read CSV file in Cplex - csv

my question is related to my previous question. I should make some change in my code. I have a number of nodes between 1 to 100 in the CSV file. I create another CSV file and generate 20 random numbers between the 100 nodes and called them demand points. Each of this demand point has specific demands which are the randomly generate numbers between 1 to 10. I want to read this demand points(the indexes) and their weights. this is the first part of my question? how can I read this?
After that, I need to have a distance between each of these demand points and all nodes. I don't how can I just read the indexes of demand points and calculate the distance between them and all the nodes.
Based on the code that I provided, I need the indexes of demand points for a lot of places. My main problem is that I don't know how should I get these indexes in Cplex through the CSV file.
The demand points with their demands picture is:
first column is demandpointindex and second column in their demands
this file has 200 rows
I tried this code for reading the demand points:
tuple demands
{
int demandpoint;
int weight;
}
{demands} demand={};
execute
{
var f=new IloOplInputFile("weight.csv");
while (!f.eof)
{
var data = f.readline().split(",");
if (ar.length==2)
demand.add(Opl.intValue(ar[0]),Opl.intValue(ar[1]));
}
f.close();
}
execute
{
writeln(demand);
}
but it's not true.
int n=100;
int p=5;
tuple demands
{
int demandpointindex;
int weight;
}
{demands} demand={};
execute
{
var f=new IloOplInputFile("weight.csv");
while (!f.eof)
{
var data = f.readline().split(",");
if (ar.length==2)
demand.add(Opl.intValue(ar[0]),Opl.intValue(ar[1]));
}
f.close();
}
execute
{
writeln(demand);
}
float d[demandpointindexes][facilities];
execute {
var f = new IloOplInputFile("test1.csv");
while (!f.eof) {
var data = f.readline().split(",");
if (data.length == 3)
d[Opl.intValue(data[0])][Opl.intValue(data[1])] = Opl.floatValue(data[2]);
}
writeln(d);
}
dvar boolean x[demandpointindexe][facilities];
...

I hope I got your explanation right. Assume you have a file weight.csv like this:
1,2,
3,7,
4,9,
Here the first item in each row is the index of a demand point, the second item is its weight. Then you can parse this as before using this scripting block:
tuple demandpoint {
int index;
int weight;
}
{demandpoint} demand={};
execute {
var f = new IloOplInputFile("weight.csv");
while (!f.eof) {
var data = f.readline().split(",");
if (data.length == 3)
demand.add(Opl.intValue(data[0]), Opl.intValue(data[1]));
}
writeln(demand);
}
Next you can create a set that contains the indeces of all the demand points:
{int} demandpoints = { d.index | d in demand };
Assume file test1.csv looks like this
1,1,0,
1,2,5,
1,3,6,
1,4,7,
3,1,1,
3,2,1.5,
3,3,0,
3,4,3.5,
4,1,1,
4,2,1.5,
4,3,1.7,
4,4,0,
Here the first item is a demand point index, the second item is a facility index and the third item is the distance between first and second item. Note that there are no lines that start with 2 since there is no demand point with index 2 in weight.csv. Also note that I assume only 4 facilities here (to keep the file short). You can read the distance between demand points and facitilies as follows:
range facilities = 1..4;
float d[demandpoints][facilities];
execute {
var f = new IloOplInputFile("test1.csv");
while (!f.eof) {
var data = f.readline().split(",");
if (data.length == 4)
d[Opl.intValue(data[0])][Opl.intValue(data[1])] = Opl.floatValue(data[2]);
}
writeln(d);
}
The full script (including a dummy objective and constraints so that it can be run) looks:
tuple demandpoint {
int index;
int weight;
}
{demandpoint} demand={};
execute {
var f = new IloOplInputFile("weight.csv");
while (!f.eof) {
var data = f.readline().split(",");
if (data.length == 3)
demand.add(Opl.intValue(data[0]), Opl.intValue(data[1]));
}
writeln(demand);
}
// Create a set that contains all the indeces of demand points
// as read from weight.csv
{int} demandpoints = { d.index | d in demand };
range facilities = 1..4;
float d[demandpoints][facilities];
execute {
var f = new IloOplInputFile("test1.csv");
while (!f.eof) {
var data = f.readline().split(",");
if (data.length == 4)
d[Opl.intValue(data[0])][Opl.intValue(data[1])] = Opl.floatValue(data[2]);
}
writeln(d);
}
minimize 0;
subject to {}
It prints
{<1 2> <3 7> <4 9>}
[[0 5 6 7]
[1 1.5 0 3.5]
[1 1.5 1.7 0]]
Be careful about how many commas you have in your csv! The code posted above assumes that each line ends with a comma. That is, each line has as many commas as it has fields. If the last field is not terminated by a comma then you have to adapt the parser.
If you have in test1.csv the distance between all the nodes then it makes sense to first read the data into an array float distance[facilities][facilities]; and then define the array d based on that as
float d[i in demandpoints][j in facilities] = distance[i][j];
Update for the more detailed specification you gave in the comments:
In order to handle the test1.csv you explained in the comments you could define a new tuple:
tuple Distance {
int demandpoint;
int facility;
float distance;
}
{Distance} distances = {};
and read/parse this exactly as you did parse the weight.csv file (with one additional field, of course).
