diff options
30 files changed, 834 insertions, 27 deletions
diff --git a/challenge-167/paulo-custodio/python/ch-2.py b/challenge-167/paulo-custodio/python/ch-2.py index 1a8fc16a90..616c4080cc 100644 --- a/challenge-167/paulo-custodio/python/ch-2.py +++ b/challenge-167/paulo-custodio/python/ch-2.py @@ -1,4 +1,4 @@ -#!/usr/bin/env perl +#!/usr/bin/env python3 # Challenge 167 # diff --git a/challenge-168/paulo-custodio/perl/ch-1.pl b/challenge-168/paulo-custodio/perl/ch-1.pl index eda2430282..ca521979bc 100644 --- a/challenge-168/paulo-custodio/perl/ch-1.pl +++ b/challenge-168/paulo-custodio/perl/ch-1.pl @@ -1,4 +1,4 @@ -#!/usr/bin/perl +#!/usr/bin/env perl # Challenge 168 # @@ -6,7 +6,7 @@ # Submitted by: Roger Bell_West # # The Perrin sequence is defined to start with [3, 0, 2]; after that, term N is -# the sum of terms N-2 and N-3. (So it continues 3, 2, 5, 5, 7, ….) +# the sum of terms N-2 and N-3. (So it continues 3, 2, 5, 5, 7, ....) # # A Perrin prime is a number in the Perrin sequence which is also a prime # number. diff --git a/challenge-168/paulo-custodio/perl/ch-2.pl b/challenge-168/paulo-custodio/perl/ch-2.pl index b3ce77932c..427fafa330 100644 --- a/challenge-168/paulo-custodio/perl/ch-2.pl +++ b/challenge-168/paulo-custodio/perl/ch-2.pl @@ -1,4 +1,4 @@ -#!/usr/bin/perl +#!/usr/bin/env perl # Challenge 168 # @@ -18,9 +18,9 @@ # # As given in the Wikipedia page, # -# HP(10) = 773, as 10 factors as 2×5 yielding HP10(1) = 25, 25 factors as 5×5 -# yielding HP10(2) = HP25(1) = 55, 55 = 5×11 implies HP10(3) = HP25(2) = HP55(1) -# = 511, and 511 = 7×73 gives HP10(4) = HP25(3) = HP55(2) = HP511(1) = 773, a +# HP(10) = 773, as 10 factors as 2x5 yielding HP10(1) = 25, 25 factors as 5x5 +# yielding HP10(2) = HP25(1) = 55, 55 = 5x11 implies HP10(3) = HP25(2) = HP55(1) +# = 511, and 511 = 7x73 gives HP10(4) = HP25(3) = HP55(2) = HP511(1) = 773, a # prime number. use Modern::Perl; diff --git a/challenge-168/paulo-custodio/python/ch-1.py b/challenge-168/paulo-custodio/python/ch-1.py new file mode 100644 index 0000000000..631b6eea96 --- /dev/null +++ b/challenge-168/paulo-custodio/python/ch-1.py @@ -0,0 +1,60 @@ +#!/usr/bin/env python3 + +# Challenge 168 +# +# Task 1: Perrin Prime +# Submitted by: Roger Bell_West +# +# The Perrin sequence is defined to start with [3, 0, 2]; after that, term N is +# the sum of terms N-2 and N-3. (So it continues 3, 2, 5, 5, 7, ....) +# +# A Perrin prime is a number in the Perrin sequence which is also a prime +# number. +# +# Calculate the first 13 Perrin Primes. +# +# f(13) = [2, 3, 5, 7, 17, 29, 277, 367, 853, 14197, 43721, 1442968193, +# 792606555396977] + +def is_prime(n): + if n <= 1: + return False + if n <= 3: + return True + if n % 2 == 0 or n % 3 == 0: + return False + i = 5 + while i * i <= n: + if n % i == 0 or n % (i + 2) == 0: + return False + i += 6 + return True + +def perrin_seq(): + t = [3, 0, 2] + def next_perrin(): + t.append(t[-2] + t[-3]) + return t.pop(0) + return next_perrin + +def perrin_prime_seq(): + it = perrin_seq() + seen = {} + def next_perrin_prime(): + while True: + n = it() + if is_prime(n) and n not