A taxicab number (also known as Hardy Ramanujan Number) is a positive integer that can be expressed as the sum of two positive cubes in more than one way. Most famous is 1729. 1729 = 1^3 + 12^3 = 9^3 + 10^3 List Y against a number if it is Taxicab number otherwise N.
📌 Challenge Details and Links
ExcelBI Excel Challenge Number: 481
Challenge Difficulty: ⭐️⭐️
📥Download Sample File
📥Link to the solutions on LinkedIn
Solving the challenge of Detect Hardy-Ramanujan Numbers with Power Query
Power Query solution 1 for Detect Hardy-Ramanujan Numbers, proposed by Aditya Kumar Darak 🇮🇳:
let
Source = Excel.CurrentWorkbook(){[Name = "data"]}[Content],
Return = Table.AddColumn(
Source,
"Answer",
each [
CR = Int64.From(Number.Power([Numbers], 1 / 3)),
G = List.Transform(
{1 .. CR},
(f) =>
[
c1 = Number.Power(f, 3),
c2 = Number.Power([Numbers] - c1, 1 / 3),
tf = Number.Mod(Number.Round(c2, 8), 1) = 0,
r = Number.From(tf)
][r]
),
R = List.Sum(G) / 2 > 1
][R]
)
in
ReturnPower Query solution 2 for Detect Hardy-Ramanujan Numbers, proposed by Alejandro Simón 🇵🇦 🇪🇸:
let
Source = Excel.CurrentWorkbook(){[Name = "Table1"]}[Content],
Sol = Table.AddColumn(
Source,
"Answer",
(z) =>
let
a = Int64.From(Number.Power(z[Numbers], 1 / 3)),
b = List.Select(
{1 .. a},
(x) =>
List.AnyTrue(
List.Transform({1 .. a}, (y) => Number.Power(x, 3) + Number.Power(y, 3) = z[Numbers])
)
),
c = if List.Count(b) > 1 then "Y" else "N"
in
c
)
in
SolPower Query solution 3 for Detect Hardy-Ramanujan Numbers, proposed by Alejandro Simón 🇵🇦 🇪🇸:
let
Source = Excel.CurrentWorkbook(){[Name = "Table1"]}[Content],
Sol = Table.AddColumn(
Source,
"Answer",
(z) =>
let
a = Int64.From(Number.Power(z[Numbers], 1 / 3)),
b = List.Transform(
{1 .. a},
(x) =>
List.Transform(
{1 .. a},
(y) => {x, y} & {Number.Power(x, 3) + Number.Power(y, 3) = z[Numbers]}
)
),
c = List.Select(List.Combine(b), each _{2} = true),
d = List.Distinct(List.Transform(c, each List.Sort(List.FirstN(_, 2)))),
e = if List.Count(d) > 1 then "Y" else "N"
in
e
)
in
SolPower Query solution 4 for Detect Hardy-Ramanujan Numbers, proposed by Luan Rodrigues:
let
Fonte = Tabela1,
res = Table.AddColumn(
Fonte,
"Personalizar",
each
let
a = List.Transform(
{1 .. Int64.From(Number.Power([Numbers], 1 / 3))},
(x) => Number.Power(([Numbers] - Number.Power(x, 3)), 1 / 3)
),
b = List.Select(a, each Number.Round(_, 2) = Int64.From(_)),
c = if (try List.Count(b) otherwise 0) = 0 then "N" else "Y"
in
c
)
in
resPower Query solution 5 for Detect Hardy-Ramanujan Numbers, proposed by Abdallah Ally:
let
Source = Excel.CurrentWorkbook(){[Name = "Table1"]}[Content],
f = (n) =>
let
a = {1 .. Number.RoundDown(Number.Power(n, 1 / 3))},
b = List.Accumulate(
a,
{},
(x, y) =>
let
c = Number.Round(Number.Power(n - Number.Power(y, 3), 1 / 3), 6),
d = if List.Contains(a, c) then x & {y, c} else x
in
List.Distinct(d)
)
in
if List.Count(b) > 2 then "Y" else "N",
g = Table.AddColumn(Source, "My Answer", each f([Numbers])),
