cleanup and added more c examples:

18 files changed, 60 insertions, 394 deletions

diff --git a/10001prime.py.orig b/10001prime.py.orig
deleted file mode 100644
index ecaa8cd..0000000
--- a/10001prime.py.orig
+++ /dev/null
@@ -1,30 +0,0 @@
-import PIL, math
-
-#Problem 7 - 10,001st prime
-
-#returns a list of every prime up to a certain number
-def listprimes(number):
-	primes = []
-	if (number <=1):
-		return [1]
-	a = [1] * number
-	a[0]=0
-	a[1]=0#1 and 0 are not prime
-	##Sieve of eratosthenes
-	for i in range(2,int(math.ceil(math.sqrt(number)))):
-		if(a[i]==1):
-			j = i**2 #cross out all the multiples of i, starting with its square
-			while(j < number):
-				a[j] = 0
-				j+=i
-	
-	for k in range(2, number):
-		#is prime
-		if(a[k]==1):
-			primes.append(k)
-	return primes
-
-out_list = listprimes(1000000)
-#print(out_list)
-#return 10,001th prime factor
-print(out_list[10000])
diff --git a/bigfactors2.py.orig b/bigfactors2.py.orig
deleted file mode 100644
index 41bc13d..0000000
--- a/bigfactors2.py.orig
+++ /dev/null
@@ -1,32 +0,0 @@
-import PIL, math
-
-#Problem 12 Highly divisible triangular number
-#finds the first number with over 500 factors
-
-def countfactors(num):
-	factors = 0
-	### we only need to know about the FIRST HALF of factors,
-	##one factor implies a second
-	root = int(math.ceil(math.sqrt(num)))
-	divs = range(1, root)
-	for d in divs:
-		if(num%d == 0):
-			factors += 2
-			
-	#Correction if the number is a perfect square
-	if (root * root == num):
-		factors-=1
-	return factors
-
-
-####   MAIN #####
-i=1
-k=1
-j = 0
-while(k < 500):
-	j += i
-	k=countfactors(j)
-	print(str(j) + " has " + str(k) + " factors")
-	i += 1
-
-print("Ding! Ding! {} has over 500 factors, wow!".format(j))
diff --git a/collatz b/collatz
index 75c5d45..37da709 100755
--- a/collatz
+++ b/collatz
diff --git a/collatz.c b/collatz.c
index 5bbb783..ec77af6 100644
--- a/collatz.c
+++ b/collatz.c
@@ -1,19 +1,13 @@
 #include <stdlib.h>
 #include <stdio.h>
 #include <math.h>
-#include <string.h>
+// In C, it's immediately obvious my python solution is inefficient
 
-// Temporary storage chain
-//long int *chain;
-// The longest sequence found so far
-//long int *seq;
-
-// Size of the above arrays
+// 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){
-    //chain = (long int *) realloc (chain, c * sizeof(long int));
-    //chain[c - 1] == seed;
     c++;
     if(seed == 1)
         return 0;
@@ -21,19 +15,17 @@ long int collatz(long int seed){
         seed = seed / 2;
     else
         seed = 3 * seed + 1;
-    //collatz(seed);
+    collatz(seed);
 }
 
