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Python code works 20 times slower than Java. Is there a way to speed up Python?

I wrote a program in Python and in Java to search for the smallest integer solution of the equation:
a^5+b^5+c^5+d^5=e^5 (expected output is 133^5+110^5+84^5+27^5=144^5)
Powers and roots are either computed directly ("direct calculation" method) or computed and stored in an array ("power lookup" method). Fifth powers are looked up like n5 = fifth_power[n]. Fifth power root is computed using a binary search in array 'fifth_power`.
I am running it on NetBeans if it matters. It takes:
30. s (Python, direct)
20. s (Python, lookup)
5.6 s (Java, direct)
0.8 s (Java, lookup)
Is there a way to boost Python performance? I am not looking for better math (sieving of some kind). I am looking for better implementation of "for each combination of a,b,c,d compute some of their powers, check if the sum is a perfect power. If it is - print the result".
Is it expected that Python runs some 20 times slower than Java?
Python 3.5
http://pastebin.com/qVthWGKm
from array import *
import math
import time
#PYTHON, BRUTEFORCE : ~30 s
millis1 = int(round(time.time() * 1000))
keep_searching = True
a=1
result=""
while(keep_searching):
a+=1
for b in range(1,a+1):
for c in range(1,b+1):
for d in range(1,c+1):
sum=math.pow(a,5)+math.pow(b,5)+math.pow(c,5)+math.pow(d,5)
root = math.pow(sum,0.2)
e = round(root)
e5 = math.pow(e,5)
if(e5==sum):
result="{}^5 + {}^5 + {}^5 + {}^5 = {}^5".format(int(a),int(b), int(c),int(d), int(e))
keep_searching = False
millis2 = int(round(time.time() * 1000))
print(result)
print("Found solution in {} ms".format(millis2-millis1))
#PYTHON, PRECOMPUTE POWERS: ~20 s
millis3 = int(round(time.time() * 1000))
#fifth_power #175 is enough
size=176
fifth_power = [None] * size
for i in range(size):
fifth_power[i]=long(math.pow(i,5))
millis4 = int(round(time.time() * 1000))
#returns value if it is a perfect power (32 returns 2)
#returns -1 if between perfect powers, -2 if greater than max value in array, -3 if smaller than min value in array
def check_perfect_power(number, min, max, fifth_power):
current=int((min+max)/2)
while(max>=min):
if(number==fifth_power[current]):
return current
elif(number>fifth_power[current]):
min=current+1
current=int((max+min)/2)
else:
max=current-1
current=int((max+min)/2)
if(min>=len(fifth_power)):
return -2
if(max<0):
return -3
return -1
keep_searching = True
a=0
result=""
while(keep_searching):
a+=1
for b in range(1,a+1):
for c in range(1,b+1):
for d in range(1,c+1):
mymax=min(int(a*1.32)+1, size-1)
e=check_perfect_power(fifth_power[a]+fifth_power[b]+fifth_power[c]+fifth_power[d], a, mymax, fifth_power)
if(e>0):
result="{}^5 + {}^5 + {}^5 + {}^5 = {}^5".format(int(a),int(b), int(c),int(d), int(e))
keep_searching = False
millis5 = int(round(time.time() * 1000))
print(result)
print("Populated in {} ms, find solution in {} ms".format(millis4-millis3,millis5-millis4))
Java 8:
http://pastebin.com/G4V3fHnD
import java.util.ArrayList;
public class Eu514 {
public static void main(String[] args) {
bruteforce(); //Solution found by bruteforce in 5600 ms.
prepopulate(); //Solution found by prepopulation in 761 ms.
}
public static void bruteforce(){ //JAVA BRUTEFORCE
Long t2 = 0L;
Long t1 = System.currentTimeMillis();
boolean keepSearching = true;
int a = 0;
long e = 0;
String solution = "";
while (keepSearching) {
a++;
for (int b = 1; b <= a; b++) {
for (int c = 1; c <= b; c++) {
for (int d = 1; d <= c; d++) {
long sum = (long) (Math.pow(a, 5) + Math.pow(b, 5) + Math.pow(c, 5) + Math.pow(d, 5)); //sum=a^5+b^5+c^5+d^5
e = Math.round(Math.pow(sum, 0.2)); //e= sum^(1/5), rounded
long e5 = (long) Math.pow(e,5); //e^5
if(e5==sum){
t2 = System.currentTimeMillis();
solution = a + "^5 + " + b + "^5 + " + c + "^5 + " + d + "^5 = " + e + "^5";
keepSearching = false;
}
}
}
}
}
long delta = ((t2-t1));
System.out.println(solution+"\nSolution found by bruteforce in "+delta+" ms.");
}
public static void prepopulate(){ //JAVA PREPOPULATE
Long t2 = 0L;
Long t1 = System.currentTimeMillis();
int size = 176;
long[] powers = populatePowers(size);
boolean keepSearching = true;
int a = 0;
int e = 0;
String solution = "";
while (keepSearching) {
a++;
for (int b = 1; b <= a; b++) {
for (int c = 1; c <= b; c++) {
for (int d = 1; d <= c; d++) {
long sum = powers[a] + powers[b] + powers[c] + powers[d];
int max = (int) Math.min(size - 1, (a * 1.32 + 1));
e = checkIfPerfectPower(sum, a, max, powers);
if (e > 0) {
t2 = System.currentTimeMillis();
solution = a + "^5 + " + b + "^5 + " + c + "^5 + " + d + "^5 = " + e + "^5";
keepSearching = false;
}
}
}
}
}
long delta = ((t2-t1));
System.out.println(solution+"\nSolution found by prepopulation in "+delta+" ms.");
}
public static long[] populatePowers(int max){
long[] powers = new long[max];
for (int i = 0; i < powers.length; i++) {
powers[i]=(long) Math.pow(i,5);
}
return powers;
}
public static int checkIfPerfectPower(long number, int min, int max, long[] arr){
int current =((min+max)/2);
while(max>=min){
if(number==arr[current]){
return current;
}else if(number>arr[current]){
min = current + 1;
current = (max + min) / 2;
}else{
max=current-1;
current=(max+min)/2;
}
}
if(min>=arr.length) return -2;
if(max<0) return -3;
return -1;
}
}

