Posts by Michelle Bennett5

Jose Scott3 said:Is there actually an app for this, or some way to block them from hitting the inbox? Or maybe we just head down to the office, find the local IT-experts, and see if we can bribe them with a bottle of high-end bourbon.

Richard Ramos said:Hey there!

So, my husband and I have our computers hooked up to the same router via Ethernet.
Every time his PC is running and I power mine on—just hitting the actual power button on the tower, not the strip—his internet drops out for maybe a minute or two.
I tried searching for a fix online, but I can't find anything that actually matches what's happening here.

Does anyone have any clue how to fix this? Maybe there's some setting somewhere that could handle it??

Michael Jackson10 says:
I haven't had a spare second until now, but here it is. At least, this is my version of a shorter way:

c=0
istaslova=""
word=input("Enter first word: ")
word2=input("Enter second word: ")
for i in range(len(word)):
--if istaslova.count(word)<min(word.count(word), word2.count(word)):
----istaslova=istaslova+word
----c=c+1

print("Common letters are:",istaslova)
print("Total count:",c)

That previous example using find() was a total disaster, because if you use find(), you have to come up with some kind of pointer logic to track where you are moving through the words...

Anyway, isn't this decent for someone who’s only had about four hours of Python classes while helping out a middle schooler nephew??
You should see me trying to mess around with Turtle 🤣
The Syntax trips me up sometimes. I've got one tab glued to a W3Schools reference and another on Replit, but Python is actually pretty sweet for learning logical structure. You can interpret things immediately or just drop in some code to check variable states. It's a bit of a pain with these indentation-only structures, and I'm constantly checking my casting since nothing is declared—I'm never quite sure what a function expects or if I just botched the Syntax... man, my last "masterpiece" was back when I was messing with Cleveland... 🤦

Besides, you didn't really define the task right in your first post: it's not "how many identical letters they have" (since that doesn't account for repeats: "aaaa" and "aaa" only share one unique letter), it's a "common subset" (where you treat every single letter as its own entity, regardless of whether the shape exists elsewhere in the set)

Benjamin Brooks2 said:c = 0
a = "Benjamin Brooks2"
b = "Michelle Bennett5 Kok"
istaslova = ""
for x in a:
for y in b:
if x == y:
if x not in istaslova:
istaslova = istaslova + x
c = c + 1
print(c)
print(istaslova)

Benjamin Brooks2 says:
I've never touched Python before, but I figure it would go something like this...

c=0
istaslova=""
sentence=input("Enter the first sentence: ")
sentence2=input("Enter the second sentence: ")
for i in range(len(sentence)):
for j in range(len(sentence2)):
for c1 in range(len(istaslova))
if sentence2.find(sentence)********* if it found the same letter in both words
if istaslova.NOTfind istaslova = istaslova + ......c=c+1****** if it DIDN'T find the letter in "istaslova" already, then add the letter

print("Duplicate letters:",istaslova)
print("Total count:",c)

Michael Jackson10 said:Nah, he's actually right. I totally misread the prompt—I thought we weren't supposed to count duplicate letters...

Whatever, just comment out that second if statement

Benjamin Brooks2 said:Indices start at 0. len is the letter count.

Michael Jackson10 says:
Python C = 0
istaslova=""
sentence=input("Enter first sentence: ")
sentence2=input("Enter second sentence: ")
for i in range(len(sentence)):
--if sentence2.find(sentence)>-1:
----if istaslova.find(sentence)==-1:
------istaslova=istaslova+sentence
------c=c+1

print("Repeating letters:",istaslova)
print("Total count:",c)

The catch is this thing counts spaces too, treating them just like any other character. You could probably just skip those when you're testing...

Benjamin Brooks2 said:int c = 0;
string sentence = "Benjamin Brooks2";
string sentence2 = "coastaltiger39";
string istaslova = "";
char[] charArr = sentence.ToCharArray();
char[] charArr2 = sentence2.ToCharArray();
foreach (char ch in charArr)
{
if (sentence2.Contains(ch))
{
if (istaslova.Contains(ch)) { }
else { istaslova = istaslova + ch; ++c; }

}
}
Console.WriteLine(istaslova);
Console.WriteLine(c);

******************************************

// Looks good now. It shows the unique letters and the total count. --> ok,2

bluecyclist64 said:You're right; I completely misread the requirements. I dropped the ball on this one.
My solution would work perfectly if we were looking for identical letters at the exact same positions, but that isn't what was asked. That's where my logic failed.

northernmason5 said:Yeah

northernmason5 said:Yeah. I don't get it. I can access it fine on one PC, but then I try on another one and it won't even let me open the folder... and the settings are identical.

bluecyclist64 said:Menščini, I think you’ve overcomplicated this because you missed the core point of the assignment.
There is absolutely no need for some complex boolean matrix here.

coastaltiger39 said:I need some help here!

I'm trying to compare two words to see how many matching letters they share.
Here’s what I mean:
apple, egg = 2 (e, p)
apple, butter = 2 (p, e)
potato, bread = 3 (p, t, o)

Tyler Smith40 said:So does that mean I can just plug any VGA cable into it?

Tyler Smith40 said:To me, they look identical.