From 95e2fe57f6e4639f6ae9f1fef368829d5090dbf6 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Ale=C5=A1=20Smodi=C5=A1?= <aless@guru.si>
Date: Tue, 18 Aug 2015 16:06:19 +0200
Subject: Exported all problems from the SQLite database into the new directory
 structure.

---
 prolog/problems/old_exams/pascal_3/en.py | 26 ++++++++++++++++++++++++++
 1 file changed, 26 insertions(+)
 create mode 100644 prolog/problems/old_exams/pascal_3/en.py

(limited to 'prolog/problems/old_exams/pascal_3/en.py')

diff --git a/prolog/problems/old_exams/pascal_3/en.py b/prolog/problems/old_exams/pascal_3/en.py
new file mode 100644
index 0000000..31a0d47
--- /dev/null
+++ b/prolog/problems/old_exams/pascal_3/en.py
@@ -0,0 +1,26 @@
+id = 179
+name = 'pascal/3'
+slug = 'pascal's triangle'
+
+description = '''\
+<p>The first five rows of the Pascal's triangle look like this:</p>
+<pre>
+        1
+       1 1
+      1 2 1
+     1 3 3 1
+    1 4 6 4 1
+</pre>
+<p>
+Each row begins and ends with 1. Every other element can be obtained as a sum of the two values above it. Write the predicate <code>pascal(I,J,N)</code> that returns the <code>J</code>-th value in the <code>I</code>-th column of the Pascal's triangle. Your solution should return exactly one answer for any input (the <code>I</code> and <code>J</code> arguments start counting with 0; you can assume that 0 ≤ <code>J</code> ≤ <code>I</code>).
+
+<pre>
+  ?- pascal(0, 0, N).
+    N = 1.
+  ?- pascal(2, 1, N).
+    N = 2.
+  ?- pascal(4, 3, N).
+    N = 4.
+</pre>'''
+
+hint = {}
-- 
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