#import "@preview/cetz:0.3.2";
#import "@preview/cetz-plot:0.1.1": plot
#import "@preview/physica:0.9.4": *
#import "@preview/plotsy-3d:0.1.0": plot-3d-parametric-surface
#import "@preview/fletcher:0.5.4" as fletcher: diagram, edge, node

#set page(paper: "a4", margin: (x: 2.6cm, y: 2.8cm), numbering: "1 : 1")
#set par(justify: true, leading: 0.52em)

#let FONT_SIZE = 18pt;
#set text(font: "FreeSerif", size: FONT_SIZE, lang: "us")
#show math.equation: set text(font: "Euler Math", size: (FONT_SIZE * 1.0), lang: "en")

#set heading(numbering: none)
#show heading.where(level: 1): it => {
  rect(inset: FONT_SIZE / 2)[#it]
}

#align(center)[
  #text(size: FONT_SIZE * 2, weight: "bold")[#underline[assignment 1]]
]

these are our solutions to the first assignment of TDT4136.

this document was created using
#link("https://typst.app/")[#text(blue.darken(5%))[typst]].

#v(42pt)

#outline(title: none)

#v(42pt)

this is the collaborative effort of group 192, Erlend Ulvund Skaarberg and Fredrik Robertsen

= part 1

== 1)

these are the costs

#table(
  ` `, `A`, `B`, `C`, `D`, `E`, `F`, `G`, `H`, `I`,
  `A`, ` `, ` `, ` `, ` `, ` `, ` `, `4`, ` `, `1`,
  `B`, ` `, ` `, `4`, ` `, ` `, ` `, ` `, `3`, `2`,
  `C`, ` `, `4`, ` `, ` `, `1`, ` `, ` `, ` `, ` `,
  `D`, ` `, ` `, ` `, ` `, `2`, `1`, ` `, ` `, ` `,
  `E`, ` `, ` `, `1`, `2`, ` `, ` `, ` `, `1`, ` `,
  `F`, ` `, ` `, ` `, `1`, ` `, ` `, ` `, ` `, ` `,
  `G`, `4`, ` `, ` `, ` `, ` `, ` `, ` `, ` `, `1`,
  `H`, ` `, `3`, ` `, ` `, `1`, ` `, ` `, ` `, ` `,
  `I`, `1`, `2`, ` `, ` `, ` `, ` `, `1`, ` `, ` `,
  columns: 10,
  inset: 10pt,
)

== 2)

=== a) BFS

- order of expansion: $A, G, I, B, C, H, E, D$
- found path: $A, I, B, C, E, D, F$
- path cost: $1 + 2 + 4 + 1 + 2 + 1 = 11$

=== b) DFS

- order of expansion: $A, G, I, B, C, E, D, F$
- found path: $A, G, I, B, C, E, D, F$
- path cost: $4 + 1 + 2 + 4 + 1 + 2 + 1 = 15$

=== c) UCS

- order of expansion: A, I, B, H, C, E, D
- found path: A, I, B, H, C, E, D, F
- path cost: $1 + 2 + 3 + 1 + 1 + 2 + 1 = 11$

= part 2

== 3)

=== a) greedy best first search

- order of expansion: $A, I, B, C, E, D$
- found path: $A, I, B, C, E, D, F$
- path cost: $1 + 2 + 4 + 1 + 2 + 1 = 11$

=== b) A#upper("*")

- order of expansion: $A, I, B, C, E, D$
- found path: $A, I, B, C, E, D, F$
- path cost: $1 + 2 + 4 + 1 + 2 + 1 = 11$

= part 3

== 4)

admissibility is needed for $A^*$ to be optimal.

our heuristics are inadmissible, as they overestimate the cost of the optimal
path from each node to the goal. if we look at the graph, the heuristics may
encode a sense of distance between each node accurately, however the problem
states that the cost of an action is the same between any two neighboring nodes
(unless there is a lift or a staircase present). thus, the heuristic estimates
are not very optimistic.

== 5)

the heuristic $h(E) = 5$ is an overestimation, as the shortest path to the goal
$F$ is trivially seen to be $3$. this is a counter-example to the statement that
the heuristic function $h(n)$ is admissible, as admissibility is defined as
$
  h(n) <= C^*(n) quad forall quad n in V
$
where $C^*(n)$ is the cost of the optimal path from the node $n$.

== 6)

- order of expansion: $A, I, B, H, E, D$
- found path: $A, I, B, H, E, D, F$
- path cost: $1 + 2 + 3 + 1 + 2 + 1 = 10$

== 7)

the new heuristic function $h'(n)$ is admissible, because it does not
overestimate the cost of getting from any node to the goal. thus we can see that
when performing $A^*$ in 6) we find the cost-optimal path from $A$ to $F$.

suppose that there is a node $n in V$ such that
$
  h'(n) > C^*(n).
$
if such a node exists, then $h'(n)$ is inadmissible. since no such node exists,
$h'(n)$ is admissible.

admissibility guarantees that $A^*$ finds the cost-optimal path, however it says
nothing about optimal efficiency, like consistency (see 8)).

== 8)

consistency of a heuristic $h(n)$ is defined as a triangle inequality
$
  h(n) <= c(n, a, n') + h(n')
$
where $c(n, a, n')$ denotes the cost of the path-action $a$ from $n$ to $n'$.

to see whether the heuristic $h'(n)$ is consistent, we must investigate each child
node of each node on the cost-optimal path to see if this identity holds.

- from $A$, it is not cheaper to get to $I$ through $G$, thus the identity holds.
- from $B$, it is not cheaper to get to $H$ through $C$, thus the identity holds.
- from $H$, it is not cheaper to get to $E$ through $C$, thus the identity holds.

these are all the nodes on the cost-optimal path with multiple children where
inconsistency could occur, but these cases are consistent, thus our heuristic
$h'(n)$ is consistent.

= bonus task

== 9)

if we let $f(n) = g(n) + h(n)$ in `Best-First-Search(problem, f)` with
a heuristic $h(n)$ that is inadmissible for at least one node on the
cost-optimal path, unless it is consistent.

however, we can also change the code implementation of $A^*$ to require
consistency of $f(n)$:

#[#set text(size: FONT_SIZE * 0.6)
  ```pseudocode
  function BEST-FIRST-SEARCH(problem,f) returns a solution node or failure
    node←NODE(STATE=problem.INITIAL)
    frontier←a priority queue ordered by f, with node as an element
    reached←a lookup table, with one entry with key problem.INITIAL and value node
    while not IS-EMPTY(frontier) do
      node←POP(frontier)
      if problem.IS-GOAL(node.STATE) then return node
      for each child in EXPAND(problem, node) do
        s←child.STATE
        if s is not in reached or child.PATH-COST < reached[s].PATH-COST then
          reached[s]←child
          add child to frontier
          // ------------
          // calculate all parts of triangle inequality
          h←f(n) - node.PATH-COST
          h'←f(n) - child.PATH-COST
          c←child.PATH-COST - node.PATH-COST
          if h > c + h' then return failure  // fail on inconsistency
          // ------------
    return failure
  ```
]