I decided to work through the CLRS Introduction to Algorithms text, and picked the printing neatly problem here.
I worked through the problem and came up with an imperative solution which was straightforward to implement in Python, but somewhat less so in Clojure.
I'm completely stumped on translating the compute-matrix function from my solution into idiomatic Clojure. Any suggestions? Here is the pseudocode for the compute-matrix function:
// 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
Additionally, I would very much appreciate feedback on the rest of my Clojure source, specifically with regards to how idiomatic it is. Have I managed to think sufficiently outside of the imperative paradigm for the Clojure code I've written thus far? Here it is:
(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: I've updated my matrix-construct function with the feedback given, so now it's actually one line shorter than my Python implementation.
;-----------------------------------------------------------------------------
; 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)))))
Instead of using let with fn's in it, try letfn.
doseq doseq -> looks like it would likely be better as a for comprehension
Your cond / zero? / = 1 code would be easier to read (and faster) with case.
My spidey-sense tells me that the loop/recurs here should be some kind of map call instead
I strongly suspect that this would be far faster with primitive arrays (and possibly cleaner in some places)
You might like to use or look at the source for Incanter
I climbed the wall and was able to think in a sufficiently Clojure-like way to translate the core compute-matrix algorithm into a workable program.
It's just one line longer than my Python implementation, although it appears to be more densely written. For sure, concepts like 'map' and 'reduce' are higher-level functions that require you to put your thinking cap on.
I believe this implementation also fixes a bug in my Python one. :)
;-----------------------------------------------------------------------------
; Compute all table entries so we can compute the optimal cost path and
; reconstruct an optimal solution.
;
(defn compute-matrix [m cost]
(letfn [; Return a function that computes 'cost(k+1,j) + c[i-1,k]'
; OR just 'c[i-1,k]' if we're on the last row.
(make-min-func [matrix i j]
(if (< j (- (count matrix) 1))
(fn [k]
(+ (cost (+ k 1) j) (get-in matrix [(- i 1) k])))
(fn [k]
(get-in matrix [(- i 1) k]))))
; Find the minimum cost for the new cost: 'cost(k+1,j)'
; added to the previous entry's cost: 'c[i-1,k]'
(min-cost [matrix i j]
(let [this-cost (make-min-func matrix i j)
rang (range (- i 1) (- j 1))
cnt (if (= rang ()) (list (- i 1)) rang)]
(apply min (map this-cost cnt))))
; Takes a matrix and indices, returns an updated matrix.
(combine [matrix indices]
(let [i (first indices)
j (nth indices 1)
opt (min-cost matrix i j)]
(assoc-in matrix [i j] opt)))]
(reduce combine m
(for [i (range 1 (count m)) j (range i (count m))] [i j]))))
Thank you Alex and Jake for your comments. They were both very helpful and have helped me on my way toward idiomatic Clojure.
My general approach to dynamic programs in Clojure is not to mess with construction of the matrix of values directly, but rather to use memorization in tandem with a fixed point combinator. Here's my example for computing edit distance:
(defn edit-distance-fp
"Computes the edit distance between two collections"
[fp coll1 coll2]
(cond
(and (empty? coll1) (empty? coll2)) 0
(empty? coll2) (count coll1)
(empty? coll1) (count coll2)
:else (let [x1 (first coll1)
xs (rest coll1)
y1 (first coll2)
ys (rest coll2)]
(min
(+ (fp fp xs ys) (if (= x1 y1) 0 1))
(inc (fp fp coll1 ys))
(inc (fp fp xs coll2))))))
The only difference from the naive recursive solution here is simply to replace the recursive calls with calls to fp.
And then I create a memoized fixed point with:
(defn memoize-recursive [f] (let [g (memoize f)] (partial g g)))
(defn mk-edit-distance [] (memoize-recursive edit-distance-fp))
And then call it with:
> (time ((mk-edit-distance)
"the quick brown fox jumped over the tawdry moon"
"quickly brown foxes moonjumped the tawdriness"))
"Elapsed time: 45.758 msecs"
23
I find memoization easier to wrap my brain around than mutating tables.
