A Cyclops number is a number that has a zero in the center (so, it needs to have odd number of digits and >=3 digits). The 0 should not appear anywhere else other than in center. Hence, 12035 is a Cyclops number but 12005 is not as there are more than one 0s. Nth Triangular number is calculated by N(N+1)/2. Hence, 1, 3, 6, 10, 15….are Triangular numbers. Find the list of first 100 Cyclops Triangular numbers i.e. which are both Cyclops as well as Triangular.
📌 Challenge Details and Links
ExcelBI Excel Challenge Number: 415
Challenge Difficulty: ⭐️
📥Download Sample File
📥Link to the solutions on LinkedIn
Solving the challenge of Cyclops and Triangular Numbers with Power Query
Power Query solution 1 for Cyclops and Triangular Numbers, proposed by Aditya Kumar Darak 🇮🇳:
let
Generate = List.Generate(
() => [a = 0, b = 0, e = false, f = 0],
each [f] < 101,
each [
a = [a] + 1,
b = a * (a + 1) / 2,
c = Text.From(b),
d = Text.Length(c),
e = Number.IsOdd(d)
and (Text.At(c, Number.RoundDown(d / 2)) = "0")
and (Text.Length(Text.Select(c, "0")) = 1),
f = [f] + Number.From(e)
],
each [Expected Number = [b], TF = [e]]
),
Table = Table.FromRecords(Generate),
Return = Table.SelectRows(Table, each [TF])[[Expected Number]]
in
ReturnPower Query solution 2 for Cyclops and Triangular Numbers, proposed by Alejandro Simón 🇵🇦 🇪🇸:
let
Source = List.Select(
List.Transform({101 .. 10000000}, each {_, Text.From(_)}),
each Text.Length(Text.Select(_{1}, "0")) = 1
),
Impar = List.Select(Source, each Number.IsOdd(Text.Length(_{1}))),
Lista = List.Transform(
List.Select(Impar, each Text.Range(_{1}, Number.RoundDown(Text.Length(_{1}) / 2), 1) = "0"),
each _{0}
),
Triang = List.Transform({1 .. 1000000}, each _ * (_ + 1) / 2),
Sol = List.FirstN(List.Intersect({Triang, Lista}), 100)
in
SolPower Query solution 3 for Cyclops and Triangular Numbers, proposed by Alexis Olson:
let
L = {1 .. 4000},
T = List.Transform(L, (n) => n * (n + 1) / 2),
IsCyclops = (n as number) =>
let
digits = Text.ToList(Number.ToText(n)),
len = List.Count(digits),
mid = if Number.IsOdd(len) then (len - 1) / 2 else null,
all_eyes = List.PositionOf(digits, "0", Occurrence.All)
in
all_eyes = {mid},
Result = List.Select(T, IsCyclops)
in
List.FirstN(Result, 100)Power Query solution 4 for Cyclops and Triangular Numbers, proposed by Ramiro Ayala Chávez:
let
T = List.Transform,
L = Text.Length,
A = Text.At,
S = List.Select,
a = List.Generate(() => [i = 1], each [i] <= 5000, each [i = [i] + 1], each [i] / 2 * ([i] + 1)),
