cs61a-part2

cs61a COMPOSING PROGRAMS CS 61A Spring 2025 question: Cats | CS自学社区 ans: [2021 Spring] CS61A 学习笔记索引 - ikventure - 博客园 - quite good ans

Tree

it add 1 to all the node including the lable

increment the leaves only

first print tree

def print_tree(t, indent=0):
    print(' '*indent, label(t)) # could adding str() to label
    for b in branches(t):
        print_tree(b, indent + 1)

fib_tree make it much easier to see the difference

numbers = tree(3, [tree(4), tree(5, [tree(6)])])
haste = tree('h', [tree('a', [tree('s'),
                              tree('t')]),
                   tree('e')])
 
>>> haste
['h', ['a', ['s'], ['t']], ['e']]
>>> print_tree(haste)
 h
  a
   s
   t
  e
  >>> label(haste)
  'h'
  >>> branches(haste)
  [['a', ['s'], ['t']], ['e']]

adding a function

 
def print_sums(t, so_far):
    so_far = so_far + label(t)
    if is_leaf(t):
        print(so_far)
    else:
        for b in branches(t):
            print_sums(b, so_far)
  >>> numbers
  [3, [4], [5, [6]]]
  >>> print_sums(numbers,0)
  7
  14
  >>>

The expression not branches(list) will return:

True → when branches(list) is an empty list ([]), meaning list has no branches (only a root).
False → when branches(list) is not empty, meaning list has at least one branch.

Example Scenarios:

tree1 = [1]          # Only root, no branches
tree2 = [1, [2], [3]]  # Root with branches
 
print(not branches(tree1))  # True (empty branches)
print(not branches(tree2))  # False (has branches)

So, is_leaf(tree) will return True only when the tree has no branches.

empty list → false (TF game)

>>> print_sums(numbers,0)
7
14
>>> numbers
[3, [4], [5, [6]]]
>>> print_sums(numbers,0)
7
14
>>> print_tree(numbers)
 3
  4
  5
   6

find total path to get the number total

how many way to get the ans

review :→

(a) → variable → 1

second(thing ) into a fuction: → into return (y)

a → y , so y = 1

g() call back parent fuction (a / 1)

lambda y:

>>> a = 3
...
... def f(g):
...     a = 2
...     return lambda y: a* g(y)
... f(lambda y : a+y )(a)
...
12
>>>
 
(a) -> into another (variable)
 
  - indent -> trace
  last step is *2

return None → ending the fuction

brute force

y = last ( variable )

search (f) → anyfn that f(self) = y , if true = return x

return None make a back to previous recurion fn 🤔
damn , so that’s why it print both case !!!!

<Each cascade frame is from a different call to cascade.

-Until the Return value appears, that call has not completed.

«Any statement can appear before or after the recursive call.

shorter version:

def cascade(n):
    print(n)
    if n>10:
        cascade(n//10)
        print(n)

longer implemenation is more clear (base case, recursive case , so first one is better)

now it is inverse cascade:

Tree recursion

>>> fib(0)
0
>>> fib(1)
1
>>> fib(3)
2
>>> fib(5)
5
>>> fib(6)
8
>>> fib(8)
21
>>>
  @trace
def fib(n):
    if n == 0:
        return 0
    elif n == 1:
        return 1
    else:
        return fib(n - 2) + fib(n - 1)

using ucb trace.py to trace the recursion function 🤩🤩🤩

ucb python file:

 
"""The UCB module contains functions specific to 61A projects at UC Berkeley."""
 
import code
import functools
import inspect
import re
import signal
import sys
 
 
def main(fn):
    """Call fn with command line arguments.  Used as a decorator.
 
    The main decorator marks the function that starts a program. For example,
 
    @main
    def my_run_function():
        # function body
 
    Use this instead of the typical __name__ == "__main__" predicate.
    """
    if inspect.stack()[1][0].f_locals['__name__'] == '__main__':
        args = sys.argv[1:] # Discard the script name from command line
        fn(*args) # Call the main function
    return fn
 
_PREFIX = ''
def trace(fn):
    """A decorator that prints a function's name, its arguments, and its return
    values each time the function is called. For example,
 
    @trace
    def compute_something(x, y):
        # function body
    """
    @functools.wraps(fn)
    def wrapped(*args, **kwds):
        global _PREFIX
        reprs = [repr(e) for e in args]
        reprs += [repr(k) + '=' + repr(v) for k, v in kwds.items()]
        log('{0}({1})'.format(fn.__name__, ', '.join(reprs)) + ':')
        _PREFIX += '    '
        try:
            result = fn(*args, **kwds)
            _PREFIX = _PREFIX[:-4]
        except Exception as e:
            log(fn.__name__ + ' exited via exception')
            _PREFIX = _PREFIX[:-4]
            raise
        # Here, print out the return value.
        log('{0}({1}) -> {2}'.format(fn.__name__, ', '.join(reprs), result))
        return result
    return wrapped
 
 
def log(message):
    """Print an indented message (used with trace)."""
    print(_PREFIX + re.sub('\n', '\n' + _PREFIX, str(message)))
 
 
def log_current_line():
    """Print information about the current line of code."""
    frame = inspect.stack()[1]
    log('Current line: File "{f[1]}", line {f[2]}, in {f[3]}'.format(f=frame))
 
 
def interact(msg=None):
    """Start an interactive interpreter session in the current environment.
 
