I have a 2d array a[3][3] and the program reads 2 sets of IDs for 2 numbers on the array.
I need to print all the possible paths from one number to another.
So far, I know how to find how many paths exist every time:
scanf("%d %d",&ai,&aj);
scanf("%d %d",&bi,&bj);
distance_i = bi - ai;
if(distance_i<0){distance_i=distance_i*-1;}
distance_j = bj - aj;
if(distance_j<0){distance_j=ap_j*-1;}
path = 1+(distance_i*distance_j);
For example, if the array a is:
1 2 3
4 5 6
7 8 9
With
input_1: 0,0
input_2: 1,2
The output must be:
there are 3 possible paths:
a) 1,2,5,8
b) 1,4,5,8
c) 1,4,7,8
But I can't find a way to print them. Any ideas?
From location[v1][h1] to location[v2][h2]
Move kinds are: DOWN, RIGHT
The width is: (v2-v1) * DOWN
The height is: (h2-h2) * RIGHT
=> all path choice action list: [width, height] = [(v2-v1) * DOWN, (h2-h2) * RIGHT]
Example: from location[0][0] to location [2][1]
action list = [DOWN, DOWN, RIGHT]
all unique path choices are (It make minus the duplicate repeated permutation from a given list):
[DOWN, DOWN, RIGHT]
[DOWN, RIGHT, DOWN]
[RIGHT, DOWN, DOWN]
You will use backtracking (Depth-First Search) to find all possible routes.
See program test here http://ideone.com/GqWLa5
#define VALID(x) ((x) >= 0 && (x) < 3)
int arr[3][3];
// to detect previous visited cells and eliminate infinite recursion
short vis[3][3] = { 0 };
int xtar, ytar; // destination cell
int xsrc, ysrc; // source cell
// to move in directions: down, up, right, and left, respectively
const int dirx[] = { 0, 0, 1, -1 };
const int diry[] = { 1, -1, 0, 0 };
// temp buffer to print paths
// max size = size of arr + zero termination char
char tmp_path[3 * 3 + 1];
void rec(int x, int y, int idx) // idx is used to fill tmp_path
{
int i;
tmp_path[idx] = arr[y][x] + '0';
if (x == xtar && y == ytar) // basic case
{
tmp_path[idx + 1] = 0; // put zero char
printf("%s\n", tmp_path); // print path
return;
}
if (vis[y][x]) return; // already visited
vis[y][x] = 1; // otherwise, mark as visited
for (i = 0; i < 4; ++i) // for each of the 4 directions
if (VALID(y + diry[i]) && VALID(x + dirx[i]))
rec(x + dirx[i], y + diry[i], idx + 1);
vis[y][x] = 0; // reset visited so that can be visited again
}
main()
{
// input xtar, ytar, xsrc, ysrc, arr
rec(xsrc, ysrc, 0);
}
Related
What am I doing wrong here?
The prototypes aren't changeable.
I need to write both of the functions and this is what I wrote.
They work at 90% of the times. The iterative way got an issue when i'm trying to search for the middle element.
And the recursive way got some issue but I don't really know what's wrong. Sometimes it finds the element and other times not.
One more thing I cannot change it's that I must avoid checking inside of the function whether array[middle] == key. The teacher wants us to do this check only when I exit the loop, and he wants the loop to only check if should I go right or left.
int *BinarySearchIter(const int SortedArray[], int key, size_t length)
{
/* variables that will set the boundries of the searched area */
int left_index = 0, right_index = 0, middle = 0;
int *result = NULL;
assert(SortedArray);
assert(length);
left_index = 0; /* first elemenet in the array */
right_index = length - 1; /* last elemenet in the array */
/* while only one elemenet is left in the searched area */
while (left_index <= right_index)
{
/* split it to half */
middle = (left_index + right_index) / 2;
/* if key greater, ignore left half, search only in right half */
if (SortedArray[middle] < key)
{
left_index = middle + 1;
}
/* if key smaller, ignore right half, search only in left half */
else
{
right_index = middle - 1;
}
}
/* if we reach here, then element is the key or was not found */
result = (int *)SortedArray + middle;
return (key == *result ? result : NULL);
}
/******************************************************************************/
int *BinarySearchRec(const int SortedArray[], int key, size_t length)
{
int left_index = 0; /* first elemenet of the array */
int right_index = length - 1; /* last elemenet in the array */
int middle = 0, isBigger = 0;
if (1 == length)
{
return (key == *SortedArray ? (int *)SortedArray : NULL);
}
middle = (left_index + right_index) / 2;
isBigger = (key > SortedArray[middle]);
return (BinarySearchRec(SortedArray+(middle + 1)*isBigger, key, isBigger * (length - length /2 ) + !isBigger*(length/2)));
}
/******************************************************************************/
I was trying to run the following test;
const int sorted_arr[] = { 2, 4, 8, 10, 12};
size_t arr_length = sizeof(sorted_arr) / sizeof(sorted_arr[0]), i = 0;
int key_to_find = 0;
int *iterative_res = NULL;
int *recursive_res = NULL;
for (i = 0; i < arr_length; ++i)
{
key_to_find = sorted_arr[i];
iterative_res = BinarySearchIter(sorted_arr, key_to_find, arr_length);
recursive_res = BinarySearchRec(sorted_arr, key_to_find, arr_length);
Print if any errors (nulls or key doesn't match any of the results)
}
And this is the output of the test:
ERRORS:
Needs to find: 8
But found:
Iterative: NULL (failure)
Recursive: NULL (failure)
Needs to find: 12
But found:
Iterative: 12 (success)
Recursive: NULL (failure)
For Iterative Function
Let's think what your code is doing. You have an array consists 5 elements and let's say you are searching for 8.