b = S(T(a, Text.From), each Number.IsOdd(L(_)) and Text.Contains(_, "0")),
c = T(
b,
each
if L(_) = 3 and A(_, 1) = "0" or L(_) = 5 and A(_, 2) = "0" or L(_) = 7 and A(_, 3) = "0" then
_
else
null
),
d = T(
c,
each
if L(_)
= 5 and A(_, 1)
= "0" or L(_)
= 5 and A(_, 3)
= "0" or L(_)
= 7 and A(_, 1)
= "0" or L(_)
= 7 and A(_, 2)
= "0" or L(_)
= 7 and A(_, 4)
= "0" or L(_)
= 7 and A(_, 5)
= "0"
then
null
else
_
),
e = List.RemoveNulls(S(d, each Text.End(_, 1) <> "0")),
f = T(List.FirstN(e, 100), Number.From),
Sol = Table.FromColumns({f}, {"Expect Answer"})
in
SolPower Query solution 5 for Cyclops and Triangular Numbers, proposed by Eric Laforce:
let
fxIsCyclop = (n as number) =>
let
digits = Text.ToList(Number.ToText(n)),
nD = List.Count(digits),
zPos = List.PositionOf(digits, "0", Occurrence.All)
in
List.Count(zPos) = 1 and Number.IsOdd(nD) and (zPos{0} = Number.RoundDown(nD / 2)),
List = List.Generate(
() => [n = 1, r = {}],
each List.Count([r]) <= 100,
each
let
_n = [n] + 1,
_t = _n * (_n + 1) / 2
in
[n = _n, r = [r] & (if fxIsCyclop(_t) then {_t} else {})]
),
Result = Table.FromColumns({List.Last(List)[r]}, {"Expect Answer"})
in
ResultPower Query solution 6 for Cyclops and Triangular Numbers, proposed by Alexandra Popoff:
let
Max_Result = 100,
Cacl_Opt = Table.FromList(
List.Generate(
() => [Output = 0, z_temp = 0, z_test = false, z_Len = 0, z_i = 1, z_count = 0],
each [z_count] <= 100,
each [
z_temp = Number.ToText(([z_i] * ([z_i] + 1)) / 2, "0"),
z_Len = Text.Length(Text.From(z_temp)),
z_test = (Text.Length(z_temp) - Text.Length(Text.Replace(z_temp, "0", "")))
= 1 and Text.End(Text.Start(z_temp, Number.RoundUp(z_Len / 2, 0)), 1)
= "0" and Number.IsOdd(z_Len),
z_count = if z_test then [z_count] + 1 else [z_count],
Output = if z_test then Number.FromText(z_temp) else 0,
z_i = [z_i] + 1
],
each [Output]
),
Splitter.SplitByNothing(),
{"Result"}
),
Keep_Result = Table.SelectRows(Cacl_Opt, each ([Result] <> 0))
in
Keep_ResultSolving the challenge of Cyclops and Triangular Numbers with Excel
Excel solution 1 for Cyclops and Triangular Numbers, proposed by Bo Rydobon 🇹🇭:
=LET(n,
SEQUENCE(
3100
),
m,
n*(n+1)/2,
TAKE(TOCOL(m/(FIND(
0,
m
)*2-1=LEN(
m
))/ISNA(
TEXTAFTER(
m,
0,
2
)
),
3),
100))Excel solution 2 for Cyclops and Triangular Numbers, proposed by Rick Rothstein:
=LET(s,
SEQUENCE(
5000
),
n,
s*(s+1)/2,
c,
LEN(
n
),
TAKE(TOCOL(IF((c>4)*ISODD(
c
)*(c-LEN(
SUBSTITUTE(
n,
0,
)
)=1)*(MID(
n,
1+c/2,
1
)="0"),
n,
1/0),
3),
100))Excel solution 3 for Cyclops and Triangular Numbers, proposed by Rick Rothstein:
=LET(s,