    On Unix:
      <Control>-D exits the interactive session and returns to normal execution.
    In Windows:
      <Control>-Z <Enter> exits the interactive session and returns to normal
      execution.
    """
    # evaluate commands in current namespace
    frame = inspect.currentframe().f_back
    namespace = frame.f_globals.copy()
    namespace.update(frame.f_locals)
 
    # exit on interrupt
    def handler(signum, frame):
        print()
        exit(0)
    signal.signal(signal.SIGINT, handler)
 
    if not msg:
        _, filename, line, _, _, _ = inspect.stack()[1]
        msg = 'Interacting at File "{0}", line {1} \n'.format(filename, line)
        msg += '    Unix:    <Control>-D continues the program; \n'
        msg += '    Windows: <Control>-Z <Enter> continues the program; \n'
        msg += '    exit() or <Control>-C exits the program'
 
    code.interact(msg, None, namespace)

with trace(ubc)

>>> fib(0)
fib(0):
fib(0) -> 0
0
>>> fib(1)
fib(1):
fib(1) -> 1
1
>>> fib(2)
fib(2):
    fib(0):
    fib(0) -> 0
    fib(1):
    fib(1) -> 1
fib(2) -> 1
1
>>> fib(3)
fib(3):
    fib(1):
    fib(1) -> 1
    fib(2):
        fib(0):
        fib(0) -> 0
        fib(1):
        fib(1) -> 1
    fib(2) -> 1
fib(3) -> 2
2

tree partitions

counting 6, but max for using number 4

spliting half

left = 6-4

right =

explore two possibilities:

use at least one 4
don’t use any 4

Decomposing problem

count_partitions(2, 4)
count_partitions(6, 3)

again , at least 3 , don’t use 3

finally , sum them all

@trace
def count_partitions(n, m):
    if n == 0:
        return 1
    elif n < 0:
        return 0
    elif m == 0:
        return 0
    else:
        with_m = count_partitions(n - m, m)
        without_m = count_partitions(n, m - 1)
        return with_m + without_m
 
>>> count_partitions(2,4)
count_partitions(2, 4):
    count_partitions(-2, 4):
    count_partitions(-2, 4) -> 0
    count_partitions(2, 3):
        count_partitions(-1, 3):
        count_partitions(-1, 3) -> 0
        count_partitions(2, 2):
            count_partitions(0, 2):
            count_partitions(0, 2) -> 1
            count_partitions(2, 1):
                count_partitions(1, 1):
                    count_partitions(0, 1):
                    count_partitions(0, 1) -> 1
                    count_partitions(1, 0):
                    count_partitions(1, 0) -> 0
                count_partitions(1, 1) -> 1
                count_partitions(2, 0):
                count_partitions(2, 0) -> 0
            count_partitions(2, 1) -> 1
        count_partitions(2, 2) -> 2
    count_partitions(2, 3) -> 2
count_partitions(2, 4) -> 2
2

(4,6)

1.  6 = 2 + 4
2.  6 = 1 + 1 + 4
3.  6 = 3 + 3
4.  6 = 1 + 2 + 3
5.  6 = 1 + 1 + 1 + 3
6.  6 = 2 + 2 + 2
7.  6 = 1 + 1 + 2 + 2
8.  6 = 1 + 1 + 1 + 1 + 2
9.  6 = 1 + 1 + 1 + 1 + 1 + 1

(162) Lists - YouTube

777-decorator

digit[3] or getitem(digits, 3) → same

containers

>>> digits=[1,8,2,8]
>>> 1 in digits
True
>>> 2 in digits
True
>>> 5 not in digits
True
>>> not(5 in digits)
True
>>> '1' == 1
False
>>> '1' in digits
False
>>> [1,8] in digits
False
>>> [1,2] in [3,[1,2],4]
True
>>> [1,2] in [3,[[1,2]],4]
False
>>>

statements

>>> def count(s, value):
...     total,index=0,0
...     while index<len(s):
...         element=s[index]
...         if element == value:
...             total = total +1
...         index=index+1
...     return total
...
>>> count([1,2,1,2,1],1)
3
# refactor into python prefer:
def count_new(s, value):
    total = 0
    for element in s:
        if element == value:
            total += 1
    return total

unpack the pairs (only work in same length sequence) for x,y in pairs: if x==y:
range:

list(range())

for _ in range(3):

just loop 3 time

list comprehensions

>>> odds =[1,3,5,7,9]
>>> [x+1 ofr x in odds]
  File "<python-input-64>", line 1
    [x+1 ofr x in odds]
     ^^^^^^^
SyntaxError: invalid syntax. Perhaps you forgot a comma?
>>> [x+1 for x in odds]
[2, 4, 6, 8, 10]
>>> [x+2 for x in odds]
[3, 5, 7, 9, 11]
>>> [x**2 for x in odds]
[1, 9, 25, 49, 81]
>>> x**2 for x in odds
  File "<python-input-68>", line 1
    x**2 for x in odds
         ^^^
SyntaxError: invalid syntax
>>> [x**2 for x in odds]
[1, 9, 25, 49, 81]
>>> (x**2 for x in odds)
<generator object <genexpr> at 0x7a5e2cddcd40>
>>> list((x**2 for x in odds))
[1, 9, 25, 49, 81]
>>> list(x**2 for x in odds)
[1, 9, 25, 49, 81]
>>> [x**2 for x in list(x**2 for x in odds)]
[1, 81, 625, 2401, 6561]
>>> [x for x in odds if 25 %x ==0]
[1, 5]
 
## other list compre:
>>> def divisor(n):
...     return [1] + [x for x in range(2,n) if n%x==0]
...
>>> divisor(1)
[1]
>>> divisor(4)
[1, 2]
>>> divisor(9)
[1, 3]
>>> list(range(2,4))
[2, 3]
>>> list(range(2,9))
[2, 3, 4, 5, 6, 7, 8]
>>>
  ... def divisor(n):
...     return [1] + [x for x in range(2,n) if x%2==0]
  ... def divisor(n):
...     return [1] + [x for x in range(2,n+1) if x%2==0]

range(2,n+1)

return s[0] + fn(s[1:]) adding the first element to the sum of the rest of the element
normal recursion using n-1 etc , this just adjust the slice number

tree recursion

Understanding the “non-first” case

The goal of the “non-first” choice is to skip the first element and find the best product from the rest of the list.

Why is this wrong?

non_first = s[1:] * max_product(s[1+2:])

Issue with Multiplication (s[1:] * something)
- s[1:] is a list, not a single number.
- You cannot multiply a list by a number like this; it’s a type error.
Incorrect Skipping Logic
- s[1+2:] skips two elements from s[1:], but the “non-first” case only means skipping the first element.
- You should not force skipping another element in this case.

Why is this correct?

non_first = max_product(s[1:])

It correctly skips only the first element
- The purpose of “non-first” is just to ignore the first element and continue processing the rest normally.
- We pass s[1:] (everything except the first element) to max_product(), letting the function decide the best product from there.
Recursive structure remains intact
- max_product(s[1:]) naturally handles choosing whether to include elements at s[1] or not.
- No need to manually skip more elements.

