SPOJ 8545. Subset Sum (Main72) with Dynamic Programming and F#

The Subset Sum (Main72) problem, officially published in SPOJ, is about computing the sum of all integers that can be obtained from the summations over any subset of the given set (of integers). A naïve solution would be to derive all the subsets of the given set, which unfortunately would result in time complexity, given that is the number of elements in the set.

This post outlines a more efficient (pseudo-polynomial) solution to this problem using Dynamic Programming and F#. Additionally, we post C# code of the solution.

Note that we have solved a similar problem in Party Schedule (PARTY) with F# blog-post.

Interpretation¶

This problem provides a set of integers , and specifies the following constraints–

The size of the given set, i.e.,

, where the value of

is bounded by:

, the following condition holds:

Given this input, we would like to find all the integers: and is the sum of the items of any subset over . Afterward, we sum all these integers, and return it as the result to the problem instance.

In essence, we reduce this problem as follows: Given , can we express it using any subset over ? If yes, we include it in the solution set for summation. Interestingly, we realize that the stated problem is a special case of a more general problem called Subset Sum, given that the sum is .

Algorithm¶

What would be the maximum possible value for ? Indeed, is not practical at all, as can be bounded by the following upper limit: , i.e., the summation of all the items in . This observation effectively reduces the search space to , for a given .

It implies that a naïve algorithm would require to iterate all the subsets over and verify whether their sum is within . Recall that, due to its exponential time complexity, it is quite impractical .

Using dynamic programming technique, a pseudo-polynomial algorithm can be derived, as the problem has an inherent optimal substructure property. That is, a solution of an instance of the problem can be expressed as the solutions of its subproblems, as described next.

We define as the function that determines whether the summation over any subset can result in the integer . So, it yields if sum can be derived over any subset, otherwise, . Also, note that, and .

To define the recurrence, we describe in terms of its smaller subproblems as follows.

In Eq. (1), the first case refers to the fact that is larger than . Consequently, can not be included in the subset to derive . Then, the case 2 of Eq. (1) expresses the problem into two subproblems as follows: we can either ignore though , or we can include it. Using any case stated in Eq. (1), if we can derive i.e. = true, we can include it in the solution set.

As we can see overlapping subproblems, we realize that we can effectively solve them using a bottom-up dynamic programming approach. What about the base cases?

Using a table– dp, we can implement the stated algorithm as follows.

	let computeSubsetSum (set:int array, n:int, sum:int):int =
	// dp[i,j] = True, if subset of [0..i-1] sum
	// equals to :
	// j-set[i] => ith item is included
	// or
	// j => ith item is not included
	let dp:bool[,] = Array2D.zeroCreate (n+1) (sum+1)

	// This dp tries answer folloowing question, given a sum
	// j, whether we can derive it by summing any subset of
	// [0..i]. It computes this answer in bottom up manner,
	// hence, if starting from (i,j) if it can reach (i,0), the
	// answer is yes. Otherwise, if sum<>0 but i=0, then answer is
	// no due to the fact that, using any subsets, the sum cannot be
	// computed.

	// Therefore.
	// base case 1 : given sum = 0, answer is true
	// sum is zero with the provided set of items.
	// Hence, storing it as 0
	for i = 0 to n do
	dp.[i,0] <- true

	// base case 2 : sum <> 0 but OPT = empty -->
	// answer is false
	for j=1 to sum do
	dp.[0,j] <- false

	for i = 1 to n do
	let v_i = set.[i-1]
	for j = 1 to sum do
	dp.[i,j] <-
	if j - v_i < 0 then
	// we can't include i th item in OPT
	dp.[i-1,j]
	else
	dp.[i-1,j]\|\|dp.[i-1,j-v_i]

	let mutable result = 0
	for j=1 to sum do
	result <- result+
	( if dp.[n,j] = true then
	j
	else
	0
	)
	result

view raw SPOJ_MAIN72.fsx hosted with ❤ by GitHub

In essence, the N^th row in the table provides the set of integers that can be derived by summing over any subset . Thereby, we compute the summation of all these integers that satisfies subsum(N,j) = true, and returns it as the result.

