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""" |
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Project Euler Problem 28: Number Spiral Diagonals |
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================================================= |
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.. module:: solutions.problem28 |
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:synopsis: My solution to problem #28. |
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The source code for this problem can be |
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`found here <https://bitbucket.org/nekedome/project-euler/src/master/solutions/problem28.py>`_. |
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Problem Statement |
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################# |
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Starting with the number :math:`1` and moving to the right in a clockwise direction a :math:`5` by :math:`5` spiral is |
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formed as follows: |
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.. math:: |
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&\\color{red}{21}, 2&2, 2&3, 2&4, &\\color{red}{25} \\\\ |
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&20, &\\color{red}{7}, &8, &\\color{red}{9}, &10 \\\\ |
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&19, &6, &\\color{red}{1}, &2, &11 \\\\ |
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&18, &\\color{red}{5}, &4, &\\color{red}{3}, &12 \\\\ |
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&\\color{red}{17}, 1&6, 1&5, 1&4, &\\color{red}{13} |
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It can be verified that the sum of the numbers on the diagonals is :math:`101`. |
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What is the sum of the numbers on the diagonals in a :math:`1001` by :math:`1001` spiral formed in the same way? |
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Solution Discussion |
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################### |
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The example :math:`5 \\times 5` matrix reveals a general algorithm to complete an :math:`n \\times n` square matrix with |
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the clockwise spiral pattern where :math:`n` is an odd integer. |
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This solution creates the matrix by storing the value of an incrementing counter in the clockwise spiral pattern. The |
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two diagonals are then summed to produce the final answer. |
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1. Mark the centre with :math:`1` |
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2. Set a `counter` to :math:`1` and `distance` to :math:`1` |
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3. Repeat the following :math:`^n / _2` times: |
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a. Mark the cells using the `counter`, moving right `distance` cells |
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b. Mark the cells using the `counter`, moving down `distance` cells |
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c. Increment `distance` |
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d. Mark the cells using the `counter`, moving left `distance` cells |
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e. Mark the cells using the `counter`, moving up `distance` cells |
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f. Increment `distance` |
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4. Mark the cells using the `counter`, moving right `distance` cells |
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Solution Implementation |
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####################### |
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.. literalinclude:: ../../solutions/problem28.py |
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:language: python |
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:lines: 57- |
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""" |
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from typing import List, Optional, Tuple |
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from lib.numbertheory import is_odd |
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def complete_line_segment(matrix: List[List[Optional[int]]], x: int, y: int, counter: int, direction: str, |
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distance: int) -> Tuple[int, int, int]: |
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""" Fill in adjacent cells in `matrix`, starting at position :math:`(x,y)`, moving in the specified `direction` for |
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the specified `distance` |
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.. note:: the initial value of `counter` is used to start, it is incremented at each step. |
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:param matrix: the two-dimensional grid to be populated with `counter` in a spiral pattern |
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:param x: the starting :math:`x` co-ordinate |
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:param y: the staring :math:`y` co-ordinate |
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:param counter: the current value of the counter to use when setting cell values |
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:param direction: the direction to move in ("R", "D", "L" or "U") |
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:param distance: the distance to move in the given direction |
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:return: the new :math:`(x,y)` position and the final value of `counter` |
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""" |
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step_coords = {"R": lambda _x, _y: (_x + 1, _y), "D": lambda _x, _y: (_x, _y + 1), |
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"L": lambda _x, _y: (_x - 1, _y), "U": lambda _x, _y: (_x, _y - 1)} |
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for i in range(distance): |
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matrix[y][x] = counter # mark the (x,y) cell in matrix with counter |
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x, y = step_coords[direction](x, y) # advance (x,y) co-ordinates in the specified direction |
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counter += 1 # get next counter value |
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return x, y, counter |
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def solve(): |
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""" Compute the answer to Project Euler's problem #28 """ |
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dimension = 1001 |
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assert is_odd(dimension), "this algorithm is only correct for an odd-valued dimension" |
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# Initialise the matrix with None values everywhere, start a position (x,y) at the centre |
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matrix = [[None for _ in range(dimension)] for _ in range(dimension)] |
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x, y = dimension // 2, dimension // 2 |
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# Iteratively fill in matrix values in a clockwise direction |
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counter = 1 |
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distance = 1 |
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for i in range(dimension // 2): |
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x, y, counter = complete_line_segment(matrix, x, y, counter, "R", distance) |
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x, y, counter = complete_line_segment(matrix, x, y, counter, "D", distance) |
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distance += 1 |
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x, y, counter = complete_line_segment(matrix, x, y, counter, "L", distance) |
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x, y, counter = complete_line_segment(matrix, x, y, counter, "U", distance) |
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distance += 1 |
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# Complete the top row of the matrix |
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complete_line_segment(matrix, x, y, counter, "R", distance) |
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# Sum the values along both diagonals |
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answer = 0 |
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for z in range(dimension): |
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answer += matrix[z][z] # top left to bottom right |
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answer += matrix[dimension - 1 - z][z] # bottom left to top right |
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answer -= 1 # we double-counted the centre cell which has a value of 1, remove one copy |
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return answer |
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expected_answer = 669171001 |
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