From 85fff24cc18f044c98a4e6fe9c55f5ac3864a8a7 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Mika=C3=ABl=20Capelle?= <capelle.mikael@gmail.com>
Date: Thu, 21 Dec 2023 21:56:38 +0100
Subject: [PATCH] 2023 day 21, version 2.

---
 src/holt59/aoc/2023/day21.py | 67 +++++++++++++++++++++++++++++++++---
 1 file changed, 62 insertions(+), 5 deletions(-)

diff --git a/src/holt59/aoc/2023/day21.py b/src/holt59/aoc/2023/day21.py
index 9618081..c44a4c1 100644
--- a/src/holt59/aoc/2023/day21.py
+++ b/src/holt59/aoc/2023/day21.py
@@ -39,8 +39,7 @@ print(f"answer 1 is {answer_1}")
 # rhombus every n steps
 #
 # we are going to find the number of cells reached for the initial rhombus, n steps
-# after and n * 2 steps after, and then interpolate the value to get a 2nd order
-# polynomial
+# after and n * 2 steps after
 #
 cycle = len(map)
 rhombus = (len(map) - 3) // 2 + 1
@@ -56,9 +55,9 @@ if logging.root.getEffectiveLevel() == logging.INFO:
     rows = [
         [
             map[i % n_rows][j % n_cols] if (i, j) not in tiles else "O"
-            for j in range(-2 * cycle, 3 * cycle + 1)
+            for j in range(-2 * cycle, 3 * cycle)
         ]
-        for i in range(-2 * cycle, 3 * cycle + 1)
+        for i in range(-2 * cycle, 3 * cycle)
     ]
 
     for i in range(len(rows)):
@@ -71,6 +70,64 @@ if logging.root.getEffectiveLevel() == logging.INFO:
 
 logging.info(f"values to fit: {values}")
 
+# version 1:
+#
+# after 3 cycles, the figure looks like the following:
+#
+#   I M D
+# I J A K D
+# H A F A L
+# C E A K B
+#   C G B
+#
+# after 4 cycles, the figure looks like the following:
+#
+#     I M D
+#   I J A K D
+# I J A B A K D
+# H A B A B A L
+# C E A B A N F
+#   C E A N F
+#     C G F
+#
+# the 'radius' of the rhombus is the number of cycles minus 1
+#
+# the 4 'corner' (M, H, L, G) are counted once, the blocks with a corner triangle (D, I,
+# C, B) are each counted radius times, the blocks with everything but one corner (J, K,
+# E, N) are each counted radius - 1 times
+#
+# there are two versions of the whole block, A and B in the above (or odd and even),
+# depending on the number of cycles, either A or B will be in the center
+#
+
+counts = [
+    [
+        sum(
+            (i, j) in tiles
+            for i in range(ci * cycle, (ci + 1) * cycle)
+            for j in range(cj * cycle, (cj + 1) * cycle)
+        )
+        for cj in range(-2, 3)
+    ]
+    for ci in range(-2, 3)
+]
+
+radius = (26501365 - rhombus) // cycle - 1
+A = counts[2][2] if radius % 2 == 0 else counts[2][1]
+B = counts[2][2] if radius % 2 == 1 else counts[2][1]
+answer_2 = (
+    (radius + 1) * A
+    + radius * B
+    + 2 * radius * (radius + 1) // 2 * A
+    + 2 * radius * (radius - 1) // 2 * B
+    + sum(counts[i][j] for i, j in ((0, 2), (-1, 2), (2, 0), (2, -1)))
+    + sum(counts[i][j] for i, j in ((0, 1), (0, 3), (-1, 1), (-1, 3))) * (radius + 1)
+    + sum(counts[i][j] for i, j in ((1, 1), (1, 3), (-2, 1), (-2, 3))) * radius
+)
+print(f"answer 2 (v1) is {answer_2}")
+
+# version 2: fitting a polynomial
+#
 # the value we are interested in (26501365) can be written as R + K * C where R is the
 # step at which we find the first rhombus, and K the repeat step, so instead of fitting
 # for X values (R, R + K, R + 2 K), we are going to fit for (0, 1, 2), giving us much
@@ -89,4 +146,4 @@ a, b, c = (y1 + y3) // 2 - y2, 2 * y2 - (3 * y1 + y3) // 2, y1
 
 n = (26501365 - rhombus) // cycle
 answer_2 = a * n * n + b * n + c
-print(f"answer 2 is {answer_2}")
+print(f"answer 2 (v2) is {answer_2}")