# Leaving Earth is a Constraint Satisfaction Problem

Mission planning with the Z3 Theorem Prover

## Leaving Earth#

Leaving Earth is a board game by Joseph Fatula that recreates the space race of the 1950s, 60s and 70s. At the start of the game, a handful of missions will be revealed and the players compete to be the first to complete each one, scoring points based on their complexity. Many of these missions are examples of actual human accomplishments - such as Venus Lander - while others - Mars Station - are more fanciful.

### Mission Planning#

The core mechanic in Leaving Earth is planning the missions using a variety of rockets, capsules and other components to perform a series of maneuvers from one location to another.

The key concept is that each maneuver has a difficulty - roughly approximating Delta-v - and each rocket can produce a certain amount of thrust, at which point it is expended.1 The amount of thrust required to perform a maneuver is based on this simple formula: thrust required = mass × difficulty. The final factor, mass, is determined by the number of components aboard the spacecraft.

A simple mission plan for a spacecraft travelling from Earth to the Lunar surface might look like the following (the numbers in parentheses indicates the mass of that component):

Earth to Earth Orbit8Saturn (20)Atlas (4), Juno (1), Probe (1)26 × 8
Earth Orbit to Lunar Orbit3Atlas (4)Juno (1), Probe (1)6 × 3
Lunar Orbit to Moon2Juno (1)Probe (1)2 × 2

As long as the rockets used in each stage of this mission can product the required thrust, then the maneuver can be completed. In the example above, a Saturn rocket produces 200 thrust, an Atlas 27 thrust and a Juno 4, so each maneuver succeeds.

An alternative route to space would be to use a two-stage launch to avoid the difficulty 8 maneuver required to go directly from the Earth’s surface to Earth Orbit.

Earth to Suborbital Flight3Soyuz (9)Soyuz (9), Atlas (4), Juno (1), Probe (1)24 × 3
Suborbital Flight to Earth Orbit5Soyuz (9)Atlas (4), Juno (1), Probe (1)15 × 5
Earth Orbit to Lunar Orbit3Atlas (4)Juno (1), Probe (1)6 × 8
Lunar Orbit to Moon2Juno (1)Probe (1)2 × 2

A Soyuz produces only 80 thrust, but using this rocket type in a two-stage configuration means that it can still reach orbit. In fact, two Soyuz rockets can carry a payload of 7 mass into orbit, whereas a single Saturn can only carry 5. And since in the game’s economy a Soyuz only costs \$8 compared to a hefty \$15 for a Saturn, the former starts to look like an attractive proposition. That is until you start planning a mission that requires moving a large amount of mass into orbit, such as transporting humans to Mars and back again.

There are many other mission planning complexities: Ion Thrusters, which are not expended but take time to travel from one location to another; slingshot maneuvers to travel to the outer reaches of the solar system; one-way maneuvers; and more. But the basic concept I presented above is enough to move on.

# Constraint Satisfaction Problems#

So how can Leaving Earth’s mission planning be reduced to a Constraint Satisfaction Problem (CSP)?

Let’s take the difficulty 8 maneuver from Earth to Earth Orbit and rewrite it as a CSP. To simplify things, we’ll assume that there are only Saturn and Soyuz rockets available.

The question we want to solve is how many Saturn and how many Soyuz rockets are needed to move a mass of, say, 2 from Earth to Earth Orbit. In other words, we want to find the values of two variables, `saturn` and `soyuz`, such that they satisfy a set of constraints.

The first two constraints are that both variables must be greater or equal to zero (we don’t allow negative numbers of rockets):

• `saturn ≥ 0`
• `soyuz ≥ 0`

Next, our Leaving Earth formula thrust required = mass × difficulty needs to be rewritten slightly as thrust produced ≥ mass × difficulty; in other words, we need to make sure that the thrust produced by the Saturn and Soyuz rockets that we use is large enough.

To write this constraint, first consider what we know:

• thrust produced = `saturn` × 200 + `soyuz` × 80
• mass = `saturn` × 20 + `soyuz` × 9 + payload
• difficulty = 8

Then bring this information together as a third and final constraint:

• `saturn × 200 + soyuz × 80 ≥ (saturn × 20 + soyuz × 9 + 2) × 8`

Let’s propose a hypothetical solution to the problem that is `saturn = 0, soyuz = 1`. To check if this is a solution we need to ensure that the three constraints are satisfied.

