各种语言版本实现欧拉猜想 Euler's sum of powers conjecture
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Euler's sum of powers conjecture
There is a conjecture in mathematics that held for over 200 years before it was disproved by the finding of a counterexample in 1966 byLander and Parkin.
- Euler's (disproved) sum of powers conjecture
- At least k positive kth powers are required to sum to a kth power, except for the trivial case of one kth power: yk = yk.
Lander and Parkin are known to have used a brute-force search on a CDC 6600 computer restricting numbers to those less than 250.
- Task
Write a program to search for an integer solution to:
x05 + x15 + x25 + x35 == y5
Where all xi's and y are distinct integers between 0 and 250 (exclusive).
Show an answer here.
Contents
[hide]- 1 360 Assembly
- 2 Ada
- 3 ALGOL 68
- 4 C
- 5 C++
- 5.1 Second version
- 6 D
- 6.1 First version
- 6.2 Second version
- 6.3 Third version
- 7 EchoLisp
- 8 Elixir
- 9 ERRE
- 10 F#
- 11 Fortran
- 12 FreeBASIC
- 13 Go
- 14 Haskell
- 15 J
- 16 Java
- 17 JavaScript
- 18 jq
- 19 Julia
- 20 Oforth
- 21 PARI/GP
- 22 Pascal
- 23 Perl 6
- 24 PHP
- 25 PowerShell
- 26 Python
- 27 Racket
- 28 REXX
- 29 Ruby
- 30 Rust
- 31 Sidef
- 32 VBScript
- 33 zkl
360 Assembly[edit]
This program could have been run in 1964. Here, for maximum compatibility, we use only the basic 360 instruction set. Macro instruction XPRNT can be replaced by a WTO.
EULERCO CSECT
USING EULERCO,R13
B 80(R15)
DC 17F'0'
DC CL8'EULERCO'
STM R14,R12,12(R13)
ST R13,4(R15)
ST R15,8(R13)
LR R13,R15
ZAP X1,=P'1'
LOOPX1 ZAP PT,MAXN do x1=1 to maxn-4
SP PT,=P'4'
CP X1,PT
BH ELOOPX1
ZAP PT,X1
AP PT,=P'1'
ZAP X2,PT
LOOPX2 ZAP PT,MAXN do x2=x1+1 to maxn-3
SP PT,=P'3'
CP X2,PT
BH ELOOPX2
ZAP PT,X2
AP PT,=P'1'
ZAP X3,PT
LOOPX3 ZAP PT,MAXN do x3=x2+1 to maxn-2
SP PT,=P'2'
CP X3,PT
BH ELOOPX3
ZAP PT,X3
AP PT,=P'1'
ZAP X4,PT
LOOPX4 ZAP PT,MAXN do x4=x3+1 to maxn-1
SP PT,=P'1'
CP X4,PT
BH ELOOPX4
ZAP PT,X4
AP PT,=P'1'
ZAP X5,PT x5=x4+1
ZAP SUMX,=P'0' sumx=0
ZAP PT,X1 x1
BAL R14,POWER5
AP SUMX,PT
ZAP PT,X2 x2
BAL R14,POWER5
AP SUMX,PT
ZAP PT,X3 x3
BAL R14,POWER5
AP SUMX,PT
ZAP PT,X4 x4
BAL R14,POWER5
AP SUMX,PT sumx=x1**5+x2**5+x3**5+x4**5
ZAP PT,X5 x5
BAL R14,POWER5
ZAP VALX,PT valx=x5**5
LOOPX5 CP X5,MAXN while x5<=maxn & valx<=sumx
BH ELOOPX5
CP VALX,SUMX
BH ELOOPX5
CP VALX,SUMX if valx=sumx
BNE NOTEQUAL
MVI BUF,C' '
MVC BUF+1(79),BUF clear buffer
MVC WC,MASK
ED WC,X1 x1
MVC BUF+0(8),WC+8
MVC WC,MASK
ED WC,X2 x2
MVC BUF+8(8),WC+8
MVC WC,MASK
ED WC,X3 x3
MVC BUF+16(8),WC+8
MVC WC,MASK
ED WC,X4 x4
MVC BUF+24(8),WC+8
MVC WC,MASK
ED WC,X5 x5
MVC BUF+32(8),WC+8
XPRNT BUF,80 output x1,x2,x3,x4,x5
B ELOOPX1
NOTEQUAL ZAP PT,X5
AP PT,=P'1'
ZAP X5,PT x5=x5+1
ZAP PT,X5
BAL R14,POWER5
ZAP VALX,PT valx=x5**5
B LOOPX5
ELOOPX5 AP X4,=P'1'
B LOOPX4
ELOOPX4 AP X3,=P'1'
B LOOPX3
ELOOPX3 AP X2,=P'1'
B LOOPX2
ELOOPX2 AP X1,=P'1'
B LOOPX1
ELOOPX1 L R13,4(0,R13)
LM R14,R12,12(R13)
XR R15,R15
BR R14
POWER5 ZAP PQ,PT ^1
MP PQ,PT ^2
MP PQ,PT ^3
MP PQ,PT ^4
MP PQ,PT ^5
ZAP PT,PQ
BR R14
MAXN DC PL8'250'
X1 DS PL8
X2 DS PL8
X3 DS PL8
X4 DS PL8
X5 DS PL8
SUMX DS PL8
VALX DS PL8
PT DS PL8
PQ DS PL8
WC DS CL17
MASK DC X'40',13X'20',X'212060' CL17
BUF DS CL80
YREGS
END
- Output:
27 84 110 133 144
Ada[edit]
with Ada.Text_IO;
procedure Sum_Of_Powers is
type Base is range 0 .. 250; -- A, B, C, D and Y are in that range
type Num is range 0 .. 4*(250**5); -- (A**5 + ... + D**5) is in that range
subtype Fit is Num range 0 .. 250**5; -- Y**5 is in that range
Modulus: constant Num := 254;
type Modular is mod Modulus;
type Result_Type is array(1..5) of Base; -- this will hold A,B,C,D and Y
type Y_Type is array(Modular) of Base;
type Y_Sum_Type is array(Modular) of Fit;
Y_Sum: Y_Sum_Type := (others => 0);
Y: Y_Type := (others => 0);
-- for I in 0 .. 250, we set Y_Sum(I**5 mod Modulus) := I**5
-- and Y(I**5 mod Modulus) := I
-- Modulus has been chosen to avoid collisions on (I**5 mod Modulus)
-- later we will compute Sum_ABCD := A**5 + B**5 + C**5 + D**5
-- and check if Y_Sum(Sum_ABCD mod modulus) = Sum_ABCD
function Compute_Coefficients return Result_Type is
Sum_A: Fit;
Sum_AB, Sum_ABC, Sum_ABCD: Num;
Short: Modular;
begin
for A in Base(0) .. 246 loop
Sum_A := Num(A) ** 5;
for B in A .. 247 loop
Sum_AB := Sum_A + (Num(B) ** 5);
for C in Base'Max(B,1) .. 248 loop -- if A=B=0 then skip C=0
Sum_ABC := Sum_AB + (Num(C) ** 5);
for D in C .. 249 loop
Sum_ABCD := Sum_ABC + (Num(D) ** 5);
Short := Modular(Sum_ABCD mod Modulus);
if Y_Sum(Short) = Sum_ABCD then
return A & B & C & D & Y(Short);
end if;
end loop;
end loop;
end loop;
end loop;
return 0 & 0 & 0 & 0 & 0;
end Compute_Coefficients;
Tmp: Fit;
ABCD_Y: Result_Type;
begin -- main program
-- initialize Y_Sum and Y
for I in Base(0) .. 250 loop
Tmp := Num(I)**5;
if Y_Sum(Modular(Tmp mod Modulus)) /= 0 then
raise Program_Error with "Collision: Change Modulus and recompile!";
else
Y_Sum(Modular(Tmp mod Modulus)) := Tmp;
Y(Modular(Tmp mod Modulus)) := I;
end if;
end loop;
-- search for a solution (A, B, C, D, Y)
ABCD_Y := Compute_Coefficients;
-- output result
for Number of ABCD_Y loop
Ada.Text_IO.Put(Base'Image(Number));
end loop;
Ada.Text_IO.New_Line;
end Sum_Of_Powers;
- Output:
27 84 110 133 144
ALGOL 68[edit]
# max number will be the highest integer we will consider #
INT max number = 250;
# Construct a table of the fifth powers of 1 : max number #
[ max number ]LONG INT fifth;
FOR i TO max number DO
LONG INT i2 = i * i;
fifth[ i ] := i2 * i2 * i
OD;
# find the first a, b, c, d, e such that a^5 + b^5 + c^5 + d^5 = e^5 #
# as the fifth powers are in order, we can use a binary search to determine #
# whether the value is in the table #
BOOL found := FALSE;
