From 93ea7fe5957b62f18e8fbd17a21696bd7de6332d Mon Sep 17 00:00:00 2001
From: mjfernez <mjfernez@gmail.com>
Date: Sun, 9 Feb 2020 15:16:26 -0500
Subject: Organized everything, update README

---
 14-Longest-Collatz-Sequence/collatz    | Bin 0 -> 16696 bytes
 14-Longest-Collatz-Sequence/collatz.c  |  31 +++++++++++++++++++++++++++++++
 14-Longest-Collatz-Sequence/collatz.py |  28 ++++++++++++++++++++++++++++
 3 files changed, 59 insertions(+)
 create mode 100755 14-Longest-Collatz-Sequence/collatz
 create mode 100644 14-Longest-Collatz-Sequence/collatz.c
 create mode 100644 14-Longest-Collatz-Sequence/collatz.py

(limited to '14-Longest-Collatz-Sequence')

diff --git a/14-Longest-Collatz-Sequence/collatz b/14-Longest-Collatz-Sequence/collatz
new file mode 100755
index 0000000..37da709
Binary files /dev/null and b/14-Longest-Collatz-Sequence/collatz differ
diff --git a/14-Longest-Collatz-Sequence/collatz.c b/14-Longest-Collatz-Sequence/collatz.c
new file mode 100644
index 0000000..ec77af6
--- /dev/null
+++ b/14-Longest-Collatz-Sequence/collatz.c
@@ -0,0 +1,31 @@
+#include <stdlib.h>
+#include <stdio.h>
+#include <math.h>
+// In C, it's immediately obvious my python solution is inefficient
+
+// For keeping track of the length of sequences
+// c is a temporary counter, s changes when c reaches a new maximum
+long int c = 0;
+long int s = 0;
+long int collatz(long int seed){
+    c++;
+    if(seed == 1)
+        return 0;
+    if(seed % 2 == 0)
+        seed = seed / 2;
+    else
+        seed = 3 * seed + 1;
+    collatz(seed);
+}
+
+int main(){
+    for(long int i = 1; i < pow(10, 6); i++){
+        c = 0;
+        collatz(i);
+        if(c > s){
+            s = c;
+        }
+    }
+    printf("The longest sequence has %d numbers \n", s);
+    return 0;
+}
diff --git a/14-Longest-Collatz-Sequence/collatz.py b/14-Longest-Collatz-Sequence/collatz.py
new file mode 100644
index 0000000..74d5e1e
--- /dev/null
+++ b/14-Longest-Collatz-Sequence/collatz.py
@@ -0,0 +1,28 @@
+import PIL
+import math
+
+# Problem 14 Longest Collatz sequence
+# Note, a little slow, takes about 15 seconds.
+# There is probably a more efficient solution out there
+chain = []
+biggo = []
+
+
+def collatz(seed):
+    chain.append(seed)
+    if(seed == 1):
+        return 0
+    if(seed % 2 == 0):
+        seed = seed/2
+    else:
+        seed = 3*seed + 1
+    collatz(seed)
+
+
+for n in range(1, pow(10, 6)):
+    collatz(n)
+    if(len(chain) > len(biggo)):
+        biggo = chain
+    chain = []
+print(biggo)
+print("Has " + str(len(biggo)) + " numbers.")
-- 
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