Then you can create the distance matrix like so:
float d[i in I][j in J] = sum (dist in distances : dist.demandpoint == i && dist.facility == j) dist.distance;
Here I and J are the sets or ranges of demand points and facilities, respectively. See above for how you can get a set of all demand points defined in the tuple set. The created matrix will have an entry for each demandpoint/distance pair. The trick in the definition d is that there are two cases:
If a pair (i,j) is defined in test1.csv then the sum will match exactly one element in distances: the one that defines the distance between two points.
If a pair (i,j) is not defined in test1.csv then the sum will not match anything and the corresponding entry in the distance matrix will thus be 0.

Related

Count no. of ones by position(place) of an array of 32-bit binary numbers

An array(10^5 size) of 32-bit binary numbers is given, we're required to count the no. of ones for every bit of those numbers.
For example:
Array : {10101,1011,1010,1}
Counts : {1's place: 3, 2's place: 2, 3's place: 1, 4's place: 2, 5's place: 1}
No bit manipulation technique seems to satisfy the constraints to me.
Well, this should be solveable with two loops: one going over the array the other one masking the right bits. Running time should be not too bad for your constraints.
Here is a rust implementation (out of my head, not throughtfully tested):
fn main() {
let mut v = vec!();
for i in 1..50*1000 {
v.push(i);
}
let r = bitcount_arr(v);
r.iter().enumerate().for_each( |(i,x)| print!("index {}:{} ",i+1,x));
}
fn bitcount_arr(input:Vec<u32>) -> [u32;32] {
let mut res = [0;32];
for num in input {
for i in 0..31 {
let mask = 1 << i;
if num & mask != 0 {
res[i] += 1;
}
}
}
res
}
This can be done with transposed addition, though the array is a bit long for it.
To transpose addition, use an array of counters, but instead of using one counter for every position we'll use one counter for every bit of the count. So a counter that tracks for each position whether the count is even/odd, a counter that tracks for each position whether the count has a 2 in it, etc.
To add an element of the array into this, only half-add operations (& to find the new carry, ^ to update) are needed, since it's only a conditional increment: (not tested)
uint32_t counters[17];
for (uint32_t elem : array) {
uint32_t c = elem;
for (int i = 0; i < 17; i++) {
uint32_t nextcarry = counters[i] & c;
counters[i] ^= c;
c = nextcarry;
}
}
I chose 17 counters because log2(10^5) is just less than 17. So even if all bits are 1, the counters won't wrap.
To read off the result for bit k, take the k'th bit of every counter.
There are slightly more efficient ways that can add several elements of the array into the counters at once using some full-adds and duplicated counters.

AS3: adding different numbers in an array to get specific result.

I got a numberArray.
It contains intergers - randomised, within a specific range.
I want to get a specific sum, but not for everything inside the numberArray,
more of trying to sum up different amount of numbers (total of 5 only) inside the numberArray and see if it'll get the specific total required. and if not, it'll randomise another number to take over one of the numbers inside the numberArray.
What's the easiest way to do this ?
doing lots of
if (numberArray[1] + numberArray[2] == specificNumber)
{
}
if (numberArray[1] + numberArray[3] == specificNumber)
{
}
etc. etc. etc.
have too many lines of codes, and it seems like there are easier codes. right now i only have 5 different numbers in the array, so it's still bearable, but if the amount of numbers are higher.... ....
Reading your question like this: For your array of random integers, find a (or all) set(s) of integers that have a given sum.
This is an NP-Complete problem - i.e. there's no known algorithm that solves it efficiently.
The fastest known way is rather complex, so we'll go with a naive solution - should be good enough if you're not doing this on every frame or the input set is huge.
This should also work with 0 or negative values in the input set.
// The sum we're looking for:
var requiredSum:int = 8;
// Our input set:
var numberArray:Array = [1, 2, 3, 4, 5, 2, 3];
// Results will be stored here:
var resultSets:Array = [];
// Go through all possible subset sizes.
// This allows subset sizes all the way up to the size of
// the input set (numberArray.length).
// You can modify it to a fixed value (say, 5), of course:
for (var subsetSize:int = 1; subsetSize <= numberArray.length; subsetSize++)
{
// We'll use the same array for all our attempts of this size:
var subset:Array = new Array(subsetSize);
findSum(numberArray, subset, 0, 0);
}
// Output results:
for (var i:int = 0; i < resultSets.length; i++)
{
trace(resultSets[i].join("+"));
}
// numberArray : Our input set
// subset : The set we're currently filling
// setIndex : The position we're at in numberArray
// subsetIndex : The position we're at in the set we're filling
function findSum(numberArray:Array, subset:Array, setIndex:int,
subsetIndex:int):void
{
// Try every value from the input set starting from our current position,
// and insert the value at the current subset index:
for (var index:int = setIndex ; index < numberArray.length; index++)
{
subset[subsetIndex] = numberArray[index];
// Have we filled the subset?