FOR a TO max number WHILE NOT found DO
FOR b FROM a TO max number WHILE NOT found DO
FOR c FROM b TO max number WHILE NOT found DO
FOR d FROM c TO max number WHILE NOT found DO
LONG INT sum = fifth[a] + fifth[b] + fifth[c] + fifth[d];
INT low := d;
INT high := max number;
WHILE low < high
AND NOT found
DO
INT e := ( low + high ) OVER 2;
IF fifth[ e ] = sum
THEN
# the value at e is a fifth power #
found := TRUE;
print( ( ( whole( a, 0 ) + "^5 + " + whole( b, 0 ) + "^5 + "
+ whole( c, 0 ) + "^5 + " + whole( d, 0 ) + "^5 = "
+ whole( e, 0 ) + "^5"
)
, newline
)
)
ELIF sum < fifth[ e ]
THEN high := e - 1
ELSE low := e + 1
FI
OD
OD
OD
OD
OD
Output:
27^5 + 84^5 + 110^5 + 133^5 = 144^5
C[edit]
The trick to speed up was the observation that for any x we have x^5=x modulo 2, 3, and 5, according to the Fermat's little theorem. Thus, based on the Chinese Remainder Theorem we have x^5==x modulo 30 for any x. Therefore, when we have computed the left sum s=a^5+b^5+c^5+d^5, then we know that the right side e^5 must be such that s==e modulo 30. Thus, we do not have to consider all values of e, but only values in the form e=e0+30k, for some starting value e0, and any k. Also, we follow the constraints 1<=a<b<c<d<e<N in the main loop.
// Alexander Maximov, July 2nd, 2015
#include <stdio.h>
#include <time.h>
typedef long long mylong;
void compute(int N, char find_only_one_solution)
{const int M = 30; /* x^5 == x modulo M=2*3*5 */
int a, b, c, d, e;
mylong s, t, max, *p5 = (mylong*)malloc(sizeof(mylong)*(N+M));
for(s=0; s < N; ++s)
p5[s] = s * s, p5[s] *= p5[s] * s;
for(max = p5[N - 1]; s < (N + M); p5[s++] = max + 1);
for(a = 1; a < N; ++a)
for(b = a + 1; b < N; ++b)
for(c = b + 1; c < N; ++c)
for(d = c + 1, e = d + ((t = p5[a] + p5[b] + p5[c]) % M); ((s = t + p5[d]) <= max); ++d, ++e)
{for(e -= M; p5[e + M] <= s; e += M); /* jump over M=30 values for e>d */
if(p5[e] == s)
{printf("%d %d %d %d %d\r\n", a, b, c, d, e);
if(find_only_one_solution) goto onexit;
}
}
onexit:
free(p5);
}
int main(void)
{
int tm = clock();
compute(250, 0);
printf("time=%d milliseconds\r\n", (int)((clock() - tm) * 1000 / CLOCKS_PER_SEC));
return 0;
}
- Output:
The fair way to measure the speed of the code above is to measure it's run time to find all possible solutions to the problem, given N (and not just a single solution, since then the time may depend on the way and the order we organize for-loops).
27 84 110 133 144time=235 milliseconds
Another test with N=1000 produces the following results:
27 84 110 133 14454 168 220 266 28881 252 330 399 432108 336 440 532 576135 420 550 665 720162 504 660 798 864time=65743 milliseconds
PS: The solution for C++ provided below is actually quite good in its design idea behind. However, with all proposed tricks to speed up, the measurements for C++ solution for N=1000 showed the execution time 81447ms (+23%) on the same environment as above for C solution (same machine, same compiler, 64-bit platform). The reason that C++ solution is a bit slower is, perhaps, the fact that the inner loops over rs have complexity ~N/2 steps in average, while with the modulo 30 trick that complexity can be reduced down to ~N/60 steps, although one "expensive" extra %-operation is still needed.
C++[edit]
The simplest brute-force find is already reasonably quick:
#include <iostream>
#include <cmath>
#include <set>
using namespace std;
bool find()
{
const auto MAX = 250;
vector<double> pow5(MAX);
for (auto i = 1; i < MAX; i++)
pow5[i] = (double)i * i * i * i * i;
for (auto x0 = 1; x0 < MAX; x0++) {
for (auto x1 = 1; x1 < x0; x1++) {
for (auto x2 = 1; x2 < x1; x2++) {
for (auto x3 = 1; x3 < x2; x3++) {
auto sum = pow5[x0] + pow5[x1] + pow5[x2] + pow5[x3];
if (binary_search(pow5.begin(), pow5.end(), sum))
{
cout << x0 << " " << x1 << " " << x2 << " " << x3 << " " << pow(sum, 1.0 / 5.0) << endl;
return true;
}
}
}
}
}
// not found
return false;
}
int main(void)
{
int tm = clock();
if (!find())
cout << "Nothing found!\n";
printf("time=%d milliseconds\r\n", (int)((clock() - tm) * 1000 / CLOCKS_PER_SEC));
return 0;
}
- Output:
133 110 84 27 144time=234 milliseconds
We can accelerate this further by creating a parallel std::set<double> p5s containing the elements of the std::vector pow5, and using it to replace the call to std::binary_search:
set<double> pow5s;
for (auto i = 1; i < MAX; i++)
{
pow5[i] = (double)i * i * i * i * i;
pow5s.insert(pow5[i]);
}
//...
if (pow5s.find(sum) != pow5s.end())
This reduces the timing to 125 ms on the same hardware.
A different, more effective optimization is to note that each successive sum is close to the previous one, and use a bidirectional linear search with memory. We also note that inside the innermost loop, we only need to search upward, so we hoist the downward linear search to the loop over x2.
bool find()
{
const auto MAX = 250;
vector<double> pow5(MAX);
for (auto i = 1; i < MAX; i++)
pow5[i] = (double)i * i * i * i * i;
auto rs = 5;
for (auto x0 = 1; x0 < MAX; x0++) {
for (auto x1 = 1; x1 < x0; x1++) {
for (auto x2 = 1; x2 < x1; x2++) {
auto s2 = pow5[x0] + pow5[x1] + pow5[x2];
while (rs > 0 && pow5[rs] > s2) --rs;
for (auto x3 = 1; x3 < x2; x3++) {
auto sum = s2 + pow5[x3];
while (rs < MAX - 1 && pow5[rs] < sum) ++rs;
if (pow5[rs] == sum)
{
cout << x0 << " " << x1 << " " << x2 << " " << x3 << " " << pow(sum, 1.0 / 5.0) << endl;
return true;
}
}
}
}
}
// not found
return false;
}
This reduces the timing to around 25 ms. We could equally well replace rs with an iterator inside pow5; the timing is unaffected.
