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authorThomas Voss <thomasvoss@live.com> 2021-10-29 23:02:39 +0200
committerThomas Voss <thomasvoss@live.com> 2021-10-29 23:02:39 +0200
commite7c9108b95e39d7ea5a29ae06d619c4727f11027 (patch)
tree237261eef3afd0720be77dbcbb9599fa66a24b67 /2015/11
Initial commit
Diffstat (limited to '2015/11')
-rw-r--r--2015/11/input1
-rw-r--r--2015/11/puzzle-17
-rw-r--r--2015/11/puzzle-29
3 files changed, 17 insertions, 0 deletions
diff --git a/2015/11/input b/2015/11/input
new file mode 100644
index 0000000..a45c477
--- /dev/null
+++ b/2015/11/input
@@ -0,0 +1 @@
+hepxcrrq
diff --git a/2015/11/puzzle-1 b/2015/11/puzzle-1
new file mode 100644
index 0000000..b0e5756
--- /dev/null
+++ b/2015/11/puzzle-1
@@ -0,0 +1,7 @@
+The starting string is "hepxcrrq"
+
+The string must contain two sets of non-overlapping duplicates, this is best done as having the last
+5 characters be "xxyzz". The first 3 characters aren't illegal, so we don't need to change them.
+This gives an end result of "hepxxyzz".
+
+QED. (I hope)
diff --git a/2015/11/puzzle-2 b/2015/11/puzzle-2
new file mode 100644
index 0000000..444ce35
--- /dev/null
+++ b/2015/11/puzzle-2
@@ -0,0 +1,9 @@
+This continues from the last puzzle, so our starting string is "hepxxyzz"
+
+Following the theme of the previous puzzle, we want to create an "xxyzz" pattern in the lowest
+amount of increments possible. Since the string currently ends in "xxyzz", we know that the next
+possible value we could have that satisfies this property is "aabcc". In order to go from "xxyzz" to
+"aabcc" however, you need to overflow which would increment the "p" to a "q". This gives an end
+result of "heqaabcc".
+
+QED.