Taxicab numbers
You are encouraged to solve this task according to the task description, using any language you may know.
A taxicab number (the definition that is being used here) is a positive integer that can be expressed as the sum of two positive cubes in more than one way.
The first taxicab number is 1729 : 13 + 123 = 93 + 103
- Task requirements
- Compute and display the lowest 25 taxicab numbers (in numeric order, and in a human-readable format).
- For each of the taxicab numbers, show the number as well as it's constituent cubes.
- Extra credit
- Show the 2,000th taxicab number, and a half dozen more
- See also
- A001235 taxicab numbers on The On-Line Encyclopedia of Integer Sequences.
- Hardy-Ramanujan Number on MathWorld.
- taxicab number on MathWorld.
- taxicab number on Wikipedia.
C
Using a priority queue to emit sum of two cubs in order. It's reasonably fast and doesn't use excessive amount of memory (the heap is only at 245 length upon the 2006th taxi). <lang c>#include <stdio.h>
- include <stdlib.h>
typedef unsigned long long xint; typedef unsigned uint; typedef struct { uint x, y; // x > y always xint value; } sum_t;
xint *cube; uint n_cubes;
sum_t *pq; uint pq_len, pq_cap;
void add_cube(void) { uint x = n_cubes++; cube = realloc(cube, sizeof(xint) * (n_cubes + 1)); cube[n_cubes] = (xint) n_cubes*n_cubes*n_cubes; if (x < 2) return; // x = 0 or 1 is useless
if (++pq_len >= pq_cap) { if (!(pq_cap *= 2)) pq_cap = 2; pq = realloc(pq, sizeof(*pq) * pq_cap); }
sum_t tmp = (sum_t) { x, 1, cube[x] + 1 }; // upheap uint i, j; for (i = pq_len; i >= 1 && pq[j = i>>1].value > tmp.value; i = j) pq[i] = pq[j];
pq[i] = tmp; }
void next_sum(void) { redo: while (!pq_len || pq[1].value >= cube[n_cubes]) add_cube();
sum_t tmp = pq[0] = pq[1]; // pq[0] always stores last seen value if (++tmp.y >= tmp.x) { // done with this x; throw it away tmp = pq[pq_len--]; if (!pq_len) goto redo; // refill empty heap } else tmp.value += cube[tmp.y] - cube[tmp.y-1];
uint i, j; // downheap for (i = 1; (j = i<<1) <= pq_len; pq[i] = pq[j], i = j) { if (j < pq_len && pq[j+1].value < pq[j].value) ++j; if (pq[j].value >= tmp.value) break; } pq[i] = tmp; }
uint next_taxi(sum_t *hist) { do next_sum(); while (pq[0].value != pq[1].value);
uint len = 1; hist[0] = pq[0]; do { hist[len++] = pq[1]; next_sum(); } while (pq[0].value == pq[1].value);
return len; }
int main(void) { uint i, l; sum_t x[10]; for (i = 1; i <= 2006; i++) { l = next_taxi(x); if (25 < i && i < 2000) continue; printf("%4u:%10llu", i, x[0].value); while (l--) printf(" = %4u^3 + %4u^3", x[l].x, x[l].y); putchar('\n'); } return 0; }</lang>
- Output:
1: 1729 = 12^3 + 1^3 = 10^3 + 9^3 2: 4104 = 15^3 + 9^3 = 16^3 + 2^3 3: 13832 = 20^3 + 18^3 = 24^3 + 2^3 4: 20683 = 27^3 + 10^3 = 24^3 + 19^3 5: 32832 = 30^3 + 18^3 = 32^3 + 4^3 6: 39312 = 33^3 + 15^3 = 34^3 + 2^3 7: 40033 = 33^3 + 16^3 = 34^3 + 9^3 8: 46683 = 30^3 + 27^3 = 36^3 + 3^3 9: 64232 = 36^3 + 26^3 = 39^3 + 17^3 10: 65728 = 33^3 + 31^3 = 40^3 + 12^3 11: 110656 = 40^3 + 36^3 = 48^3 + 4^3 12: 110808 = 45^3 + 27^3 = 48^3 + 6^3 13: 134379 = 43^3 + 38^3 = 51^3 + 12^3 14: 149389 = 50^3 + 29^3 = 53^3 + 8^3 15: 165464 = 48^3 + 38^3 = 54^3 + 20^3 16: 171288 = 54^3 + 24^3 = 55^3 + 17^3 17: 195841 = 57^3 + 22^3 = 58^3 + 9^3 18: 216027 = 59^3 + 22^3 = 60^3 + 3^3 19: 216125 = 50^3 + 45^3 = 60^3 + 5^3 20: 262656 = 60^3 + 36^3 = 64^3 + 8^3 21: 314496 = 66^3 + 30^3 = 68^3 + 4^3 22: 320264 = 66^3 + 32^3 = 68^3 + 18^3 23: 327763 = 58^3 + 51^3 = 67^3 + 30^3 24: 373464 = 60^3 + 54^3 = 72^3 + 6^3 25: 402597 = 61^3 + 56^3 = 69^3 + 42^3 2000:1671816384 = 1168^3 + 428^3 = 944^3 + 940^3 2001:1672470592 = 1124^3 + 632^3 = 1187^3 + 29^3 2002:1673170856 = 1034^3 + 828^3 = 1164^3 + 458^3 2003:1675045225 = 1153^3 + 522^3 = 1081^3 + 744^3 2004:1675958167 = 1096^3 + 711^3 = 1159^3 + 492^3 2005:1676926719 = 1188^3 + 63^3 = 1095^3 + 714^3 2006:1677646971 = 990^3 + 891^3 = 1188^3 + 99^3
C#
<lang csharp>using System; using System.Collections.Generic; using System.Linq; using System.Text;
namespace TaxicabNumber {
class Program
{
static void Main(string[] args)
{
IDictionary<long, IList<Tuple<int, int>>> taxicabNumbers = GetTaxicabNumbers(2006);
PrintTaxicabNumbers(taxicabNumbers);
Console.ReadKey();
}
private static IDictionary<long, IList<Tuple<int, int>>> GetTaxicabNumbers(int length)
{
SortedList<long, IList<Tuple<int, int>>> sumsOfTwoCubes = new SortedList<long, IList<Tuple<int, int>>>();
for (int i = 1; i < int.MaxValue; i++)
{
for (int j = 1; j < int.MaxValue; j++)
{
long sum = (long)(Math.Pow((double)i, 3) + Math.Pow((double)j, 3));
if (!sumsOfTwoCubes.ContainsKey(sum))
{
sumsOfTwoCubes.Add(sum, new List<Tuple<int, int>>());
}
sumsOfTwoCubes[sum].Add(new Tuple<int, int>(i, j));
if (j >= i)
{
break;
}
}
// Found that you need to keep going for a while after the length, because higher i values fill in gaps
if (sumsOfTwoCubes.Count(t => t.Value.Count >= 2) >= length * 1.1)
{
break;
}
}
IDictionary<long, IList<Tuple<int, int>>> values = (from t in sumsOfTwoCubes where t.Value.Count >= 2 select t)
.Take(2006)
.ToDictionary(u => u.Key, u => u.Value);
return values;
}
private static void PrintTaxicabNumbers(IDictionary<long, IList<Tuple<int, int>>> values)
{
int i = 1;
foreach (long taxicabNumber in values.Keys)
{
StringBuilder output = new StringBuilder().AppendFormat("{0,10}\t{1,4}", i, taxicabNumber);
foreach (Tuple<int, int> numbers in values[taxicabNumber])
{
output.AppendFormat("\t= {0}^3 + {1}^3", numbers.Item1, numbers.Item2);
}
if (i <= 25 || (i >= 2000 && i <= 2006))
{
Console.WriteLine(output.ToString());
}
i++;
}
}
}
}</lang>
Clojure
<lang clojure>(ns test-project-intellij.core
(:gen-class))
(defn cube [x]
"Cube a number through triple multiplication" (* x x x))
(defn sum3 i j
" [i j] -> i^3 + j^3" (+ (cube i) (cube j)))
(defn next-pair i j
" Generate next [i j] pair of sequence (producing lower triangle pairs) " (if (< j i) [i (inc j)] [(inc i) 1]))
- Pair sequence generator [1 1] [2 1] [2 2] [3 1] [3 2] [3 3] ...
(def pairs-seq (iterate next-pair [1 1]))
(defn dict-inc [m pair]
" Add pair to pair map m, with the key of the map based upon the cubic sum (sum3) and the value appends the pair " (update-in m [(sum3 pair)] (fnil #(conj % pair) [])))
(defn enough? [m n-to-generate]
" Checks if we have enough taxi numbers (i.e. if number in map >= count-needed "
(->> m ; hash-map of sum of cube of numbers [key] and their pairs as value
(filter #(if (> (count (second %)) 1) true false)) ; filter out ones which don't have more than 1 entry
(count) ; count the item remaining
(<= n-to-generate))) ; true iff count-needed is less or equal to the nubmer filtered
(defn find-taxi-numbers [n-to-generate]
" Generates 1st n-to-generate taxi numbers"
(loop [m {} ; Hash-map containing cube of pairs (key) and set of pairs that produce sum (value)
p pairs-seq ; select pairs from our pair sequence generator (i.e. [1 1] [2 1] [2 2] ...)
num-tried 0 ; Since its expensve to count how many taxi numbers we have found
check-after 1] ; we only check if we have enough numbers every time (num-tried equals check-after)
; num-tried increments by 1 each time we try the next pair and
; check-after doubles if we don't have enough taxi numbers
(if (and (= num-tried check-after) (enough? m n-to-generate)) ; check if we found enough taxi numbers
(sort-by first (into [] (filter #(> (count (second %)) 1) m))) ; sort the taxi numbers and this is the result
(if (= num-tried check-after) ; Check if we need to increase our count between checking
(recur (dict-inc m (first p)) (rest p) (inc num-tried) (* 2 check-after)) ; increased count between checking
(recur (dict-inc m (first p)) (rest p) (inc num-tried) check-after))))) ; didn't increase the count
- Generate 1st 2006 taxi numbers
(def result (find-taxi-numbers 2006))
- Show First 25
(defn show-result [n sample]
" Prints one line of result "
(print (format "%4d:%10d" n (first sample)))
(doseq [q (second sample)
:let [[i j] q]]
(print (format " = %4d^3 + %4d^3" i j)))
(println))
- 1st 25 taxi numbers
(doseq [n (range 1 26)
:let [sample (nth result (dec n))]] (show-result n sample))
- taxi numbers from 2000th to 2006th
(doseq [n (range 2000 2007)
:let [sample (nth result (dec n))]] (show-result n sample))
}</lang>
- Output:
1: 1729 = 10^3 + 9^3 = 12^3 + 1^3 2: 4104 = 15^3 + 9^3 = 16^3 + 2^3 3: 13832 = 20^3 + 18^3 = 24^3 + 2^3 4: 20683 = 24^3 + 19^3 = 27^3 + 10^3 5: 32832 = 30^3 + 18^3 = 32^3 + 4^3 6: 39312 = 33^3 + 15^3 = 34^3 + 2^3 7: 40033 = 33^3 + 16^3 = 34^3 + 9^3 8: 46683 = 30^3 + 27^3 = 36^3 + 3^3 9: 64232 = 36^3 + 26^3 = 39^3 + 17^3 10: 65728 = 33^3 + 31^3 = 40^3 + 12^3 11: 110656 = 40^3 + 36^3 = 48^3 + 4^3 12: 110808 = 45^3 + 27^3 = 48^3 + 6^3 13: 134379 = 43^3 + 38^3 = 51^3 + 12^3 14: 149389 = 50^3 + 29^3 = 53^3 + 8^3 15: 165464 = 48^3 + 38^3 = 54^3 + 20^3 16: 171288 = 54^3 + 24^3 = 55^3 + 17^3 17: 195841 = 57^3 + 22^3 = 58^3 + 9^3 18: 216027 = 59^3 + 22^3 = 60^3 + 3^3 19: 216125 = 50^3 + 45^3 = 60^3 + 5^3 20: 262656 = 60^3 + 36^3 = 64^3 + 8^3 21: 314496 = 66^3 + 30^3 = 68^3 + 4^3 22: 320264 = 66^3 + 32^3 = 68^3 + 18^3 23: 327763 = 58^3 + 51^3 = 67^3 + 30^3 24: 373464 = 60^3 + 54^3 = 72^3 + 6^3 25: 402597 = 61^3 + 56^3 = 69^3 + 42^3 2000:1671816384 = 944^3 + 940^3 = 1168^3 + 428^3 2001:1672470592 = 1124^3 + 632^3 = 1187^3 + 29^3 2002:1673170856 = 1034^3 + 828^3 = 1164^3 + 458^3 2003:1675045225 = 1081^3 + 744^3 = 1153^3 + 522^3 2004:1675958167 = 1096^3 + 711^3 = 1159^3 + 492^3 2005:1676926719 = 1095^3 + 714^3 = 1188^3 + 63^3 2006:1677646971 = 990^3 + 891^3 = 1188^3 + 99^3
D
High Level Version
<lang d>void main() /*@safe*/ {
import std.stdio, std.range, std.algorithm, std.typecons, std.string;
auto iCubes = iota(1u, 1201u).map!(x => tuple(x, x ^^ 3));
bool[Tuple!(uint, uint)][uint] sum2cubes;
foreach (i, immutable i3; iCubes)
foreach (j, immutable j3; iCubes[i .. $])
sum2cubes[i3 + j3][tuple(i, j)] = true;
const taxis = sum2cubes.byKeyValue.filter!(p => p.value.length > 1)
.array.schwartzSort!(p => p.key).release;
foreach (/*immutable*/ const r; [[0, 25], [2000 - 1, 2000 + 6]]) {
foreach (immutable i, const t; taxis[r[0] .. r[1]])
writefln("%4d: %10d =%-(%s =%)", i + r[0] + 1, t.key,
t.value.keys.sort().map!q{"%4d^3 + %4d^3".format(a[])});
writeln;
}
}</lang>
- Output:
1: 1729 = 1^3 + 12^3 = 9^3 + 10^3 2: 4104 = 2^3 + 16^3 = 9^3 + 15^3 3: 13832 = 2^3 + 24^3 = 18^3 + 20^3 4: 20683 = 10^3 + 27^3 = 19^3 + 24^3 5: 32832 = 4^3 + 32^3 = 18^3 + 30^3 6: 39312 = 2^3 + 34^3 = 15^3 + 33^3 7: 40033 = 9^3 + 34^3 = 16^3 + 33^3 8: 46683 = 3^3 + 36^3 = 27^3 + 30^3 9: 64232 = 17^3 + 39^3 = 26^3 + 36^3 10: 65728 = 12^3 + 40^3 = 31^3 + 33^3 11: 110656 = 4^3 + 48^3 = 36^3 + 40^3 12: 110808 = 6^3 + 48^3 = 27^3 + 45^3 13: 134379 = 12^3 + 51^3 = 38^3 + 43^3 14: 149389 = 8^3 + 53^3 = 29^3 + 50^3 15: 165464 = 20^3 + 54^3 = 38^3 + 48^3 16: 171288 = 17^3 + 55^3 = 24^3 + 54^3 17: 195841 = 9^3 + 58^3 = 22^3 + 57^3 18: 216027 = 3^3 + 60^3 = 22^3 + 59^3 19: 216125 = 5^3 + 60^3 = 45^3 + 50^3 20: 262656 = 8^3 + 64^3 = 36^3 + 60^3 21: 314496 = 4^3 + 68^3 = 30^3 + 66^3 22: 320264 = 18^3 + 68^3 = 32^3 + 66^3 23: 327763 = 30^3 + 67^3 = 51^3 + 58^3 24: 373464 = 6^3 + 72^3 = 54^3 + 60^3 25: 402597 = 42^3 + 69^3 = 56^3 + 61^3 2000: 1671816384 = 428^3 + 1168^3 = 940^3 + 944^3 2001: 1672470592 = 29^3 + 1187^3 = 632^3 + 1124^3 2002: 1673170856 = 458^3 + 1164^3 = 828^3 + 1034^3 2003: 1675045225 = 522^3 + 1153^3 = 744^3 + 1081^3 2004: 1675958167 = 492^3 + 1159^3 = 711^3 + 1096^3 2005: 1676926719 = 63^3 + 1188^3 = 714^3 + 1095^3 2006: 1677646971 = 99^3 + 1188^3 = 891^3 + 990^3
Run-time: about 2.9 seconds with dmd compiler.