For comparison with the C code, we also check the timing of an exhaustive search up to MAX=1000. (Don't try this in Python.) This takes 87.2 seconds on the same hardware, comparable to the results found by the C code authors, and supports their conclusion that the mod-30 trick used in the C solution leads to better scalability than the iterator optimizations.
Fortunately, we can incorporate the same trick into the above code, by inserting a forward jump to a feasible solution mod 30 into the loop over x3:
for (auto x3 = 1; x3 < x2; x3++)
{
// go straight to the first appropriate x3, mod 30
if (int err30 = (x0 + x1 + x2 + x3 - rs) % 30)
x3 += 30 - err30;
if (x3 >= x2)
break;
auto sum = s2 + pow5[x3];
With this refinement, the exhaustive search up to MAX=1000 takes 16.9 seconds.
Thanks, C guys!
Second version[edit]
We can create a more efficient method by using the idea (taken from the EchoLisp listing below) of precomputing difference between pairs of fifth powers. If we combine this with the above idea of using linear search with memory, this still requires asymptotically O(N^4) time (because of the linear search within diffs), but is at least twice as fast as the solution above using the mod-30 trick. Exhaustive search up to MAX=1000 took 6.2 seconds for me (64-bit on 3.4GHz i7). It is not clear how it can be combined with the mod-30 trick.
The asymptotic behavior can be improved to O(N^3 ln N) by replacing the linear search with an increasing-increment "hunt" (and the outer linear search, which is also O(N^4), with a call to std::upper_bound). With this replacement, the first solution was found in 0.05 seconds; exhaustive search up to MAX=1000 took 2.80 seconds; and the second nontrivial solution (discarding multiples of the first solution), at y==2615, was found in 94.6 seconds.
template<class C_, class LT_> C_ Unique(const C_& src, const LT_& less)
{
C_ retval(src);
std::sort(retval.begin(), retval.end(), less);
retval.erase(unique(retval.begin(), retval.end()), retval.end());
return retval;
}
template<class I_, class P_> I_ HuntFwd(const I_& hint, const I_& end, const P_& less)// if less(x) is false, then less(x+1) must also be false
{
I_ retval(hint);
int step = 1;
// expanding phase
while (end - retval > step)
{
I_ test = retval + step;
if (!less(test))
break;
retval = test;
step <<= 1;
}
// contracting phase
while (step > 1)
{
step >>= 1;
if (end - retval <= step)
continue;
I_ test = retval + step;
if (less(test))
retval = test;
}
if (retval != end && less(retval))
++retval;
return retval;
}
bool DPFind(int how_many)
{
const int MAX = 1000;
vector<double> pow5(MAX);
for (int i = 1; i < MAX; i++)
pow5[i] = (double)i * i * i * i * i;
vector<pair<double, int>> diffs;
for (int i = 2; i < MAX; ++i)
{
for (int j = 1; j < i; ++j)
diffs.emplace_back(pow5[i] - pow5[j], j);
}
auto firstLess = [](const pair<double, int>& lhs, const pair<double, int>& rhs) { return lhs.first < rhs.first; };
diffs = Unique(diffs, firstLess);
for (int x4 = 4; x4 < MAX - 1; ++x4)
{
for (int x3 = 3; x3 < x4; ++x3)
{
// if (133 * x3 == 110 * x4) continue;// skip duplicates of first solution
const auto s2 = pow5[x4] + pow5[x3];
auto pd = upper_bound(diffs.begin() + 1, diffs.end(), make_pair(s2, 0), firstLess) - 1;
for (int x2 = 2; x2 < x3; ++x2)
{
const auto sum = s2 + pow5[x2];
pd = HuntFwd(pd, diffs.end(), [&](decltype(pd) it){ return it->first < sum; });
if (pd != diffs.end() && pd->first == sum && pd->second < x3)// find each solution only once
{
const double y = pow(pd->first + pow5[pd->second], 0.2);
cout << x4 << " " << x3 << " " << x2 << " " << pd->second << " -> " << static_cast<int>(y + 0.5) << "\n";
if (--how_many <= 0)
return true;
}
}
}
}
return false;
}
Thanks, EchoLisp guys!
D[edit]
First version[edit]
import std.stdio, std.range, std.algorithm, std.typecons;
auto eulersSumOfPowers() {
enum maxN = 250;
auto pow5 = iota(size_t(maxN)).map!(i => ulong(i) ^^ 5).array.assumeSorted;
foreach (immutable x0; 1 .. maxN)
foreach (immutable x1; 1 .. x0)
foreach (immutable x2; 1 .. x1)
foreach (immutable x3; 1 .. x2) {
immutable powSum = pow5[x0] + pow5[x1] + pow5[x2] + pow5[x3];
if (pow5.contains(powSum))
return tuple(x0, x1, x2, x3, pow5.countUntil(powSum));
}
assert(false);
}
void main() {
writefln("%d^5 + %d^5 + %d^5 + %d^5 == %d^5", eulersSumOfPowers[]);
}
- Output:
133^5 + 110^5 + 84^5 + 27^5 == 144^5
Run-time about 0.64 seconds. A Range-based Haskell-like solution is missing because of Issue 14833.
Second version[edit]
void main() {
import std.stdio, std.range, std.algorithm, std.typecons;
enum uint MAX = 250;
uint[ulong] p5;
Tuple!(uint, uint)[ulong] sum2;
foreach (immutable i; 1 .. MAX) {
p5[ulong(i) ^^ 5] = i;
foreach (immutable j; i .. MAX)
sum2[ulong(i) ^^ 5 + ulong(j) ^^ 5] = tuple(i, j);
}
const sk = sum2.keys.sort().release;
foreach (p; p5.keys.sort())
foreach (immutable s; sk) {
if (p <= s)
break;
if (p - s in sum2) {
writeln(p5[p], " ", tuple(sum2[s][], sum2[p - s][]));
return; // Finds first only.
}
}
}
- Output:
144 Tuple!(uint, uint, uint, uint)(27, 84, 110, 133)
Run-time about 0.10 seconds.
Third version[edit]
This solution is a brutal translation of the iterator-based C++ version, and it should be improved to use more idiomatic D Ranges.
import core.stdc.stdio, std.typecons, std.math, std.algorithm, std.range;
alias Pair = Tuple!(double, int);
alias PairPtr = Pair*;
// If less(x) is false, then less(x + 1) must also be false.
