All possible ways to split a number
$begingroup$
I am trying to find a way to find (if it is possible) how many ways there are to split a number of n digits considering that the "splits" can occur everywhere and the subsets don't have to be the same length. So for example, if I have a 5 digits number "12345" I can split it into 16(?) ways:
1)12345
2)1-2-3-4-5
3)1-2345
4)12-345
5)123-45
6)1-23-45 etc.
So I am looking for all possible strategies to split the number. I am looking for both a formula and an algorithm. I am guessing that it is 2^(n-1) but I am not sure and I'd like to know why, if this is the case
combinatorics algorithms combinations
$endgroup$
add a comment |
$begingroup$
I am trying to find a way to find (if it is possible) how many ways there are to split a number of n digits considering that the "splits" can occur everywhere and the subsets don't have to be the same length. So for example, if I have a 5 digits number "12345" I can split it into 16(?) ways:
1)12345
2)1-2-3-4-5
3)1-2345
4)12-345
5)123-45
6)1-23-45 etc.
So I am looking for all possible strategies to split the number. I am looking for both a formula and an algorithm. I am guessing that it is 2^(n-1) but I am not sure and I'd like to know why, if this is the case
combinatorics algorithms combinations
$endgroup$
add a comment |
$begingroup$
I am trying to find a way to find (if it is possible) how many ways there are to split a number of n digits considering that the "splits" can occur everywhere and the subsets don't have to be the same length. So for example, if I have a 5 digits number "12345" I can split it into 16(?) ways:
1)12345
2)1-2-3-4-5
3)1-2345
4)12-345
5)123-45
6)1-23-45 etc.
So I am looking for all possible strategies to split the number. I am looking for both a formula and an algorithm. I am guessing that it is 2^(n-1) but I am not sure and I'd like to know why, if this is the case
combinatorics algorithms combinations
$endgroup$
I am trying to find a way to find (if it is possible) how many ways there are to split a number of n digits considering that the "splits" can occur everywhere and the subsets don't have to be the same length. So for example, if I have a 5 digits number "12345" I can split it into 16(?) ways:
1)12345
2)1-2-3-4-5
3)1-2345
4)12-345
5)123-45
6)1-23-45 etc.
So I am looking for all possible strategies to split the number. I am looking for both a formula and an algorithm. I am guessing that it is 2^(n-1) but I am not sure and I'd like to know why, if this is the case
combinatorics algorithms combinations
combinatorics algorithms combinations
asked Dec 8 '18 at 16:38
Craig MontevecchiCraig Montevecchi
447
447
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Your problem can be solved using permutation with repetition in general.
Permutations with repetition
Ordered arrangements of the elements of a set S of length n where repetition is allowed are called n-tuples, but have sometimes been referred to as permutations with repetition although they are not permutations in general. They are also called words over the alphabet S in some contexts. If the set S has k elements, the number of n-tuples over S is:
The general formula is $ k^n $ where $k$ is the number of elements of S and $n$ the length of the tuple to form.
There is no restriction on how often an element can appear in an n-tuple, but if restrictions are placed on how often an element can appear, this formula is no longer valid.
In particular your problem to split 5 figures is equivalent to arrange 4 elements ($n=4$) from a set S of 2 elements ($k=2$) i.e.
${"-", "Nothing"}$
in order to get 4 tuples;
$("-/Nothing","-/Nothing","-/Nothing","-/Nothing")$ where slash $/$ means logical operator "or".
Therefore you will get the amount that you have inferred above which:
${2^{4}} = 16$
possible splits.
An algorithm in python 3.5 to solve this problem
import itertools
def getSplits(myciphers):
comb=
for split in [p for p in itertools.product([0,1], repeat=4)]:
tsplit=
tsplit.append(myciphers[0])
for c,s in zip(myciphers[1:],split):
if s == 1:
tsplit.append("-")
tsplit.append(c)
comb.append("".join(tsplit))
return comb
# Test Case 1
myciphers="12345"
print(getSplits(myciphers))
# Test Case 2
myciphers="ABCD"
print(getSplits(myciphers))
The output of this program is as follows:
rcolomina@workstation-rig:~$ python boo.py
['12345', '1234-5', '123-45', '123-4-5', '12-345', '12-34-5', '12-3-45', '12-3-4-5', '1-2345', '1-234-5', '1-23-45', '1-23-4-5', '1-2-345', '1-2-34-5', '1-2-3-45', '1-2-3-4-5']
['ABCD', 'ABCD', 'ABC-D', 'ABC-D', 'AB-CD', 'AB-CD', 'AB-C-D', 'AB-C-D', 'A-BCD', 'A-BCD', 'A-BC-D', 'A-BC-D', 'A-B-CD', 'A-B-CD', 'A-B-C-D', 'A-B-C-D']
$endgroup$
add a comment |
$begingroup$
When you have a sequence of $n$ numbers, you can $n-1$ slots to make cuts. So we need to know all the ways we can choose $0$ cuts out if it, then $1$ cut, $2$ cuts, etc
The total number of ways to choose the cuts is $$sum_{k=0}^{n-1} { {n-1}choose{k}} =2^{n-1} $$
To prove this equality, use a subtle trick: Set $2^{n-1} = (1+1)^{n-1}$ and use the Binomial Theorem.
