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Most efficient way to fill a vector from back to front
(2 answers)
Is it possible to collect an iterator to populate a collection backwards?
(2 answers)
Closed 3 years ago.
I've got a vector:
let v = Vec<T>::with_capacity(10);
I want to populate it with data, in reverse order:
let i = 10;
while i > 0 {
i -= 1;
v[i] = non_pure_function(i);
}
Unfortunately, this panics; when allocating with Vec::with_capacity, the actual initial length of the vector is still 0.
How can I best accomplish this? What if I have no constructor for T?
How can I Populate a vector in reverse?
You can do it with VecDeque:
use std::collections::VecDeque;
fn main() {
let mut v = VecDeque::new();
for i in 0..5 {
v.push_front(i);
}
for item in v {
print!("{}", item);
}
}
Playground
Instead of using VecDeque you can solve this problem with using array and the iterator:
Declare an array containing Option types.
Set array cells with the indexes.
Iterate over your array.
Filter your array to eliminate None values
Map the unwrapped values
Collect them into vector.
fn main() {
let mut v = [None; 10];
let mut i = 5i32;
while i > 0 {
i -= 1;
v[i as usize] = Some(4i32 - i);
}
let reverse_populated: Vec<i32> = v
.iter()
.filter(|x| x.is_some())
.map(|x| x.unwrap())
.collect();
for item in reverse_populated {
print!("{:?}", item);
}
}
Playground
Both output will be: 43210
Related
I have got a code that generates all possible correct strings of balanced brackets. So if the input is n = 4 there should be 4 brackets in the string and thus the answers the code will give are: {}{} and
{{}}.
Now, what I would like to do is print the number of possible strings. For example, for n = 4 the outcome would be 2.
Given my code, is this possible and how would I make that happen?
Just introduce a counter.
// Change prototype to return the counter
int findBalanced(int p,int n,int o,int c)
{
static char str[100];
// The counter
static int count = 0;
if (c == n) {
// Increment it on every printout
count ++;
printf("%s\n", str);
// Just return zero. This is not used anyway and will give
// Correct result for n=0
return 0;
} else {
if (o > c) {
str[p] = ')';
findBalanced(p + 1, n, o, c + 1);
}
if (o < n) {
str[p] = '(';
findBalanced(p + 1, n, o + 1, c);
}
}
// Return it
return count;
}
What you're looking for is the n-th Catalan number. You'll need to implement binomial coefficient to calculate it, but that's pretty much it.
Given matrices A and B the tropical product is defined to be the usual matrix product with multiplication traded out for addition and addition traded out for minimum. That is, it returns a new matrix C such that,
C_ij = minimum(A_ij, B_ij, A_i1 + B_1j, A_i2 + B_12,..., A_im + B_mj)
Given the underlying adjacency matrix A_g of a graph g, the nth "power" with respect to the tropical product represents the connections between nodes reachable in at most n steps. That is, C_ij = (A**n)_ij has value m if nodes i and j are separated by m<=n edges.
In general, given some graph with N nodes. The diameter of the graph can only be at most N; and, given a graph with diameter k, A**n = A**k for all n>k and the matrix D_ij = A**k is called the "distance matrix" entries representing the distances between all nodes in the graph.
I have written a tropical product function in chapel and I want to write a function that takes an adjacency matrix and returns the resulting distance matrix. I have tried the following approaches to no avail. Guidance in getting past these errors would be greatly appreciated!
proc tropicLimit(A:[] real,B:[] real) {
var R = tropic(A,B);
if A == R {
return A;
} else {
tropicLimit(R,B);
}
}
which threw a domain mismatch error so I made the following edit:
proc tropicLimit(A:[] real,B:[] real) {
var R = tropic(A,B);
if A.domain == R.domain {
if && reduce (A == R) {
return R;
} else {
tropicLimit(R,B);
}
} else {
tropicLimit(R,B);
}
}
which throws
src/MatrixOps.chpl:602: error: control reaches end of function that returns a value
proc tropicLimit(A:[] real,B:[] real) {
var R = tropic(A,B);
if A.domain == R.domain {
if && reduce (A == R) { // Line 605 is this one
} else {
tropicLimit(R,B);
}
} else {
tropicLimit(R,B);
}
return R;
}
Brings me back to this error
src/MatrixOps.chpl:605: error: halt reached - Sparse arrays can't be zippered with anything other than their domains and sibling arrays (CS layout)
I also tried using a for loop with a break condition but that didn't work either
proc tropicLimit(B:[] real) {
var R = tropic(B,B);
for n in B.domain.dim(2) {
var S = tropic(R,B);
if S.domain != R.domain {
R = S; // Intended to just reassign the handle "R" to the contents of "S" i.o.w. destructive update of R
} else {
break;
}
}
return R;
}
Any suggestions?
src/MatrixOps.chpl:605: error: halt reached - Sparse arrays can't be zippered with anything other than their domains and sibling arrays (CS layout)
I believe you are encountering a limitation of zippering sparse arrays in the current implementation, documented in #6577.