PairPtr huntForward(Pred)(PairPtr hint, const PairPtr end, const Pred less) pure nothrow @nogc {
PairPtr result = hint;
int step = 1;
// Expanding phase.
while (end - result > step) {
PairPtr test = result + step;
if (!less(test))
break;
result = test;
step <<= 1;
}
// Contracting phase.
while (step > 1) {
step >>= 1;
if (end - result <= step)
continue;
PairPtr test = result + step;
if (less(test))
result = test;
}
if (result != end && less(result))
++result;
return result;
}
bool dPFind(int how_many) nothrow {
enum MAX = 1_000;
double[MAX] pow5;
foreach (immutable i; 1 .. MAX)
pow5[i] = double(i) ^^ 5;
Pair[] diffs0; // Will contain (MAX-1) * (MAX-2) / 2 pairs.
foreach (immutable i; 2 .. MAX)
foreach (immutable j; 1 .. i)
diffs0 ~= Pair(pow5[i] - pow5[j], j);
// Remove pairs with duplicate first items.
diffs0.length -= diffs0.sort!q{ a[0] < b[0] }.uniq.copy(diffs0).length;
auto diffs = diffs0.assumeSorted!q{ a[0] < b[0] };
foreach (immutable x4; 4 .. MAX - 1) {
foreach (immutable x3; 3 .. x4) {
immutable s2 = pow5[x4] + pow5[x3];
auto pd0 = diffs[1 .. $].upperBound(Pair(s2, 0));
PairPtr pd = &pd0[0] - 1;
foreach (immutable x2; 2 .. x3) {
immutable sum = s2 + pow5[x2];
const PairPtr endPtr = &diffs[$ - 1] + 1;
// This lambda heap-allocates.
pd = huntForward(pd, endPtr, (in PairPtr p) pure => (*p)[0] < sum);
if (pd != endPtr && (*pd)[0] == sum && (*pd)[1] < x3) { // Find each solution only once.
immutable y = ((*pd)[0] + pow5[(*pd)[1]]) ^^ 0.2;
printf("%d %d %d %d : %d\n", x4, x3, x2, (*pd)[1], cast(int)(y + 0.5));
if (--how_many <= 0)
return true;
}
}
}
}
return false;
}
void main() nothrow {
if (!dPFind(100))
printf("Search finished.\n");
}
- Output:
133 110 27 84 : 144133 110 84 27 : 144266 220 54 168 : 288266 220 168 54 : 288399 330 81 252 : 432399 330 252 81 : 432532 440 108 336 : 576532 440 336 108 : 576665 550 135 420 : 720665 550 420 135 : 720798 660 162 504 : 864798 660 504 162 : 864Search finished.
Run-time about 7.1 seconds.
EchoLisp[edit]
To speed up things, we search for x0, x1, x2 such as x0^5 + x1^5 + x2^5 = a difference of 5-th powers.
(define dim 250)
;; speed up n^5
(define (p5 n) (* n n n n n))
(remember 'p5) ;; memoize
;; build vector of all y^5 - x^5 diffs - length 30877
(define all-y^5-x^5
(for*/vector
[(x (in-range 1 dim)) (y (in-range (1+ x) dim))]
(- (p5 y) (p5 x))))
;; sort to use vector-search
(begin (vector-sort! < all-y^5-x^5) 'sorted)
;; find couple (x y) from y^5 - x^5
(define (x-y y^5-x^5)
(for*/fold (x-y null)
[(x (in-range 1 dim)) (y (in-range (1+ x ) dim))]
(when
(= (- (p5 y) (p5 x)) y^5-x^5)
(set! x-y (list x y))
(break #t)))) ; stop on first
;; search
(for*/fold (sol null)
[(x0 (in-range 1 dim)) (x1 (in-range (1+ x0) dim)) (x2 (in-range (1+ x1) dim))]
(set! sol (+ (p5 x0) (p5 x1) (p5 x2)))
(when
(vector-search sol all-y^5-x^5) ;; x0^5 + x1^5 + x2^5 = y^5 - x3^5 ???
(set! sol (append (list x0 x1 x2) (x-y sol))) ;; found
(break #t))) ;; stop on first
→ (27 84 110 133 144) ;; time 2.8 sec
Elixir[edit]
defmodule Euler do
def sum_of_power(max \\ 250) do
{p5, sum2} = setup(max)
sk = Enum.sort(Dict.keys(sum2))
Enum.reduce(Enum.sort(Dict.keys(p5)), Map.new, fn p,map ->
sum(sk, p5, sum2, p, map)
end)
end
def setup(max) do
Enum.reduce(1..max, {%{}, %{}}, fn i,{p5,sum2} ->
i5 = i*i*i*i*i
add = for j <- i..max, do: {i5 + j*j*j*j*j, [i,j]}
{Dict.put(p5, i5, i), Enum.into(add, sum2)}
end)
end
def sum([], _, _, _, map), do: map
def sum([s|_], _, _, p, map) when p<=s, do: map
def sum([s|t], p5, sum2, p, map) do
if sum2[p - s],
do: sum(t, p5, sum2, p, Dict.put(map, Enum.sort(sum2[s] ++ sum2[p-s]), p5[p])),
else: sum(t, p5, sum2, p, map)
end
end
Enum.each(Euler.sum_of_power, fn {k,v} ->
IO.puts Enum.map_join(k, " + ", fn i -> "#{i}**5" end) <> " = #{v}**5"
end)
- Output:
27**5 + 84**5 + 110**5 + 133**5 = 144**5
ERRE[edit]
PROGRAM EULERO
CONST MAX=250
!$DOUBLE
FUNCTION POW5(X)
POW5=X*X*X*X*X
END FUNCTION
!$INCLUDE="PC.LIB"
BEGIN
CLS
FOR X0=1 TO MAX DO
FOR X1=1 TO X0 DO
FOR X2=1 TO X1 DO
FOR X3=1 TO X2 DO
LOCATE(3,1) PRINT(X0;X1;X2;X3)
SUM=POW5(X0)+POW5(X1)+POW5(X2)+POW5(X3)
S1=INT(SUM^0.2#+0.5#)
IF SUM=POW5(S1) THEN PRINT(X0,X1,X2,X3,S1) END IF
END FOR
END FOR
END FOR
END FOR
END PROGRAM
- Output:
133 110 84 27 144
F#[edit]
//Find 4 integers whose 5th powers sum to the fifth power of an integer (Quickly!) - Nigel Galloway: April 23rd., 2015
let G =
let GN = Array.init<float> 250 (fun n -> (float n)**5.0)
let rec gng (n, i, g, e) =
match (n, i, g, e) with
| (250,_,_,_) -> "No Solution Found"
| (_,250,_,_) -> gng (n+1, n+1, n+1, n+1)
| (_,_,250,_) -> gng (n, i+1, i+1, i+1)
| (_,_,_,250) -> gng (n, i, g+1, g+1)
| _ -> let l = GN.[n] + GN.[i] + GN.[g] + GN.[e]
match l with
| _ when l > GN.[249] -> gng(n,i,g+1,g+1)
| _ when l = round(l**0.2)**5.0 -> sprintf "%d**5 + %d**5 + %d**5 + %d**5 = %d**5" n i g e (int (l**0.2))
| _ -> gng(n,i,g,e+1)
gng (1, 1, 1, 1)
- Output:
"27**5 + 84**5 + 110**5 + 133**5 = 144**5"
Fortran[edit]
program sum_of_powers
implicit none
integer, parameter :: maxn = 249
integer, parameter :: dprec = selected_real_kind(15)
integer :: i, x0, x1, x2, x3, y
real(dprec) :: n(maxn), sumx
n = (/ (real(i, dprec)**5, i = 1, maxn) /)
outer: do x0 = 1, maxn
do x1 = 1, maxn
do x2 = 1, maxn
do x3 = 1, maxn
sumx = n(x0)+ n(x1)+ n(x2)+ n(x3)
y = 1
do while(y <= maxn .and. n(y) <= sumx)
if(n(y) == sumx) then
write(*,*) x0, x1, x2, x3, y
exit outer
end if
y = y + 1
end do
end do
end do
end do
end do outer
end program
- Output:
27 84 110 133 144
FreeBASIC[edit]
' version 14-09-2015
' compile with: fbc -s console
' some constants calculated when the program is compiled
Const As UInteger max = 250
Const As ULongInt pow5_max = CULngInt(max) * max * max * max * max
' limit x1, x2, x3
Const As UInteger limit_x1 = (pow5_max / 4) ^ 0.2
Const As UInteger limit_x2 = (pow5_max / 3) ^ 0.2
Const As UInteger limit_x3 = (pow5_max / 2) ^ 0.2
' ------=< MAIN >=------
Dim As ULongInt pow5(max), ans1, ans2, ans3
Dim As UInteger x1, x2, x3, x4, x5 , m1, m2
Cls : Print
For x1 = 1 To max
pow5(x1) = CULngInt(x1) * x1 * x1 * x1 * x1
Next x1
For x1 = 1 To limit_x1
For x2 = x1 +1 To limit_x2
m1 = x1 + x2
ans1 = pow5(x1) + pow5(x2)
If ans1 > pow5_max Then Exit For
For x3 = x2 +1 To limit_x3
ans2 = ans1 + pow5(x3)
If ans2 > pow5_max Then Exit For
m2 = (m1 + x3) Mod 30
If m2 = 0 Then m2 = 30
For x4 = x3 +1 To max -1
ans3 = ans2 + pow5(x4)
If ans3 > pow5_max Then Exit For
For x5 = x4 + m2 To max Step 30
If ans3 < pow5(x5) Then Exit For
If ans3 = pow5(x5) Then
Print x1; "^5 + "; x2; "^5 + "; x3; "^5 + "; _
x4; "^5 = "; x5; "^5"
Exit For, For
EndIf
Next x5
Next x4
Next x3
Next x2
Next x1
Print "done"
' empty keyboard buffer
While Inkey <> "" : Var _key_ = Inkey : Wend
Print : Print "hit any key to end program"
Sleep
End
- Output:
27^5 + 84^5 + 110^5 + 133^5 = 144^5
Go[edit]
package main
import (
"fmt"
"log"
)
func main() {
fmt.Println(eulerSum())
}
func eulerSum() (x0, x1, x2, x3, y int) {
var pow5 [250]int
for i := range pow5 {
pow5[i] = i * i * i * i * i
}
for x0 = 4; x0 < len(pow5); x0++ {
for x1 = 3; x1 < x0; x1++ {
for x2 = 2; x2 < x1; x2++ {
for x3 = 1; x3 < x2; x3++ {
sum := pow5[x0] +
pow5[x1] +
pow5[x2] +
pow5[x3]
for y = x0 + 1; y < len(pow5); y++ {
if sum == pow5[y] {
return
}
}
}
}
}
}
log.Fatal("no solution")
return
}
- Output:
133 110 84 27 144
Haskell[edit]
import Data.List
import Data.List.Ordered
main :: IO ()
main = print $ head [(x0,x1,x2,x3,x4) |
-- choose x0, x1, x2, x3
-- so that 250 < x3 < x2 < x1 < x0
x3 <- [1..250-1],
x2 <- [1..x3-1],
x1 <- [1..x2-1],
x0 <- [1..x1-1],
let p5Sum = x0^5 + x1^5 + x2^5 + x3^5,
-- lazy evaluation of powers of 5
let p5List = [i^5|i <- [1..]],
-- is sum a power of 5 ?
member p5Sum p5List,
-- which power of 5 is sum ?
let Just x4 = elemIndex p5Sum p5List ]
- Output:
(27,84,110,133,144)
J[edit]
require 'stats'
(#~ (= <.)@((+/"1)&.:(^&5)))1+4 comb 248
27 84 110 133
Explanation:
1+4 comb 248finds all the possibilities for our four arguments. Then,
(#~ (= <.)@((+/"1)&.:(^&5)))discards the cases we are not interested in. (It only keeps the case(s) where the fifth root of the sum of the fifth powers is an integer.)
Only one possibility remains.
Here's a significantly faster approach (about 100 times faster), based on the echolisp implementation:
find5=:3 :0
y=. 250
n=. i.y
p=. n^5
a=. (#~ 0&<),-/~p
s=. /:~a
l=. (i.*:y)(#~ 0&<),-/~p
c=. 3 comb <.5%:(y^5)%4
t=. +/"1 c{p
x=. (t e. s)#t
|.,&<&~./|:(y,y)#:l#~a e. x
)
Use:
find5''
┌─────────────┬───┐
│27 84 110 133│144│
└─────────────┴───┘
Note that this particular implementation is a bit hackish, since it relies on the solution being unique for the range of numbers being considered. If there were more solutions it would take a little extra code (though not much time) to untangle them.
Java[edit]
Tested with Java 6.
public class eulerSopConjecture
{
static final int MAX_NUMBER = 250;
public static void main( String[] args )
{
boolean found = false;
long[] fifth = new long[ MAX_NUMBER ];
for( int i = 1; i <= MAX_NUMBER; i ++ )
{
long i2 = i * i;
fifth[ i - 1 ] = i2 * i2 * i;
} // for i
for( int a = 0; a < MAX_NUMBER && ! found ; a ++ )
{
for( int b = a; b < MAX_NUMBER && ! found ; b ++ )
{
for( int c = b; c < MAX_NUMBER && ! found ; c ++ )
{
for( int d = c; d < MAX_NUMBER && ! found ; d ++ )
{
long sum = fifth[a] + fifth[b] + fifth[c] + fifth[d];
int e = java.util.Arrays.binarySearch( fifth, sum );
found = ( e >= 0 );
if( found )
{
// the value at e is a fifth power
System.out.print( (a+1) + "^5 + "
+ (b+1) + "^5 + "
+ (c+1) + "^5 + "
+ (d+1) + "^5 = "
+ (e+1) + "^5"
);
} // if found;;
} // for d
} // for c
} // for b
} // for a
} // main
} // eulerSopConjecture
Output:
27^5 + 84^5 + 110^5 + 133^5 = 144^5
JavaScript[edit]
var eulers_sum_of_powers = function (iMaxN) {
var aPow5 = [];
var oPow5ToN = {};
for (var iP = 0; iP <= iMaxN; iP ++) {
var iPow5 = Math.pow(iP, 5);
aPow5.push(iPow5);
oPow5ToN[iPow5] = iP;
}
for (var i0 = 1; i0 <= iMaxN; i0 ++) {
for (var i1 = 1; i1 <= i0; i1 ++) {
for (var i2 = 1; i2 <= i1; i2 ++) {
for (var i3 = 1; i3 <= i2; i3 ++) {
var iPow5Sum = aPow5[i0] + aPow5[i1] + aPow5[i2] + aPow5[i3];
if (typeof oPow5ToN[iPow5Sum] != 'undefined') {
return {
i0: i0,
i1: i1,
i2: i2,
i3: i3,
iSum: oPow5ToN[iPow5Sum]
};
}
}
}
}
}
};
var oResult = eulers_sum_of_powers(250);
console.log(oResult.i0 + '^5 + ' + oResult.i1 + '^5 + ' + oResult.i2 + '^5 + ' + oResult.i3 + '^5 = ' + oResult.iSum + '^5');
- Output:
133^5 + 110^5 + 84^5 + 27^5 = 144^5Thisthat verify: a^5 + b^5 + c^5 + d^5 = x^5
var N=1000, first=falseOr thisthat verify: a^5 + b^5 + c^5 + d^5 = x^5
var ns={}, npv=[]
for (var n=0; n<=N; n++) {
var np=Math.pow(n,5); ns[np]=n; npv.push(np)
}
loop:
for (var a=1; a<=N; a+=1)
for (var b=a+1; b<=N; b+=1)
for (var c=b+1; c<=N; c+=1)
for (var d=c+1; d<=N; d+=1) {
var x = ns[ npv[a]+npv[b]+npv[c]+npv[d] ]
if (!x) continue
print( [a, b, c, d, x] )
if (first) break loop
}
function print(c) {
var e='<sup>5</sup>', ep=e+' + '