Result = Table.AddColumn(g, "Check", each [Answer Expected] = [My Answer])
in
ResultPower Query solution 6 for Detect Hardy-Ramanujan Numbers, proposed by Ramiro Ayala Chávez:
let
S = Excel.CurrentWorkbook(){[Name = "Table1"]}[Content],
Sol = Table.AddColumn(
S,
"Answer Expected",
each
let
A = [Numbers],
B = {1 .. Number.RoundDown(Number.Power(A, 1 / 3))},
C = List.TransformMany(B, (x) => B, (x, y) => {x, y}),
D = List.Select(C, each _{0} <> _{1} and _{0} < _{1}),
E = List.Transform(D, each Number.Power(_{0}, 3) + Number.Power(_{1}, 3)),
F = List.Select(E, each _ = A),
G = if List.Count(F) <> 0 then "Y" else "N"
in
G
)
in
SolSolving the challenge of Detect Hardy-Ramanujan Numbers with Excel
Excel solution 1 for Detect Hardy-Ramanujan Numbers, proposed by Bo Rydobon 🇹🇭:
=MAP(A2:A10,
LAMBDA(a,
LET(q,
1/3,
b,
(a-SEQUENCE(
a^q
)^3)^q,
IF(
SUM(
N(
INT(
b
)=b
)
)>2,
"Y",
"N"
))))Excel solution 2 for Detect Hardy-Ramanujan Numbers, proposed by Rick Rothstein:
=IF(MAP(A2:A10,
LAMBDA(x,
SUM(0+(x=SEQUENCE(
500
)^3+SEQUENCE(
,
500
)^3))>2)),
"Y",
"N")Excel solution 3 for Detect Hardy-Ramanujan Numbers, proposed by John V.:
=MAP(
A2:A10,
LAMBDA(
x,
LET(
s,
ROW(
1:500
)^3,
IF(
SUM(
N(
s+TOROW(
s
)=x
)
)>1,
"Y",
"N"
)
)
)
)
✅=MAP(A2:A10,
LAMBDA(x,
LET(s,
SEQUENCE(x^(1/3))^3,
IF(
SUM(
N(
s+TOROW(
s
)=x
)
)>1,
"Y",
"N"
))))Excel solution 4 for Detect Hardy-Ramanujan Numbers, proposed by Kris Jaganah:
=MAP(A2:A10,
LAMBDA(x,
LET(a,
SEQUENCE(x^(1/3)),
IF(
SUM(
N(
a^3+TOROW(
a
)^3=x
)
)>3,
"Y",
"N"
))))Excel solution 5 for Detect Hardy-Ramanujan Numbers, proposed by Julian Poeltl:
=MAP(A2:A10,
LAMBDA(N,
LET(M,
MAX(N^(1/3)),
S,
SEQUENCE(
M
)^3,
IF(SUM(--(N=(S+TOROW(
S
))))>2,
"Y",
"N"))))Excel solution 6 for Detect Hardy-Ramanujan Numbers, proposed by Timothée BLIOT:
=MAP(A2:A10,
LAMBDA(z,
IF(SUM(--(z=MAKEARRAY(
999,
999,
LAMBDA(
x,
y,
x^3+y^3
)
)))>1,
"Y",
"N")))Excel solution 7 for Detect Hardy-Ramanujan Numbers, proposed by Oscar Mendez Roca Farell:
=MAP(A2:A10,
LAMBDA(a,
LET(c,
1/3,
s,
(a-SEQUENCE(
a^c
)^3)^c,
IF(
OR(
s=INT(
s
)
),
"Y",
"N"
))))Excel solution 8 for Detect Hardy-Ramanujan Numbers, proposed by Sunny Baggu:
=IF(
MAP(
A2:A10,
LAMBDA(a,
LET(
b,
SEQUENCE(a ^ (1 / 3)) ^ 3,
c,
TOROW(
b
),
SUM(N((b + c) = a)) > 1
)
)
),
"Y",
"N"
)Excel solution 9 for Detect Hardy-Ramanujan Numbers, proposed by Abdallah Ally:
=MAP(A2:A10,
LAMBDA(x,
LET(a,
x,
b,
SEQUENCE(a^(1/3)),
c,
DROP(
REDUCE(
"",
b,
LAMBDA(
x,
y,
IF(
OR(
a-y^3=FILTER(
b,
b<>y
)^3
),
VSTACK(
x,
y
),
x
)
)
),
1
),
IF(
COUNT(
UNIQUE(
c
)
)>2,
"Y",
"N"
))))Excel solution 10 for Detect Hardy-Ramanujan Numbers, proposed by Tyler Cameron:
=LET(z,
500,
a,
MUNIT(
z
),
b,
SEQUENCE(
z
)^3,
MAP(A2:A10,
LAMBDA(x,
IF(SUM(--(IF(
a,
"",
b+TOROW(
b
)
)=x))>2,
"Y",
"N"))))Solving the challenge of Detect Hardy-Ramanujan Numbers with Python
Python solution 1 for Detect Hardy-Ramanujan Numbers, proposed by Konrad Gryczan, PhD:
Code longer because of time measuring.