Your matrix-lookup and matrix-set functions can be simplified. You can use assoc-in and get-in for manipulating nested associative structures.
(defn matrix-lookup [matrix row col]
(get-in matrix [row col]))
(defn matrix-set [matrix row col value]
(assoc-in matrix [row col] value))
Alex Miller mentioned using primitive arrays. If you end up needing to go that direction you can start by looking at int-array, aset-int, and aget. Look at the clojure.core documentation to find out more.
Related
This is the code to calculate the failure function (how many steps we have to go back) in Scheme, when we use the Knuth-Morris-Pratt algorithm:
(define (compute-failure-function p)
(define n-p (string-length p))
(define sigma-table (make-vector n-p 0))
(let loop
((i-p 2)
(k 0))
(cond
((>= i-p n-p)
(vector-set! sigma-table (- n-p 1) k))
((eq? (string-ref p k)
(string-ref p (- i-p 1)))
(vector-set! sigma-table i-p (+ k 1))
(loop (+ i-p 1) (+ k 1)))
((> k 0)
(loop i-p (vector-ref sigma-table k)))
(else ; k=0
(vector-set! sigma-table i-p 0)
(loop (+ i-p 1) k))))
(vector-set! sigma-table 0 -1)
(lambda (q)
(vector-ref sigma-table q)))
But I do not understand the part when k > 0. Can someone explain it please?
I see you're confused with the syntax of a named let. This post does a good job explaining how it works, but perhaps an example with more familiar syntax will make things clearer. Take this code in Python, it adds all integers from 1 to 10:
sum = 0
n = 1
while n <= 10:
sum += n
n += 1
print(sum)
=> 55
Now let's try to write it in a recursive fashion, I'll call my function loop. This is completely equivalent:
def loop(n, sum):
if n > 10:
return sum
else:
return loop(n + 1, n + sum)
loop(1, 0)
=> 55
In the above example, the loop function implements an iteration, the parameter n is used to keep track of the current position, and the parameter sum accumulates the answer. Now let's write the exact same code, but in Scheme:
(let loop ((n 1) (sum 0))
(cond ((> n 10) sum)
(else (loop (+ n 1) (+ n sum)))))
=> 55
Now we've defined a local procedure called loop which is then automatically called with the initial values 1 and 0 for its parameters n and sum. When the base case of the recursion is reached, we return sum, otherwise we keep calling this procedure, passing updated values for the parameters. It's exactly the same as in the Python code! Don't be confused by the syntax.
In your algorithm, i-p and k are the iteration variables, which are initialized to 2 and 0 respectively. Depending on which condition is true, the iteration continues when we call loop again with updated values for i-p and k, or it ends when the case (>= i-p n-p) is reached, at this point the loop exits and the computed value is in the variable sigma-table. The procedure ends by returning a new function, referred to as the "failure function".
I am trying to solve the Maximum subarray problem on hacker rank. This is a standard DP problem and I write an O(n) solution:
(defn dp
[v]
(let [n (count v)]
(loop [i 1 f (v 0) best f]
(if (< i n)
(let [fi (max (v i) (+ f (v i)))]
(recur (inc i) fi (max fi best)))
best))))
(defn positive-only
[v]
(reduce + (filterv #(> % 0) v)))
(defn line->ints
[line]
(->>
(clojure.string/split line #" ")
(map #(Integer. %))
(into [])
))
(let [T (Integer. (read-line))]
(loop [test 0]
(when (< test T)
(let [_ (read-line)
x (read-line)
v (line->ints x)
a (dp-array v)
b (let [p (positive-only v)]
(if (= p 0) (reduce max v) p))]
(printf "%d %d\n" a b))
(recur (inc test)))))
To my surprise, I got time-limited-exceed for a large test case. I downloaded the input file, and found that the above version needs about 3 seconds to run.