if (subsetIndex == subset.length - 1)
{
var sum:int = 0;
for (var i:int = 0; i < subset.length; i++)
{
sum += subset[i];
}
if (sum == requiredSum)
{
// Clone the array before adding it to our results,
// since we'll be modifying it if we find more:
resultSets.push(subset.concat());
}
}
else
{
// Recursion takes care of combining our subset so far
// with every possible value for the remaining subset indices:
findSum(numberArray, subset, index + 1, subsetIndex + 1);
}
}
}
Output for the values used in the above code:
3+5
5+3
1+2+5
1+3+4
1+4+3
1+5+2
2+3+3
2+4+2
3+2+3
1+2+3+2
1+2+2+3
If we only need to know IF a sum exists, there's no need for the result set - we just return true/false, and break out of the recursive algorithm completely when a sum has been found:
var requiredSum:int = 8;
var numberArray:Array = [1, 2, 3, 4, 5, 2, 3];
// Go through all possible subset sizes:
for (var subsetSize:int = 1; subsetSize <= numberArray.length; subsetSize++)
{
// We'll use the same array for all our attempts of this size:
var subset:Array = new Array(subsetSize);
if (findSum(numberArray, subset, 0, 0))
{
trace("Found our sum!");
// If we found our sum, no need to look for more sets:
break;
}
}
// numberArray : Our input set
// subset : The set we're currently filling
// setIndex : The position we're at in numberArray
// subsetIndex : The position we're at in the set we're filling
// RETURNS : True if the required sum was found, otherwise false.
function findSum(numberArray:Array, subset:Array, setIndex:int,
subsetIndex:int):Boolean
{
// Try every value from the input set starting from our current position,
// and insert the value at the current subset index:
for (var index:int = setIndex ; index < numberArray.length; index++)
{
subset[subsetIndex] = numberArray[index];
// Have we filled the subset?
if (subsetIndex == subset.length - 1)
{
var sum:int = 0;
for (var i:int = 0; i < subset.length; i++)
{
sum += subset[i];
}
// Return true if we found our sum, false if not:
return sum == requiredSum;
}
else
{
if (findSum(numberArray, subset, index + 1, subsetIndex + 1))
{
// If the "inner" findSum found a sum, we're done, so return
// - otherwise stay in the loop and keep looking:
return true;
}
}
}
// We found no subset with our required sum this time around:
return false;
}
ETA: How this works... As mentioned, it's the naive solution - in other words, we're simply checking every single permutation of numberArray, summing each permutation, and checking if it's the sum we want.
The most complicated part is making all the permutations. The way this code does it is through recursion - i.e., the findSum() function filling a slot then calling itself to fill the next one, until all slots are filled and it can check the sum. We'll use the numberArray [1, 5, 4, 2] as an example here:
Go through all subset sizes in a loop - i.e., start by making all [a], then all [a,b], [a,b,c], [a,b,c,d]... etc.
For each subset size:
Fill slot 1 of the subset...
... with each value of numberArray - [1, ?, ?], [5, ?, ?], [4, ?, ?]...
If all slots in subset have been filled, check if the sum matches and skip step 4.
(Recursively) call findSum to:
Fill slot 2 of the subset...
... with each remaining value of numberArray - [1, 5, ?], [1, 4, ?], [1, 2, ?]
If all slots in subset have been filled, check if the sum matches and skip step 4.
(Recursively) call findSum to:
Fill slot 3 of the subset
... with each remaining value of numberArray - [1, 5, 4], [1, 5, 2]
If all slots in subset have been filled, check if the sum matches and skip step 4.
(Recursively) call findSum (this goes on "forever", or until all slots are filled and we "skip step 4")
Go to 2.4.4.1. to try next value for slot 3.
Go to 2.4.1 to try next value for slot 2.
Go to 2.1 to try next value for slot 1.
This way, we go through every permutation of size 1, 2, 3, 4...
There's more optimization that could be done here, since the code never checks that it actually has enough values left in the input set to fill the remaining slots - i.e. it does some loops and calls to findSum() that are unneeded. This is only a matter of efficiency, however - the result is still correct.
I would do something like the following:
shuffle array
take random amount of numbers from the array
sum them up
if the sum is not the total sum you want, repeat
hm, not sure what you want to do at the end when a "conclusion" or "no conclusion" is reached, but you could generate a Power set from your set of numbers then for each subset add up all the numbers in it to see if you get your desired sum.
(This would be a 'brute force' approach and could be slow if you have many numbers.)
Possibly useful for how to create a Power set:
Calculating all of the subsets of a set of numbers

Removal of every 'kth' person from a circle. Find the last remaining person

As part of a recent job application I was asked to code a solution to this problem.
Given,
n = number of people standing in a circle.
k = number of people to count over each time
Each person is given a unique (incrementing) id. Starting with the first person (the lowest id), they begin counting from 1 to k.
The person at k is then removed and the circle closes up. The next remaining person (following the eliminated person) resumes counting at 1. This process repeats until only one person is left, the winner.
The solution must provide:
the id of each person in the order they are removed from the circle
the id of the winner.
Performance constraints:
Use as little memory as possible.
Make the solution run as fast as possible.
I remembered doing something similar in my CS course from years ago but could not recall the details at the time of this test. I now realize it is a well known, classic problem with multiple solutions. (I will not mention it by name yet as some may just 'wikipedia' an answer).
I've already submitted my solution so I'm absolutely not looking for people to answer it for me. I will provide it a bit later once/if others have provided some answers.
My main goal for asking this question is to see how my solution compares to others given the requirements and constraints.
