Derivative of the l2-norm of a multivariate complex matrix Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Gradient And Hessian Of General 2-NormLinear Algebra ResponseMinimizing $ell^infty$ norm of complex vectorHow to find the $S_kappa$ elements in the product $ sum_kappa=1^K S_kappa Q_kappa(r, theta,phi) = 1$?What is the derivative of a Radial Basis Interpolation function?First derivative of matrix expressionFirst derivative of RVM related matrix expressionDifferential of Matrix InverseHow to compute this derivative of a square root of a sum?Finding the amount partitions with gives sizes of a multisetI need help with a problem involving the nth derivative of arcsin x
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Derivative of the l2-norm of a multivariate complex matrix
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Gradient And Hessian Of General 2-NormLinear Algebra ResponseMinimizing $ell^infty$ norm of complex vectorHow to find the $S_kappa$ elements in the product $ sum_kappa=1^K S_kappa Q_kappa(r, theta,phi) = 1$?What is the derivative of a Radial Basis Interpolation function?First derivative of matrix expressionFirst derivative of RVM related matrix expressionDifferential of Matrix InverseHow to compute this derivative of a square root of a sum?Finding the amount partitions with gives sizes of a multisetI need help with a problem involving the nth derivative of arcsin x
$begingroup$
I have a 4x1 vector with complex elements $mathbfc=left[c_1, dots, c_Nright]^ T $ and the following $ell2$-norm function $f(mathbfc) = |textbfA.textbfc|_2^2 = sum_m=1^Mleft|mathbfa_m^Hmathbfcright|^2=sum_m=1^Mmathbfc^Hmathbfa_mmathbfa_m^Hmathbfc$ , where the elements of c are complex and have the form $c_i = r_icdot exp(jPhi_i)$
I am trying to find the derivatives the aforementioned function given $r_i$ and $Phi_i$
I tried this by decomposing my complex numbers and sums but it takes too long and I cannot shorten my formulas at the end. I think there is an alternative way as this post states but I am unable to find it. Any help or hints are much appreciated. Thank you in advance.
linear-algebra derivatives complex-numbers norm
asked Mar 27 at 3:08
SuperKogitoSuperKogito
1035
$endgroup$
add a comment |
$begingroup$
I have a 4x1 vector with complex elements $mathbfc=left[c_1, dots, c_Nright]^ T $ and the following $ell2$-norm function $f(mathbfc) = |textbfA.textbfc|_2^2 = sum_m=1^Mleft|mathbfa_m^Hmathbfcright|^2=sum_m=1^Mmathbfc^Hmathbfa_mmathbfa_m^Hmathbfc$ , where the elements of c are complex and have the form $c_i = r_icdot exp(jPhi_i)$
I am trying to find the derivatives the aforementioned function given $r_i$ and $Phi_i$
I tried this by decomposing my complex numbers and sums but it takes too long and I cannot shorten my formulas at the end. I think there is an alternative way as this post states but I am unable to find it. Any help or hints are much appreciated. Thank you in advance.
linear-algebra derivatives complex-numbers norm
asked Mar 27 at 3:08
SuperKogitoSuperKogito
1035
$endgroup$
add a comment |
$begingroup$
I have a 4x1 vector with complex elements $mathbfc=left[c_1, dots, c_Nright]^ T $ and the following $ell2$-norm function $f(mathbfc) = |textbfA.textbfc|_2^2 = sum_m=1^Mleft|mathbfa_m^Hmathbfcright|^2=sum_m=1^Mmathbfc^Hmathbfa_mmathbfa_m^Hmathbfc$ , where the elements of c are complex and have the form $c_i = r_icdot exp(jPhi_i)$
I am trying to find the derivatives the aforementioned function given $r_i$ and $Phi_i$
I tried this by decomposing my complex numbers and sums but it takes too long and I cannot shorten my formulas at the end. I think there is an alternative way as this post states but I am unable to find it. Any help or hints are much appreciated. Thank you in advance.