1. `saturn ≥ 0`: 0 is greater or equal to 0
2. `soyuz ≥ 0`: 1 is greater or equal to 0
3. `saturn × 200 + soyuz × 80 ≥ (saturn × 20 + soyuz × 9 + 2) × 8`: 80 ≥ 88

Unfortunately, we do not satisfy the third constraints, so this is not a valid solution; a single Soyuz rocket is not powerful enough to carry this payload into Earth Orbit. What if we try again with two Soyuz rockets?

1. `saturn ≥ 0`: 0 is greater or equal to 0
2. `soyuz ≥ 0`: 2 is greater or equal to 0
3. `saturn × 200 + soyuz × 80 ≥ (saturn × 20 + soyuz × 9 + 2) × 8`: 160 ≥ 160

The final constraint is now satisfied, and this two-engine spacecraft is able to boost a payload of mass 2 into Earth Orbit!

Finally, if we used a single Saturn rocket, we also have a valid solution:

1. `saturn ≥ 0`: 1 is greater or equal to 0
2. `soyuz ≥ 0`: 0 is greater or equal to 0
3. `saturn × 200 + soyuz × 80 ≥ (saturn × 20 + soyuz × 9 + 2) × 8`: 200 ≥ 176

In fact, any combination of two or more Saturn and Soyuz rockets will satisfy this CSP (e.g. `saturn = 2, soyuz = 0`, `saturn = 1, soyuz = 1` and `saturn = 5, soyuz = 3` are all valid solutions).

Now if the payload size increased from 2 to, say, 10, and we allowed all the different rocket types modelled in Leaving Earth, then the set of solutions is perhaps not so obvious. This is where a CSP solver comes in handy.

## Z3 theorem prover#

Z3 is a toolkit for solving CSPs with bindings for several languages, including Python, which is the language I’ll be using here.

### Solving a single maneuver#

Rewriting the CSP above in Z3 for a payload of 10 and a maneuver of difficulty 5 would look like this:

``````from z3 import solve, Int

saturn = Int("saturn rockets")
soyuz = Int("soyuz rockets")

thrust = saturn*200 + soyuz*80
mass = saturn*20 + soyuz*9 + payload
difficulty = 5

solve(saturn>=0, soyuz>=0, thrust >= mass * difficulty)
``````

Z3 will return the following result:

``````[saturn rockets = 0, soyuz rockets = 2]
``````

In other words, 2 Soyuz rockets can transport a payload of 10 to Suborbital Flight. This certainly is a solution, but is it the optimal solution? For this, we have to decide what defines optimal. Is it the number of rockets? The total mass of the rockets? Or is it the cost of the rockets used? If we are thinking about a game of Leaving Earth, it’s probably the cost.

Let’s rewrite the function but with a cost optimization and let’s add all the remaining rockets in as variables.

``````from z3 import Optimize, Int

saturn = Int("saturn rockets")
soyuz = Int("soyuz rockets")
atlas = Int("atlas rockets")
juno = Int("juno rockets")

thrust = saturn*200 + soyuz*80 + atlas*27 + juno*4
mass = saturn*20 + soyuz*9 + atlas*4 + juno*1 + payload
difficulty = 5

cost = saturn*15 + soyuz*8 + atlas*5 + juno*1

solver = Optimize()
solver.minimize(cost)

solver.check()
solver.model()
``````

And this returns:

``````[saturn rockets = 1, soyuz rockets = 0, atlas rockets = 0, juno rockets = 0]
``````

And of course you can get a mix of rockets with the right conditions, such as by setting `payload` to `21`, giving:

``````[saturn rockets = 1, soyuz rockets = 0, atlas rockets = 1, juno rockets = 0]
``````

### Solving longer missions#

When it’s just a single maneuver, the constraints end up being quite simple; however, missions in Leaving Earth inevitably have multiple steps. Fortunately, this is not difficult to express in Z3. Let’s take the mission from Earth Orbit to the Moon. We have a maneuver of difficulty 3, followed by a maneuver of difficulty 2.

