Here is what I want to get:
$ python ex33.py
At the top i is 0
Numbers now: [0]
At the bottom i is 1
At the top i is 1
Numbers now: [0, 1]
At the bottom i is 2
At the top i is 2
Numbers now: [0, 1, 2]
At the bottom i is 3
At the top i is 3
Numbers now: [0, 1, 2, 3]
At the bottom i is 4
At the top i is 4
Numbers now: [0, 1, 2, 3, 4]
At the bottom i is 5
At the top i is 5
Numbers now: [0, 1, 2, 3, 4, 5]
At the bottom i is 6
The numbers:
0
1
2
3
4
5
This was a while loop:
1 i = 0
2 numbers = []
3
4 while i < 6:
5 print "At the top i is %d" % i
6 numbers.append(i)
7
8 i = i + 1
9 print "Numbers now: ", numbers
10 print "At the bottom i is %d" % i
11
12
13 print "The numbers: "
14
15 for num in numbers:
16 print num
Now Mr Shaw asks me to build this into a function. His exact question is: Convert this while- loop to a function that you can call, and replace 6 in the test (i < 6) with a variable. I am not sure I have build a function (since those mostly start with a def? Or do they always start with a def?)
I am not sure if I understand completely what he is asking, but this is what I did:
i = 0
numbers = []
for i in range (0, 6):
print "At the top i is %d" % i
numbers.append(i)
i = i + 1
print "Numbers now: ", numbers
print "At the bottom i is %d" % i
print "The numbers: "
for num in numbers:
print num
Which actually makes me kind of proud for how for I have gotten with this:
At the top i is 0
Numbers now: [0]
At the bottom i is 1
The numbers:
0
At the top i is 1
Numbers now: [0, 1]
At the bottom i is 2
The numbers:
0
1
At the top i is 2
Numbers now: [0, 1, 2]
At the bottom i is 3
The numbers:
0
1
2
At the top i is 3
Numbers now: [0, 1, 2, 3]
At the bottom i is 4
The numbers:
0
1
2
3
At the top i is 4
Numbers now: [0, 1, 2, 3, 4]
At the bottom i is 5
The numbers:
0
1
2
3
4
At the top i is 5
Numbers now: [0, 1, 2, 3, 4, 5]
At the bottom i is 6
The numbers:
0
1
2
3
4
5
Where I am stuck is that "The numbers" is something I only need to get at the end. And that, I can't get my head around. What am I missing (except amazing coding skills and a well functioning brain)?
You have not made a function. While it's not always necessary to use def to make a function, that's the usual way. And I think you've misunderstood what your teacher wanted you to change in your while loop. He wanted you to replace the 6 in the condition i < 6 with a variable (a parameter of the function). You've replaced the while with a for, but still kept the 6 as a constant. Using for loops instead of while loops is often a good idea, but in this case it's now what was being asked for.
As for why your numbers output is being repeated, I suspect it's because that code is inside your main loop now, while it wasn't before. This is not entirely clear though, since you seem to have lost the indentation of the original code when you copied it into Stack Overflow. Since indentation is significant in Python, this makes it hard to know exactly the old code ran.
what you are missing is that in python block of code are defined by the indentation level, you put the part that print the whole list inside the for that fill the list so each time that you put a number in the list also print if afterward. To put that outside reduce the indentation level of that part like this
i = 0
numbers = []
for i in range (0, 6):
print "At the top i is %d" % i
numbers.append(i)
i = i + 1
print "Numbers now: ", numbers
print "At the bottom i is %d" % i
#this part is now outside the for-loop
print "The numbers: "
for num in numbers:
print num
About your other question, yes functions are defined by the key world def you can check the documentation about the details, but in your case you can transform you code into a function very easy like this
def my_function():
i = 0
numbers = []
for i in range (0, 6):
print "At the top i is %d" % i
numbers.append(i)
i = i + 1
print "Numbers now: ", numbers
print "At the bottom i is %d" % i
print "The numbers: "
for num in numbers:
print num
that is, declare that you are creating a new function with the def key world, give it a name, in this case I called my_function, and put your previous code inside it by just give it a extra indentation level as show above.
Functions have the property that they can take a number of arguments and operate according to that, that let you define some behavior and make it more general.