Removing some unknowns from the equation, I believe this distilled code snippet demonstrates the issue you are encountering:
use LayoutCS;
var dom = {1..10, 1..10};
var Adom: sparse subdomain(dom) dmapped CS();
var Bdom: sparse subdomain(dom) dmapped CS();
var A: [Adom] real;
var B: [Bdom] real;
Adom += (1,1);
Bdom += (1,1);
A[1,1] = 1.0;
B[1,1] = 2.0;
writeln(A.domain == B.domain); // true
var willThisWork = && reduce (A == B);
// dang.chpl:19: error: halt reached - Sparse arrays can't be zippered with
// anything other than their domains and sibling arrays (CS layout)
As a work-around, I would suggest looping over the sparse indices after confirming the domains are equal and performing a && reduce. This is something you could wrap in a helper function, e.g.
proc main() {
var dom = {1..10, 1..10};
var Adom: sparse subdomain(dom) dmapped CS();
var Bdom: sparse subdomain(dom) dmapped CS();
var A: [Adom] real;
var B: [Bdom] real;
Adom += (1,1);
Bdom += (1,1);
A[1,1] = 1.0;
B[1,1] = 2.0;
if A.domain == B.domain {
writeln(equal(A, B));
}
}
/* Some day, this should be A.equals(B) ! */
proc equal(A: [], B: []) {
// You could also return 'false' if domains do not match
assert(A.domain == B.domain);
var s = true;
forall (i,j) in A.domain with (&& reduce s) {
s &&= (A[i,j] == B[i,j]);
}
return s;
}
src/MatrixOps.chpl:602: error: control reaches end of function that returns a value
This error is a result of not returning something in every condition. I believe you intended to do:
proc tropicLimit(A:[] real,B:[] real) {
var R = tropic(A,B);
if A.domain == R.domain {
if && reduce (A == R) {
return R;
} else {
return tropicLimit(R,B);
}
} else {
return tropicLimit(R,B);
}
}
I was under the impression that the assignment to entry in nil map error would only happen if we would want to assign to a double map, that is, when a map on a deeper level is trying to be assigned while the higher one doesn't exist, e.g.:
var mm map[int]map[int]int
mm[1][2] = 3
But it also happens for a simple map (though with struct as a key):
package main
import "fmt"
type COO struct {
x int
y int
}
var neighbours map[COO][]COO
func main() {
for i := 0; i < 30; i++ {
for j := 0; j < 20; j++ {
var buds []COO
if i < 29 {
buds = append(buds, COO{x: i + 1, y: j})
}
if i > 0 {
buds = append(buds, COO{x: i - 1, y: j})
}
if j < 19 {
buds = append(buds, COO{x: i, y: j + 1})
}
if j > 0 {
buds = append(buds, COO{x: i, y: j - 1})
}
neighbours[COO{x: i, y: j}] = buds // <--- yields error
}
}
fmt.Println(neighbours)
}
What could be wrong?
You need to initialize neighbours: var neighbours = make(map[COO][]COO)
See the second section in: https://blog.golang.org/go-maps-in-action
You'll get a panic whenever you try to insert a value into a map that hasn't been initialized.
In Golang, everything is initialized to a zero value, it's the default value for uninitialized variables.
So, as it has been conceived, a map's zero value is nil. When trying to use an non-initialized map, it panics. (Kind of a null pointer exception)
Sometimes it can be useful, because if you know the zero value of something you don't have to initialize it explicitly:
var str string
str += "42"
fmt.Println(str)
// 42 ; A string zero value is ""
var i int
i++
fmt.Println(i)
// 1 ; An int zero value is 0
var b bool
b = !b
fmt.Println(b)
// true ; A bool zero value is false
If you have a Java background, that's the same thing: primitive types have a default value and objects are initialized to null;
Now, for more complex types like chan and map, the zero value is nil, that's why you have to use make to instantiate them. Pointers also have a nil zero value. The case of arrays and slice is a bit more tricky:
var a [2]int
fmt.Println(a)
// [0 0]
var b []int
fmt.Println(b)
// [] ; initialized to an empty slice
The compiler knows the length of the array (it cannot be changed) and its type, so it can already instantiate the right amount of memory. All of the values are initialized to their zero value (unlike C where you can have anything inside your array). For the slice, it is initialized to the empty slice [], so you can use append normally.