(Note the requirements carefully as I think they may invalidate some of the 'classic' solutions.)
Manuel Gonzalez noticed correctly that this is the general form of the famous Josephus problem.
If we are only interested in the survivor f(N,K) of a circle of size N and jumps of size K, then we can solve this with a very simple dynamic programming loop (In linear time and constant memory). Note that the ids start from 0:
int remaining(int n, int k) {
int r = 0;
for (int i = 2; i <= n; i++)
r = (r + k) % i;
return r;
}
It is based on the following recurrence relation:
f(N,K) = (f(N-1,K) + K) mod N
This relation can be explained by simulating the process of elimination, and after each elimination re-assigning new ids starting from 0. The old indices are the new ones with a circular shift of k positions. For a more detailed explanation of this formula, see http://blue.butler.edu/~phenders/InRoads/MathCounts8.pdf.
I know that the OP asks for all the indices of the eliminated items in their correct order. However, I believe that the above insight can be used for solving this as well.
You can do it using a boolean array.
Here is a pseudo code:
Let alive be a boolean array of size N. If alive[i] is true then ith person is alive else dead. Initially it is true for every 1>=i<=N
Let numAlive be the number of persons alive. So numAlive = N at start.
i = 1 # Counting starts from 1st person.
count = 0;
# keep looping till we've more than 1 persons.
while numAlive > 1 do
if alive[i]
count++
end-if
# time to kill ?
if count == K
print Person i killed
numAlive --
alive[i] = false
count = 0
end-if
i = (i%N)+1 # Counting starts from next person.
end-while
# Find the only alive person who is the winner.
while alive[i] != true do
i = (i%N)+1
end-while
print Person i is the winner
The above solution is space efficient but not time efficient as the dead persons are being checked.
To make it more efficient time wise you can make use of a circular linked list. Every time you kill a person you delete a node from the list. You continue till a single node is left in the list.
The problem of determining the 'kth' person is called the Josephus Problem.
Armin Shams-Baragh from Ferdowsi University of Mashhad published some formulas for the Josephus Problem and the extended version of it.
The paper is available at: http://www.cs.man.ac.uk/~shamsbaa/Josephus.pdf
This is my solution, coded in C#. What could be improved?
public class Person
{
public Person(int n)
{
Number = n;
}
public int Number { get; private set; }
}
static void Main(string[] args)
{
int n = 10; int k = 4;
var circle = new List<Person>();
for (int i = 1; i <= n; i++)
{
circle.Add(new Person(i));
}
var index = 0;
while (circle.Count > 1)
{
index = (index + k - 1) % circle.Count;
var person = circle[index];
circle.RemoveAt(index);
Console.WriteLine("Removed {0}", person.Number);
}
Console.ReadLine();
}
Console.WriteLine("Winner is {0}", circle[0].Number);
Essentially the same as Ash's answer, but with a custom linked list:
using System;
using System.Linq;
namespace Circle
{
class Program
{
static void Main(string[] args)
{
Circle(20, 3);
}
static void Circle(int k, int n)
{
// circle is a linked list representing the circle.
// Each element contains the index of the next member
// of the circle.
int[] circle = Enumerable.Range(1, k).ToArray();
circle[k - 1] = 0; // Member 0 follows member k-1
int prev = -1; // Used for tracking the previous member so we can delete a member from the list
int curr = 0; // The member we're currently inspecting
for (int i = 0; i < k; i++) // There are k members to remove from the circle
{
// Skip over n members
for (int j = 0; j < n; j++)
{
prev = curr;
curr = circle[curr];
}
Console.WriteLine(curr);
circle[prev] = circle[curr]; // Delete the nth member
curr = prev; // Start counting again from the previous member
}
}
}
}
Here is a solution in Clojure:
(ns kthperson.core
(:use clojure.set))
(defn get-winner-and-losers [number-of-people hops]
(loop [people (range 1 (inc number-of-people))
losers []
last-scan-start-index (dec hops)]
(if (= 1 (count people))
{:winner (first people) :losers losers}
(let [people-to-filter (subvec (vec people) last-scan-start-index)
additional-losers (take-nth hops people-to-filter)
remaining-people (difference (set people)
(set additional-losers))
new-losers (concat losers additional-losers)
index-of-last-removed-person (* hops (count additional-losers))]
(recur remaining-people
new-losers
(mod last-scan-start-index (count people-to-filter)))))))
Explanation:
start a loop, with a collection of people 1..n
if there is only one person left, they are the winner and we return their ID, as well as the IDs of the losers (in order of them losing)
we calculate additional losers in each loop/recur by grabbing every N people in the remaining list of potential winners
a new, shorter list of potential winners is determined by removing the additional losers from the previously-calculated potential winners.
rinse & repeat (using modulus to determine where in the list of remaining people to start counting the next time round)
This is a variant of the Josephus problem.
General solutions are described here.
Solutions in Perl, Ruby, and Python are provided here. A simple solution in C using a circular doubly-linked list to represent the ring of people is provided below. None of these solutions identify each person's position as they are removed, however.