2 4 8 10 12
^
In the first iteration the values are like this:
left_index = 0, right_index = 4, middle_index = 2
In the if statement the program checks for (SortedArray[middle_index] < key) and that is wrong because 8 is not bigger than 8. So it executes the else statement. The mistake is here.
In the else statement you are discarding the middle element. But in this case middle element is the key that you are looking for.
So first thing that you need to change just like #Eric Postpischil said is changing your else statement to this:
else {
right_index = middle;
}
Let's continue with second iteration:
2 4 8
^
left_index = 0, right_index = 2, middle_index = 1
After the second iteration, third iteration is only going to consist 1 element.
8
left_index = 2, right_index = 2, middle_index = 2
At this point the program stucks in the infinite loop because it always executes the else statement. Left and right indexes always stay same.
So the second change that you need to do is changing your while condition to this:
while (left_index < right_index)
At this point when the left index and right index are equal, the loop will break. Than the last thing that you need to do is updating your middle index again. Or you can use left_index or right_index (which one is doesn't matter they are equal).
At the end, your function should look like this:
const int *BinarySearchIter(const int SortedArray[], int key, size_t length) {
int left_index = 0, right_index = length - 1;
while (left_index < right_index) {
int middle = (left_index + right_index) / 2;
if (SortedArray[middle] < key) {
left_index = middle + 1;
} else {
right_index = middle;
}
}
return SortedArray[right_index] == key ? SortedArray + right_index : NULL;
}
For Recursive Function
Only thing you need to change is this:
return BinarySearchRec(SortedArray+(middle + 1)*isBigger, key, (length / 2) + !isBigger * (length % 2));
The reasons are same as iterative function. You need to include the middle element. For lower part of the array, the array length needs to be equal to ceil(length / 2), and for upper part of the array, it needs to be equal to floor(length / 2).
You can also do this:
const int *BinarySearchRec(const int SortedArray[], int key, size_t length) {
if (1 == length) return (key == *SortedArray ? SortedArray : NULL);
int middle = (length + 1) / 2;
if (key >= SortedArray[middle]) return BinarySearchRec(SortedArray + middle, key, length / 2);
else return BinarySearchRec(SortedArray, key, middle);
}
I have removed all the storylines for this question.
Q. You are given N numbers. You have to find 2 equal sum sub-sequences, with maximum sum. You don't necessarily need to use all numbers.
Eg 1:-
5
1 2 3 4 1
Sub-sequence 1 : 2 3 // sum = 5
Sub-sequence 2 : 4 1 // sum = 5
Possible Sub-sequences with equal sum are
{1,2} {3} // sum = 3
{1,3} {4} // sum = 4
{2,3} {4,1} // sum = 5
Out of which 5 is the maximum sum.
Eg 2:-
6
1 2 4 5 9 1
Sub-sequence 1 : 2 4 5 // sum = 11
Sub-sequence 2 : 1 9 1 // sum = 11
The maximum sum you can get is 11
Constraints:
5 <= N <= 50
1<= number <=1000
sum of all numbers is <= 1000
Important: Only <iostream> can be used. No STLs.
N numbers are unsorted.
If array is not possible to split, print 0.
Number of function stacks is limited. ie your recursive/memoization solution won't work.
Approach 1:
I tried a recursive approach something like the below:
#include <iostream>
using namespace std;
bool visited[51][1001][1001];
int arr[51];
int max_height=0;
int max_height_idx=0;
int N;
void recurse( int idx, int sum_left, int sum_right){
if(sum_left == sum_right){
if(sum_left > max_height){
max_height = sum_left;
max_height_idx = idx;
}
}
if(idx>N-1)return ;
if(visited[idx][sum_left][sum_right]) return ;
recurse( idx+1, sum_left+arr[idx], sum_right);
recurse( idx+1, sum_left , sum_right+arr[idx]);
recurse( idx+1, sum_left , sum_right);
visited[idx][sum_left][sum_right]=true;
/*
We could reduce the function calls, by check the visited condition before calling the function.
This could reduce stack allocations for function calls. For simplicity I have not checking those conditions before function calls.
Anyways, this recursive solution would get time out. No matter how you optimize it.
Btw, there are T testcases. For simplicity, removed that constraint.
*/
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cin>>N;
for(int i=0; i<N; i++)
cin>>arr[i];
recurse(0,0,0);
cout<< max_height <<"\n";
}
NOTE: Passes test-cases. But time out.
Approach 2:
I also tried, taking advantage of constraints.
Every number has 3 possible choice:
1. Be in sub-sequence 1
2. Be in sub-sequence 2
3. Be in neither of these sub-sequences
So
1. Be in sub-sequence 1 -> sum + 1*number
2. Be in sub-sequence 2 -> sum + -1*number
3. None -> sum
Maximum sum is in range -1000 to 1000.
So dp[51][2002] could be used to save the maximum positive sum achieved so far (ie till idx).
CODE:
#include <iostream>
using namespace std;
int arr[51];
int N;
int dp[51][2002];
int max3(int a, int b, int c){
return max(a,max(b,c));
}
int max4(int a, int b, int c, int d){
return max(max(a,b),max(c,d));
}
int recurse( int idx, int sum){
if(sum==0){
// should i perform anything here?
}
if(idx>N-1){
return 0;
}
if( dp[idx][sum+1000] ){
return dp[idx][sum+1000];
}
return dp[idx][sum+1000] = max3 (
arr[idx] + recurse( idx+1, sum + arr[idx]),
0 + recurse( idx+1, sum - arr[idx]),
0 + recurse( idx+1, sum )
) ;
/*
This gives me a wrong output.
4
1 3 5 4
*/
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cin>>N;
for(int i=0; i<N; i++)
cin>>arr[i];
cout<< recurse(0,0) <<"\n";
}
The above code gives me wrong answer. Kindly help me with solving/correcting this memoization.