in seen: + seen[n] = True + return n + return next_perrin_prime + +import sys + +if len(sys.argv) != 2: + raise ValueError("usage: ch-1.py n") +n = int(sys.argv[1]) +pp = [] +it = perrin_prime_seq() +while len(pp) < n: + pp.append(it()) +print(", ".join(map(str, sorted(pp)))) diff --git a/challenge-168/paulo-custodio/python/ch-2.py b/challenge-168/paulo-custodio/python/ch-2.py new file mode 100644 index 0000000000..75f56e23b6 --- /dev/null +++ b/challenge-168/paulo-custodio/python/ch-2.py @@ -0,0 +1,64 @@ +#!/usr/bin/env python3 + +# Challenge 168 +# +# Task 2: Home Prime +# Submitted by: Mohammad S Anwar +# +# You are given an integer greater than 1. +# +# Write a script to find the home prime of the given number. +# +# In number theory, the home prime HP(n) of an integer n greater than 1 is the +# prime number obtained by repeatedly factoring the increasing concatenation of +# prime factors including repetitions. +# +# Further information can be found on Wikipedia and OEIS. +# Example +# +# As given in the Wikipedia page, +# +# HP(10) = 773, as 10 factors as 2x5 yielding HP10(1) = 25, 25 factors as 5x5 +# yielding HP10(2) = HP25(1) = 55, 55 = 5x11 implies HP10(3) = HP25(2) = HP55(1) +# = 511, and 511 = 7x73 gives HP10(4) = HP25(3) = HP55(2) = HP511(1) = 773, a +# prime number. + +def is_prime(n): + if n <= 1: + return False + if n <= 3: + return True + if n % 2 == 0 or n % 3 == 0: + return False + i = 5 + while i * i <= n: + if n % i == 0 or n % (i + 2) == 0: + return False + i += 6 + return True + +def prime_factors(n): + factors = [] + p = 0 + while n > 1: + p += 1 + while not is_prime(p): + p += 1 + while n % p == 0: + factors.append(p) + n //= p + return factors + +def home_prime(n): + if n <= 1: + raise ValueError("Input must be greater than 1") + while not is_prime(n): + factors = prime_factors(n) + n = int(''.join(map(str, factors))) + return n + +import sys + +if len(sys.argv) != 2: + raise ValueError("usage: script.py n") +print(home_prime(int(sys.argv[1]))) diff --git a/challenge-169/paulo-custodio/perl/ch-1.pl b/challenge-169/paulo-custodio/perl/ch-1.pl index 313305f986..058a608817 100644 --- a/challenge-169/paulo-custodio/perl/ch-1.pl +++ b/challenge-169/paulo-custodio/perl/ch-1.pl @@ -1,4 +1,4 @@ -#!/usr/bin/perl +#!/usr/bin/env perl # Challenge 169 # @@ -9,7 +9,7 @@ # # Brilliant numbers are numbers with two prime factors of the same length. # -# The number should have exactly two prime factors, i.e. it’s the product of two +# The number should have exactly two prime factors, i.e. it's the product of two # primes of the same length. # # For example, diff --git a/challenge-169/paulo-custodio/perl/ch-2.pl b/challenge-169/paulo-custodio/perl/ch-2.pl index e049f4f458..bf1625cd34 100644 --- a/challenge-169/paulo-custodio/perl/ch-2.pl +++ b/challenge-169/paulo-custodio/perl/ch-2.pl @@ -1,4 +1,4 @@ -#!/usr/bin/perl +#!/usr/bin/env perl # Challenge 169 # diff --git a/challenge-169/paulo-custodio/python/ch-1.py b/challenge-169/paulo-custodio/python/ch-1.py new file mode 100644 index 0000000000..37ef60de29 --- /dev/null +++ b/challenge-169/paulo-custodio/python/ch-1.py @@ -0,0 +1,76 @@ +#!