## Tuesday, April 30, 2024

### From IP to CP

Someone asked on OR Stack Exchange how to convert an integer programming model into a constraint programming model. I think you can reasonably say that it involves a "paradigm shift", for a couple of reasons.

The first paradigm shift has to do with how you frame the problem, mainly in terms of the decision variables. Math programmers are trained to turn discrete decisions with a logical flavor into binary variables. Discrete quantities, such as how many bins of a certain type to use or how many workers to assign to a task, are expressed as general integer variables, but most other things end up turning into a slew of binary variables. The problem being solved in the ORSE question illustrates this nicely.

The problem is as follows. You have $N$ participants in a tournament involving some kind of racing. Importantly, $N$ is guaranteed to be an even number. There is one track with two lanes, and races are spread over $N-1$ days. Every participant races head to head with every other participant exactly once, and nobody races twice in the same day. For whatever reason, the left lane is preferable to the right lane, and so there is a "fairness" constraint that nobody is assigned the left lane on more than $M$ consecutive days. For some reason, the author also imposed a second fairness constraint that nobody be assigned to the right lane on more than $M$ consecutive days. Dimensions for the author's problem were $N=20$ and $M=2.$

The model has to assign participant pairs (races) to days and also make lane assignments. To decide against whom I must race on a given day, someone building an IP model will use binary variables to select my opponent. Similarly, they will use binary variables to select my lane assignment each day. So the author of the question had in his IP model a variable array opp[Competitors][Competitors][Tracks][Days] taking value 1 "if competitor 'c1' races with 'c2' on track 't' on day 'd'".

CP models are more flexible in their use of variables, and in particular general integer variables. So to decide my opponent on a given day, I can just an integer variable array indexed by day where the value is the index number of my opponent on the given day. Similarly, I could (and would) use an integer variable indexed by day to indicate my lane assignment that day, although in this case that variable does turn out to be binary, since there are only two lanes.

The second paradigm shift has to do with constraints, and it ties to what solver you are using. IP models have a very limited constraint "vocabulary". They all understand linear equalities and inequalities, and some understand some combination of SOS1, SOS2, second order cone and implication constraints. That's pretty much it. CP solvers have a richer "vocabulary" of constraints, but with the caveat that not many of those constraints are universal. I would wager that every CP solver has the "all different" constraint, and they must have the usual arithmetic comparisons ($=,\neq,\lt,\le,\gt,\ge$). Beyond that, it pays to check in advance.

I wrote a CP model (in Java) using IBM's CP Optimizer (CPO) to solve the scheduling problem. Details of the model can be sussed out from the Java code, but I will mention a few pertinent details here.

• I did use an integer variable array to determine, for each combination of participant and day, the participant's opponent that day, as well as an integer array giving the lane assignment (0 or 1) for each combination of participant and day.
• To make sure that, on any day, the opponent of X's opponent is X I used CPO's  inverse constraint. The constraint inverse(f, g) says that f[g[x]] = x and g[f[x]] = x for any x in the domain of the inner function.
• To ensure that nobody raced the same opponent twice, I used allDiff, which is CPO's version of the all different constraint.
• We have to do something to force opponents in a race to be in different lanes. Let $x_{i,d}$ and $y_{i,d}$ denote respectively the opponent and lane assignment for participant $i$ on day $d.$ In mathematical terms, the constraint we want is $y_{x_{i,d},d} \neq y_{i,d}.$ Indexing a variable with another variable is impossible in an IP model. In CPO, I used the element constraint to do just that.

I added an objective function, namely to minimize the difference between the most and fewest times any participant gets assigned the preferred left lane. I also added one constraint to mitigate symmetry. Since any solution remains a solution (with the same objective value) under any permutation of the participant indices, I froze the first day's schedule as $1\ v.\ N$, $2\ v.\ N-1$, $3\ v.\ N-2$ etc.

On my decent but not screamingly fast PC, CPO found a feasible solution almost instantly and a solution with objective value 1 in under a second. In that solution, every participant gets the left lane either nine or ten times out of the 19 racing days. It's not hard to prove that 1 is the optimal value (you cannot have everybody get exactly the same number of left lane assignments), but don't tell CPO that -- it was still chugging along trying when it hit my five minute time limit.

My Java code is available from my repository under a Creative Commons 4.0 open source license.

## Sunday, April 21, 2024

### Where Quadratic, Positive Definite and Binary Meet

A comment by Rob Pratt (of SAS) on OR Stack Exchange pointed out two things that are glaringly obvious in hindsight but that somehow I keep forgetting. Both pertain to an expression of the form $x'Qx + c'x,$ either in an objective function or in a second order cone constraint, where $x$ is a vector of variables and $Q$ and $c$ are parameters.

The first observation does not depend on the nature of the $x$ variables. We can without loss of generality assume that $Q$ is symmetric. If it is not, replace $Q$ with the symmetric matrix $\hat{Q} = \frac{1}{2}\left(Q + Q'\right),$ which is symmetric. A wee bit of algebra should convince you that $x'\hat{Q}x = x'Qx.$

The second observation is specific to the case where the $x$ variables are binary (which was the case in the ORSE question which drew the comment from Rob). When minimizing an objective function of the form $x'Qx + c'x$ or when using it in a second order cone constraint of the form $x'Qx + c'x \le 0,$ you want the $Q$ matrix to be positive definite. When $x$ is binary, this can be imposed easily.

