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+# the number of ways to change amount a using n kinds of coins:
+
+# the number of ways to change amount a using all but the first
+# kind of coin
+
+# +
+
+# the number of ways to change a - d using all n kinds of coins,
+# where d is the denomination of the first kind of coin.
+
+# c(a, [n = list of coins]) = c(a, n[:1]) + c(a-n[0], n)
+
+# using the following base rules:
+
+# if a = 0, result = 1
+# if a < 0, result = 0
+# if n = 0, result = 0
+
+(defn first-denomination [kinds-of-coins]
+  (cond
+    (= kinds-of-coins 1) 1
+    (= kinds-of-coins 2) 5
+    (= kinds-of-coins 3) 10
+    (= kinds-of-coins 4) 25
+    (= kinds-of-coins 5) 50))
+
+(defn cc [amount kind-of-coins]
+  (cond
+    (= amount 0) 1
+    (or
+      (< amount 0)
+      (= kind-of-coins 0)) 0
+    (+
+     (cc amount (- kind-of-coins 1))
+     (cc (- amount (first-denomination kind-of-coins)) kind-of-coins))))
+
+(defn count-change [amount] 
+  (cc amount 5))
+
+# (cc 11.0 5)
+# (cc 11.0 4) (cc -39.0 5)
+# (cc 11.0 3) (cc -14.0 4)
+# (cc 11.0 2) (cc 1.0 3)
+# (cc 11.0 1) (cc 6.0 2) (cc 1.0 2) (cc -9.0 3)
+# (cc 11.0 0) (cc 10.0 1) (cc 6.0 1) (cc 1.0 2) (cc 1.0 1) (cc -4.0 2)
+# (cc 10.0 0) (cc 9.0 1) (cc 6.0 0) (cc 5.0 1) (cc 1.0 1) (cc -4.0 2) (cc 1.0 0) (cc 0.0 1)
+# (cc 9.0 0) (cc 8.0 1) (cc 5.0 0) (cc 4.0 1) (cc 1.0 0) (cc 0.0 1) 1
+# (cc 8.0 0) (cc 7.0 1) (cc 4.0 0) (cc 3.0 1) 1 1
+# (cc 7.0 0) (cc 6.0 1) (cc 3.0 0) (cc 2.0 1) 1 1
+# (cc 6.0 0) (cc 5.0 1) (cc 2.0 0) (cc 1.0 1) 1 1
+# (cc 5.0 0) (cc 4.0 1) (cc 1.0 0) (cc 0.0 1) 1 1
+# (cc 4.0 0) (cc 3.0 1) 1 1 1
+# (cc 3.0 0) (cc 2.0 1) 1 1 1
+# (cc 2.0 0) (cc 1.0 1) 1 1 1 
+# (cc 1.0 0) 1 1 1 1
+# 1 1 1 1 
+# 4
+
+(import ./dot :as dot)
+
+(defn add-to [graph counter amount kind-of-coins parent]
+  (def label (string/format "(cc %d %d)" amount kind-of-coins))
+  (def node_name (string/format "node_%d" counter))
+  (def clr
+    (cond (= kind-of-coins 0) "lightgray"
+          (= amount 0) "cyan"
+          "lightgreen"))
+  (cond
+    (= kind-of-coins 2) (dot/add-node (get (get graph :subgraphs) 1) node_name :label label :shape "box" :color clr)
+    (= kind-of-coins 1) (dot/add-node (get (get graph :subgraphs) 0) node_name :label label :shape "box" :color clr)
+    (do
+      (dot/add-node graph node_name :label label :shape "box" :color clr)))
+
+  (dot/add-relation graph parent node_name)
+  node_name)
+
+(defn count-change2 [amount denom]
+  (def graph (dot/create-graph :digraph "change"))