linear-algebra derivatives complex-numbers norm
asked Mar 27 at 3:08
SuperKogitoSuperKogito
1035
$endgroup$
I have a 4x1 vector with complex elements $mathbfc=left[c_1, dots, c_Nright]^ T $ and the following $ell2$-norm function $f(mathbfc) = |textbfA.textbfc|_2^2 = sum_m=1^Mleft|mathbfa_m^Hmathbfcright|^2=sum_m=1^Mmathbfc^Hmathbfa_mmathbfa_m^Hmathbfc$ , where the elements of c are complex and have the form $c_i = r_icdot exp(jPhi_i)$
I am trying to find the derivatives the aforementioned function given $r_i$ and $Phi_i$
I tried this by decomposing my complex numbers and sums but it takes too long and I cannot shorten my formulas at the end. I think there is an alternative way as this post states but I am unable to find it. Any help or hints are much appreciated. Thank you in advance.
linear-algebra derivatives complex-numbers norm
linear-algebra derivatives complex-numbers norm
asked Mar 27 at 3:08
SuperKogitoSuperKogito
1035
asked Mar 27 at 3:08
SuperKogitoSuperKogito
1035
asked Mar 27 at 3:08
SuperKogitoSuperKogito
1035
asked Mar 27 at 3:08
SuperKogitoSuperKogito
1035
asked Mar 27 at 3:08
SuperKogitoSuperKogito
1035
1035
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Given the real vectors $(r,phi)$, define the complex vectors $(p,c,b)$ as
$$eqalign
p &= exp(jphi) &implies dp = j,podot dphi cr
c &= rodot p &implies dc = rodot dp + podot dr cr
b &= A^HAc cr
$$
where $odot$ denotes the elementwise/Hadamard product.
Further, a colon will denote the trace/Frobenius product, i.e.
$,,A:B = rm Tr(A^TB)$
(NB: The exp() function is applied elementwise)
Calculate the differential of the real function $(f)$ in terms of the real variables $(r,phi)$.
$$eqalign
f &= (Ac)^* : Ac cr
df
&= (Ac)^* : (A , dc) + (Ac) : (A , dc)^* cr
&= b^*:dc + b:dc^* cr
&= b^*:(rodot dp + podot dr) + b:(rodot dp + podot dr)^* cr
&= (rodot b^*):dp + (r^*odot b):dp^* + (podot b^*):dr + (p^*odot b):dr^* cr
&= (rodot b^*):(j,podot dphi) + (r^*odot b):(j,podot dphi)^*
+ (podot b^*):dr + (p^*odot b):dr^* cr
&= (j,p)odot(rodot b^*):dphi + (j,p)^*odot(r^*odot b):dphi^*
+ (podot b^*):dr + (p^*odot b):dr^* cr
&= 2,mathcal Re(j,podot rodot b^*):dphi + 2,mathcal Re(podot b^*):dr cr
$$
The fact that $(dr^*=dr,,,dphi^*=dphi)$ and $mathcal Re(z)=tfrac12(z+z^*)$ allows terms to be combined in that last line.
In this form, the gradients can be identified as
$$eqalign
fracpartial fpartialphi
&= 2,mathcal Re(j,podot rodot b^*) cr
&= 2rodot mathcal Im(p^*odot b) cr
fracpartial fpartial r &= 2,mathcal Re(p^*odot b) cr
$$
answered Mar 27 at 14:30
greggreg
9,3911825
$endgroup$
$begingroup$
Thank you for your answer but what do you mean exactly by "the real and imaginary components of a variable are treated independently" and the c you are using here is the vector, I assume? or is it an element of the vector? And how is the $mathbfc*$ relevant here? Anyway my variables are $r_i$ & $Phi_i$ so my derivatives are $fracpartial fpartial Phi_i$ & $fracpartial fpartial r_i$ and not $fracpartial fpartial mathbfc$
$endgroup$
– SuperKogito
Mar 27 at 15:29
$begingroup$
Sorry. the answer has been re-worked to find the gradients with respect to $(r,phi)$ instead of $(c,c^*)$.
$endgroup$
– greg
Mar 27 at 16:35
$begingroup$
Thank you so much @greg . This is exactly what I needed. Just one question: Is the $A^*$ the Conjugate transpose or just the Conjugate?
