Q1: Run, K, Run
An increasing run of an integer is a sequence of consecutive digits in which each digit is larger than the last. For example, the number 123444345 has four increasing runs: 1234, 4, 4 and 345. Each run can be indexed from the end of the number, starting with index 0. In the previous example, the 0th run is 345, the first run is 4, the second run is 4 and the third run is 1234.
Implement get_k_run_starter, which takes in integers n and k and returns the 0th digit of the kth increasing run within n. The 0th digit is the leftmost number in the run. You may assume that there are at least k+1 increasing runs in n.
由於源代碼本身已提供while框架，按照框架解答：
- 首先定義i=0，即當前片段索引值為0；
- 從右至左依次取數：當前位為current = n % 10，下一位為next = (n // 10) % 10, current > next時，繼續取下一位；current <= next時，current即為該片段的leftmost number；
- 將final賦值為current，即為第i個run的leftmost；
- 當 i == k時，這個final即為結果。
def get_k_run_starter(n, k):
"""
>>> get_k_run_starter(123444345, 0) # example from description
3
>>> get_k_run_starter(123444345, 1)
4
>>> get_k_run_starter(123444345, 2)
4
>>> get_k_run_starter(123444345, 3)
1
>>> get_k_run_starter(123412341234, 1)
1
>>> get_k_run_starter(1234234534564567, 0)
4
>>> get_k_run_starter(1234234534564567, 1)
3
>>> get_k_run_starter(1234234534564567, 2)
2
"""
i = 0
final = None
while ____________________________:
while ____________________________:
____________________________
final = ____________________________
i = ____________________________
n = ____________________________
return final
Q1 solution:
i = 0
final = None
while i < k+1:
while (n // 10) % 10 < n % 10 and n // 10:
n = n // 10
final = n % 10
i = i + 1
n = n // 10
return final
Q2: High Score
For the purposes of this problem, a score function is a pure function which takes a single number s as input and outputs another number, referred to as the score of s. Complete the best_k_segmenter function, which takes in a positive integer k and a score function score.
best_k_segmenter returns a function that takes in a single number n as input and returns the best k-segment of n, where a k-segment is a set of consecutive digits obtained by segmenting n into pieces of size k and the best segment is the segment with the highest score as determined by score. The segmentation is right to left.
For example, consider 1234567. Its 2-segments are 1, 23, 45 and 67 (a segment may be shorter than k if k does not divide the length of the number; in this case, 1 is the leftover, since the segmenation is right to left). Given the score function lambda x: -x, the best 2-segment is 1. With lambda x: x, the best segment is 67.
按照框架：
- n, seg = partitioner(n)，說明 partitioner函數傳入待分段的數字n，返回（n-seg片段）和seg片段，使用//和%操作符即可；
- 函數目的是找到score函數map在各seg后的最大值，注意初始best_seg=None，不能直接使用score函數，先判斷是否為None，再比較score(best_seg)與score(seg)。
（否則可能出現TypeError: unsupported operand type(s) for //: 'NoneType' and 'int'）
def best_k_segmenter(k, score):
"""
>>> largest_digit_getter = best_k_segmenter(1, lambda x: x)
>>> largest_digit_getter(12345)
5
>>> largest_digit_getter(245351)
5
>>> largest_pair_getter = best_k_segmenter(2, lambda x: x)
>>> largest_pair_getter(12345)
45
>>> largest_pair_getter(245351)
53
>>> best_k_segmenter(1, lambda x: -x)(12345)
1
>>> best_k_segmenter(3, lambda x: (x // 10) % 10)(192837465)
192
"""
partitioner = lambda x: (________________, ________________)
def best_getter(n):
assert n > 0
best_seg = None
while __________________:
n, seg = partitioner(n)
if __________________:
best_seg = seg
return best_seg
return best_getter
Q2 solution:
partitioner = lambda x: (x // (10 ** k), x % (10 ** k))
def best_getter(n):
assert n > 0
best_seg = None
while n:
n, seg = partitioner(n)
if not best_seg or score(seg) > score(best_seg):
best_seg = seg
return best_seg
return best_getter
Q3: It's Always a Good Prime
Implement div_by_primes_under, which takes in an integer n and returns an n-divisibility checker. An n-divisibility-checker is a function that takes in an integer k and returns whether k is divisible by any integers between 2 and n, inclusive. Equivalently, it returns whether k is divisible by any primes less than or equal to n.