/usr/bin/env python3 + +# Challenge 169 +# +# Task 1: Brilliant Numbers +# Submitted by: Mohammad S Anwar +# +# Write a script to generate first 20 Brilliant Numbers. +# +# Brilliant numbers are numbers with two prime factors of the same length. +# +# The number should have exactly two prime factors, i.e. it's the product of two +# primes of the same length. +# +# For example, +# +# 24287 = 149 x 163 +# 24289 = 107 x 227 +# +# Therefore 24287 and 24289 are 2-brilliant numbers. +# These two brilliant numbers happen to be consecutive as there are no even +# brilliant numbers greater than 14. +# +# +# Output +# +# 4, 6, 9, 10, 14, 15, 21, 25, 35, 49, 121, 143, 169, 187, 209, 221, 247, 253, +# 289, 299 + +import sys + +def is_prime(n): + if n <= 1: + return False + if n <= 3: + return True + if n % 2 == 0 or n % 3 == 0: + return False + i = 5 + while i * i <= n: + if n % i == 0 or n % (i + 2) == 0: + return False + i += 6 + return True + +def prime_factors(n): + factors = [] + p = 0 + while n > 1: + p += 1 + while not is_prime(p): + p += 1 + while n % p == 0: + factors.append(p) + n //= p + return factors + +def is_brillant(n): + factors = prime_factors(n) + return len(factors) == 2 and is_prime(factors[0]) and is_prime(factors[1]) and len(str(factors[0])) == len(str(factors[1])) + +def brillant_seq(): + n = 1 + while True: + n += 1 + if is_brillant(n): + yield n + +if len(sys.argv) != 2: + raise ValueError("usage: ch-1.py n") +n = int(sys.argv[1]) +brillant = [] +it = brillant_seq() +while len(brillant) < n: + brillant.append(next(it)) +print(", ".join(map(str, brillant))) diff --git a/challenge-169/paulo-custodio/python/ch-2.py b/challenge-169/paulo-custodio/python/ch-2.py new file mode 100644 index 0000000000..aca0a73fd8 --- /dev/null +++ b/challenge-169/paulo-custodio/python/ch-2.py @@ -0,0 +1,95 @@ +#!/usr/bin/env python3 + +# Challenge 169 +# +# Task 2: Achilles Numbers +# Submitted by: Mohammad S Anwar +# +# Write a script to generate first 20 Achilles Numbers. Please checkout wikipedia +# for more information. +# +# An Achilles number is a number that is powerful but imperfect (not a +# perfect power). Named after Achilles, a hero of the Trojan war, who was +# also powerful but imperfect. +# +# A positive integer n is a powerful number if, for every prime factor p of +# n, p^2 is also a divisor. +# +# A number is a perfect power if it has any integer roots (square root, cube +# root, etc.). +# +# For example 36 factors to (2, 2, 3, 3) - every prime factor (2, 3) also has its +# square as a divisor (4, 9). But 36 has an integer square root, 6, so the number +# is a perfect power. +# +# But 72 factors to (2, 2, 2, 3, 3); it similarly has 4 and 9 as divisors, but it +# has no integer roots. This is an Achilles number. +# +# Output +# +# 72, 108, 200, 288, 392, 432, 500, 648, 675, 800, 864, 968, 972, 1125, +# 1152, 1323, 1352, 1372, 1568, 1800 + +def is_prime(n): + if n <= 1: + return False + if n <= 3: + return True + if n % 2 == 0 or n % 3 == 0: + return False + i = 5 + while i * i <= n: + if n % i == 0 or n % (i + 2) == 0: + return False + i += 6 + return True + +def prime_factors(n): + factors = [] + p = 0 + while n > 1: + p += 1 + while not is_prime(p): + p += 1 + while n % p == 0: + factors.append(p) + n //= p + return factors + +def is_powerfull(n): + factors = prime_factors(n) + for factor in factors: + if n % (factor * factor) != 0: + return False + return True + +def is_perfect(n): + for e in range(2, n + 1): + for b in range(1, n + 1): + power = b ** e + if power == n: + return True + if power > n: + break + return False + +def is_aquilles(n): + return is_powerfull(n) and not is_perfect(n) + +def aquilles_seq(): + n = 1 + while True: + n += 1 + if is_aquilles(n): + yield n + +import sys + +if len(sys.argv) != 2: + raise ValueError("usage: ch-2.py n") +n = int(sys.argv[1]) +aquilles = [] +it = aquilles_seq() +while len(aquilles) < n: + aquilles.append(next(it)) +print(", ".join(map(str, aquilles))) diff --git a/challenge-170/paulo-custodio/perl/ch-1.pl b/challenge-170/paulo-custodio/perl/ch-1.pl index 4d36966785..3d07569f3b 100644 --- a/challenge-170/paulo-custodio/perl/ch-1.pl +++ b/challenge-170/paulo-custodio/perl/ch-1.pl @@ -1,4 +1,4 @@ -#!/usr/bin/perl +#!/usr/bin/env perl # Challenge 170 # @@ -14,9 +14,9 @@ # # P(0) = 1 (1) # P(1) = 2 (1x2) -# P(2) = 6 (1x2×3) -# P(3) = 30 (1x2×3×5) -# P(4) = 210 (1x2×3×5×7) +# P(2) = 6 (1x2x3) +# P(3) = 30 (1x2x3x5) +# P(4) = 210 (1x2x3x5x7) use Modern::Perl; diff --git a/challenge-170/paulo-custodio/perl/ch-2.pl b/challenge-170/paulo-custodio/perl/ch-2.pl index dfff30bef8..92e0d5b3c2 100644 --- a/challenge-170/paulo-custodio/perl/ch-2.pl +++ b/challenge-170/paulo-custodio/perl/ch-2.pl @@ -1,4 +1,4 @@ -#!/usr/bin/perl +#!/usr/bin/env perl # Challenge 170 # @@ -61,7 +61,7 @@ sub parse_input { sub kronecker_product { my($a, $b) = @_; my $wa = @{$a->[0]}; my $ha = @$a; - my $wb = @{$b->[0]}; my $hb = @$a; + my $wb = @{$b->[0]}; my $hb = @$b; my @prod; for my $ar (0..$ha-1) { for my $ac (0..$wa-1) { diff --git a/challenge-170/paulo-custodio/python/ch-1.py b/challenge-170/paulo-custodio/python/ch-1.py new file mode 100644 index 0000000000..1b2a973b87 --- /dev/null +++ b/challenge-170/paulo-custodio/python/ch-1.py @@ -0,0 +1,52 @@ +#!/usr/bin/env python3 + +# Challenge 170 +# +# Task 1: Primorial Numbers +# Submitted by: Mohammad S Anwar +# +# Write a script to generate first 10 Primorial Numbers. +# +# Primorial numbers are those formed by multiplying successive prime numbers. +# +# +# For example, +# +# P(0) = 1 (1) +# P(1) = 2 (1x2) +# P(2) = 6 (1x2x3) +# P(3) = 30 (1x2x3x5) +# P(4) = 210 (1x2x3x5x7) + +import sys + +def is_prime(n): + if n == 2 or n == 3: + return True + if n <= 1 or n % 2 == 0 or n % 3 == 0: + return False + i = 5 + while i * i <= n: + if n % i == 0 or n % (i + 2) == 0: + return False + i += 6 + return True + +def next_prime(p): + p += 1 + while not is_prime(p): + p += 1 + return p + +def primorial_numbers(n): + primorial = [1] + p = 1 + while len(primorial) < n: + p = next_prime(p) + primorial.append(primorial[-1] * p) + return primorial + +if len(sys.argv) != 2: + raise ValueError("usage: ch-1.py n") +n = int(sys.argv[1]) +print(", ".join(map(str, primorial_numbers(n)))) diff --git a/challenge-170/paulo-custodio/python/ch-2.py b/challenge-170/paulo-custodio/python/ch-2.py new file mode 100644 index 0000000000..ed5be6a669 --- /dev/null +++ b/challenge-170/paulo-custodio/python/ch-2.py @@ -0,0 +1,80 @@ +#!