User-person's solutions for challenge 54.

2 files changed, 182 insertions, 0 deletions

diff --git a/challenge-054/user-person/python/ch-1.py b/challenge-054/user-person/python/ch-1.py
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+++ b/challenge-054/user-person/python/ch-1.py
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+#!/usr/bin/env python3
+
+###########################################################################
+# script name: ch-1.py                                                    #
+# Wed Apr  1 04:40:14 2020 | 1585730414                                   #
+#                                                                         #
+# https://github.com/user-person                                          #
+#                                                                         #
+# https://perlweeklychallenge.org/blog/perl-weekly-challenge-054/         #
+#                                                                         #
+# kth Permutation Sequence                                                #
+# Write a script to accept two integers n (>=1) and k (>=1).              #
+# It should print the kth permutation of n integers.                      #
+# For more information, please follow the wiki page.                      #
+# https://en.wikipedia.org/wiki/Permutation#k-permutations_of_n           #
+# For example, n=3 and k=4,                                               #
+# the possible permutation sequences are listed below:                    #
+#                                                                         #
+#   123                                                                   #
+#   132                                                                   #
+#   213                                                                   #
+#   231                                                                   #
+#   312                                                                   #
+#   321                                                                   #
+# The script should print the 4th permutation sequence 231.               #
+#                                                                         #
+###########################################################################
+
+from itertools import permutations
+import os
+import re
+import sys
+
+def msgExit():
+    print(os.path.basename(sys.argv[0]),  " requires 2 arguments. The first argument to determine the sequence ( 1 >= ),\nThe second argument to determine which entry to print. Both arguments should be >= 1 .\n");
+    exit(1)
+
+def printSmooshed(ints):
+    for num in ints:
+        print(num,end='')
+    print()
+    
+if len(sys.argv) != 3: 
+          msgExit()
+          
+n = int(sys.argv[1])
+k = int(sys.argv[2])
+
+if re.search(r'\A\d+\Z', str(n)) and re.search(r'\A\d+\Z', str(k)):
+     if n < 1 or k < 1:
+         msgExit()
+else:
+    msgExit()              
+
+kth = 1;
+failure = True
+
+for i in permutations(range(1,n)):
+    if kth == k:
+        printSmooshed(i)
+        failure = False
+        break
+    else:
+        kth += 1
+
+if failure:
+    print('There is no ' +  str(k) + "-th number in the '" + str(n) + " sequence'.")
diff --git a/challenge-054/user-person/python/ch-2.py b/challenge-054/user-person/python/ch-2.py
new file mode 100755
index 0000000000..fe8dc75fdd
--- /dev/null
+++ b/challenge-054/user-person/python/ch-2.py
@@ -0,0 +1,115 @@
+#!/usr/bin/env python3
+
+###########################################################################
+# script name: ch-2.py                                                    #
+# Wed Apr  1 13:45:26 2020 | 1585763126                                   #
+#                                                                         #
+# https://github.com/user-person                                          #
+#                                                                         #
+# https://perlweeklychallenge.org/blog/perl-weekly-challenge-054/         #
+#                                                                         #
+# Collatz Conjecture                                                      #
+# Contributed by Ryan Thompson                                            #
+#                                                                         #
+# It is thought that the following sequence will always reach 1:          #
+# $n = $n / 2 when $n is even                                             #
+# $n = 3*$n + 1 when $n is odd                                            #
+#                                                                         #
+# For example, if we start at 23, we get the following sequence:          #
+# 23 -> 70 -> 35 -> 106 -> 53 -> 160 -> 80 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1   #
+#                                                                         #
+# Write a function that finds the Collatz sequence for any positive       #
+# integer.                                                                #
+# Notice how the sequence itself may go far above the original starting   #
+# number.                                                                 #
+#                                                                         #
+# Extra Credit                                                            #
+# Have your script calculate the sequence length for all starting numbers #
+# up to 1000000 (1e6), and output the starting number                     #
+# and sequence length for the longest 20 sequences.perl                   #
+#                                                                         #
+###########################################################################
+
+import os
+import re
+import sys
+
+def msgExit():
+    print(os.path.basename(sys.argv[0]),  "requires one positive integer as an argument (or extracredit).")
+    exit(1)
+
+high = {}
+
+def checkHighKeys(key,keyCount,highKeys):
+    if len(highKeys) == 20:
+        highKeys = {k: v for k, v in sorted(highKeys.items(), key=lambda item: (item[1],item[0]))}
+        if highKeys[list(highKeys)[0]] < keyCount:
+            del highKeys[list(highKeys)[0]]
+            highKeys[key] = keyCount
+        elif highKeys[list(highKeys)[0]] == keyCount:
+            if list(highKeys)[0] < key:
+                del highKeys[list(highKeys)[0]]
+                highKeys[key] = keyCount
+    else:
+        highKeys[key] = keyCount
+
+    return highKeys
+
+if len(sys.argv) != 2:
+    msgExit()
+
+n = sys.argv[1]
+extraCredit = False
+ecPattern = re.compile(r'-{0,2}extra[-~`!@#$%^&*=+|\\;:",.?/ ]?credit', flags=re.IGNORECASE)
+
+if  ecPattern.search(str(n)):
+    n = 2
+    extraCredit = True
+    sys.stderr.flush()
+    sys.stderr.write('Allow time for calculations.\n')
+    sys.stderr.write("        They finish when '#'s reach this point-> |\n")
+
+elif re.search(r'\A\d+\Z',str(n)):
+    n = int(n)
+    if n < 1:
+        msgExit()
+    if n == 1:
+        print(1)
+        exit()
+else:
+    msgExit()
+    
+MAX = 1000000
+i = n
+
+while i <= MAX:
+    count = 0
+
+    while n != 1:
+        if extraCredit:
+            count += 1
+        else:
+            print(int(n),'-> ', end='',flush=True)
+
+        if n % 2 == 0:          # EVEN
+            n /= 2
+        else:                   # ODD
+            n = 3*n + 1
+
+    if extraCredit:
+        high = checkHighKeys(i, count, high)
+        i += 1
+        n = i
+
+        if n % 20000 == 0:
+            sys.stderr.write('#')
+            sys.stderr.flush()
+    else:
+        print(1)
+        break
+
+if extraCredit:
+    sys.stderr.write('\n\n')
+    high = {k: v for k, v in sorted(high.items(), key=lambda item: (item[1],item[0]),reverse=True)}
+    for thisKey in high.keys():
+        print('Starting number:', thisKey, 'with sequence length:',  high[thisKey])