Recently I came across a nice Ted-Ed video presenting a Fish Riddle.

I thought it would be fun to try solving it using Julia’s award winning JuMP package. Before we get started, please watch the above video-you might want to pause at 2:24 if you want to solve it yourself.

To attempt this problem in Julia, you will have to install the `JuMP`

package.

julia> Pkg.add("JuMP")

`JuMP`

provides an algebraic modeling language for dealing with mathematical optimization problems. Basically, that allows you to focus on describing your problem in a simple syntax and it would then take care of transforming that description in a form that can be handled by any number of *solvers*. Those solvers can deal with several types of optimization problems, and some solvers are more generic than others. It is important to pick the right solver for the problem that you are attempting.

The problem premises are:

1. There are 50 creatures in total. That includes sharks outside the tanks and fish

2. Each SECTOR has anywhere from 1 to 7 sharks, with no two sectors having the same number of sharks.

3. Each tank has an equal number of fish

4. In total, there are 13 or fewer tanks

5. SECTOR ALPHA has 2 sharks and 4 tanks

6. SECTOR BETA has 4 sharsk and 2 tanks

We want to find the number of tanks in sector GAMMA!

Here we identify the problem as mixed integer non-linear program (MINLP). We know that because the problem involves an integer number of fish tanks, sharks, and number of fish inside each tank. It also non-linear (quadratic to be exact) because it involves multiplying two two of the problem variables to get the total number or creatures. Looking at the table of solvers in the JuMP manual. pick the `Bonmin`

solver from `AmplNLWriter`

package. This is an open source solver, so installation should be hassle free.

julia> Pkg.add("AmplNLWriter")

We are now ready to write some code.

using JuMP, AmplNLWriter # Solve model m = Model(solver=BonminNLSolver()) # Number of fish in each tank @variable(m, n>=1, Int) # Number of sharks in each sector @variable(m, s[i=1:3], Int) # Number of tanks in each sector @variable(m, nt[i=1:3]>=0, Int) @constraints m begin # Constraint 2 sharks[i=1:3], 1 <= s[i] <= 7 numfish[i=1:3], 1 <= nt[i] # Missing uniqueness in restriction # Constraint 4 sum(nt) <= 13 # Constraint 5 s[1] == 2 nt[1] == 4 # Constraint 6 s[2] == 4 nt[2] == 2 end # Constraints 1 & 3 @NLconstraint(m, s[1]+s[2]+s[3]+n*(nt[1]+nt[2]+nt[3]) == 50) # Solve it status = solve(m) sharks_in_each_sector=getvalue(s) fish_in_each_tank=getvalue(n) tanks_in_each_sector=getvalue(nt) @printf("We have %d fishes in each tank.\n", fish_in_each_tank) @printf("We have %d tanks in sector Gamma.\n",tanks_in_each_sector[3]) @printf("We have %d sharks in sector Gamma.\n",sharks_in_each_sector[3])

In that representation we could not capture the restriction that “no two sectors having the same number of sharks”. We end up with the following output:

We have 4 fishes in each tank. We have 4 tanks in sector Gamma. We have 4 sharks in sector Gamma.

Since the problem domain is limited, we can possible fix that by adding a constrain that force the number of sharks in sector Gamma to be greater than 4.

@constraint(m,s[3]>=5)

This will result in an answer that that does not violate any of the stated constraints.

We have 3 fishes in each tank. We have 7 tanks in sector Gamma. We have 5 sharks in sector Gamma.

However, this seems like a bit of kludge. The proper way go about it is represent the number of sharks in the each sector as binary array, with only one value set to 1.

# Number of sharks in each sector @variable(m, s[i=1:3,j=1:7], Bin)

We will have to modify our constraint block accordingly

@constraints m begin # Constraint 2 sharks[i=1:3], sum(s[i,:]) == 1 u_sharks[j=1:7], sum(s[:,j]) <=1 # uniquness # Constraint 4 sum(nt) <= 13 # Constraint 5 s[1,2] == 1 nt[1] == 4 # Constraint 6 s[2,4] == 1 nt[2] == 2 end

We invent a new variable array `st`

to capture the number of sharks in each sector. This simply obtained by multiplying the binary array by the vector

@variable(m,st[i=1:3],Int) @constraint(m, st.==s*collect(1:7))

We rewrite our last constraint as

# Constraints 1 & 3 @NLconstraint(m, st[1]+st[2]+st[3]+n*(nt[1]+nt[2]+nt[3]) == 50)

After the model has been *solved*, we extract our output for the number of sharks.

sharks_in_each_sector=getvalue(st)

…and we get the correct output.

This problem might have been an overkill for using a full blown mixed integer non-linear optimizer. It can be solved by a simple table as shown in the video. However, we might not alway find ourselves in such a fortunate position. We could have also use mixed integer quadratic programming solver such as Gurobi which would be more efficient for that sort of problem. Given the small problem size, efficiency hardly matters here.