Conclusion¶

Full source code of the solution can be downloaded from this gist. For C# source code, please visit following gist. Please leave a comment if you have any question/suggestion regarding this post.

Enough said… now, it’s time for a and a new problem. See you soon; till then, happy problem-solving!

Interpretation¶

The problem definition enforces following rules to perform the exchange. Consider, a Bytelandian gold coin —

It can be exchanged to three other coins, i.e.,

coins. Thus, coin

yields

value in bytelandian gold coins.

Alternatively,

coin can be exchanged for

dollars.

Our objective is to derive an algorithm that maximizes the dollars exchanged from the gold coin .

Algorithm¶

From the above interpretation, it is evident that the maximum achievable dollars, (from the exchange of coin ) can be computed as follows.

It effectively demonstrates an optimal substructure and therefore, hints to a dynamic programming (DP) technique to solve it. That is, for a coin , the optimal value of dollar is given by the following function.

We employ a top-down DP approach, as it seems more efficient than a bottom-up approach in this context. It is due to the fact that a bottom-up approach generally requires an OPT table to persist results of smaller subproblems. As in this case, the value of can be very large (i.e., , a bottom-up DP would require a very large array, and performs more computations. Hence, for the overlapping subproblems, we employ memoization.

	let computeMaxDollars (n:int) (memo:Dictionary<int64,int64>)=

	let rec computeMaxDollars' (ni:int64) =
	if ni = 0L \|\| ni = 1L then // base case
	ni
	else
	match memo\|> Memo.tryFind ni with
	\| Some (nx) -> nx // found in memo. Returning Result.
	\| None ->
	let f = computeMaxDollars'
	let nx =
	(ni/2L, ni/3L, ni/4L)
	\|> (fun (x,y,z) -> (f x) + (f y) + (f z))
	\|> (fun nx -> Math.Max(ni,nx))

	memo\|> Memo.add ni nx \|> ignore // storing the result in memo
	nx
	computeMaxDollars' (n\|>int64)

view raw spoj_COINS.fsx hosted with ❤ by GitHub

The following code snippet outlines the implementation of Memo.

	module Memo =
	let empty () = new Dictionary<int64,int64>()

	let add k v (memo:Dictionary<int64,int64>) =
	memo.[k] <- v; memo

	let tryFind k (memo:Dictionary<int64,int64>) =
	match memo.TryGetValue(k) with
	\| true, v -> Some(v)
	\| false,_ -> None

view raw spoj_COINS.fsx hosted with ❤ by GitHub

Full source code of the solution can be downloaded from this gist. Please leave a comment if you have any question/suggestion regarding this post.
Happy problem-solving!

Project Euler 06. Sum square difference with F#

Problem Definition

Available at Sum square difference

The sum of the squares of the first ten natural numbers is,

1² + 2² + ... + 10² = 385

The square of the sum of the first ten natural numbers is,

(1 + 2 + ... + 10)² = 55² = 3025

Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025-385 = 2640.Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

Implementation

	open System

	let squareOfSum n =
	let i = n*(n+1)/2
	i*i

	let sumOfSquare n =
	[1..n]
	\|> List.fold (fun acc x -> acc + x*x) 0

	let solveEuler6 N =
	(squareOfSum N)- (sumOfSquare N)

view raw euler06.fs hosted with ❤ by GitHub

Project Euler 04. Largest Palindrome Product with F#

Problem Definition

Available at Largest Palindrome Product

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 * 99.

Find the largest palindrome made from the product of two 3-digit numbers.