$endgroup$
– SuperKogito
Apr 1 at 13:45
$begingroup$
In my post, the symbols $(A^T,A^*,A^H)$ denote the $($transpose, complex conjugate, hermitian conjugate$)$ of $A$, respectively.
$endgroup$
– greg
Apr 3 at 21:58
$begingroup$
I figured so, but I wanted to be sure. Again thank you so much, you are a hero sir :)
$endgroup$
– SuperKogito
Apr 3 at 22:00
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Given the real vectors $(r,phi)$, define the complex vectors $(p,c,b)$ as
$$eqalign
p &= exp(jphi) &implies dp = j,podot dphi cr
c &= rodot p &implies dc = rodot dp + podot dr cr
b &= A^HAc cr
$$
where $odot$ denotes the elementwise/Hadamard product.
Further, a colon will denote the trace/Frobenius product, i.e.
$,,A:B = rm Tr(A^TB)$
(NB: The exp() function is applied elementwise)
Calculate the differential of the real function $(f)$ in terms of the real variables $(r,phi)$.
$$eqalign
f &= (Ac)^* : Ac cr
df
&= (Ac)^* : (A , dc) + (Ac) : (A , dc)^* cr
&= b^*:dc + b:dc^* cr
&= b^*:(rodot dp + podot dr) + b:(rodot dp + podot dr)^* cr
&= (rodot b^*):dp + (r^*odot b):dp^* + (podot b^*):dr + (p^*odot b):dr^* cr
&= (rodot b^*):(j,podot dphi) + (r^*odot b):(j,podot dphi)^*
+ (podot b^*):dr + (p^*odot b):dr^* cr
&= (j,p)odot(rodot b^*):dphi + (j,p)^*odot(r^*odot b):dphi^*
+ (podot b^*):dr + (p^*odot b):dr^* cr
&= 2,mathcal Re(j,podot rodot b^*):dphi + 2,mathcal Re(podot b^*):dr cr
$$
The fact that $(dr^*=dr,,,dphi^*=dphi)$ and $mathcal Re(z)=tfrac12(z+z^*)$ allows terms to be combined in that last line.
In this form, the gradients can be identified as
$$eqalign
fracpartial fpartialphi
&= 2,mathcal Re(j,podot rodot b^*) cr
&= 2rodot mathcal Im(p^*odot b) cr
fracpartial fpartial r &= 2,mathcal Re(p^*odot b) cr
$$
answered Mar 27 at 14:30
greggreg
9,3911825
$endgroup$
$begingroup$
Thank you for your answer but what do you mean exactly by "the real and imaginary components of a variable are treated independently" and the c you are using here is the vector, I assume? or is it an element of the vector? And how is the $mathbfc*$ relevant here? Anyway my variables are $r_i$ & $Phi_i$ so my derivatives are $fracpartial fpartial Phi_i$ & $fracpartial fpartial r_i$ and not $fracpartial fpartial mathbfc$
$endgroup$
– SuperKogito
Mar 27 at 15:29
$begingroup$
Sorry. the answer has been re-worked to find the gradients with respect to $(r,phi)$ instead of $(c,c^*)$.
$endgroup$
– greg
Mar 27 at 16:35
$begingroup$
Thank you so much @greg . This is exactly what I needed. Just one question: Is the $A^*$ the Conjugate transpose or just the Conjugate?
$endgroup$
– SuperKogito
Apr 1 at 13:45
$begingroup$
In my post, the symbols $(A^T,A^*,A^H)$ denote the $($transpose, complex conjugate, hermitian conjugate$)$ of $A$, respectively.
$endgroup$
– greg
Apr 3 at 21:58
$begingroup$
I figured so, but I wanted to be sure. Again thank you so much, you are a hero sir :)
$endgroup$
– SuperKogito
Apr 3 at 22:00
add a comment |
$begingroup$
Given the real vectors $(r,phi)$, define the complex vectors $(p,c,b)$ as
$$eqalign
p &= exp(jphi) &implies dp = j,podot dphi cr
c &= rodot p &implies dc = rodot dp + podot dr cr
b &= A^HAc cr
$$
where $odot$ denotes the elementwise/Hadamard product.
Further, a colon will denote the trace/Frobenius product, i.e.
