MAKE THE GIVEN CODE SHORTER WITH LESS LOOPS def cost_to_hike_naive(m: list[list[int]], start_point: tuple[int, int], end_point: tuple[int, int]) -> int: """ Given an elevation map and a start and end point, calculate the cost it would take to hike from to . If the start and end points are the same, then return 0. Some definitions and rules: 1. You can only hike to either a vertically or horizontally adjacent location (you cannot travel diagonally). 2. You must only travel in the direction of the . More explicitly, this means that any move you make MUST take you closer to the end point, so you cannot travel in the other direction. 3. We define the cost to travel between two adjacent blocks as the absolute difference in elevation between those blocks. 4. You will calculate the naive route here, so at every position, you will have two choices, to either move closer to the end point along a row, or along a column, and you will chose the position that costs the least at that point (you do not need to look ahead to see if the costlier route now will result in an overall cheaper route). 5. If two choices along the route have the exact same cost, you will choose the direction that comes first clockwise. For example, you will choose to move horizontally right before vertically down, vertically down before horizontally left, horizontally left before vertically up, and vertically up before horizontally right. Preconditions: the start and end points will be a valid position on the map. Tips to get started: since there is no guarantee where the start and end points will be in relation to each other, it may be easier to use your rotate function from last week to rotate the map such that the end point is always to the bottom right of the start point. That is, rotate the map such that start_point[0] <= end_point[0] and start_point[1] <= end_point[1]. Keep in mind that if you choose to rotate the map, the start and end point coordinates will change as well. If you choose to do it this way, then you can code your function to only consider moving horizontally right, or vertically down. Not rotating is fine as well, but you would have to consider moves in all 4 directions, depending on the start and end coordinates. If you choose to rotate the map, you may also find the crop function you implmented last week useful, as that will guarantee that the start and end points will be at corners of the map. >>> sample_map = [[1, 3], ... [0, 4]] >>> cost_to_hike_naive(sample_map,(0, 0),(1, 1)) 5 >>> cost_to_hike_naive(sample_map,(1, 1),(0, 0)) 3
MAKE THE GIVEN CODE SHORTER WITH LESS LOOPS def cost_to_hike_naive(m: list[list[int]], start_point: tuple[int, int], end_point: tuple[int, int]) -> int: """ Given an elevation map and a start and end point, calculate the cost it would take to hike from to . If the start and end points are the same, then return 0. Some definitions and rules: 1. You can only hike to either a vertically or horizontally adjacent location (you cannot travel diagonally). 2. You must only travel in the direction of the . More explicitly, this means that any move you make MUST take you closer to the end point, so you cannot travel in the other direction. 3. We define the cost to travel between two adjacent blocks as the absolute difference in elevation between those blocks. 4. You will calculate the naive route here, so at every position, you will have two choices, to either move closer to the end point along a row, or along a column, and you will chose the position that costs the least at that point (you do not need to look ahead to see if the costlier route now will result in an overall cheaper route). 5. If two choices along the route have the exact same cost, you will choose the direction that comes first clockwise. For example, you will choose to move horizontally right before vertically down, vertically down before horizontally left, horizontally left before vertically up, and vertically up before horizontally right. Preconditions: the start and end points will be a valid position on the map. Tips to get started: since there is no guarantee where the start and end points will be in relation to each other, it may be easier to use your rotate function from last week to rotate the map such that the end point is always to the bottom right of the start point. That is, rotate the map such that start_point[0] <= end_point[0] and start_point[1] <= end_point[1]. Keep in mind that if you choose to rotate the map, the start and end point coordinates will change as well. If you choose to do it this way, then you can code your function to only consider moving horizontally right, or vertically down. Not rotating is fine as well, but you would have to consider moves in all 4 directions, depending on the start and end coordinates. If you choose to rotate the map, you may also find the crop function you implmented last week useful, as that will guarantee that the start and end points will be at corners of the map. >>> sample_map = [[1, 3], ... [0, 4]] >>> cost_to_hike_naive(sample_map,(0, 0),(1, 1)) 5 >>> cost_to_hike_naive(sample_map,(1, 1),(0, 0)) 3
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
Related questions
Question
MAKE THE GIVEN CODE SHORTER WITH LESS LOOPS
def cost_to_hike_naive(m: list[list[int]], start_point: tuple[int, int],
end_point: tuple[int, int]) -> int:
"""
Given an elevation map <m> and a start and end point, calculate the cost it
would take to hike from <start_point> to <end_point>. If the start and end
points are the same, then return 0.
Some definitions and rules:
1. You can only hike to either a vertically or horizontally adjacent
location (you cannot travel diagonally).
2. You must only travel in the direction of the <end_point>. More
explicitly, this means that any move you make MUST take you closer
to the end point, so you cannot travel in the other direction.