Also open to iterative approach for the same.
Idea of your second approach is correct, it's basically a reduction to the knapsack problem. However, it looks like your code lacks clear contract: what the recurse function is supposed to do.
Here is my suggestion: int recurse(int idx, int sum) distributes elements on positions idx..n-1 into three multisets A, B, C such that sum+sum(A)-sum(B)=0 and returns maximal possible sum(A), -inf otherwise (here -inf is some hardcoded constant which serves as a "marker" of no answer; there are some restrictions on it, I suggest -inf == -1000).
Now you're to write a recursive backtracking using that contract and then add memoization. Voila—you've got a dynamic programming solution.
In recursive backtracking we have two distinct situations:
There are no more elements to distribute, no choices to make: idx == n. In that case, we should check that our condition holds (sum + sum(A) - sum(B) == 0, i.e. sum == 0) and return the answer. If sum == 0, then the answer is 0. However, if sum != 0, then there is no answer and we should return something which will never be chosen as the answer, unless there are no answer for the whole problem. As we modify returning value of recurse and do not want extra ifs, it cannot be simply zero or even -1; it should be a number which, when modified, still remains "the worst answer ever". The biggest modification we can make is to add all numbers to the resulting value, hence we should choose something less or equal to negative maximal sum of numbers (i.e. -1000), as existing answers are always strictly positive, and that fictive answer will always be non-positive.
There is at least one remaining element which should be distributed to either A, B or C. Make the choice and choose the best answer among three options. Answers are calculated recursively.
Here is my implementation:
const int MAXN = 50;
const int MAXSUM = 1000;
bool visited[MAXN + 1][2 * MAXSUM + 1]; // should be filled with false
int dp[MAXN + 1][2 * MAXSUM + 1]; // initial values do not matter
int recurse(int idx, int sum){
// Memoization.
if (visited[idx][sum + MAXSUM]) {
return dp[idx][sum + MAXSUM];
}
// Mark the current state as visited in the beginning,
// it's ok to do before actually computing it as we're
// not expect to visit it while computing.
visited[idx][sum + MAXSUM] = true;
int &answer = dp[idx][sum + MAXSUM];
// Backtracking search follows.
answer = -MAXSUM; // "Answer does not exist" marker.
if (idx == N) {
// No more choices to make.
if (sum == 0) {
answer = 0; // Answer exists.
} else {
// Do nothing, there is no answer.
}
} else {
// Option 1. Current elemnt goes to A.
answer = max(answer, arr[idx] + recurse(idx + 1, sum + arr[idx]));
// Option 2. Current element goes to B.
answer = max(answer, recurse(idx + 1, sum - arr[idx]));
// Option 3. Current element goes to C.
answer = max(answer, recurse(idx + 1, sum));
}
return answer;
}
Here is a simple dynamic programming based solution for anyone interested, based on the idea suggested by Codeforces user lemelisk here. Complete post here. I haven't tested this code completely though.
#include <iostream>
using namespace std;
#define MAXN 20 // maximum length of array
#define MAXSUM 500 // maximum sum of all elements in array
#define DIFFSIZE (2*MAXSUM + 9) // possible size of differences array (-maxsum, maxsum) + some extra
int dp[MAXN][DIFFSIZE] = { 0 };
int visited[DIFFSIZE] = { 0 }; // visited[diff] == 1 if the difference 'diff' can be reached
int offset = MAXSUM + 1; // offset so that indices in dp table don't become negative
// 'diff' replaced by 'offset + diff' below everywhere
int max(int a, int b) {
return (a > b) ? a : b;
}
int max_3(int a, int b, int c) {
return max(a, max(b, c));
}
int main() {
int a[] = { 1, 2, 3, 4, 6, 7, 5};
int n = sizeof(a) / sizeof(a[0]);
int *arr = new int[n + 1];
int sum = 0;
for (int i = 1; i <= n; i++) {
arr[i] = a[i - 1]; // 'arr' same as 'a' but with 1-indexing for simplicity
sum += arr[i];
} // 'sum' holds sum of all elements of array
for (int i = 0; i < MAXN; i++) {
for (int j = 0; j < DIFFSIZE; j++)
dp[i][j] = INT_MIN;
}
/*
dp[i][j] signifies the maximum value X that can be reached till index 'i' in array such that diff between the two sets is 'j'
In other words, the highest sum subsets reached till index 'i' have the sums {X , X + diff}
See http://codeforces.com/blog/entry/54259 for details
*/
// 1 ... i : (X, X + diff) can be reached by 1 ... i-1 : (X - a[i], X + diff)
dp[0][offset] = 0; // subset sum is 0 for null set, difference = 0 between subsets
visited[offset] = 1; // initially zero diff reached
for (int i = 1; i <= n; i++) {
for (int diff = (-1)*sum; diff <= sum; diff++) {
if (visited[offset + diff + arr[i]] || visited[offset + diff - arr[i]] || visited[offset + diff]) {
// if difference 'diff' is reachable, then only update, else no need
dp[i][offset + diff] = max_3
(
dp[i - 1][offset + diff],
dp[i - 1][offset + diff + arr[i]] + arr[i],
dp[i - 1][offset + diff - arr[i]]
);
visited[offset + diff] = 1;
}
}
/*
dp[i][diff] = max {
dp[i - 1][diff] : not taking a[i] in either subset
dp[i - 1][diff + arr[i]] + arr[i] : putting arr[i] in first set, thus reducing difference to 'diff', increasing X to X + arr[i]
dp[i - 1][diff - arr[i]] : putting arr[i] in second set
initialization: dp[0][0] = 0
*/
// O(N*SUM) algorithm
}
cout << dp[n][offset] << "\n";
return 0;
}
Output:
14
State is not updated in Approach 1. Change the last line of recurse
visited[idx][sum_left][sum_right];
to
visited[idx][sum_left][sum_right] = 1;
Also memset the visited array to false before calling recurse from main.