$endgroup$
add a comment |
$begingroup$
Look at the cuts as a binary number where each digit i of the number indicates if we cut between digit i and i+1 in the real number.
for example:
12345 would be 0000.
12-345 would be 0100.
Now,for a number of size n our binary number would be of size n-1,how many unique values can a n-1 digits binary number represent?
$$2^{(n-1)}$$.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
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active
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$begingroup$
Your problem can be solved using permutation with repetition in general.
Permutations with repetition
Ordered arrangements of the elements of a set S of length n where repetition is allowed are called n-tuples, but have sometimes been referred to as permutations with repetition although they are not permutations in general. They are also called words over the alphabet S in some contexts. If the set S has k elements, the number of n-tuples over S is:
The general formula is $ k^n $ where $k$ is the number of elements of S and $n$ the length of the tuple to form.
There is no restriction on how often an element can appear in an n-tuple, but if restrictions are placed on how often an element can appear, this formula is no longer valid.
In particular your problem to split 5 figures is equivalent to arrange 4 elements ($n=4$) from a set S of 2 elements ($k=2$) i.e.
${"-", "Nothing"}$
in order to get 4 tuples;
$("-/Nothing","-/Nothing","-/Nothing","-/Nothing")$ where slash $/$ means logical operator "or".
Therefore you will get the amount that you have inferred above which:
${2^{4}} = 16$
possible splits.
An algorithm in python 3.5 to solve this problem
import itertools
def getSplits(myciphers):
comb=
for split in [p for p in itertools.product([0,1], repeat=4)]:
tsplit=
tsplit.append(myciphers[0])
for c,s in zip(myciphers[1:],split):
if s == 1:
tsplit.append("-")
tsplit.append(c)
comb.append("".join(tsplit))
return comb
# Test Case 1
myciphers="12345"
print(getSplits(myciphers))
# Test Case 2
myciphers="ABCD"
print(getSplits(myciphers))
The output of this program is as follows:
rcolomina@workstation-rig:~$ python boo.py
['12345', '1234-5', '123-45', '123-4-5', '12-345', '12-34-5', '12-3-45', '12-3-4-5', '1-2345', '1-234-5', '1-23-45', '1-23-4-5', '1-2-345', '1-2-34-5', '1-2-3-45', '1-2-3-4-5']
['ABCD', 'ABCD', 'ABC-D', 'ABC-D', 'AB-CD', 'AB-CD', 'AB-C-D', 'AB-C-D', 'A-BCD', 'A-BCD', 'A-BC-D', 'A-BC-D', 'A-B-CD', 'A-B-CD', 'A-B-C-D', 'A-B-C-D']
$endgroup$
add a comment |
$begingroup$
Your problem can be solved using permutation with repetition in general.
Permutations with repetition
Ordered arrangements of the elements of a set S of length n where repetition is allowed are called n-tuples, but have sometimes been referred to as permutations with repetition although they are not permutations in general. They are also called words over the alphabet S in some contexts. If the set S has k elements, the number of n-tuples over S is:
The general formula is $ k^n $ where $k$ is the number of elements of S and $n$ the length of the tuple to form.
There is no restriction on how often an element can appear in an n-tuple, but if restrictions are placed on how often an element can appear, this formula is no longer valid.
In particular your problem to split 5 figures is equivalent to arrange 4 elements ($n=4$) from a set S of 2 elements ($k=2$) i.e.
${"-", "Nothing"}$
in order to get 4 tuples;
$("-/Nothing","-/Nothing","-/Nothing","-/Nothing")$ where slash $/$ means logical operator "or".
Therefore you will get the amount that you have inferred above which:
${2^{4}} = 16$
possible splits.