document.write(c[0], ep, c[1], ep, c[2], ep, c[3], e, ' = ', c[4], e, '<br>')
}
var N=1000, first=falseOr thisthat verify: a^5 + b^5 + c^5 = x^5 - d^5
var npv=[], M=30 // x^5 == x modulo M (=2*3*5)
for (var n=0; n<=N; n+=1) npv[n]=Math.pow(n, 5)
var mx=1+npv[N]; while(n<=N+M) npv[n++]=mx
loop:
for (var a=1; a<=N; a+=1)
for (var b=a+1; b<=N; b+=1)
for (var c=b+1; c<=N; c+=1)
for (var t=npv[a]+npv[b]+npv[c], d=c+1, x=t%M+d; (n=t+npv[d])<mx; d+=1, x+=1) {
while (npv[x]<=n) x+=M; x-=M // jump over M=30 values for x>d
if (npv[x] != n) continue
print( [a, b, c, d, x] )
if (first) break loop;
}
function print(c) {
var e='<sup>5</sup>', ep=e+' + '
document.write(c[0], ep, c[1], ep, c[2], ep, c[3], e, ' = ', c[4], e, '<br>')
}
var N=1000, first=falseOr thisthat verify: a^5 + b^5 = x^5 - (c^5 + d^5)
var dxs={}, pow=Math.pow
for (var d=1; d<=N; d+=1)
for (var dp=pow(d,5), x=d+1; x<=N; x+=1)
dxs[pow(x,5)-dp]=[d,x]
loop:
for (var a=1; a<N; a+=1)
for (var ap=pow(a,5), b=a+1; b<N; b+=1)
for (var abp=ap+pow(b,5), c=b+1; c<N; c+=1) {
var dx = dxs[ abp+pow(c,5) ]
if (!dx || c >= dx[0]) continue
print( [a, b, c].concat( dx ) )
if (first) break loop
}
function print(c) {
var e='<sup>5</sup>', ep=e+' + '
document.write(c[0], ep, c[1], ep, c[2], ep, c[3], e, ' = ', c[4], e, '<br>')
}
var N=1000, first=false
var is={}, ipv=[], ijs={}, ijpv=[], pow=Math.pow
for (var i=1; i<=N; i+=1) {
var ip=pow(i,5); is[ip]=i; ipv.push(ip)
for (var j=i+1; j<=N; j+=1) {
var ijp=ip+pow(j,5); ijs[ijp]=[i,j]; ijpv.push(ijp)
}
}
ijpv.sort( function (a,b) {return a - b } )
loop:
for (var i=0, ei=ipv.length; i<ei; i+=1)
for (var xp=ipv[i], j=0, je=ijpv.length; j<je; j+=1) {
var cdp = ijpv[j]
if (cdp >= xp) break
var cd = ijs[xp-cdp]
if (!cd) continue
var ab = ijs[cdp]
if (ab[1] >= cd[0]) continue
print( [].concat(ab, cd, is[xp]) )
if (first) break loop
}
function print(c) {
var e='<sup>5</sup>', ep=e+' + '
document.write(c[0], ep, c[1], ep, c[2], ep, c[3], e, ' = ', c[4], e, '<br>')
}
- Output:
275 + 845 + 1105 + 133 = 1445 545 + 1685 + 2205 + 266 = 2885 815 + 2525 + 3305 + 399 = 4325 1085 + 3365 + 4405 + 532 = 5765 1355 + 4205 + 5505 + 665 = 7205 1625 + 5045 + 6605 + 798 = 8645
jq[edit]
This version finds all non-decreasing solutions within the specified bounds, using a brute-force but not entirely blind approach.
# Search for y in 1 .. maxn (inclusive) for a solution to SIGMA (xi ^ 5) = y^5
# and for each solution with x0<=x1<=...<x3, print [x0, x1, x3, x3, y]
#
def sum_of_powers_conjecture(maxn):
def p5: . as $in | (.*.) | ((.*.) * $in);
def fifth: log / 5 | exp;
# return the fifth root if . is a power of 5
def integral_fifth_root: fifth | if . == floor then . else false end;
(maxn | p5) as $uber
| range(1; maxn) as $x0
| ($x0 | p5) as $s0
| if $s0 < $uber then range($x0; ($uber - $s0 | fifth) + 1) as $x1
| ($s0 + ($x1 | p5)) as $s1
| if $s1 < $uber then range($x1; ($uber - $s1 | fifth) + 1) as $x2
| ($s1 + ($x2 | p5)) as $s2
| if $s2 < $uber then range($x2; ($uber - $s2 | fifth) + 1) as $x3
| ($s2 + ($x3 | p5)) as $sumx
| ($sumx | integral_fifth_root)
| if . then [$x0,$x1,$x2,$x3,.] else empty end
else empty
end
else empty
end
else empty
end ;
The task:
sum_of_powers_conjecture(249)
- Output:
$ jq -c -n -f Euler_sum_of_powers_conjecture_fifth_root.jq
[27,84,110,133,144]
Julia[edit]
const lim = 250
const pwr = 5
const p = [i^pwr for i in 1:lim]
x = zeros(Int, pwr-1)
y = 0
for a in combinations(1:lim, pwr-1)
b = searchsorted(p, sum(p[a]))
0 < length(b) || continue
x = a
y = b[1]
break
end
if y == 0
println("No solution found for power = ", pwr, " and limit = ", lim, ".")
else
s = [@sprintf("%d^%d", i, pwr) for i in x]
s = join(s, " + ")
println("A solution is ", s, " = ", @sprintf("%d^%d", y, pwr), ".")
end
- Output:
A solution is 27^5 + 84^5 + 110^5 + 133^5 = 144^5.
Oforth[edit]
: eulerSum
{
| i j k l ip jp kp |
250 loop: i [
i 5 pow ->ip
i 1 + 250 for: j [
j 5 pow ip + ->jp
j 1 + 250 for: k [
k 5 pow jp + ->kp
k 1 + 250 for: l [
kp l 5 pow + 0.2 powf dup asInteger == ifTrue: [ [ i, j, k, l ] println ]
]
]
]
]
}
- Output:
>eulerSum[27, 84, 110, 133]
PARI/GP[edit]
Naive script:
forvec(v=vector(4,i,[0,250]), if(ispower(v[1]^5+v[2]^5+v[3]^5+v[4]^5,5,&n), print(n" "v)), 2)
- Output:
144 [27, 84, 110, 133]
Naive + caching (setbinop):
{
v2=setbinop((x,y)->[min(x,y),max(x,y),x^5+y^5],[0..250]); \\ sums of two fifth powers
for(i=2,#v2,
for(j=1,i-1,
if(v2[i][2]<v2[j][2] && ispower(v2[i][3]+v2[j][3],5,&n) && #(v=Set([v2[i][1],v2[i][2],v2[j][1],v2[j][2]]))==4,
print(n" "v)
)
)
)
}
- Output:
144 [27, 84, 110, 133]
Pascal[edit]
slightly improved.Reducing calculation time by temporary sum and early break.
program Pot5Test;
{$IFDEF FPC} {$MODE DELPHI}{$ELSE]{$APPTYPE CONSOLE}{$ENDIF}
type
tTest = double;//UInt64;{ On linux 32Bit double is faster than Uint64 }
var
Pot5 : array[0..255] of tTest;
res,tmpSum : tTest;
x0,x1,x2,x3, y : NativeUint;//= Uint32 or 64 depending on OS xx-Bit
i : byte;
BEGIN
For i := 1 to 255 do
Pot5[i] := (i*i*i*i)*Uint64(i);
For x0 := 1 to 250-3 do
For x1 := x0+1 to 250-2 do
For x2 := x1+1 to 250-1 do
Begin
//set y here only, because pot5 is strong monoton growing,
//therefor the sum is strong monoton growing too.
y := x2+2;// aka x3+1
tmpSum := Pot5[x0]+Pot5[x1]+Pot5[x2];
For x3 := x2+1 to 250 do
Begin
res := tmpSum+Pot5[x3];
while (y< 250) AND (res > Pot5[y]) do
inc(y);
IF y > 250 then BREAK;
if res = Pot5[y] then
writeln(x0,'^5+',x1,'^5+',x2,'^5+',x3,'^5 = ',y,'^5');
end;
end;
END.
