Gena Playing Hanoi
Problem Statement :
The Tower of Hanoi is a famous game consisting of rods and a number of discs of incrementally different diameters. The puzzle starts with the discs neatly stacked on one rod, ordered by ascending size with the smallest disc at the top. The game's objective is to move the entire stack to another rod, obeying the following rules: Only one disk can be moved at a time. In one move, remove the topmost disk from one rod and move it to another rod. No disk may be placed on top of a smaller disk. Gena has a modified version of the Tower of Hanoi. This game of Hanoi has rods and disks ordered by ascending size. Gena has already made a few moves following the rules above. Given the state of Gena's Hanoi, determine the minimum number of moves needed to restore the tower to its original state with all disks on rod . Note: Gena's rods are numbered from to . The radius of a disk is its index in the input array, so disk is the smallest disk with a radius of , and disk is the largest with a radius of . Function Description Complete the hanoi function below. hanoi has the following parameters: int posts[n]: is the location of the disk with radius Returns int: the minimum moves to reset the game to its initial state Input Format The first line contains a single integer, , the number of disks. The second line contains space-separated integers, where the integer is the index of the rod where the disk with diameter is located.
Solution :
Solution in C :
In C :
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
typedef struct _QueueElement {
int move;
int state;
} QueueElement;
int N;
#define QUEUE_SIZE (1024*1024*16)
QueueElement queue[QUEUE_SIZE];
char enqueued[QUEUE_SIZE];
int queueCount = 0;
int queueStart = 0;
int emptyQueue()
{
return (queueCount == 0);
}
void pushQueue(int move, int state)
{
int index = queueStart + queueCount;
if (index >= QUEUE_SIZE) {
index -= QUEUE_SIZE;
}
if (enqueued[state] != move) {
queue[index].move = move;
queue[index].state = state;
enqueued[state] = move;
queueCount += 1;
}
}
int popQueue(int *move)
{
int res = queue[queueStart].state;
*move = queue[queueStart].move;
queueStart += 1;
if (queueStart >= QUEUE_SIZE) {
queueStart -= QUEUE_SIZE;
}
queueCount -= 1;
return res;
}
void genMoves(int state, int *moves, int *movesCount)
{
int rod;
int i = 0;
int r[4] = { -1, -1, -1, -1 };
int tmp = state;
while (tmp) {
rod = tmp & 0x3;
if (r[rod] < 0) {
r[rod] = i;
}
tmp >>= 2;
i += 1;
}
*movesCount = 0;
if (r[0] >= 0) {
if (r[0] < r[1] || r[1] < 0) {
moves[*movesCount] = state | (1 << (r[0] * 2));
*movesCount += 1;
}
if (r[0] < r[2] || r[2] < 0) {
moves[*movesCount] = state | (2 << (r[0] * 2));
*movesCount += 1;
}
if (r[0] < r[3] || r[3] < 0) {
moves[*movesCount] = state | (3 << (r[0] * 2));
*movesCount += 1;
}
}
if (r[1] >= 0) {
if (r[1] < r[0] || r[0] < 0) {
moves[*movesCount] = state & (~(1 << (r[1] * 2)));
*movesCount += 1;
}
if (r[1] < r[2] || r[2] < 0) {
moves[*movesCount] = state & (~(1 << (r[1] * 2)));
moves[*movesCount] |= (2 << (r[1] * 2));
*movesCount += 1;
}
if (r[1] < r[3] || r[3] < 0) {
moves[*movesCount] = state | (3 << (r[1] * 2));
*movesCount += 1;
}
}
if (r[2] >= 0) {
if (r[2] < r[0] || r[0] < 0) {
moves[*movesCount] = state & (~(2 << (r[2] * 2)));
*movesCount += 1;
}
if (r[2] < r[1] || r[1] < 0) {
moves[*movesCount] = state & (~(2 << (r[2] * 2)));
moves[*movesCount] |= (1 << (r[2] * 2));
*movesCount += 1;
}
if (r[2] < r[3] || r[3] < 0) {
moves[*movesCount] = state | (3 << (r[2] * 2));
*movesCount += 1;
}
}
if (r[3] >= 0) {
if (r[3] < r[0] || r[0] < 0) {
moves[*movesCount] = state & (~(3 << (r[3] * 2)));
*movesCount += 1;
}
if (r[3] < r[1] || r[1] < 0) {
moves[*movesCount] = state & (~(3 << (r[3] * 2)));
moves[*movesCount] |= (1 << (r[3] * 2));
*movesCount += 1;
}
if (r[3] < r[2] || r[2] < 0) {
moves[*movesCount] = state & (~(3 << (r[3] * 2)));
moves[*movesCount] |= (2 << (r[3] * 2));
*movesCount += 1;
}
}
}
int main()
{
int i;
for (i = 0; i < QUEUE_SIZE; ++i) {
enqueued[i] = -1;
}
scanf("%d", &N);