Example Breakdown

Example: `s = [10, 3, 1, 9, 2]`

Let’s see how each approach behaves.

Your incorrect approach

non_first = s[1:] * max_product(s[1+2:])  # Incorrect!

s[1:] = [3, 1, 9, 2]
max_product(s[1+2:]) = max_product([9, 2])
But s[1:] * max_product([9, 2]) is invalid because s[1:] is a list, and we cannot multiply a list by a number.

Correct approach

non_first = max_product(s[1:])

We correctly pass [3, 1, 9, 2] into max_product(), letting it handle the computation recursively.

Final Takeaway

The “first” case involves multiplying because we actively pick an element.
The “non-first” case just moves forward without multiplying, letting recursion handle it.
Don’t manually skip extra elements for “non-first”; just call max_product(s[1:]).

Does this explanation help? 😊

# Official (alternative) version
return max(s[0] * max_product(s[2:]), s[1] * max_product(s[3:]))

🔍 Key Difference:

Instead of just skipping the first element, the official version tries both:

Taking s[0] and skipping s[1]
Taking s[1] and skipping s[2]

✅ Why it’s valid:

This version is exploring more options, not just one recursive path. It’s still selecting non-consecutive elements, but now it considers:

What if we start by skipping the first element, and then pick s[1]?

It might give a better product in some cases.

🧠 Summary:

Your version is recursive and safe: max(s[0]*..., max_product(s[1:]))
The alternative tries both s[0] and s[1] as starting points.

Both are valid; the official one explores a broader search space.

Want to see an example where this difference matters?

official ans→ no need using +1, just [0:] and [2:] & [1] and [3] 🔥

def sums(n, m):
    """Return lists that sum to n containing positive numbers up to m that
    have no adjacent repeats.
 
    >>> sums(5, 1)
    []
    >>> sums(5, 2)
    [[2, 1, 2]]
    >>> sums(5, 3)
    [[1, 3, 1], [2, 1, 2], [2, 3], [3, 2]]
    >>> sums(5, 5)
    [[1, 3, 1], [1, 4], [2, 1, 2], [2, 3], [3, 2], [4, 1], [5]]
    >>> sums(6, 3)
    [[1, 2, 1, 2], [1, 2, 3], [1, 3, 2], [2, 1, 2, 1], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]
    """
    if n < 0:
        return []
    if n == 0:
        sums_to_zero = []     # The only way to sum to zero using positives
        return [sums_to_zero] # Return a list of all the ways to sum to zero
    result = []
    for k in range(1, m + 1):
        result = result + [ ___ for rest in ___ if rest == [] or ___ ]
    return result

📌 Line in Question:

result = result + [[k] + rest for rest in sums(n - k, m) if rest == [] or rest[0] != k]

🧠 First, let’s understand the goal

We’re trying to build all possible lists of numbers that:

Add up to n
Use only numbers 1 to m
Have no adjacent duplicates

🧱 Break the line into smaller parts:

Part 1: The loop

for rest in sums(n - k, m)

This is the recursive part.

We already chose a number k (in a loop from 1 to m)
So we subtract k from the target: n - k
And we ask: What are all the valid lists that sum to n - k?
Those results are stored in rest

So rest is a valid list after we’ve already picked k

Part 2: The condition

if rest == [] or rest[0] != k

This is the adjacent repeat check.

We don’t want to allow something like [2, 2, 1]
So we check:
- If rest is empty ([]) → OK, there’s nothing to compare
- Otherwise: make sure rest[0] != k, so k isn’t repeated

✅ Example:

Say k = 2 and rest = [2, 1], then:

rest[0] == k → ❌ skip it (would give [2, 2, 1], which has adjacent 2s)

Say rest = [1, 2]:

rest[0] != k → ✅ keep it (result would be [2, 1, 2])

Part 3: The result building

[[k] + rest ...]

Once we know rest is valid:

We add k to the front of it → [k] + rest
That’s one of our valid result lists

Part 4: Append to final list

result = result + ...

We’re collecting all valid combinations, so we keep adding to result

✅ Final Visual Summary:

For every number k from 1 to m:

Get all valid sublists that sum to n - k
For each one:
- If it doesn’t start with k, then:
  - Add k in front
  - Save it in the final list

🧪 Tiny Example:

sums(3, 2) — we want to find all valid lists of 1s and 2s that add up to 3, no adjacent repeats.

Here’s what would happen:

Try k = 1
- n - k = 2
- Call sums(2, 2) → maybe gives [[2], [1, 1]]
- Check: can we put 1 in front?
  - [1, 1] ❌ skip (1 == 1)
  - [2] ✅ keep → [1, 2]
Try k = 2
- n - k = 1
- Call sums(1, 2) → maybe gives [[1]]
- Can we put 2 in front?
  - [1] ✅ → [2, 1]

✅ Final result: [[1, 2], [2, 1]]

Let me know if you’d like this rewritten as a for-loop version — sometimes that’s easier to read than list comprehensions.

