindocument

beginalgorithm
captionPHORIZONTAL Algorithmlabelalg:phorizontal
KwIn$R$: relation; $Pr$: set of simple predicates
KwOut$F_R$: set of horizontal fragments of $R$
Begin
$Pr' leftarrow $COM_MIN($R, Pr$) ;
determine the set $M$ of minterm predicates ;
determine the set $I$ of implications among $p_i in Pr'$ ;
ForEach$m_i in M$
If $m_i$ is contradictory according to $I$$M leftarrow M - m_i$

$F_R = sigma_m_i R, m_i in M, forall m_i in M$ ;

endalgorithm
enddocument


enter image description here






share|improve this answer























  • Thank you. I was hoping I could keep both, but apparently that is not possible.

    – ozsu
    Mar 30 at 16:02















5














Referring to the answert by @egreg in End word is hidden in algorithm block , the vlined option suppresses the printing of “end”. Just remove the option:



documentclassarticle
usepackage[longend,ruled]algorithm2e

begindocument

beginalgorithm
captionPHORIZONTAL Algorithmlabelalg:phorizontal
KwIn$R$: relation; $Pr$: set of simple predicates
KwOut$F_R$: set of horizontal fragments of $R$
Begin
$Pr' leftarrow $COM_MIN($R, Pr$) ;
determine the set $M$ of minterm predicates ;
determine the set $I$ of implications among $p_i in Pr'$ ;
ForEach$m_i in M$
If $m_i$ is contradictory according to $I$$M leftarrow M - m_i$

$F_R = sigma_m_i R, m_i in M, forall m_i in M$ ;

endalgorithm
enddocument


enter image description here






share|improve this answer























  • Thank you. I was hoping I could keep both, but apparently that is not possible.

    – ozsu
    Mar 30 at 16:02













5












5








5







Referring to the answert by @egreg in End word is hidden in algorithm block , the vlined option suppresses the printing of “end”. Just remove the option:



documentclassarticle
usepackage[longend,ruled]algorithm2e

begindocument

beginalgorithm
captionPHORIZONTAL Algorithmlabelalg:phorizontal
KwIn$R$: relation; $Pr$: set of simple predicates
KwOut$F_R$: set of horizontal fragments of $R$
Begin
$Pr' leftarrow $COM_MIN($R, Pr$) ;
determine the set $M$ of minterm predicates ;
determine the set $I$ of implications among $p_i in Pr'$ ;
ForEach$m_i in M$
If $m_i$ is contradictory according to $I$$M leftarrow M - m_i$

$F_R = sigma_m_i R, m_i in M, forall m_i in M$ ;

endalgorithm
enddocument


enter image description here






share|improve this answer













Referring to the answert by @egreg in End word is hidden in algorithm block , the vlined option suppresses the printing of “end”. Just remove the option:



documentclassarticle
usepackage[longend,ruled]algorithm2e

begindocument

beginalgorithm
captionPHORIZONTAL Algorithmlabelalg:phorizontal
KwIn$R$: relation; $Pr$: set of simple predicates
KwOut$F_R$: set of horizontal fragments of $R$
Begin
$Pr' leftarrow $COM_MIN($R, Pr$) ;
determine the set $M$ of minterm predicates ;
determine the set $I$ of implications among $p_i in Pr'$ ;
ForEach$m_i in M$
If $m_i$ is contradictory according to $I$$M leftarrow M - m_i$

$F_R = sigma_m_i R, m_i in M, forall m_i in M$ ;

endalgorithm
enddocument


enter image description here







share|improve this answer












share|improve this answer



share|improve this answer










answered Mar 30 at 12:50









subham sonisubham soni

4,99183188




4,99183188












  • Thank you. I was hoping I could keep both, but apparently that is not possible.

    – ozsu
    Mar 30 at 16:02

















  • Thank you. I was hoping I could keep both, but apparently that is not possible.

    – ozsu
    Mar 30 at 16:02
















Thank you. I was hoping I could keep both, but apparently that is not possible.

– ozsu
Mar 30 at 16:02





Thank you. I was hoping I could keep both, but apparently that is not possible.

– ozsu
Mar 30 at 16:02

















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