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);
}
}
Related
How does this code work? (leetcode 95 question) I don't understand how the 2 recursions work inside the for loop. Does the 2nd inner for loop end when the recursive function returns NULL? Or would it continue executing the 3rd inner for loop?
class Solution {
public:
vector<TreeNode*> generateTrees(int n) {
if(n == 0) {
return {};
}
vector<TreeNode*> ans = generateT(1,n);
return ans;
}
vector<TreeNode*> generateT(int l, int r) {
if(l > r) return {nullptr};
vector<TreeNode*> ans;
for(int i=l; i <= r; ++i) {
for(TreeNode*left: generateT(l, i-1)) {
for(TreeNode* right:generateT(i+1, r)) {
ans.push_back(new TreeNode(i));
ans.back()->left = left;
ans.back()->right = right;
}
}
}
return ans;
}
};
Problem statement:
Given an integer n, return all the structurally unique BST's (binary search trees), which has exactly n nodes of unique values from 1 to n. Return the answer in any order.
Does the 2nd inner for loop end when the recursive function returns NULL?
No. The recursive function is not returing NULL, it is returning vector of nullptr.
Or would it continue executing the 3rd inner for loop?
Of course, it will.
How does this code work? I don't understand how the 2 recursions work inside the loop.
I suppose the following snippet is the cause of confusion, so commented the case when nullptr provided by outer loop.
vector<TreeNode*> generateT(int l, int r) {
if(l > r) return { nullptr };
vector<TreeNode*> ans;
for ( int i = l; i <= r; i++ ) {
// if l = 0, i = 0
for ( TreeNode* left :generateT(l, i-1) ) // if l = 0, i = -1, returns { nullptr } (vector of nullptr)
for (TreeNode* right :generateT(i+1, r)) { // now this snippet will execute
auto node = new TreeNode(i);
ans.push_back(node);
node->left = left; // the nullptr we have from the outer loop, will provide null value for this
node->right = right;
}
}
return ans;
}
Visually, for a combination of node where,
a
\
b
/ \
null c
/
null
the above pattern occurs the provided { nullptr } from outer loop will come in handy setting left node.
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.
I'm working on binary search tree homework and am asked to convert a recursive method to an iterative method. Here is the recursive method and below that is my iterative method. This method should return node containing the kth key. My method keeps giving me a NullPointerException and I'm not sure why. Thank you.
Provided code:
public Key select(int k) {
Node node = select(root, k);
if (node==null) {
return null;
} else {
return node.key;
}
}
// Return Node containing kth key (zero based)
private Node select(Node node, int k) {
if (node == null) return null;
int t = size(node.left);
if (t > k)
return select(node.left, k);
else if (t < k)
return select(node.right, k - t - 1);
else
return node;
}
My Code:
public Key selectI(int k) {
return selectI(root, k);
}
private Key selectI(Node node, int k) {
Node curr = node;
while (curr != null) {
int t = size(node.left);
if (t > k) {
curr = node.left;
} else if (t < k) {
curr = node.right;
k = (k - (t - 1));
} else
return curr.key;
}
return null;
}
The problem seems to be that you are not updating the value for k. This is normally done recursively, but you have to do it mathematically if you are going to make an iterative function. When you pass to the left (t > k) you continue searching for the node with the size of k. When you pass to the right (t < k) you are searching for a node with size k = (k - (t - 1)). Eventually t and k will either be equal or zero, in which case you've found the node you're looking for!
Also make sure that you are constantly updating the size of the current node you are looking at. You don't want to only look at the size of the tree, this ruins the mathematical relationship between your t and k values, which will cause the program to run until there are no more nodes to look at!
This is a bit more intricate than a simple left-recursion or tail-call recursion. So I'm wondering how I can eliminate this kind of recursion. I'm already keeping my own stack as you can see below, so the function needs to no params or return values. However, it's still calling itself up (or down) to a certain level and I want to turn this into a loop, but been scratching my head over this for some time now.