Heap-Based Version
<lang d>import std.stdio, std.string, std.container;
struct CubeSum {
ulong x, y, value;
this(in ulong x_, in ulong y_) pure nothrow @safe @nogc {
this.x = x_;
this.y = y_;
this.value = x_ ^^ 3 + y_ ^^ 3;
}
}
final class Taxi {
BinaryHeap!(Array!CubeSum, "a.value > b.value") pq; CubeSum last; ulong n = 0;
this() {
last = nextSum();
}
CubeSum nextSum() {
while (pq.empty || pq.front.value >= n ^^ 3)
pq.insert(CubeSum(++n, 1));
auto s = pq.front;
pq.removeFront;
if (s.x > s.y + 1)
pq.insert(CubeSum(s.x, s.y + 1));
return s; }
CubeSum[] nextTaxi() {
CubeSum s;
typeof(return) train;
while ((s = nextSum).value != last.value)
last = s;
train ~= last;
do {
train ~= s;
} while ((s = nextSum).value == last.value);
last = s;
return train; }
}
void main() {
auto taxi = new Taxi;
foreach (immutable i; 1 .. 2007) {
const t = taxi.nextTaxi;
if (i > 25 && i < 2000)
continue;
writef("%4d: %10d", i, t[0].value);
foreach (const s; t)
writef(" = %4d^3 + %4d^3", s.x, s.y);
writeln;
}
}</lang>
- Output:
1: 1729 = 10^3 + 9^3 = 12^3 + 1^3 2: 4104 = 15^3 + 9^3 = 16^3 + 2^3 3: 13832 = 20^3 + 18^3 = 24^3 + 2^3 4: 20683 = 24^3 + 19^3 = 27^3 + 10^3 5: 32832 = 30^3 + 18^3 = 32^3 + 4^3 6: 39312 = 33^3 + 15^3 = 34^3 + 2^3 7: 40033 = 33^3 + 16^3 = 34^3 + 9^3 8: 46683 = 30^3 + 27^3 = 36^3 + 3^3 9: 64232 = 39^3 + 17^3 = 36^3 + 26^3 10: 65728 = 40^3 + 12^3 = 33^3 + 31^3 11: 110656 = 40^3 + 36^3 = 48^3 + 4^3 12: 110808 = 45^3 + 27^3 = 48^3 + 6^3 13: 134379 = 51^3 + 12^3 = 43^3 + 38^3 14: 149389 = 50^3 + 29^3 = 53^3 + 8^3 15: 165464 = 48^3 + 38^3 = 54^3 + 20^3 16: 171288 = 54^3 + 24^3 = 55^3 + 17^3 17: 195841 = 57^3 + 22^3 = 58^3 + 9^3 18: 216027 = 59^3 + 22^3 = 60^3 + 3^3 19: 216125 = 50^3 + 45^3 = 60^3 + 5^3 20: 262656 = 60^3 + 36^3 = 64^3 + 8^3 21: 314496 = 66^3 + 30^3 = 68^3 + 4^3 22: 320264 = 68^3 + 18^3 = 66^3 + 32^3 23: 327763 = 67^3 + 30^3 = 58^3 + 51^3 24: 373464 = 60^3 + 54^3 = 72^3 + 6^3 25: 402597 = 69^3 + 42^3 = 61^3 + 56^3 2000: 1671816384 = 1168^3 + 428^3 = 944^3 + 940^3 2001: 1672470592 = 1124^3 + 632^3 = 1187^3 + 29^3 2002: 1673170856 = 1164^3 + 458^3 = 1034^3 + 828^3 2003: 1675045225 = 1153^3 + 522^3 = 1081^3 + 744^3 2004: 1675958167 = 1159^3 + 492^3 = 1096^3 + 711^3 2005: 1676926719 = 1095^3 + 714^3 = 1188^3 + 63^3 2006: 1677646971 = 990^3 + 891^3 = 1188^3 + 99^3
Run-time: about 0.31 seconds with ldc2 compiler. It's faster than the Java solution.
Low Level Heap-Based Version
<lang d>struct Taxicabs {
alias CubesSumT = uint; // Or ulong.
static struct Sum {
CubesSumT value;
uint x, y;
}
// The cubes can be pre-computed if CubesSumT is a BigInt. private uint nCubes; private Sum[] pq; private uint pq_len;
private void addCube() pure nothrow @safe {
nCubes = nCubes ? nCubes + 1 : 2;
if (nCubes < 2)
return; // 0 or 1 is useless.
pq_len++;
if (pq_len >= pq.length)
pq.length = (pq.length == 0) ? 2 : (pq.length * 2);
immutable tmp = Sum(CubesSumT(nCubes - 2) ^^ 3 + 1,
nCubes - 2, 1);
// Upheap.
uint i = pq_len;
for (; i >= 1 && pq[i >> 1].value > tmp.value; i >>= 1)
pq[i] = pq[i >> 1];
pq[i] = tmp; }
private void nextSum() pure nothrow @safe {
while (!pq_len || pq[1].value >= (nCubes - 1) ^^ 3)
addCube();
Sum tmp = pq[0] = pq[1]; //pq[0] always stores last seen value.
tmp.y++;
if (tmp.y >= tmp.x) { // Done with this x; throw it away.
tmp = pq[pq_len];
pq_len--;
if (!pq_len)
return nextSum(); // Refill empty heap.
} else
tmp.value += tmp.y ^^ 3 - (tmp.y - 1) ^^ 3;
// Downheap.
uint i = 1;
while (true) {
uint j = i << 1;
if (j > pq_len)
break;
if (j < pq_len && pq[j + 1].value < pq[j].value)
j++;
if (pq[j].value >= tmp.value)
break;
pq[i] = pq[j];
i = j;
}
pq[i] = tmp; }
Sum[] nextTaxi(size_t N)(ref Sum[N] hist)
pure nothrow @safe {
do {
nextSum();
} while (pq[0].value != pq[1].value);
uint len = 1;
hist[0] = pq[0];
do {
hist[len] = pq[1];
len++;
nextSum();
} while (pq[0].value == pq[1].value);
return hist[0 .. len]; }
}
void main() nothrow {
import core.stdc.stdio;
Taxicabs t; Taxicabs.Sum[3] x;
foreach (immutable uint i; 1 .. 2007) {
const triples = t.nextTaxi(x);
if (i > 25 && i < 2000)
continue;
printf("%4u: %10lu", i, triples[0].value);
foreach_reverse (const s; triples)
printf(" = %4u^3 + %4u^3", s.x, s.y);
'\n'.putchar;
}
}</lang>
- Output:
1: 1729 = 12^3 + 1^3 = 10^3 + 9^3 2: 4104 = 15^3 + 9^3 = 16^3 + 2^3 3: 13832 = 20^3 + 18^3 = 24^3 + 2^3 4: 20683 = 27^3 + 10^3 = 24^3 + 19^3 5: 32832 = 30^3 + 18^3 = 32^3 + 4^3 6: 39312 = 33^3 + 15^3 = 34^3 + 2^3 7: 40033 = 33^3 + 16^3 = 34^3 + 9^3 8: 46683 = 30^3 + 27^3 = 36^3 + 3^3 9: 64232 = 36^3 + 26^3 = 39^3 + 17^3 10: 65728 = 33^3 + 31^3 = 40^3 + 12^3 11: 110656 = 40^3 + 36^3 = 48^3 + 4^3 12: 110808 = 45^3 + 27^3 = 48^3 + 6^3 13: 134379 = 43^3 + 38^3 = 51^3 + 12^3 14: 149389 = 50^3 + 29^3 = 53^3 + 8^3 15: 165464 = 48^3 + 38^3 = 54^3 + 20^3 16: 171288 = 54^3 + 24^3 = 55^3 + 17^3 17: 195841 = 57^3 + 22^3 = 58^3 + 9^3 18: 216027 = 59^3 + 22^3 = 60^3 + 3^3 19: 216125 = 50^3 + 45^3 = 60^3 + 5^3 20: 262656 = 60^3 + 36^3 = 64^3 + 8^3 21: 314496 = 66^3 + 30^3 = 68^3 + 4^3 22: 320264 = 66^3 + 32^3 = 68^3 + 18^3 23: 327763 = 58^3 + 51^3 = 67^3 + 30^3 24: 373464 = 60^3 + 54^3 = 72^3 + 6^3 25: 402597 = 61^3 + 56^3 = 69^3 + 42^3 2000: 1671816384 = 1168^3 + 428^3 = 944^3 + 940^3 2001: 1672470592 = 1124^3 + 632^3 = 1187^3 + 29^3 2002: 1673170856 = 1034^3 + 828^3 = 1164^3 + 458^3 2003: 1675045225 = 1153^3 + 522^3 = 1081^3 + 744^3 2004: 1675958167 = 1096^3 + 711^3 = 1159^3 + 492^3 2005: 1676926719 = 1188^3 + 63^3 = 1095^3 + 714^3 2006: 1677646971 = 990^3 + 891^3 = 1188^3 + 99^3
Run-time: about 0.08 seconds with ldc2 compiler.
DCL
We invoke external utility SORT which I suppose technically speaking is not a formal part of the language but is darn handy at times; <lang DCL>$ close /nolog sums_of_cubes $ on control_y then $ goto clean $ open /write sums_of_cubes sums_of_cubes.txt $ i = 1 $ loop1: $ write sys$output i $ j = 1 $ loop2: $ sum = i * i * i + j * j * j $ if sum .lt. 0 $ then $ write sys$output "overflow at ", j $ goto next_i $ endif $ write sums_of_cubes f$fao( "!10SL,!10SL,!10SL", sum, i, j ) $ j = j + 1 $ if j .le. i then $ goto loop2 $ next_i: $ i = i + 1 $ if i .le. 1289 then $ goto loop1 ! cube_root of 2^31-1 $ close sums_of_cubes $ sort sums_of_cubes.txt sorted_sums_of_cubes.txt $ close /nolog sorted_sums_of_cubes $ open sorted_sums_of_cubes sorted_sums_of_cubes.txt $ count = 0 $ read sorted_sums_of_cubes prev_prev_line ! need to detect when there are more than just 2 different sums, e.g. 456 $ prev_prev_sum = f$element( 0, ",", f$edit( prev_prev_line, "collapse" )) $ read sorted_sums_of_cubes prev_line $ prev_sum = f$element( 0,",", f$edit( prev_line, "collapse" )) $ loop3: $ read /end_of_file = done sorted_sums_of_cubes line $ sum = f$element( 0, ",", f$edit( line, "collapse" )) $ if sum .eqs. prev_sum $ then $ if sum .nes. prev_prev_sum then $ count = count + 1 $ int_sum = f$integer( sum ) $ i1 = f$integer( f$element( 1, ",", prev_line )) $ j1 = f$integer( f$element( 2, ",", prev_line )) $ i2 = f$integer( f$element( 1, ",", line )) $ j2 = f$integer( f$element( 2, ",", line )) $ if count .le. 25 .or. ( count .ge. 2000 .and. count .le. 2006 ) then - $ write sys$output f$fao( "!4SL:!11SL =!5SL^3 +!5SL^3 =!5SL^3 +!5SL^3", count, int_sum, i1, j1, i2, j2 ) $ endif $ prev_prev_line = prev_line $ prev_prev_sum = prev_sum $ prev_line = line $ prev_sum = sum $ goto loop3 $ done: $ close sorted_sums_of_cubes $ exit $ $ clean: $ close /nolog sorted_sums_of_cubes $ close /nolog sums_of_cubes</lang>
- Output:
$ @taxicab_numbers 1: 1729 = 10^3 + 9^3 = 12^3 + 1^3 2: 4104 = 15^3 + 9^3 = 16^3 + 2^3 3: 13832 = 20^3 + 18^3 = 24^3 + 2^3 4: 20683 = 24^3 + 19^3 = 27^3 + 10^3 5: 32832 = 30^3 + 18^3 = 32^3 + 4^3 6: 39312 = 33^3 + 15^3 = 34^3 + 2^3 7: 40033 = 33^3 + 16^3 = 34^3 + 9^3 8: 46683 = 30^3 + 27^3 = 36^3 + 3^3 9: 64232 = 36^3 + 26^3 = 39^3 + 17^3 10: 65728 = 33^3 + 31^3 = 40^3 + 12^3 11: 110656 = 40^3 + 36^3 = 48^3 + 4^3 12: 110808 = 45^3 + 27^3 = 48^3 + 6^3 13: 134379 = 43^3 + 38^3 = 51^3 + 12^3 14: 149389 = 50^3 + 29^3 = 53^3 + 8^3 15: 165464 = 48^3 + 38^3 = 54^3 + 20^3 16: 171288 = 54^3 + 24^3 = 55^3 + 17^3 17: 195841 = 57^3 + 22^3 = 58^3 + 9^3 18: 216027 = 59^3 + 22^3 = 60^3 + 3^3 19: 216125 = 50^3 + 45^3 = 60^3 + 5^3 20: 262656 = 60^3 + 36^3 = 64^3 + 8^3 21: 314496 = 66^3 + 30^3 = 68^3 + 4^3 22: 320264 = 66^3 + 32^3 = 68^3 + 18^3 23: 327763 = 58^3 + 51^3 = 67^3 + 30^3 24: 373464 = 60^3 + 54^3 = 72^3 + 6^3 25: 402597 = 61^3 + 56^3 = 69^3 + 42^3 2000: 1671816384 = 944^3 + 940^3 = 1168^3 + 428^3 2001: 1672470592 = 1124^3 + 632^3 = 1187^3 + 29^3 2002: 1673170856 = 1034^3 + 828^3 = 1164^3 + 458^3 2003: 1675045225 = 1081^3 + 744^3 = 1153^3 + 522^3 2004: 1675958167 = 1096^3 + 711^3 = 1159^3 + 492^3 2005: 1676926719 = 1095^3 + 714^3 = 1188^3 + 63^3 2006: 1677646971 = 990^3 + 891^3 = 1188^3 + 99^3
EchoLisp
Using the heap library, and a heap to store the taxicab numbers. For taxi tuples - decomposition in more than two sums - we use the group function which transforms a list ( 3 5 5 6 8 ...) into ((3) (5 5) (6) ...). <lang scheme> (require '(heap compile))
(define (scube a b) (+ (* a a a) (* b b b))) (compile 'scube "-f") ; "-f" means : no bigint, no rational used
- is n - a^3 a cube b^3?