sums(5, 3):
    sums(4, 3):
        sums(3, 3):
            sums(2, 3):
                sums(1, 3):
                    sums(0, 3):
                    sums(0, 3) -> [[]]
                    sums(-1, 3):
                    sums(-1, 3) -> []
                    sums(-2, 3):
                    sums(-2, 3) -> []
                sums(1, 3) -> [[1]]
                sums(0, 3):
                sums(0, 3) -> [[]]
                sums(-1, 3):
                sums(-1, 3) -> []
            sums(2, 3) -> [[2]]
            sums(1, 3):
                sums(0, 3):
                sums(0, 3) -> [[]]
                sums(-1, 3):
                sums(-1, 3) -> []
                sums(-2, 3):
                sums(-2, 3) -> []
            sums(1, 3) -> [[1]]
            sums(0, 3):
            sums(0, 3) -> [[]]
        sums(3, 3) -> [[1, 2], [2, 1], [3]]
        sums(2, 3):
            sums(1, 3):
                sums(0, 3):
                sums(0, 3) -> [[]]
                sums(-1, 3):
                sums(-1, 3) -> []
                sums(-2, 3):
                sums(-2, 3) -> []
            sums(1, 3) -> [[1]]
            sums(0, 3):
            sums(0, 3) -> [[]]
            sums(-1, 3):
            sums(-1, 3) -> []
        sums(2, 3) -> [[2]]
        sums(1, 3):
            sums(0, 3):
            sums(0, 3) -> [[]]
            sums(-1, 3):
            sums(-1, 3) -> []
            sums(-2, 3):
            sums(-2, 3) -> []
        sums(1, 3) -> [[1]]
    sums(4, 3) -> [[1, 2, 1], [1, 3], [3, 1]]
    sums(3, 3):
        sums(2, 3):
            sums(1, 3):
                sums(0, 3):
                sums(0, 3) -> [[]]
                sums(-1, 3):
                sums(-1, 3) -> []
                sums(-2, 3):
                sums(-2, 3) -> []
            sums(1, 3) -> [[1]]
            sums(0, 3):
            sums(0, 3) -> [[]]
            sums(-1, 3):
            sums(-1, 3) -> []
        sums(2, 3) -> [[2]]
        sums(1, 3):
            sums(0, 3):
            sums(0, 3) -> [[]]
            sums(-1, 3):
            sums(-1, 3) -> []
            sums(-2, 3):
            sums(-2, 3) -> []
        sums(1, 3) -> [[1]]
        sums(0, 3):
        sums(0, 3) -> [[]]
    sums(3, 3) -> [[1, 2], [2, 1], [3]]
    sums(2, 3):
        sums(1, 3):
            sums(0, 3):
            sums(0, 3) -> [[]]
            sums(-1, 3):
            sums(-1, 3) -> []
            sums(-2, 3):
            sums(-2, 3) -> []
        sums(1, 3) -> [[1]]
        sums(0, 3):
        sums(0, 3) -> [[]]
        sums(-1, 3):
        sums(-1, 3) -> []
    sums(2, 3) -> [[2]]
sums(5, 3) -> [[1, 3, 1], [2, 1, 2], [2, 3], [3, 2]]
[[1, 3, 1], [2, 1, 2], [2, 3], [3, 2]]

Let’s take an easy example:
👉 sums(3, 2)
So:

We want all combinations of 1s and 2s
That sum to 3
And don’t have adjacent repeats

🔁 Step-by-step with simple loops

Let’s write the logic more clearly using regular loops:

def sums(n, m):
    if n < 0:
        return []
    if n == 0:
        return [[]]  # Only one way to sum to zero
 
    result = []
    for k in range(1, m + 1):  # Try every number from 1 to m
        sublists = sums(n - k, m)  # Recursive call
        for rest in sublists:
            if rest == [] or rest[0] != k:  # No repeats
                result.append([k] + rest)   # Build the new list
    return result

🧠 What happens to `rest`?

Let’s trace sums(3, 2) step by step.

First call: `sums(3, 2)`

Try k = 1:

Call sums(2, 2)

Second call: `sums(2, 2)`

Try k = 1 again:

Call sums(1, 2)

Third call: `sums(1, 2)`

Try k = 1:

Call sums(0, 2) → returns [[]]
Now we loop over rest in [[]]
- rest = [] → ✅ pass the check
- Add [1] + [] → [1] ✅

Now sums(1, 2) → has [ [1] ] so far.

Try k = 2:

Call sums(-1, 2) → returns []
So done here.

Return: sums(1, 2) → [[1]]

Back to `sums(2, 2)`

Now we’re still handling k = 1, and sums(1, 2) returned [[1]]

We loop:

rest = [1]
Check: is rest[0] != k? → 1 == 1 ❌ skip it!

Now try k = 2:

Call sums(0, 2) → returns [[]]

Loop:

rest = [] → ✅
Add [2] + [] → [2]

Return: sums(2, 2) → [[2]]

Back to `sums(3, 2)`

We’re at k = 1, and sums(2, 2) gave [[2]]

Loop:

rest = [2] → 2 != 1 ✅
Add [1] + [2] → [1, 2]

Now try k = 2:

Call sums(1, 2) → returned [[1]]

Loop:

rest = [1] → 1 != 2 ✅
Add [2] + [1] → [2, 1]

✅ Final result of `sums(3, 2)`:

[[1, 2], [2, 1]]

🔄 How `rest` changes

It’s the result of the recursive call each time:

sums(0, 2) → [[]]
sums(1, 2) → [[1]]
sums(2, 2) → [[2]]
sums(3, 2) builds on that to form [1, 2], [2, 1]

Each time you add a number k, you get smaller problems, and rest is the partial solution to that smaller problem.

TL;DR:

rest is a partial solution from a recursive call
You’re trying to build on rest by adding a new number k in front
But you skip it if rest[0] == k to avoid adjacent repeats

Awesome — let’s visualize the recursive calls of sums(3, 2) as a tree diagram 🌲, so you can clearly see how the recursion works, how rest is built, and how adjacent repeats are avoided.

📌 Problem: `sums(3, 2)`

We want to find all the non-repeating adjacent sequences of numbers 1 to 2 that sum to 3.

🌲 Tree of Recursive Calls

We’ll show it like:

current_call (n)
├── try k = 1
│   └── recursive call: sums(n - 1, m)
└── try k = 2
    └── recursive call: sums(n - 2, m)

Root: `sums(3, 2)`

sums(3, 2)
├── k = 1 → sums(2, 2)
│   ├── k = 1 → sums(1, 2)
│   │   ├── k = 1 → sums(0, 2) ✅ returns [[]]
│   │   └── k = 2 → sums(-1, 2) ❌ invalid
│   │   → builds: [1] + [] = [1] ✅
│   │   → return: [[1]]
│   └── k = 2 → sums(0, 2) ✅ returns [[]]
│       → builds: [2] + [] = [2] ✅
│   → Now from `sums(2, 2)` we have: [[2]], because [1] is invalid (adj repeat with prev 1)
│   → Back to `sums(3, 2)`, build: [1] + [2] = [1, 2] ✅
├── k = 2 → sums(1, 2)
│   ├── k = 1 → sums(0, 2) ✅ returns [[]]
│   │   → builds: [1] + [] = [1] ✅
│   └── k = 2 → sums(-1, 2) ❌ invalid
│   → Return: [[1]]
│   → Back to `sums(3, 2)`, build: [2] + [1] = [2, 1] ✅

✅ Final Result:

From root sums(3, 2), we got:

[1, 2]
[2, 1]

So:

sums(3, 2) → [[1, 2], [2, 1]]

🔁 And how `rest` changes:

Call	Returned `rest`	Built `[k] + rest`	Keep or Skip
`sums(0, 2)`	`[[]]`	`[1]`, `[2]`	✅
`sums(1, 2)`	`[[1]]`	`[1, 1]`, `[2, 1]`	❌, ✅
`sums(2, 2)`	`[[2]]`	`[1, 2]`	✅

Let me know if you want to go one level deeper (e.g., sums(4, 2) tree) or see it as a visual chart!