3. We define the cost to travel between two adjacent blocks as the
absolute difference in elevation between those blocks.
4. You will calculate the naive route here, so at every position, you
will have two choices, to either move closer to the end point along
a row, or along a column, and you will chose the position that costs
the least at that point (you do not need to look ahead to see if the
costlier route now will result in an overall cheaper route).
5. If two choices along the route have the exact same cost, you will
choose the direction that comes first clockwise. For example, you
will choose to move horizontally right before vertically down,
vertically down before horizontally left, horizontally left before
vertically up, and vertically up before horizontally right.
Preconditions: the start and end points will be a valid position on the map.
Tips to get started:
since there is no guarantee where the start and end points will be in
relation to each other, it may be easier to use your rotate function from
last week to rotate the map such that the end point is always to the
bottom right of the start point. That is, rotate the map such that
start_point[0] <= end_point[0] and start_point[1] <= end_point[1].
Keep in mind that if you choose to rotate the map, the start and end
point coordinates will change as well. If you choose to do it this way,
then you can code your function to only consider moving horizontally
right, or vertically down. Not rotating is fine as well, but you would
have to consider moves in all 4 directions, depending on the start and
end coordinates. If you choose to rotate the map, you may also find the
crop function you implmented last week useful, as that will guarantee
that the start and end points will be at corners of the map.
>>> sample_map = [[1, 3],
... [0, 4]]
>>> cost_to_hike_naive(sample_map,(0, 0),(1, 1))
5
>>> cost_to_hike_naive(sample_map,(1, 1),(0, 0))
3
"""
![File Edit View Navigate Code Refactor Run Iools VCS Window Help
lab9.py - lab9.py
O X
C Def Cost_to_hike
Def Cost_to_hike x
Def Cost to_hike X
Answered: def su X
C Def Cost_to_hike X
C Def Cost_to_hike
LAB RESTRICTIO
+
Downloads , lab9.py
Doctest cost_to_hike_naive v
A bartleby.com/questions-and-answers/def-substring-with-largest-sumstring-str-greater-str-wwwwin-given-a-string-return-the-substring-with/891a6ba1-1038-410... 2
O Project -
e lab6.py x
fe lab9.py
e pydev_execfile.py x
I Apps
E Reading list
> Downloads D:\Downloads
> uli External Libraries
100
col -= 1
A 3 X6 ^ v
2,559 /5,000 E
= bartl
Question Asking Tips
101
elif col < col_end:
E Scratches and Consoles
I
U x?
X2
- Ask one academic question at a time
102
c += abs(m[row][col + 1] - m[row][coi])
enu cooruimates. in you tnoose to TOlale uie Tmap, you may diso inu the
- Keep questions specific and include all
details
103
col += 1
Engineering
calculated by ...
crop function you implmented last week useful, as that will guarantee
that the start and end points will be at corners of the map.
104
else:
def sub
Contact Support for textbook, writing, or
account help
e sub...
105
if row < row_end:
>>> sample_map = [[1, 3],
.. [0, 4]]
>>> cost_to_hike_naive(sample_map,(0, 0).(1, 1))
106
a = abs(m[row + 1][col] - m[row][col])
107
elif row > row_end:
Like you, we deeply value honesty and
108
a = abs(m[row - 1][col] - m[row][col])
integrity. Learn about our honor code.
5
Question
*Response times may vary by subject and
109
elif row == row_end:
>>> cost_to_hike_naive(sample_map,(1, 1),(0, 0))
question complexity. Median response time is
110
a = 0
34 minutes for paid subscribers and may be
Do not add a
longer for promotional offers.
111
if col < col_end:
break or con
**Limited time offer.
112
b = abs(m[row][col + 1] - m[row][col])
A 0/2 Im
113
elif col > col_end:
114
b = abs(m[row][col - 1] - m[row][col])
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Ask an Expert
115
elif col == col_end:
116
b = 0
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if a <= b:
Get on-demand help from a live
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118
C += a
tutor, at no additional cost**
119
if row < row_end:
120
row += 1
121
elif row > row_end:
122
row -= 1
8:23 PM
PC
123
elif a > b:
2021-11-21
124
C += b
125
if col < col_end:
126
col += 1
127
elif col > col_end:
128
col -= 1
129
return c
130
cost_to_hike_naive()
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Debug
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a Event Log
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近](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F67b13aff-bfb5-4d63-8e56-0718b4a6f235%2Febce44fa-24f9-4364-a437-6891fa938521%2Fbuw057k_processed.png&w=3840&q=75)
Transcribed Image Text:File Edit View Navigate Code Refactor Run Iools VCS Window Help
lab9.py - lab9.py
O X
C Def Cost_to_hike
Def Cost_to_hike x
Def Cost to_hike X
Answered: def su X
C Def Cost_to_hike X
C Def Cost_to_hike
LAB RESTRICTIO
+
Downloads , lab9.py
Doctest cost_to_hike_naive v
A bartleby.com/questions-and-answers/def-substring-with-largest-sumstring-str-greater-str-wwwwin-given-a-string-return-the-substring-with/891a6ba1-1038-410... 2
O Project -
e lab6.py x
fe lab9.py
e pydev_execfile.py x
I Apps
E Reading list
> Downloads D:\Downloads
> uli External Libraries
100
col -= 1
A 3 X6 ^ v
2,559 /5,000 E
= bartl
Question Asking Tips
101
elif col < col_end:
E Scratches and Consoles
I
U x?