- if yes return b, else #f
(define (taxi? n a (b 0)) (set! b (cbrt (- n (* a a a)))) ;; cbrt is ∛ (when (and (< b a) (integer? b)) b)) (compile 'taxi? "-f")
- |-------------------
looking for taxis
|#
- remove from heap until heap-top >= a
- when twins are removed, it is a taxicab number
- push it
- at any time (top stack) = last removed
(define (clean-taxi H limit: a min-of-heap: htop) (when (and htop (> a htop)) (when (!= (stack-top S) htop) (pop S)) (push S htop) (heap-pop H) (clean-taxi H a (heap-top H)))) (compile 'clean-taxi "-f")
- loop on a and b, b <=a , until n taxicabs found
(define (taxicab (n 2100)) (for ((a (in-naturals))) (clean-taxi H (* a a a) (heap-top H)) #:break (> (stack-length S) n) (for ((b a)) (heap-push H (scube a b)))))
- |------------------
printing taxis
|#
- string of all decompositions
(define (taxi->string i n) (string-append (format "%d. %d " (1+ i) n) (for/string ((a (cbrt n))) #:when (taxi? n a) (format " = %4d^3 + %4d^3" a (taxi? n a)))))
(define (taxi-print taxis (nfrom 0) (nto 26)) (for ((i (in-naturals nfrom)) (taxi (sublist taxis nfrom nto))) (writeln (taxi->string i (first taxi))))) </lang>
- Output:
<lang scheme> (define S (stack 'S)) ;; to push taxis (define H (make-heap < )) ;; make min heap of all scubes
(taxicab 2100) (define taxis (group (stack->list S))) (taxi-print taxis )
1. 1729 = 10^3 + 9^3 = 12^3 + 1^3 2. 4104 = 15^3 + 9^3 = 16^3 + 2^3 3. 13832 = 20^3 + 18^3 = 24^3 + 2^3 4. 20683 = 24^3 + 19^3 = 27^3 + 10^3
- | ... |#
24. 373464 = 60^3 + 54^3 = 72^3 + 6^3 25. 402597 = 61^3 + 56^3 = 69^3 + 42^3 26. 439101 = 69^3 + 48^3 = 76^3 + 5^3
(taxi-print taxis 1999 2006) 2000. 1671816384 = 944^3 + 940^3 = 1168^3 + 428^3 2001. 1672470592 = 1124^3 + 632^3 = 1187^3 + 29^3 2002. 1673170856 = 1034^3 + 828^3 = 1164^3 + 458^3 2003. 1675045225 = 1081^3 + 744^3 = 1153^3 + 522^3 2004. 1675958167 = 1096^3 + 711^3 = 1159^3 + 492^3 2005. 1676926719 = 1095^3 + 714^3 = 1188^3 + 63^3 2006. 1677646971 = 990^3 + 891^3 = 1188^3 + 99^3
- extra bonus
- print all taxis which are triplets
(define (taxi-tuples taxis (nfrom 0) (nto 2000)) (for ((i (in-naturals nfrom)) (taxi (sublist taxis nfrom nto))) #:when (> (length taxi) 1) ;; filter for tuples is here (writeln (taxi->string i (first taxi)))))
(taxi-tuples taxis)
455. 87539319 = 414^3 + 255^3 = 423^3 + 228^3 = 436^3 + 167^3 535. 119824488 = 428^3 + 346^3 = 492^3 + 90^3 = 493^3 + 11^3 588. 143604279 = 423^3 + 408^3 = 460^3 + 359^3 = 522^3 + 111^3 655. 175959000 = 525^3 + 315^3 = 552^3 + 198^3 = 560^3 + 70^3 888. 327763000 = 580^3 + 510^3 = 661^3 + 339^3 = 670^3 + 300^3 1299. 700314552 = 828^3 + 510^3 = 846^3 + 456^3 = 872^3 + 334^3 1398. 804360375 = 920^3 + 295^3 = 927^3 + 198^3 = 930^3 + 15^3 1515. 958595904 = 856^3 + 692^3 = 984^3 + 180^3 = 986^3 + 22^3 1660. 1148834232 = 846^3 + 816^3 = 920^3 + 718^3 = 1044^3 + 222^3 1837. 1407672000 = 1050^3 + 630^3 = 1104^3 + 396^3 = 1120^3 + 140^3 </lang>
Elixir
<lang elixir>defmodule Taxicab do
def numbers(n \\ 1200) do
(for i <- 1..n, j <- i..n, do: {i,j})
|> Enum.group_by(fn {i,j} -> i*i*i + j*j*j end)
|> Enum.filter(fn {_,v} -> length(v)>1 end)
|> Enum.sort
end
end
nums = Taxicab.numbers |> Enum.with_index Enum.each(nums, fn {x,i} ->
if i in 0..24 or i in 1999..2005 do
IO.puts "#{i+1} : #{inspect x}"
end
end)</lang>
- Output:
1 : {1729, [{9, 10}, {1, 12}]}
2 : {4104, [{9, 15}, {2, 16}]}
3 : {13832, [{18, 20}, {2, 24}]}
4 : {20683, [{19, 24}, {10, 27}]}
5 : {32832, [{18, 30}, {4, 32}]}
6 : {39312, [{15, 33}, {2, 34}]}
7 : {40033, [{16, 33}, {9, 34}]}
8 : {46683, [{27, 30}, {3, 36}]}
9 : {64232, [{26, 36}, {17, 39}]}
10 : {65728, [{31, 33}, {12, 40}]}
11 : {110656, [{36, 40}, {4, 48}]}
12 : {110808, [{27, 45}, {6, 48}]}
13 : {134379, [{38, 43}, {12, 51}]}
14 : {149389, [{29, 50}, {8, 53}]}
15 : {165464, [{38, 48}, {20, 54}]}
16 : {171288, [{24, 54}, {17, 55}]}
17 : {195841, [{22, 57}, {9, 58}]}
18 : {216027, [{22, 59}, {3, 60}]}
19 : {216125, [{45, 50}, {5, 60}]}
20 : {262656, [{36, 60}, {8, 64}]}
21 : {314496, [{30, 66}, {4, 68}]}
22 : {320264, [{32, 66}, {18, 68}]}
23 : {327763, [{51, 58}, {30, 67}]}
24 : {373464, [{54, 60}, {6, 72}]}
25 : {402597, [{56, 61}, {42, 69}]}
2000 : {1671816384, [{940, 944}, {428, 1168}]}
2001 : {1672470592, [{632, 1124}, {29, 1187}]}
2002 : {1673170856, [{828, 1034}, {458, 1164}]}
2003 : {1675045225, [{744, 1081}, {522, 1153}]}
2004 : {1675958167, [{711, 1096}, {492, 1159}]}
2005 : {1676926719, [{714, 1095}, {63, 1188}]}
2006 : {1677646971, [{891, 990}, {99, 1188}]}
FreeBASIC
<lang freebasic>' version 11-10-2016 ' compile with: fbc -s console
' Brute force
' adopted from "Sorting algorithms/Shell" sort Task Sub shellsort(s() As String)
' sort from lower bound to the highter bound Dim As UInteger lb = LBound(s) Dim As UInteger ub = UBound(s) Dim As Integer done, i, inc = ub - lb
Do inc = inc / 2.2 If inc < 1 Then inc = 1
Do
done = 0
For i = lb To ub - inc
If s(i) > s(i + inc) Then
Swap s(i), s(i + inc)
done = 1
End If
Next
Loop Until done = 0
Loop Until inc = 1
End Sub
' ------=< MAIN >=------
Dim As UInteger x, y, count, c, sum Dim As UInteger cube(1290) Dim As String result(), str1, str2, str3 Dim As String buf11 = Space(11), buf5 = Space(5) ReDim result(900000) ' ~1291*1291\2
' set up the cubes Print : Print " Calculate cubes" For x = 1 To 1290
cube(x) = x*x*x
Next
' combine and store Print : Print " Combine cubes" For x = 1 To 1290
For y = x To 1290 sum = cube(x)+cube(y) RSet buf11, Str(sum) : str1 = buf11 RSet buf5, Str(x) : str2 = buf5 RSet buf5, Str(y) : Str3 = buf5 result(count)=buf11 + " = " + str2 + " ^ 3 + " + str3 + " ^ 3" count = count +1 Next
Next
count= count -1 ReDim Preserve result(count) ' trim the array
Print : Print " Sort (takes some time)" shellsort(result()) ' sort
Print : Print " Find the Taxicab numbers" c = 1 ' start at index 1 For x = 0 To count -1
' find sums that match
If Left(result(x), 11) = Left(result(x + 1), 11) Then
result(c) = result(x)
y = x +1
Do ' merge the other solution(s)
result(c) = result(c) + Mid(result(y), 12)
y = y +1
Loop Until Left(result(x), 11) <> Left(result(y), 11)
x = y -1 ' let x point to last match result
c = c +1
End If
Next
c = c -1 Print : Print " "; c; " Taxicab numbers found" ReDim Preserve result(c) ' trim the array again
cls Print : Print " Print first 25 numbers" : Print For x = 1 To 25
Print result(x)
Next
Print : Print " The 2000th to the 2006th" : Print For x = 2000 To 2006
Print result(x)
Next
' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End</lang>
- Output:
Print first 25 numbers
1729 = 1 ^ 3 + 12 ^ 3 = 9 ^ 3 + 10 ^ 3
4104 = 2 ^ 3 + 16 ^ 3 = 9 ^ 3 + 15 ^ 3
13832 = 2 ^ 3 + 24 ^ 3 = 18 ^ 3 + 20 ^ 3
20683 = 10 ^ 3 + 27 ^ 3 = 19 ^ 3 + 24 ^ 3
32832 = 4 ^ 3 + 32 ^ 3 = 18 ^ 3 + 30 ^ 3
39312 = 2 ^ 3 + 34 ^ 3 = 15 ^ 3 + 33 ^ 3
40033 = 9 ^ 3 + 34 ^ 3 = 16 ^ 3 + 33 ^ 3
46683 = 3 ^ 3 + 36 ^ 3 = 27 ^ 3 + 30 ^ 3
64232 = 17 ^ 3 + 39 ^ 3 = 26 ^ 3 + 36 ^ 3
65728 = 12 ^ 3 + 40 ^ 3 = 31 ^ 3 + 33 ^ 3
110656 = 4 ^ 3 + 48 ^ 3 = 36 ^ 3 + 40 ^ 3
110808 = 6 ^ 3 + 48 ^ 3 = 27 ^ 3 + 45 ^ 3
134379 = 12 ^ 3 + 51 ^ 3 = 38 ^ 3 + 43 ^ 3
149389 = 8 ^ 3 + 53 ^ 3 = 29 ^ 3 + 50 ^ 3
165464 = 20 ^ 3 + 54 ^ 3 = 38 ^ 3 + 48 ^ 3
171288 = 17 ^ 3 + 55 ^ 3 = 24 ^ 3 + 54 ^ 3
195841 = 9 ^ 3 + 58 ^ 3 = 22 ^ 3 + 57 ^ 3
216027 = 3 ^ 3 + 60 ^ 3 = 22 ^ 3 + 59 ^ 3
216125 = 5 ^ 3 + 60 ^ 3 = 45 ^ 3 + 50 ^ 3
262656 = 8 ^ 3 + 64 ^ 3 = 36 ^ 3 + 60 ^ 3
314496 = 4 ^ 3 + 68 ^ 3 = 30 ^ 3 + 66 ^ 3
320264 = 18 ^ 3 + 68 ^ 3 = 32 ^ 3 + 66 ^ 3
327763 = 30 ^ 3 + 67 ^ 3 = 51 ^ 3 + 58 ^ 3
373464 = 6 ^ 3 + 72 ^ 3 = 54 ^ 3 + 60 ^ 3
402597 = 42 ^ 3 + 69 ^ 3 = 56 ^ 3 + 61 ^ 3
The 2000th to the 2006th
1671816384 = 428 ^ 3 + 1168 ^ 3 = 940 ^ 3 + 944 ^ 3
1672470592 = 29 ^ 3 + 1187 ^ 3 = 632 ^ 3 + 1124 ^ 3
1673170856 = 458 ^ 3 + 1164 ^ 3 = 828 ^ 3 + 1034 ^ 3
1675045225 = 522 ^ 3 + 1153 ^ 3 = 744 ^ 3 + 1081 ^ 3
1675958167 = 492 ^ 3 + 1159 ^ 3 = 711 ^ 3 + 1096 ^ 3
1676926719 = 63 ^ 3 + 1188 ^ 3 = 714 ^ 3 + 1095 ^ 3
1677646971 = 99 ^ 3 + 1188 ^ 3 = 891 ^ 3 + 990 ^ 3
Go
<lang go>package main
import ( "container/heap" "fmt" )
type CubeSum struct { x, y uint16 value uint64 }
func (c *CubeSum) fixvalue() { c.value = cubes[c.x] + cubes[c.y] }
type CubeSumHeap []*CubeSum
func (h CubeSumHeap) Len() int { return len(h) } func (h CubeSumHeap) Less(i, j int) bool { return h[i].value < h[j].value } func (h CubeSumHeap) Swap(i, j int) { h[i], h[j] = h[j], h[i] } func (h *CubeSumHeap) Push(x interface{}) { (*h) = append(*h, x.(*CubeSum)) } func (h *CubeSumHeap) Pop() interface{} { x := (*h)[len(*h)-1] *h = (*h)[:len(*h)-1] return x }
type TaxicabGen struct { n int h CubeSumHeap }
var cubes []uint64 // cubes[i] == i*i*i func cubesExtend(i int) { for n := uint64(len(cubes)); n <= uint64(i); n++ { cubes = append(cubes, n*n*n) } }
func (g *TaxicabGen) min() CubeSum { for len(g.h) == 0 || g.h[0].value > cubes[g.n] { g.n++ cubesExtend(g.n) heap.Push(&g.h, &CubeSum{uint16(g.n), 1, cubes[g.n] + 1}) } // Note, we use g.h[0] to "peek" at the min heap entry. c := *(g.h[0]) if c.y+1 <= c.x { // Instead of Pop and Push we modify in place and fix. g.h[0].y++ g.h[0].fixvalue() heap.Fix(&g.h, 0) } else { heap.Pop(&g.h) } return c }
// Originally this was just: type Taxicab [2]CubeSum // and we always returned two sums. Now we return all the sums. type Taxicab []CubeSum
func (t Taxicab) String() string { //return fmt.Sprintf("%12d =%5d³ +%5d³ =%5d³ +%5d³", // t[0].value, t[0].x, t[0].y, t[1].x, t[1].y)
// Likely more efficient to use a bytes.Buffer here. s := fmt.Sprintf("%12d", t[0].value) for _, p := range t { s = s + fmt.Sprintf(" =%5d³ +%5d³", p.x, p.y) } return s }
func (g *TaxicabGen) Next() Taxicab { a, b := g.min(), g.min() for a.value != b.value { a, b = b, g.min() } //return Taxicab{a,b}
// Originally this just returned Taxicab{a,b} and we didn't look // further into the heap. Since we start by looking at the next // pair, that is okay until the first Taxicab number with four // ways of expressing the cube, which doesn't happen until the // 97,235th Taxicab: // 6963472309248 = 16630³ + 13322³ = 18072³ + 10200³ // = 18948³ + 5436³ = 19083³ + 2421³ // Now we return all ways so we need to peek into the heap. t := Taxicab{a, b} for g.h[0].value == b.value { t = append(t, g.min()) } return t }
func main() { const ( low = 25 mid = 2e3 high = 4e4 ) var tg TaxicabGen firstn := 3 // To show the first tripple, quadruple, etc for i := 1; i <= high+6; i++ { t := tg.Next() switch { case len(t) >= firstn: firstn++ fallthrough case i <= low || (mid <= i && i <= mid+6) || i >= high: //fmt.Printf("h:%-4d ", len(tg.h)) fmt.Printf("%5d: %v\n", i, t) } } }</lang> On a modern 64 bit processor this takes ~89 msec for the first 2000 and ~9.2 sec for the first 40,000.