I'm trying to write a program in C that will solve the following cryptarithm:
one + one = two
seven is prime
nine is a perfect square
Namely, I need to find the numerical values for the words one, two, seven and nine where each letter (o, n, e, t, w, s, v, i) is assigned a numerical value and the complete number also meets all of the above conditions.
I was thinking along the lines of creating an int array for each of the words and then 1) checking if each word meets the condition (e.g is a prime for "seven") and then 2) checking if each integer in the array is consistant with the value of the other words, where the other words also are found to meet their respective conditions.
I can't really see this working though as I would have to continuously convert the int array to a single int throughout every iteration and then I'm not sure how I can simultaneously match each element in the array with the other words.
Perhaps knowing the MIN and MAX numerical range that must be true for each of the words would be useful?
Any ideas?
For a brute-force (ish) method, I'd start with the prime seven, and use the Sieve of Eratosthenes to get all the prime numbers up to 99999. You could discard all answers where the 2nd and 4th digit aren't the same. After that you could move on to the square nine, because three of the digits are determined by the prime seven. That should narrow down the possibilities nicely, and then you can just use the answer of #pmg to finish it off :-).
Update: The following C# program seems to do it
bool[] poss_for_seven = new bool[100000]; // this will hold the possibilities for `seven`
for (int seven = 0; seven < poss_for_seven.Length; seven++)
poss_for_seven[seven] = (seven > 9999); // `seven` must have 5 digits
// Sieve of Eratosthenes to make `seven` prime
for (int seven = 2; seven < poss_for_seven.Length; seven++) {
for (int j = 2 * seven; j < poss_for_seven.Length; j += seven) {
poss_for_seven[j] = false;
}
}
// look through the array poss_for_seven[], considering each possibility in turn
for (int seven = 10000; seven < poss_for_seven.Length; seven++) {
if (poss_for_seven[seven]) {
int second_digit = ((seven / 10) % 10);
int fourth_digit = ((seven / 1000) % 10);
if (second_digit == fourth_digit) {
int e = second_digit;
int n = (seven % 10); // NB: `n` can't be zero because otherwise `seven` wouldn't be prime
for (int i = 0; i < 10; i++) {
int nine = n * 1000 + i * 100 + n * 10 + e;
int poss_sqrt = (int)Math.Floor(Math.Sqrt(nine) + 0.1); // 0.1 in case of of rounding error
if (poss_sqrt * poss_sqrt == nine) {
int o = ((2 * e) % 10); // since 2 * `one` = `two`, we now know `o`
int one = o * 100 + n * 10 + e;
int two = 2 * one;
int t = ((two / 100) % 10);
int w = ((two / 10) % 10);
// turns out that `one`=236, `two`=472, `nine` = 3136.
// look for solutions where `s` != `v` with `s` and `v' different from `o`, `n`, `e`,`t`, `w` and `i`
int s = ((seven / 10000) % 10);
int v = ((seven / 100) % 10);
if (s != v && s != o && s != n && s != e && s != t && s != w && s != i && v != o && v != n && v != e && v != t && v != w && v != i) {
System.Diagnostics.Trace.WriteLine(seven + "," + nine + "," + one + "," + two);
}
}
}
}
}
}
It seems that nine is always equal to 3136, so that one = 236 and two = 472. However, there are 21 possibiliites for seven. If one adds the constraint that no two digits can take the same value (which is what the C# code above does), then it reduces to just one possibility (although a bug in my code meant this answer originally had 3 possibilities):
seven,nine,one,two
56963,3136,236,472
I just found the time to build a c program to solve your cryptarithm.
I think that tackling the problem mathematicaly, prior to starting the brute force programming, will heavily increase the speed of the output.
Some math (number theory):
Since ONE + ONE = TWO, O cant be larget than 4, because ONE + ONE would result 4 digits. Also O cant be 0. TWO end with O and is an even number, because it is 2 * ONE.
Applying these 3 filters to O, the possible values remain O= {2,4}
Hence E can be {1,2,6,7} because (E+E) modulus 10 must be = O. More specificaly, O=2 implicates E={1,6} and O=4 implicates E={2,7}
Now lets filter N. Given that SEVEN is prime, N must be an odd number. Also N cant be 5, because all that ends with 5 is divisible by 5. Hence N={1,3,7,9}
Now that we have reduced the possibilites for the most ocurring characters (O,E,N), we are ready to hit this cryptarith with all of our brutality, having iterations drastically reduced.