SEQUENCE(
5000
),
n,
s*(s+1)/2,
c,
LEN(
n
),
x,
n*(c>4)*ISODD(
c
)*(c-LEN(
SUBSTITUTE(
n,
0,
)
)=1)*(-MID(
n,
1+c/2,
1
)=0),
TAKE(
FILTER(
x,
x
),
100
))Excel solution 4 for Cyclops and Triangular Numbers, proposed by John V.:
=TOCOL(LET(n,
ROW(
1:3023
),
MAP((n^2+n)/2,
LAMBDA(
x,
x/AND(
LEN(
TEXTSPLIT(
x,
0
)
)*{2,
2}=LEN(
x
)-1
)
))),
2)Excel solution 5 for Cyclops and Triangular Numbers, proposed by محمد حلمي:
=LET(n,
SEQUENCE(
3040
),
j,
n*(n+1)/2,
FILTER(j,
MAP(j,
LAMBDA(a,
LET(i,
LEN(
a
)/2,
IFERROR(
1-FIND(
0,
LEFT(
a,
i
)
)^0,
1
)*
IFERROR(
1-FIND(
0,
RIGHT(
a,
i
)
)^0,
1
)*(MID(
a,
i+0.5,
1
)="0"))))))Excel solution 6 for Cyclops and Triangular Numbers, proposed by Kris Jaganah:
=LET(a,
SEQUENCE(
999
),
b,
SORT(
--TOCOL(
a&0&TOROW(
a
)
)
),
c,
LEN(
b
),
d,
MOD((-1+(1+(8*b))^0.5)/2,
1)=0,
e,
FIND(
0,
b
)=(c/2)+0.5,
f,
(c-LEN(
SUBSTITUTE(
b,
"0",
""
)
))=1,
g,
MOD(
c,
2
),
TAKE(
FILTER(
b,
d*e*f*g
),
100
))Excel solution 7 for Cyclops and Triangular Numbers, proposed by Julian Poeltl:
=TAKE(LET(N,
SEQUENCE(
100000,
,
14
),
T,
N*(N+1)/2,
LT,
LEN(
T
),
CZ,
LT-LEN(
SUBSTITUTE(
T,
"0",
""
)
),
MZ,
RIGHT(
LEFT(
T,
LT/2+1
),
1
)="0",
FILTER(T,
(CZ=1)*MZ*ISODD(
LT
))),
100)Excel solution 8 for Cyclops and Triangular Numbers, proposed by Timothée BLIOT:
=LET(A,
SEQUENCE(
10^4
),
B,
A*(A+1)/2,
C,
LEN(
B
),
TAKE(FILTER(B,
(MID(
B,
FLOOR(
C/2,
1
)+1,
1
)="0")*(ISODD(
C
))*(LEN(
SUBSTITUTE(
B,
"0",
""
)
)=C-1)),
100))Excel solution 9 for Cyclops and Triangular Numbers, proposed by Hussein SATOUR:
=LET(a,
SEQUENCE(
457990,
,
10
),
b,
FILTER(a,
(ISEVEN(
LEN(
a
)
))*(ISERR(
FIND(
0,
a
)
))),
c,
--REPLACE(
b,
LEN(
b
)/2+1,
0,
0
),
d,
INT(
SQRT(
c*2
)
),
FILTER(c,
c*2=d*(d+1)))Excel solution 10 for Cyclops and Triangular Numbers, proposed by Sunny Baggu:
=LET(
_s,
SEQUENCE(
3073
),
_n,
_s * (_s + 1) / 2,
_cri,
(ISODD(
LEN(
_n
)
)) * (LEN(
_n
) >= 3) *
(MID(
_n,
ROUNDUP(
LEN(
_n
) / 2,
0
),
1
) = "0") *
(1 = MAP(
_n,
LAMBDA(
a,
SUM(
N(
MID(
a,
SEQUENCE(
LEN(
a
)
),
1
) = "0"
)
)
)
)),
TAKE(
FILTER(
_n,
_cri
),
100
)
)Excel solution 11 for Cyclops and Triangular Numbers, proposed by LEONARD OCHEA 🇷🇴:
=LET(s,
SEQUENCE(
3100
),
TAKE(TOCOL(MAP(s*(s+1)/2,
LAMBDA(a,
LET( m,
(LEN(
a
)+1)/2,
a/AND(
INT(
m
)=m,
--MID(
a,
m+{-1,
1}*SEQUENCE(
m-1
),
1
),
--MID(
a,
m,
1
)=0
)))),
3),
100))Excel solution 12 for Cyclops and Triangular Numbers, proposed by Charles Roldan:
=LET(
First,
LAMBDA(n,
f,
SCAN(0,
SEQUENCE(
n
),
LAMBDA(
g,
g(
g
)
)(LAMBDA(
g,
LAMBDA(
n,
[_],
IF(
f(
n + 1
),
n + 1,
g(
g
)(
n + 1
)