Now, for structs, it is the same as for arrays. Go creates a struct with all its fields initialized to zero values. It makes a deep initialization, example here:
type Point struct {
x int
y int
}
type Line struct {
a Point
b Point
}
func main() {
var line Line
// the %#v format prints Golang's deep representation of a value
fmt.Printf("%#v\n", line)
}
// main.Line{a:main.Point{x:0, y:0}, b:main.Point{x:0, y:0}}
Finally, the interface and func types are also initialized to nil.
That's really all there is to it. When working with complex types, you just have to remember to initialize them. The only exception is for arrays because you can't do make([2]int).
In your case, you have map of slice, so you need at least two steps to put something inside: Initialize the nested slice, and initialize the first map:
var buds []COO
neighbours := make(map[COO][]COO)
neighbours[COO{}] = buds
// alternative (shorter)
neighbours := make(map[COO][]COO)
// You have to use equal here because the type of neighbours[0] is known
neighbours[COO{}] = make([]COO, 0)
I am working on creating a struct that represents a Matrix with methods to manipulate the data within the type. There are two methods that I use as an example to set a single row or column to a specific value. Here is a snippet of my code:
type Matrix struct {
Height, Width int
data [][]int
}
func NewMatrix(nrows, ncols int) (mat *Matrix) {
mat = new(Matrix)
mat.Height = nrows
mat.Width = ncols
mat.data = make([][]int, nrows)
for i := range mat.data {
mat.data[i] = make([]int, ncols)
for j := range mat.data[i]{
mat.data[i][j] = 0
}
}
return
}
func (mat *Matrix) SetCol(col, val int) {
for i := 0; i < mat.Height; i++ {
mat.data[i][col] = val
}
}
func (mat *Matrix) SetRow(row, val int) {
for i := 0; i < mat.Width; i++ {
mat.data[row][i] = val
}
}
When I use this Matrix type and manipulating the data attribute like so:
mat := NewMatrix(33,33)
mat.SetCol(2, 3)
mat.SetRow(2, 3)
I am finding that the data attribute of the Matrix instance is being set within the method SetCol but once it returns from this method the data appears to be the empty matrix that I initialized it to.
Why is the data attribute that I am manipulating in the method not persisting past the lifetime of the method call? How can I fix this?
Edit
I found out that the issue was that I was instantiating a new instance of a Matrix on each iteration in a loop which is why the matrix always appeared to be empty after I manipulated it with SetCol and SetRow. So the question is not really valid.
Project Euler problem 14:
The following iterative sequence is
defined for the set of positive
integers:
n → n/2 (n is even) n → 3n + 1 (n is
odd)
Using the rule above and starting with
13, we generate the following
sequence: 13 → 40 → 20 → 10 → 5 → 16 →
8 → 4 → 2 → 1
It can be seen that this sequence
(starting at 13 and finishing at 1)
contains 10 terms. Although it has not
been proved yet (Collatz Problem), it
is thought that all starting numbers
finish at 1.
Which starting number, under one
million, produces the longest chain?
My first instinct is to create a function to calculate the chains, and run it with every number between 1 and 1 million. Obviously, that takes a long time. Way longer than solving this should take, according to Project Euler's "About" page. I've found several problems on Project Euler that involve large groups of numbers that a program running for hours didn't finish. Clearly, I'm doing something wrong.
How can I handle large groups of numbers quickly?
What am I missing here?
Have a read about memoization. The key insight is that if you've got a sequence starting A that has length 1001, and then you get a sequence B that produces an A, you don't to repeat all that work again.
This is the code in Mathematica, using memoization and recursion. Just four lines :)
f[x_] := f[x] = If[x == 1, 1, 1 + f[If[EvenQ[x], x/2, (3 x + 1)]]];
Block[{$RecursionLimit = 1000, a = 0, j},
Do[If[a < f[i], a = f[i]; j = i], {i, Reverse#Range#10^6}];
Print#a; Print[j];
]
Output .... chain length´525´ and the number is ... ohhhh ... font too small ! :)
BTW, here you can see a plot of the frequency for each chain length
Starting with 1,000,000, generate the chain. Keep track of each number that was generated in the chain, as you know for sure that their chain is smaller than the chain for the starting number. Once you reach 1, store the starting number along with its chain length. Take the next biggest number that has not being generated before, and repeat the process.
