_Addendum: This is mostly about the story behind our solution to N. To see more about the math, see [here](https://codeforces.com/blog/ekzhang)_↵
↵
When I was younger, fairytales always captivated me, with their nice and clean stories. All through my life I've liked the idea of having a fairytale ending to a life arc, but I was frequently disappointed. Many times after an unsuccessful experience, one simply gets more unsucessful experiences and then time runs out...↵
↵
At ICPC recently, I was lucky enough to finally attain my fairytale ending to my competitive programming career. I figured I would tell the fairytale, to close a career that started about 6 years ago.↵
↵
On my World Finals team, my role is to help the team get hard math problems, as I haven't actively processed algorithms in years. So imagine my delight when my teammate ([user:ekzhang,2021-10-06]) points out that problem N reduces to finding a vector so that the (l2)-norm of x is r, and $Ax = b$ for a matrix $A$ and vector $b$. This has to be doable!↵
↵
The first observation is that *finding the min-norm solution $x_{\text{min}}$ to $Ax=b$* is useful. Once we have the minimum-norm solution, it's sufficient to just find an $e$ in the null space of $Ax = b$, and consider $x_c = x_{\text{min}} + c \cdot e$, choosing $c$ so that the norm of $x_c$ is equal to $r$. Note that if we replace $x_{\text{min}}$ with any such $x$, it may not be possible to find a $c$ so that the norm of $x_c$ is equal to $r$. ↵
↵
When I hear min-norm solution to $Ax = b$ my ears perk up -- a well known result in machine learning is that stochastic gradient descent on $Ax - b$ (more precisely, $(Ax - b)^{\top}(Ax - b)$)) converges to the min-norm solution. Could this be my chance to use preconditioned gradient descent?↵
↵
As it turns out this is not to be. I try using gradient descent, and the precision after a finite number of operations is simply not high enough, especially given the massive precision demanded by the problem.↵
↵
It's here that I stop trying to overcomplicate things. I rewrite $Ax = b$ as $Cx = d$, where $C$ is a matrix composed of orthogonal rows, via Gram-Schmidt orthogonalization (or QR decomposition). As such, if we write ↵
$$ C = \begin{bmatrix} -- & e_1 & -- \\ ↵
-- & \vdots & -- \\ ↵
-- & e_n & -- \end{bmatrix}$$↵
↵
Now, to find the min-norm solution of $Cx = d$, note that we can simply take $x_{\text{min}} = d_1e_1 + \ldots + d_ne_n$. Note that any other solution to $Cx = d$ is just $x_{\text{min}}$ added to vectors in $e_{n+1}, e_{n+2}, \ldots e_d$, each of which only add norm to $x$. And now I knew I was almost done! We just need to find an $e_{n+1}$ that's orthogonal to each of $e_1, e_2, \ldots e_n$, and then find a $c$ so that the norm of $x_{\text{min}} + ce_{n+1}$ is equal to $r$. ↵
↵
Inspired by this, and wanting to get the coveted first solve award, I begin coding this solution, typing the fastest that I have in a while. After about 20 minutes, at the three-hour mark, I have a mostly working solution that seems to check out on the sample data, and I excitedly submit, to get↵
↵
__${\color{red}{\textbf{Wrong Answer__}}}$. ↵
↵
After this, there's a fairly obvious typo I fix, and I resubmit again, getting↵
↵
__${\color{red}{\textbf{Runtime Error__}}}$. ↵
↵
At the last few programming contests I've participated in, run-time error has been a pretty common occurence. This is a fairly rare error to have, and one that is usually not too hard to fix. Calculating the value of $c$ in the above uses a square root operation, and due to floating point precision issues sometimes this number is slightly below zero, so I patch that issue. Resubmit! ↵
↵
__${\color{red}{\textbf{Runtime Error__}}}$ ↵
↵
At this point, I try running my new code on the samples again, and get something quite unusual: my code prints the answer, but throws a run-time error in the middle. Frustrated, I step away from the problem briefly and let my other teammate ([user:halohalo,2021-10-06]) try looking for a run-time error. He quickly finds something amiss. Namely, in order to compute $e_{n+1}$ my code calls a subroutine which essentially runs Gram-Schmidt orthogonalization on $Ax = b$, but with a random vector appended to the end of $A$. The last value of $b$ doesn't matter for this component (in fact the value of $b$ doesn't matter at all for computing $e_{n+1}$) but the way I had it implemented, $b$ was one shorter than it was supposed to be, and was thus leading to out-of-bounds errors. We fix this, and are greeted by: ↵