Heres the C code:
#include <stdio.h>
#include <math.h>
#define O 0
#define N 1
#define E 2
#define T 3
#define W 4
#define S 5
#define V 6
#define I 7
bool isPerfectSquare(int number);
bool isPrime(int number);
void printSolutions(int countSolutions);
int filterNoRepeat(int unfilteredCount);
int solutions[1000][8]; // solution holder
int possibilitiesO[2] = {2,4};
int possibilitiesN[4] = {1,3,7,9};
int possibilitiesE[4] = {1,6,2,7};
void main() {
int countSolutions = 0;
int numberOne;
// iterate to fill up the solutions array by: one + one = two
for(int o=0;o<2;o++) {
for(int n=0;n<4;n++) {
for(int e=2*o;e<2*o+2;e++) { // following code is iterated 2*4*2 = 16 times
numberOne = 100*possibilitiesO[o] + 10*possibilitiesN[n] + possibilitiesE[e];
int w = ((2*numberOne)/10)%10;
int t = ((2*numberOne)/100)%10;
// check if NINE is a perfect square
for(int i=0;i<=9;i++) { // i can be anything ----- 10 iterations
int numberNine = 1000*possibilitiesN[n] + 100*i + 10*possibilitiesN[n] + possibilitiesE[e];
if(isPerfectSquare(numberNine)) {
// check if SEVEN is prime
for(int s=1;s<=9;s++) { // s cant be 0 ------ 9 iterations
for(int v=0;v<=9;v++) { // v can be anything other than s ------- 10 iterations
if(v==s) continue;
int numberSeven = 10000*s + 1000*possibilitiesE[e] + 100*v + 10*possibilitiesE[e] + possibilitiesN[n];
if(isPrime(numberSeven)) { // store solution
solutions[countSolutions][O] = possibilitiesO[o];
solutions[countSolutions][N] = possibilitiesN[n];
solutions[countSolutions][E] = possibilitiesE[e];
solutions[countSolutions][T] = t;
solutions[countSolutions][W] = w;
solutions[countSolutions][S] = s;
solutions[countSolutions][V] = v;
solutions[countSolutions][I] = i;
countSolutions++;
}
}
}
}
}
}
}
}
// 16 * 9 * 10 * 10 = 14400 iterations in the WORST scenario, conditions introduced reduce MOST of these iterations to 1 if() line
// iterations consumed by isPrime() function are not taken in count in the aproximation above.
// filter solutions so that no two letter have the same digit
countSolutions = filterNoRepeat(countSolutions);
printSolutions(countSolutions); // voila!
}
bool isPerfectSquare(int number) { // check if given number is a perfect square
double root = sqrt((double)number);
if(root==floor(root)) return true;
else return false;
}
bool isPrime(int number) { // simple algoritm to determine if given number is prime, check interval from sqrt(number) to number/2 with a step of +2
int startValue = sqrt((double)number);
if(startValue%2==0) startValue--; // make it odd
for(int k=startValue;k<number/2;k+=2) {
if(number%k==0) return false;
}
return true;
}
void printSolutions(int countSolutions) {
for(int k=0;k<countSolutions;k++) {
int one = 100*solutions[k][O] + 10*solutions[k][N] + solutions[k][E];
int two = 100*solutions[k][T] + 10*solutions[k][W] + solutions[k][O];
int seven = 10000*solutions[k][S] + 1000*solutions[k][E] + 100*solutions[k][V] + 10*solutions[k][E] + solutions[k][N];
int nine = 1000*solutions[k][N] + 100*solutions[k][I] + 10*solutions[k][N] + solutions[k][E];
printf("ONE: %d, TWO: %d, SEVEN: %d, NINE %d\n",one,two,seven,nine);
}
}
int filterNoRepeat(int unfilteredCount) {
int nrSol = 0;
for(int k=0;k<unfilteredCount;k++) {
bool isValid = true;
for(int i=0;i<7;i++) { // if two letters match, solution is not valid
for(int j=i+1;j<8;j++) {
if(solutions[k][i]==solutions[k][j]) {
isValid = false;
break;
}
}
if(!isValid) break;
}
if(isValid) { // store solution
for(int i=0;i<8;i++) {
solutions[nrSol][i] = solutions[k][i];
}
nrSol++;
}
}
return nrSol;
}
You can try the code yourself if you are still interested in this :P. The result is one single solution: ONE: 236, TWO: 472, SEVEN: 56963, NINE: 3136
This solution is the same as Stochastically's solutions, confirming the correctness of both algorithms i think :).
Thanks for providing this nice cryptarithm and have a nice day!
Brute force FTW!
#define ONE ((o*100) + (n*10) + e)
#define TWO ((t*100) + (w*10) + o)
#define SEVEN ((s*10000) + (e*1010) + (v*100) + n)
#define NINE ((n*1010) + (i*100) + e)
for (o = 1; o < 10; o++) { /* 1st digit cannot be zero (one) */
for (n = 1; n < 10; n++) { /* 1st digit cannot be zero (nine) */
if (n == o) continue;
for (e = 0; n < 10; n++) {
if (e == n) continue;
if (e == o) continue;
/* ... */
if (ONE + ONE == TWO) /* whatever */;
/* ... */
}
}
}
In reading in Chapter 14 of Jon Bentley's "Programming Pearls", 2nd Edition, I understand that heaps use a one-based array and the easiest approach in C is to declare x[n+1] and waste element x[0] (page 148).
On page 157, Jon listed the complete heapsort pseudo code:
for i = [2, n]
siftup(i)
for (i = n; i >= 2; i--)
swap(1, i)
siftdown(i - 1)
Here is an implementation in C. However, the array index starts with 0, instead of 1.
void heapSort(int numbers[], int array_size)
{
int i, temp;
// Qiang: shouldn't the stop-condition be i >= 1?
for (i = (array_size / 2)-1; i >= 0; i--)
siftDown(numbers, i, array_size);
for (i = array_size-1; i >= 1; i--)
{
// Qiang: shouldn't the swap be done with numbmers[1], instead of numbers[0]?
temp = numbers[0];
numbers[0] = numbers[i];
numbers[i] = temp;
siftDown(numbers, 0, i-1);
}
}
void siftDown(int numbers[], int root, int bottom)
{
int done, maxChild, temp;
done = 0;
while ((root*2 <= bottom) && (!done))
{
if (root*2 == bottom)
maxChild = root * 2;
else if (numbers[root * 2] > numbers[root * 2 + 1])
maxChild = root * 2;
else
maxChild = root * 2 + 1;
if (numbers[root] < numbers[maxChild])
{
temp = numbers[root];
numbers[root] = numbers[maxChild];
numbers[maxChild] = temp;
root = maxChild;
}
else
done = 1;
}
}
My worry is, if the array starts with index 0, then the following properties will not hold (as written on page 148 in Jon's book):
leftchild(i) = 2*i
rightchild(i) = 2*i+1
parent(i) = i/2
It looks to me that the properties here only hold when the i starts with 1.