- Output:
1: 1729 = 12³ + 1³ = 10³ + 9³
2: 4104 = 16³ + 2³ = 15³ + 9³
3: 13832 = 24³ + 2³ = 20³ + 18³
4: 20683 = 27³ + 10³ = 24³ + 19³
5: 32832 = 32³ + 4³ = 30³ + 18³
6: 39312 = 34³ + 2³ = 33³ + 15³
7: 40033 = 34³ + 9³ = 33³ + 16³
8: 46683 = 36³ + 3³ = 30³ + 27³
9: 64232 = 36³ + 26³ = 39³ + 17³
10: 65728 = 40³ + 12³ = 33³ + 31³
11: 110656 = 48³ + 4³ = 40³ + 36³
12: 110808 = 48³ + 6³ = 45³ + 27³
13: 134379 = 51³ + 12³ = 43³ + 38³
14: 149389 = 53³ + 8³ = 50³ + 29³
15: 165464 = 54³ + 20³ = 48³ + 38³
16: 171288 = 55³ + 17³ = 54³ + 24³
17: 195841 = 58³ + 9³ = 57³ + 22³
18: 216027 = 60³ + 3³ = 59³ + 22³
19: 216125 = 60³ + 5³ = 50³ + 45³
20: 262656 = 64³ + 8³ = 60³ + 36³
21: 314496 = 68³ + 4³ = 66³ + 30³
22: 320264 = 66³ + 32³ = 68³ + 18³
23: 327763 = 58³ + 51³ = 67³ + 30³
24: 373464 = 72³ + 6³ = 60³ + 54³
25: 402597 = 69³ + 42³ = 61³ + 56³
455: 87539319 = 436³ + 167³ = 423³ + 228³ = 414³ + 255³
2000: 1671816384 = 1168³ + 428³ = 944³ + 940³
2001: 1672470592 = 1187³ + 29³ = 1124³ + 632³
2002: 1673170856 = 1164³ + 458³ = 1034³ + 828³
2003: 1675045225 = 1081³ + 744³ = 1153³ + 522³
2004: 1675958167 = 1096³ + 711³ = 1159³ + 492³
2005: 1676926719 = 1188³ + 63³ = 1095³ + 714³
2006: 1677646971 = 990³ + 891³ = 1188³ + 99³
40000: 976889700163 = 8659³ + 6894³ = 9891³ + 2098³
40001: 976942087381 = 7890³ + 7861³ = 8680³ + 6861³
40002: 976946344920 = 9476³ + 5014³ = 9798³ + 3312³
40003: 976962998375 = 9912³ + 1463³ = 8415³ + 7250³
40004: 976974757064 = 9365³ + 5379³ = 9131³ + 5997³
40005: 977025552984 = 9894³ + 2040³ = 9792³ + 3366³
40006: 977104161000 = 9465³ + 5055³ = 9920³ + 970³
Haskell
<lang haskell>import Data.List (sortBy, groupBy, tails, transpose) import Data.Ord (comparing)
-- TAXICAB NUMBERS ---------------------------------------------------- taxis :: Int -> (Int, ((Int, Int), (Int, Int))) taxis nCubes =
filter ((> 1) . length) $ groupBy ((. fst) . (==) . fst) $ sortBy (comparing fst) [ (fst x + fst y, (x, y)) | (x:t) <- tails $ ((,) =<< (^ 3)) <$> [1 .. nCubes] , y <- t ]
-- Taxicab numbers composed from first 1200 cubes xs :: [(Int, [(Int, ((Int, Int), (Int, Int)))])] xs = zip [1 ..] (taxis 1200)
-- PRETTY PRINTING ---------------------------------------------------- taxiRow :: (Int, [(Int, ((Int, Int), (Int, Int)))]) -> [String] taxiRow (n, [(a, ((axc, axr), (ayc, ayr))), (b, ((bxc, bxr), (byc, byr)))]) =
[show n, ". ", show a, eq, op, show axr, cb, leq, show axc, cl] ++
[ad, op, show ayr, cb, leq, show ayc, cl, or, op, show bxr, cb] ++
[leq, show bxc, cl, ad, op, show byr, cb, leq, show byc, cl]
where
cb = "^3"
eq = " = "
leq = "="
ad = " + "
or = " or "
op = "("
cl = ")"
-- OUTPUT ------------------------------------------------------------- main :: IO () main =
mapM_ putStrLn $
concat <$>
transpose
((\col ->
let w = maximum (length <$> col)
in (justifyRight w ' ' <$> col)) <$>
transpose (taxiRow <$> (take 25 xs ++ take 6 (drop 1999 xs))))
where
justifyRight n c s = drop (length s) (replicate n c ++ s)</lang>
- Output:
1. 1729 = ( 1^3= 1) + ( 12^3= 1728) or ( 9^3= 729) + ( 10^3= 1000) 2. 4104 = ( 2^3= 8) + ( 16^3= 4096) or ( 9^3= 729) + ( 15^3= 3375) 3. 13832 = ( 2^3= 8) + ( 24^3= 13824) or ( 18^3= 5832) + ( 20^3= 8000) 4. 20683 = ( 10^3= 1000) + ( 27^3= 19683) or ( 19^3= 6859) + ( 24^3= 13824) 5. 32832 = ( 4^3= 64) + ( 32^3= 32768) or ( 18^3= 5832) + ( 30^3= 27000) 6. 39312 = ( 2^3= 8) + ( 34^3= 39304) or ( 15^3= 3375) + ( 33^3= 35937) 7. 40033 = ( 9^3= 729) + ( 34^3= 39304) or ( 16^3= 4096) + ( 33^3= 35937) 8. 46683 = ( 3^3= 27) + ( 36^3= 46656) or ( 27^3= 19683) + ( 30^3= 27000) 9. 64232 = ( 17^3= 4913) + ( 39^3= 59319) or ( 26^3= 17576) + ( 36^3= 46656) 10. 65728 = ( 12^3= 1728) + ( 40^3= 64000) or ( 31^3= 29791) + ( 33^3= 35937) 11. 110656 = ( 4^3= 64) + ( 48^3= 110592) or ( 36^3= 46656) + ( 40^3= 64000) 12. 110808 = ( 6^3= 216) + ( 48^3= 110592) or ( 27^3= 19683) + ( 45^3= 91125) 13. 134379 = ( 12^3= 1728) + ( 51^3= 132651) or ( 38^3= 54872) + ( 43^3= 79507) 14. 149389 = ( 8^3= 512) + ( 53^3= 148877) or ( 29^3= 24389) + ( 50^3= 125000) 15. 165464 = ( 20^3= 8000) + ( 54^3= 157464) or ( 38^3= 54872) + ( 48^3= 110592) 16. 171288 = ( 17^3= 4913) + ( 55^3= 166375) or ( 24^3= 13824) + ( 54^3= 157464) 17. 195841 = ( 9^3= 729) + ( 58^3= 195112) or ( 22^3= 10648) + ( 57^3= 185193) 18. 216027 = ( 3^3= 27) + ( 60^3= 216000) or ( 22^3= 10648) + ( 59^3= 205379) 19. 216125 = ( 5^3= 125) + ( 60^3= 216000) or ( 45^3= 91125) + ( 50^3= 125000) 20. 262656 = ( 8^3= 512) + ( 64^3= 262144) or ( 36^3= 46656) + ( 60^3= 216000) 21. 314496 = ( 4^3= 64) + ( 68^3= 314432) or ( 30^3= 27000) + ( 66^3= 287496) 22. 320264 = ( 18^3= 5832) + ( 68^3= 314432) or ( 32^3= 32768) + ( 66^3= 287496) 23. 327763 = ( 30^3= 27000) + ( 67^3= 300763) or ( 51^3= 132651) + ( 58^3= 195112) 24. 373464 = ( 6^3= 216) + ( 72^3= 373248) or ( 54^3= 157464) + ( 60^3= 216000) 25. 402597 = ( 42^3= 74088) + ( 69^3= 328509) or ( 56^3= 175616) + ( 61^3= 226981) 2000. 1671816384 = (428^3= 78402752) + (1168^3=1593413632) or (940^3=830584000) + ( 944^3= 841232384) 2001. 1672470592 = ( 29^3= 24389) + (1187^3=1672446203) or (632^3=252435968) + (1124^3=1420034624) 2002. 1673170856 = (458^3= 96071912) + (1164^3=1577098944) or (828^3=567663552) + (1034^3=1105507304) 2003. 1675045225 = (522^3=142236648) + (1153^3=1532808577) or (744^3=411830784) + (1081^3=1263214441) 2004. 1675958167 = (492^3=119095488) + (1159^3=1556862679) or (711^3=359425431) + (1096^3=1316532736) 2005. 1676926719 = ( 63^3= 250047) + (1188^3=1676676672) or (714^3=363994344) + (1095^3=1312932375) 2006. 1677646971 = ( 99^3= 970299) + (1188^3=1676676672) or (891^3=707347971) + ( 990^3= 970299000)
J
<lang J>cubes=: 3^~1+i.100 NB. first 100 cubes triples=: /:~ ~. ,/ (+ , /:~@,)"0/~cubes NB. ordered pairs of cubes (each with their sum) candidates=: ;({."#. <@(0&#`({.@{.(;,)<@}."1)@.(1<#))/. ])triples
NB. we just want the first 25 taxicab numbers 25{.(,.~ <@>:@i.@#) candidates ┌──┬──────┬────────────┬─────────────┐ │1 │1729 │1 1728 │729 1000 │ ├──┼──────┼────────────┼─────────────┤ │2 │4104 │8 4096 │729 3375 │ ├──┼──────┼────────────┼─────────────┤ │3 │13832 │8 13824 │5832 8000 │ ├──┼──────┼────────────┼─────────────┤ │4 │20683 │1000 19683 │6859 13824 │ ├──┼──────┼────────────┼─────────────┤ │5 │32832 │64 32768 │5832 27000 │ ├──┼──────┼────────────┼─────────────┤ │6 │39312 │8 39304 │3375 35937 │ ├──┼──────┼────────────┼─────────────┤ │7 │40033 │729 39304 │4096 35937 │ ├──┼──────┼────────────┼─────────────┤ │8 │46683 │27 46656 │19683 27000 │ ├──┼──────┼────────────┼─────────────┤ │9 │64232 │4913 59319 │17576 46656 │ ├──┼──────┼────────────┼─────────────┤ │10│65728 │1728 64000 │29791 35937 │ ├──┼──────┼────────────┼─────────────┤ │11│110656│64 110592 │46656 64000 │ ├──┼──────┼────────────┼─────────────┤ │12│110808│216 110592 │19683 91125 │ ├──┼──────┼────────────┼─────────────┤ │13│134379│1728 132651 │54872 79507 │ ├──┼──────┼────────────┼─────────────┤ │14│149389│512 148877 │24389 125000 │ ├──┼──────┼────────────┼─────────────┤ │15│165464│8000 157464 │54872 110592 │ ├──┼──────┼────────────┼─────────────┤ │16│171288│4913 166375 │13824 157464 │ ├──┼──────┼────────────┼─────────────┤ │17│195841│729 195112 │10648 185193 │ ├──┼──────┼────────────┼─────────────┤ │18│216027│27 216000 │10648 205379 │ ├──┼──────┼────────────┼─────────────┤ │19│216125│125 216000 │91125 125000 │ ├──┼──────┼────────────┼─────────────┤ │20│262656│512 262144 │46656 216000 │ ├──┼──────┼────────────┼─────────────┤ │21│314496│64 314432 │27000 287496 │ ├──┼──────┼────────────┼─────────────┤ │22│320264│5832 314432 │32768 287496 │ ├──┼──────┼────────────┼─────────────┤ │23│327763│27000 300763│132651 195112│ ├──┼──────┼────────────┼─────────────┤ │24│373464│216 373248 │157464 216000│ ├──┼──────┼────────────┼─────────────┤ │25│402597│74088 328509│175616 226981│ └──┴──────┴────────────┴─────────────┘</lang>
Explanation:
First, generate 100 cubes.
Then, form a 3 column table of unique rows: sum, small cube, large cube
Then, gather rows where the first entry is the same. Keep the ones with at least two such entries (sorted by ascending order of sum).
Then, place an counting index (starting from 1) in front of each row, so the columns are now: counting index, sum, small cube, large cube.
Note that the cube root of the 25th entry is slightly smaller than 74, so testing against the first 100 cubes is more than sufficient.
Note that here we have elected to show the constituent cubes as themselves rather than as expressions involving their cube roots.
Extra credit:
<lang J> x:each 7 {. 1999 }. (,.~ <@>:@i.@#) ;({."#. <@(0&#`({.@{.(;,)<@}."1)@.(1<#))/. ])/:~~.,/(+,/:~@,)"0/~3^~1+i.10000 ┌────┬──────────┬────────────────────┬────────────────────┬┐ │2000│1671816384│78402752 1593413632 │830584000 841232384 ││ ├────┼──────────┼────────────────────┼────────────────────┼┤ │2001│1672470592│24389 1672446203 │252435968 1420034624││ ├────┼──────────┼────────────────────┼────────────────────┼┤ │2002│1673170856│96071912 1577098944 │567663552 1105507304││ ├────┼──────────┼────────────────────┼────────────────────┼┤ │2003│1675045225│142236648 1532808577│411830784 1263214441││ ├────┼──────────┼────────────────────┼────────────────────┼┤ │2004│1675958167│119095488 1556862679│359425431 1316532736││ ├────┼──────────┼────────────────────┼────────────────────┼┤ │2005│1676926719│250047 1676676672 │363994344 1312932375││ ├────┼──────────┼────────────────────┼────────────────────┼┤ │2006│1677646971│970299 1676676672 │707347971 970299000 ││ └────┴──────────┴────────────────────┴────────────────────┴┘</lang>
The extra blank box at the end is because when tackling this large of a data set, some sums can be achieved by three different pairs of cubes.