Implementation

We have to find out a palindrome p such that —

where both a and b are three digit number. To do so, we first define a function that checks whether a number (e.g., p) is a palindrome.

	let ispalindrom (x:int):bool =
	let s = x.ToString()
	s = (s \|> (fun x -> new string(x.ToCharArray() \|> Array.rev)))

view raw ispalindrom.fs hosted with ❤ by GitHub

Then, we iterate over all the tuples (a,b) of three digit numbers e.g., [100..999] that satisfy the following equation.

Finally, we check if a*b is a palindrome and get the largest palindrome, as outlined below.

	seq{
	for x in 100..999 do
	for y in x..999 do
	if ispalindrom (xy) then yield (xy)}
	\|> Seq.max

view raw euler04.fs hosted with ❤ by GitHub

Would/did you solve it differently? Please let me know your opinion in the comment section below.

Happy problem solving …

problem solving

SPOJ 6219. Edit Distance (EDIST) with F#

This problem can be solved using dynamic programming with memoization technique. In essence, it is about computing the Edit Distance, also known as, Levenshtein Distance between two given strings.

Definition

Edit Distance—a.k.a “Lavenshtein Distance”–is the minimum number of edit operations required to transform one word into another. The allowable edit operations are letter insertion, letter deletion and letter substitution.

Implementation

Using Dynamic Programming, we can compute the edit distance between two string sequences. But for that, we need to derive a recursive definition of Edit Distance. We denote the distance between two strings as D, which can be defined using a recurrence as follows.

Case 1 : Both and are empty strings, denoted as :

Case 2 : Either or is :

Case 3 : Both and are not :

Here , where is the last character of and contains rest of the characters. Same goes for . We define edit distance between and using a recurrence and in term of and .

can be defined as the minimum, or the least expensive one of the following three alternatives stated in the above equation.

Substitution: If , then the overall distance is simply . Otherwise, we need a substitution operation that replaces with , and thus, the overall distance will be .
Insertion: Second possibility is to convert to by inserting in . In this case, the distance will be . Here, +1 is the cost of the insert operation.
Deletion: Last alternative is to convert to by deleting from that costs +1. Then the distance become .

As this is a ternary recurrence, it would result in an exponential run-time, which is quite impractical. However, using the dynamic programming with memoization, this recurrence can be solved using a 2D array. The code to solve this problem is outline below.

	let computeEditDistance (source:string,target:string) =
	let height,width = (source.Length, target.Length)
	let grid: int [,] = Array2D.zeroCreate<int> (height+1) (width+1) // 2D Array for memoization

	for h = 0 to height do
	for w = 0 to width do
	grid.[h,w] <-
	match h,w with
	\| h,0 -> h // case 1 and 2
	\| 0, w -> w
	\| h, w ->
	let s,t = source.[h-1],target.[w-1]
	let substitution = grid.[h-1,w-1]+(if s = t then 0 else 1)
	let insertion = grid.[h,w-1] + 1
	let deletion = grid.[h-1,w] + 1
	min (insertion, deletion, substitution) // case 3
	grid.[height,width]

view raw gistfile1.fs hosted with ❤ by GitHub

As shown in line 14, the distance grid.[h,w] can be computed locally by taking the min of the three alternatives stated in the recurrence (computed in line 11,12, 13). By obtaining the locally optimum solutions, we eventually get the edit distance from grid.[s.length, t.length].

Complexity: Run-time complexity: . Lets denote the lengths of both strings as . Then, the complexity become . Space complexity is also same.

Complete source code is outlined in the next page.

Pages: 12

Tag: problemsolving

SPOJ 8545. Subset Sum (Main72) with Dynamic Programming and F#

Interpretation¶

Algorithm¶

Conclusion¶

See Also

SPOJ 346. Bytelandian Gold Coins (COINS) with Dynamic Programming and F#

Interpretation¶

Algorithm¶

Project Euler 06. Sum square difference with F#

Problem Definition

Implementation

Project Euler 04. Largest Palindrome Product with F#

Problem Definition

SPOJ 6219. Edit Distance (EDIST) with F#

Definition

Implementation