I thought the bottleneck is in (v i) (getting the i-th element in vector v). So I changed the data structure from vector to an array:
(defn dp-array
[v0]
(let [v (into-array v0)
n (int (alength v))]
(loop [i 1
f (aget v 0)
best f]
(if (< i n)
(let [fi (max (aget v i) (+ f (aget v i)))]
(recur (inc i) fi (max fi best)))
best))))
This array version is even slower. On the same input, it costs 33 seconds, much slower than the vector version. I think the slowness is due to boxing and unboxing. I tried to add type hints, but encountered run-time errors. Could anyone help me improve dp-array function? Thanks!
Also, great appreciate if anyone knows how to improve the vector version.
UPDATE:
Finally I managed to get my clojure program accepted, not by optimizing over the dynamic programming function, but by changing (Integer. str) to (Integer/parseInt str). In this way, reflection is avoided in converting from string to integer.
I also replace into-array by int-array. But the speed of both versions are still on par with each other. I would expect the array version be faster than the vector version.
The Clojure compiler can't infer the type of v in the array version of the dp-array function whose argument v0 has unknown type. This causes costs to reflections when evaluating the following alength and aget. In order to avoid these unnecessary reflections, you have to replace into-array with long-array.
I wonder how this can be done in Clojure idiomatically and efficiently:
1) Given a vector containing n integers in it: [A0 A1 A2 A3 ... An]
2) Increase the last x items by 1 (let's say x is 100) so the vector will become: [A0 A1 A2 A3 ... (An-99 + 1) (An-98 + 1)... (An-1 + 1) (An + 1)]
One naive implementation looks like:
(defn inc-last [x nums]
(let [n (count nums)]
(map #(if (>= % (- n x)) (inc %2) %2)
(range n)
nums)))
(inc-last 2 [1 2 3 4])
;=> [1 2 4 5]
In this implementation, basically you just map the entire vector to another vector by examine each item to see if it needs to be increased.
However, this is an O(n) operation while I only want to change the last x items in the vector. Ideally, this should be done in O(x) instead of O(n).
I am considering using some functions like split-at/concat to implement it like below:
(defn inc-last [x nums]
(let [[nums1 nums2] (split-at x nums)]
(concat nums1 (map inc nums2))))
However, I am not sure if this implementation is O(n) or O(x). I am new to Clojure and not really sure what the time complexity will be for operations like concat/split-at on persistent data structures in Clojure.
So my questions are:
1) What the time complexity here in second implementation?
2) If it is still O(n), is there any idiomatic and efficient implementation that takes only O(x) in Clojure for solving this problem?
Any comment is appreciated. Thanks.
Update:
noisesmith's answer told me that split-at will convert the vector into a list, which was a fact I did not realised previously. Since I will do random access for the result (call nth after processing the vector), I would like to have an efficient solution (O(x) time) while keeping the vector instead of list otherwise nth will slow down my program as well.
Concat and split-at both turn the input into a seq, effectively a linked-list representation, O(x) time. Here is how to do it with a vector for O(n) performance.
user> (defn inc-last-n
[n x]
(let [count (count x)
update (fn [x i] (update-in x [i] inc))]
(reduce update x (range (- count n) count))))
#'user/inc-last-n
user> (inc-last-n 3 [0 1 2 3 4 5 6])
[0 1 2 3 5 6 7]
This will fail on input that is not associative (like seq / lazy-seq) because there is no O(1) access time in non-associative types.
inc-last is an implementation using a transient, which allows to get a modifiable "in place" vector in constant time and return a persistent! vector also in constant time, which allows to make the updates in O(x). The original implementation used an imperative doseq loop but, as mentioned in the comments, transient operations can return a new object, so it's better to keep doing things in a functional way.
I added a doall to the call to inc-last-2 since it returns a lazy seq, but inc-last and inc-last-3 returns a vector so the doall is needed to be able to compare them all.
According to some quick tests I made, inc-last and inc-last-3 don't actually differ much in performance, not even for huge vectors (10000000 elements). For the inc-last-2 implementation though, there's quite a difference even for a vector of 1000 elements, modifying only the last 10, it's ~100x slower. For smaller vectors or when the n is close to (count nums) the difference is not really that much.
