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Shaun Setlock
2020-08-02 21:06:44 -04:00
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problems/011_problem/011_problem.py Normal file
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# # Problem 11:
#
# [Euler Project #11](https://projecteuler.net/problem=11)
#
#
# > In the 20×20 grid below, four numbers along a diagonal line have been marked in red.
# >
# > 08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08 <br>
# > 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00 <br>
# > 81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65 <br>
# > 52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91 <br>
# > 22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80 <br>
# > 24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50 <br>
# > 32 98 81 28 64 23 67 10 **26** 38 40 67 59 54 70 66 18 38 64 70 <br>
# > 67 26 20 68 02 62 12 20 95 **63** 94 39 63 08 40 91 66 49 94 21 <br>
# > 24 55 58 05 66 73 99 26 97 17 **78** 78 96 83 14 88 34 89 63 72 <br>
# > 21 36 23 09 75 00 76 44 20 45 35 **14** 00 61 33 97 34 31 33 95 <br>
# > 78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92 <br>
# > 16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57 <br>
# > 86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58 <br>
# > 19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40 <br>
# > 04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66 <br>
# > 88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69 <br>
# > 04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36 <br>
# > 20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16 <br>
# > 20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54 <br>
# > 01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48 <br>
# >
# > The product of these numbers is 26 × 63 × 78 × 14 = 1788696.
# >
# > What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20×20 grid?
#
#
# ---
import os
import pprint
import time # Typically imported for sleep function, to slow down execution in terminal.
import typing
import decorators # Typically imported to compute execution duration of functions.
import math
import numpy
def check_horizontal(table,solution):
'''
Args:
table: list of lists representing the input table
solution: a dictionary storing the product, and three tuples of cell coordinates
'''
# Let's scan the horizontals first.
# Acceptable cells would have a root position (on the LHS) which may reside anywhere in the table,
# except for the last three columns.
# create a list of tuples to store cell coordinates while generating the products
cells=[(0,0),(0,0),(0,0),(0,0)]
# loop over row dimension
for row in range(len(table)):
# loop over column dimension
for column in range(len(table[row])-4):
# calculate product
product = 1
for i in range(4):
product *= table[row][column+i]
cells[i]=(row,column+i)
#print("product=",product)
if product>solution['product']:
solution['product']=product
solution['cella']=cells[0]
solution['cellb']=cells[1]
solution['cellc']=cells[2]
solution['celld']=cells[3]
return solution
def check_vertical(table,solution):
'''
Args:
table: list of lists representing the input table
solution: a dictionary storing the product, and three tuples of cell coordinates
'''
# Let's scan the verticals next.
# Acceptable cells would have a root position (at the top of column) which may reside anywhere in the table,
# except for the last three rows.
# create a list of tuples to store cell coordinates while generating the products
cells=[(0,0),(0,0),(0,0),(0,0)]
# loop over row dimension
count = 0
for row in range(len(table)-4):
# loop over column dimension
for column in range(len(table[row])):
# convert data type from string to int
product = 1
for i in range(4):
product *= table[row+i][column]
cells[i]=(row,column+i)
#print("product=",product)
count+=1
#print("count=",count)
if product>solution['product']:
solution['product']=product
solution['cella']=cells[0]
solution['cellb']=cells[1]
solution['cellc']=cells[2]
solution['celld']=cells[3]
return solution
def check_upwards_diagonals(table,solution):
'''
Args:
table: list of lists representing the input table
solution: a dictionary storing the product, and three tuples of cell coordinates
'''
# Let's scan the upward (left to right) diagonals next.
# Acceptable cells would have a root position (at the bottom-left of the diagonal) which may reside anywhere in the table,
# except for the first three rows, and the last three columns.
# create a list of tuples to store cell coordinates while generating the products
cells=[(0,0),(0,0),(0,0),(0,0)]
# loop over row dimension
count = 0
for row in range(len(table)-3):
# loop over column dimension
#print("row=",row)
for column in range(len(table[row])-3):
# convert data type from string to int
product = 1
for i in range(4):
product *= table[row+3-i][column+i]
cells[i]=(row+3-i,column+i)
#print("cell=[",row+3-i,"][",column+i,"]")
#print("product=",product)
count += 1
if product>solution['product']:
solution['product']=product
solution['cella']=cells[0]
solution['cellb']=cells[1]
solution['cellc']=cells[2]
solution['celld']=cells[3]
return solution
def check_downwards_diagonals(table,solution):
'''
Args:
table: list of lists representing the input table
solution: a dictionary storing the product, and three tuples of cell coordinates
'''