X2
- Ask one academic question at a time
102
c += abs(m[row][col + 1] - m[row][coi])
enu cooruimates. in you tnoose to TOlale uie Tmap, you may diso inu the
- Keep questions specific and include all
details
103
col += 1
Engineering
calculated by ...
crop function you implmented last week useful, as that will guarantee
that the start and end points will be at corners of the map.
104
else:
def sub
Contact Support for textbook, writing, or
account help
e sub...
105
if row < row_end:
>>> sample_map = [[1, 3],
.. [0, 4]]
>>> cost_to_hike_naive(sample_map,(0, 0).(1, 1))
106
a = abs(m[row + 1][col] - m[row][col])
107
elif row > row_end:
Like you, we deeply value honesty and
108
a = abs(m[row - 1][col] - m[row][col])
integrity. Learn about our honor code.
5
Question
*Response times may vary by subject and
109
elif row == row_end:
>>> cost_to_hike_naive(sample_map,(1, 1),(0, 0))
question complexity. Median response time is
110
a = 0
34 minutes for paid subscribers and may be
Do not add a
longer for promotional offers.
111
if col < col_end:
break or con
**Limited time offer.
112
b = abs(m[row][col + 1] - m[row][col])
A 0/2 Im
113
elif col > col_end:
114
b = abs(m[row][col - 1] - m[row][col])
Live Chat with an Expert
Ask an Expert
115
elif col == col_end:
116
b = 0
A Start chat now
Submit your question
117
if a <= b:
Get on-demand help from a live
Answers in as fast as 30 minutes*
118
C += a
tutor, at no additional cost**
119
if row < row_end:
120
row += 1
121
elif row > row_end:
122
row -= 1
8:23 PM
PC
123
elif a > b:
2021-11-21
124
C += b
125
if col < col_end:
126
col += 1
127
elif col > col_end:
128
col -= 1
129
return c
130
cost_to_hike_naive()
> Run
E TODO
O Problems
Debug
2 Terminal
S Python Packages
2 Python Console
a Event Log
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近
![File Edit View Navigate Code Refactor Run Iools VCS Window Help
lab9.py - lab9.py
O X
C Def Cost_to_hike
Def Cost_to_hike x
Def Cost to_hike X
Answered: def su X
C Def Cost_to_hike X
C Def Cost_to_hike
LAB RESTRICTIO
+
Downloads , lab9.py
Doctest cost_to_hike_naive v
A bartleby.com/questions-and-answers/def-substring-with-largest-sumstring-str-greater-str-wwwwin-given-a-string-return-the-substring-with/891a6ba1-1038-410... 2
O Project -
e lab6.py x
fe lab9.py
e pydev_execfile.py x
I Apps
E Reading list
> Downloads D:\Downloads
> uli External Libraries
79
A 3 X6 ^ v
2,559 /5,000 E
= bartl
Question Asking Tips
80
row = start_point[0]
E Scratches and Consoles
I
U x X2
- Ask one academic question at a time
81
col = start_point[1]
enu cooruimates. in you tnoose to TOlale uie Tmap, you may diso inu the
- Keep questions specific and include all
details
82
row_end = end_point[0]
Engineering
calculated by ...
crop function you implmented last week useful, as that will guarantee
that the start and end points will be at corners of the map.
83
col_end = end_point[1]
def sub
Contact Support for textbook, writing, or
account help
e sub...
84
C = 0
>>> sample_map = [[1, 3],
.. [0, 4]]
>>> cost_to_hike_naive(sample_map,(0, 0).(1, 1))
85
a = 0
86
b = 0
Like you, we deeply value honesty and
87
if start_point == end_point:
integrity. Learn about our honor code.
5
Question
*Response times may vary by subject and
88
return 0
>>> cost_to_hike_naive(sample_map,(1, 1),(0, 0))
question complexity. Median response time is
89
while row != row_end or col != col_end:
34 minutes for paid subscribers and may be
Do not add a
longer for promotional offers.
