Day 10 of 2020 solved in Python

2020/Day 10/Solution.py

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+input_file = "input.txt"
+
+adapters = []
+
+with open(input_file) as adapter_ratings:
+    line = adapter_ratings.readline()
+
+    while line and line != "\n":
+        adapters.append(int(line))
+        line = adapter_ratings.readline()
+
+adapters.sort()
+computer_adapter = max(adapters) + 3
+
+# List of couples separated by three, which must be present to have a working chain
+fixed_points = []
+
+one_diffs = 0
+three_diffs = 1  # As the computer adapter is always the highest one we got plus three
+
+# The charging outlet is 0, compute the difference now
+if adapters[0] == 1:
+    one_diffs += 1
+elif adapters[0] == 3:
+    fixed_points.append((adapters[0], 0))
+    three_diffs += 1
+
+for index in range(len(adapters)-1):
+    current_difference = adapters[index+1] - adapters[index]
+    if current_difference == 1:
+        one_diffs += 1
+    elif current_difference == 3:
+        if (adapters[index], index) not in fixed_points:
+            fixed_points.append((adapters[index], index))
+        fixed_points.append((adapters[index+1], index+1))
+        three_diffs += 1
+
+print(f"The product of one and three differences is : {one_diffs}*{three_diffs} = {one_diffs*three_diffs}")
+
+# The last adapter is always needed as it is the only link possible to the computer adapter
+if (adapters[-1], len(adapters)-1) not in fixed_points:
+    fixed_points.append((adapters[-1], len(adapters)-1))
+
+# Compute the distance separating each fixed point. If they are not successive, it means that there are possible
+# permutations between those two fixed points. Store the count of numbers we can choose from
+
+separation_distance = []
+for index in range(len(fixed_points) - 1):
+    if fixed_points[index + 1][1] - fixed_points[index][1] > 1:
+        separation_distance.append(fixed_points[index + 1][1] - fixed_points[index][1] - 1)
+
+# Distance between 0 and the first fixed point
+separation_distance.insert(0, fixed_points[0][1])
+
+total_combinations = 1
+
+# This would probably not work in a general case other than this puzzle, as I only take into account small possible
+# separations, which is all I have got.
+for separation in separation_distance:
+    # If we have three numbers, it means that we have at least a separation of 4 jolts between the two fixed points,
+    # which is not possible to cross if we do not choose any number between those two. So, do not count this choice.
+    if separation == 3:
+        # Number of subsets of a set is 2^n where n is the number of elements in the set.
+        total_combinations *= 2**3 - 1
+    else:
+        total_combinations *= 2**separation
+
+print(f"The number of combinations is : {total_combinations}")

2020/Day 10/input.txt

@@ -0,0 +1,110 @@
+160
+34
+123
+159
+148
+93
+165
+56
+179
+103
+171
+44
+110
+170
+147
+98
+25
+37
+137
+71
+5
+6
+121
+28
+19
+134
+18
+7
+66
+90
+88
+181
+89
+41
+156
+46
+8
+61
+124
+9
+161
+72
+13
+172
+111
+59
+105
+51
+109
+27
+152
+117
+52
+68
+95
+164
+116
+75
+78
+180
+81
+47
+104
+12
+133
+175
+16
+149
+135
+99
+112
+38
+67
+53
+153
+2
+136
+113
+17
+145
+106
+31
+45
+169
+146
+168
+26
+36
+118
+62
+65
+142
+130
+1
+140
+84
+94
+141
+122
+22
+48
+102
+60
+178
+127
+73
+74
+87
+182
+35