What strikes me is that the analysis part in the implementation used an array starting with index 1, while the implementation part used an array starting with index 0 and didn't waste the first element.
Am I missing anything here?
Edited
With help from interjay, I realized the error contained in the original implementation, which could be shown with a testing input array of {66,4,23,4,78,6,44,11,22,1,99}.
Changed the original siftDown() function a little bit to adjust the relationship between the index of parent and those of its children:
void siftDown(int numbers[], int root, int bottom)
{
int done, maxChild, temp;
done = 0;
while ((root*2 + 1 <= bottom) && (!done))
{
if (root*2 + 1 == bottom ||
numbers[root * 2 + 1] > numbers[root * 2 + 2])
maxChild = root * 2 + 1;
else
maxChild = root * 2 + 2;
if (numbers[root] < numbers[maxChild])
{
temp = numbers[root];
numbers[root] = numbers[maxChild];
numbers[maxChild] = temp;
root = maxChild;
}
else
done = 1;
}
}
Credits go to interjay, :-)
Afterword:
It looks the same error doesn't appear in the implementations in wikibooks and algorithmist. Hooray!
The heap elements can be stored starting with index 0 or index 1, the decision on which to use is up to you.
If the root element is at index 1, the mathematical relations between parent and child indices are simple as you've shown above, and for that reason many books choose to teach it that way.
If the root is at index 0, you'd get these relations instead:
leftchild(i) = 2*i+1
rightchild(i) = 2*i+2
parent(i) = (i-1) / 2
It doesn't matter which one you pick as long as you are consistent.
The C code you've shown seems incorrect to me. It starts with array index 0, but uses the parent/child relations appropriate for starting with index 1.
A reusable implementation of heapsort would want to start at a root index of 0 so the user could use a normal (0 based) array with it. You wouldn't want to require the user to allocate an extra member and start the array at index 1 just so they can use your heapsort function. You do need to use the modified parent/child calculations that #interjay shows.
Replying to little old thread, thought my small contribution might helps future visitors.
Experts please validate and correct my logic if I missed any scenarios.
Considered Qiang Xu link and interjay zero based index logic.
And here is the C# code and tested with the below inputs.
//-----------------------------------------------------------------------------------------------------------------------------------------------
// Input Arrays :
int[] ErrCaseArry = new int[] { 66, 4, 23, 4, 78, 6, 44, 11, 22, 1, 99};
int[] GenCaseArry = new int[] { 30, 20, 40, 10, 90, 160, 140, 100, 80, 70 };
int[] NearlySortedArry = new int[] { 1, 2, 3, 4, 6, 5 };
int[] FewSortedArry1 = new int[] { 3, 2, 1, 4, 5, 6 };
int[] FewSortedArry2 = new int[] { 6, 2, 3, 1, 5, 4 };
int[] ReversedArry1 = new int[] { 6, 5, 4, 3, 2, 1 };
int[] FewDuplsArry2 = new int[] { 1, 3, 1, 2, 1, 3 };
int[] MoreDuplsArry3 = new int[] { 1, 1, 2, 2, 1, 2 };
//-----------------------------------------------------------------------------------------------------------------------------------------------
public void HeapSort(int[] listToSort)
{
int LastChildIndex = listToSort.Length -1;
int parentElementIndex = ((LastChildIndex - 1)/ 2);
//1. Use this loop to Construct Heap Array (Max/Min) by using Heapify function on every node.
while (parentElementIndex >= 0) // (N - 1) / 2 to 0
{
Heapify(listToSort, parentElementIndex, LastChildIndex); // (N - 1) / 2 & Lenght - 1
parentElementIndex--;
}
//-----------------------------------------------------------------------------------------------------------------------------------------------
AppendArrayToResultString("Max Heap\t", listToSort);
//2. Heap sort algorithm takes largest element off the heap and places it at the end of an array.
// This phase continue until all the elements are placed in the array that are in sorted order.
int sortedElementIndex = listToSort.Length - 1;
//-----------------------------------------------------------------------------------------------------------------------------------------------
// In this loop get Largest Element to Zero'th postion and move to end. and reduce the loop count from Heapify Array. So that elements gets sorted from right.
while (sortedElementIndex >= 0) // (N - 1) to 1
{
// Swap the elements (root(maximum value)) of the heap with the last element of the heap
Swap(ref listToSort[0], ref listToSort[sortedElementIndex]);
// sortedElementIndex-- : Decrease the size of the heap by one so that the previous max value will stay in its proper placement
sortedElementIndex--;
if (sortedElementIndex == -1) break;
// Since largest elemented from 0 to last, Re Heapify and get the remaining largest element and place it in 0 position.
Heapify(listToSort, 0, (sortedElementIndex)); // 0 to (N - 1)
}
//-----------------------------------------------------------------------------------------------------------------------------------------------
}
//Heapify() function maintain the heap property (Max Heap or Min Heap). Can be recursive or can use iteration loop like while/for.
void Heapify(int[] listToSort, int parentIndext, int lastChildIndext)
{
//bool doneFlag = false;
int largestElementIndex = 0;
int leftChildIndex = parentIndext * 2 + 1;
int rightChildIndex = parentIndext * 2 + 2;
while (leftChildIndex <= lastChildIndext) //&& !doneFlag)
{
// If leftChild is larger than rightChild or it is the last child and there is no rightChild for this parent.
// Then consider leftChild as largestElement else consider rightChild as largestElement.
if (leftChildIndex == lastChildIndext || listToSort[leftChildIndex] > listToSort[rightChildIndex])
{
largestElementIndex = leftChildIndex;
}
else
{
largestElementIndex = rightChildIndex;
}
//-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
// If largestElement is larger than parent then swap them and make parent as largestElement to continue the loop.
if (listToSort[parentIndext] < listToSort[largestElementIndex])
{
// Make largestElement as parent. And continue finding if childs (left and right) are bigger than element in largestIndex position.