For example: say that you want to print the number 0 to 9 you can do
for num in range(0,10):
print num
Now say that you want to print the number 0-5, 0-20, and 10-20 you can write a similar code for all of them:
for num in range(0,6):
print num
for num in range(0,21):
print num
for num in range(10,21):
print num
but wait there is pattern here all have the same exact code in every case save for the arguments of range, here is when a function come to play, we can define a function that take as arguments 2 numbers and our function do the job of print all the number in between; that is something like this
def my_function(star,stop):
for num in range(star,stop+1): # the +1 is to include the stop number
print num
(this is fun_test.py in the example below)
then we can call our function like this
my_function(0,5)
my_function(0,20)
my_function(10,20)
or if we open the file in a python interpreter or in interactive mode (that is $ python -i ex33.py) we can call with any pair of numbers of our desire
$ python -i fun_test.py
>>> my_function(8,17)
8
9
10
11
12
13
14
15
16
17
>>> my_function(0,5)
0
1
2
3
4
5
>>>
as you can see a function let us re-use a piece of code as many times as we want. The arguments of a function is the part of it that is variable while its code is the behavior we want according to the variable part (if any).
With this little explanation I think that you can modify your function to the requirements that your teacher (?) ask you
Related
from __future__ import print_function
i = 0
for i in range (0, 100)
print(i, end = " ")
i += 1
if i % 10 == 0
print(" ")
print()
I just start my python learning several days ago.
It is about the for_loop training.
As my codes tell, it should print out numbers from 0 to 98 increasing by 2 till 98. As 10 digits printed out, it should print another 10 digits in the next line and keep going on. But it do not work properly. However, it works from changing function to
range(0,100) `
range(0,100)
But do not work when I change the function to
range(0, 100, 2)`
range(0, 100, 2)
Hint: When you use range(0, 100, 2), i is always even and i += 1 would make it odd, hence never satisfying i % 10 == 0. Try using different variable for digit counter or modifying the operands for addition and modulus.
sticks = int(raw_input());
stickList= map(int,raw_input().split()) ;
stickList = sorted(stickList);
for i in xrange(0,len(stickList)):
stickList[i] = stickList[i]-stickList[0];
print stickList;
Given Input is :
6
5 4 4 2 2 8
Why the output is this: [0, 2, 4, 4, 5, 8]
instead of [0,0,2,2,3,6]
That is because you are changing the value in source stickList in for loop.
After first iteration in loop stickList[0] will become 0 for remaining iterations.
As ShadowRanger mentioned reversed list will do,
stickList = map(int, "5 4 4 2 2 8".split())
stickList.sort()
for i in reversed(xrange(len(stickList))):
stickList[i] -= stickList[0]
print stickList
Given a string of digits, I wish to find the number of ways of breaking up the string into individual numbers so that each number is under 26.
For example, "8888888" can only be broken up as "8 8 8 8 8 8 8". Whereas "1234567" can be broken up as "1 2 3 4 5 6 7", "12 3 4 5 6 7" and "1 23 4 5 6 7".
I'd like both a recurrence relation for the solution, and some code that uses dynamic programming.
This is what I've got so far. It only covers the base cases which are a empty string should return 1 a string of one digit should return 1 and a string of all numbers larger than 2 should return 1.
int countPerms(vector<int> number, int currentPermCount)
{
vector< vector<int> > permsOfNumber;
vector<int> working;
int totalPerms=0, size=number.size();
bool areAllOverTwo=true, forLoop = true;
if (number.size() <=1)
{
//TODO: print out permetations
return 1;
}
for (int i = 0; i < number.size()-1; i++) //minus one here because we dont care what the last digit is if all of them before it are over 2 then there is only one way to decode them
{
if (number.at(i) <= 2)
{
areAllOverTwo = false;
}
}
if (areAllOverTwo) //if all the nubmers are over 2 then there is only one possable combination 3456676546 has only one combination.
{
permsOfNumber.push_back(number);
//TODO: write function to print out the permetions
return 1;
}
do
{
//TODO find all the peremtions here
} while (forLoop);
return totalPerms;
}
Assuming you either don't have zeros, or you disallow numbers with leading zeros), the recurrence relations are:
N(1aS) = N(S) + N(aS)
N(2aS) = N(S) + N(aS) if a < 6.
