Decidí trabajar a través del texto Introducción a Algoritmos de CLRS, y elegí claramente el problema de impresión here.Idiomático Clojure para resolver el algoritmo de programación dinámica
Trabajé a través del problema y se me ocurrió una solución imperativa que fue fácil de implementar en Python, pero algo menos en Clojure.
Estoy completamente perplejo al traducir la función de matriz de cálculo de mi solución a Clojure idiomática. ¿Alguna sugerencia? Aquí está el pseudocódigo para la función de cálculo de matriz:
// n is the dimension of the square matrix.
// c is the matrix.
function compute-matrix(c, n):
// Traverse through the left-lower triangular matrix and calculate values.
for i=2 to n:
for j=i to n:
// This is our minimum value sentinal.
// If we encounter a value lower than this, then we store the new
// lowest value.
optimal-cost = INF
// Index in previous column representing the row we want to point to.
// Whenever we update 't' with a new lowest value, we need to change
// 'row' to point to the row we're getting that value from.
row = 0
// This iterates through each entry in the previous column.
// Note: we have a lower triangular matrix, meaning data only
// exists in the left-lower half.
// We are on column 'i', but because we're in a left-lower triangular
// matrix, data doesn't start until row (i-1).
//
// Similarly, we go to (j-1) because we can't choose a configuration
// where the previous column ended on a word who's index is larger
// than the word index this column starts on - the case which occurs
// when we go for k=(i-1) to greater than (j-1)
for k=(i-1) to (j-1):
// When 'j' is equal to 'n', we are at the last cell and we
// don't care how much whitespace we have. Just take the total
// from the previous cell.
// Note: if 'j' < 'n', then compute normally.
if (j < n):
z = cost(k + 1, j) + c[i-1, k]
else:
z = c[i-1, k]
if z < optimal-cost:
row = k
optimal-cost = z
c[i,j] = optimal-cost
c[i,j].row = row
Además, me gustaría mucho recibir sus comentarios como en el resto de mi fuente Clojure, específicamente en lo que respecta a la forma idiomática que es. ¿He logrado pensar lo suficiente fuera del paradigma imperativo para el código de Clojure que he escrito hasta ahora? Aquí está:
(ns print-neatly)
;-----------------------------------------------------------------------------
; High-order function which returns a function that computes the cost
; for i and j where i is the starting word index and j is the ending word
; index for the word list "word-list."
;
(defn make-cost [word-list max-length]
(fn [i j]
(let [total (reduce + (map #(count %1) (subvec word-list i j)))
result (- max-length (+ (- j i) total))]
(if (< result 0)
nil
(* result result result)))))
;-----------------------------------------------------------------------------
; initialization function for nxn matrix
;
(defn matrix-construct [n cost-func]
(let [; Prepend nil to our collection.
append-empty
(fn [v]
(cons nil v))
; Like append-empty; append cost-func for first column.
append-cost
(fn [v, index]
(cons (cost-func 0 index) v))
; Define an internal helper which calls append-empty N times to create
; a new vector consisting of N nil values.
; ie., [nil[0] nil[1] nil[2] ... nil[N]]
construct-empty-vec
(fn [n]
(loop [cnt n coll()]
(if (neg? cnt)
(vec coll)
(recur (dec cnt) (append-empty coll)))))
; Construct the base level where each entry is the basic cost function
; calculated for the base level. (ie., starting and ending at the
; same word)
construct-base
(fn [n]
(loop [cnt n coll()]
(if (neg? cnt)
(vec coll)
(recur (dec cnt) (append-cost coll cnt)))))]
; The main matrix-construct logic, which just creates a new Nx1 vector
; via construct-empty-vec, then prepends that to coll.
; We end up with a vector of N entries where each entry is a Nx1 vector.
(loop [cnt n coll()]
(cond
(zero? cnt) (vec coll)
(= cnt 1) (recur (dec cnt) (cons (construct-base n) coll))
:else (recur (dec cnt) (cons (construct-empty-vec n) coll))))))
;-----------------------------------------------------------------------------
; Return the value at a given index in a matrix.
;
(defn matrix-lookup [matrix row col]
(nth (nth matrix row) col))
;-----------------------------------------------------------------------------
; Return a new matrix M with M[row,col] = value
; but otherwise M[i,j] = matrix[i,j]
;
(defn matrix-set [matrix row col value]
(let [my-row (nth matrix row)
my-cel (assoc my-row col value)]
(assoc matrix row my-cel)))
;-----------------------------------------------------------------------------
; Print the matrix out in a nicely formatted fashion.
;
(defn matrix-print [matrix]
(doseq [j (range (count matrix))]
(doseq [i (range (count matrix))]
(let [el (nth (nth matrix i) j)]
(print (format "%1$8.8s" el)))) ; 1st item max 8 and min 8 chars
(println)))
;-----------------------------------------------------------------------------
; Main
;-----------------------------------------------------------------------------
;-----------------------------------------------------------------------------
; Grab all arguments from the command line.
;
(let [line-length (Integer. (first *command-line-args*))
words (vec (rest *command-line-args*))
cost (make-cost words line-length)
matrix (matrix-construct (count words) cost)]
(matrix-print matrix))
EDIT: He actualizado mi función matricial-construcción con la información dada, por lo que ahora es en realidad una línea más corta que mi implementación de Python.
;-----------------------------------------------------------------------------
; Initialization function for nxn matrix
;
(defn matrix-construct [n cost-func]
(letfn [; Build an n-length vector of nil
(construct-empty-vec [n]
(vec (repeat n nil)))
; Short-cut so we can use 'map' to apply the cost-func to each
; element in a range.
(my-cost [j]
(cost-func 0 j))
; Construct the base level where each entry is the basic cost function
; calculated for the base level. (ie., starting and ending at the
; same word)
(construct-base-vec [n]
(vec (map my-cost (range n))))]
; The main matrix-construct logic, which just creates a new Nx1 vector
; via construct-empty-vec, then prepends that to coll.
; We end up with a vector of N entries where each entry is a Nx1 vector.
(let [m (repeat (- n 1) (construct-empty-vec n))]
(vec (cons (construct-base-vec n) m)))))
1. Gran sugerencia. No tenía idea de que había algo como letfn. 2. Luché para comprender la construcción de la matriz usando doseq. Con suerte, hay más sugerencias para ejemplos prácticos. :) 4. Había usado algunos mapas, pero terminé usando '_' para ignorar algunos valores y pensé que era 'impuro', por falta de una palabra mejor. Sin embargo, podría ser más idiomático, como lo he visto en algunos ejemplos recientemente. 5. ¿Qué quiere decir con 'matrices primitivas'? ¡Gracias por los comentarios! (respuesta abreviada para encajar en las restricciones de comentarios) – GrooveStomp