Project2

Let’s analyze the provided implementation of the choose function (which seems to be equivalent to the pick function from your earlier question) and compare it to the approach we discussed. Then, I’ll explain why the author initializes count = -1 and explore an alternative approach, highlighting the differences.

Provided Implementation:

def choose(paragraphs, select, k):
    """Return the Kth paragraph from PARAGRAPHS for which SELECT called on the
    paragraph returns true. If there are fewer than K such paragraphs, return
    the empty string.
    """
    count = -1
    for p in paragraphs:
        if select(p):
            count += 1
            if count == k:
                return p
    return ""

How It Works:

Initialization: count = -1
- Starts the counter at -1 instead of 0.
Loop: Iterates over paragraphs directly (using p instead of indices).
Condition: When select(p) is True:
- Increments count (from -1 to 0 on the first valid paragraph, then 1, etc.).
- Checks if count == k. If true, returns the current paragraph p.
Default Return: If the loop ends (i.e., not enough valid paragraphs), returns ''.

Testing with Example:

ps = ['short', 'really long', 'tiny']
s = lambda p: len(p) <= 5
k = 2
- 'short': len = 5, True, count = -1 + 1 = 0 (not 2, continue)
- 'really long': len = 11, False, count stays 0
- 'tiny': len = 4, True, count = 0 + 1 = 1 (not 2, continue)
- End: count = 1, fewer than k = 2 valid paragraphs, returns ''.

This matches the expected behavior.

Alternative Approach (from Earlier):

Here’s the version we discussed previously:

def pick(paragraphs, select, k):
    count = 0
    for i in range(len(paragraphs)):
        if select(paragraphs[i]):
            if count == k:
                return paragraphs[i]
            count += 1
    return ''

Differences:

Starting Value of count:
- Provided (count = -1): Starts at -1 and increments before checking against k.
- Alternative (count = 0): Starts at 0 and increments after checking against k.
Order of Operations:
- Provided: Increments count first, then checks if it equals k. This means count represents the index of the current valid paragraph after incrementing.
- Alternative: Checks count == k first, then increments. Here, count represents how many valid paragraphs have been seen before the current one.
Loop Style:
- Provided: Uses for p in paragraphs (direct iteration over elements).
- Alternative: Uses for i in range(len(paragraphs)) (iteration over indices).

Why `count = -1`?

The choice of count = -1 is a subtle but intentional design decision:

Logic: By starting at -1, the first valid paragraph increments count to 0, the second to 1, and so on. When count == k, it’s naturally aligned with the 0-based index of the kth valid paragraph (e.g., k = 0 is the 0th valid paragraph, k = 1 is the 1st, etc.).
Simplicity: This allows the increment and comparison to happen in one clean step inside the if select(p) block, avoiding an extra condition or adjustment elsewhere.
Example:
- k = 0 (first valid paragraph):
  - 'short': count = -1 + 1 = 0, 0 == 0, returns 'short'.
- k = 1:
  - 'short': count = -1 + 1 = 0, 0 != 1, continue
  - 'tiny': count = 0 + 1 = 1, 1 == 1, returns 'tiny'.

If it started at 0 and incremented first, you’d need to adjust the comparison (e.g., count - 1 == k), which adds complexity.

Another Way to Do It:

Here’s an alternative that avoids the -1 trick by using a different structure:

def choose(paragraphs, select, k):
    valid_paragraphs = [p for p in paragraphs if select(p)]  # Filter valid paragraphs
    return valid_paragraphs[k] if k < len(valid_paragraphs) else ''

How It Works:

List Comprehension: Creates a list of all paragraphs where select(p) is True.
Indexing: Returns the kth element if it exists, otherwise ''.

Differences:

Efficiency: Less efficient for large lists because it processes the entire list upfront, even if k is small (e.g., k = 0). The other versions stop as soon as they find the kth match.
Readability: More concise and arguably easier to understand at a glance.
Memory: Uses more memory by storing the filtered list, whereas the loop versions don’t.

Example:

ps = ['short', 'really long', 'tiny']
s = lambda p: len(p) <= 5
k = 2
- valid_paragraphs = ['short', 'tiny']
- len(valid_paragraphs) = 2, k = 2 >= 2, returns ''.

Why `count = -1` vs. `count = 0`?

count = -1: Aligns the increment with the index naturally, making the logic flow smoothly without extra adjustments. It’s a common pattern when you want the counter to match the current item’s position after processing.
count = 0: Requires incrementing after the check, which works but separates the “found it” logic from the counter update. It’s more intuitive if you think of count as “how many I’ve seen so far.”

Both are correct, but count = -1 is slightly more elegant here because it avoids an extra line or condition. The list comprehension alternative trades elegance for simplicity but sacrifices early termination.

Does this clarify the differences and reasoning? Let me know if you’d like to explore further!

def about(topic):
    """Return a select function that returns whether a paragraph contains one
    of the words in TOPIC.
 
    >>> about_dogs = about(['dog', 'dogs', 'pup', 'puppy'])
    >>> choose(['Cute Dog!', 'That is a cat.', 'Nice pup!'], about_dogs, 0)
    'Cute Dog!'
    >>> choose(['Cute Dog!', 'That is a cat.', 'Nice pup.'], about_dogs, 1)
    'Nice pup.'
    """
    assert all([lower(x) == x for x in topic]), 'topics should be lowercase.'
    # BEGIN PROBLEM 2
    "*** YOUR CODE HERE ***"
    def select(paragraph):
        words = set(split(lower(remove_punctuation(paragraph))))
        for x in topic:
            if x in words:
                return True
        return False
    return select
    # END PROBLEM 2
   # other method :
    def about2(text):
            text1 = remove_punctuation(text)
            text2 = split(text1)
            text2_low = [lower(x) for x in text2]
 
            subject_low = [lower(x) for x in subject]
 
            for i in subject_low:
                for j in text2_low:
                    if i == j:
                        return True
            return False
    return about2

"""Return the accuracy (percentage of words typed correctly) of TYPED
when compared to the prefix of REFERENCE that was typed.