)
)
)))),
B,
LAMBDA(
f,
g,
LAMBDA(
n,
f(
g(
n
)
)
)
),
Triangle,
LAMBDA(
n,
n * (
n + 1
) / 2
),
isCyclops,
LAMBDA(n,
LET(f,
LAMBDA(
k,
IFERROR(
SEARCH(
"0",
n,
k + 1
),
)
),
AND(f(
HSTACK(
0,
f(
0
)
)
) * 2 = {1,
0} * (LEN(
n
) + 1)))),
Triangle(
First(
100,
B(
isCyclops,
Triangle
)
)
)
)Excel solution 13 for Cyclops and Triangular Numbers, proposed by Charles Roldan:
=LET(
First,
LAMBDA(f,
LAMBDA(n,
SCAN(0,
SEQUENCE(
n
),
LAMBDA(
g,
g(
g
)
)(LAMBDA(
g,
LAMBDA(
n,
[_],
IF(
f(
n + 1
),
n + 1,
g(
g
)(
n + 1
)
)
)
))))),
Triangle,
LAMBDA(
n,
SERIESSUM(
n,
1,
1,
{1,
1}
) / 2
),
isCyclops,
LAMBDA(
n,
AND(
HSTACK(
2 * IFERROR(
SEARCH(
"0",
n
),
),
LEN(
SUBSTITUTE(
n,
"0",
)
)
) = LEN(
n
) + {1,
-1}
)
),
B,
LAMBDA(
f,
g,
LAMBDA(
n,
f(
g(
n
)
)
)
),
B(
Triangle,
First(
B(
isCyclops,
Triangle
)
)
)
)(100)Excel solution 14 for Cyclops and Triangular Numbers, proposed by Giorgi Goderdzishvili:
= 1
while True:
triangular_number = n * (n + 1) // 2
if triangular_number == num:
return True
elif triangular_number > num:
return False
n += 1
def is_cyclops(
num
):
t = str(
num
)
mdl = len(
t
) // 2
cyc_check = all(
[t[mdl] == "0",
t.count(
'0'
) == 1]
)
if len(
t
) % 2 == 0:
return False
elif cyc_check:
return True
else:
return False
# Output
fin = []
counter = 1
N = 100
while len(
fin
)Excel solution 15 for Cyclops and Triangular Numbers, proposed by LUIS FLORENTINO COUTO CORTEGOSO:
=LET(a,
ROW(
1:10000
),
t,
a*(a+1)/2,
l,
LEN(
t
),
b,
TOCOL(t/((ISODD(
l
))*((l-LEN(
SUBSTITUTE(
t,
"0",
""
)
))=1)*(MID(
t,
INT(
l/2
)+1,
1
)="0")),
3),
TAKE(
b,
100
))
other formula:
=LET(s,
ROW(
1:9
),
g,
LAMBDA(y,
LET(n,
((1+8*y)^0.5-1)/2,
n=INT(
n
))),
f,
LAMBDA(
x,
TOCOL(
TOROW(
s
)&TOCOL(
x&TOROW(
s
)
)
)
),
a,
f(
0
),
b,
f(
a
),
c,
f(
b
),
d,
VSTACK(
a,
b,
c
),
TAKE(
SORT(
TOCOL(
d/MAP(
d,
g
),
3
)
),
100
))Excel solution 16 for Cyclops and Triangular Numbers, proposed by Ernesto Vega Castillo:
=TAKE(LET(f,
LET(sec,
SEQUENCE(
3500
),
num,
sec*(sec+1)/2,
cic,
LEN(
num
),
IFERROR(IF((cic>1)*ISODD(
cic
)*(cic-LEN(
SUBSTITUTE(
num,
0,
""
)
)=1)*(MID(
num,
1+cic/2,
1
)="0"),
num,
1/0),
"")),
FILTER(
f,
f<>""
)),
100)Excel solution 17 for Cyclops and Triangular Numbers, proposed by Tyler Cameron:
=LET(a,SEQUENCE(4000),b,a*(a+1)/2,c,MAP(b,LAMBDA(x,IF(AND(ISODD(LEN(x)),--MID(x,ROUNDUP(LEN(x)/2,0),1)=0,SUM(--(--MID(x,SEQUENCE(LEN(x)),1)=0))=1),x,""))),INDEX(FILTER(c,c<>""),SEQUENCE(100)))Excel solution 18 for Cyclops and Triangular Numbers, proposed by Alexandra Popoff:
= LAMBDA(Max_Resul,
[Max_Seq],
LET(
z_Max_Seq,