from array import *
import time
import numpy as np
#PYTHON, BRUTEFORCE : ~30 s
millis1 = int(round(time.time() * 1000))
keep_searching = True
a = 1
result = ""
while(keep_searching):
a += 1
a_pow = a ** 5
for b in xrange(1, a+1):
b_pow = b ** 5
for c in xrange(1, b+1):
c_pow = c ** 5
for d in xrange(1, c+1):
d_pow = d ** 5
sum_pow = a_pow + b_pow + c_pow + d_pow
root = sum_pow ** 0.2
e = round(root)
e5 = e ** 5
if(e5 == sum_pow):
result="{}^5 + {}^5 + {}^5 + {}^5 = {}^5".format(a, b, c, d, e)
keep_searching = False
millis2 = int(round(time.time() * 1000))
print(result)
print("Found solution in {} ms".format(millis2-millis1))
Python 2.7, with some code optimizations
133^5 + 110^5 + 84^5 + 27^5 = 144.0^5
Found solution in 8333 ms
It could be a little different from CPU to CPU.

What about improving the java code?
int size = 200;
long[] pow5 = new long[size];
for (int i = 1; i < size; ++i)
{
long sqr = i * i;
pow5[i] = sqr * sqr * i;
}
for (int a = 1; a < size; ++a)
{
for (int b = 1; b <= a; ++b)
{
for (int c = 1; c <= b; ++c)
{
int e = a + 1;
for (int d = 1; d <= c; ++d)
{
long sum = pow5[a] + pow5[b] + pow5[c] + pow5[d];
while(pow5[e] < sum){ e++; }
if (pow5[e] == sum)
{
System.out.println(a + "^5 + " + b + "^5 + " + c + "^5 + " + d + "^5 = " + e + "^5");
return;
}
}
}
}
}

Switching between codes in my program

I have three programs calculating PI and I want to refactor it into one. I can't get it working using an if or switch command. It is something to do with case 3.
public class ProgramPI {
public static void main(String[] args) {
// TODO Auto-generated method stub
int z = Integer.parseInt(JOptionPane.showInputDialog("Wybierz metode liczenia PI 1=Leibnitz 2=Wallis 3=MonteCarlo:"));{
switch(z){
case 1:
double pi=0;
double mianownik=1;
double i = Integer.parseInt(JOptionPane.showInputDialog("Obliczamy PI metoda Leibnitza podaj n: "));
for (int x = 0; x < i; x++){
// obliczanie ciagu
if (x % 2 == 0) {
pi = pi + (1 / mianownik);
} else {
pi=pi - (1 / mianownik);
}
mianownik = mianownik + 2;
}
pi = pi * 4;
System.out.println("Przy n= " + i + " Pi = " + pi);
break;
case 2:
double termCount = 2;
double polpi = 1;
double termin;
// double termino2 =0;
double n = Integer.parseInt(JOptionPane.showInputDialog("Liczymy Pi wzorem Wallisa, podaj n:"));
for (double j = 0; j <= n; j++)
{
termin = (termCount / (termCount-1));
// System.out.println(termin);
polpi = polpi * termin;
termin = (termCount / (termCount+1));
// System.out.println(termin+" "+ mediopi);
polpi = polpi * termin;
termCount= termCount + 2;
}
System.out.println("Ilosc wyrazen = " + termCount);
polpi = polpi * 2;
System.out.println("Przy n= " + n + " Pi wynosi: " + polpi);
break;
case 3:
static double podajPi(int liczbaStrzalow) {
int wKole = 0;
// generowanie punktów
for (int i = 0; i < liczbaStrzalow; i++) {
double x = Math.random();
double y = Math.random();
// Sprawdzanie, czy punkt jest w kole
// r = 1, więc r * r = 1
if ((x * x + y * y) <= 1) {
wKole++;
}
}
// obliczanie stosunku liczby punktów w kole do wszystkich punktów
double s = (double) wKole / liczbaStrzalow;
return 4 * s;
}
System.out.println("Pi = " + Math.PI);
// obliczanie Pi dla 00, 1000, 10 000, 100 000 oraz 1000 000 punktów
// można to zrobić za pomocą pętli
for (int i = 100; i <= 1000000; i *= 10) {
double pi = podajPi(i);
System.out.println("Dla " + i + " punktów Pi = " + pi + "\n"
+ "błąd = " + (pi - Math.PI));
}
}
{
}
}

