Add notes and assignment solution

7 files changed, 257 insertions, 8 deletions

diff --git a/sem7/pp/larger.hs b/sem7/pp/larger.hs
new file mode 100644
index 0000000..5618faf
--- /dev/null
+++ b/sem7/pp/larger.hs
@@ -0,0 +1,29 @@
+import Data.Maybe
+
+data LTree = LLeaf String | LNode LTree LTree
+data LTreeRoot = LTreeRoot LTree Float
+
+treemerge :: LTreeRoot -> LTreeRoot -> LTreeRoot
+treemerge (LTreeRoot ta la) (LTreeRoot tb lb) = LTreeRoot (LNode ta tb) (la + lb)
+
+data Bit = Zero | One
+
+instance Show Bit where
+  show Zero = "0"
+  show One = "1"
+
+createBitList :: String -> Maybe [Bit]
+createBitList [] = Just []
+createBitList (c:rest) | c == '0' = createBitList rest >>= Just . (:) Zero
+                       | c == '1' = createBitList rest >>= Just . (:) One
+                       | otherwise = Nothing
+
+maketable :: LTree -> [([Bit], String)]
+maketable = recur []
+            where recur prefix (LLeaf val) = [(reverse prefix, val)]
+                  recur prefix (LNode left right) =
+                    recur (Zero : prefix) left ++
+                    recur (One : prefix) right
+
+
+testTree = LNode (LLeaf "Hej") (LNode (LLeaf "Med") (LLeaf "Dig"))
diff --git a/sem7/pp/lec6.hs b/sem7/pp/lec6.hs
index 4598d7a..31f666c 100644
--- a/sem7/pp/lec6.hs
+++ b/sem7/pp/lec6.hs
@@ -94,11 +94,3 @@ filterWithFold func = foldr (\new pre -> if func new
                                             else pre
                             ) []
 