#include <stdio.h>
#include <stdlib.h>
/* remove every k-th soldier from a circle of n */
#define n 40
#define k 3
struct man {
int pos;
struct man *next;
struct man *prev;
};
int main(int argc, char *argv[])
{
/* initialize the circle of n soldiers */
struct man *head = (struct man *) malloc(sizeof(struct man));
struct man *curr;
int i;
curr = head;
for (i = 1; i < n; ++i) {
curr->pos = i;
curr->next = (struct man *) malloc(sizeof(struct man));
curr->next->prev = curr;
curr = curr->next;
}
curr->pos = n;
curr->next = head;
curr->next->prev = curr;
/* remove every k-th */
while (curr->next != curr) {
for (i = 0; i < k; ++i) {
curr = curr->next;
}
curr->prev->next = curr->next;
curr->next->prev = curr->prev;
}
/* announce last person standing */
printf("Last person standing: #%d.\n", curr->pos);
return 0;
}
Here's my answer in C#, as submitted. Feel free to criticize, laugh at, ridicule etc ;)
public static IEnumerable<int> Move(int n, int k)
{
// Use an Iterator block to 'yield return' one item at a time.
int children = n;
int childrenToSkip = k - 1;
LinkedList<int> linkedList = new LinkedList<int>();
// Set up the linked list with children IDs
for (int i = 0; i < children; i++)
{
linkedList.AddLast(i);
}
LinkedListNode<int> currentNode = linkedList.First;
while (true)
{
// Skip over children by traversing forward
for (int skipped = 0; skipped < childrenToSkip; skipped++)
{
currentNode = currentNode.Next;
if (currentNode == null) currentNode = linkedList.First;
}
// Store the next node of the node to be removed.
LinkedListNode<int> nextNode = currentNode.Next;
// Return ID of the removed child to caller
yield return currentNode.Value;
linkedList.Remove(currentNode);
// Start again from the next node
currentNode = nextNode;
if (currentNode== null) currentNode = linkedList.First;
// Only one node left, the winner
if (linkedList.Count == 1) break;
}
// Finally return the ID of the winner
yield return currentNode.Value;
}

Generating combination of letters

Given a set of letters, say from A.. F, how can one generate a combination of these letters for a specific length. i.e for length 4, generate all string containing these letters {AAAA, ABCD, ...} (duplicates included). I am not able to understand how to come out with a code that does it.This is pertaining to the Mastermind game that I am trying to simulate. Is there any algorithm to perform this generation.
regards,
darkie
There is an algorithm called Heap's Algorithm for generating permutations. This might suit your purposes. I found an example implementation here
I'm not sure what the name would be of such an algorithm, but it is a recursive one. That is, have a method that figures out one character, and simply keep calling itself until you're at the desired length of string that you want, then start filling in your array. Here's some sample C# code that should help:
public void GetPermutations()
{
string currentPrefix = ""; // Just a starting point
int currentLength = 1; // one-based
int desiredLength = 4; // one-based
string alphabet = "ABCDEF"; // Characters to build permutations from
List<string> permutations = new List<string>();
FillPermutations(currentPrefix, currentLength, alphabet, desiredLength, permutations);
}
public void FillPermutations(string currentPrefix, int currentLength, string alphabet, int desiredLength, List<string> permutations)
{
// If we're not at the desired depth yet, keep calling this function recursively
// until we attain what we want.
for (int i = 0; i < alphabet.Length; i++)
{
string currentPermutation = currentPrefix + alphabet[i].ToString();
if (currentLength < desiredLength)
{
// Increase current length by one and recurse. Current permutation becomes new prefix
int newCurrentLength = currentLength + 1;
FillPermutations(currentPermutation, newCurrentLength, alphabet, desiredLength, permutations);
}
else
{
// We're at the desired length, so add this permutation to the list
permutations.Add(currentPermutation);
}
}
}

How to implement three stacks using a single array

I came across this problem in an interview website. The problem asks for efficiently implement three stacks in a single array, such that no stack overflows until there is no space left in the entire array space.
For implementing 2 stacks in an array, it's pretty obvious: 1st stack grows from LEFT to RIGHT, and 2nd stack grows from RIGHT to LEFT; and when the stackTopIndex crosses, it signals an overflow.
Thanks in advance for your insightful answer.
You can implement three stacks with a linked list:
You need a pointer pointing to the next free element. Three more pointers return the last element of each stack (or null, if the stack is empty).
When a stack gets another element added, it has to use the first free element and set the free pointer to the next free element (or an overflow error will be raised). Its own pointer has to point to the new element, from there back to the next element in the stack.
When a stack gets an element removed it will hand it back into the list of free elements. The own pointer of the stack will be redirected to the next element in the stack.
A linked list can be implemented within an array.
How (space) efficent is this?
It is no problem to build a linked list by using two cells of an array for each list element (value + pointer). Depending on the specification you could even get pointer and value into one array element (e.g. the array is long, value and pointer are only int).
Compare this to the solution of kgiannakakis ... where you lose up to 50% (only in the worst case). But I think that my solution is a bit cleaner (and maybe more academic, which should be no disadvantage for an interview question ^^).
See Knuth, The Art of Computer Programming, Volume 1, Section 2.2.2. titled "Sequential allocation". Discusses allocating multiple queues/stacks in a single array, with algorithms dealing with overflows, etc.
We can use long bit array representing to which stack the i-th array cell belongs to.