An algorithm in python 3.5 to solve this problem
import itertools
def getSplits(myciphers):
comb=
for split in [p for p in itertools.product([0,1], repeat=4)]:
tsplit=
tsplit.append(myciphers[0])
for c,s in zip(myciphers[1:],split):
if s == 1:
tsplit.append("-")
tsplit.append(c)
comb.append("".join(tsplit))
return comb
# Test Case 1
myciphers="12345"
print(getSplits(myciphers))
# Test Case 2
myciphers="ABCD"
print(getSplits(myciphers))
The output of this program is as follows:
rcolomina@workstation-rig:~$ python boo.py
['12345', '1234-5', '123-45', '123-4-5', '12-345', '12-34-5', '12-3-45', '12-3-4-5', '1-2345', '1-234-5', '1-23-45', '1-23-4-5', '1-2-345', '1-2-34-5', '1-2-3-45', '1-2-3-4-5']
['ABCD', 'ABCD', 'ABC-D', 'ABC-D', 'AB-CD', 'AB-CD', 'AB-C-D', 'AB-C-D', 'A-BCD', 'A-BCD', 'A-BC-D', 'A-BC-D', 'A-B-CD', 'A-B-CD', 'A-B-C-D', 'A-B-C-D']
$endgroup$
add a comment |
$begingroup$
Your problem can be solved using permutation with repetition in general.
Permutations with repetition
Ordered arrangements of the elements of a set S of length n where repetition is allowed are called n-tuples, but have sometimes been referred to as permutations with repetition although they are not permutations in general. They are also called words over the alphabet S in some contexts. If the set S has k elements, the number of n-tuples over S is:
The general formula is $ k^n $ where $k$ is the number of elements of S and $n$ the length of the tuple to form.
There is no restriction on how often an element can appear in an n-tuple, but if restrictions are placed on how often an element can appear, this formula is no longer valid.
In particular your problem to split 5 figures is equivalent to arrange 4 elements ($n=4$) from a set S of 2 elements ($k=2$) i.e.
${"-", "Nothing"}$
in order to get 4 tuples;
$("-/Nothing","-/Nothing","-/Nothing","-/Nothing")$ where slash $/$ means logical operator "or".
Therefore you will get the amount that you have inferred above which:
${2^{4}} = 16$
possible splits.
An algorithm in python 3.5 to solve this problem
import itertools
def getSplits(myciphers):
comb=
for split in [p for p in itertools.product([0,1], repeat=4)]:
tsplit=
tsplit.append(myciphers[0])
for c,s in zip(myciphers[1:],split):
if s == 1:
tsplit.append("-")
tsplit.append(c)
comb.append("".join(tsplit))
return comb
# Test Case 1
myciphers="12345"
print(getSplits(myciphers))
# Test Case 2
myciphers="ABCD"
print(getSplits(myciphers))
The output of this program is as follows:
rcolomina@workstation-rig:~$ python boo.py
['12345', '1234-5', '123-45', '123-4-5', '12-345', '12-34-5', '12-3-45', '12-3-4-5', '1-2345', '1-234-5', '1-23-45', '1-23-4-5', '1-2-345', '1-2-34-5', '1-2-3-45', '1-2-3-4-5']
['ABCD', 'ABCD', 'ABC-D', 'ABC-D', 'AB-CD', 'AB-CD', 'AB-C-D', 'AB-C-D', 'A-BCD', 'A-BCD', 'A-BC-D', 'A-BC-D', 'A-B-CD', 'A-B-CD', 'A-B-C-D', 'A-B-C-D']
$endgroup$
Your problem can be solved using permutation with repetition in general.
Permutations with repetition
Ordered arrangements of the elements of a set S of length n where repetition is allowed are called n-tuples, but have sometimes been referred to as permutations with repetition although they are not permutations in general. They are also called words over the alphabet S in some contexts. If the set S has k elements, the number of n-tuples over S is:
The general formula is $ k^n $ where $k$ is the number of elements of S and $n$ the length of the tuple to form.
There is no restriction on how often an element can appear in an n-tuple, but if restrictions are placed on how often an element can appear, this formula is no longer valid.
In particular your problem to split 5 figures is equivalent to arrange 4 elements ($n=4$) from a set S of 2 elements ($k=2$) i.e.
${"-", "Nothing"}$
in order to get 4 tuples;
$("-/Nothing","-/Nothing","-/Nothing","-/Nothing")$ where slash $/$ means logical operator "or".
Therefore you will get the amount that you have inferred above which:
${2^{4}} = 16$
possible splits.