Here's the simplified test case, replacing all "real logic" with printf("dostuff at level #n") messages. This is in Go but the problem is applicable to most languages. Use of loops and goto's would be perfectly acceptable (but I played with this and it gets convoluted, out-of-hand and seemingly unworkable to begin with); however, additional helper functions should be avoided. I guess I should to turn this into some kind of simple state machine, but... which? ;)
As for the practicality, this is to run at about 20 million times per second (stack depth can range from 1 through 25 max later on). This is a case where maintaining my own stack is bound to be more stable / faster than the function call stack. (There are no other function calls in this function, only calculations.) Also, no garbage generated = no garbage collected.
So here goes:
func testRecursion () {
var root *TMyTreeNode = makeSomeDeepTreeStructure()
// rl: current recursion level
// ml: max recursion level
var rl, ml = 0, root.MaxDepth
// node: "the stack"
var node = make([]*TMyTreeNode, ml + 1)
// the recursive and the non-recursive / iterative test functions:
var walkNodeRec, walkNodeIt func ();
walkNodeIt = func () {
log.Panicf("YOUR ITERATIVE / NON-RECURSIVE IDEAS HERE")
}
walkNodeRec = func () {
log.Printf("ENTER LEVEL %v", rl)
if (node[rl].Level == ml) || (node[rl].ChildNodes == nil) {
log.Printf("EXIT LEVEL %v", rl)
return
}
log.Printf("PRE-STUFF LEVEL %v", rl)
for i := 0; i < 3; i++ {
switch i {
case 0:
log.Printf("PRECASE %v.%v", rl, i)
node[rl + 1] = node[rl].ChildNodes[rl + i]; rl++; walkNodeRec(); rl--
log.Printf("POSTCASE %v.%v", rl, i)
case 1:
log.Printf("PRECASE %v.%v", rl, i)
node[rl + 1] = node[rl].ChildNodes[rl + i]; rl++; walkNodeRec(); rl--
log.Printf("POSTCASE %v.%v", rl, i)
case 2:
log.Printf("PRECASE %v.%v", rl, i)
node[rl + 1] = node[rl].ChildNodes[rl + i]; rl++; walkNodeRec(); rl--
log.Printf("POSTCASE %v.%v", rl, i)
}
}
}
// test recursion for reference:
if true {
rl, node[0] = 0, root
log.Printf("\n\n=========>RECURSIVE ML=%v:", ml)
walkNodeRec()
}
// test non-recursion, output should be identical
if true {
rl, node[0] = 0, root
log.Printf("\n\n=========>ITERATIVE ML=%v:", ml)
walkNodeIt()
}
}
UPDATE -- after some discussion here, and further thinking:
I just made up the following pseudo-code which in theory should do what I need:
curLevel = 0
for {
cn = nextsibling(curLevel, coords)
lastnode[curlevel] = cn
if cn < 8 {
if isleaf {
process()
} else {
curLevel++
}
} else if curLevel == 0 {
break
} else {
curLevel--
}
}
Of course the tricky part will be filling out nextsibling() for my custom use-case. But just as a general solution to eliminating inner recursion while maintaining the depth-first traversal order I need, this rough outline should do so in some form or another.
I'm not really sure I understand what it is you want to do since your recursion code looks a little strange. However if I understand the structure of your TMyTreeNode then this is what I would do for a non recursive version.
// root is our root node
q := []*TMyTreeNode{root}
processed := make(map[*TMyTreeNode]bool
for {
l := len(q)
if l < 1 {
break // our queue is empty
}
curr := q[l - 1]
if !processed[curr] && len(curr.childNodes) > 0 {
// do something with curr
processed[curr] = true
q = append(q, curr.childNodes...)
continue // continue on down the tree.
} else {
// do something with curr
processed[curr] = true
q := q[:l-2] // pop current off the queue
}
}
NOTE: This will go arbitrarily deep into the structure. If that's not what you want it will need some modifications.
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);
}