Suppose that $x$ is binary and $Q$ is symmetric but not positive definite. The following argument uses the euclidean 2-norm. Let $$\Lambda = \max_{\parallel y \parallel = 1} -y'Qy,$$ so that $y'Qy \ge -\Lambda$ for any unit vector $y.$ Under the assumption that $Q$ is not positive definite, $\Lambda \ge 0.$ Choose some $\lambda > \Lambda$ and set $\hat{Q} = Q + \lambda I,$ where $I$ is the identity matrix of appropriate dimension. For any nonzero vector $y,$

\begin{align*} y'\hat{Q}y & =y'Qy+\lambda y'Iy\\ & =\parallel y\parallel^{2}\left(\frac{y'}{\parallel y\parallel}Q\frac{y}{\parallel y\parallel}+\lambda\right)\\ & \ge\parallel y\parallel^{2}\left(-\Lambda+\lambda\right)\\ & >0. \end{align*}

So $\hat{Q}$ is positive definite. Of course, $x'\hat{Q}x \neq x'Qx,$ but this is where the assumption that $x$ is binary sneaks in. For $x_i$ binary we have $x_i^2 = x_i.$ So

\begin{align*} x'\hat{Q}x & =x'Qx+\lambda x'Ix\\ & =x'Qx+\lambda\sum_{i}x_{i}^{2}\\ & =x'Qx+\lambda e'x \end{align*}

where $e=(1,\dots,1).$ That means the original expression $x'Qx + c'x$ is equal to $x'\hat{Q}x+(c-\lambda e)'x,$ giving us an equivalent expression with a positive definite quadratic term.

## Thursday, April 11, 2024

### Finding Duplicated Records in R

Someone asked a question about finding which records (rows) in their data frame are duplicated by other records. If you just want to know which records are duplicates, base R has a duplicated() function that will do just that. It occurred to me, though, that the questioner might have wanted to know not just which records were duplicates but also which records were the corresponding "originals". Here's a bit of R code that creates a small data frame with duplicated rows and then identifies original/duplicate pairs by row number.

library(dplyr)

# Create source data.
df <- data.frame(a = c(3, 1, 1, 2, 3, 1, 3), b = c("c", "a", "a", "b", "c", "a", "c"))

# Find the indices of duplicated rows.
dup <- df |> duplicated() |> which()

# Split the source data into two data frames.
df1 <- df[-dup, ]  # originals (rows 1, 2 and 4)
df2 <- df[dup, ]   # duplicates (rows 3, 5, 6 and 7)

# The row names are the row indices in the original data frame df. Assign them to columns.
df1$Original <- row.names(df1) df2$Duplicate <- row.names(df2)

# Perform an inner join to find the original/duplicate pairings. The "NULL" value for "by"
# (which is actually the default and can be omitted) means rows of df1 and df2 are paired
# based on identical values in all columns they have in common (i.e., all the original
# columns of df).
inner_join(df1, df2, by = NULL) |> select(Original, Duplicate)

# Result:
#   Original Duplicate
# 1        1         5
# 2        1         7
# 3        2         3
# 4        2         6

The key here is that the inner_join function pairs rows from each data frame (originals and duplicates) based on matching values in the "by" columns. The default value of "by" (NULL) tells it to match by all the columns the two data frames have in common -- which in the is case is all the columns in the source data frame. The resulting data frame will have the columns from the source data frame (here "a" and "b") plus the columns unique to each data frame ("Original" and "Duplicate"). We use the select() command to drop the source columns and just keep the indices of the original and duplicate rows.

## Monday, April 8, 2024

### File Access in RStudio

I've been spending a fair bit of time in RStudio Desktop recently, much of it related to my work with INFORMS Pro Bono Analytics. I really like RStudio as a development environment for R code, including Shiny apps. It does, however, come with the occasional quirk. One of those has to do with how RStudio accesses the file system.

I tripped over this a couple of times recently when I wanted to open an R file that I had dropped in the /tmp directory on my Linux Mint system. The Files tab in RStudio appeared to be limited to the directory tree under my home directory. There was no way to browse to system directories like /tmp. Similarly, there is a way to set the default working directory (Tools > Global Options... > General > Basic > R Sessions). RStudio does not let you type in a directory name (perhaps a defense against typos?), and the Browse... button will not leave your home tree.

Initially I decided this was not important enough to worry about, but then I saw a post on the Posit Community forum by someone who was stuck trying to work from home due to a related issue. So I did a little experimentation and found a workaround, at least for the first problem (accessing files in places like /tmp). If I run setwd("/tmp") in the Console tab (which sets the working directory for the current R session), then click the More menu in the Files tab and select Go To Working Directory, the Files tab now browses /tmp, and I can navigate up to the system root directory and then down to anywhere within reason.

Changing the default starting directory is not something I actually care to do, but I'll document it here in case a reader might wish to do so. You can go to the IDE configuration directory (~/.config/rstudio on Linux and OS X, %appdata%\RStudio on Windows), open the rstudio-prefs.json file in a text editor, and change the value of the "initial_working_directory" entry to whatever starting directory you want. Save it, (re)start RStudio Desktop, and hopefully you begin in the right place.