N(a) = 1
N(aS) = N(S) otherwise
Here, a refers to a single digit, and S to a number. The first line of the recurrence relation says that if your string starts with a 1, then you can either have it on its own, or join it with the next digit. The second line says that if you start with a 2 you can either have it on its own, or join it with the next digit assuming that gives a number less than 26. The third line is the termination condition: when you're down to 1 digit, the result is 1. The final line says if you haven't been able to match one of the previous rules, then the first digit can't be joined to the second, so it must stand on its own.
The recurrence relations can be implemented fairly directly as an iterative dynamic programming solution. Here's code in Python, but it's easy to translate into other languages.
def N(S):
a1, a2 = 1, 1
for i in xrange(len(S) - 2, -1, -1):
if S[i] == '1' or S[i] == '2' and S[i+1] < '6':
a1, a2 = a1 + a2, a1
else:
a1, a2 = a1, a1
return a1
print N('88888888')
print N('12345678')
Output:
1
3
An interesting observation is that N('1' * n) is the n+1'st fibonacci number:
for i in xrange(1, 20):
print i, N('1' * i)
Output:
1 1
2 2
3 3
4 5
5 8
6 13
7 21
8 34
9 55
If I understand correctly, there are only 25 possibilities. My first crack at this would be to initialize an array of 25 ints all to zero and when I find a number less than 25, set that index to 1. Then I would count up all the 1's in the array when I was finished looking at the string.
What do you mean by recurrence? If you're looking for a recursive function, you would need to find a good way to break the string of numbers down recursively. I'm not sure that's the best approach here. I would just go through digit by digit and as you said if the digit is 2 or less, then store it and test appending the next digit... i.e. 10*digit + next. I hope that helped! Good luck.
Another way to think about it is that, after the initial single digit possibility, for every sequence of contiguous possible pairs of digits (e.g., 111 or 12223) of length n we multiply the result by:
1 + sum, i=1 to floor (n/2), of (n-i) choose i
For example, with a sequence of 11111, we can have
i=1, 1 1 1 11 => 5 - 1 = 4 choose 1 (possibilities with one pair)
i=2, 1 11 11 => 5 - 2 = 3 choose 2 (possibilities with two pairs)
This seems directly related to Wikipedia's description of Fibonacci numbers' "Use in Mathematics," for example, in counting "the number of compositions of 1s and 2s that sum to a given total n" (http://en.wikipedia.org/wiki/Fibonacci_number).
Using the combinatorial method (or other fast Fibonacci's) could be suitable for strings with very long sequences.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Generating a sequence using prime numbers 2, 3, and 5 only, and then displaying an nth term (C++)
I've been brainstorming over this forever, and I just can't figure this out. I need to solve the following problem:
Generate the following sequence and display the nth term in the
sequence
2,3,4,5,6,8,9,10,12,15, etc..... Sequence only has Prime numbers
2,3,5
I need to use basic C++, such as while, for, if, etc. Nothing fancy. I can't use arrays simply because I don't know much about them yet, and I want to understand the code for the solution.
I'm not asking for a complete solution, but I am asking for guidance to get through this... please.
My problem is that I can't figure out how to check if the number if the number in the sequence is divisible by any other prime numbers other than 2, 3, and 5.
Also let's say I'm checking the number like this:
for(int i=2; i<n; i++){
if(i%2==0){
cout<<i<<", ";
}else if(i%3==0){
cout<<i<<", ";
}else if(i%5==0){
cout<<i<<", ";
}
It doesn't work simply due to the fact that it'll produce numbers such as 14, which can be divided by prime number 7. So I need to figure out how to ensure that that sequence is only divisible by 2, 3, and 5..... I've found lots of material online with solutions for the problem, but the solutions they have are far too advance, and I can't use them (also most of them are in other languages... not C++). I'm sure there's a simpler way.
The problem with your code is that you just check one of the prime factors, not all of them.
Take your example of 14. Your code only checks if 2,3 or 5 is a factor of 14, which is not exactly what you need. Indeed, you find that 2 is a factor of 14, but the other factor is 7, as you said. What you are missing is to further check if 7 has as only factors 2,3 and 5 (which is not the case). What you need to do is to eliminate all the factors 2,3 and 5 and see what is remaining.
Let's take two examples: 60 and 42
For 60
Start with factors 2
60 % 2 = 0, so now check 60 / 2 = 30.
