Add Lua solution to challenge 101

2 files changed, 152 insertions, 0 deletions

diff --git a/challenge-101/paulo-custodio/lua/ch-1.lua b/challenge-101/paulo-custodio/lua/ch-1.lua
new file mode 100644
index 0000000000..168993f879
--- /dev/null
+++ b/challenge-101/paulo-custodio/lua/ch-1.lua
@@ -0,0 +1,108 @@
+#!/usr/bin/env lua
+
+--[[
+You are given an array @A of items (integers say, but they can be anything).
+
+Your task is to pack that array into an MxN matrix spirally counterclockwise, 
+as tightly as possible.
+
+‘Tightly’ means the absolute value |M-N| of the difference has to be as small 
+as possible.
+--]]
+
+-- find out m,n
+function smallest_rect(n)
+    local low = 1
+    local high = n
+    for i=1, math.floor(math.sqrt(n)) do
+        if math.floor(math.fmod(n, i)) == 0 then
+            low, high = i, math.floor(n/i)
+        end
+    end
+    return low, high
+end
+
+-- find max width of elements, convert to int
+function max_width(numbers)
+    local num_width = 1
+    for i=1, #numbers do
+        if #numbers[i] < num_width then
+            num_width = #numbers[i]
+            numbers[i] = tonumber(numbers[i])
+        end
+    end
+    return num_width
+end
+
+-- build empty rectangle
+function build_empty_rectangle(m, n)
+    local rect = {}
+    for r=1, m do
+        rect[r] = {}
+        for c=1, n do 
+            rect[r][c] = ""
+        end
+    end
+    return rect
+end
+
+-- build spiral rectangle
+function spiral(numbers)
+    local m, n = smallest_rect(#numbers)
+    local num_width = max_width(numbers)
+    local rect = build_empty_rectangle(m, n)
+    local r, c = m, 1
+    local i = 1
+    while i <= #numbers do
+        -- go East
+        while c <= n do
+            if rect[r][c] ~= "" then break end
+            rect[r][c] = string.format("%"..tostring(num_width+1).."d",
+                                       numbers[i])
+            i = i + 1
+            c = c + 1
+        end
+        c = c - 1 
+        r = r - 1
+        -- go North
+        while r >= 1 do
+            if rect[r][c] ~= "" then break end
+            rect[r][c] = string.format("%"..tostring(num_width+1).."d",
+                                       numbers[i])
+            i = i + 1
+            r = r - 1
+        end
+        r = r + 1 
+        c = c - 1
+        -- go West
+        while c >= 1 do
+            if rect[r][c] ~= "" then break end
+            rect[r][c] = string.format("%"..tostring(num_width+1).."d",
+                                       numbers[i])
+            i = i + 1
+            c = c - 1
+        end
+        c = c + 1 
+        r = r + 1
+        -- go South
+        while r <= m do
+            if rect[r][c] ~= "" then break end
+            rect[r][c] = string.format("%"..tostring(num_width+1).."d",
+                                       numbers[i])
+            i = i + 1
+            r = r + 1
+        end
+        r = r - 1 
+        c = c + 1
+    end
+    
+    -- print result
+    for r=1, m do
+        for c=1, n do 
+            io.write(rect[r][c])
+        end
+        io.write("\n")
+    end
+end
+
+spiral(arg)
diff --git a/challenge-101/paulo-custodio/lua/ch-2.lua b/challenge-101/paulo-custodio/lua/ch-2.lua
new file mode 100644
index 0000000000..a748e1bd02
--- /dev/null
+++ b/challenge-101/paulo-custodio/lua/ch-2.lua
@@ -0,0 +1,44 @@
+#!/usr/bin/env lua
+
+--[[
+TASK #2 › Origin-containing Triangle
+Submitted by: Stuart Little
+You are given three points in the plane, as a list of six co-ordinates: 
+A=(x1,y1), B=(x2,y2) and C=(x3,y3).
+
+Write a script to find out if the triangle formed by the given three 
+co-ordinates contain origin (0,0).
+
+Print 1 if found otherwise 0.
+--]]
+
+function sign(x1,y1,x2,y2,x3,y3)
+    return (x1 - x3) * (y2 - y3) - (x2 - x3) * (y1 - y3)
+end
+
+function point_in_triangle(xp,yp, x1,y1,x2,y2,x3,y3)
+    local d1 = sign(xp,yp, x1,y1, x2,y2)
+    local d2 = sign(xp,yp, x2,y2, x3,y3)
+    local d3 = sign(xp,yp, x3,y3, x1,y1)
+
+    local has_neg
+    if (d1 < 0) or (d2 < 0) or (d3 < 0) then 
+        has_neg = true
+    else
+        has_neg = false
+    end
+    
+    local has_neg = (d1 < 0) or (d2 < 0) or (d3 < 0)
+    local has_pos = (d1 > 0) or (d2 > 0) or (d3 > 0)
+
+    if has_neg and has_pos then
+        return 0
+    else
+        return 1
+    end
+end
+
+io.write(point_in_triangle(0,0, 
+                           tonumber(arg[1]), tonumber(arg[2]),
+                           tonumber(arg[3]), tonumber(arg[4]),
+                           tonumber(arg[5]), tonumber(arg[6])), "\n")