0.012 sec
import pandas as pd
import itertools
import time
path = '481 Taxicab Numbers.xlsx'
input = pd.read_excel(path, usecols="A")
test = pd.read_excel(path, usecols="B")
start_time = time.time()
def is_taxicab(num):
croot = round(num**(1/3) + 0.5)
seq = range(1, croot)
comb = [x for x in itertools.combinations(seq, 2) if x[0]**3 + x[1]**3 == num]
if len(comb) >= 2:
return "Y"
else:
return "N"
result = input.copy()
result['Answer Expected'] = result['Numbers'].apply(is_taxicab)
end_time = time.time()
execution_time = end_time - start_time
print("Execution time:", execution_time, "seconds")
# Execution time: 0.012981891632080078 seconds
print(result['Answer Expected'].equals(test['Answer Expected'])) # True
Solving the challenge of Detect Hardy-Ramanujan Numbers with Python in Excel
Python in Excel solution 1 for Detect Hardy-Ramanujan Numbers, proposed by Abdallah Ally:
import pandas as pd
from itertools import product
def is_taxicab_number(number):
numbers = []
values = range(1, int(number ** (1/3)) + 1)
for num1, num2 in product(values, values):
if num1 != num2 and num1 ** 3 + num2 ** 3 == number:
numbers.extend([num1, num2])
return 'Y' if len(set(numbers)) > 2 else 'N'
file_path = 'Excel_Challenge_481 - Taxicab Numbers.xlsx'
df = pd.read_excel(file_path)
# Perform data wrangling
df['My Answer'] = df['Numbers'].map(is_taxicab_number)
df['check'] = df['Answer Expected'] == df['My Answer']
df
Solving the challenge of Detect Hardy-Ramanujan Numbers with R
R solution 1 for Detect Hardy-Ramanujan Numbers, proposed by Konrad Gryczan, PhD:
library(tidyverse)
library(readxl)
library(tictoc)
path = "Excel/481 Taxicab Numbers.xlsx"
input = read_excel(path, range = "A1:A10")
test = read_excel(path, range = "B1:B10")
tic()
is_taxicab = function(number) {
x = ceiling(number^(1/3))
df = tibble(a = 1:x, b = 1:x) %>%
expand.grid() %>%
filter(a <= b,
a^3 + b^3 == number)
check = ifelse(nrow(df) >= 2, "Y", "N")
return(check)
}
result = input %>%
mutate(`Answer Expected` = map_chr(Numbers, is_taxicab))
toc()
identical(result$`Answer Expected`, test$`Answer Expected`)
# [1] TRUE
R solution 2 for Detect Hardy-Ramanujan Numbers, proposed by Anil Kumar Goyal:
Fast enough
df <- data.frame(
Numbers = c(4104L,8435L,13832L,28939L,
60625L,65728L,314496L,1392000L,87539319L)
)
check_hardy <- function(n){
map_lgl(seq(floor(n^(1/3))), ~ dplyr::near(
round((n - .x^3)^(1/3)),
(n - .x^3)^(1/3)
)) %>% {sum(.) > 1}
}
df %>%
mutate(ans = map_lgl(Numbers, check_hardy))
Solving the challenge of Detect Hardy-Ramanujan Numbers with Excel VBA
Excel VBA solution 1 for Detect Hardy-Ramanujan Numbers, proposed by Vasin Nilyok:
Sub TaxicabNum()
LastRow = Cells(Rows.Count, 1).End(xlUp).Row
For r = 2 To LastRow
QNum = Cells(r, 1)
LoopNum = Fix(WorksheetFunction.Power(Cells(r, 1) - 1, 1 / 3))
For a = 1 To LoopNum
LeftOne = WorksheetFunction.Power(a, 3)
For b = a + 1 To LoopNum
GoTo nextb
Else
GoTo CheckRight
End If
nextb:
Next b
Next a
CheckRight:
For c = a + 1 To LoopNum
RightOne = WorksheetFunction.Power(c, 3)
For d = c + 1 To LoopNum
GoTo nextd
Else
If AnsLeft = AnsRight Then
Cells(r, 3) = "Y"
GoTo NextQNum
End If
End If
nextd:
Next d
Next c
Cells(r, 3) = "N"
NextQNum:
Next r
End Sub
&&&