>>> accuracy('Cute Dog!', 'Cute Dog.')
50.0
>>> accuracy('A Cute Dog!', 'Cute Dog.')
0.0
>>> accuracy('cute Dog.', 'Cute Dog.')
50.0
>>> accuracy('Cute Dog. I say!', 'Cute Dog.')
50.0
>>> accuracy('Cute', 'Cute Dog.')
100.0
>>> accuracy('', 'Cute Dog.')
0.0
unit reference/typed
length
saniterize
"""

    a = len(typed_words)
    b = len(source_words)
    if  a == 0 and b == 0 :
        return 100.0
    elif a == 0 or b == 0 :
        return 0.0
 
    count = 0
    for i in range(0, a):
        if i+1 > b:
            break
        elif typed_words[i] == source_words[i]:
            count += 1
    return count * 100 / a
 
def accuracy(typed, reference):
    typed_words = split(typed)
    reference_words = split(reference)
    # BEGIN PROBLEM 3
    "*** YOUR CODE HERE ***"
    if len(typed_words) == 0 and len(source_words) == 0:
        return 100.0
    elif len(typed_words) == 0:
        return 0.0
 
    correct = 0
    for x, y in zip(typed_words, source_words):
        if x == y:
            correct += 1
 
    return correct / len(typed_words) * 100.0
 
    # END PROBLEM 3

return correct / len(typed_words) * 100.0

Let’s dive into comparing these two implementations of the accuracy function, analyze their differences, determine which one is “better” based on the problem requirements, and explain how zip works in the first solution. We’ll evaluate correctness, readability, and efficiency, and then tie it all back to the problem’s intent.

Problem Recap:

Function: accuracy(typed, source)
Purpose: Calculate the percentage of words in typed that correctly match the corresponding words in the prefix of source.
Inputs:
- typed: A string that may contain typos.
- source: A reference string without errors.
Output: A float (0.0 to 100.0) representing the percentage of correctly typed words.
Assumptions: split is a helper function (not shown) that splits a string into a list of words (e.g., 'Cute Dog!' → ['Cute', 'Dog!']).

Test Cases:

accuracy('Cute Dog!', 'Cute Dog.') → 50.0 (1/2 words correct)
accuracy('A Cute Dog!', 'Cute Dog.') → 0.0 (no matches in prefix)
accuracy('cute Dog.', 'Cute Dog.') → 50.0 (1/2 correct, case-sensitive)
accuracy('Cute Dog. I say!', 'Cute Dog.') → 50.0 (1/2 correct in prefix)
accuracy('Cute', 'Cute Dog.') → 100.0 (1/1 correct)
accuracy('', 'Cute Dog.') → 0.0
accuracy('', '') → 100.0

Solution 1: Using `zip`

def accuracy(typed, source):
    typed_words = split(typed)
    reference_words = split(source)  # Note: 'reference_words' should be 'source_words'
    if len(typed_words) == 0 and len(reference_words) == 0:
        return 100.0
    elif len(typed_words) == 0:
        return 0.0
 
    correct = 0
    for x, y in zip(typed_words, reference_words):
        if x == y:
            correct += 1
 
    return correct / len(typed_words) * 100.0

How It Works:

Base Cases:
- If both typed and source are empty → 100.0.
- If typed is empty but source isn’t → 0.0.
Main Logic:
- Uses zip to pair words from typed_words and reference_words (should be source_words—a naming inconsistency).
- Counts matches with correct.
- Returns (correct / len(typed_words)) * 100.

How `zip` Works:

zip(typed_words, source_words) creates an iterator of tuples, pairing elements from both lists up to the length of the shorter list.
Example:
- typed_words = ['Cute', 'Dog!']
- source_words = ['Cute', 'Dog.']
- zip → [('Cute', 'Cute'), ('Dog!', 'Dog.')]
- If typed_words = ['A', 'Cute', 'Dog!'] and source_words = ['Cute', 'Dog.'], zip → [('A', 'Cute'), ('Cute', 'Dog.')].
for x, y in zip(...) unpacks each pair, comparing x (from typed) and y (from source).

Pros:

Concise: zip handles pairing automatically.
Readable: The intent (comparing corresponding words) is clear.

Cons:

Naming Bug: Uses reference_words internally but source_words should be consistent with the parameter source.
Incomplete Base Case: Doesn’t check len(source_words) == 0 when typed_words isn’t empty, which could lead to division by zero or incorrect behavior if source is empty but typed isn’t.

Solution 2: Using Index-Based Loop

def accuracy(typed, source):
    typed_words = split(typed)
    source_words = split(source)
    a = len(typed_words)
    b = len(source_words)
    if a == 0 and b == 0:
        return 100.0
    elif a == 0 or b == 0:
        return 0.0
 
    count = 0
    for i in range(0, a):
        if i + 1 > b:
            break
        elif typed_words[i] == source_words[i]:
            count += 1
    return count * 100 / a

How It Works:

Base Cases:
- Both empty → 100.0.
- Either empty → 0.0 (covers typed empty or source empty).
Main Logic:
- Loops over indices of typed_words (up to a).
- Checks if index i exceeds source_words length (b) to avoid index errors.
- Compares words at matching indices, increments count for matches.
- Returns (count / a) * 100.

Pros:

Robust Base Case: Handles all empty scenarios explicitly.
Prefix Focus: Stops comparing when source_words runs out, aligning with “prefix of source” requirement.

Cons:

Verbose: Manual index checking is less elegant than zip.
Potential Confusion: i + 1 > b could be written as i >= b for clarity (since i is 0-based).