30 % 2 = 0, so now check 30 / 2 = 15.
15 % 2 = 1, so no more factors of 2.
Go on with factors 3
15 % 3 = 0, so now check 15 / 3 = 5.
5 % 3 = 2, so no more factors of 3.
Finish with factors 5
5 % 5 = 0, so now check 5 / 5 = 1
1 % 5 = 1, so no more factors of 5.
We end up with 1, so this number is part of the sequence.
For 42
Again, start with factors 2
42 % 2 = 0, so now check 42 / 2 = 21.
21 % 2 = 1, so no more factors of 2.
Go on with factors 3
21 % 3 = 0, so now check 21 / 3 = 7.
7 % 3 = 1, so no more factors of 3.
Finish with factors 5
7 % 5 = 2, so no more factors of 5.
We end up with 7 (something different from 1), so this number is NOT part of the sequence.
So in your implementation, you should probably nest 3 while loops in your for loop to reflect this reasoning.
Store the next i value in temporary variable and then divide it by 2 as long as you can (eg. as long as i%2 == 0). Then divide by 3 as long as you can. Then by 5. And then check, what is left.
What about this?
bool try_hamming(int n)
{
while(n%2 == 0)
{
n = n/2;
}
while(n%3 == 0)
{
n = n/3;
}
while(n%5 == 0)
{
n = n/5;
}
return n==1;
}
This should return true when n is a hamming nummer and false other wise. So the main function could look something like this
#include<iostream>
using namespace std;
main()
{
for(int i=2;i<100;++i)
{
if(try_hamming(i) )
cout<< i <<",";
}
cout<<end;
}
this schould print out all Hamming numbers less then 100
How to effectively generate permutations of a number (or chars in word), if i need some char/digit on specified place?
e.g. Generate all numbers with digit 3 at second place from the beginning and digit 1 at second place from the end of the number. Each digit in number has to be unique and you can choose only from digits 1-5.
4 3 2 1 5
4 3 5 1 2
2 3 4 1 5
2 3 5 1 4
5 3 2 1 4
5 3 4 1 2
I know there's a next_permutation function, so i can prepare an array with numbers {4, 2, 5} and post this in cycle to this function, but how to handle the fixed positions?
Generate all permutations of 2 4 5 and insert 3 and 1 in your output routine. Just remember the positions were they have to be:
int perm[3] = {2, 4, 5};
const int N = sizeof(perm) / sizeof(int);
std::map<int,int> fixed; // note: zero-indexed
fixed[1] = 3;
fixed[3] = 1;
do {
for (int i=0, j=0; i<5; i++)
if (fixed.find(i) != fixed.end())
std::cout << " " << fixed[i];
else
std::cout << " " << perm[j++];
std::cout << std::endl;
} while (std::next_permutation(perm, perm + N));
outputs
2 3 4 1 5
2 3 5 1 4
4 3 2 1 5
4 3 5 1 2
5 3 2 1 4
5 3 4 1 2
I've read the other answers and I believe they are better than mine for your specific problem. However I'm answering in case someone needs a generalized solution to your problem.
I recently needed to generate all permutations of the 3 separate continuous ranges [first1, last1) + [first2, last2) + [first3, last3). This corresponds to your case with all three ranges being of length 1 and separated by only 1 element. In my case the only restriction is that distance(first3, last3) >= distance(first1, last1) + distance(first2, last2) (which I'm sure could be relaxed with more computational expense).
My application was to generate each unique permutation but not its reverse. The code is here:
http://howardhinnant.github.io/combinations.html
And the specific applicable function is combine_discontinuous3 (which creates combinations), and its use in reversible_permutation::operator() which creates the permutations.
This isn't a ready-made packaged solution to your problem. But it is a tool set that could be used to solve generalizations of your problem. Again, for your exact simple problem, I recommend the simpler solutions others have already offered.
Remember at which places you want your fixed numbers. Remove them from the array.
Generate permutations as usual. After every permutation, insert your fixed numbers to the spots where they should appear, and output.
If you have a set of digits {4,3,2,1,5} and you know that 3 and 1 will not be permutated, then you can take them out of the set and just generate a powerset for {4, 2, 5}. All you have to do after that is just insert 1 and 3 in their respective positions for each set in the power set.
I posted a similar question and in there you can see the code for a powerset.