 int main(){
     for(long int i = 1; i < pow(10, 6); i++){
         c = 0;
-        //chain = (long int *) malloc(c * sizeof(long int));
         collatz(i);
-        printf("%d \n", i);
         if(c > s){
             s = c;
         }
     }
-    printf("Has %d numbers \n", s);
+    printf("The longest sequence has %d numbers \n", s);
     return 0;
 }
diff --git a/collatz.py.orig b/collatz.py.orig
deleted file mode 100644
index 404a2ed..0000000
--- a/collatz.py.orig
+++ /dev/null
@@ -1,24 +0,0 @@
-import PIL, 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.")
diff --git a/fib b/fib
new file mode 100755
index 0000000..e9b1b3b
--- /dev/null
+++ b/fib
diff --git a/fib.c b/fib.c
new file mode 100644
index 0000000..915fccd
--- /dev/null
+++ b/fib.c
@@ -0,0 +1,33 @@
+#include <stdlib.h> 
+#include <stdio.h>
+#include <math.h>
+
+const long int MAX = 4 * pow(10, 6);
+int *fib;
+
+int main(){
+    // Intialize sequence with first two terms
+    fib = (int *) malloc (2 * sizeof(int));
+    fib[0] = 1;
+    fib[1] = 1;
+    
+    // loop counter
+    int k = 1;
+    // sum of even terms
+    int sum = 0;
+    //next term in sequence
+    int next;
+    while(1){
+        next = fib[k] + fib[k - 1];
+        printf("%d\n", next);
+        if(next > MAX)
+            break;
+        if(next % 2 == 0)
+            sum += next;
+        k++;
+        fib = (int *) realloc (fib, (k + 1) * sizeof(int));
+        fib[k] = next;
+    }
+    printf("The sum of the even terms is %d\n", sum);
+    return 0;
+}
diff --git a/fib.py.orig b/fib.py.orig
deleted file mode 100644
index 77b32fa..0000000
--- a/fib.py.orig
+++ /dev/null
@@ -1,33 +0,0 @@
-import PIL, math
-MAX = 4*10**6
-
-#Problem 2 Even Fibonnacci numbers
-fib = [1, 1]
-
-k=1;
-while (True):
-	n = fib[k] + fib[k-1]
-	if(n>MAX):
-		break
-	else:
-		fib.append(n);
-		k+=1
-
-c=0
-for i in fib:
-	num = str(i)
-	print(num + " ", end='')
-	c+=1
-	if(c%10==0):
-		print()
-	
-print()
-
-print("The sum is: " + str(sum(fib)))
-
-s=0
-for i in fib:
-	if (i%2==0):
-		s+=i
-
-print("The sum of the even terms is: " + str(s))
diff --git a/largestprime b/largestprime
new file mode 100755
index 0000000..c2afe6f
--- /dev/null
+++ b/largestprime
diff --git a/largestprime.c b/largestprime.c
new file mode 100644
index 0000000..bf98414
--- /dev/null
+++ b/largestprime.c
@@ -0,0 +1,21 @@
+#include <stdio.h>
+#include <stdlib.h>
+
+long lpf(long num){
+    long factor = 2;
+    while(num > factor){
+        if(num % factor == 0){
+            num = num / factor;
+            factor = 2;
+        }
+        else{
+            factor ++;
+        }
+    }
+    return factor;
+}
+
+int main(){
+	printf("%d\n", lpf(600851475143));
+	return 0;
+}
diff --git a/largestprime.py b/largestprime.py
index fa53dc5..387de01 100644
--- a/largestprime.py
+++ b/largestprime.py
@@ -1,47 +1,6 @@
 import PIL
 import math
-import numpy as np
 # Problem 3 Largest Prime Factor
-
-###INEFFICIENT METHODS###
-# def isPrime(number):
-#	if (number <=1):
-#		return False
-#	for i in range(2,number):
-#		if(number%i==0):
-#			return False
-#	return True
-
-# def prime_factors(number):
-#primes = []
-# for i in range(number, 2, -1):
-# if(number%i==0):
-# if (isPrime(i)):
-# return i
-# return primes
-
-
-def prime_factors(number):
-    primes = []
-    if (number <= 1):
-        return [1]
-    a = np.ones(number)
-    a[0] = 0
-    a[1] = 0  # not prime
-    for i in range(2, int(math.ceil(math.sqrt(number)))):
-        if(a[i] == 1):
-            j = i**2  # cross out all the multiples
-            while(j < number):
-                a[j] = False
-                j += i
-
-    for k in range(2, number):
-        # is prime		is a factor
-        if(a[k] == 1) and (number % k == 0):
-            primes.append(k)
-    return primes
-
-
 def lpf(number):
     factor = 2
     while (number > factor):
@@ -54,4 +13,4 @@ def lpf(number):
 