(Thanks to Michał Marczyk for his useful comments)
(def x (vec (range 1000)))
(defn inc-last [n x]
(let [x (transient x)
l (count x)]
(->>
(range (- l n) l)
(reduce #(assoc! %1 %2 (inc (%1 %2))) x)
persistent!)))
(defn inc-last-2 [x nums]
(let [n (count nums)]
(map #(if (>= % (- n x)) (inc %2) %2)
(range n)
nums)))
(defn inc-last-3 [n x]
(let [l (count x)]
(reduce #(assoc %1 %2 (inc (%1 %2))) x (range (- l n) l))))
(time
(dotimes [i 100]
(inc-last 50 x)))
(time
(dotimes [i 100]
(doall (inc-last-2 10 x))))
(time
(dotimes [i 100]
(inc-last-3 50 x)))
;=> "Elapsed time: 49.7965 msecs"
;=> "Elapsed time: 1751.964501 msecs"
;=> "Elapsed time: 67.651 msecs"
I was wondering if someone could help me with the performance of this code snippet in Clojure 1.3. I am trying to implement a simple function that takes two vectors and does a sum of products.
So let's say the vectors are X (size 10,000 elements) and B (size 3 elements), and the sum of products are stored in a vector Y, mathematically it looks like this:
Y0 = B0*X2 + B1*X1 + B2*X0
Y1 = B0*X3 + B1*X2 + B2*X1
Y2 = B0*X4 + B1*X3 + B2*X2
and so on ...
For this example, the size of Y will end up being 9997, which corresponds to (10,000 - 3). I've set up the function to accept any size of X and B.
Here's the code: It basically takes (count b) elements at a time from X, reverses it, maps * onto B and sums the contents of the resulting sequence to produce an element of Y.
(defn filt [b-vec x-vec]
(loop [n 0 sig x-vec result []]
(if (= n (- (count x-vec) (count b-vec)))
result
(recur (inc n) (rest sig) (conj result (->> sig
(take (count b-vec))
(reverse)
(map * b-vec)
(apply +)))))))
Upon letting X be (vec (range 1 10001)) and B being [1 2 3], this function takes approximately 6 seconds to run. I was hoping someone could suggest improvements to the run time, whether it be algorithmic, or perhaps a language detail I might be abusing.
Thanks!
P.S. I have done (set! *warn-on-reflection* true) but don't get any reflection warning messages.
You are using count many times unnecessary. Below code calculate count one time only
(defn filt [b-vec x-vec]
(let [bc (count b-vec) xc (count x-vec)]
(loop [n 0 sig x-vec result []]
(if (= n (- xc bc))
result
(recur (inc n) (rest sig) (conj result (->> sig
(take bc)
(reverse)
(map * b-vec)
(apply +))))))))
(time (def b (filt [1 2 3] (range 10000))))
=> "Elapsed time: 50.892536 msecs"
If you really want top performance for this kind of calculation, you should use arrays rather than vectors. Arrays have a number of performance advantages:
They support O(1) indexed lookup and writes - marginally better than vectors which are O(log32 n)
They are mutable, so you don't need to construct new arrays all the time - you can just create a single array to serve as the output buffer
They are represented as Java arrays under the hood, so benefit from the various array optimisations built into the JVM
You can use primitive arrays (e.g. of Java doubles) which are much faster than if you use boxed number objects
Code would be something like:
(defn filt [^doubles b-arr
^doubles x-arr]
(let [bc (count b-arr)
xc (count x-arr)
rc (inc (- xc bc))
result ^doubles (double-array rc)]
(dotimes [i rc]
(dotimes [j bc]
(aset result i (+ (aget result i) (* (aget x-arr (+ i j)) (aget b-arr j))))))
result))
To follow on to Ankur's excellent answer, you can also avoid repeated calls to the reverse function, which gets us even a little more performance.