if(
isomitted(
Max_Seq
),
10000,
Max_Seq
),
z_Input,
SEQUENCE(
z_Max_Seq,
1,
1,
1
),
z_Numb_Temp,
TEXT(z_Input * (z_Input + 1) / 2,
"0"),
z_Numb,
FILTER(z_Numb_Temp,
(LEN(
z_Numb_Temp
) > 2) * (ISODD(
LEN(
z_Numb_Temp
)
))),
z_Test,
SCAN(
0,
z_Numb,
LAMBDA(
x_Count,
z_i,
IF(
x_Count >= Max_Resul,
Max_Resul,
x_Count + N(
LET(
x_temp,
z_i,
x_Len,
LEN(
INDEX(
x_temp,
1,
1
)
),
x_Seq,
SEQUENCE(
x_Len,
1,
1,
1
),
x_Arr,
BYROW(
SEQUENCE(
x_Len,
1,
1,
1
),
LAMBDA(
y_i,
RIGHT(
LEFT(
x_temp,
y_i
),
1
)
)
),
AND(
INDEX(
x_Arr,
ROUNDUP(
x_Len / 2,
0
),
1
) = "0",
IFERROR(
ROWS(
FILTER(
x_Arr,
x_Arr = "0"
)
),
0
) = 1
)
)
)
)
)
),
z_Out,
VALUE(FILTER(z_Numb,
(z_Test - VSTACK(
0,
DROP(
z_Test,
-1
)
)) = 1)),
IF(
ROWS(
z_Out
) < Max_Resul,
"z_Max Seq too low",
z_Out
)
))Solving the challenge of Cyclops and Triangular Numbers with Python
Python solution 1 for Cyclops and Triangular Numbers, proposed by Cristobal Salcedo Beltran:
Code:
import pandas as pd
def is_cyclops(n):
s = str(n)
length = len(s)
if length % 2 == 0 or length < 3:
return False
middle_index = length // 2
return s[middle_index] == '0' and '0' not in s[:middle_index] + s[middle_index+1:]
num_range = pd.DataFrame({'id': range(1, 10000)})
num_range['Expected Number'] = num_range['id'] * (num_range['id'] + 1) // 2
cyclops_triangular_numbers = num_range[num_range['Expected Number'].apply(is_cyclops)].drop('id', axis=1)
print(cyclops_triangular_numbers.head(100))
Python solution 2 for Cyclops and Triangular Numbers, proposed by Cristobal Salcedo Beltran:
Code:
from pyspark.sql import SparkSession
from pyspark.sql.functions import col, udf
from pyspark.sql.types import BooleanType, LongType
spark = SparkSession.builder.appName("CyclopsTriangularNumbers").getOrCreate()
def is_cyclops(n):
s = str(n)
length = len(s)
if length % 2 == 0 or length < 3:
return False
middle_index = length // 2
return s[middle_index] == '0' and '0' not in s[:middle_index] + s[middle_index+1:]
is_cyclops_udf = udf(is_cyclops, BooleanType())
num_range = spark.range(1, 10000)
triangular_numbers = num_range.withColumn("Expected Number", (col("id") * (col("id") + 1) / 2).cast(LongType()))
cyclops_triangular_numbers = triangular_numbers.filter(is_cyclops_udf("Expected Number")).drop("id")
cyclops_triangular_numbers.show(100)
Solving the challenge of Cyclops and Triangular Numbers with Python in Excel
Python in Excel solution 1 for Cyclops and Triangular Numbers, proposed by Abdallah Ally:
(Influenced by Konrad Gryczan, PhD link he posted)
import pandas as pd