Review the Disc 01 is_prime problem for a reminder about prime numbers.
You can also choose to do the no lambda version, which is the same problem, just with defining functions with def instead of lambda.
Hint: If struggling, here is a partially filled out line for after the if statement. checker = (lambda f, i: lambda x: __________)(checker, i)
判斷一個數k能否被i整除：k % i == 0
def div_by_primes_under(n):
"""
>>> div_by_primes_under(10)(11)
False
>>> div_by_primes_under(10)(121)
False
>>> div_by_primes_under(10)(12)
True
>>> div_by_primes_under(5)(1)
False
"""
checker = lambda x: False
i = ____________________________
while ____________________________:
if not checker(i):
checker = ____________________________
i = ____________________________
return ____________________________
def div_by_primes_under_no_lambda(n):
"""
>>> div_by_primes_under_no_lambda(10)(11)
False
>>> div_by_primes_under_no_lambda(10)(121)
False
>>> div_by_primes_under_no_lambda(10)(12)
True
>>> div_by_primes_under_no_lambda(5)(1)
False
"""
def checker(x):
return False
i = ____________________________
while ____________________________:
if not checker(i):
def outer(____________________________):
def inner(____________________________):
return ____________________________
return ____________________________
checker = ____________________________
i = ____________________________
return ____________________________
(2023/4/9 updates, from 2023sp lab03 solution)
當初學這門課的時候，這道題應該是覺得很難沒有做出來，這次按照評論好心人提醒找到了解答。
個人理解：
1、傳入參數n，找到[2,n]的所有質數；
2、傳入參數k，判斷k是否能被1中的質數整除。
使用pythontutor可以查看調用過程。
Q3 Solution:
def div_by_primes_under(n):
checker = lambda x: False
i = 2
while i <= n:
if not checker(i):
checker = (lambda f, i: lambda x: x % i == 0 or f(x))(checker, i)
i = i + 1
return checker
def div_by_primes_under_no_lambda(n):
def checker(x):
return False
i = 2
while i <= n:
if not checker(i):
def outer(f, i):
def inner(x):
return x % i == 0 or f(x)
return inner
checker = outer(checker, i)
i = i + 1
return checker
Q4: My Last Three Brain Cells
A k-memory function takes in a single input, prints whether that input was seen exactly k function calls ago, and returns a new k-memory function. For example, a 2-memory function will display “Found" if its input was seen exactly two function calls ago, and otherwise will display “Not found".
Implement three_memory, which is a three-memory function. You may assume that the value None is never given as an input to your function, and that in the first two function calls the function will display “Not found" for any valid inputs given.
根據題意，當前輸入i與前三個輸入x,y,z中的x相同即可，再將f(x,y,z)替換為最新的三個f(y,z,i)。
def three_memory(n):
"""
>>> f = three_memory('first')
>>> f = f('first')
Not found
>>> f = f('second')
Not found
>>> f = f('third')
Not found
>>> f = f('second') # 'second' was not input three calls ago
Not found
>>> f = f('second') # 'second' was input three calls ago
Found
>>> f = f('third') # 'third' was input three calls ago
Found
>>> f = f('third') # 'third' was not input three calls ago
Not found
"""
def f(x, y, z):
def g(i):
if ____________________________:
____________________________
else:
____________________________
return ____________________________
return ____________________________
return f(None, None, n)
Q4 solution:
def f(x, y, z):
def g(i):
if i == x:
print('Found')
else:
print('Not found')
return f(y,z,i)
return g
return f(None, None, n)