+  (dot/add-subgraph graph (dot/create-subgraph "rank_1" :rankdir "TB"))
+  (dot/add-subgraph graph (dot/create-subgraph "rank_2" :rankdir "LR"))
+  (var counter 0)
+  (defn cc [amount kind-of-coins parent]
+    (def name (add-to graph counter amount kind-of-coins parent))
+    (set counter (+ counter 1))
+    (cond
+      (= amount 0) 1
+      (or
+          (< amount 0)
+          (= kind-of-coins 0)) 0
+      (+
+        (cc amount (- kind-of-coins 1) name)
+        (cc (- amount (first-denomination kind-of-coins)) kind-of-coins name))))
+
+  (def result (cc amount denom "test"))
+  (dot/write-graph graph "exercise_1_14.gv")
+  result)
+
+(print (count-change2 11 1))
+
+# Using the graph we can find the following relation for (cc n 1)
+# f(0,1) = 1
+# f(1,1) = 3 = f(0,1) + 2
+# f(2,1) = 5 = f(1,1) + 2
+# f(3,1) = 7 = f(2,1) + 2
+# f(n,1) = f(n-1, 1) + 2
+# f(n,1) = 2*n + 1
+
+# graphing (cc n 2)
+# we can see that every iteration of (cc n 2) splits into two trees:
+# (cc n 1) and (cc (n-5) 2).
+# Starting with f(1,2) = f(-4, 2) + f(1, 1).
+# we know that f(1,1) = 3 steps, f(-4, 2) = 1 step, and accounting
+# for the original f(1,2), we get a total of 5 steps.
+# f(6, 2) = f(1,2) + f(6,1) = 5 + 6 + 1 = 11
+# in general we can say:
+# f(n, 2) = f(n, 1) + f(n-x_2, 2) + 1
+# given x_2 = 5
+# example f(1, 2) = f(1, 1) + f(-4, 2) + 1 = 3 + 1 + 1 = 5
+# example f(6, 2) = f(6, 1) + f(1, 2) + 1 = 13 + 5 + 1 = 19
+# the closed form of this is:
+# f(n, 2) = ceil(n/5) + sum_i=0^floor(n/5){ f(n-x_2*i, 1) } + (ceil(n/5) - floor(n/5))
+# f(6, 2) = 2 + [f(6, 1) + f(1, 1)] + (2-1) = 2 + 13 + 3 + 1 = 19
+# f(11, 2) = 3 + [f(11, 1) + f(6, 1) + f(1,1)] + (3-2) = 3 + 23 + 13 + 3 + 1 = 43
+# f(n, 2) = 2*ceil(n/5) - floor(n/5) + (ceil(n/5) * (f(n, 1) + f(n%x_2, 1)) / 2)
+# f(n, 2) = ceil(n/x_2) * [(f(n,1) + f(n%x_2, 1)) / 2 + 2] - floor(n%x_2)
+# f(n, 2) = ceil(n/x_2) * [(2*n + 1 + 2*(n%x_2) + 1) / 2 + 2] - floor(n%x_2)
+# f(n, 2) = ceil(n/x_2) * [n + (n<F2><F3><F4>-x_2*floor(n/x_2)) + 3] - floor(n/x_2)
+# f(11, 2) = 2*3 - 2 + 3 * (23 + 3) / 2 = 6 - 2 + 39 = 43
+# f(11, 2) = 3 * [(23 + 3) / 2 + 2] - 2 = 3 * 45 - 2 = 43
+# f(11, 2) = 3 * [11 + 1 + 3] - 2 = 3 * [15] - 2 = 43
+
+# graphing (cc n 3)
+# we can see that every iteration of (cc n 3) splits into two trees:
+# (cc n 2) + (cc (n-x_3),3)
+# an recursive form of this definition would be:
+# f(n, 3) = f(n, 2) + f(n-x_2, 3) + 1
+# test:
+# f(1, 3) = f(1, 2) + f(-9, 3) + 1 = 5 + 1 + 1 = 7
+# f(11, 3) = f(11, 2) + f(1, 3) + 1 = 43 + 1 + 7 = 51
+