from math import sqrt
file_path = 'Excel_Challenge_415 - Triangular Cyclops Numbers.xlsx'
df = pd.read_excel(file_path)
# Create a function to generate the
def triangular_cyclops(n):
numbers = []
number = 100
while len(numbers) != n:
number += 1
num_str = str(number)
if len(num_str) % 2:
single_zero = num_str.count('0') == 1
middle_zero = num_str[int(len(num_str) / 2)] == '0'
odd_length = (len(num_str) % 2) == 1
triangular = int(sqrt(number * 8 + 1)) == sqrt(number * 8 + 1)
else:
continue
if all([single_zero, middle_zero, odd_length, triangular]):
numbers.append(number)
return numbers
# Add results column
df['My Answer'] = pd.Series(triangular_cyclops(100))
print(f'n{df.head()}nn{df.tail()}')
Python in Excel solution 2 for Cyclops and Triangular Numbers, proposed by Abdallah Ally:
import pandas as pd
from math import sqrt
file_path = 'Excel_Challenge_415 - Triangular Cyclops Numbers.xlsx'
df = pd.read_excel(file_path)
# Create a function to generate the
def triangular_cyclops(n):
numbers = []
number = 100
while len(numbers) != n:
number += 1
num_str = str(number)
if len(num_str) % 2:
single_zero = num_str.count('0') == 1
middle_zero = num_str[int(len(num_str) / 2)] == '0'
odd_length = (len(num_str) % 2) == 1
triangular = False
for i in range(1, int(sqrt(2 * number)) + 2):
if i * (i + 1) == 2 * number:
triangular = True
break
else:
continue
if all([single_zero, middle_zero, odd_length, triangular]):
numbers.append(number)
return numbers
df['My Answer'] = pd.Series(triangular_cyclops(100))
print(f'n{df.head()}n{df.tail()})
Python in Excel solution 3 for Cyclops and Triangular Numbers, proposed by JvdV -:
=PY() in ms365:
import regex as re
[n for n in [int(i*(i+1)/2) for i in range(3024)] if re.match(r'(.(?=[^0]*0[^0]*$)0?(?1)?[^0])$',str(n))]
Solving the challenge of Cyclops and Triangular Numbers with R
R solution 1 for Cyclops and Triangular Numbers, proposed by Konrad Gryczan, PhD:
library(tidyverse)
library(readxl)
test = read_excel("Excel/415 Triangular Cyclops Numbers.xlsx", range = "A1:A101" )
range = 1:1e7
is_triangular <- function(x) {
n <- (-1 + sqrt(1 + 8 * x)) / 2
n == floor(n)
}
r = data.frame(n = range) %>%
mutate(nchar = nchar(n)) %>%
filter(nchar %% 2 == 1) %>%
mutate(zeroes = str_count(n, "0"),
central = substr(n, nchar/2+1, nchar/2+1)) %>%
filter(zeroes == 1,
central == "0") %>%
mutate(triangular = is_triangular(n)) %>%
filter(triangular == TRUE) %>%
head(100) %>%
mutate(n = as.numeric(n))
https://github.com/kgryczan/excelbi_puzzles/blob/main/Excel/415%20Challenge.R
&&