/usr/bin/env python3 + +# Challenge 170 +# +# Task 2: Kronecker Product +# Submitted by: Mohammad S Anwar +# +# You are given 2 matrices. +# +# Write a script to implement Kronecker Product on the given 2 matrices. +# +# For more information, please refer wikipedia page. +# +# For example, +# +# A = [ 1 2 ] +# [ 3 4 ] +# +# B = [ 5 6 ] +# [ 7 8 ] +# +# A x B = [ 1 x [ 5 6 ] 2 x [ 5 6 ] ] +# [ [ 7 8 ] [ 7 8 ] ] +# [ 3 x [ 5 6 ] 4 x [ 5 6 ] ] +# [ [ 7 8 ] [ 7 8 ] ] +# +# = [ 1x5 1x6 2x5 2x6 ] +# [ 1x7 1x8 2x7 2x8 ] +# [ 3x5 3x6 4x5 4x6 ] +# [ 3x7 3x8 4x7 4x8 ] +# +# = [ 5 6 10 12 ] +# [ 7 8 14 16 ] +# [ 15 18 20 24 ] +# [ 21 24 28 32 ] + +import sys +import re + +def parse_matrix(line): + mx = [] + while re.match(r'^\s*$', line): + line = sys.stdin.readline() + line, count = re.subn(r'^\s*\w+\s*=\s*', '', line) + if count == 0: + raise ValueError("assignment expected") + while found := re.match(r'^\s*\[([\d\s]+)\]\s*$', line): + row = [int(x) for x in found.group(1).split()] + mx.append(row) + line = sys.stdin.readline() + return mx, line + +def parse_input(): + line = sys.stdin.readline() + a, line = parse_matrix(line) + b, line = parse_matrix(line) + return a, b + +def print_matrix(mx): + for row in mx: + print("[ "+ " ".join([str(x) for x in row]) +" ]") + +def kronecker_product(a, b): + wa = len(a[0]) + ha = len(a) + wb = len(b[0]) + hb = len(b) + prod = [[0 for _ in range(wa*wb)] for _ in range(ha*hb)] + for ar in range(ha): + for ac in range(wa): + for br in range(hb): + for bc in range(wb): + tr = ar*hb + br + tc = ac*wb + bc + prod[tr][tc] = a[ar][ac] * b[br][bc] + return prod + +a, b = parse_input() +p = kronecker_product(a, b) +print_matrix(p) diff --git a/challenge-171/paulo-custodio/perl/ch-1.pl b/challenge-171/paulo-custodio/perl/ch-1.pl index b67f270336..735f7b2082 100644 --- a/challenge-171/paulo-custodio/perl/ch-1.pl +++ b/challenge-171/paulo-custodio/perl/ch-1.pl @@ -1,4 +1,4 @@ -#!/usr/bin/perl +#!/usr/bin/env perl # Challenge 171 # diff --git a/challenge-171/paulo-custodio/perl/ch-2.pl b/challenge-171/paulo-custodio/perl/ch-2.pl index 45c3da0abf..96387742bc 100644 --- a/challenge-171/paulo-custodio/perl/ch-2.pl +++ b/challenge-171/paulo-custodio/perl/ch-2.pl @@ -1,4 +1,4 @@ -#!/usr/bin/perl +#!/usr/bin/env perl # Challenge 171 # diff --git a/challenge-171/paulo-custodio/python/ch-1.py b/challenge-171/paulo-custodio/python/ch-1.py new file mode 100644 index 0000000000..a84a80698e --- /dev/null +++ b/challenge-171/paulo-custodio/python/ch-1.py @@ -0,0 +1,54 @@ +#!