Java
<lang java>import java.util.PriorityQueue; import java.util.ArrayList; import java.util.List; import java.util.Iterator;
class CubeSum implements Comparable<CubeSum> { public long x, y, value;
public CubeSum(long x, long y) { this.x = x; this.y = y; this.value = x*x*x + y*y*y; }
public String toString() { return String.format("%4d^3 + %4d^3", x, y); }
public int compareTo(CubeSum that) { return value < that.value ? -1 : value > that.value ? 1 : 0; } }
class SumIterator implements Iterator<CubeSum> { PriorityQueue<CubeSum> pq = new PriorityQueue<CubeSum>(); long n = 0;
public boolean hasNext() { return true; } public CubeSum next() { while (pq.size() == 0 || pq.peek().value >= n*n*n) pq.add(new CubeSum(++n, 1));
CubeSum s = pq.remove(); if (s.x > s.y + 1) pq.add(new CubeSum(s.x, s.y+1));
return s; } }
class TaxiIterator implements Iterator<List<CubeSum>> { Iterator<CubeSum> sumIterator = new SumIterator(); CubeSum last = sumIterator.next();
public boolean hasNext() { return true; } public List<CubeSum> next() { CubeSum s; List<CubeSum> train = new ArrayList<CubeSum>();
while ((s = sumIterator.next()).value != last.value) last = s;
train.add(last);
do { train.add(s); } while ((s = sumIterator.next()).value == last.value); last = s;
return train; } }
public class Taxi { public static final void main(String[] args) { Iterator<List<CubeSum>> taxi = new TaxiIterator();
for (int i = 1; i <= 2006; i++) { List<CubeSum> t = taxi.next(); if (i > 25 && i < 2000) continue;
System.out.printf("%4d: %10d", i, t.get(0).value); for (CubeSum s: t) System.out.print(" = " + s); System.out.println(); } } }</lang>
- Output:
1: 1729 = 10^3 + 9^3 = 12^3 + 1^3 2: 4104 = 15^3 + 9^3 = 16^3 + 2^3 3: 13832 = 20^3 + 18^3 = 24^3 + 2^3 4: 20683 = 24^3 + 19^3 = 27^3 + 10^3 5: 32832 = 30^3 + 18^3 = 32^3 + 4^3 6: 39312 = 33^3 + 15^3 = 34^3 + 2^3 7: 40033 = 34^3 + 9^3 = 33^3 + 16^3 8: 46683 = 30^3 + 27^3 = 36^3 + 3^3 9: 64232 = 36^3 + 26^3 = 39^3 + 17^3 10: 65728 = 33^3 + 31^3 = 40^3 + 12^3 11: 110656 = 40^3 + 36^3 = 48^3 + 4^3 12: 110808 = 45^3 + 27^3 = 48^3 + 6^3 13: 134379 = 43^3 + 38^3 = 51^3 + 12^3 14: 149389 = 50^3 + 29^3 = 53^3 + 8^3 15: 165464 = 48^3 + 38^3 = 54^3 + 20^3 16: 171288 = 54^3 + 24^3 = 55^3 + 17^3 17: 195841 = 57^3 + 22^3 = 58^3 + 9^3 18: 216027 = 59^3 + 22^3 = 60^3 + 3^3 19: 216125 = 50^3 + 45^3 = 60^3 + 5^3 20: 262656 = 60^3 + 36^3 = 64^3 + 8^3 21: 314496 = 66^3 + 30^3 = 68^3 + 4^3 22: 320264 = 66^3 + 32^3 = 68^3 + 18^3 23: 327763 = 58^3 + 51^3 = 67^3 + 30^3 24: 373464 = 60^3 + 54^3 = 72^3 + 6^3 25: 402597 = 61^3 + 56^3 = 69^3 + 42^3 2000: 1671816384 = 1168^3 + 428^3 = 944^3 + 940^3 2001: 1672470592 = 1124^3 + 632^3 = 1187^3 + 29^3 2002: 1673170856 = 1164^3 + 458^3 = 1034^3 + 828^3 2003: 1675045225 = 1153^3 + 522^3 = 1081^3 + 744^3 2004: 1675958167 = 1159^3 + 492^3 = 1096^3 + 711^3 2005: 1676926719 = 1095^3 + 714^3 = 1188^3 + 63^3 2006: 1677646971 = 990^3 + 891^3 = 1188^3 + 99^3
JavaScript
<lang JavaScript>var n3s=[], s3s={} for (var n=1,e=1200; n<e; n+=1) n3s[n]=n*n*n for (var a=1; a<e-1; a+=1) { var a3 = n3s[a] for (var b=a; b<e; b+=1) { var b3 = n3s[b] var s3 = a3+b3, abs = s3s[s3] if (!abs) s3s[s3] = abs = [] abs.push([a,b]) } }
var i=0
for (var s3 in s3s) {
var abs = s3s[s3]
if (abs.length < 2) continue
i+=1
if (abs.length == 2 && i > 25 && i < 2000) continue
if (i > 2006) break
document.write(i, ': ', s3)
for (var ab of abs) document.write( ' = ', ab[0], '3+', ab[1], '3')
document.write('
')
}</lang>
- Output:
1: 1729 = 13+123 = 93+103 2: 4104 = 23+163 = 93+153 3: 13832 = 23+243 = 183+203 4: 20683 = 103+273 = 193+243 5: 32832 = 43+323 = 183+303 6: 39312 = 23+343 = 153+333 7: 40033 = 93+343 = 163+333 8: 46683 = 33+363 = 273+303 9: 64232 = 173+393 = 263+363 10: 65728 = 123+403 = 313+333 11: 110656 = 43+483 = 363+403 12: 110808 = 63+483 = 273+453 13: 134379 = 123+513 = 383+433 14: 149389 = 83+533 = 293+503 15: 165464 = 203+543 = 383+483 16: 171288 = 173+553 = 243+543 17: 195841 = 93+583 = 223+573 18: 216027 = 33+603 = 223+593 19: 216125 = 53+603 = 453+503 20: 262656 = 83+643 = 363+603 21: 314496 = 43+683 = 303+663 22: 320264 = 183+683 = 323+663 23: 327763 = 303+673 = 513+583 24: 373464 = 63+723 = 543+603 25: 402597 = 423+693 = 563+613 455: 87539319 = 1673+4363 = 2283+4233 = 2553+4143 535: 119824488 = 113+4933 = 903+4923 = 3463+4283 588: 143604279 = 1113+5223 = 3593+4603 = 4083+4233 655: 175959000 = 703+5603 = 1983+5523 = 3153+5253 888: 327763000 = 3003+6703 = 3393+6613 = 5103+5803 1299: 700314552 = 3343+8723 = 4563+8463 = 5103+8283 1398: 804360375 = 153+9303 = 1983+9273 = 2953+9203 1515: 958595904 = 223+9863 = 1803+9843 = 6923+8563 1660: 1148834232 = 2223+10443 = 7183+9203 = 8163+8463 1837: 1407672000 = 1403+11203 = 3963+11043 = 6303+10503 2000: 1671816384 = 4283+11683 = 9403+9443 2001: 1672470592 = 293+11873 = 6323+11243 2002: 1673170856 = 4583+11643 = 8283+10343 2003: 1675045225 = 5223+11533 = 7443+10813 2004: 1675958167 = 4923+11593 = 7113+10963 2005: 1676926719 = 633+11883 = 7143+10953 2006: 1677646971 = 993+11883 = 8913+9903
jq
<lang jq># Output: an array of the form [i^3 + j^3, [i, j]] sorted by the sum.
- Only cubes of 1 to ($in-1) are considered; the listing is therefore truncated
- as it might not capture taxicab numbers greater than $in ^ 3.
def sum_of_two_cubes:
def cubed: .*.*.; . as $in | (cubed + 1) as $limit | [range(1;$in) as $i | range($i;$in) as $j
| [ ($i|cubed) + ($j|cubed), [$i, $j] ] ] | sort | map( select( .[0] < $limit ) );
- Output a stream of triples [t, d1, d2], in order of t,
- where t is a taxicab number, and d1 and d2 are distinct
- decompositions [i,j] with i^3 + j^3 == t.
- The stream includes each taxicab number once only.
def taxicabs0:
sum_of_two_cubes as $sums
| range(1;$sums|length) as $i
| if $sums[$i][0] == $sums[$i-1][0]
and ($i==1 or $sums[$i][0] != $sums[$i-2][0])
then [$sums[$i][0], $sums[$i-1][1], $sums[$i][1]]
else empty
end;
- Output a stream of $n taxicab triples: [t, d1, d2] as described above,
- without repeating t.
def taxicabs:
# If your jq includes until/2 then the following definition # can be omitted: def until(cond; next): def _until: if cond then . else (next|_until) end; _until; . as $n | [10, ($n / 10 | floor)] | max as $increment | [20, ($n / 2 | floor)] | max | [ ., [taxicabs0] ] | until( .[1] | length >= $m; (.[0] + $increment) | [., [taxicabs0]] ) | .[1][0:$n] ;</lang>
The task <lang jq>2006 | taxicabs as $t | (range(0;25), range(1999;2006)) as $i | "\($i+1): \($t[$i][0]) ~ \($t[$i][1]) and \($t[$i][2])"</lang>
- Output:
<lang sh>$ jq -n -r -f Taxicab_numbers.jq 1: 1729 ~ [1,12] and [9,10] 2: 4104 ~ [2,16] and [9,15] 3: 13832 ~ [2,24] and [18,20] 4: 20683 ~ [10,27] and [19,24] 5: 32832 ~ [4,32] and [18,30] 6: 39312 ~ [2,34] and [15,33] 7: 40033 ~ [9,34] and [16,33] 8: 46683 ~ [3,36] and [27,30] 9: 64232 ~ [17,39] and [26,36] 10: 65728 ~ [12,40] and [31,33] 11: 110656 ~ [4,48] and [36,40] 12: 110808 ~ [6,48] and [27,45] 13: 134379 ~ [12,51] and [38,43] 14: 149389 ~ [8,53] and [29,50] 15: 165464 ~ [20,54] and [38,48] 16: 171288 ~ [17,55] and [24,54] 17: 195841 ~ [9,58] and [22,57] 18: 216027 ~ [3,60] and [22,59] 19: 216125 ~ [5,60] and [45,50] 20: 262656 ~ [8,64] and [36,60] 21: 314496 ~ [4,68] and [30,66] 22: 320264 ~ [18,68] and [32,66] 23: 327763 ~ [30,67] and [51,58] 24: 373464 ~ [6,72] and [54,60] 25: 402597 ~ [42,69] and [56,61] 2000: 1671816384 ~ [428,1168] and [940,944] 2001: 1672470592 ~ [29,1187] and [632,1124] 2002: 1673170856 ~ [458,1164] and [828,1034] 2003: 1675045225 ~ [522,1153] and [744,1081] 2004: 1675958167 ~ [492,1159] and [711,1096] 2005: 1676926719 ~ [63,1188] and [714,1095] 2006: 1677646971 ~ [99,1188] and [891,990]</lang>
Kotlin
<lang scala>// version 1.0.6
import java.util.PriorityQueue
class CubeSum(val x: Long, val y: Long) : Comparable<CubeSum> {
val value: Long = x * x * x + y * y * y
override fun toString() = String.format("%4d^3 + %3d^3", x, y)
override fun compareTo(other: CubeSum) = value.compareTo(other.value)
}
class SumIterator : Iterator<CubeSum> {
private val pq = PriorityQueue<CubeSum>() private var n = 0L override fun hasNext() = true
override fun next(): CubeSum {
while (pq.size == 0 || pq.peek().value >= n * n * n)
pq.add(CubeSum(++n, 1))
val s: CubeSum = pq.remove()
if (s.x > s.y + 1) pq.add(CubeSum(s.x, s.y + 1))
return s
}
}
class TaxiIterator : Iterator<MutableList<CubeSum>> {
private val sumIterator = SumIterator() private var last: CubeSum = sumIterator.next()
override fun hasNext() = true
override fun next(): MutableList<CubeSum> {
var s: CubeSum = sumIterator.next()
val train = mutableListOf<CubeSum>()
while (s.value != last.value) {
last = s
s = sumIterator.next()
}
train.add(last)
do {
train.add(s)
s = sumIterator.next()
}
while (s.value == last.value)
last = s
return train
}
}
fun main(args: Array<String>) {
val taxi = TaxiIterator()
for (i in 1..2006) {
val t = taxi.next()
if (i in 26 until 2000) continue
print(String.format("%4d: %10d", i, t[0].value))
for (s in t) print(" = $s")
println()
}
}</lang>
- Output:
1: 1729 = 10^3 + 9^3 = 12^3 + 1^3 2: 4104 = 15^3 + 9^3 = 16^3 + 2^3 3: 13832 = 20^3 + 18^3 = 24^3 + 2^3 4: 20683 = 24^3 + 19^3 = 27^3 + 10^3 5: 32832 = 30^3 + 18^3 = 32^3 + 4^3 6: 39312 = 33^3 + 15^3 = 34^3 + 2^3 7: 40033 = 33^3 + 16^3 = 34^3 + 9^3 8: 46683 = 30^3 + 27^3 = 36^3 + 3^3 9: 64232 = 39^3 + 17^3 = 36^3 + 26^3 10: 65728 = 40^3 + 12^3 = 33^3 + 31^3 11: 110656 = 40^3 + 36^3 = 48^3 + 4^3 12: 110808 = 45^3 + 27^3 = 48^3 + 6^3 13: 134379 = 51^3 + 12^3 = 43^3 + 38^3 14: 149389 = 50^3 + 29^3 = 53^3 + 8^3 15: 165464 = 48^3 + 38^3 = 54^3 + 20^3 16: 171288 = 54^3 + 24^3 = 55^3 + 17^3 17: 195841 = 57^3 + 22^3 = 58^3 + 9^3 18: 216027 = 59^3 + 22^3 = 60^3 + 3^3 19: 216125 = 50^3 + 45^3 = 60^3 + 5^3 20: 262656 = 60^3 + 36^3 = 64^3 + 8^3 21: 314496 = 66^3 + 30^3 = 68^3 + 4^3 22: 320264 = 68^3 + 18^3 = 66^3 + 32^3 23: 327763 = 67^3 + 30^3 = 58^3 + 51^3 24: 373464 = 60^3 + 54^3 = 72^3 + 6^3 25: 402597 = 69^3 + 42^3 = 61^3 + 56^3 2000: 1671816384 = 1168^3 + 428^3 = 944^3 + 940^3 2001: 1672470592 = 1124^3 + 632^3 = 1187^3 + 29^3 2002: 1673170856 = 1164^3 + 458^3 = 1034^3 + 828^3 2003: 1675045225 = 1153^3 + 522^3 = 1081^3 + 744^3 2004: 1675958167 = 1159^3 + 492^3 = 1096^3 + 711^3 2005: 1676926719 = 1095^3 + 714^3 = 1188^3 + 63^3 2006: 1677646971 = 990^3 + 891^3 = 1188^3 + 99^3
PARI/GP
<lang parigp>taxicab(n)=my(t); for(k=sqrtnint((n-1)\2,3)+1, sqrtnint(n,3), if(ispower(n-k^3, 3), if(t, return(1), t=1))); 0; cubes(n)=my(t); for(k=sqrtnint((n-1)\2,3)+1, sqrtnint(n,3), if(ispower(n-k^3, 3, &t), print(n" = \t"k"^3\t+ "t"^3"))) select(taxicab, [1..402597]) apply(cubes, %);</lang>
- Output:
%1 = [1729, 4104, 13832, 20683, 32832, 39312, 40033, 46683, 64232, 65728, 110656, 110808, 134379, 149389, 165464, 171288, 195841, 216027, 216125, 262656, 314496, 320264, 327763, 373464, 402597] 1729 = 10^3 + 9^3 1729 = 12^3 + 1^3 4104 = 15^3 + 9^3 4104 = 16^3 + 2^3 13832 = 20^3 + 18^3 13832 = 24^3 + 2^3 20683 = 24^3 + 19^3 20683 = 27^3 + 10^3 32832 = 30^3 + 18^3 32832 = 32^3 + 4^3 39312 = 33^3 + 15^3 39312 = 34^3 + 2^3 40033 = 33^3 + 16^3 40033 = 34^3 + 9^3 46683 = 30^3 + 27^3 46683 = 36^3 + 3^3 64232 = 36^3 + 26^3 64232 = 39^3 + 17^3 65728 = 33^3 + 31^3 65728 = 40^3 + 12^3 110656 = 40^3 + 36^3 110656 = 48^3 + 4^3 110808 = 45^3 + 27^3 110808 = 48^3 + 6^3 134379 = 43^3 + 38^3 134379 = 51^3 + 12^3 149389 = 50^3 + 29^3 149389 = 53^3 + 8^3 165464 = 48^3 + 38^3 165464 = 54^3 + 20^3 171288 = 54^3 + 24^3 171288 = 55^3 + 17^3 195841 = 57^3 + 22^3 195841 = 58^3 + 9^3 216027 = 59^3 + 22^3 216027 = 60^3 + 3^3 216125 = 50^3 + 45^3 216125 = 60^3 + 5^3 262656 = 60^3 + 36^3 262656 = 64^3 + 8^3 314496 = 66^3 + 30^3 314496 = 68^3 + 4^3 320264 = 66^3 + 32^3 320264 = 68^3 + 18^3 327763 = 58^3 + 51^3 327763 = 67^3 + 30^3 373464 = 60^3 + 54^3 373464 = 72^3 + 6^3 402597 = 61^3 + 56^3 402597 = 69^3 + 42^3
Pascal
Brute force: Create all combinations x³+ y³ | y < x one by on and test if there is a combination v < x and v> w > y with the same cube-sum. Combinations to check = n*(n-1)/2.The mean distance of one Combination m is m/2 from m³+1³ to m³+(m-1)³. searchSameSum checks one half of this distance == m/4.So O(n) ~ n³ /8 checks are needed. searchSameSum takes most of the time (>95% ), sorting is neglectable. [[1]]C-Version is ~6 times faster aka 43 vs 247 ms for max = 1290^3. Here limit set to 1190 to just reach the goal of element 2006 ;-) so 200ms are possible. Its impressive, that over all one check takes ~3.5 cpu-cycles on i4330 3.5Ghz
<lang pascal>program taxiCabNo; uses
sysutils;
type
tPot3 = Uint32;
tPot3Sol = record
p3Sum : tPot3;
i1,j1,
i2,j2 : Word;
end;
tpPot3 = ^tPot3;
tpPot3Sol = ^tPot3Sol;
var //1290^3 = 2'146'689'000 < 2^31-1 //1190 is the magic number of the task ;-)
pot3 : array[0..1190{1290}] of tPot3;//
AllSol : array[0..3000] of tpot3Sol;
AllSolHigh : NativeInt;
procedure SolOut(const s:tpot3Sol;no: NativeInt); begin
with s do writeln(no:5,p3Sum:12,' = ',j1:5,'^3 +',i1:5,'^3 =',j2:5,'^3 +',i2:5,'^3');
end;
procedure InsertAllSol;
var
tmp: tpot3Sol; p :tpPot3Sol; p3Sum: tPot3; i: NativeInt;
Begin
i := AllSolHigh;
IF i > 0 then
Begin
p := @AllSol[i];
tmp := p^;
p3Sum := p^.p3Sum;
//search the right place for insertion
repeat
dec(i);
dec(p);
IF (p^.p3Sum <= p3Sum) then
BREAK;
until (i<=0);
IF p^.p3Sum = p3Sum then
EXIT;
//free the right place by moving one place up
inc(i);
inc(p);
IF i<AllSolHigh then
Begin
move(p^,AllSol[i+1],SizeOf(AllSol[0])*(AllSolHigh-i));
p^ := tmp;
end;
end;
inc(AllSolHigh);
end;
function searchSameSum(var sol:tpot3Sol):boolean; //try to find a new combination for the same sum //within the limits given by lo and hi var
Sum, SumLo: tPot3; hi,lo: NativeInt;
Begin
with Sol do Begin Sum := p3Sum; lo:= i1; hi:= j1; end;
repeat
//Move hi down
dec(hi);
SumLo := Sum-Pot3[hi];
//Move lo up an check until new combination found or implicite lo> hi
repeat
inc(lo)
until (SumLo<=Pot3[lo]);
//found?