$,,A:B = rm Tr(A^TB)$
(NB: The exp() function is applied elementwise)
Calculate the differential of the real function $(f)$ in terms of the real variables $(r,phi)$.
$$eqalign
f &= (Ac)^* : Ac cr
df
&= (Ac)^* : (A , dc) + (Ac) : (A , dc)^* cr
&= b^*:dc + b:dc^* cr
&= b^*:(rodot dp + podot dr) + b:(rodot dp + podot dr)^* cr
&= (rodot b^*):dp + (r^*odot b):dp^* + (podot b^*):dr + (p^*odot b):dr^* cr
&= (rodot b^*):(j,podot dphi) + (r^*odot b):(j,podot dphi)^*
+ (podot b^*):dr + (p^*odot b):dr^* cr
&= (j,p)odot(rodot b^*):dphi + (j,p)^*odot(r^*odot b):dphi^*
+ (podot b^*):dr + (p^*odot b):dr^* cr
&= 2,mathcal Re(j,podot rodot b^*):dphi + 2,mathcal Re(podot b^*):dr cr
$$
The fact that $(dr^*=dr,,,dphi^*=dphi)$ and $mathcal Re(z)=tfrac12(z+z^*)$ allows terms to be combined in that last line.
In this form, the gradients can be identified as
$$eqalign
fracpartial fpartialphi
&= 2,mathcal Re(j,podot rodot b^*) cr
&= 2rodot mathcal Im(p^*odot b) cr
fracpartial fpartial r &= 2,mathcal Re(p^*odot b) cr
$$
answered Mar 27 at 14:30
greggreg
9,3911825
$endgroup$
$begingroup$
Thank you for your answer but what do you mean exactly by "the real and imaginary components of a variable are treated independently" and the c you are using here is the vector, I assume? or is it an element of the vector? And how is the $mathbfc*$ relevant here? Anyway my variables are $r_i$ & $Phi_i$ so my derivatives are $fracpartial fpartial Phi_i$ & $fracpartial fpartial r_i$ and not $fracpartial fpartial mathbfc$
$endgroup$
– SuperKogito
Mar 27 at 15:29
$begingroup$
Sorry. the answer has been re-worked to find the gradients with respect to $(r,phi)$ instead of $(c,c^*)$.
$endgroup$
– greg
Mar 27 at 16:35
$begingroup$
Thank you so much @greg . This is exactly what I needed. Just one question: Is the $A^*$ the Conjugate transpose or just the Conjugate?
$endgroup$
– SuperKogito
Apr 1 at 13:45
$begingroup$
In my post, the symbols $(A^T,A^*,A^H)$ denote the $($transpose, complex conjugate, hermitian conjugate$)$ of $A$, respectively.
$endgroup$
– greg
Apr 3 at 21:58
$begingroup$
I figured so, but I wanted to be sure. Again thank you so much, you are a hero sir :)
$endgroup$
– SuperKogito
Apr 3 at 22:00
add a comment |
$begingroup$
Given the real vectors $(r,phi)$, define the complex vectors $(p,c,b)$ as
$$eqalign
p &= exp(jphi) &implies dp = j,podot dphi cr
c &= rodot p &implies dc = rodot dp + podot dr cr
b &= A^HAc cr
$$
where $odot$ denotes the elementwise/Hadamard product.
Further, a colon will denote the trace/Frobenius product, i.e.
$,,A:B = rm Tr(A^TB)$
(NB: The exp() function is applied elementwise)
Calculate the differential of the real function $(f)$ in terms of the real variables $(r,phi)$.
$$eqalign
f &= (Ac)^* : Ac cr
df
&= (Ac)^* : (A , dc) + (Ac) : (A , dc)^* cr
&= b^*:dc + b:dc^* cr
&= b^*:(rodot dp + podot dr) + b:(rodot dp + podot dr)^* cr
&= (rodot b^*):dp + (r^*odot b):dp^* + (podot b^*):dr + (p^*odot b):dr^* cr
&= (rodot b^*):(j,podot dphi) + (r^*odot b):(j,podot dphi)^*
+ (podot b^*):dr + (p^*odot b):dr^* cr
&= (j,p)odot(rodot b^*):dphi + (j,p)^*odot(r^*odot b):dphi^*
+ (podot b^*):dr + (p^*odot b):dr^* cr
&= 2,mathcal Re(j,podot rodot b^*):dphi + 2,mathcal Re(podot b^*):dr cr
$$
The fact that $(dr^*=dr,,,dphi^*=dphi)$ and $mathcal Re(z)=tfrac12(z+z^*)$ allows terms to be combined in that last line.