/usr/bin/env python3 + +# Challenge 171 +# +# Task 1: Abundant Number +# Submitted by: Mohammad S Anwar +# +# Write a script to generate first 20 Abundant Odd Numbers. +# +# According to wikipedia, +# +# A number n for which the sum of divisors s(n) > 2n, or, equivalently, the +# sum of proper divisors (or aliquot sum) s(n) > n. +# +# +# For example, 945 is the first Abundant Odd Number. +# +# Sum of divisors: +# 1 + 3 + 5 + 7 + 9 + 15 + 21 + 27 + 35 + 45 + 63 + 105 + 135 + 189 + 315 = 975 + +import sys +import math +from functools import reduce + +def divisors(n): + div_low = [] + div_high = [] + for i in range(1, int(math.sqrt(n)) + 1): + if n % i == 0: + div_low.append(i) + if n // i != i: + div_high.insert(0, n // i) + return div_low + div_high + +def proper_divisors(n): + div = divisors(n) + return div[:-1] + +def is_abundant(n): + return sum(proper_divisors(n)) > n + +def abundant_numbers(N): + abundant = [] + n = 1 + while len(abundant) < N: + if is_abundant(n): + abundant.append(n) + n += 1 + return abundant + +if len(sys.argv) != 2: + raise ValueError("usage: script.py n") +N = int(sys.argv[1]) +print(", ".join(map(str, abundant_numbers(N)))) diff --git a/challenge-171/paulo-custodio/python/ch-2.py b/challenge-171/paulo-custodio/python/ch-2.py new file mode 100644 index 0000000000..f780acf173 --- /dev/null +++ b/challenge-171/paulo-custodio/python/ch-2.py @@ -0,0 +1,32 @@ +#!/usr/bin/env python3 + +# Challenge 171 +# +# Task 2: First-class Function +# Submitted by: Mohammad S Anwar +# +# Create sub compose($f, $g) which takes in two parameters $f and $g as +# subroutine refs and returns subroutine ref i.e. compose($f, $g)->($x) = +# $f->($g->($x)) +# +# e.g. +# +# $f = (one or more parameters function) +# $g = (one or more parameters function) +# +# $h = compose($f, $g) +# $f->($g->($x,$y, ..)) == $h->($x, $y, ..) for any $x, $y, ... + +import sys + +def compose(f, g): + return lambda x: f(g(x)) + +def times3(x): + return 3 * x + +def times5(x): + return 5 * x + +h = compose(times3, times5) +print(h(int(int(sys.argv[1])))) diff --git a/challenge-172/paulo-custodio/perl/ch-1.pl b/challenge-172/paulo-custodio/perl/ch-1.pl index 6753187f1a..1a28191ff2 100644 --- a/challenge-172/paulo-custodio/perl/ch-1.pl +++ b/challenge-172/paulo-custodio/perl/ch-1.pl @@ -1,4 +1,4 @@ -#!/usr/bin/perl +#!/usr/bin/env perl # Challenge 172 # diff --git a/challenge-172/paulo-custodio/perl/ch-2.pl b/challenge-172/paulo-custodio/perl/ch-2.pl index a7a6852c2e..1f4bf144c9 100644 --- a/challenge-172/paulo-custodio/perl/ch-2.pl +++ b/challenge-172/paulo-custodio/perl/ch-2.pl @@ -1,4 +1,4 @@ -#!/usr/bin/perl +#!/usr/bin/env perl # Challenge 172 # @@ -40,4 +40,4 @@ sub five_number_summary { return ($n[0], lower_quartile(@n), median(@n), upper_quartile(@n), $n[-1]); } -say join ", ", five_number_summary(@ARGV); +say join ", ", map {sprintf("%.1f", $_)} five_number_summary(@ARGV); diff --git a/challenge-172/paulo-custodio/python/ch-1.py b/challenge-172/paulo-custodio/python/ch-1.py new file mode 100644 index 0000000000..7bc35bc760 --- /dev/null +++ b/challenge-172/paulo-custodio/python/ch-1.py @@ -0,0 +1,90 @@ +#!