↵
__${\color{red}{\textbf{Wrong Answer__}}}$↵
↵
At this point [user:ekzhang,2021-10-06] decided to reimplement my program from scratch. I worked on preparing large tests, hoping to find an error. I find it, a large test where the variable $n$ is equal to $500$ and where $d$ is $490$. In this case, the linear system is actually overdetermined, and we have $499$ equations in $490$ variables. In these cases, naive Gram-Schmidt will start trying to nomalize vectors which (up to floating point error) are zero, leading to potentially unusual consequences. In this case, my program explodes. ↵
↵
His program passes the samples, but doesn't pass the large case either. There are now about $10$ minutes left in the contest, so he tries a hail mary adding one line of code to the beginning of the program:↵
$n = min(n, d + 1)$. In fact, I thought that I had expressed this implicitly later in the program. ↵
↵
His solution still fails the weird case, but is much closer to right. With nothing else to try, and with 4 minutes left and 15 wrong submissions, I add this line of code to the beginning of my program, and submit. ↵
↵
__${\color{green}{\textbf{Correct__}}}$↵
↵
When I see the green__${\color{green}{\textbf{Correct__}}}$ on the screen, it's probably the happiest moment I've ever had in any contest ever. ↵
↵
You couldn't have scripted a better ending to my ICPC career (I'm not doing it in the future), one where I solved the hardest math problem on the set. ↵
↵
Originally we thought that Latvia may have solved the problem before us, but then when we see their last submission is after ours (at 298, with 2 minutes left) we know that we've won the first to solve award for the problem. ↵
↵
There were many things that could have gone better in solving this problem. We probably didn't need to make these wrong submits, and I could've made the $n = min(n, d + 1)$ check earlier (never hurts to check twice!) But at the end we had our shiny Correct, and that's all that mattered to me in the end.↵
↵
I'd like to thank everyone who helped me (far too numerous to list in the post) for a great 6 years of competitive programming. Bon voyage! ↵
↵
↵
When I was younger, fairytales always captivated me, with their nice and clean stories. All through my life I've liked the idea of having a fairytale ending to a life arc, but I was frequently disappointed. Many times after an unsuccessful experience, one simply gets more unsucessful experiences and then time runs out...↵
↵
At ICPC recently, I was lucky enough to finally attain my fairytale ending to my competitive programming career. I figured I would tell the fairytale, to close a career that started about 6 years ago.↵
↵
On my World Finals team, my role is to help the team get hard math problems, as I haven't actively processed algorithms in years. So imagine my delight when my teammate ([user:ekzhang,2021-10-06]) points out that problem N reduces to finding a vector so that the (l2)-norm of x is r, and $Ax = b$ for a matrix $A$ and vector $b$. This has to be doable!↵
↵
The first observation is that *finding the min-norm solution $x_{\text{min}}$ to $Ax=b$* is useful. Once we have the minimum-norm solution, it's sufficient to just find an $e$ in the null space of $Ax = b$, and consider $x_c = x_{\text{min}} + c \cdot e$, choosing $c$ so that the norm of $x_c$ is equal to $r$. Note that if we replace $x_{\text{min}}$ with any such $x$, it may not be possible to find a $c$ so that the norm of $x_c$ is equal to $r$. ↵
↵
When I hear min-norm solution to $Ax = b$ my ears perk up -- a well known result in machine learning is that stochastic gradient descent on $Ax - b$ (more precisely, $(Ax - b)^{\top}(Ax - b)$)) converges to the min-norm solution. Could this be my chance to use preconditioned gradient descent?↵
↵
As it turns out this is not to be. I try using gradient descent, and the precision after a finite number of operations is simply not high enough, especially given the massive precision demanded by the problem.↵
↵