- output
27^5+84^5+110^5+133^5 = 144^5real 0m1.091s {Uint64; Linux 32}real 0m0.761s {double; Linux 32}real 0m0.511s{Uint64; Linux 64}Perl 6[edit]
constant MAX = 250;
my %p5{Int};
my %sum2{Int};
for 1..MAX -> $i {
%p5{$i**5} = $i;
for 1..MAX -> $j {
%sum2{$i**5 + $j**5} = ($i, $j);
}
}
my @sk = %sum2.keys.sort;
for %p5.keys.sort -> $p {
for @sk -> $s {
next if $p <= $s;
if %sum2{$p - $s} {
say (%sum2{$s}[],%sum2{$p-$s}[] X~ '⁵').join(' + ') ~ " = %p5{$p}⁵";
exit;
}
}
}
- Output:
84⁵ + 27⁵ + 133⁵ + 110⁵ = 144⁵
PHP[edit]
<?php
function eulers_sum_of_powers () {
$max_n = 250;
$pow_5 = array();
$pow_5_to_n = array();
for ($p = 1; $p <= $max_n; $p ++) {
$pow5 = pow($p, 5);
$pow_5 [$p] = $pow5;
$pow_5_to_n[$pow5] = $p;
}
foreach ($pow_5 as $n_0 => $p_0) {
foreach ($pow_5 as $n_1 => $p_1) {
if ($n_0 < $n_1) continue;
foreach ($pow_5 as $n_2 => $p_2) {
if ($n_1 < $n_2) continue;
foreach ($pow_5 as $n_3 => $p_3) {
if ($n_2 < $n_3) continue;
$pow_5_sum = $p_0 + $p_1 + $p_2 + $p_3;
if (isset($pow_5_to_n[$pow_5_sum])) {
return array($n_0, $n_1, $n_2, $n_3, $pow_5_to_n[$pow_5_sum]);
}
}
}
}
}
}
list($n_0, $n_1, $n_2, $n_3, $y) = eulers_sum_of_powers();
echo "$n_0^5 + $n_1^5 + $n_2^5 + $n_3^5 = $y^5";
?>
- Output:
133^5 + 110^5 + 84^5 + 27^5 = 144^5
PowerShell[edit]
Brute Force Search
This is a very slow algorithm.
#EULER.PS1
$max = 250
for($x0=1;$x0 -lt $max;$x0++){
for($x1=1;$x1 -lt $x0;$x1++){
for($x2=1;$x2 -lt $x1;$x2++){
for($x3=1;$x3 -lt $x2;$x3++){
$sum=[Math]::pow($x0,5)+`
[Math]::pow($x1,5)+`
[Math]::pow($x2,5)+`
[Math]::pow($x3,5)
$S1=[Math]::truncate([Math]::pow($sum,0.2))
if($sum -eq [Math]::pow($S1,5)){
Write-host $x0 " " $x1 " " $x2 " " $x3 "-" $S1
return
}
}
}
}
}
- Output:
133 110 84 27 - 144
Python[edit]
def eulers_sum_of_powers():
max_n = 250
pow_5 = [n**5 for n in range(max_n)]
pow5_to_n = {n**5: n for n in range(max_n)}
for x0 in range(1, max_n):
for x1 in range(1, x0):
for x2 in range(1, x1):
for x3 in range(1, x2):
pow_5_sum = sum(pow_5[i] for i in (x0, x1, x2, x3))
if pow_5_sum in pow5_to_n:
y = pow5_to_n[pow_5_sum]
return (x0, x1, x2, x3, y)
print("%i**5 + %i**5 + %i**5 + %i**5 == %i**5" % eulers_sum_of_powers())
- Output:
133**5 + 110**5 + 84**5 + 27**5 == 144**5
The above can be written as:
from itertools import combinations
def eulers_sum_of_powers():
max_n = 250
pow_5 = [n**5 for n in range(max_n)]
pow5_to_n = {n**5: n for n in range(max_n)}
for x0, x1, x2, x3 in combinations(range(1, max_n), 4):
pow_5_sum = sum(pow_5[i] for i in (x0, x1, x2, x3))
if pow_5_sum in pow5_to_n:
y = pow5_to_n[pow_5_sum]
return (x0, x1, x2, x3, y)
print("%i**5 + %i**5 + %i**5 + %i**5 == %i**5" % eulers_sum_of_powers())
- Output:
27**5 + 84**5 + 110**5 + 133**5 == 144**5
It's much faster to cache and look up sums of two fifth powers, due to the small allowed range:
MAX = 250
p5, sum2 = {}, {}
for i in range(1, MAX):
p5[i**5] = i
for j in range(i, MAX):
sum2[i**5 + j**5] = (i, j)
sk = sorted(sum2.keys())
for p in sorted(p5.keys()):
for s in sk:
if p <= s: break
if p - s in sum2:
print(p5[p], sum2[s] + sum2[p-s])
exit()
- Output:
144 (27, 84, 110, 133)
Racket[edit]
#lang racket
(define MAX 250)
(define pow5 (make-vector MAX))
(for ([i (in-range 1 MAX)])
(vector-set! pow5 i (expt i 5)))
(define pow5s (list->set (vector->list pow5)))
(let/ec break
(for* ([x0 (in-range 1 MAX)]
[x1 (in-range 1 x0)]
[x2 (in-range 1 x1)]
[x3 (in-range 1 x2)])
(define sum (+ (vector-ref pow5 x0)
(vector-ref pow5 x1)
(vector-ref pow5 x2)
(vector-ref pow5 x3)))
(when (set-member? pow5s sum)
(displayln (list x0 x1 x2 x3 (inexact->exact (round (expt sum 1/5)))))
(break))))
- Output:
(133 110 84 27 144)
REXX[edit]
Programming note: the 3rd argument can be specified which causes an attempt to find N solutions.
The starting and ending (low and high) values can also be specified (to limit or expand the search range).
If any of the arguments are omitted, they default to the Rosetta Code task's specifications.
The method used is:
- precompute all powers of five (within the confines of allowed integers)
- precompute all (positive) differences between two applicable 5th powers
- see if any of the sums of any three 5th powers are equal to any of those (above) differences
- {thanks to the real nifty idea (↑↑↑) from userID G. Brougnard}
- see if the sum of any four 5th powers is equal to any 5th power
- (this is needed as the fourth number d isn't known yet).
- {all of the above utilizes REXX's sparse stemmed array hashing which eliminates the need for sorting.}
By implementing (userID) G. Brougnard's idea of differences of two 5th powers,
the time used for computation was cut by over a factor of seventy.