In this form, the gradients can be identified as
$$eqalign
fracpartial fpartialphi
&= 2,mathcal Re(j,podot rodot b^*) cr
&= 2rodot mathcal Im(p^*odot b) cr
fracpartial fpartial r &= 2,mathcal Re(p^*odot b) cr
$$
answered Mar 27 at 14:30
greggreg
9,3911825
$endgroup$
Given the real vectors $(r,phi)$, define the complex vectors $(p,c,b)$ as
$$eqalign
p &= exp(jphi) &implies dp = j,podot dphi cr
c &= rodot p &implies dc = rodot dp + podot dr cr
b &= A^HAc cr
$$
where $odot$ denotes the elementwise/Hadamard product.
Further, a colon will denote the trace/Frobenius product, i.e.
$,,A:B = rm Tr(A^TB)$
(NB: The exp() function is applied elementwise)
Calculate the differential of the real function $(f)$ in terms of the real variables $(r,phi)$.
$$eqalign
f &= (Ac)^* : Ac cr
df
&= (Ac)^* : (A , dc) + (Ac) : (A , dc)^* cr
&= b^*:dc + b:dc^* cr
&= b^*:(rodot dp + podot dr) + b:(rodot dp + podot dr)^* cr
&= (rodot b^*):dp + (r^*odot b):dp^* + (podot b^*):dr + (p^*odot b):dr^* cr
&= (rodot b^*):(j,podot dphi) + (r^*odot b):(j,podot dphi)^*
+ (podot b^*):dr + (p^*odot b):dr^* cr
&= (j,p)odot(rodot b^*):dphi + (j,p)^*odot(r^*odot b):dphi^*
+ (podot b^*):dr + (p^*odot b):dr^* cr
&= 2,mathcal Re(j,podot rodot b^*):dphi + 2,mathcal Re(podot b^*):dr cr
$$
The fact that $(dr^*=dr,,,dphi^*=dphi)$ and $mathcal Re(z)=tfrac12(z+z^*)$ allows terms to be combined in that last line.
In this form, the gradients can be identified as
$$eqalign
fracpartial fpartialphi
&= 2,mathcal Re(j,podot rodot b^*) cr
&= 2rodot mathcal Im(p^*odot b) cr
fracpartial fpartial r &= 2,mathcal Re(p^*odot b) cr
$$
answered Mar 27 at 14:30
greggreg
9,3911825
edited Mar 30 at 18:32
answered Mar 27 at 14:30
greggreg
9,3911825
answered Mar 27 at 14:30
greggreg
9,3911825
answered Mar 27 at 14:30
greggreg
9,3911825
9,3911825
$begingroup$
Thank you for your answer but what do you mean exactly by "the real and imaginary components of a variable are treated independently" and the c you are using here is the vector, I assume? or is it an element of the vector? And how is the $mathbfc*$ relevant here? Anyway my variables are $r_i$ & $Phi_i$ so my derivatives are $fracpartial fpartial Phi_i$ & $fracpartial fpartial r_i$ and not $fracpartial fpartial mathbfc$
$endgroup$
– SuperKogito
Mar 27 at 15:29
$begingroup$
Sorry. the answer has been re-worked to find the gradients with respect to $(r,phi)$ instead of $(c,c^*)$.
$endgroup$
– greg
Mar 27 at 16:35
$begingroup$
Thank you so much @greg . This is exactly what I needed. Just one question: Is the $A^*$ the Conjugate transpose or just the Conjugate?
$endgroup$
– SuperKogito
Apr 1 at 13:45
$begingroup$
In my post, the symbols $(A^T,A^*,A^H)$ denote the $($transpose, complex conjugate, hermitian conjugate$)$ of $A$, respectively.