IF SumLo = Pot3[lo] then
BREAK;
until lo>=hi;
IF lo<hi then Begin sol.i2:= lo; sol.j2:= hi; searchSameSum := true; end else searchSameSum := false;
end;
procedure Search; var
i,j: LongInt;
Begin
AllSolHigh := 0;
For j := 2 to High(pot3)-1 do
Begin
For i := 1 to j-1 do
Begin
with AllSol[AllSolHigh] do
Begin
p3Sum:= pot3[i]+pot3[j];
i1:= i;
j1:= j;
end;
IF searchSameSum(AllSol[AllSolHigh]) then
BEGIN
InsertAllSol;
IF AllSolHigh>High(AllSol) then EXIT;
end;
end;
end;
end;
var
i: LongInt;
Begin
For i := Low(pot3) to High(pot3) do
pot3[i] := i*i*i;
AllSolHigh := 0;
Search;
For i := 0 to 24 do SolOut(AllSol[i],i+1);
For i := 1999 to 2005 do SolOut(AllSol[i],i+1);
writeln('count of solutions ',AllSolHigh);
end. </lang>
1 1729 = 12^3 + 1^3 = 10^3 + 9^3
2 4104 = 16^3 + 2^3 = 15^3 + 9^3
3 13832 = 24^3 + 2^3 = 20^3 + 18^3
......
24 373464 = 72^3 + 6^3 = 60^3 + 54^3
25 402597 = 69^3 + 42^3 = 61^3 + 56^3
2000 1671816384 = 1168^3 + 428^3 = 944^3 + 940^3
2001 1672470592 = 1187^3 + 29^3 = 1124^3 + 632^3
...
2005 1676926719 = 1188^3 + 63^3 = 1095^3 + 714^3
2006 1677646971 = 1188^3 + 99^3 = 990^3 + 891^3
count of solutions 2050
//checks 196438017
real 0m0.196s
Perl
Uses segmentation so memory use is constrained as high values are searched for. Also has parameter to look for Ta(3) and Ta(4) numbers (which is when segmentation is really needed). By default shows the first 25 numbers; with one argument shows that many; with two arguments shows results in the range.
<lang perl>my($beg, $end) = (@ARGV==0) ? (1,25) : (@ARGV==1) ? (1,shift) : (shift,shift);
my $lim = 1e14; # Ought to be dynamic as should segment size my @basis = map { $_*$_*$_ } (1 .. int($lim ** (1.0/3.0) + 1)); my $paira = 2; # We're looking for Ta(2) and larger
my ($segsize, $low, $high, $i) = (500_000_000, 0, 0, 0);
while ($i < $end) {
$low = $high+1;
die "lim too low" if $low > $lim;
$high = $low + $segsize - 1;
$high = $lim if $high > $lim;
foreach my $p (_find_pairs_segment(\@basis, $paira, $low, $high,
sub { sprintf("%4d^3 + %4d^3", $_[0], $_[1]) }) ) {
$i++;
next if $i < $beg;
last if $i > $end;
my $n = shift @$p;
printf "%4d: %10d = %s\n", $i, $n, join(" = ", @$p);
}
}
sub _find_pairs_segment {
my($p, $len, $start, $end, $formatsub) = @_; my $plen = $#$p;
my %allpairs;
foreach my $i (0 .. $plen) {
my $pi = $p->[$i];
next if ($pi+$p->[$plen]) < $start;
last if (2*$pi) > $end;
foreach my $j ($i .. $plen) {
my $sum = $pi + $p->[$j];
next if $sum < $start;
last if $sum > $end;
push @{ $allpairs{$sum} }, $i, $j;
}
# If we wanted to save more memory, we could filter and delete every entry
# where $n < 2 * $p->[$i+1]. This can cut memory use in half, but is slow.
}
my @retlist;
foreach my $list (grep { scalar @$_ >= $len*2 } values %allpairs) {
my $n = $p->[$list->[0]] + $p->[$list->[1]];
my @pairlist;
while (@$list) {
push @pairlist, $formatsub->(1 + shift @$list, 1 + shift @$list);
}
push @retlist, [$n, @pairlist];
}
@retlist = sort { $a->[0] <=> $b->[0] } @retlist;
return @retlist;
}</lang>
- Output:
1: 1729 = 1^3 + 12^3 = 9^3 + 10^3 2: 4104 = 2^3 + 16^3 = 9^3 + 15^3 3: 13832 = 2^3 + 24^3 = 18^3 + 20^3 4: 20683 = 10^3 + 27^3 = 19^3 + 24^3 5: 32832 = 4^3 + 32^3 = 18^3 + 30^3 6: 39312 = 2^3 + 34^3 = 15^3 + 33^3 7: 40033 = 9^3 + 34^3 = 16^3 + 33^3 8: 46683 = 3^3 + 36^3 = 27^3 + 30^3 9: 64232 = 17^3 + 39^3 = 26^3 + 36^3 10: 65728 = 12^3 + 40^3 = 31^3 + 33^3 11: 110656 = 4^3 + 48^3 = 36^3 + 40^3 12: 110808 = 6^3 + 48^3 = 27^3 + 45^3 13: 134379 = 12^3 + 51^3 = 38^3 + 43^3 14: 149389 = 8^3 + 53^3 = 29^3 + 50^3 15: 165464 = 20^3 + 54^3 = 38^3 + 48^3 16: 171288 = 17^3 + 55^3 = 24^3 + 54^3 17: 195841 = 9^3 + 58^3 = 22^3 + 57^3 18: 216027 = 3^3 + 60^3 = 22^3 + 59^3 19: 216125 = 5^3 + 60^3 = 45^3 + 50^3 20: 262656 = 8^3 + 64^3 = 36^3 + 60^3 21: 314496 = 4^3 + 68^3 = 30^3 + 66^3 22: 320264 = 18^3 + 68^3 = 32^3 + 66^3 23: 327763 = 30^3 + 67^3 = 51^3 + 58^3 24: 373464 = 6^3 + 72^3 = 54^3 + 60^3 25: 402597 = 42^3 + 69^3 = 56^3 + 61^3
With arguments 2000 2006:
2000: 1671816384 = 428^3 + 1168^3 = 940^3 + 944^3 2001: 1672470592 = 29^3 + 1187^3 = 632^3 + 1124^3 2002: 1673170856 = 458^3 + 1164^3 = 828^3 + 1034^3 2003: 1675045225 = 522^3 + 1153^3 = 744^3 + 1081^3 2004: 1675958167 = 492^3 + 1159^3 = 711^3 + 1096^3 2005: 1676926719 = 63^3 + 1188^3 = 714^3 + 1095^3 2006: 1677646971 = 99^3 + 1188^3 = 891^3 + 990^3
Perl 6
This uses a pretty simple search algorithm that doesn't necessarily return the Taxicab numbers in order. Assuming we want all the Taxicab numbers within some range S to N, we'll search until we find N values. When we find the Nth value, we continue to search up to the cube root of the largest Taxicab number found up to that point. That ensures we will find all of them inside the desired range without needing to search arbitrarily or use magic numbers. Defaults to returning the Taxicab numbers from 1 to 25. Pass in a different start and end value if you want some other range. <lang perl6>sub MAIN ($start = 1, $end = 25) {
my %taxi; my $taxis = 0; my $terminate = 0;
for 1 .. * -> $c1 {
display( %taxi, $start, $end ) and exit if 0 < $terminate < $c1;
my $c = $c1 ** 3;
for 1 .. $c1 -> $c2 {
my $this = $c2 ** 3 + $c;
%taxi{$this}.push: [$c2, $c1];
$taxis++ if %taxi{$this}.elems == 2;
$terminate = %taxi.grep( { $_.value.elems > 1 } ).sort( +*.key )[*-1].key**(1/3)
if $taxis == $end and !$terminate;
}
}
}
sub display (%this_stuff, $start, $end) {
my $i = $start;
printf "%4d %10d =>\t%s\n", $i++, $_.key,
($_.value.map({ sprintf "%4d³ + %-s", $_[0], "$_[1]\³" })).join: ",\t"
for %this_stuff.grep( { $_.value.elems > 1 } ).sort( +*.key )[$start-1..$end-1];
1;
}</lang>
- Output:
With no passed parameters (default)
1 1729 => 9³ + 10³, 1³ + 12³ 2 4104 => 9³ + 15³, 2³ + 16³ 3 13832 => 18³ + 20³, 2³ + 24³ 4 20683 => 19³ + 24³, 10³ + 27³ 5 32832 => 18³ + 30³, 4³ + 32³ 6 39312 => 15³ + 33³, 2³ + 34³ 7 40033 => 16³ + 33³, 9³ + 34³ 8 46683 => 27³ + 30³, 3³ + 36³ 9 64232 => 26³ + 36³, 17³ + 39³ 10 65728 => 31³ + 33³, 12³ + 40³ 11 110656 => 36³ + 40³, 4³ + 48³ 12 110808 => 27³ + 45³, 6³ + 48³ 13 134379 => 38³ + 43³, 12³ + 51³ 14 149389 => 29³ + 50³, 8³ + 53³ 15 165464 => 38³ + 48³, 20³ + 54³ 16 171288 => 24³ + 54³, 17³ + 55³ 17 195841 => 22³ + 57³, 9³ + 58³ 18 216027 => 22³ + 59³, 3³ + 60³ 19 216125 => 45³ + 50³, 5³ + 60³ 20 262656 => 36³ + 60³, 8³ + 64³ 21 314496 => 30³ + 66³, 4³ + 68³ 22 320264 => 32³ + 66³, 18³ + 68³ 23 327763 => 51³ + 58³, 30³ + 67³ 24 373464 => 54³ + 60³, 6³ + 72³ 25 402597 => 56³ + 61³, 42³ + 69³
With passed parameters 2000 2006:
2000 1671816384 => 940³ + 944³, 428³ + 1168³ 2001 1672470592 => 632³ + 1124³, 29³ + 1187³ 2002 1673170856 => 828³ + 1034³, 458³ + 1164³ 2003 1675045225 => 744³ + 1081³, 522³ + 1153³ 2004 1675958167 => 711³ + 1096³, 492³ + 1159³ 2005 1676926719 => 714³ + 1095³, 63³ + 1188³ 2006 1677646971 => 891³ + 990³, 99³ + 1188³
Python
(Magic number 1201 found by trial and error) <lang python>from collections import defaultdict from itertools import product from pprint import pprint as pp
cube2n = {x**3:x for x in range(1, 1201)} sum2cubes = defaultdict(set) for c1, c2 in product(cube2n, cube2n): if c1 >= c2: sum2cubes[c1 + c2].add((cube2n[c1], cube2n[c2]))
taxied = sorted((k, v) for k,v in sum2cubes.items() if len(v) >= 2)
- pp(len(taxied)) # 2068
for t in enumerate(taxied[:25], 1):
pp(t)
print('...') for t in enumerate(taxied[2000-1:2000+6], 2000):
pp(t)</lang>
- Output:
(1, (1729, {(12, 1), (10, 9)}))
(2, (4104, {(16, 2), (15, 9)}))
(3, (13832, {(20, 18), (24, 2)}))
(4, (20683, {(27, 10), (24, 19)}))
(5, (32832, {(30, 18), (32, 4)}))
(6, (39312, {(33, 15), (34, 2)}))
(7, (40033, {(33, 16), (34, 9)}))
(8, (46683, {(30, 27), (36, 3)}))
(9, (64232, {(36, 26), (39, 17)}))
(10, (65728, {(33, 31), (40, 12)}))
(11, (110656, {(48, 4), (40, 36)}))
(12, (110808, {(48, 6), (45, 27)}))
(13, (134379, {(51, 12), (43, 38)}))
(14, (149389, {(50, 29), (53, 8)}))
(15, (165464, {(54, 20), (48, 38)}))
(16, (171288, {(54, 24), (55, 17)}))
(17, (195841, {(57, 22), (58, 9)}))
(18, (216027, {(60, 3), (59, 22)}))
(19, (216125, {(60, 5), (50, 45)}))
(20, (262656, {(64, 8), (60, 36)}))
(21, (314496, {(66, 30), (68, 4)}))
(22, (320264, {(66, 32), (68, 18)}))
(23, (327763, {(58, 51), (67, 30)}))
(24, (373464, {(72, 6), (60, 54)}))
(25, (402597, {(69, 42), (61, 56)}))
...
(2000, (1671816384, {(1168, 428), (944, 940)}))
(2001, (1672470592, {(1187, 29), (1124, 632)}))
(2002, (1673170856, {(1164, 458), (1034, 828)}))
(2003, (1675045225, {(1153, 522), (1081, 744)}))
(2004, (1675958167, {(1159, 492), (1096, 711)}))
(2005, (1676926719, {(1188, 63), (1095, 714)}))
(2006, (1677646971, {(990, 891), (1188, 99)}))
Although, for this task it's simply faster to look up the cubes in the sum when we need to print them, because we can now store and sort only the sums: <lang python>cubes, crev = [x**3 for x in range(1,1200)], {}
- for cube root lookup
for x,x3 in enumerate(cubes): crev[x3] = x + 1
sums = sorted(x+y for x in cubes for y in cubes if y < x)
idx = 0 for i in range(1, len(sums)-1):
if sums[i-1] != sums[i] and sums[i] == sums[i+1]:
idx += 1
if idx > 25 and idx < 2000 or idx > 2006: continue
n,p = sums[i],[]
for x in cubes:
if n-x < x: break
if n-x in crev:
p.append((crev[x], crev[n-x]))
print "%4d: %10d"%(idx,n),
for x in p: print " = %4d^3 + %4d^3"%x,
print</lang>
- Output:
Output trimmed to reduce clutter.
