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authorThomas Voss <mail@thomasvoss.com> 2022-11-30 12:23:17 +0100
committerThomas Voss <mail@thomasvoss.com> 2022-11-30 12:23:17 +0100
commitf3e3e3e354b87f4ce27bd74bf14cec05d06974b6 (patch)
tree112352e9c3ba5d5d924036def2b5f853d3f8dc28 /2015/11
parent4742dd73046c53cff7fe54deee58358c3b193206 (diff)
Lots of cleanup and stuff
Diffstat (limited to '2015/11')
-rw-r--r--2015/11/puzzle-111
-rw-r--r--2015/11/puzzle-213
2 files changed, 13 insertions, 11 deletions
diff --git a/2015/11/puzzle-1 b/2015/11/puzzle-1
index b0e5756..913a194 100644
--- a/2015/11/puzzle-1
+++ b/2015/11/puzzle-1
@@ -1,7 +1,8 @@
-The starting string is "hepxcrrq"
+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".
+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)
+QED. (I hope)
diff --git a/2015/11/puzzle-2 b/2015/11/puzzle-2
index 444ce35..089865d 100644
--- a/2015/11/puzzle-2
+++ b/2015/11/puzzle-2
@@ -1,9 +1,10 @@
-This continues from the last puzzle, so our starting string is "hepxxyzz"
+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".
+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.