It's here that I stop trying to overcomplicate things. I rewrite $Ax = b$ as $Cx = d$, where $C$ is a matrix composed of orthogonal rows, via Gram-Schmidt orthogonalization (or QR decomposition). As such, if we write ↵
$$ C = \begin{bmatrix} -- & e_1 & -- \\ ↵
-- & \vdots & -- \\ ↵
-- & e_n & -- \end{bmatrix}$$↵
↵
Now, to find the min-norm solution of $Cx = d$, note that we can simply take $x_{\text{min}} = d_1e_1 + \ldots + d_ne_n$. Note that any other solution to $Cx = d$ is just $x_{\text{min}}$ added to vectors in $e_{n+1}, e_{n+2}, \ldots e_d$, each of which only add norm to $x$. And now I knew I was almost done! We just need to find an $e_{n+1}$ that's orthogonal to each of $e_1, e_2, \ldots e_n$, and then find a $c$ so that the norm of $x_{\text{min}} + ce_{n+1}$ is equal to $r$. ↵
↵
Inspired by this, and wanting to get the coveted first solve award, I begin coding this solution, typing the fastest that I have in a while. After about 20 minutes, at the three-hour mark, I have a mostly working solution that seems to check out on the sample data, and I excitedly submit, to get↵
↵
↵
After this, there's a fairly obvious typo I fix, and I resubmit again, getting↵
↵
↵
At the last few programming contests I've participated in, run-time error has been a pretty common occurence. This is a fairly rare error to have, and one that is usually not too hard to fix. Calculating the value of $c$ in the above uses a square root operation, and due to floating point precision issues sometimes this number is slightly below zero, so I patch that issue. Resubmit! ↵
↵
↵
At this point, I try running my new code on the samples again, and get something quite unusual: my code prints the answer, but throws a run-time error in the middle. Frustrated, I step away from the problem briefly and let my other teammate ([user:halohalo,2021-10-06]) try looking for a run-time error. He quickly finds something amiss. Namely, in order to compute $e_{n+1}$ my code calls a subroutine which essentially runs Gram-Schmidt orthogonalization on $Ax = b$, but with a random vector appended to the end of $A$. The last value of $b$ doesn't matter for this component (in fact the value of $b$ doesn't matter at all for computing $e_{n+1}$) but the way I had it implemented, $b$ was one shorter than it was supposed to be, and was thus leading to out-of-bounds errors. We fix this, and are greeted by: ↵
↵
↵
At this point [user:ekzhang,2021-10-06] decided to reimplement my program from scratch. I worked on preparing large tests, hoping to find an error. I find it, a large test where the variable $n$ is equal to $500$ and where $d$ is $490$. In this case, the linear system is actually overdetermined, and we have $499$ equations in $490$ variables. In these cases, naive Gram-Schmidt will start trying to nomalize vectors which (up to floating point error) are zero, leading to potentially unusual consequences. In this case, my program explodes. ↵
↵
His program passes the samples, but doesn't pass the large case either. There are now about $10$ minutes left in the contest, so he tries a hail mary adding one line of code to the beginning of the program:↵
$n = min(n, d + 1)$. In fact, I thought that I had expressed this implicitly later in the program. ↵
↵
His solution still fails the weird case, but is much closer to right. With nothing else to try, and with 4 minutes left and 15 wrong submissions, I add this line of code to the beginning of my program, and submit. ↵
↵
↵
When I see the green
↵
You couldn't have scripted a better ending to my ICPC career (I'm not doing it in the future), one where I solved the hardest math problem on the set. ↵
↵
Originally we thought that Latvia may have solved the problem before us, but then when we see their last submission is after ours (at 298, with 2 minutes left) we know that we've won the first to solve award for the problem. ↵
↵
There were many things that could have gone better in solving this problem. We probably didn't need to make these wrong submits, and I could've made the $n = min(n, d + 1)$ check earlier (never hurts to check twice!) But at the end we had our shiny Correct, and that's all that mattered to me in the end.↵
↵
I'd like to thank everyone who helped me (far too numerous to list in the post) for a great 6 years of competitive programming. Bon voyage! ↵
↵