We can take values by modulo 3 (00 - empty, 01 - A, 10 - B, 11 - C). It would take N/2 bits or N/4 bytes of additional memory for N sized array.
For example for 1024 long int elements (4096 bytes) it would take only 256 bytes or 6%.
This bit array map can be placed in the same array at the beginning or at the end, just shrinking the size of the given array by constant 6%!
First stack grows from left to right.
Second stack grows from right to left.
Third stack starts from the middle. Suppose odd sized array for simplicity. Then third stack grows like this:
* * * * * * * * * * *
5 3 1 2 4
First and second stacks are allowed to grow maximum at the half size of array. The third stack can grow to fill in the whole array at a maximum.
Worst case scenario is when one of the first two arrays grows at 50% of the array. Then there is a 50% waste of the array. To optimise the efficiency the third array must be selected to be the one that grows quicker than the other two.
This is an interesting conundrum, and I don't have a real answer but thinking slightly outside the box...
it could depend on what each element in the stack consists of. If it's three stacks of true/false flags, then you could use the first three bits of integer elements. Ie. bit 0 is the value for the first stack, bit 1 is the value for the second stack, bit 2 is the value for the third stack. Then each stack can grow independently until the whole array is full for that stack. This is even better as the other stacks can also continue to grow even when the first stack is full.
I know this is cheating a bit and doesn't really answer the question but it does work for a very specific case and no entries in the stack are wasted. I am watching with interest to see whether anyone can come up with a proper answer that works for more generic elements.
Split array in any 3 parts (no matter if you'll split it sequentially or interleaved). If one stack grows greater than 1/3 of array you start filling ends of rest two stacks from the end.
aaa bbb ccc
1 2 3
145 2 3
145 2 6 3
145 2 6 3 7
145 286 3 7
145 286 397
The worse case is when two stacks grows up to 1/3 boundary and then you have 30% of space waste.
Assuming that all array positions should be used to store values - I guess it depends on your definition of efficiency.
If you do the two stack solution, place the third stack somewhere in the middle, and track both its bottom and top, then most operations will continue to be efficient, at a penalty of an expensive Move operation (of the third stack towards wherever free space remains, moving to the half way point of free space) whenever a collision occurs.
It's certainly going to be quick to code and understand. What are our efficiency targets?
A rather silly but effective solution could be:
Store the first stack elements at i*3 positions: 0,3,6,...
Store the second stack elements at i*3+1 positions: 1,4,7...
And third stack elements at i*3+2 positions.
The problem with this solution is that the used memory will be always three times the size of the deepest stack and that you can overflow even when there are available positions at the array.
Make a HashMap with keys to the begin and end positions e.g. < "B1" , 0 >, <"E1" , n/3 >
for each Push(value) add a condition to check if position of Bx is previous to Ex or there is some other "By" in between. -- lets call it condition (2)
with above condition in mind,
if above (2) is true // if B1 and E1 are in order
{ if ( S1.Push()), then E1 ++ ;
else // condition of overflow ,
{ start pushing at end of E2 or E3 (whichever has a space) and update E1 to be E2-- or E3-- ; }
}
if above (2) is false
{ if ( S1.Push()), then E1 -- ;
else // condition of overflow ,
{ start pushing at end of E2 or E3 (whichever has a space) and update E1 to be E2-- or E3-- ; }
}
Assume you only has integer index. if it's treated using FILO (First In Last Out) and not referencing individual, and only using an array as data. Using it's 6 space as stack reference should help:
[head-1, last-1, head-2, last-2, head-3, last-3, data, data, ... ,data]
you can simply using 4 space, because head-1 = 0 and last-3 = array length. If using FIFO (First In First Out) you need to re-indexing.
nb: I’m working on improving my English.
first stack grows at 3n,
second stack grows at 3n+1,
third grows at 3n+2
for n={0...N}
Yet another approach (as additional to linked-list) is to use map of stacks. In that case you'll have to use additional log(3^n)/log(2) bits for building map of data distribution in your n-length array. Each of 3-value part of map says which stack is owns next element.
Ex. a.push(1); b.push(2); c.push(3); a.push(4); a.push(5); will give you image
aacba
54321
appropriate value of map is calculated while elements is pushed onto stack (with shifting contents of array)
map0 = any
map1 = map0*3 + 0
map2 = map1*3 + 1
map3 = map2*3 + 2
map4 = map3*3 + 0
map5 = map4*3 + 0 = any*3^5 + 45
and length of stacks 3,1,1
Once you'll want to do c.pop() you'll have to reorganize your elements by finding actual position of c.top() in original array through walking in cell-map (i.e. divide by 3 while mod by 3 isn't 2) and then shift all contents in array back to cover that hole. While walking through cell-map you'll have to store all position you have passed (mapX) and after passing that one which points to stack "c" you'll have to divide by 3 yet another time and multiply it by 3^(amount positions passed-1) and add mapX to get new value of cells-map.
Overhead for that fixed and depends on size of stack element (bits_per_value):
(log(3n)/log(2)) / (nlog(bits_per_value)/log(2)) = log(3n) / (nlog(bits_per_value)) = log(3) / log(bits_per_value)
So for bits_per_value = 32 it will be 31.7% space overhead and with growing bits_per_value it will decay (i.e. for 64 bits it will be 26.4%).