This will give you the list of numbers and chain length. Take the greatest chain length, and that's your answer.
I'll make some code to clarify.
public static long nextInChain(long n) {
if (n==1) return 1;
if (n%2==0) {
return n/2;
} else {
return (3 * n) + 1;
}
}
public static void main(String[] args) {
long iniTime=System.currentTimeMillis();
HashSet<Long> numbers=new HashSet<Long>();
HashMap<Long,Long> lenghts=new HashMap<Long, Long>();
long currentTry=1000000l;
int i=0;
do {
doTry(currentTry,numbers, lenghts);
currentTry=findNext(currentTry,numbers);
i++;
} while (currentTry!=0);
Set<Long> longs = lenghts.keySet();
long max=0;
long key=0;
for (Long aLong : longs) {
if (max < lenghts.get(aLong)) {
key = aLong;
max = lenghts.get(aLong);
}
}
System.out.println("number = " + key);
System.out.println("chain lenght = " + max);
System.out.println("Elapsed = " + ((System.currentTimeMillis()-iniTime)/1000));
}
private static long findNext(long currentTry, HashSet<Long> numbers) {
for(currentTry=currentTry-1;currentTry>=0;currentTry--) {
if (!numbers.contains(currentTry)) return currentTry;
}
return 0;
}
private static void doTry(Long tryNumber,HashSet<Long> numbers, HashMap<Long, Long> lenghts) {
long i=1;
long n=tryNumber;
do {
numbers.add(n);
n=nextInChain(n);
i++;
} while (n!=1);
lenghts.put(tryNumber,i);
}
Suppose you have a function CalcDistance(i) that calculates the "distance" to 1. For instance, CalcDistance(1) == 0 and CalcDistance(13) == 9. Here is a naive recursive implementation of this function (in C#):
public static int CalcDistance(long i)
{
if (i == 1)
return 0;
return (i % 2 == 0) ? CalcDistance(i / 2) + 1 : CalcDistance(3 * i + 1) + 1;
}
The problem is that this function has to calculate the distance of many numbers over and over again. You can make it a little bit smarter (and a lot faster) by giving it a memory. For instance, lets create a static array that can store the distance for the first million numbers:
static int[] list = new int[1000000];
We prefill each value in the list with -1 to indicate that the value for that position is not yet calculated. After this, we can optimize the CalcDistance() function:
public static int CalcDistance(long i)
{
if (i == 1)
return 0;
if (i >= 1000000)
return (i % 2 == 0) ? CalcDistance(i / 2) + 1 : CalcDistance(3 * i + 1) + 1;
if (list[i] == -1)
list[i] = (i % 2 == 0) ? CalcDistance(i / 2) + 1: CalcDistance(3 * i + 1) + 1;
return list[i];
}
If i >= 1000000, then we cannot use our list, so we must always calculate it. If i < 1000000, then we check if the value is in the list. If not, we calculate it first and store it in the list. Otherwise we just return the value from the list. With this code, it took about ~120ms to process all million numbers.
This is a very simple example of memoization. I use a simple list to store intermediate values in this example. You can use more advanced data structures like hashtables, vectors or graphs when appropriate.
Minimize how many levels deep your loops are, and use an efficient data structure such as IList or IDictionary, that can auto-resize itself when it needs to expand. If you use plain arrays they need to be copied to larger arrays as they expand - not nearly as efficient.
This variant doesn't use an HashMap but tries only to not repeat the first 1000000 numbers. I don't use an hashmap because the biggest number found is around 56 billions, and an hash map could crash.
I have already done some premature optimization. Instead of / I use >>, instead of % I use &. Instead of * I use some +.
void Main()
{
var elements = new bool[1000000];
int longestStart = -1;
int longestRun = -1;
long biggest = 0;
for (int i = elements.Length - 1; i >= 1; i--) {
if (elements[i]) {
continue;
}
elements[i] = true;
int currentStart = i;
int currentRun = 1;
long current = i;
while (current != 1) {
if (current > biggest) {
biggest = current;
}
if ((current & 1) == 0) {
current = current >> 1;
} else {
current = current + current + current + 1;
}
currentRun++;
if (current < elements.Length) {
elements[current] = true;
}
}
if (currentRun > longestRun) {
longestStart = i;
longestRun = currentRun;
}
}
Console.WriteLine("Longest Start: {0}, Run {1}", longestStart, longestRun);
Console.WriteLine("Biggest number: {0}", biggest);
}