-
--- Problems part of larger thing
-data LTreeMid = LLeaf Char | LNode LTreeMid LTreeMid
-data LTree = LTree LTreeMid Float
-
-treemerge :: LTree -> LTree -> LTree
-treemerge (LTree ta la) (LTree tb lb) = LTree (LNode ta tb) (la + lb)
-
diff --git a/sem7/pp/prolog/lec1.dl b/sem7/pp/prolog/lec1.dl
new file mode 100644
index 0000000..84154c2
--- /dev/null
+++ b/sem7/pp/prolog/lec1.dl
@@ -0,0 +1,33 @@
+#lang datalog
+
+% It finds whether a parent is a father or a mother depending on the gender of the parent.
+
+
+
+% Opgave 2
+
+% In the first program, a person is only happy if he/she is both rich and famous.
+% In the second program, a person is happy if he/she is either rich or famous.
+
+
+% Opgave 3
+
+% There is one program containing 3 clauses.
+% The first two clauses are called facts each only containing a single atom.
+% The first two facts each contain a atom with the single term beyonce which is a constant
+
+% Last line is a clause with a head and a body containing two atoms.
+% The body uses the predicates created in the last two lines, while the clause introduces the new predicate happy.
+% The clause uses the variable Person, which is used once in all 3 atoms.
+
+
+% Opgave 4
+
+destinct(red, green).
+destinct(green, blue).
+destinct(red, blue).
+
+% It is symmetric
+destinct(X, Y) :- destinct(Y, X).
+
+colouring(X, Y, XC, YC) :- neighbour(X, Y), destinct(XC, YC).
diff --git a/sem7/pp/prolog/lec3.pl b/sem7/pp/prolog/lec3.pl
new file mode 100644
index 0000000..d434134
--- /dev/null
+++ b/sem7/pp/prolog/lec3.pl
@@ -0,0 +1,110 @@
+
+/* Opgave 1
+
+{
+    loves(rose, jack).
+    loves(jack, rose).
+    loves(caledon, rose).
+    happy(rose).
+    happy(jack).
+}
+{
+    loves(rose, jack).
+    loves(caledon, rose).
+}
+{
+    loves(jack, rose).
+    loves(caledon, rose).
+}
+{
+    loves(rose, jack).
+    loves(jack, rose).
+}
+{
+    loves(rose, jack).
+    loves(jack, rose).
+    loves(caledon, rose).
+    happy(jack).
+}
+{
+    loves(rose, jack).
+    loves(jack, rose).
+    loves(caledon, rose).
+    happy(rose).
+}
+{
+    loves(rose, jack).
+    loves(jack, rose).
+    loves(caledon, rose).
+}
+{
+    loves(rose, jack).
+    loves(jack, rose).
+    happy(jack).
+}
+{
+    loves(rose, jack).
+    loves(jack, rose).
+    happy(rose).
+}
+{
+    loves(rose, jack).
+    loves(jack, rose).
+    happy(rose).
+    happy(jack)
+}
+
+Well okay i feel stupid
+
+We say that the universe U_p = {rose, jack, caledon}.
+We will then way that the base is:
+
+U_b = { loves(x, y) | x, y \in U_p } \cup { happy(x) | x \in U_p }
+
+Then all the interpretations are.
+
+I = { S | S \subseteq U_b }
+
+*/
+
+
+/* Opgave 2
+
+loves(rose, jack).
+happy(rose)
+
+happy(rose) <= loves(rose, jack),loves(jack,rose)
+We know rose is happy, we do not need to check the predicates.
+For some reason, kind of TODO.
+
+*/
+
+/* Opgave 3
+
+I_4 og I_5 er modeller for P, hvor I_4 lige har en extra happy(caledon). 
+
+Her er I_4 minimal fordi ingen anden model for P er mindre.
+
+*/
+
+/* Opgave 4
+
+M_1 = T_P(Ø) = {god(odin),son(odin,thor),son(odin,baldr),son(thor,mothi),son(thor,magni)}
+M_2 = T_P(M_1) = M_1 \cup {demigod(thor),demigod(baldr)}
+M_3 = T_P(M_2) = M_2 \cup {mortal(mothi),mortal(magni)}
+
+*/
+
+/* Opgave 5
+
+B -+-> A
+D -+-> C
+B -+-> A
+C ---> A
+C -+-> B
+D ---> B
+
+D, C -> A, B
+D -> C -> B -> A
+
+Det er stratifyable.
diff --git a/sem7/pp/prolog/lec4.pl b/sem7/pp/prolog/lec4.pl
new file mode 100644
index 0000000..36d3880
--- /dev/null
+++ b/sem7/pp/prolog/lec4.pl
@@ -0,0 +1,50 @@
+
+last([X], X).
+last([_|XS], X) :- last(XS, X).
+
+/*
+last([1,2,3], X) # Bruger regel 2
+last([2,3], Y) # Bruger regel 2
+last([3], Z) # Bruger regel 1, så Z = 3
+# Så Y = Z = 3
+# Så X = Y = Z = 3
+
+
+Okay det er også bare fint at skrive
+last([1,2,3],X)
+     |
+last([2,3],X)
+     |
+last([3], X)
+*/
+
+attach([], E, [E]).
+attach([X|XS], E, [X|R]) :- attach(XS, E, R).
+
+
+/*
+Proof search of attach
+
+attach([1,2], a, L)
+        |      X = 1, XS = [2], E = a, L = [1|L1]
+attach([2], a, L1)
+        |      X1 = 2, XS1 = [], E1 = a, L1 = [2|L2]
+attach([], a, L2)
+        |      E2 = a, L2 = [a]
+*/
+
+nat(zero).
+nat(succ(X)) :- nat(X).
+
+after(succ(X), X).
+
+leq(zero, Y) :- nat(Y).
+leq(succ(X), succ(Y)) :- leq(X, Y), nat(X), nat(Y).
+
+add(X, zero, X) :- nat(X).
+add(X, succ(Y), succ(R)) :- add(X, Y, R), nat(X), nat(Y), nat(R).
+
+sub(X, Y, R) :- add(R, Y, X).
+
+min(X, Y, X) :- leq(X, Y).
+min(succ(X), Y, Y) :- leq(Y, X).
diff --git a/sem7/pp/prolog/notes_lec2.md b/sem7/pp/prolog/notes_lec2.md
new file mode 100644
index 0000000..49511e8
--- /dev/null
+++ b/sem7/pp/prolog/notes_lec2.md
@@ -0,0 +1,22 @@
+# Logic Programming
+
+## Herbrand
+
+The *herbrand universe* denotes all the constants in the program.
+Thus all the things we can talk about.
+
+While the *herbrand base* is the set of possible facts.
+So all the things that one can build using predicates and constants.
+So any model is the subset of the *herbrand base*.
+Just because something is in the *herbrand base* does not mean that it holds.
+
+The *herbrand universe* is what we can talk about, while the *herbrand base* is what we can say about it.
+
+## Minimal Model
+
+A minimal model for some program P, does not contain a smaller Herbrand model for P.
+
+Compute finite model with *least fixed point*, which calculates all things that are true.
+Has something to do with *minimal model*, probably the saaaaaame??. TODO
+
+
diff --git a/sem7/pp/prolog/shell.nix b/sem7/pp/prolog/shell.nix
new file mode 100644
index 0000000..45b9e32
--- /dev/null
+++ b/sem7/pp/prolog/shell.nix
@@ -0,0 +1,13 @@
+{ pkgs ? import <nixpkgs> {} }:
+let
+    pkgs_old = import (builtins.fetchTarball {
+        name = "nixpkgs-racket";
+        url = "https://github.com/nixos/nixpkgs/archive/1bf0327ef6d7e0959f3250758b4d33b4066a732b.tar.gz";
+        sha256 = "1pz4xaimpb1y8xmqz9c8a2g1nsr77jc7nxi6m8v4ph8q1r3c7pz9";
+    }) {};
+in
+pkgs.mkShell {
+    buildInputs = with pkgs; [
+        pkgs_old.racket swiProlog
+    ];
+}