Key Differences:

Handling source Shorter than typed:
- Zip: Automatically stops at the shorter list. If source_words is shorter, it ignores extra typed_words, which aligns with “prefix of source”.
- Index: Explicitly breaks when i >= b, achieving the same result.
Base Case Completeness:
- Zip: Misses len(source_words) == 0 when typed_words isn’t empty (e.g., 'Cute' vs. ''). This should return 0.0, but it attempts division by zero or incorrect logic.
- Index: Covers all cases with a == 0 or b == 0.
Style:
- Zip: More Pythonic, leveraging built-in tools.
- Index: More manual, but explicit.

Which is Better?

Correctness: Solution 2 (Index) is more robust due to its complete base case handling. Solution 1 (Zip) has a bug when source is empty but typed isn’t, potentially causing runtime errors or incorrect results unless fixed.
Readability: Solution 1 (Zip) wins for conciseness and elegance, assuming the base case is corrected.
Efficiency: Both are O(min(n, m)) where n = len(typed_words), m = len(source_words), so no significant difference.

Fixed Zip Version:

def accuracy(typed, source):
    typed_words = split(typed)
    source_words = split(source)
    if len(typed_words) == 0 and len(source_words) == 0:
        return 100.0
    if len(typed_words) == 0 or len(source_words) == 0:
        return 0.0
    correct = 0
    for x, y in zip(typed_words, source_words):
        if x == y:
            correct += 1
    return correct / len(typed_words) * 100.0

Adds len(source_words) == 0 check.
Now fully correct and more readable than the index version.

Conclusion:

Best Choice: The fixed zip version is better because it’s:
- Correct: Handles all edge cases.
- Readable: Uses Python’s zip for a clean, idiomatic solution.
- Maintainable: Less prone to manual errors like index mismanagement.
Why zip Works Well Here: It naturally aligns with comparing two sequences positionally up to the shorter length, perfectly matching the “prefix of source” requirement.

Let me know if you’d like to explore further or test more edge cases!

def wpm(typed, elapsed):
    """Return the words-per-minute (WPM) of the TYPED string.
 
    Arguments:
        typed: an entered string
        elapsed: an amount of time in seconds
 
    >>> wpm('hello friend hello buddy hello', 15)
    24.0
    >>> wpm('0123456789',60)
    2.0
    """
    assert elapsed > 0, "Elapsed time must be positive"
    # BEGIN PROBLEM 4
    "*** YOUR CODE HERE ***"
    length = len(typed)
    rate = 60 / elapsed
    return length * rate / 5
    # END PROBLEM 4

Let’s dive into these two implementations of the autocorrect function, break down their concepts, and compare how they solve the same problem in different ways. I’ll explain the logic behind each approach, highlight their strengths and weaknesses, and clarify why they both work (or don’t) for the given test cases. This will help you understand the problem-solving strategies and trade-offs involved.

Problem Recap

Goal: Return the word from word_list (or valid_words) with the smallest difference from typed_word, as measured by diff_function. If the smallest difference exceeds limit, return typed_word. If multiple words tie for the smallest difference, return the first one in the list.
Inputs:
- typed_word: A string (e.g., "hwllo").
- word_list/valid_words: A list of strings (e.g., ["butter", "hello", "potato"]).
- diff_function: A function returning a number representing the difference between two words.
- limit: A number (e.g., 20).
Test Cases:
- autocorrect("hwllo", ["butter", "hello", "potato"], ten_diff, 20) → 'butter'
- autocorrect("tosting", ["testing", "asking", "fasting"], first_diff, 10) → 'testing'

Solution 1: Iterative with Early Initialization

def autocorrect(typed_word, word_list, diff_function, limit):
    min_diff = limit + 1
    best_word = typed_word
 
    for word in word_list:
        if word == typed_word:
            return typed_word
 
        diff = diff_function(typed_word, word, limit)
 
        if min_diff > abs(diff):
            min_diff = abs(diff)
            best_word = word
 
    return typed_word if min_diff > limit else best_word

Concept Explained:

Initialization:
- min_diff = limit + 1: Starts higher than limit so that if no word is better, typed_word is returned by default.
- best_word = typed_word: Pre-sets the fallback return value, avoiding extra logic later.
Early Exit:
- if word == typed_word: If the exact word is in word_list, return it immediately. This optimizes for the case where no correction is needed.
Loop Logic:
- Computes diff for each word using diff_function.
- Uses abs(diff) to handle potential negative differences (though test cases suggest non-negative outputs).
- Updates min_diff and best_word if the new diff is smaller (using > ensures the first tied word is kept).
Final Decision:
- Ternary expression: Returns typed_word if min_diff > limit, otherwise best_word.

How It Works with Test Cases:

"hwllo", ["butter", "hello", "potato"], ten_diff, 20:
- min_diff = 21, best_word = "hwllo".
- "hwllo" not in list, no early return.
- diff(butter) = 10, 21 > 10, min_diff = 10, best_word = "butter".
- diff(hello) = 10, 10 > 10 is false (tie), no update.
- diff(potato) = 10, no update.
- 10 ≤ 20, return "butter".
"tosting", ["testing", "asking", "fasting"], first_diff, 10:
- min_diff = 11, best_word = "tosting".
- "tosting" not in list".
- diff(testing) = 0, 11 > 0, min_diff = 0, best_word = "testing".
- diff(asking) = 1, 0 > 1 is false.
- diff(fasting) = 1, no update.
- 0 ≤ 10, return "testing".

Key Concept:

Single Pass with Default: It iterates once, updating the best candidate only when a strictly smaller difference is found, and defaults to typed_word if the limit is exceeded. The abs(diff) adds flexibility for diff functions that might return negatives.

Solution 2: List Comprehension with Index Lookup

def autocorrect(typed_word, valid_words, diff_function, limit):
    if typed_word in valid_words:
        return typed_word
    diff_list = [diff_function(typed_word, word, limit) for word in valid_words]
    if min(diff_list) > limit:
        return typed_word
    else:
        return valid_words[diff_list.index(min(diff_list))]

Concept Explained:

Early Exit:
- if typed_word in valid_words: Similar to Solution 1, returns typed_word if it’s already valid.
Compute All Differences:
- diff_list = [diff_function(...)]: Uses a list comprehension to create a list of differences for all words in valid_words.
Find Minimum and Decide:
- min(diff_list): Finds the smallest difference.
- If min > limit, returns typed_word.
- Otherwise, uses diff_list.index(min(diff_list)) to find the first word with that minimum difference and returns it from valid_words.