In this approach, any stack can grow as long as there is any free space in the array.
We sequentially allocate space to the stacks and we link new blocks to the previous block. This means any new element in a stack keeps a pointer to the previous top element of that particular stack.
int stackSize = 300;
int indexUsed = 0;
int[] stackPointer = {-1,-1,-1};
StackNode[] buffer = new StackNode[stackSize * 3];
void push(int stackNum, int value) {
int lastIndex = stackPointer[stackNum];
stackPointer[stackNum] = indexUsed;
indexUsed++;
buffer[stackPointer[stackNum]]=new StackNode(lastIndex,value);
}
int pop(int stackNum) {
int value = buffer[stackPointer[stackNum]].value;
int lastIndex = stackPointer[stackNum];
stackPointer[stackNum] = buffer[stackPointer[stackNum]].previous;
buffer[lastIndex] = null;
indexUsed--;
return value;
}
int peek(int stack) { return buffer[stackPointer[stack]].value; }
boolean isEmpty(int stackNum) { return stackPointer[stackNum] == -1; }
class StackNode {
public int previous;
public int value;
public StackNode(int p, int v){
value = v;
previous = p;
}
}
This code implements 3 stacks in single array. It takes care of empty spaces and fills the empty spaces in between the data.
#include <stdio.h>
struct stacknode {
int value;
int prev;
};
struct stacknode stacklist[50];
int top[3] = {-1, -1, -1};
int freelist[50];
int stackindex=0;
int freeindex=-1;
void push(int stackno, int value) {
int index;
if(freeindex >= 0) {
index = freelist[freeindex];
freeindex--;
} else {
index = stackindex;
stackindex++;
}
stacklist[index].value = value;
if(top[stackno-1] != -1) {
stacklist[index].prev = top[stackno-1];
} else {
stacklist[index].prev = -1;
}
top[stackno-1] = index;
printf("%d is pushed in stack %d at %d\n", value, stackno, index);
}
int pop(int stackno) {
int index, value;
if(top[stackno-1] == -1) {
printf("No elements in the stack %d\n", value, stackno);
return -1;
}
index = top[stackno-1];
freeindex++;
freelist[freeindex] = index;
value = stacklist[index].value;
top[stackno-1] = stacklist[index].prev;
printf("%d is popped put from stack %d at %d\n", value, stackno, index);
return value;
}
int main() {
push(1,1);
push(1,2);
push(3,3);
push(2,4);
pop(3);
pop(3);
push(3,3);
push(2,3);
}
Another solution in PYTHON, please let me know if that works as what you think.
class Stack(object):
def __init__(self):
self.stack = list()
self.first_length = 0
self.second_length = 0
self.third_length = 0
self.first_pointer = 0
self.second_pointer = 1
def push(self, stack_num, item):
if stack_num == 1:
self.first_pointer += 1
self.second_pointer += 1
self.first_length += 1
self.stack.insert(0, item)
elif stack_num == 2:
self.second_length += 1
self.second_pointer += 1
self.stack.insert(self.first_pointer, item)
elif stack_num == 3:
self.third_length += 1
self.stack.insert(self.second_pointer - 1, item)
else:
raise Exception('Push failed, stack number %d is not allowd' % stack_num)
def pop(self, stack_num):
if stack_num == 1:
if self.first_length == 0:
raise Exception('No more element in first stack')
self.first_pointer -= 1
self.first_length -= 1
self.second_pointer -= 1
return self.stack.pop(0)
elif stack_num == 2:
if self.second_length == 0:
raise Exception('No more element in second stack')
self.second_length -= 1
self.second_pointer -= 1
return self.stack.pop(self.first_pointer)
elif stack_num == 3:
if self.third_length == 0:
raise Exception('No more element in third stack')
self.third_length -= 1
return self.stack.pop(self.second_pointer - 1)
def peek(self, stack_num):
if stack_num == 1:
return self.stack[0]
elif stack_num == 2:
return self.stack[self.first_pointer]
elif stack_num == 3:
return self.stack[self.second_pointer]
else:
raise Exception('Peek failed, stack number %d is not allowd' % stack_num)
def size(self):
return len(self.items)
s = Stack()
# push item into stack 1
s.push(1, '1st_stack_1')
s.push(1, '2nd_stack_1')
s.push(1, '3rd_stack_1')
#
## push item into stack 2
s.push(2, 'first_stack_2')
s.push(2, 'second_stack_2')
s.push(2, 'third_stack_2')
#
## push item into stack 3
s.push(3, 'FIRST_stack_3')
s.push(3, 'SECOND_stack_3')
s.push(3, 'THIRD_stack_3')
#
print 'Before pop out: '
for i, elm in enumerate(s.stack):
print '\t\t%d)' % i, elm
#
s.pop(1)
s.pop(1)
#s.pop(1)
s.pop(2)
s.pop(2)
#s.pop(2)
#s.pop(3)
s.pop(3)
s.pop(3)
#s.pop(3)
#
print 'After pop out: '
#
for i, elm in enumerate(s.stack):
print '\t\t%d)' % i, elm
May be this can help you a bit...i wrote it by myself
:)
// by ashakiran bhatter
// compile: g++ -std=c++11 test.cpp
// run : ./a.out
// sample output as below
// adding: 1 2 3 4 5 6 7 8 9
// array contents: 9 8 7 6 5 4 3 2 1
// popping now...