$endgroup$
– greg
Apr 3 at 21:58
$begingroup$
I figured so, but I wanted to be sure. Again thank you so much, you are a hero sir :)
$endgroup$
– SuperKogito
Apr 3 at 22:00
add a comment |
$begingroup$
Thank you for your answer but what do you mean exactly by "the real and imaginary components of a variable are treated independently" and the c you are using here is the vector, I assume? or is it an element of the vector? And how is the $mathbfc*$ relevant here? Anyway my variables are $r_i$ & $Phi_i$ so my derivatives are $fracpartial fpartial Phi_i$ & $fracpartial fpartial r_i$ and not $fracpartial fpartial mathbfc$
$endgroup$
– SuperKogito
Mar 27 at 15:29
$begingroup$
Sorry. the answer has been re-worked to find the gradients with respect to $(r,phi)$ instead of $(c,c^*)$.
$endgroup$
– greg
Mar 27 at 16:35
$begingroup$
Thank you so much @greg . This is exactly what I needed. Just one question: Is the $A^*$ the Conjugate transpose or just the Conjugate?
$endgroup$
– SuperKogito
Apr 1 at 13:45
$begingroup$
In my post, the symbols $(A^T,A^*,A^H)$ denote the $($transpose, complex conjugate, hermitian conjugate$)$ of $A$, respectively.
$endgroup$
– greg
Apr 3 at 21:58
$begingroup$
I figured so, but I wanted to be sure. Again thank you so much, you are a hero sir :)
$endgroup$
– SuperKogito
Apr 3 at 22:00
$begingroup$
Thank you for your answer but what do you mean exactly by "the real and imaginary components of a variable are treated independently" and the c you are using here is the vector, I assume? or is it an element of the vector? And how is the $mathbfc*$ relevant here? Anyway my variables are $r_i$ & $Phi_i$ so my derivatives are $fracpartial fpartial Phi_i$ & $fracpartial fpartial r_i$ and not $fracpartial fpartial mathbfc$
$endgroup$
– SuperKogito
Mar 27 at 15:29
$begingroup$
Thank you for your answer but what do you mean exactly by "the real and imaginary components of a variable are treated independently" and the c you are using here is the vector, I assume? or is it an element of the vector? And how is the $mathbfc*$ relevant here? Anyway my variables are $r_i$ & $Phi_i$ so my derivatives are $fracpartial fpartial Phi_i$ & $fracpartial fpartial r_i$ and not $fracpartial fpartial mathbfc$
$endgroup$
– SuperKogito
Mar 27 at 15:29
$begingroup$
Sorry. the answer has been re-worked to find the gradients with respect to $(r,phi)$ instead of $(c,c^*)$.
$endgroup$
– greg
Mar 27 at 16:35
$begingroup$
Sorry. the answer has been re-worked to find the gradients with respect to $(r,phi)$ instead of $(c,c^*)$.
$endgroup$
– greg
Mar 27 at 16:35
$begingroup$
Thank you so much @greg . This is exactly what I needed. Just one question: Is the $A^*$ the Conjugate transpose or just the Conjugate?
$endgroup$
– SuperKogito
Apr 1 at 13:45
$begingroup$
Thank you so much @greg . This is exactly what I needed. Just one question: Is the $A^*$ the Conjugate transpose or just the Conjugate?
$endgroup$
– SuperKogito
Apr 1 at 13:45
$begingroup$
In my post, the symbols $(A^T,A^*,A^H)$ denote the $($transpose, complex conjugate, hermitian conjugate$)$ of $A$, respectively.
$endgroup$
– greg
Apr 3 at 21:58
$begingroup$
In my post, the symbols $(A^T,A^*,A^H)$ denote the $($transpose, complex conjugate, hermitian conjugate$)$ of $A$, respectively.
$endgroup$
– greg
Apr 3 at 21:58
$begingroup$
I figured so, but I wanted to be sure. Again thank you so much, you are a hero sir :)
$endgroup$
– SuperKogito
Apr 3 at 22:00
$begingroup$
I figured so, but I wanted to be sure. Again thank you so much, you are a hero sir :)
$endgroup$
– SuperKogito
Apr 3 at 22:00
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