graph TD E((Earth)) -->|3| ESo(Suborbital Flight) ESo -->|5| Eo E -->|8| Eo(Earth Orbit) subgraph Eo ==>|3| Lo(Lunar Orbit) Lo ==>|2| L(Moon) end Eo -->|3| Ipt(Inner Planets Transfer) Ipt -->|5| Hfb(Mercury Fly-by) Hfb -->|2| Ho(Mercury Orbit) Ho -->|2| H(Mercury) Ipt -->|4| Mo(Mars Orbit) Ipt -->|1| Mfb(Mars Fly-by) Mfb -->|3| Mo Mo -->|0| M(Mars) Ipt -->|3| Vo(Venus Orbit) Ipt -->|2| Vfb(Venus Fly-by) Vfb -->|1| Vo Vo -->|0| V(Venus)

The key here is to express this mission backwards; in other words, we’ll make sure we can travel from Lunar Orbit to the Moon first and then ensure we can travel from Earth Orbit to Lunar Orbit.

And crucially, the payload gets larger and larger as we move backwards in the mission by the number of rockets that will be expended in the future (remember that once a rocket is used, it is removed from the spacecraft):

0Lunar Orbit to Moon25
1Earth Orbit to Lunar Orbit35 + mass of rockets used in Stage 0

In Z3, this would be:

``````from z3 import Optimize, IntVector, Int
solver = Optimize()
maneuvers = [3, 2]
maneuvers.reverse() # plan the mission backwards

saturn = IntVector("saturn rockets", len(maneuvers))
soyuz = IntVector("soyuz rockets", len(maneuvers))
atlas = IntVector("atlas rockets", len(maneuvers))
juno = IntVector("juno rockets", len(maneuvers))

cost = 0 # running total of the cost
for i, difficulty in enumerate(maneuvers):

thrust = saturn[i]*200 + soyuz[i]*80 + atlas[i]*27 + juno[i]*4
thruster_mass = saturn[i]*20 + soyuz[i]*9 + atlas[i]*4 + juno[i]*1

cost += saturn[i]*15 + soyuz[i]*8 + atlas[i]*5 + juno[i]*1

solver.minimize(cost)
solver.check()
solver.model()
``````

Removing all the unused rockets, we are left with the following:

``````[atlas rockets0 = 1, soyuz rockets1 = 1]
``````

This is saying that for maneuver `0`, we need an Atlas and for maneuver `1`, we need a Soyuz. Remember that we planned this backwards, so `0` is the last maneuver (i.e. Lunar Orbit to the Moon) and `1` is the maneuver that came before it (i.e. Earth Obit to Lunar Orbit).

The amazing thing is that by simply changing the line `maneuvers = [3, 2]`, we can solve any one-way mission in Leaving Earth that uses expendable rockets.

Earth to the Moon? `maneuvers = [8, 3, 2]`

Earth to the Moon with a two stage launch? `maneuvers = [3, 5, 3, 2]`

Earth to Mercury with a two stage launch? `maneuvers = [3, 5, 3, 5, 2, 2]`

### Ion Thrusters#

Ion Thrusters work a little differently: they can’t be used for maneuvers travelling to or from celestial bodies; they are not expended when they are used; they produce 5 thrust for every year the journey takes. As an example, if a spacecraft composed of a Probe (mass 1) and an Ion Thruster (mass 1) travels from Earth Orbit to Lunar Orbit, then the journey will take 2 years to complete. The same formula - thrust required = mass × difficulty - applies, giving (1+1) × 3 = 6. Ion Thrusters generate 5 thrust per year so one year of travel would only provide 5 thrust and we need 6, hence two years.

Since Ion Thrusters are not expended, they are very efficient as long as you aren’t in a hurry, but remember Leaving Earth is a race to complete missions, so players must be mindful about how long travel takes!