{
int state = 0;
int tmp;
for (i = 0; i < N; ++i) {
scanf("%d", &tmp);
state |= (tmp - 1) << (i * 2);
}
pushQueue(0, state);
}
{
int state;
int move;
int moves[6];
int movesCount;
while (! emptyQueue()) {
state = popQueue(&move);
if (! state) {
break;
}
genMoves(state, moves, &movesCount);
for (i = 0; i < movesCount; ++i) {
pushQueue(move + 1, moves[i]);
}
}
printf("%d\n", move);
}
return 0;
}
Solution in C++ :
In C++ :
#include <bits/stdc++.h>
using namespace std;
#define sz(x) ((int) (x).size())
#define forn(i,n) for (int i = 0; i < int(n); ++i)
#define forab(i,a,b) for (int i = int(a); i < int(b); ++i)
typedef long long ll;
typedef long long i64;
typedef long double ld;
const int inf = int(1e9) + int(1e5);
const ll infl = ll(2e18) + ll(1e10);
int d[1 << 20];
int main() {
cout.precision(10);
cout.setf(ios::fixed);
#ifdef LOCAL
assert(freopen("c.in", "r", stdin));
#else
#endif
int n;
cin >> n;
int mask = 0;
forn (i, n) {
int k;
cin >> k;
mask |= ((k - 1) << (2 * i));
}
forn (i, 1 << 20)
d[i] = inf;
d[mask] = 0;
queue<int> o;
o.push(mask);
while (!o.empty()) {
int mask = o.front();
o.pop();
if (mask == 0)
break;
vector<int> b[4];
forn (i, n)
b[(mask >> (2 * i)) & 3].push_back(i);
forn (i, 4)
b[i].push_back(inf);
forn (i, 4)
forn (j, 4) {
if (b[i].front() >= b[j].front())
continue;
int id = b[i].front();
int to = mask ^ ((i ^ j) << (2 * id));
if (d[to] < inf)
continue;
d[to] = d[mask] + 1;
o.push(to);
}
}
cout << d[0] << '\n';
#ifdef LOCAL
cerr << "Time: " << double(clock()) / CLOCKS_PER_SEC << '\n';
#endif
}
Solution in Java :
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static int[] readIntArray3(Scanner in, int size) {
int[] arr = new int[size];
for (int i = 0; i < size; i++) {
arr[i] = in.nextInt();
}
return arr;
}
public static void main(String[] args) throws Exception {
Scanner in = new Scanner(System.in);
int cases = 1;//in.nextInt();
for (int testcase = 0; testcase < cases; testcase++) {
int n = in.nextInt();
int[] arr = readIntArray3(in, n);
genahanoi(n, arr);
}
}
public static void genahanoi(int n, int[] locs) {
int[] powers = new int[]{1,4,16,64,256,1024,4096,16384,65536,262144};
int score = 0;
for (int i = 0; i < n; i++) {
score += powers[i]*(locs[i]-1);
}
Set<Integer> allTests = new HashSet();
Set<Integer> currentTests = new HashSet();
currentTests.add(score);
allTests.add(score);
int currentMoves = 0;
// printScore(score, n);
while (!currentTests.contains(0)) {
Set<Integer> nextTests = new HashSet();
for (int test : currentTests) {
// System.out.print("Looking at position "); printScore(test, n);
int[] tops = new int[]{-1,-1,-1,-1};
for (int i = n-1; i >=0; i--) {
int loc = (test>>(2*i))%4;
tops[loc] = i;
}
for (int j = 0; j < 4; j++) {
if (tops[j] >= 0) {
for (int k = 0; k < 4; k++) {
if ( k != j) {
if (tops[k] == -1 || tops[k] > tops[j]) {
int newTest = test - (j<<(2*tops[j])) + (k<<(2*tops[j]));
if (!allTests.contains(newTest)) {
// System.out.print("Found new position "); printScore(newTest, n);
nextTests.add(newTest);
} else {
// System.out.print("Already seen position "); printScore(newTest, n);
}
}
}
}
}
}
}
currentTests = nextTests;
allTests.addAll(currentTests);
currentMoves++;
}
System.out.println(currentMoves);
}
}
Solution in Python :
In Python3 :
#!/bin/python3
import sys
def generate(move, N):
next = []
i = 0
available = [1, 2, 3, 4]
while (i < N):
if len(available) == 0:
break
elif (move[i] in available):
available.remove(move[i])
else:
i = i + 1
continue
for m in available:
amove = list(move)
amove[i] = m
next.append(tuple(amove))
i = i + 1
return set(next);
N = int(input().strip())
a = [int(a_temp) for a_temp in input().strip().split(' ')]
initial = [];
i = 0
while (i < N):
initial.append(1)
i = i + 1
moves = set();
moves.add(tuple(initial));
current = set(moves)
gen = 0
while True:
next = set()
for move in current:
next.update(generate(move, N))
gen = gen + 1
next.difference_update(moves)
moves.update(next)
if (tuple(a) in next):
print(gen)
break
current = set(next)
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