Swap(ref listToSort[parentIndext], ref listToSort[largestElementIndex]);
// Repeat to continue sifting down the child now
parentIndext = largestElementIndex;
leftChildIndex = ((parentIndext * 2) + 1);
rightChildIndex = ((parentIndext * 2) + 2);
}
else
{
//doneFlag = true;
break; // Trying to avoid extra flag condition check. Or return.
}
}
}
//-----------------------------------------------------------------------------------------------------------------------------------------------
void Swap(ref int num1, ref int num2)
{
int temp = num1;
num1 = num2;
num2 = temp;
}
In a general binary search, we are looking for a value which appears in the array. Sometimes, however, we need to find the first element which is either greater or less than a target.
Here is my ugly, incomplete solution:
// Assume all elements are positive, i.e., greater than zero
int bs (int[] a, int t) {
int s = 0, e = a.length;
int firstlarge = 1 << 30;
int firstlargeindex = -1;
while (s < e) {
int m = (s + e) / 2;
if (a[m] > t) {
// how can I know a[m] is the first larger than
if(a[m] < firstlarge) {
firstlarge = a[m];
firstlargeindex = m;
}
e = m - 1;
} else if (a[m] < /* something */) {
// go to the right part
// how can i know is the first less than
}
}
}
Is there a more elegant solution for this kind of problem?
One way of thinking about this problem is to think about doing a binary search over a transformed version of the array, where the array has been modified by applying the function
f(x) = 1 if x > target
0 else
Now, the goal is to find the very first place that this function takes on the value 1. We can do that using a binary search as follows:
int low = 0, high = numElems; // numElems is the size of the array i.e arr.size()
while (low != high) {
int mid = (low + high) / 2; // Or a fancy way to avoid int overflow
if (arr[mid] <= target) {
/* This index, and everything below it, must not be the first element
* greater than what we're looking for because this element is no greater
* than the element.
*/
low = mid + 1;
}
else {
/* This element is at least as large as the element, so anything after it can't
* be the first element that's at least as large.
*/
high = mid;
}
}
/* Now, low and high both point to the element in question. */
To see that this algorithm is correct, consider each comparison being made. If we find an element that's no greater than the target element, then it and everything below it can't possibly match, so there's no need to search that region. We can recursively search the right half. If we find an element that is larger than the element in question, then anything after it must also be larger, so they can't be the first element that's bigger and so we don't need to search them. The middle element is thus the last possible place it could be.
Note that on each iteration we drop off at least half the remaining elements from consideration. If the top branch executes, then the elements in the range [low, (low + high) / 2] are all discarded, causing us to lose floor((low + high) / 2) - low + 1 >= (low + high) / 2 - low = (high - low) / 2 elements.
If the bottom branch executes, then the elements in the range [(low + high) / 2 + 1, high] are all discarded. This loses us high - floor(low + high) / 2 + 1 >= high - (low + high) / 2 = (high - low) / 2 elements.
Consequently, we'll end up finding the first element greater than the target in O(lg n) iterations of this process.
Here's a trace of the algorithm running on the array 0 0 1 1 1 1.
Initially, we have
0 0 1 1 1 1
L = 0 H = 6
So we compute mid = (0 + 6) / 2 = 3, so we inspect the element at position 3, which has value 1. Since 1 > 0, we set high = mid = 3. We now have
0 0 1
L H
We compute mid = (0 + 3) / 2 = 1, so we inspect element 1. Since this has value 0 <= 0, we set mid = low + 1 = 2. We're now left with L = 2 and H = 3:
0 0 1
L H
Now, we compute mid = (2 + 3) / 2 = 2. The element at index 2 is 1, and since 1 ≥ 0, we set H = mid = 2, at which point we stop, and indeed we're looking at the first element greater than 0.
You can use std::upper_bound if the array is sorted (assuming n is the size of array a[]):
int* p = std::upper_bound( a, a + n, x );
if( p == a + n )
std::cout << "No element greater";
else
std::cout << "The first element greater is " << *p
<< " at position " << p - a;
After many years of teaching algorithms, my approach for solving binary search problems is to set the start and the end on the elements, not outside of the array. This way I can feel what's going on and everything is under control, without feeling magic about the solution.
The key point in solving binary search problems (and many other loop-based solutions) is a set of good invariants. Choosing the right invariant makes problem-solving a cake. It took me many years to grasp the invariant concept although I had learned it first in college many years ago.
Even if you want to solve binary search problems by choosing start or end outside of the array, you can still achieve it with a proper invariant. That being said, my choice is stated above to always set a start on the first element and end on the last element of the array.
So to summarize, so far we have:
int start = 0;
int end = a.length - 1;
Now the invariant. The array right now we have is [start, end]. We don't know anything yet about the elements. All of them might be greater than the target, or all might be smaller, or some smaller and some larger. So we can't make any assumptions so far about the elements. Our goal is to find the first element greater than the target. So we choose the invariants like this:
Any element to the right of the end is greater than the target. Any
element to the left of the start is smaller than or equal to the
target.
We can easily see that our invariant is correct at the start (ie before going into any loop). All the elements to the left of the start (no elements basically) are smaller than or equal to the target, same reasoning for the end.
With this invariant, when the loop finishes, the first element after the end will be the answer (remember the invariant that the right side of the end are all greater than the target?). So answer = end + 1.
Also, we need to note that when the loop finishes, the start will be one more than the end. ie start = end + 1. So equivalently we can say start is the answer as well (invariant was that anything to the left of the start is smaller than or equal to the target, so start itself is the first element larger than the target).