(defn filt [b-vec x-vec]
(let [bc (count b-vec) xc (count x-vec) bb-vec (reverse b-vec)]
(loop [n 0 sig x-vec result []]
(if (= n (- xc bc))
result
(recur (inc n) (rest sig) (conj result (->> sig
(take bc)
(map * bb-vec)
(apply +))))))))
I'm sorry for deleting the original question, here it is:
We have a bag or an array of n integers, we need to find the product of each of the (n-1) subsets. e.g:
S = {1, 0, 3, 6}
ps[1] = 0*3*6 = 0;
ps[2] = 1*3*6 = 18; etc.
After discussions, we need to take care of the three cases and they are illustrated in the following:
1. S is a set (contains one zero element)
for i=1 to n
if s[i]=0
sp[i] = s[1] * s[2] * ...* s[i-1] * s[i+1] *.....*s[n]
else
sp[i] = 0;
2. S is a bag (contains more than one zero element)
for i=1 to n
sp[i] = 0;
3. S is a set (contains no zero elements)
product = 1
for i=1 to n
product *= s[i];
for i=1 to n
sp[i] = product / s[i];
Thanks.
If the set is very large, it may be convenient to:
compute the product P of all the elements beforehand, and then
for each element x, obtain a (n-1) product as P/x
If the set contains zero (i.e. P=0, x=0), you must deal with it as a special case.
EDIT. Here is a solution in Scheme, taking into account andand's answer. I'm a complete beginner - can someone help me improve the following code (make it more efficient, more readable, more lisp-ish)? (Feel free to edit my answer.)
#!/usr/bin/env guile !#
(use-modules (ice-9 pretty-print))
(define (count-zeros l)
(cond ((null? l) 0)
((= 0 (car l)) (+ 1 (count-zeros (cdr l))))
(else (count-zeros (cdr l)))))
(define (non-zero-product l)
(define (non-zero-product-loop l product)
(cond ((null? l) product)
((= 0 (car l)) (non-zero-product-loop (cdr l) product))
(else (non-zero-product-loop (cdr l) (* (car l) product)))))
(non-zero-product-loop l 1))
(define (n-1-products l)
(let ((nzeros (count-zeros l)))
(cond ((> nzeros 1)
(map (lambda (x) 0) l))
((= 1 nzeros)
(map (lambda (x) (if (= 0 x) (non-zero-product l) 0)) l))
(else
(map (lambda (x) (/ (non-zero-product l) x)) l)))))
(pretty-print (n-1-products '(1 2 3 4 5)))
(pretty-print (n-1-products '(0 1 2 3 4)))
(pretty-print (n-1-products '(0 1 2 3 0)))
You need to explicitly consider the three cases:
1) No zeros: Precompute the product of all of the elements and divide out the desired set element from that product.
2) One zero: Precompute the product of the non-zero elements. The answer is always 0 except when you remove the single zero element, in which case it's the precomputed product.
3) More than one zero: The answer is always 0.
This assumes that you have a data type that can contain the products... that is, you need to be careful that your product doesn't exceed the maximum value of the type you're using to store it.
For the actual implementation, always precompute the product of the non-zero elements and keep track of how many zeros there are. If the "set" is dynamic (its values change) you'll need to update the product and the zero count. When asked for a particular subset product, consider the various cases and act accordingly.
Set product = 1;
for item in set:
if item index == argument index
ignore
else
product *= item
If I understand your question, this is the trivial solution. It should be easy to implement in any programming language.
You can solve this in O(N), using O(1) additional space (not counting the O(N) output array), without even using division. Here's the algorithm in Java.
static int[] products(int... nums) {
final int N = nums.length;
int[] prods = new int[N];
int pi = 1;
for (int i = 0; i < N; i++) {
prods[i] = pi;
pi *= nums[i];
}
int pj = 1;
for (int j = N-1; j >= 0; j--) {
prods[j] *= pj;
pj *= nums[j];
}
return prods;
}
//...
System.out.println(
Arrays.toString(products(1, 2, 3, 4, 5))
); // prints "[120, 60, 40, 30, 24]"
See also
Interview Q: given an array of numbers, return array of products of all other numbers (no division)
Assuming you can use Python:
You can use the combinations method from the itertools module to lazily generate the various subsets of the set in question. Once you have that, you can use reduce and operator.mul to generate the product of each one.