1: 1729 = 1^3 + 12^3 = 9^3 + 10^3 2: 4104 = 2^3 + 16^3 = 9^3 + 15^3 3: 13832 = 2^3 + 24^3 = 18^3 + 20^3 4: 20683 = 10^3 + 27^3 = 19^3 + 24^3 5: 32832 = 4^3 + 32^3 = 18^3 + 30^3 ... 2004: 1675958167 = 492^3 + 1159^3 = 711^3 + 1096^3 2005: 1676926719 = 63^3 + 1188^3 = 714^3 + 1095^3 2006: 1677646971 = 99^3 + 1188^3 = 891^3 + 990^3
Using heapq module
A priority queue that holds cube sums. When consecutive sums come out with the same value, they are taxis. <lang python>from heapq import heappush, heappop
def cubesum():
h,n = [],1
while True:
while not h or h[0][0] > n**3: # could also pre-calculate cubes
heappush(h, (n**3 + 1, n, 1))
n += 1
(s, x, y) = heappop(h)
yield((s, x, y))
y += 1
if y < x: # should be y <= x?
heappush(h, (x**3 + y**3, x, y))
def taxis():
out = [(0,0,0)]
for s in cubesum():
if s[0] == out[-1][0]:
out.append(s)
else:
if len(out) > 1: yield(out)
out = [s]
n = 0 for x in taxis():
n += 1
if n >= 2006: break
if n <= 25 or n >= 2000:
print(n, x)</lang>
- Output:
(1, [(1729, 10, 9), (1729, 12, 1)]) (2, [(4104, 15, 9), (4104, 16, 2)]) (3, [(13832, 20, 18), (13832, 24, 2)]) (4, [(20683, 24, 19), (20683, 27, 10)]) (5, [(32832, 30, 18), (32832, 32, 4)]) (6, [(39312, 33, 15), (39312, 34, 2)]) (7, [(40033, 33, 16), (40033, 34, 9)]) (8, [(46683, 30, 27), (46683, 36, 3)]) (9, [(64232, 36, 26), (64232, 39, 17)]) (10, [(65728, 33, 31), (65728, 40, 12)]) (11, [(110656, 40, 36), (110656, 48, 4)]) (12, [(110808, 45, 27), (110808, 48, 6)]) (13, [(134379, 43, 38), (134379, 51, 12)]) (14, [(149389, 50, 29), (149389, 53, 8)]) (15, [(165464, 48, 38), (165464, 54, 20)]) (16, [(171288, 54, 24), (171288, 55, 17)]) (17, [(195841, 57, 22), (195841, 58, 9)]) (18, [(216027, 59, 22), (216027, 60, 3)]) (19, [(216125, 50, 45), (216125, 60, 5)]) (20, [(262656, 60, 36), (262656, 64, 8)]) (21, [(314496, 66, 30), (314496, 68, 4)]) (22, [(320264, 66, 32), (320264, 68, 18)]) (23, [(327763, 58, 51), (327763, 67, 30)]) (24, [(373464, 60, 54), (373464, 72, 6)]) (25, [(402597, 61, 56), (402597, 69, 42)]) (2000, [(1671816384, 944, 940), (1671816384, 1168, 428)]) (2001, [(1672470592, 1124, 632), (1672470592, 1187, 29)]) (2002, [(1673170856, 1034, 828), (1673170856, 1164, 458)]) (2003, [(1675045225, 1081, 744), (1675045225, 1153, 522)]) (2004, [(1675958167, 1096, 711), (1675958167, 1159, 492)]) (2005, [(1676926719, 1095, 714), (1676926719, 1188, 63)])
Racket
This is the straighforward implementation, so it finds only the first 25 values in a sensible amount of time. <lang Racket>#lang racket
(define (cube x) (* x x x))
- floor of cubic root
(define (cubic-root x)
(let ([aprox (inexact->exact (round (expt x (/ 1 3))))])
(if (> (cube aprox) x)
(- aprox 1)
aprox)))
(let loop ([p 1] [n 1])
(let ()
(define pairs
(for*/list ([j (in-range 1 (add1 (cubic-root (quotient n 2))))]
[k (in-value (cubic-root (- n (cube j))))]
#:when (= n (+ (cube j) (cube k))))
(cons j k)))
(if (>= (length pairs) 2)
(begin
(printf "~a: ~a" p n)
(for ([pair (in-list pairs)])
(printf " = ~a^3 + ~a^3" (car pair) (cdr pair)))
(newline)
(when (< p 25)
(loop (add1 p) (add1 n))))
(loop p (add1 n)))))</lang>
- Output:
1: 1729 = 1^3 + 12^3 = 9^3 + 10^3 2: 4104 = 2^3 + 16^3 = 9^3 + 15^3 3: 13832 = 2^3 + 24^3 = 18^3 + 20^3 4: 20683 = 10^3 + 27^3 = 19^3 + 24^3 5: 32832 = 4^3 + 32^3 = 18^3 + 30^3 6: 39312 = 2^3 + 34^3 = 15^3 + 33^3 7: 40033 = 9^3 + 34^3 = 16^3 + 33^3 8: 46683 = 3^3 + 36^3 = 27^3 + 30^3 9: 64232 = 17^3 + 39^3 = 26^3 + 36^3 10: 65728 = 12^3 + 40^3 = 31^3 + 33^3 11: 110656 = 4^3 + 48^3 = 36^3 + 40^3 12: 110808 = 6^3 + 48^3 = 27^3 + 45^3 13: 134379 = 12^3 + 51^3 = 38^3 + 43^3 14: 149389 = 8^3 + 53^3 = 29^3 + 50^3 15: 165464 = 20^3 + 54^3 = 38^3 + 48^3 16: 171288 = 17^3 + 55^3 = 24^3 + 54^3 17: 195841 = 9^3 + 58^3 = 22^3 + 57^3 18: 216027 = 3^3 + 60^3 = 22^3 + 59^3 19: 216125 = 5^3 + 60^3 = 45^3 + 50^3 20: 262656 = 8^3 + 64^3 = 36^3 + 60^3 21: 314496 = 4^3 + 68^3 = 30^3 + 66^3 22: 320264 = 18^3 + 68^3 = 32^3 + 66^3 23: 327763 = 30^3 + 67^3 = 51^3 + 58^3 24: 373464 = 6^3 + 72^3 = 54^3 + 60^3 25: 402597 = 42^3 + 69^3 = 56^3 + 61^3
REXX
<lang rexx>/*REXX program displays the specified first (lowest) taxicab numbers (for three ranges).*/ parse arg L.1 H.1 L.2 H.2 L.3 H.3 . /*obtain optional arguments from the CL*/
if L.1== | L.1=="," then L.1= 1 /*L1 is the low part of 1st range. */ if H.1== | H.1=="," then H.1= 25 /*H1 " " high " " " " */
if L.2== | L.2=="," then L.2= 454 /*L2 " " low " " 2nd " */ if H.2== | H.2=="," then H.2= 456 /*H2 " " high " " " " */
if L.3== | L.3=="," then L.3=2000 /*L3 " " low " " 3rd " */ if H.3== | H.3=="," then H.3=2006 /*H3 " " high " " " " */
mx=max(L.1, H.1, L.2, H.2, L.3, H.3) /*find how many taxicab numbers needed.*/ mx=mx+mx%10; ww=length(mx)*3 /*cushion; compensate for the triples.*/ w=ww%2; numeric digits max(9,ww) /*prepare to use some larger numbers. */ @.=.; #=0; @@.=0; @and=" ──and── "; $.= /*set some REXX vars and handy literals*/
/* [↓] generate extra taxicab numbers.*/
do j=1 until #>=mx; C=j**3 /*taxicab numbers may not be in order. */
!.j=C /*use memoization for cube calculation.*/
do k=1 for j-1; s=C+!.k /*define a whole bunch of cube sums. */
if @.s==. then do /*Cube not defined? Then process it. */
@.s=j; b.s=k /*define @.S and B.S≡sum of 2 cubes*/
iterate /* ··· and then go and do another.*/
end /* [↑] define one cube sum at a time. */
has=@@.s /*has this number been defined before? */
if \has then $.s=right(s,ww) '───►' r(@.s," +")r(b.s) @and r(j,' +')r(k)
else $.s=$.s @and r(j,' +')r(k) /*◄─ build a display string. [↑] */
@@.s=1 /*mark taxicab number as a sum of cubes*/
if has then iterate /*S is a triple (or sometimes better).*/
#=#+1; #.#=s /*bump taxicab counter; define taxicab#*/
end /*k*/ /* [↑] build the cubes one─at─a─time. */
end /*j*/ /* [↑] complete with overage numbers. */
call bSort # /*use a bubble sort on taxicab numbers.*/
do range=1 for 3; call tell L.range, H.range; say; end /*range*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ r: return right(arg(1), w)'^3'arg(2) /*right─justify a number, append "^3" */ tell: do t=arg(1) to arg(2); _=#.t; say right(t, 9)':' $._; end; return /*──────────────────────────────────────────────────────────────────────────────────────*/ bSort: procedure expose @.; parse arg #; m=#-1 /*a bubble sort (in ascending order). */
do m=m for m by -1 until ok; ok=1
do j=1 for m; k=j+1; if @.j<=@.k then iterate
_=@.j; @.j=@.k; @.k=_; ok=0 /*swap 2 elements.*/
end /*j*/
end /*m*/
return</lang>
output using the default inputs:
1: 1729 ───► 10^3 + 9^3 ──and── 12^3 + 1^3
2: 4104 ───► 15^3 + 9^3 ──and── 16^3 + 2^3
3: 13832 ───► 20^3 + 18^3 ──and── 24^3 + 2^3
4: 20683 ───► 24^3 + 19^3 ──and── 27^3 + 10^3
5: 32832 ───► 30^3 + 18^3 ──and── 32^3 + 4^3
6: 39312 ───► 33^3 + 15^3 ──and── 34^3 + 2^3
7: 40033 ───► 33^3 + 16^3 ──and── 34^3 + 9^3
8: 46683 ───► 30^3 + 27^3 ──and── 36^3 + 3^3
9: 64232 ───► 36^3 + 26^3 ──and── 39^3 + 17^3
10: 65728 ───► 33^3 + 31^3 ──and── 40^3 + 12^3
11: 110656 ───► 40^3 + 36^3 ──and── 48^3 + 4^3
12: 110808 ───► 45^3 + 27^3 ──and── 48^3 + 6^3
13: 134379 ───► 43^3 + 38^3 ──and── 51^3 + 12^3
14: 149389 ───► 50^3 + 29^3 ──and── 53^3 + 8^3
15: 165464 ───► 48^3 + 38^3 ──and── 54^3 + 20^3
16: 171288 ───► 54^3 + 24^3 ──and── 55^3 + 17^3
17: 195841 ───► 57^3 + 22^3 ──and── 58^3 + 9^3
18: 216027 ───► 59^3 + 22^3 ──and── 60^3 + 3^3
19: 216125 ───► 50^3 + 45^3 ──and── 60^3 + 5^3
20: 262656 ───► 60^3 + 36^3 ──and── 64^3 + 8^3
21: 314496 ───► 66^3 + 30^3 ──and── 68^3 + 4^3
22: 320264 ───► 66^3 + 32^3 ──and── 68^3 + 18^3
23: 327763 ───► 58^3 + 51^3 ──and── 67^3 + 30^3
24: 373464 ───► 60^3 + 54^3 ──and── 72^3 + 6^3
25: 402597 ───► 61^3 + 56^3 ──and── 69^3 + 42^3
454: 87483968 ───► 363^3 + 341^3 ──and── 440^3 + 132^3
455: 87539319 ───► 414^3 + 255^3 ──and── 423^3 + 228^3 ──and── 436^3 + 167^3
456: 87579037 ───► 370^3 + 333^3 ──and── 444^3 + 37^3
2000: 1671816384 ───► 944^3 + 940^3 ──and── 1168^3 + 428^3
2001: 1672470592 ───► 1124^3 + 632^3 ──and── 1187^3 + 29^3
2002: 1673170856 ───► 1034^3 + 828^3 ──and── 1164^3 + 458^3
2003: 1675045225 ───► 1081^3 + 744^3 ──and── 1153^3 + 522^3
2004: 1675958167 ───► 1096^3 + 711^3 ──and── 1159^3 + 492^3
2005: 1676926719 ───► 1095^3 + 714^3 ──and── 1188^3 + 63^3
2006: 1677646971 ───► 990^3 + 891^3 ──and── 1188^3 + 99^3
Ruby
<lang ruby>def taxicab_number(nmax=1200)
[*1..nmax].repeated_combination(2).group_by{|x,y| x**3 + y**3}.select{|k,v| v.size>1}.sort
end
t = [0] + taxicab_number
[*1..25, *2000...2007].each do |i|
puts "%4d: %10d" % [i, t[i][0]] + t[i][1].map{|a| " = %4d**3 + %4d**3" % a}.join
end</lang>
- Output:
1: 1729 = 1**3 + 12**3 = 9**3 + 10**3 2: 4104 = 2**3 + 16**3 = 9**3 + 15**3 3: 13832 = 2**3 + 24**3 = 18**3 + 20**3 4: 20683 = 10**3 + 27**3 = 19**3 + 24**3 5: 32832 = 4**3 + 32**3 = 18**3 + 30**3 6: 39312 = 2**3 + 34**3 = 15**3 + 33**3 7: 40033 = 9**3 + 34**3 = 16**3 + 33**3 8: 46683 = 3**3 + 36**3 = 27**3 + 30**3 9: 64232 = 17**3 + 39**3 = 26**3 + 36**3 10: 65728 = 12**3 + 40**3 = 31**3 + 33**3 11: 110656 = 4**3 + 48**3 = 36**3 + 40**3 12: 110808 = 6**3 + 48**3 = 27**3 + 45**3 13: 134379 = 12**3 + 51**3 = 38**3 + 43**3 14: 149389 = 8**3 + 53**3 = 29**3 + 50**3 15: 165464 = 20**3 + 54**3 = 38**3 + 48**3 16: 171288 = 17**3 + 55**3 = 24**3 + 54**3 17: 195841 = 9**3 + 58**3 = 22**3 + 57**3 18: 216027 = 3**3 + 60**3 = 22**3 + 59**3 19: 216125 = 5**3 + 60**3 = 45**3 + 50**3 20: 262656 = 8**3 + 64**3 = 36**3 + 60**3 21: 314496 = 4**3 + 68**3 = 30**3 + 66**3 22: 320264 = 18**3 + 68**3 = 32**3 + 66**3 23: 327763 = 30**3 + 67**3 = 51**3 + 58**3 24: 373464 = 6**3 + 72**3 = 54**3 + 60**3 25: 402597 = 42**3 + 69**3 = 56**3 + 61**3 2000: 1671816384 = 428**3 + 1168**3 = 940**3 + 944**3 2001: 1672470592 = 29**3 + 1187**3 = 632**3 + 1124**3 2002: 1673170856 = 458**3 + 1164**3 = 828**3 + 1034**3 2003: 1675045225 = 522**3 + 1153**3 = 744**3 + 1081**3 2004: 1675958167 = 492**3 + 1159**3 = 711**3 + 1096**3 2005: 1676926719 = 63**3 + 1188**3 = 714**3 + 1095**3 2006: 1677646971 = 99**3 + 1188**3 = 891**3 + 990**3
Scala
<lang scala>import scala.math.pow
implicit class Pairs[A, B]( p:List[(A, B)]) {
def collectPairs: Map[A, List[B]] = p.groupBy(_._1).mapValues(_.map(_._2)).filterNot(_._2.size<2)
}
// Make a sorted List of Taxi Cab Numbers. Limit it to the cube of 1200 because we know it's high enough. val taxiNums = {