Getting same random numbers every time

I'm trying to write a program that generates a random school timetable with random positions and a random amount of hours for each teacher, but with a fixed total amount of time per day. For now the program is written to work with two days, and I'm encountering an issue: the random-generated values for the amount of time between the two days are the same:
import java.util.Random;
public class randomTimetable {
public static void main(String[] args) {
String newLine = System.getProperty("line.separator");
System.out.println("For each day (x + y + ... n) >= 5 and" +newLine +"(x && y && ... n) <= 2" +newLine);
createTimetable();
}
private static void createTimetable() {
String x_g1 = "x";
String y_g1 = "y";
String z_g1 = "z";
String m_g1 = "m";
String[] arrayTimetablePosition1={x_g1, y_g1, z_g1, m_g1};
String newLine = System.getProperty("line.separator");
System.out.println("Work In Progress" +newLine +"Total subjects = 5" +newLine +"Day 1");
Random rand = new Random();
int min = 0;
int max = 2;
int x1 = rand.nextInt(max - min + 1) + min;
int y1 = rand.nextInt(max - min + 1) + min;
int z1 = rand.nextInt(max - min + 1) + min;
int m1 = rand.nextInt(max - min + 1) + min;
while((x1 + y1 + z1 + m1) != 5) {
x1 = rand.nextInt(max - min + 1) + min;
y1 = rand.nextInt(max - min + 1) + min;
z1 = rand.nextInt(max - min + 1) + min;
m1 = rand.nextInt(max - min + 1) + min;
}
System.out.println("x1 = " +x1 +newLine +"y1 = " +y1 +newLine +"z1 = " +z1 +newLine +"m1 = " +m1 +newLine);
System.out.println("Total subjects = 5" +newLine +"Day 2");
int x2 = rand.nextInt(max - min + 1) + min;
int y2 = rand.nextInt(max - min + 1) + min;
int z2 = rand.nextInt(max - min + 1) + min;
int m2 = rand.nextInt(max - min + 1) + min;
while((x2 + y2 + z2 + m2) != 5 && (x1 == x2 || y1 == y2 || z1 == z2 || m1 == m2)) {
x2 = rand.nextInt(max - min + 1) + min;
y2 = rand.nextInt(max - min + 1) + min;
z2 = rand.nextInt(max - min + 1) + min;
m2 = rand.nextInt(max - min + 1) + min;
}
System.out.println("x2 = " +x1 +newLine +"y2 = " +y1 +newLine +"z2 = " +z1 +newLine +"m2 = " +m1 +newLine);
}
}
specifically the value of x1 is the same of x2, the one of y1 is the same of y2 and so on.

You're Random construction is fine - you're using the default constructor, which automatically uses the time as a seed:
public Random() { this(System.currentTimeMillis()); }
But you have a copy/paste error in your last debug-print statement. You're label says x2, but you're printing x1, etc.
System.out.println("x2 = " +x1 +newLine +"y2 = " +y1 +newLine +"z2 = " +z1 +newLine +"m2 = " +m1 +newLine);

I cannot see any initialization of the pseudorandom number generator.
You need to set the seed of the PRNG.

Looks like you are using the same seed. see:
http://docs.oracle.com/javase/6/docs/api/java/util/Random.html

LOL, it's a copy-paste error
last line should read
System.out.println("x2 = " +x2 +newLine +"y2 = " +y2 +newLine +"z2 = " +z2 +newLine +"m2 = " +m2 +newLine);
but it sounds like a classic random-seed problem. That's pretty funny.