90
if row != row_end and col == col_end:
break or con
**Limited time offer.
91
if row > row_end:
A 0/2 Im
92
c += abs(m[row - 1][col] - m[row][coi])
93
row -= 1
Live Chat with an Expert
Ask an Expert
94
elif row < row_end:
95
c += abs(m[row + 1][col] - m[row][coi])
A Start chat now
Submit your question
96
row +=
Get on-demand help from a live
Answers in as fast as 30 minutes*
97
elif row == row_end and col != col_end:
tutor, at no additional cost**
98
if col > col_end:
99
c += abs(m[row][col - 1] - m[row][col]D
100
col -= 1
101
elif col < col_end:
8:22 PM
PC
102
c += abs(m[row][col + 1] - m[row][col]D
2021-11-21
103
col += 1
104
else:
105
if row < row_end:
106
a = abs(m[row + 1][col] - m[row][col])
107
elif row > row_end:
108
a = abs(m[row - 1][col] - m[row][col])
109
olif nOw -- nOw ond:
cost_to_hike_naive()
> Run
E TODO
O Problems
Debug
2 Terminal
Python Packages
2 Python Console
A Event Log
O Tests passed: 3 (3 minutes ago)
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8:22 PM
Address
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2021-11-21
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1. Structure](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F67b13aff-bfb5-4d63-8e56-0718b4a6f235%2Febce44fa-24f9-4364-a437-6891fa938521%2Fqzeqsaa_processed.png&w=3840&q=75)
Transcribed Image Text:File Edit View Navigate Code Refactor Run Iools VCS Window Help
lab9.py - lab9.py
O X
C Def Cost_to_hike
Def Cost_to_hike x
Def Cost to_hike X
Answered: def su X
C Def Cost_to_hike X
C Def Cost_to_hike
LAB RESTRICTIO
+
Downloads , lab9.py
Doctest cost_to_hike_naive v
A bartleby.com/questions-and-answers/def-substring-with-largest-sumstring-str-greater-str-wwwwin-given-a-string-return-the-substring-with/891a6ba1-1038-410... 2
O Project -
e lab6.py x
fe lab9.py
e pydev_execfile.py x
I Apps
E Reading list
> Downloads D:\Downloads
> uli External Libraries
79
A 3 X6 ^ v
2,559 /5,000 E
= bartl
Question Asking Tips
80
row = start_point[0]
E Scratches and Consoles
I
U x X2
- Ask one academic question at a time
81
col = start_point[1]
enu cooruimates. in you tnoose to TOlale uie Tmap, you may diso inu the
- Keep questions specific and include all
details
82
row_end = end_point[0]
Engineering
calculated by ...
crop function you implmented last week useful, as that will guarantee
that the start and end points will be at corners of the map.
83
col_end = end_point[1]
def sub
Contact Support for textbook, writing, or
account help
e sub...
84
C = 0
>>> sample_map = [[1, 3],
.. [0, 4]]
>>> cost_to_hike_naive(sample_map,(0, 0).(1, 1))
85
a = 0
86
b = 0
Like you, we deeply value honesty and
87
if start_point == end_point:
integrity. Learn about our honor code.
5
Question
*Response times may vary by subject and
88
return 0
>>> cost_to_hike_naive(sample_map,(1, 1),(0, 0))
question complexity. Median response time is
89
while row != row_end or col != col_end:
34 minutes for paid subscribers and may be
Do not add a
longer for promotional offers.
90
if row != row_end and col == col_end:
break or con
**Limited time offer.
91
if row > row_end:
A 0/2 Im
92
c += abs(m[row - 1][col] - m[row][coi])
93
row -= 1
Live Chat with an Expert
Ask an Expert
94
elif row < row_end:
95
c += abs(m[row + 1][col] - m[row][coi])
A Start chat now
Submit your question
96
row +=
Get on-demand help from a live
Answers in as fast as 30 minutes*
97
elif row == row_end and col != col_end:
tutor, at no additional cost**
98
if col > col_end:
99
c += abs(m[row][col - 1] - m[row][col]D
100
col -= 1
101
elif col < col_end:
8:22 PM
PC
102
c += abs(m[row][col + 1] - m[row][col]D
2021-11-21
103
col += 1
104
else:
105
if row < row_end:
106
a = abs(m[row + 1][col] - m[row][col])
107
elif row > row_end:
108
a = abs(m[row - 1][col] - m[row][col])
109
olif nOw -- nOw ond:
cost_to_hike_naive()
> Run
E TODO
O Problems
Debug
2 Terminal
Python Packages
2 Python Console
A Event Log
O Tests passed: 3 (3 minutes ago)
37:72 LF UTE-8 4 spaces Python 3.9
8:22 PM
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