From e556b7afce94442cdeb67f98eff7896f176aa5fd Mon Sep 17 00:00:00 2001 From: fredrikr79 Date: Wed, 5 Nov 2025 11:28:37 +0100 Subject: [PATCH] ex5: report --- assignment5/report.md | 150 +++ assignment5/report.pdf | 2287 ++++++++++++++++++++++++++++++++++++++++ 2 files changed, 2437 insertions(+) create mode 100644 assignment5/report.md create mode 100644 assignment5/report.pdf diff --git a/assignment5/report.md b/assignment5/report.md new file mode 100644 index 0000000..93ad182 --- /dev/null +++ b/assignment5/report.md @@ -0,0 +1,150 @@ +--- +title: "assignment 5 - prolog" +date: 2025-11-05 +author: fredrik robertsen +margin: + x: 3.5cm + y: 4cm +fontsize: 14pt +mainfont: libertinus serif +lineheight: 1.5 +pagesize: a4 +--- + +## task 1 + +the following code is my solution: + +```prolog +payment(Sum, Coins) :- + payment_acc(0, Sum, Coins). + +payment_acc(Acc, Sum, []) :- + Acc #= Sum. + +payment_acc(Acc, Sum, [coin(Count, Value, Available)|Tail]) :- + Count in 0..Available, + NewAcc #= Acc + Count * Value, + payment_acc(NewAcc, Sum, Tail). +``` + +the predicate `payment/2` has two arguments (hence `/2`). the first, `Sum`, is +a target cost we are attempting to make from a collection of `Coins`, a list of +`coin/3` items, specifying the value of the coin type and how many of such coins +we have available. + +this is a classic constraint satisfaction problem which is solved elegantly in +prolog, even with my messy first-timer code above. + +we are essentially making a search in the state graph of our problem space. this +is what prolog does when it performs a combination of inference/deduction, DFS +and backtracking to find values that fit our variables, given the stated +predicates. + +in this program we create a helper function which `acc`umulates a sum accross +the recursion. this is similar to how we would solve it in oz, except we would +need to implement the search explicitly. prolog does a lot of heavy lifting in +that regard, since we are essentially only recursively creating a long string of +CNF logic statements that constrain our problem until a solution can be found. + +note that i am also using patternmatching to destructure the arguments of the +`payment_acc` predicate. + +## task 2 + +### subtask 1 + +this is my solution code: + +```prolog +% arity 4 predicate +plan(Cabin1, Cabin2, Path, TotalDistance) :- + plan(Cabin1, Cabin2, [Cabin1], Path, TotalDistance). + +% base case +plan(Cabin1, Cabin2, Visited, Path, Distance) :- + not(Cabin1 = Cabin2), + distance(Cabin1, Cabin2, Distance, 1), + append(Visited, [Cabin2], Path). + +% recursive case +plan(Cabin1, Cabin2, Visited, Path, TotalDistance) :- + not(Cabin1 = Cabin2), + distance(Cabin1, CabinX, Distance, 1), + \+ member(CabinX, Visited), + append(Visited, [CabinX], NewVisited), + plan(CabinX, Cabin2, NewVisited, Path, SubDistance), + TotalDistance is Distance + SubDistance. +``` + +as you can see, i learned that you can overload predicates with differing +arities, such that the previous `payment_acc` could've only been named +`payment`. we can also do multiple definitions to clearly state the different +branches of a recursive algorithm, such as the base case and the recursive step. + +we can then use a `plan/5` auxiliary function to carry a log of what cabins we +have already `Visited`. thus, our base case becomes the case where we have +a direct connection between the first and last cabin, such that we can easily +read the distance from the predicate. then just make sure to mark `Cabin2` as +visited. + +in the recursive step we assume there is a `CabinX` that lies between our start +and end cabins. this cabin cannot have been visited previously, lest we enter an +infinite cycle -- `\+ member(CabinX, Visited)`. we can then visit `CabinX` and +continue our search recursively from there -- `plan(CabinX, Cabin2, ...)`. +lastly, we can calculate the `TotalDistance` as the sum of the total distance from `CabinX` +to `Cabin2` and the total distance from `Cabin1` to `CabinX`. + +``` +Cabin1 --> CabinX --> Cabin2 + L Distance J L SubDistance J + L TotalDistance J +``` + +### subtask 2 + +my initial solution was: + +```prolog +bestplan(Cabin1, Cabin2, ShortestPath, ShortestDistance) :- + findall([Path, Distance], (plan(Cabin1, Cabin2, Path, Distance)), Solutions), + shortestpath(Solutions, [ShortestPath, ShortestDistance]). + +% takes a list of [Path, Distance] pairs +shortestpath([[Path, Distance]|[]], Solution) :- Solution = [Path, Distance]. +shortestpath([[Path, Distance]|Tail], [ShortestPath, ShortestDistance]) :- + shortestpath(Tail, [TailPath, TailDistance]), + (Distance < TailDistance -> + (ShortestPath = Path, ShortestDistance = Distance) + ; + (ShortestPath = TailPath, ShortestDistance = TailDistance) + ). +``` + +but i cleaned it up to this: + +```prolog +bestplan(Cabin1, Cabin2, ShortestPath, ShortestDistance) :- + findall([Path, Distance], (plan(Cabin1, Cabin2, Path, Distance)), Solutions), + sort(2, @=<, Solutions, [ShortestSolution|_]), + [ShortestPath, ShortestDistance] = ShortestSolution. +``` + +the main idea is a bit naive: we are just picking out the shortest path from +_all_ the paths we find using the `plan` predicate from the last subtask. it +sounds like it wouldn't be particularly performant, but i think prolog does +a lot of work such that it isn't too terrible. + +anyway, i initially just expanded all the paths found by `plan`, then used +a homemade `shortestpath/2` predicate that works on a list of `[Path, Distance]` +pairs to tail recurse and find the shortest such pair, keeping a running +shortest distance result. this is similar to oz, taking the head of the list, +checking if it is smaller than the current accumulated value, then carry on the +smaller value of the two. the base case is when we only have a single path and +distance: we return that as a solution. + +but this entire process can be shortened to simply sorting the solutions based +on the distance and then picking the first one of that list. the code above +performs such a sort and only takes the first solution in the sorted list, +sorting from low to high based on the key `2`, such that we are sorting the +distance. diff --git a/assignment5/report.pdf b/assignment5/report.pdf new file mode 100644 index 0000000..2f099f3 --- /dev/null +++ b/assignment5/report.pdf @@ -0,0 +1,2287 @@ +%PDF-1.7 +% + +1 0 obj +<< + /Type /Pages + /Count 5 + /Kids [154 0 R 156 0 R 158 0 R 160 0 R 162 0 R] +>> +endobj + +2 0 obj +<< + /Type /Outlines + /First 3 0 R + /Last 4 0 R + /Count 2 +>> +endobj + +3 0 obj +<< + /Parent 2 0 R + /Next 4 0 R + /Title (task 1) + /Dest 150 0 R +>> +endobj + +4 0 obj +<< + /Parent 2 0 R + /Prev 3 0 R + /First 5 0 R + /Last 6 0 R + /Count -2 + /Title (task 2) + /Dest 153 0 R +>> +endobj + +5 0 obj +<< + /Parent 4 0 R + /Next 6 0 R + /Title (subtask 1) + /Dest 151 0 R +>> +endobj + +6 0 obj +<< + /Parent 4 0 R + /Prev 5 0 R + /Title (subtask 2) + /Dest 152 0 R +>> +endobj + +7 0 obj +<< + 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