Let’s add support for Ion Thrusters to the code above. First we need to add the Ion Thrusters’ mass and cost once but add their thrust at each stage. Second, we need to introduce a new variable for the number of years each maneuver takes.

``````from z3 import Optimize, IntVector, Int
solver = Optimize()
maneuvers = [3, 5, 2]
maneuvers.reverse() # plan the mission backwards

saturn = IntVector("saturn rockets", len(maneuvers))
soyuz = IntVector("soyuz rockets", len(maneuvers))
atlas = IntVector("atlas rockets", len(maneuvers))
juno = IntVector("juno rockets", len(maneuvers))
ion = Int("ion thrusters")
year = IntVector("years travelled", len(maneuvers))

cost = 0 # running total of the cost
years = 0 # running total of journey time

payload += ion * 1 # add the weight of the ion thrusters to the payload once
cost += ion * 10 # add the cost of the ion thrusters to the total once

for i, difficulty in enumerate(maneuvers):

thrust = saturn[i]*200 + soyuz[i]*80 + atlas[i]*27 + juno[i]*4 + ion*year[i]*5
thruster_mass = saturn[i]*20 + soyuz[i]*9 + atlas[i]*4 + juno[i]*1

years += year[i]
cost += saturn[i]*15 + soyuz[i]*8 + atlas[i]*5 + juno[i]*1

solver.minimize(cost)
solver.check()
solver.model()
``````

The code above is for a journey from Earth Orbit to Mercury Orbit. Because Ion Thrusters can’t be used to reach space from Earth, the code would need to be more complex to support the full journey, but this is good enough for now.

``````[ion thrusters = 1,
years travelled0 = 3,
years travelled1 = 9,
years travelled2 = 4]
``````
StageManeuverDifficultyTime
2Earth Orbit to Inner Planets Transfer34
1Inner Planets Transfer to Mercury Fly-by59
0Mercury Fly-by to Mercury Orbit23

For \$10 in game money, a single Ion Thruster will carry a payload all the way to Mercury! The only problem is that it will take 16 years … but what if we ask Z3 to optimize for the journey time?

``````solver.minimize(cost)
solver.minimize(years)
``````

Now cost is still the most important factor but years will also be minimized and the new results are that the journey will take 13 years. You can also flip the order of the two minimize statements so that the critical factor is the journey time and cost is only of secondary importance!

``````[juno rockets0 = 5, saturn rockets1 = 1, saturn rockets2 = 1]
``````

Now suddenly, Z3 wants us to use expendable rockets for each maneuver because they don’t have the additional time cost of the Ion Thrusters.2 The monetary cost is significantly higher, however, at \$35!

### Controlling costs#

In order to controlling spiralling program costs, we can add an additional constraint on how much is spent on a mission. Let’s say we are budgeting \$25 on our mission to Mercury from Earth Orbit. We still want it completed as fast as possible, but it has to fall below this budget.

``````solver.add(cost <= 25)
solver.minimize(years)
solver.minimize(cost)
``````

Z3 finds a nice balance between rockets and Ion Thrusters, with the Ion Thrusters used at every stage but an Atlas also providing an initial boost:

``````[ion thrusters = 2, atlas rockets2 = 1,
years travelled0 = 2, years travelled1 = 4, years travelled2 = 1]
``````
StageManeuverDifficultyTimeRockets
2Earth Orbit to Inner Planets Transfer31Atlas
1Inner Planets Transfer to Mercury Fly-by54
0Mercury Fly-by to Mercury Orbit22

## Leaving Earth Solver#

I put all of this and more together into a Python package to solve many (but not all) Leaving Earth missions, including rules from Outer Planets and Stations. It’s a command-line tool that works like this:

bosth/leaving-earth-solver

Leaving Earth Solver

Python
2
2

### Usage#

Use `les --help` to get a list of arguments and parameters:

``````Usage: les [OPTIONS] ORIGIN DESTINATION [PAYLOAD]

Options:
--version                       Show the version and exit.
-v, --verbose                   Verbose mode
-j, --juno RANGE                Number of Juno rockets
-a, --atlas RANGE               Number of Atlas rockets
-s, --soyuz RANGE               Number of Soyuz rockets
-p, --proton RANGE              Number of Proton rockets
-n, --saturn RANGE              Number of Saturn rockets
-i, --ion RANGE                 Number of Ion thrusters
-c, --cost RANGE                Cost of mission
-m, --minimize [time|cost|mass]
Minimization goal
--single-stage                  Check a single stage configuration for
launches from Earth (by default only a two-
stage configuration will be attempted)
--aerobraking / --no-aerobraking
Use aerobraking
--rendezvous / --no-rendezvous  If rendezvous technology is available, Ion
thrusters will be detached when no longer
needed
-t, --time RANGE                Number of time tokens
-y, --year INTEGER RANGE        Year that journey starts  [1956<=x<=1986]
--help                          Show this message and exit.
``````