/usr/bin/env python3 + +# Challenge 172 +# +# Task 1: Prime Partition +# Submitted by: Mohammad S Anwar +# +# You are given two positive integers, $m and $n. +# +# Write a script to find out the Prime Partition of the given number. No +# duplicates allowed. +# +# For example, +# +# Input: $m = 18, $n = 2 +# Output: 5, 13 or 7, 11 +# +# Input: $m = 19, $n = 3 +# Output: 3, 5, 11 + +import sys +from itertools import count +from functools import reduce + +def is_prime(n): + if n <= 1: + return False + if n <= 3: + return True + if n % 2 == 0 or n % 3 == 0: + return False + for i in range(5, int(n**0.5) + 1, 6): + if n % i == 0 or n % (i + 2) == 0: + return False + return True + +def next_prime(n): + if n <= 1: + return 2 + p = n + while True: + p += 1 + if is_prime(p): + return p + +def primes(n): + p = 2 + prime_list = [] + while p <= n: + prime_list.append(p) + p = next_prime(p) + return prime_list + +def odometer_increment(limit, n): + i = len(n) - 1 + while True: + n[i] += 1 + if n[i] >= limit: + n[i] = 0 + i -= 1 + if i < 0: + n.insert(0, 0) + i = 0 + else: + break + return n + +def has_duplicates(n): + seen = set() + for num in n: + if num in seen: + return True + seen.add(num) + return False + +def prime_partition(m, n): + prime_list = primes(m) + idx = [0] * n + while len(idx) == n: + if not has_duplicates(idx): + terms = [prime_list[i] for i in idx] + if sum(terms) == m: + return terms + idx = odometer_increment(len(prime_list), idx) + return [] + +if len(sys.argv) != 3: + raise ValueError("usage: ch-1.py m n") +m, n = map(int, sys.argv[1:]) +print(", ".join(map(str, prime_partition(m, n)))) diff --git a/challenge-172/paulo-custodio/python/ch-2.py b/challenge-172/paulo-custodio/python/ch-2.py new file mode 100644 index 0000000000..d6de8fcc5b --- /dev/null +++ b/challenge-172/paulo-custodio/python/ch-2.py @@ -0,0 +1,32 @@ +#!/usr/bin/env python3 + +# Challenge 172 +# +# Task 2: Five-number Summary +# Submitted by: Mohammad S Anwar +# +# You are given an array of integers. +# +# Write a script to compute the five-number summary of the given set of integers. +# +# You can find the definition and example in the wikipedia page. + +def median(n): + if len(n) % 2 == 0: + i = len(n) // 2 + return (n[i - 1] + n[i]) / 2 + else: + return n[len(n) // 2] + +def lower_quartile(n): + return median(n[:len(n) // 2]) + +def upper_quartile(n): + return median(n[len(n) // 2:]) + +def five_number_summary(n): + n.sort() + return (n[0], lower_quartile(n), median(n), upper_quartile(n), n[-1]) + +import sys +print(", ".join([f"{x:.1f}" for x in five_number_summary(list(map(float, sys.argv[1:])))])) diff --git a/challenge-172/paulo-custodio/t/test-2.yaml b/challenge-172/paulo-custodio/t/test-2.yaml index 7271a491ad..c840c8ef46 100644 --- a/challenge-172/paulo-custodio/t/test-2.yaml +++ b/challenge-172/paulo-custodio/t/test-2.yaml @@ -2,4 +2,4 @@ cleanup: args: 0 0 1 2 63 61 27 13 input: - output: 0, 0.5, 7.5, 44, 63 + output: 0.0, 0.5, 7.5, 44.0, 63.0 diff --git a/challenge-173/paulo-custodio/perl/ch-1.pl b/challenge-173/paulo-custodio/perl/ch-1.pl index caba1bf96d..859df10ea8 100644 --- a/challenge-173/paulo-custodio/perl/ch-1.pl +++ b/challenge-173/paulo-custodio/perl/ch-1.pl @@ -1,4 +1,4 @@ -#!/usr/bin/perl +#!/usr/bin/env perl # Challenge 173 # diff --git a/challenge-173/paulo-custodio/perl/ch-2.pl b/challenge-173/paulo-custodio/perl/ch-2.pl |