Refactor

lua/src/lib/math.lua

@@ -16,41 +16,9 @@ local function n_choose_r(n, r)
     end
 
     -- slow path for big numbers// slow path for larger numbers
-    local factors = {}
+    local factors = factor_gen(n, r)
     local answer
     local tmp
-
-    -- collect factors of final number
-    for i = 2,n,1
-    do
-        factors[i] = 1
-    end
-
-    -- negative factor values indicate need to divide
-    for i = 2,r,1
-    do
-        factors[i] = factors[i] - 1
-    end
-    for i = 2,(n - r),1
-    do
-        factors[i] = factors[i] - 1
-    end
-
-    -- this loop reduces to prime factors only
-    for i = n,2,-1
-    do
-        for j = 2,(i - 1),1
-        do
-            if i % j == 0
-            then
-                factors[j] = factors[j] + factors[i]
-                factors[i / j] = factors[i / j] + factors[i]
-                factors[i] = 0
-                break
-            end
-        end
-    end
-
     local i = 2
     local j = 2
     answer = 1;
@@ -93,6 +61,44 @@ local function n_choose_r(n, r)
     return answer
 end
 
+local function factor_gen(n, r)
+    local factors = {}
+
+    -- collect factors of final number
+    for i = 2,n,1
+    do
+        factors[i] = 1
+    end
+
+    -- negative factor values indicate need to divide
+    for i = 2,r,1
+    do
+        factors[i] = factors[i] - 1
+    end
+    for i = 2,(n - r),1
+    do
+        factors[i] = factors[i] - 1
+    end
+
+    -- this loop reduces to prime factors only
+    for i = n,2,-1
+    do
+        for j = 2,(i - 1),1
+        do
+            if i % j == 0
+            then
+                factors[j] = factors[j] + factors[i]
+                factors[i / j] = factors[i / j] + factors[i]
+                factors[i] = 0
+                break
+            end
+        end
+    end
+
+    return factors
+end
+
 return {
     factorial = factorial,
+    n_choose_r = n_choose_r,
 }

lua/src/p0015.lua

@@ -3,8 +3,8 @@
 -- Turns out this is easy, if you think sideways a bit
 --
 -- You can only go down or right. If we say right=1, then you can only have 20 1s, since otherwise you go off the grid.
--- You also can't have fewer than 20 1s, since then you go off the grid the other way. This means you can look at it as a
--- bit string, and the number of 40-bit strings with 20 1s is 40c20.
+-- You also can't have fewer than 20 1s, since then you go off the grid the other way. This means you can look at it as
+-- a bit string, and the number of 40-bit strings with 20 1s is 40c20.
 --
 -- Problem:
 --