// array contents: 8 7 6 5 4 3 2 1
#include <iostream>
#include <cstdint>
#define MAX_LEN 9
#define LOWER 0
#define UPPER 1
#define FULL -1
#define NOT_SET -1
class CStack
{
private:
int8_t array[MAX_LEN];
int8_t stack1_range[2];
int8_t stack2_range[2];
int8_t stack3_range[2];
int8_t stack1_size;
int8_t stack2_size;
int8_t stack3_size;
int8_t stack1_cursize;
int8_t stack2_cursize;
int8_t stack3_cursize;
int8_t stack1_curpos;
int8_t stack2_curpos;
int8_t stack3_curpos;
public:
CStack();
~CStack();
void push(int8_t data);
void pop();
void print();
};
CStack::CStack()
{
stack1_range[LOWER] = 0;
stack1_range[UPPER] = MAX_LEN/3 - 1;
stack2_range[LOWER] = MAX_LEN/3;
stack2_range[UPPER] = (2 * (MAX_LEN/3)) - 1;
stack3_range[LOWER] = 2 * (MAX_LEN/3);
stack3_range[UPPER] = MAX_LEN - 1;
stack1_size = stack1_range[UPPER] - stack1_range[LOWER];
stack2_size = stack2_range[UPPER] - stack2_range[LOWER];
stack3_size = stack3_range[UPPER] - stack3_range[LOWER];
stack1_cursize = stack1_size;
stack2_cursize = stack2_size;
stack3_cursize = stack3_size;
stack1_curpos = stack1_cursize;
stack2_curpos = stack2_cursize;
stack3_curpos = stack3_cursize;
}
CStack::~CStack()
{
}
void CStack::push(int8_t data)
{
if(stack3_cursize != FULL)
{
array[stack3_range[LOWER] + stack3_curpos--] = data;
stack3_cursize--;
} else if(stack2_cursize != FULL) {
array[stack2_range[LOWER] + stack2_curpos--] = data;
stack2_cursize--;
} else if(stack1_cursize != FULL) {
array[stack1_range[LOWER] + stack1_curpos--] = data;
stack1_cursize--;
} else {
std::cout<<"\tstack is full...!"<<std::endl;
}
}
void CStack::pop()
{
std::cout<<"popping now..."<<std::endl;
if(stack1_cursize < stack1_size)
{
array[stack1_range[LOWER] + ++stack1_curpos] = 0;
stack1_cursize++;
} else if(stack2_cursize < stack2_size) {
array[stack2_range[LOWER] + ++stack2_curpos] = 0;
stack2_cursize++;
} else if(stack3_cursize < stack3_size) {
array[stack3_range[LOWER] + ++stack3_curpos] = 0;
stack3_cursize++;
} else {
std::cout<<"\tstack is empty...!"<<std::endl;
}
}
void CStack::print()
{
std::cout<<"array contents: ";
for(int8_t i = stack1_range[LOWER] + stack1_curpos + 1; i <= stack1_range[UPPER]; i++)
std::cout<<" "<<static_cast<int>(array[i]);
for(int8_t i = stack2_range[LOWER] + stack2_curpos + 1; i <= stack2_range[UPPER]; i++)
std::cout<<" "<<static_cast<int>(array[i]);
for(int8_t i = stack3_range[LOWER] + stack3_curpos + 1; i <= stack3_range[UPPER]; i++)
std::cout<<" "<<static_cast<int>(array[i]);
std::cout<<"\n";
}
int main()
{
CStack stack;
std::cout<<"adding: ";
for(uint8_t i = 1; i < 10; i++)
{
std::cout<<" "<<static_cast<int>(i);
stack.push(i);
}
std::cout<<"\n";
stack.print();
stack.pop();
stack.print();
return 0;
}
Perhaps you can implement N number of stacks or queues in the single array. My defination of using single array is that we are using single array to store all the data of all the stacks and queues in the single array, anyhow we can use other N array to keep track of indices of all elements of particular stack or queue.
solution :
store data sequentially to in the array during the time of insertion in any of the stack or queue. and store it's respective index to the index keeping array of that particular stack or queue.
for eg : you have 3 stacks (s1, s2, s3) and you want to implement this using a single array (dataArray[]). Hence we will make 3 other arrays (a1[], a2[], a3[]) for s1, s2 and s3 respectively which will keep track of all of their elements in dataArray[] by saving their respective index.
insert(s1, 10) at dataArray[0] a1[0] = 0;
insert(s2, 20) at dataArray[1] a2[0] = 1;
insert(s3, 30) at dataArray[2] a3[0] = 2;
insert(s1, 40) at dataArray[3] a1[1] = 3;
insert(s3, 50) at dataArray[4] a3[1] = 4;
insert(s3, 60) at dataArray[5] a3[2] = 5;
insert(s2, 30) at dataArray[6] a2[1] = 6;
and so on ...
now we will perform operation in dataArray[] by using a1, a2 and a3 for respective stacks and queues.
to pop an element from s1
return a1[0]
shift all elements to left
do similar approach for other operations too and you can implement any number of stacks and queues in the single array.