 
 out = lpf(600851475143)
-print out
+print(out)
diff --git a/largestprime.py.orig b/largestprime.py.orig
deleted file mode 100644
index fa39904..0000000
--- a/largestprime.py.orig
+++ /dev/null
@@ -1,55 +0,0 @@
-import PIL, math
-import numpy as np
-#Problem 3 Largest Prime Factor
-
-###INEFFICIENT METHODS###
-#def isPrime(number):
-#	if (number <=1):
-#		return False
-#	for i in range(2,number):
-#		if(number%i==0):
-#			return False
-#	return True
-
-#def prime_factors(number):
-	#primes = []
-	#for i in range(number, 2, -1):
-		#if(number%i==0):
-			#if (isPrime(i)):
-				#return i
-	#return primes
-
-
-
-def prime_factors(number):
-	primes = []
-	if (number <=1):
-		return [1]
-	a = np.ones(number)
-	a[0]=0
-	a[1]=0#not prime
-	for i in range(2,int(math.ceil(math.sqrt(number)))):
-		if(a[i]==1):
-			j = i**2 #cross out all the multiples
-			while(j < number):
-				a[j] = False
-				j+=i
-	
-	for k in range(2, number):
-		#is prime		is a factor
-		if(a[k]==1) and (number%k==0):
-			primes.append(k)
-	return primes
-	
-def lpf(number):
-	factor = 2
-	while (number > factor):
-		if(number % factor == 0):
-			number = number/factor
-			factor = 2
-		else:
-			factor +=1
-	return factor
-			
-out = lpf(600851475143)
-print out
diff --git a/palindrome.py.orig b/palindrome.py.orig
deleted file mode 100644
index 9690553..0000000
--- a/palindrome.py.orig
+++ /dev/null
@@ -1,40 +0,0 @@
-import PIL, math
-
-#Problem 4 - Palindrome Products
-
-def isPalindrome(number):
-	numchar = str(number)
-	middle = len(numchar)/2
-	face = numchar[:middle]
-	ref = numchar[len(numchar):middle-1:-1]
-	if(face == ref):
-		return True
-	else:
-		return False
-
-def findProduct(number):
-	for i in range(999,100,-1):
-		for j in range(999,100,-1):
-			if(i*j==number):
-				return [i,j]
-			
-def findMaxPalindrome():
-	large = 0
-	for i in range(999,100,-1):
-		for j in range(999,100,-1):
-			test = i*j
-			if(isPalindrome(test) and test > large):
-				large = test
-	return large
-
-answer = findMaxPalindrome()
-print(answer)
-print("The factors are: " + str(findProduct(answer)))
-#x = input("Type a palindromic number: ")
-
-#if(isPalindrome(x)):
-	#print "The factors are: " + str(findProduct(x))
-#else:
-	#print "not a palindrome"
-
-			
diff --git a/productofdigits.py.orig b/productofdigits.py.orig
deleted file mode 100644
index ddd5edc..0000000
--- a/productofdigits.py.orig
+++ /dev/null
@@ -1,18 +0,0 @@
-import PIL, math
-import numpy as np
-
-#Problem 8 - largest Product in a series
-bigBoy=7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450
-
-def parseNum(x, r):
-	search = str(x)
-	maxi = 0
-	for ch in range(0,len(search)-r):
-		prod = int(search[ch])
-		for dig in range(ch+1, ch+r):
-			prod*=int(search[dig])
-		if (prod > maxi):
-			maxi = prod
-	return maxi
-			
-print(parseNum(bigBoy, 13))
diff --git a/smallmult.py.orig b/smallmult.py.orig
deleted file mode 100644
index d672ef0..0000000
--- a/smallmult.py.orig
+++ /dev/null
@@ -1,26 +0,0 @@
-#Problem 5 smallest multiple
-#2520 is the smallest number that can be divided by 
-#each of the numbers from 1 to 10 without any remainder.
-#What is the smallest positive number that is evenly divisible 
-#by all of the numbers from 1 to 20?
-
-import PIL, math
-
-def isDivisible(num, divisors):
-	divisible = True
-	for j in divisors:
-		if(num%j !=0):
-			divisible = False
-			break
-	return divisible
-
-divs = range(1,21)
-
-found = False
-start = 2520
-while (not found):
-	start+=20
-	found = isDivisible(start,divs)
-	
-print(start)
-print(str(isDivisible(start, divs)))
diff --git a/sumexp.py.orig b/sumexp.py.orig
deleted file mode 100644
index 90dd2da..0000000
--- a/sumexp.py.orig
+++ /dev/null
@@ -1,10 +0,0 @@
-import PIL, math
-#Problem 16 Power digit sum
-n = 2**1000
-strn = str(n)
-s = 0
-
-for i in strn:
-	s += int(i)
-	
-print(s)
diff --git a/sumofprimes.py.orig b/sumofprimes.py.orig
deleted file mode 100644
index 3799e94..0000000
--- a/sumofprimes.py.orig
+++ /dev/null
@@ -1,44 +0,0 @@
-import PIL, math
-
-#Problem 10 sum of primes
-###INEFFICIENT###
-#def isPrime(number):
-#	if (number <=1):
-#		return False
-#	for i in range(2,number):
-#		if(number%i==0):
-#			return False
-#	return True
-
-#def prime_factors(number):
-	#primes = []
-	#for i in range(number, 2, -1):
-		#if(number%i==0):
-			#if (isPrime(i)):
-				#return i
-	#return primes
-
-
-
-def prime_factors(number):
-	primes = []
-	if (number <=1):
-		return [1]
-	a = [1] * number
-	a[0]=0
-	a[1]=0#not prime
-	for i in range(2,int(math.ceil(math.sqrt(number)))):
-		if(a[i]==1):
-			j = i**2 #cross out all the multiples
-			while(j < number):
-				a[j] = False
-				j+=i
-	
-	for k in range(2, number):
-		#is prime
-		if(a[k]==1):
-			primes.append(k)
-	return primes
-
-out = prime_factors(2000000)
-print(str(sum(out)))
diff --git a/sumsq.py.orig b/sumsq.py.orig
deleted file mode 100644
index e5e8988..0000000
--- a/sumsq.py.orig
+++ /dev/null
@@ -1,27 +0,0 @@
-#The sum of the squares of the first ten natural numbers is,
-#12 + 22 + ... + 102 = 385
-#The square of the sum of the first ten natural numbers is,
-#(1 + 2 + ... + 10)2 = 552 = 3025
-#Hence the difference between the sum of the squares 
-#of the first ten natural numbers and the square of the sum is 
-#Find the difference between the sum of the squares of the first 
-#one hundred natural numbers and the square of the sum.
-
-import PIL, math
-
-nums = range(1,11)
-numsq = range(1,11)
-
-for i in nums:
-    i = i*i
-
-print(nums)
-print(numsq)
-
-sum1 = (sum(nums)**2)
-sum2 = (sum(numsq))
-
-print("The sum squared is: " + str(sum1))
-print("The sum of the squares is: " + str(sum2))
-
-print("The difference is: " + str(abs(sum2-sum1)))