The basic pattern is to provide the origin and destination and `les` will assume a payload of size 1:

``````les Eo L
``````

The results are a JSON description of the entire mission:

``````{
"components": {
"juno": 5
},
"mass": 5,
"cost": 5,
"time": 0,
"plan": [
{
"origin": "Earth orbit",
"destination": "Lunar fly-by",
"difficulty": 1,
"components": {
"juno": 2
},
"thrust": 8
},
{
"origin": "Lunar fly-by",
"destination": "Lunar orbit",
"difficulty": 2,
"components": {
"juno": 2
},
"thrust": 8
},
{
"origin": "Lunar orbit",
"destination": "Moon",
"difficulty": 2,
"components": {
"juno": 1
},
"thrust": 4
}
]
}
``````

If you don’t have Saturn technology (or don’t want Saturn rockets to be part of the solution), you can force them to `0`:

``````les Eo L --saturn 0
``````

`les` understands the following format for the numbers provided to it:

ParameterRange
1exactly 1
1+1 or more
1-31 to 3
-33 or fewer

### More examples#

To find the optimal solution for a trip that taking 5 mass from Earth (`E`) to the Moon that uses no Saturn rockets, between 1 and 3 Soyuz rockets and where we are able to detach the Ion Thruster before landing on the Moon (this avoids moving unnecessary mass):

``````les E L 5 --saturn 0 --soyuz 1-3 --rendezvous
``````

To find the optimal solution for a trip taking 10 mass from Earth to Mars (`M`) in 5 years or less with aerobraking technology available:

``````les E M 10 --aerobraking -t -5
``````

To find the optimal solution for the fastest trip carrying 10 mass from Earth Orbit to Jupiter Fly-by (`Jfb`) that leaves in 1960.

``````les Eo Jfb 10 --year 1960 --minimize time
``````

Result:

``````{
"start": 1960,
"end": 1963,
"components": {
"juno": 3,
"soyuz": 1,
"ion": 1,
"atlas": 1
},
"mass": 17,
"cost": 26,
"time": 3,
"plan": [
{
"origin": "Earth orbit",
"destination": "Inner planets transfer",
"difficulty": 3,
"components": {
"soyuz": 1,
"ion": 1
},
"aerobraking": false,
"year": 1960,
"time": 1,
"thrust": 85,
"slingshot": false
},
{
"origin": "Inner planets transfer",
"destination": "Venus fly-by",
"difficulty": 2,
"components": {
"juno": 1,
"atlas": 1,
"ion": 1
},
"aerobraking": false,
"time": 1,
"year": 1961,
"thrust": 36,
"slingshot": false
},
{
"origin": "Venus fly-by",
"destination": "Jupiter fly-by",
"difficulty": 1,
"components": {
"juno": 2,
"ion": 1
},
"aerobraking": false,
"year": 1962,
"time": 1,
"thrust": 13,
"slingshot": true
}
]
}
``````

### How it works#

The core concept is identical to what I described above but `les` also finds the optimal paths from the origin to the destination. In most cases there are multiple paths between two points and the optimal path depends on a number of external factors. Even travelling to the Moon from Earth presents many options:

graph TD E((Earth)) -->|3| ESo(Suborbital Flight) ESo -->|5| Eo(Earth Orbit) E -->|8| Eo Eo -->|3| Lo(Lunar Orbit) Lo -->|2| L((fa:fa-ban Moon)) Eo -->|1| Lfb(Lunar Fly-by) Lfb -->|2| Lo Lfb -->|4| L

By default, `les` will try to predict a few plausible paths and solve for each of those, selecting the best solution at the end.

`les` is aware of which maneuvers support Ion Thrusters and can detatch these opportunistically when they will no longer be needed to complete a mission. Conversely, it will carry unused Ion Thrusters from one location to the next if they will be used at a future stage.

It also can make use of aerobraking, slingshot maneuvers and Proton rockets from the expansions. It doesn’t support the reusable shuttles, however, and it won’t increase or decrease the difficulty of maneuvers to change travel times. The biggest missing feature, however, is probably not being able to plan return missions (e.g. going to the Moon and back again).

1. When a rocket is expended, its card is discarded and it is no longer part of the spacecraft. ↩︎

2. Leaving Earth players will note that some maneuvers take a minimum amount of time even with rockets, but I am ignoring that for the moment. ↩︎