How It Works with Test Cases:

"hwllo", ["butter", "hello", "potato"], ten_diff, 20:
- "hwllo" not in list".
- diff_list = [10, 10, 10].
- min(diff_list) = 10, 10 ≤ 20.
- diff_list.index(10) = 0 (first occurrence), return "butter".
"tosting", ["testing", "asking", "fasting"], first_diff, 10:
- "tosting" not in list".
- diff_list = [0, 1, 1].
- min(diff_list) = 0, 0 ≤ 10.
- diff_list.index(0) = 0, return "testing".

Key Concept:

Batch Processing: Computes all differences upfront, then processes them with built-in functions (min and index). It relies on Python’s list operations to handle ties (first occurrence by default).

Comparison of Concepts

1. Approach to Finding the Minimum

Solution 1 (Iterative):
- Tracks the minimum difference incrementally in a single loop.
- Updates best_word only when a strictly smaller difference is found (>).
- Pros: Memory-efficient (no extra list), stops updating once the minimum stabilizes.
- Cons: Slightly more manual logic.
Solution 2 (List Comprehension):
- Computes all differences into a list, then finds the minimum and its index.
- Pros: Cleaner, leverages Python’s built-ins.
- Cons: Uses more memory (stores diff_list), less efficient for large lists.

2. Handling Ties

Both satisfy “return the one closest to the front”:
- Solution 1: Uses < (not <=), so ties don’t overwrite best_word, keeping the first.
- Solution 2: index(min()) returns the first occurrence of the minimum naturally.

3. Limit Check

Solution 1: Integrates it into the return logic with a pre-set default (typed_word).
Solution 2: Explicitly checks min(diff_list) > limit, more declarative.

4. Early Exit

Both check if typed_word is in the list, optimizing for exact matches.

5. Use of `abs`

Solution 1: Applies abs(diff) to handle potential negative differences.
Solution 2: Assumes non-negative differences (test cases support this).
Note: Test cases (ten_diff, first_diff) return non-negative values, so abs isn’t strictly necessary but adds robustness.

Which is Better?

Correctness:
- Both work for the test cases.
- Solution 1’s abs(diff) is overkill here but safer for untested diff_functions.
Efficiency:
- Solution 1: O(n) time, O(1) space (n = len(word_list)).
- Solution 2: O(n) time for list creation, O(n) space for diff_list, plus O(n) for index, totaling O(n) time but higher memory.
- Winner: Solution 1 for large lists.
Readability:
- Solution 1: More explicit, but requires understanding the initialization trick (limit + 1).
- Solution 2: More concise and Pythonic, easier to read at a glance.
- Winner: Solution 2 for clarity.
Maintainability:
- Solution 2’s list comprehension is easier to modify (e.g., for debugging or adding conditions).

Verdict:

Solution 2 is better for small lists and readability, aligning with Python’s style.
Solution 1 is preferable for performance with large lists or memory constraints.

Conceptual Takeaway

Iterative (Solution 1): “Build the answer as you go.” Good for control and efficiency.
Declarative (Solution 2): “Compute everything, then decide.” Good for simplicity and leveraging Python’s tools.

Both embody the core idea: minimize difference, respect the limit, and prioritize the first tie. Your choice depends on context (e.g., list size, coding style preference). Does this clarify the concepts? Let me know if you want to explore further!

extra info about diff

>>> def length_diff(w1, w2, limit):
...     return min(limit + 1, abs(len(w2) - len(w1)))
>>> length_diff('mellow', 'cello', 10)
1
>>> length_diff('hippo', 'hippopotamus', 5)
6

Recent Writing

20250402-cs61a-part2

cs61a-part2

Tree

Example Scenarios:

Understanding the “non-first” case

Why is this wrong?

Why is this correct?

Example Breakdown

Example: s = [10, 3, 1, 9, 2]

Your incorrect approach

Correct approach

Final Takeaway

🔍 Key Difference:

✅ Why it’s valid:

🧠 Summary:

📌 Line in Question:

🧠 First, let’s understand the goal

🧱 Break the line into smaller parts:

Part 1: The loop

Part 2: The condition

✅ Example:

Part 3: The result building

Part 4: Append to final list

✅ Final Visual Summary:

🧪 Tiny Example:

🔁 Step-by-step with simple loops

🧠 What happens to rest?

First call: sums(3, 2)

Second call: sums(2, 2)

Third call: sums(1, 2)

Back to sums(2, 2)

Back to sums(3, 2)

✅ Final result of sums(3, 2):

🔄 How rest changes

TL;DR:

📌 Problem: sums(3, 2)

🌲 Tree of Recursive Calls

Root: sums(3, 2)

✅ Final Result:

🔁 And how rest changes:

Project2

Provided Implementation:

How It Works:

Testing with Example:

Alternative Approach (from Earlier):

Differences:

Why count = -1?

Another Way to Do It:

How It Works:

Differences:

Example:

Why count = -1 vs. count = 0?

Problem Recap:

Test Cases:

Solution 1: Using zip

How It Works:

How zip Works:

Pros:

Cons:

Solution 2: Using Index-Based Loop

How It Works:

Pros:

Cons:

Key Differences:

Which is Better?

Fixed Zip Version:

Conclusion:

Problem Recap

Solution 1: Iterative with Early Initialization

Concept Explained:

How It Works with Test Cases:

Key Concept:

Solution 2: List Comprehension with Index Lookup

Concept Explained:

How It Works with Test Cases:

Key Concept:

Comparison of Concepts

1. Approach to Finding the Minimum

2. Handling Ties

Example: `s = [10, 3, 1, 9, 2]`

🧠 What happens to `rest`?

First call: `sums(3, 2)`

Second call: `sums(2, 2)`

Third call: `sums(1, 2)`

Back to `sums(2, 2)`

Back to `sums(3, 2)`

✅ Final result of `sums(3, 2)`:

🔄 How `rest` changes

📌 Problem: `sums(3, 2)`

Root: `sums(3, 2)`

🔁 And how `rest` changes:

Why `count = -1`?

Why `count = -1` vs. `count = 0`?

Solution 1: Using `zip`

How `zip` Works:

5. Use of `abs`