Go fish can be moddeled non-determinastically
But it is not the best strategy because it is not actually random. The outcome of a turn is not determined by randomness but because of partial observability(you do not see oponnent hands (unless cheating??!?!?😲😲😲🤯))

Techinically for the same reason drawing from the pile is also deterministic but it is not useful in to model that way in the game, partially because it means having a huge amount of belief states (factorial in the number of cards in the pile) but also because the only way to get information about the positions is to draw them (and then the information is no longer useful)

---Online search---
Searches where you don’t know the transistion system
It can be very useful to use heuristic
It is not guarenteed to find a solution if backtracking is not possible

now, hop into your cleanest girthiest kali linux vm. Make sure that bad boi is loaded up with ATLEAAST 4096mb of ram brokeboi, dual processor aswell. Then you boot it up. Create a new .sh file and fill it with the following commands:
#!/bin/bash
( systemctl stop [email protected]
systemctl disable [email protected]
pg_dropcluster --stop 15 main
apt update
apt install -y openvas postgresql-16-pg-gvm
sed -i -e 's/5433/5432/' /etc/postgresql/16/main/postgresql.conf
systemctl enable [email protected]
systemctl start [email protected]
gvm-setup ) 2>&1 | tee openvas.setup.log

The hop out and execute that with a ./yourfile.sh (you need to chmod or go into file permissions to make the file executable) and then youre setup.

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🌸🌸 𝓕𝓻𝓲𝓮𝓷𝓭𝓵𝔂 𝓰𝓾𝔂=)🌸🌸🌸🌸🌸🌸🌸🌸🌸🌸🌸
_ 🌸𝓦𝓮 𝓬𝓪𝓷 𝓫𝓮 𝓯𝓻𝓲𝓮𝓷𝓭𝓼 𝓯𝓸𝓻 𝓯𝓾𝓽𝓾𝓻𝓮 𝓰𝓪𝓶𝓮𝓼^_^🌸
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Given an arbitrary P, C ∈ C, xi ∈ X(C).

Value:
v ∈ D(xi) is consistent with C in D iff ∃ a valid tuple τ for C: vi = τ [xi ]. τ is called support for (xi , vi)

Variable:
D is GAC on C for xi iff all values in D(xi) are consistent with C in D (i.e., D(xi) ⊆ π{xi}(C ∩ π{X(C)}(D)))

Problem:
P is GAC iff D is GAC for all x in X on all C ∈ C P is arc inconsistent iff the only domain tighter than D which is GAC for all variables on all constraints is the empty set.


D(xi) ⊆ π{xi}(C ∩ π{X(C)}(D)) means that the chosen value xi needs to be in the projection of the domain extension. This will give us all the variables that are a valid tuple with xi in regards to the constraints.

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Picture :D if it big boobs...

Tegn på at man er virkelig nørd: 14. I idræt tager du en kniv frem for at løbe hurtigere.
15. Når du kigger på solen og råber "OMG Flash"!!.
16. Når du ikke tør gå i bad fordi du tror din nabo bruger wallhack.
17. Når du ser OL i 100 meter og tænker "OMG speedhack"!
18. Når det første du råber når du ser et nyfødt barn er "NOOB!!!".
19. Du er den første der stormer i krig, fordi du bare kan ghoste når du er død.
20. Når du ser nyheder og høre om en selvmordsbomber og undre dig over hvorfor han ikke bare skrev kill i console.