So everything being said, here is the code.
public static int find(int a[], int target) {
int st = 0;
int end = a.length - 1;
while(st <= end) {
int mid = (st + end) / 2; // or elegant way of st + (end - st) / 2;
if (a[mid] <= target) {
st = mid + 1;
} else { // mid > target
end = mid - 1;
}
}
return st; // or return end + 1
}
A few extra notes about this way of solving binary search problems:
This type of solution always shrinks the size of subarrays by at least 1. This is obvious in the code. The new start or end are either +1 or -1 in the mid. I like this approach better than including the mid in both or one side, and then reason later why the algo is correct. This way it's more tangible and more error-free.
The condition for the while loop is st <= end. Not st < end. That means the smallest size that enters the while loop is an array of size 1. And that totally aligns with what we expect. In other ways of solving binary search problems, sometimes the smallest size is an array of size 2 (if st < end), and honestly I find it much easier to always address all array sizes including size 1.
So hope this clarifies the solution for this problem and many other binary search problems. Treat this solution as a way to professionally understand and solve many more binary search problems without ever wobbling whether the algorithm works for edge cases or not.
How about the following recursive approach:
public static int minElementGreaterThanOrEqualToKey(int A[], int key,
int imin, int imax) {
// Return -1 if the maximum value is less than the minimum or if the key
// is great than the maximum
if (imax < imin || key > A[imax])
return -1;
// Return the first element of the array if that element is greater than
// or equal to the key.
if (key < A[imin])
return imin;
// When the minimum and maximum values become equal, we have located the element.
if (imax == imin)
return imax;
else {
// calculate midpoint to cut set in half, avoiding integer overflow
int imid = imin + ((imax - imin) / 2);
// if key is in upper subset, then recursively search in that subset
if (A[imid] < key)
return minElementGreaterThanOrEqualToKey(A, key, imid + 1, imax);
// if key is in lower subset, then recursively search in that subset
else
return minElementGreaterThanOrEqualToKey(A, key, imin, imid);
}
}
public static int search(int target, int[] arr) {
if (arr == null || arr.length == 0)
return -1;
int lower = 0, higher = arr.length - 1, last = -1;
while (lower <= higher) {
int mid = lower + (higher - lower) / 2;
if (target == arr[mid]) {
last = mid;
lower = mid + 1;
} else if (target < arr[mid]) {
higher = mid - 1;
} else {
lower = mid + 1;
}
}
return (last > -1 && last < arr.length - 1) ? last + 1 : -1;
}
If we find target == arr[mid], then any previous element would be either less than or equal to the target. Hence, the lower boundary is set as lower=mid+1. Also, last is the last index of 'target'. Finally, we return last+1 - taking care of boundary conditions.
My implementation uses condition bottom <= top which is different from the answer by templatetypedef.
int FirstElementGreaterThan(int n, const vector<int>& values) {
int B = 0, T = values.size() - 1, M = 0;
while (B <= T) { // B strictly increases, T strictly decreases
M = B + (T - B) / 2;
if (values[M] <= n) { // all values at or before M are not the target
B = M + 1;
} else {
T = M - 1;// search for other elements before M
}
}
return T + 1;
}
Hhere is a modified binary search code in JAVA with time complexity O(logn) that :
returns index of element to be searched if element is present
returns index of next greater element if searched element is not present in array
returns -1 if an element greater than the largest element of array is searched
public static int search(int arr[],int key) {
int low=0,high=arr.length,mid=-1;
boolean flag=false;
while(low<high) {
mid=(low+high)/2;
if(arr[mid]==key) {
flag=true;
break;
} else if(arr[mid]<key) {
low=mid+1;
} else {
high=mid;
}
}
if(flag) {
return mid;
}
else {
if(low>=arr.length)
return -1;
else
return low;
//high will give next smaller
}
}
public static void main(String args[]) throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
//int n=Integer.parseInt(br.readLine());
int arr[]={12,15,54,221,712};
int key=71;
System.out.println(search(arr,key));
br.close();
}
kind =0 : exact match
kind=1 : just grater than x
kind=-1 : just smaller than x;
It returns -1 if no match is found.
#include <iostream>
#include <algorithm>
using namespace std;
int g(int arr[], int l , int r, int x, int kind){
switch(kind){
case 0: // for exact match
if(arr[l] == x) return l;
else if(arr[r] == x) return r;
else return -1;
break;
case 1: // for just greater than x
if(arr[l]>=x) return l;
else if(arr[r]>=x) return r;
else return -1;
break;
case -1: // for just smaller than x
if(arr[r]<=x) return r;
else if(arr[l] <= x) return l;
else return -1;
break;
default:
cout <<"please give "kind" as 0, -1, 1 only" << ednl;
}
}
int f(int arr[], int n, int l, int r, int x, int kind){
if(l==r) return l;
if(l>r) return -1;
int m = l+(r-l)/2;
while(m>l){
if(arr[m] == x) return m;
if(arr[m] > x) r = m;
if(arr[m] < x) l = m;
m = l+(r-l)/2;
}
int pos = g(arr, l, r, x, kind);
return pos;
}
int main()
{
int arr[] = {1,2,3,5,8,14, 22, 44, 55};
int n = sizeof(arr)/sizeof(arr[0]);
sort(arr, arr+n);
int tcs;
cin >> tcs;
while(tcs--){
int l = 0, r = n-1, x = 88, kind = -1; // you can modify these values
cin >> x;
int pos = f(arr, n, l, r, x, kind);
// kind =0: exact match, kind=1: just grater than x, kind=-1: just smaller than x;
cout <<"position"<< pos << " Value ";
if(pos >= 0) cout << arr[pos];
cout << endl;
}
return 0;
}