(1 to 1200).toList // Start with a sequential list of integers
.combinations(2).toList // Find all two number combinations
.map {
case a :: b :: nil => ((pow(a, 3) + pow(b, 3)).toInt, (a, b))
case _ => 0 ->(0, 0)
} // Turn the list into the sum of two cubes and
// remember what we started with, eg. 28->(1,3)
.collectPairs // Only keep taxi cab numbers with a duplicate
.toList.sortBy(_._1) // Sort the results
}
def output() : Unit = {
println( "%20s".format( "Taxi Cab Numbers" ) ) println( "%20s%15s%15s".format( "-"*20, "-"*15, "-"*15 ) )
taxiNums.take(25) foreach {
case (p, a::b::Nil) => println( "%20d\t(%d\u00b3 + %d\u00b3)\t\t(%d\u00b3 + %d\u00b3)".format(p,a._1,a._2,b._1,b._2) )
}
taxiNums.slice(1999,2007) foreach {
case (p, a::b::Nil) => println( "%20d\t(%d\u00b3 + %d\u00b3)\t(%d\u00b3 + %d\u00b3)".format(p,a._1,a._2,b._1,b._2) )
}
} </lang>
- Output:
Taxi Cab Numbers
--------------------------------------------------
1729 (1³ + 12³) (9³ + 10³)
4104 (2³ + 16³) (9³ + 15³)
13832 (2³ + 24³) (18³ + 20³)
20683 (10³ + 27³) (19³ + 24³)
32832 (4³ + 32³) (18³ + 30³)
39312 (2³ + 34³) (15³ + 33³)
40033 (9³ + 34³) (16³ + 33³)
46683 (3³ + 36³) (27³ + 30³)
64232 (17³ + 39³) (26³ + 36³)
65728 (12³ + 40³) (31³ + 33³)
110656 (4³ + 48³) (36³ + 40³)
110808 (6³ + 48³) (27³ + 45³)
134379 (12³ + 51³) (38³ + 43³)
149389 (8³ + 53³) (29³ + 50³)
165464 (20³ + 54³) (38³ + 48³)
171288 (17³ + 55³) (24³ + 54³)
195841 (9³ + 58³) (22³ + 57³)
216027 (3³ + 60³) (22³ + 59³)
216125 (5³ + 60³) (45³ + 50³)
262656 (8³ + 64³) (36³ + 60³)
314496 (4³ + 68³) (30³ + 66³)
320264 (18³ + 68³) (32³ + 66³)
327763 (30³ + 67³) (51³ + 58³)
373464 (6³ + 72³) (54³ + 60³)
402597 (42³ + 69³) (56³ + 61³)
1671816384 (428³ + 1168³) (940³ + 944³)
1672470592 (29³ + 1187³) (632³ + 1124³)
1673170856 (458³ + 1164³) (828³ + 1034³)
1675045225 (522³ + 1153³) (744³ + 1081³)
1675958167 (492³ + 1159³) (711³ + 1096³)
1676926719 (63³ + 1188³) (714³ + 1095³)
1677646971 (99³ + 1188³) (891³ + 990³)
Sidef
<lang ruby>var (start=1, end=25) = ARGV.map{.to_i}...;
func display (h, start, end) {
var i = start
h.grep { |_,v| v.len > 1 }.keys.sort_by{.to_i}.ft(start-1, end-1).each {|n|
printf("%4d %10d =>\t%s\n", i++, n,
h{n}.map{ "%4d³ + %-s" % (.first, "#{.last}³") }.join(",\t"))
}
}
var taxi = Hash() var taxis = 0 var terminate = 0
Inf.times { |c1|
if (terminate>0 && terminate>c1) {
display(taxi, start, end)
break
}
var c = c1**3
c1.times { |c2|
var this = (c2**3 + c)
taxi{this} := [] << [c2, c1]
++taxis if (taxi{this}.len == 2);
if (taxis == end && terminate.is_zero) {
terminate = taxi.grep{|_,v| v.len > 1 }.keys.map{.to_i}.max.root(3)
}
}
}</lang>
- Output:
1 1729 => 9³ + 10³, 1³ + 12³ 2 4104 => 9³ + 15³, 2³ + 16³ 3 13832 => 18³ + 20³, 2³ + 24³ 4 20683 => 19³ + 24³, 10³ + 27³ 5 32832 => 18³ + 30³, 4³ + 32³ 6 39312 => 15³ + 33³, 2³ + 34³ 7 40033 => 16³ + 33³, 9³ + 34³ 8 46683 => 27³ + 30³, 3³ + 36³ 9 64232 => 26³ + 36³, 17³ + 39³ 10 65728 => 31³ + 33³, 12³ + 40³ 11 110656 => 36³ + 40³, 4³ + 48³ 12 110808 => 27³ + 45³, 6³ + 48³ 13 134379 => 38³ + 43³, 12³ + 51³ 14 149389 => 29³ + 50³, 8³ + 53³ 15 165464 => 38³ + 48³, 20³ + 54³ 16 171288 => 24³ + 54³, 17³ + 55³ 17 195841 => 22³ + 57³, 9³ + 58³ 18 216027 => 22³ + 59³, 3³ + 60³ 19 216125 => 45³ + 50³, 5³ + 60³ 20 262656 => 36³ + 60³, 8³ + 64³ 21 314496 => 30³ + 66³, 4³ + 68³ 22 320264 => 32³ + 66³, 18³ + 68³ 23 327763 => 51³ + 58³, 30³ + 67³ 24 373464 => 54³ + 60³, 6³ + 72³ 25 402597 => 56³ + 61³, 42³ + 69³
With passed parameters 2000 and 2006:
2000 1671816384 => 940³ + 944³, 428³ + 1168³ 2001 1672470592 => 632³ + 1124³, 29³ + 1187³ 2002 1673170856 => 828³ + 1034³, 458³ + 1164³ 2003 1675045225 => 744³ + 1081³, 522³ + 1153³ 2004 1675958167 => 711³ + 1096³, 492³ + 1159³ 2005 1676926719 => 714³ + 1095³, 63³ + 1188³ 2006 1677646971 => 891³ + 990³, 99³ + 1188³
Tcl
<lang tcl>package require Tcl 8.6
proc heappush {heapName item} {
upvar 1 $heapName heap
set idx [lsearch -bisect -index 0 -integer $heap [lindex $item 0]]
set heap [linsert $heap [expr {$idx + 1}] $item]
} coroutine cubesum apply {{} {
yield
set h {}
set n 1
while true {
while {![llength $h] || [lindex $h 0 0] > $n**3} { heappush h [list [expr {$n**3 + 1}] $n 1] incr n } set h [lassign $h item] yield $item lassign $item s x y if {[incr y] < $x} { heappush h [list [expr {$x**3 + $y**3}] $x $y] }
}
}} coroutine taxis apply {{} {
yield set out Template:0 0 0 while true {
set s [cubesum] if {[lindex $s 0] == [lindex $out end 0]} { lappend out $s } else { if {[llength $out] > 1} {yield $out} set out [list $s] }
}
}}
- Put a cache in front for convenience
variable taxis {} proc taxi {n} {
variable taxis
while {$n > [llength $taxis]} {lappend taxis [taxis]}
return [lindex $taxis [expr {$n-1}]]
}
set 3 "\u00b3" for {set n 1} {$n <= 25} {incr n} {
puts ${n}:[join [lmap t [taxi $n] {format " %d = %d$3 + %d$3" {*}$t}] ","]
} for {set n 2000} {$n <= 2006} {incr n} {
puts ${n}:[join [lmap t [taxi $n] {format " %d = %d$3 + %d$3" {*}$t}] ","]
}</lang>
- Output:
1: 1729 = 10³ + 9³, 1729 = 12³ + 1³ 2: 4104 = 15³ + 9³, 4104 = 16³ + 2³ 3: 13832 = 20³ + 18³, 13832 = 24³ + 2³ 4: 20683 = 24³ + 19³, 20683 = 27³ + 10³ 5: 32832 = 30³ + 18³, 32832 = 32³ + 4³ 6: 39312 = 33³ + 15³, 39312 = 34³ + 2³ 7: 40033 = 33³ + 16³, 40033 = 34³ + 9³ 8: 46683 = 30³ + 27³, 46683 = 36³ + 3³ 9: 64232 = 36³ + 26³, 64232 = 39³ + 17³ 10: 65728 = 33³ + 31³, 65728 = 40³ + 12³ 11: 110656 = 40³ + 36³, 110656 = 48³ + 4³ 12: 110808 = 45³ + 27³, 110808 = 48³ + 6³ 13: 134379 = 43³ + 38³, 134379 = 51³ + 12³ 14: 149389 = 50³ + 29³, 149389 = 53³ + 8³ 15: 165464 = 48³ + 38³, 165464 = 54³ + 20³ 16: 171288 = 54³ + 24³, 171288 = 55³ + 17³ 17: 195841 = 57³ + 22³, 195841 = 58³ + 9³ 18: 216027 = 59³ + 22³, 216027 = 60³ + 3³ 19: 216125 = 50³ + 45³, 216125 = 60³ + 5³ 20: 262656 = 60³ + 36³, 262656 = 64³ + 8³ 21: 314496 = 66³ + 30³, 314496 = 68³ + 4³ 22: 320264 = 66³ + 32³, 320264 = 68³ + 18³ 23: 327763 = 58³ + 51³, 327763 = 67³ + 30³ 24: 373464 = 60³ + 54³, 373464 = 72³ + 6³ 25: 402597 = 61³ + 56³, 402597 = 69³ + 42³ 2000: 1671816384 = 944³ + 940³, 1671816384 = 1168³ + 428³ 2001: 1672470592 = 1124³ + 632³, 1672470592 = 1187³ + 29³ 2002: 1673170856 = 1034³ + 828³, 1673170856 = 1164³ + 458³ 2003: 1675045225 = 1081³ + 744³, 1675045225 = 1153³ + 522³ 2004: 1675958167 = 1096³ + 711³, 1675958167 = 1159³ + 492³ 2005: 1676926719 = 1095³ + 714³, 1676926719 = 1188³ + 63³ 2006: 1677646971 = 990³ + 891³, 1677646971 = 1188³ + 99³
zkl
An array of bytes is used to hold n, where array[n³+m³]==n. <lang zkl>fcn taxiCabNumbers{
const HeapSZ=0d5_000_000;
iCubes:=[1..120].apply("pow",3);
sum2cubes:=Data(HeapSZ).fill(0); // BFheap of 1 byte zeros
taxiNums:=List();
foreach i,i3 in ([1..].zip(iCubes)){
foreach j,j3 in ([i+1..].zip(iCubes[i,*])){
ij3:=i3+j3;
if(z:=sum2cubes[ij3]){ taxiNums.append(T(ij3, z,(ij3-z.pow(3)).toFloat().pow(1.0/3).round().toInt(), i,j)); } else sum2cubes[ij3]=i;
}
}
taxiNums.sort(fcn([(a,_)],[(b,_)]){ a<b })
}</lang> <lang zkl>fcn print(n,taxiNums){
[n..].zip(taxiNums).pump(Console.println,fcn([(n,t)]){
"%4d: %10,d = %2d\u00b3 + %2d\u00b3 = %2d\u00b3 + %2d\u00b3".fmt(n,t.xplode())
})
} taxiNums:=taxiCabNumbers(); // 63 pairs taxiNums[0,25]:print(1,_);</lang>
- Output:
1: 1,729 = 1³ + 12³ = 9³ + 10³ 2: 4,104 = 2³ + 16³ = 9³ + 15³ 3: 13,832 = 2³ + 24³ = 18³ + 20³ 4: 20,683 = 10³ + 27³ = 19³ + 24³ 5: 32,832 = 4³ + 32³ = 18³ + 30³ 6: 39,312 = 2³ + 34³ = 15³ + 33³ 7: 40,033 = 9³ + 34³ = 16³ + 33³ 8: 46,683 = 3³ + 36³ = 27³ + 30³ 9: 64,232 = 17³ + 39³ = 26³ + 36³ 10: 65,728 = 12³ + 40³ = 31³ + 33³ 11: 110,656 = 4³ + 48³ = 36³ + 40³ 12: 110,808 = 6³ + 48³ = 27³ + 45³ 13: 134,379 = 12³ + 51³ = 38³ + 43³ 14: 149,389 = 8³ + 53³ = 29³ + 50³ 15: 165,464 = 20³ + 54³ = 38³ + 48³ 16: 171,288 = 17³ + 55³ = 24³ + 54³ 17: 195,841 = 9³ + 58³ = 22³ + 57³ 18: 216,027 = 3³ + 60³ = 22³ + 59³ 19: 216,125 = 5³ + 60³ = 45³ + 50³ 20: 262,656 = 8³ + 64³ = 36³ + 60³ 21: 314,496 = 4³ + 68³ = 30³ + 66³ 22: 320,264 = 18³ + 68³ = 32³ + 66³ 23: 327,763 = 30³ + 67³ = 51³ + 58³ 24: 373,464 = 6³ + 72³ = 54³ + 60³ 25: 402,597 = 42³ + 69³ = 56³ + 61³
Using a binary heap: <lang zkl>fcn cubeSum{
heap,n:=Heap(fcn([(a,_)],[(b,_)]){ a<=b }), 1; // heap cnt maxes out @ 244
while(1){
while(heap.empty or heap.top[0]>n.pow(3)){ # could also pre-calculate cubes
heap.push(T(n.pow(3) + 1, n,1)); n+=1;
}
s,x,y:= sxy:=heap.pop();
vm.yield(sxy);
y+=1;
if(y<x) # should be y <= x?
heap.push(T(x.pow(3) + y.pow(3), x,y));
}
} fcn taxis{
out:=List(T(0,0,0));
foreach s in (Utils.Generator(cubeSum)){
if(s[0]==out[-1][0]) out.append(s);
else{
if(out.len()>1) vm.yield(out); out.clear(s)
} }
} n:=0; foreach x in (Utils.Generator(taxis)){
n += 1; if(n >= 2006) break; if(n <= 25 or n >= 2000) println(n,": ",x);
}</lang> And a quickie heap implementation: <lang zkl>class Heap{ // binary heap
fcn init(lteqFcn='<=){
var [const, private] heap=List().pad(64,Void); // a power of 2
var cnt=0, cmp=lteqFcn;
}
fcn push(v){
// Resize the heap if it is too small to hold another item
if (cnt==heap.len()) heap.pad(cnt*2,Void);
index:=cnt; cnt+=1; while(index){ // Find out where to put the element
parent:=(index - 1)/2; if(cmp(heap[parent],v)) break; heap[index] = heap[parent]; index = parent;
}
heap[index] = v;
}
fcn pop{ // Remove the biggest element and return it
if(not cnt) return(Void);
v,temp:=heap[0], heap[cnt-=1];
// Reorder the elements
index:=0; while(1){ // Find the child to swap with
swap:=index*2 + 1; if (swap>=cnt) break; // If there are no children, the heap is reordered other:=swap + 1; if(other<cnt and cmp(heap[other],heap[swap])) swap = other; if(cmp(temp,heap[swap])) break; // If the bigger child is less than or equal to its parent, the heap is reordered
heap[index]=heap[swap]; index = swap;
}
heap[index] = temp;
v
}
var [proxy] top=fcn { if(cnt==0) Void else heap[0] };
var [proxy] empty=fcn{ (not cnt) };
}</lang>
- Output:
1: L(L(1729,10,9),L(1729,12,1)) ... 23: L(L(327763,67,30),L(327763,58,51)) 24: L(L(373464,60,54),L(373464,72,6)) 25: L(L(402597,61,56),L(402597,69,42)) 2000: L(L(1671816384,944,940),L(1671816384,1168,428)) 2001: L(L(1672470592,1124,632),L(1672470592,1187,29)) 2002: L(L(1673170856,1034,828),L(1673170856,1164,458)) 2003: L(L(1675045225,1153,522),L(1675045225,1081,744)) 2004: L(L(1675958167,1096,711),L(1675958167,1159,492)) 2005: L(L(1676926719,1188,63),L(1676926719,1095,714))