# Let's scan the downward (left to right) diagonals next.
# Acceptable cells would have a root position (at the top-left of the diagonal) which may reside anywhere in the table,
# except for the last three rows, and the last three columns.
# create a list of tuples to store cell coordinates while generating the products
cells=[(0,0),(0,0),(0,0),(0,0)]
# loop over row dimension
count = 0
for row in range(len(table)-3):
# loop over column dimension
#print("row=",row)
for column in range(len(table[row])-3):
# convert data type from string to int
product = 1
for i in range(4):
product *= table[row+i][column+i]
cells[i]=(row+i,column+i)
#print("cell=[",row+i,"][",column+i,"]")
#print("product=",product)
count += 1
#print("count=",count)
if product>solution['product']:
solution['product']=product
solution['cella']=cells[0]
solution['cellb']=cells[1]
solution['cellc']=cells[2]
solution['celld']=cells[3]
return solution
table = [ ['08', '02', '22', '97', '38', '15', '00', '40', '00', '75', '04', '05', '07', '78', '52', '12', '50', '77', '91', '08'],
['49', '49', '99', '40', '17', '81', '18', '57', '60', '87', '17', '40', '98', '43', '69', '48', '04', '56', '62', '00'],
['81', '49', '31', '73', '55', '79', '14', '29', '93', '71', '40', '67', '53', '88', '30', '03', '49', '13', '36', '65'],
['52', '70', '95', '23', '04', '60', '11', '42', '69', '24', '68', '56', '01', '32', '56', '71', '37', '02', '36', '91'],
['22', '31', '16', '71', '51', '67', '63', '89', '41', '92', '36', '54', '22', '40', '40', '28', '66', '33', '13', '80'],
['24', '47', '32', '60', '99', '03', '45', '02', '44', '75', '33', '53', '78', '36', '84', '20', '35', '17', '12', '50'],
['32', '98', '81', '28', '64', '23', '67', '10', '26', '38', '40', '67', '59', '54', '70', '66', '18', '38', '64', '70'],
['67', '26', '20', '68', '02', '62', '12', '20', '95', '63', '94', '39', '63', '08', '40', '91', '66', '49', '94', '21'],
['24', '55', '58', '05', '66', '73', '99', '26', '97', '17', '78', '78', '96', '83', '14', '88', '34', '89', '63', '72'],
['21', '36', '23', '09', '75', '00', '76', '44', '20', '45', '35', '14', '00', '61', '33', '97', '34', '31', '33', '95'],
['78', '17', '53', '28', '22', '75', '31', '67', '15', '94', '03', '80', '04', '62', '16', '14', '09', '53', '56', '92'],
['16', '39', '05', '42', '96', '35', '31', '47', '55', '58', '88', '24', '00', '17', '54', '24', '36', '29', '85', '57'],
['86', '56', '00', '48', '35', '71', '89', '07', '05', '44', '44', '37', '44', '60', '21', '58', '51', '54', '17', '58'],
['19', '80', '81', '68', '05', '94', '47', '69', '28', '73', '92', '13', '86', '52', '17', '77', '04', '89', '55', '40'],
['04', '52', '08', '83', '97', '35', '99', '16', '07', '97', '57', '32', '16', '26', '26', '79', '33', '27', '98', '66'],
['88', '36', '68', '87', '57', '62', '20', '72', '03', '46', '33', '67', '46', '55', '12', '32', '63', '93', '53', '69'],
['04', '42', '16', '73', '38', '25', '39', '11', '24', '94', '72', '18', '08', '46', '29', '32', '40', '62', '76', '36'],
['20', '69', '36', '41', '72', '30', '23', '88', '34', '62', '99', '69', '82', '67', '59', '85', '74', '04', '36', '16'],
['20', '73', '35', '29', '78', '31', '90', '01', '74', '31', '49', '71', '48', '86', '81', '16', '23', '57', '05', '54'],
['01', '70', '54', '71', '83', '51', '54', '69', '16', '92', '33', '48', '61', '43', '52', '01', '89', '19', '67', '48']
]
# Firstly, this data input creates a list-of-lists which have cells of the ```string``` data type.
# loop over row dimension
for row in range(len(table)):
# loop over column dimension
for column in range(len(table[row])):
# convert data type from string to int
table[row][column] = int(table[row][column])
# create a dictionary that stores information about the solution
solution = {
'product':1,
'cella':(0,0),
'cellb':(0,0),
'cellc':(0,0),
'celld':(0,0)
}
# We know brute force will work...
# Can we can scan through this list-of-lists, following the accepted pattern, and store the largest product?
solution = check_horizontal(table,solution)
print(solution)
solution = check_vertical(table,solution)
print(solution)
solution = check_upwards_diagonals(table,solution)
print(solution)
solution = check_downwards_diagonals(table,solution)
print('Final Answer!')
print(solution)