Double function not returning Double value?

public class Nitin12assignA6 {
public static void main(String args[]) throws IOException {
series ob = new series();
ob.input();
ob.findSum();
ob.display();
}
}
class series {
int x, n;
double sum;
series() {
x = n = 0;
sum = 0.0f;
}
void input() throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
System.out.print("\nEnter the value of x = ");
x = Integer.parseInt(in.readLine());
System.out.print("Enter the value of n = ");
n = Integer.parseInt(in.readLine());
}
void display() {
System.out.println("The sum of Series upto " + n + " terms is: " + sum);
System.out.println();
}
long fact(int num) {
if(num == 1) {
return 1;
}
return num * fact(num - 1);
}
int power(int num, int exp) {
if(exp == 1) {
return 1;
}
return num * power(num, exp - 1);
}
double term(int numr, long denom) {
return(numr / denom);
}
void findSum() {
int u = 2, l = 4;
sum = 1.0f;
for(int i = 0; i < n; i++) {
if(l % 8 == 0) {
sum += term(power(x, u), fact(l));
// Test start
System.out.println("add" + sum + " power " + power(x, u) + " fact " + fact(l) + " x " + x + " u " + u
+ " l " + l);
System.out.println("term " + term(power(x, u), fact(l)));
System.out.println("test term " + term(5, 2)); // printing 2.0
// instead of 2.5
// Test end
} else {
sum -= term(power(x, u), fact(l));
// Test start
System.out.println("minus" + sum + " power " + power(x, u) + " fact " + fact(l) + " x " + x + " u " + u
+ " l " + l);
System.out.println("term " + term(power(x, u), fact(l)));
// Test end
}
u += 2;
l += 4;
}
}
}
//double term(int numr,long denom)
//{return (numr/denom);}
This function is not returning double value. Please help... This program is find a sum a series upto n terms. Example - test term is returning 2.0 instead of 2.5... Thanks.

You need to cast explicitly to double before devision.
double term(int numr, long denom) {
return ((double)numr / denom);
}

The casting to double is done on the value resulting from the division, so you can read
double term(int numr,long denom) {
return (double) (numr/denom);
}
So numr/denom is evaluated first, and since int/long type is long, the result would be 2L. Then it is casted to double. If you want it to be evaluated a double, do
double term(int numr,long denom) {
return ((double)numr/denom);
}

try
double term(int numr, long denom) {
return ((double) numr/denom);
}
your version had int/long which will always return a non floating point number.
See this:
int a = 2;
int b = 4;
double c = a/b; // 0.0
double d = ((double) a/b); // 0.5

double term(int numr, long denom) {
return (numr / denom);
}
The return value is double but the devision (numr / denom) creates a long which then is converted to double.

This may help.
double term(int numr, long denom) {
return ((double)numr / denom);
}

Printing the digits of a number in reverse order

Can someone please help me solve this? The number should be variable and not constant.
The output should be:
Timestamping In 6 Digit
8
5
6
3
0
1
Average In 6 Digit
9
8
7
6
5
2
class Timestamp1 extends Average1
{
public static void main (String args[]) {
int i = 103658;
int j = 10;
int k = i % j;
System.out.println(" Timestamping In 6 Digit " );
System.out.println(" " + k);
int o = 10365;
int p = 10;
int q = o % p;
System.out.println(" " + q);
int l = 1036;
int m = 10;
int n = l % m;
System.out.println(" " + n);
int r = 103;
int s = 10;
int t = r % s;
System.out.println(" " + t);
int u = 10;
int v = 10;
int w = u % v;
System.out.println(" " + w);
int x = 1;
int y = 10;
int z = x % y;
System.out.println(" " + z);
class Average1 extends Timestamp1 {
public void main() {
int i = 256789;
int j = 10;
int k = i % j;
System.out.println(" Average In 6 Digit ");
System.out.println(" " + k);
int o = 25678;
int p = 10;
int q = o % p;
System.out.println(" " + q);
int l = 2567;
int m = 10;
int n = l % m;
System.out.println(" " + n);
int r = 256;
int s = 10;
int t = r % s;
System.out.println(" " + t);
int u = 25;
int v = 10;
int w = u % v;
System.out.println(" " + w);
int x = 2;
int y = 10;
int z = x % y;
System.out.println(" " + z);
}
}
}
}

If you just want to reverse a number then here you go
public String reverseNum(Integer i)
{
return new StringBuilder(i.toString()).reverse();
}

What do you really want to do? Print two integer numbers reversed? Then you should say so.
public static String reverseDigits(int i) {
StringBuilder sb = new StringBuilder();
sb.append(i);
sb.reverse();
return sb.toString();
}
...
System.out.println(reverseDigits(1234567));
...