In essence, the new formula being solved is: aⁿ + bⁿ + cⁿ == xⁿ ─ dⁿ
which lends itself to algorithm optimization by (only) having to:
- [the right side of the above equation] pre-compute all possible differences between any two applicable integer powers of five (there are 30,135 unique differences)
- [the left side of the above equation] sum any applicable three integer powers of five
- [the == part of the above equation] see if any of the above sums match any of the ≈30k differences
/*REXX program finds unique positive integer solution(s) for the equation: ─────
────────────────────────────────────── aⁿ + bⁿ + cⁿ + dⁿ == xⁿ where n=5*/
say center(' 'strip(translate(sourceLine(2), ,"*─/"))' ', 79, "─") /*title.*/
parse arg L H N . /*get optional LOW, HIGH, #solutions.*/
if L=='' | L==',' then L= 0 + 1 /*Not specified? Then use the default.*/
if H=='' | H==',' then H=250 - 1 /* " " " " " " */
if N=='' | N==',' then N= 1 /* " " " " " " */
numeric digits 1000 /*be able to handle the next expression*/
numeric digits max(9,length(3*H**5)) /* " " " " 3* [H to 5th power]*/
aH=H-3; bH=H-2; cH=H-1 /*calculate the upper DO loop limits.*/
!.=0 /* [↓] define values of 5th powers. */
do pow=1 for H; @.pow=pow**5; _=@.pow; !._=1; $._=pow; end
?.=!.
do j=4 to cH; do k=j+1 to H; _=@.k-@.j; ?._=1; end; end
/*define [↑] 5th power differences.*/
#=0 /*#: is the number of solutions found.*/ /* [↓] for N=∞ solutions.*/
do a=L to aH; s0= @.a /*traipse through possible A values. */ /*◄──done 246 times.*/
do b=a+1 to bH; s1=s0+@.b /* " " " B " */ /*◄──done 30,381 times.*/
do c=b+1 to cH; s2=s1+@.c /* " " " C " */ /*◄──done 2,511,496 times.*/
if ?.s2 then do d=c+1 to H; s=s2+@.d /*find appropriate solution.*/
if !.s then call results /*A solution? Then show it.*/
end /*d*/ /* [↑] !.S is a boolean.*/
end /*c*/
end /*b*/
end /*a*/
if #==0 then say "Didn't find a solution."; signal done
/*────────────────────────────────────────────────────────────────────────────*/
results: _=left('',5); #=#+1 /*_: used as a spacer; bump # counter.*/
say _ 'solution' #":" _ 'a='a _ "b="||b _ 'c='c _ "d="d _ 'x='$.s
if #<N then return /*return, keep searching for more sols.*/
done: exit # /*stick a fork in it, we're all done. */
output when using the default inputs:
─────────────────── aⁿ + bⁿ + cⁿ + dⁿ == xⁿ where n=5 ─────────────────── solution 1: a=27 b=84 c=110 d=133 x=144
Ruby[edit]
Brute force:
power5 = (1..250).each_with_object({}){|i,h| h[i**5]=i}
result = power5.keys.repeated_combination(4).select{|a| power5[a.inject(:+)]}
puts result.map{|a| a.map{|i| "#{power5[i]}**5"}.join(' + ') + " = #{power5[a.inject(:+)]}**5"}
- Output:
27**5 + 84**5 + 110**5 + 133**5 = 144**5
Faster version:
p5, sum2, max = {}, {}, 250
(1..max).each do |i|
p5[i**5] = i
(i..max).each{|j| sum2[i**5 + j**5] = [i,j]}
end
result = {}
sk = sum2.keys.sort
p5.keys.sort.each do |p|
sk.each do |s|
break if p <= s
result[(sum2[s] + sum2[p-s]).sort] = p5[p] if sum2[p - s]
end
end
result.each{|k,v| puts k.map{|i| "#{i}**5"}.join(' + ') + " = #{v}**5"}
The output is the same above.
Rust[edit]
const MAX_N : u64 = 250;
fn eulers_sum_of_powers() -> (usize, usize, usize, usize, usize) {
let pow5: Vec<u64> = (0..MAX_N).map(|i| i.pow(5)).collect();
let pow5_to_n = |pow| pow5.binary_search(&pow);
for x0 in 1..MAX_N as usize {
for x1 in 1..x0 {
for x2 in 1..x1 {
for x3 in 1..x2 {
let pow_sum = pow5[x0] + pow5[x1] + pow5[x2] + pow5[x3];
if let Ok(n) = pow5_to_n(pow_sum) {
return (x0, x1, x2, x3, n)
}
}
}
}
}
panic!();
}
fn main() {
let (x0, x1, x2, x3, y) = eulers_sum_of_powers();
println!("{}^5 + {}^5 + {}^5 + {}^5 == {}^5", x0, x1, x2, x3, y)
}
- Output:
133^5 + 110^5 + 84^5 + 27^5 == 144^5
Sidef[edit]
define MAX = 250;
var p5 = Hash();
var sum2 = Hash();
MAX.times { |i|
p5{i**5} = i;
MAX.times { |j|
sum2{i**5 + j**5} = [i, j];
}
}
var sk = sum2.keys.map{.to_n}.sort;
p5.keys.map{.to_n}.sort.each { |p|
sk.each { |s|
next if (p <= s);
if (sum2.exists(p - s)) {
sum2{s} + sum2{p-s} -> map{|n| "#{n}⁵" }
-> join(' + ') + " = #{p5{p}}⁵"
-> say;
goto :END;
}
}
} @:END;
- Output:
84⁵ + 27⁵ + 133⁵ + 110⁵ = 144⁵
VBScript[edit]
Max=250
For X0=1 To Max
For X1=1 To X0
For X2=1 To X1
For X3=1 To X2
Sum=fnP5(X0)+fnP5(X1)+fnP5(X2)+fnP5(X3)
S1=Int(Sum^0.2)
If Sum=fnP5(S1) Then
WScript.StdOut.Write X0 & " " & X1 & " " & X2 & " " & X3 & " " & S1
WScript.Quit
End If
Next
Next
Next
Next
Function fnP5(n)
fnP5 = n ^ 5
End Function
- Output:
133 110 84 27 144
zkl[edit]
Uses two look up tables for efficiency. Counts from 0 for ease of coding.
pow5s:=[1..249].apply("pow",5); // (1^5, 2^5, 3^5 .. 249^5)
pow5r:=pow5s.enumerate().apply("reverse").toDictionary(); // [144^5:144, ...]
foreach x0,x1,x2,x3 in (249,x0,x1,x2){
sum:=pow5s[x0] + pow5s[x1] + pow5s[x2] + pow5s[x3];
if(pow5r.holds(sum))
println("%d^5 + %d^5 + %d^5 + %d^5 = %d^5"
.fmt(x3+1,x2+1,x1+1,x0+1,pow5r[sum]+1));
}- Output:
27^5 + 84^5 + 110^5 + 133^5 = 144^5
Using the Python technique of caching double sums results in a 5x speed up [to the first/only solution]; actually the speed up is close to 25x but creating the caches dominates the runtime to the first solution.
p5,sum2:=D(),D();
foreach i in ([1..249]){
p5[i.pow(5)]=i;
foreach j in ([i..249]){ sum2[i.pow(5) + j.pow(5)]=T(i,j) } // 31,125 keys
}
sk:=sum2.keys.apply("toInt").copy().sort(); // RW list sorts faster than a RO one
foreach p,s in (p5.keys.apply("toInt"),sk){
if(p<=s) break;
if(sum2.holds(p - s)){
println("%d^5 + %d^5 + %d^5 + %d^5 = %d^5"
.fmt(sum2[s].xplode(),sum2[p - s].xplode(),p5[p]));
break(2); // or get permutations
}
}
Note: dictionary keys are always strings and copying a read only list creates a read write list.
- Output:
27^5 + 84^5 + 110^5 + 133^5 = 144^5
0 0
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