From be8b528b5cc878f7576a5b7d98bb457e9099662c Mon Sep 17 00:00:00 2001
From: Martin Henz <henz@comp.nus.edu.sg>
Date: Thu, 11 Jul 2024 02:08:46 +0800
Subject: [PATCH] fixes #986 (#1044)

---
 xml/chapter4/section1/subsection6.xml | 39 +++++++++++++++++++++++++++
 1 file changed, 39 insertions(+)

diff --git a/xml/chapter4/section1/subsection6.xml b/xml/chapter4/section1/subsection6.xml
index 69e7c334a..ab112127d 100644
--- a/xml/chapter4/section1/subsection6.xml
+++ b/xml/chapter4/section1/subsection6.xml
@@ -387,6 +387,45 @@ function f(x) {
 	    </SNIPPET>
 	  </LI>
 	</OL>
+	<SOLUTION>
+	  Part (a)
+	  <SNIPPET>
+	    <JAVASCRIPT>
+// solution provided by GitHub user LucasGdosR
+
+// The Fibonacci function receives n as an argument
+// It applies the fib function recursively, passing n as an argument,
+// as well as the initial arguments (k = 1, fib1 = 1, fib2 = 1)
+(n => (fib => fib(fib, n, 2, 1, 1))
+      // The fib function is then defined as ft,
+      // with parameters n, k, fib1, and fib2
+      // Establish the base cases: n === 1 or n === 2
+      ((ft, n, k, fib1, fib2) => n === 1
+                  ? 1
+                  : n === 2
+                  ? 1
+                  :
+                  // Iterate until k equals n. Notice k starts at 2, and gets incremented every iteration
+                  k === n
+                  // When k reaches n, return the accumulated fib2
+                  ? fib2
+                  // Otherwise, accumulate the sum as the new fib2
+                  : ft(ft, n, k + 1, fib2, fib1 + fib2)));
+	    </JAVASCRIPT>
+	  </SNIPPET>
+	  Part (b)
+	  <SNIPPET>
+	    <JAVASCRIPT>
+// solution provided by GitHub user LucasGdosR
+
+function f(x) {
+    return ((is_even, is_odd) => is_even(is_even, is_odd, x))
+           ((is_ev, is_od, n) => n === 0 ? true : is_od(is_ev, is_od, n - 1),
+           (is_ev, is_od, n) => n === 0 ? false : is_ev(is_ev, is_od, n - 1));
+}
+	    </JAVASCRIPT>
+	  </SNIPPET>
+	</SOLUTION>
       </EXERCISE>
     </JAVASCRIPT>
   </SPLIT>