Apologies for being a little late today… Took a little bit to

understand

some of what was going on in the solutions for the Preferable Pairs

quiz,

and I’m not certain I explained it well enough, but… here it is.

This “Preferable Pairs” quiz is, as was noted, very similar to the

Stable

Roommates problem, though the goal as originally stated is slightly

different from the typical presentation of the Stable Roommates problem.

Still, the minor goal difference does not change that this is an

NP-complete problem, and as such, large input sets would require

some

sort of approximation, heuristics or other techniques to keep the run

time

reasonable.

Fortunately, I had no intention of testing the solutions provided on

large

sets. My test involved a set of 20 people, to be grouped into 10 pairs

according to their preferences. Here are the results of my test on the

solutions provided.

```
Solution Time Score
Andrea F. 0.159 438
Dustin Barker 0.020 589
Eric Ivancich 4.671 311
Steven Hahn -DNF-
Eric M. 0.211 311
Matthew M. 0.022 589
Thomas ML 0.114 311
```

I neglected to post my own solution, but it’s nearly identical to

Dustin’s

both in algorithm, results and performance (but mine is much uglier).

Also,

apologies to Steven… I tried to wait, really I did, but it just kept

going, and going…

Andrea, Dustin and myself made straightforward attempts that were fast

but

not optimal. While the numbers presented above don’t show a huge

disparity

in performance, I suspect that would be a different story if we were

attempting to solve larger datasets, of thousands of people as opposed

to

twenty. For the tested dataset, I believe there might be a few grumbling

people, forced to play with someone they didn’t like very much, but the

tournament would go on.

Eric Ivancich provided a genetic algorithm that is notably slower (at

this

sample size), and isn’t guaranteed to get the most optimal answer, but

it

happened to do so here. It would be interesting to compare its

performance

against the optimal solution algorithms for larger data sets.

Eric M. and Thomas ML provided algorithms that find the optimal

solution. There is a fair bit of setup code in both to get the data into

a

convenient form for the optimization algorithm to work. I’m going to

skip

over that and jump right into Thomas’ `optimize`

method. As input to

`optimize`

, the `pairings`

argument looks like this for the sample data:

```
[
[["David", "Helen"], 0],
[["David", "Vicki"], 1],
[["Helen", "Vicki"], 2],
[["Joseph", "Vicki"], 4],
[["Helen", "Joseph"], 5],
[["David", "Joseph"], 8]
]
```

Basically, the input is an ordered list of all pairs with the

corresponding

score (as calculated according to the metric as described by the quiz).

It

is a combined score, and reflects the preferences of both people in the

pair. For example, the pair of Helen and Joseph has a score of 5,

because

Helen scores Joseph as 4, while Joseph scores Helen as 1: so, 4 + 1 ==

5.

Before we conquer the whole of the algorithm, let’s look at a simplified

version of `optimize`

to get a feel for the general structure. The input

here (for argument `pairings`

) is similar to the above, but without the

scores.

```
def optimize(pairings, names, pos, posmax)
bestpairs = nil
while pos < posmax
pair = pairings[pos]
pos += 1
if names & pair == pair
names1 = names - pair
if names1.size < 2
bestpairs = [pair]
bestpairs << names1 unless names1.empty?
return bestpairs
elsif (rv = optimize(pairings, names1, pos, posmax)
bestpairs = rv
bestpairs << pair
end
end
end
return bestpairs
end
```

`bestpairs`

starts off empty, but by the end should be an array of pairs

chosen as the solution. Each iteration of the loop grabs the next pair

from

`pairings`

and checks to see if it is still usable by set intersection:

```
if names & pair == pair
```

`names`

will contain the names of people still available; that is, those

that have not already been chosen previously. The `&`

operator treats

the

two arrays as sets and performs set intersections. If the result of

intersection is equal to one of the arguments, that argument must be a

subset of the other, and in this context, is safe to choose for the

solution.

When a pair is chosen, those names are removed by Array difference (not

quite the same as set difference, but close enough for this case). So it

is

with:

```
names1 = names - pair
```

That we remove the chosen pair from the list of remaining people. If

that is

the last possible pair to be made (there are fewer than 2 names

remaining),

we finish up by setting and returning the `bestpairs`

array

```
bestpairs = [pair]
bestpairs << names1 unless names1.empty?
return bestpairs
```

In the case that there is at least one more pair remaining, we continue

onto

the recursive part of this solution:

```
elsif (rv = optimize(pairings, names1, pos, posmax)
bestpairs = rv
bestpairs << pair
```

We know that `bestpairs`

, as returned by the recursive call to

`optimize`

,

contains the best pairs from the subset of `names1`

, which we got above

after removing `pair`

. So we make sure to concatenate `pair`

onto

`bestpairs`

before it is returned to the caller.

As simplified, this amounts to a greedy algorithm, since the pairs were

initially sorted according to score, and the simplified algorithm, at

each

level, simply takes the first pair possible.

Now let’s go back to the complete `optimize`

method.

```
def optimize(pairings, names, pos, posmax, maxweight=nil)
bestpairs = nil
maxweight ||= pairings.size ** 2 + 1
while pos < posmax
pair, weight1 = pairings[pos]
break unless weight1 * (names.size / 2).floor < maxweight
pos += 1
if names & pair == pair
names1 = names - pair
if names1.size < 2
bestpairs = [pair]
bestpairs << names1 unless names1.empty?
return [weight1, bestpairs]
elsif (rv = optimize(pairings, names1, pos, posmax,
```

maxweight - weight1))

maxweight, bestpairs = rv

maxweight += weight1

bestpairs << pair

end

end

end

return bestpairs && [maxweight, bestpairs]

end

The algorithm structure is basically the same: recursive, greedily

selecting

pairs and removing those names from the set of names available… The

major

difference here is the inclusion of the weights (i.e. scores) and

possible

rejection of pairs with respect to those weights.

As recursive calls to `optimize`

are made, the `maxweight`

value is

updated

via this code:

```
maxweight, bestpairs = rv
maxweight += weight1
```

and checked via this code:

```
break unless weight1 * (names.size / 2).floor < maxweight
```

So a pair may be skipped if it’s weight (scaled by the number of names)

exceeds the current maxweight, determined the the current set of

choices.

When a possible solution is found, the `maxweight`

variable represents

the

total score for that solution. But the algorithm does not stop

immediately;

it keeps checking pairs and solutions of pairs, rejecting those

solutions

and partial solutions whose total score (i.e. `maxweight`

) would exceed

that

previously known `maxweight`

.

In the end, the solution reported is the collection of pairs with the

lowest

total score, and is returned via the original invocation of `optimize`

.

–

Matthew M. [email protected]