An algorithm in python 3.5 to solve this problem
import itertools
def getSplits(myciphers):
comb=
for split in [p for p in itertools.product([0,1], repeat=4)]:
tsplit=
tsplit.append(myciphers[0])
for c,s in zip(myciphers[1:],split):
if s == 1:
tsplit.append("-")
tsplit.append(c)
comb.append("".join(tsplit))
return comb
# Test Case 1
myciphers="12345"
print(getSplits(myciphers))
# Test Case 2
myciphers="ABCD"
print(getSplits(myciphers))
The output of this program is as follows:
rcolomina@workstation-rig:~$ python boo.py
['12345', '1234-5', '123-45', '123-4-5', '12-345', '12-34-5', '12-3-45', '12-3-4-5', '1-2345', '1-234-5', '1-23-45', '1-23-4-5', '1-2-345', '1-2-34-5', '1-2-3-45', '1-2-3-4-5']
['ABCD', 'ABCD', 'ABC-D', 'ABC-D', 'AB-CD', 'AB-CD', 'AB-C-D', 'AB-C-D', 'A-BCD', 'A-BCD', 'A-BC-D', 'A-BC-D', 'A-B-CD', 'A-B-CD', 'A-B-C-D', 'A-B-C-D']
answered Dec 8 '18 at 17:32
Rubén ColominaRubén Colomina
436
436
add a comment |
add a comment |
$begingroup$
When you have a sequence of $n$ numbers, you can $n-1$ slots to make cuts. So we need to know all the ways we can choose $0$ cuts out if it, then $1$ cut, $2$ cuts, etc
The total number of ways to choose the cuts is $$sum_{k=0}^{n-1} { {n-1}choose{k}} =2^{n-1} $$
To prove this equality, use a subtle trick: Set $2^{n-1} = (1+1)^{n-1}$ and use the Binomial Theorem.
$endgroup$
add a comment |
$begingroup$
When you have a sequence of $n$ numbers, you can $n-1$ slots to make cuts. So we need to know all the ways we can choose $0$ cuts out if it, then $1$ cut, $2$ cuts, etc
The total number of ways to choose the cuts is $$sum_{k=0}^{n-1} { {n-1}choose{k}} =2^{n-1} $$
To prove this equality, use a subtle trick: Set $2^{n-1} = (1+1)^{n-1}$ and use the Binomial Theorem.
$endgroup$
add a comment |
$begingroup$
When you have a sequence of $n$ numbers, you can $n-1$ slots to make cuts. So we need to know all the ways we can choose $0$ cuts out if it, then $1$ cut, $2$ cuts, etc
The total number of ways to choose the cuts is $$sum_{k=0}^{n-1} { {n-1}choose{k}} =2^{n-1} $$
To prove this equality, use a subtle trick: Set $2^{n-1} = (1+1)^{n-1}$ and use the Binomial Theorem.
$endgroup$
When you have a sequence of $n$ numbers, you can $n-1$ slots to make cuts. So we need to know all the ways we can choose $0$ cuts out if it, then $1$ cut, $2$ cuts, etc
The total number of ways to choose the cuts is $$sum_{k=0}^{n-1} { {n-1}choose{k}} =2^{n-1} $$
To prove this equality, use a subtle trick: Set $2^{n-1} = (1+1)^{n-1}$ and use the Binomial Theorem.
answered Dec 8 '18 at 17:05
WaveXWaveX
2,6522722
2,6522722
add a comment |
add a comment |
$begingroup$
Look at the cuts as a binary number where each digit i of the number indicates if we cut between digit i and i+1 in the real number.
for example:
12345 would be 0000.
12-345 would be 0100.
Now,for a number of size n our binary number would be of size n-1,how many unique values can a n-1 digits binary number represent?
$$2^{(n-1)}$$.
$endgroup$
add a comment |
$begingroup$
Look at the cuts as a binary number where each digit i of the number indicates if we cut between digit i and i+1 in the real number.
for example:
12345 would be 0000.
12-345 would be 0100.
Now,for a number of size n our binary number would be of size n-1,how many unique values can a n-1 digits binary number represent?
$$2^{(n-1)}$$.
$endgroup$
add a comment |
$begingroup$
Look at the cuts as a binary number where each digit i of the number indicates if we cut between digit i and i+1 in the real number.
for example:
12345 would be 0000.
12-345 would be 0100.
Now,for a number of size n our binary number would be of size n-1,how many unique values can a n-1 digits binary number represent?
$$2^{(n-1)}$$.
$endgroup$
Look at the cuts as a binary number where each digit i of the number indicates if we cut between digit i and i+1 in the real number.
for example:
12345 would be 0000.
12-345 would be 0100.
Now,for a number of size n our binary number would be of size n-1,how many unique values can a n-1 digits binary number represent?
$$2^{(n-1)}$$.
answered Dec 8 '18 at 17:03
AdddisonAdddison
846
846
add a comment |
add a comment |
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