How many different ways do we have to group them? [duplicate]












-1
















This question already has an answer here:




  • Partition of an Integer + Number of partitions

    3 answers




I want to group them in all possible ways.



For example: If we have 5 stones, we will have 7 different ways to group them.



X   X   X   X   X

XX   X   X   X

XXX   X   X

XXX   XX

XX   XX   X

XXXX   X

XXXXX


I've been trying to get the algorithm out all day but I can not get it, do you have any idea how to do this?



Thanks






algorithm math







share|improve this question













marked as duplicate by Rory Daulton, James K Polk, Mark Dickinson, Community♦ Nov 17 '18 at 23:10


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.



















  • Can you show some attempts and what was not working with these?

    – Willem Van Onsem
    Nov 17 '18 at 19:43






  • 1





    You can here use a form of dynamic programming for this.

    – Willem Van Onsem
    Nov 17 '18 at 19:43











  • This question has been asked here many times before, usually in a different context. Look at this web site for more on this topic of partitions.

    – Rory Daulton
    Nov 17 '18 at 20:03











  • Yes, I currently have a formula that works for me in some cases but not very large numbers. The formula that I am implementing recursively is import math n = 10 p = [0,1,2] for i in range (3, n + 1): p.append ((sum (p) + 1-2) / 2 + 2) print (p) The idea is to add the previous possibilities and add the current one.

    – Akai shur
    Nov 17 '18 at 20:04
















-1
















This question already has an answer here:




  • Partition of an Integer + Number of partitions

    3 answers




I want to group them in all possible ways.



For example: If we have 5 stones, we will have 7 different ways to group them.



X   X   X   X   X

XX   X   X   X

XXX   X   X

XXX   XX

XX   XX   X

XXXX   X

XXXXX


I've been trying to get the algorithm out all day but I can not get it, do you have any idea how to do this?



Thanks






algorithm math







share|improve this question













marked as duplicate by Rory Daulton, James K Polk, Mark Dickinson, Community♦ Nov 17 '18 at 23:10


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.



















  • Can you show some attempts and what was not working with these?

    – Willem Van Onsem
    Nov 17 '18 at 19:43






  • 1





    You can here use a form of dynamic programming for this.

    – Willem Van Onsem
    Nov 17 '18 at 19:43











  • This question has been asked here many times before, usually in a different context. Look at this web site for more on this topic of partitions.

    – Rory Daulton
    Nov 17 '18 at 20:03











  • Yes, I currently have a formula that works for me in some cases but not very large numbers. The formula that I am implementing recursively is import math n = 10 p = [0,1,2] for i in range (3, n + 1): p.append ((sum (p) + 1-2) / 2 + 2) print (p) The idea is to add the previous possibilities and add the current one.

    – Akai shur
    Nov 17 '18 at 20:04














-1












-1








-1









This question already has an answer here:




  • Partition of an Integer + Number of partitions

    3 answers




I want to group them in all possible ways.



For example: If we have 5 stones, we will have 7 different ways to group them.



X   X   X   X   X

XX   X   X   X

XXX   X   X

XXX   XX

XX   XX   X

XXXX   X

XXXXX


I've been trying to get the algorithm out all day but I can not get it, do you have any idea how to do this?



Thanks






algorithm math







share|improve this question















This question already has an answer here:




  • Partition of an Integer + Number of partitions

    3 answers




I want to group them in all possible ways.



For example: If we have 5 stones, we will have 7 different ways to group them.



X   X   X   X   X

XX   X   X   X

XXX   X   X

XXX   XX

XX   XX   X

XXXX   X

XXXXX


I've been trying to get the algorithm out all day but I can not get it, do you have any idea how to do this?



Thanks





This question already has an answer here:




  • Partition of an Integer + Number of partitions

    3 answers







algorithm math




algorithm math






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 17 '18 at 19:41









Akai shurAkai shur

6910




6910




marked as duplicate by Rory Daulton, James K Polk, Mark Dickinson, Community♦ Nov 17 '18 at 23:10


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Rory Daulton, James K Polk, Mark Dickinson, Community♦ Nov 17 '18 at 23:10


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • Can you show some attempts and what was not working with these?

    – Willem Van Onsem
    Nov 17 '18 at 19:43






  • 1





    You can here use a form of dynamic programming for this.

    – Willem Van Onsem
    Nov 17 '18 at 19:43











  • This question has been asked here many times before, usually in a different context. Look at this web site for more on this topic of partitions.

    – Rory Daulton
    Nov 17 '18 at 20:03











  • Yes, I currently have a formula that works for me in some cases but not very large numbers. The formula that I am implementing recursively is import math n = 10 p = [0,1,2] for i in range (3, n + 1): p.append ((sum (p) + 1-2) / 2 + 2) print (p) The idea is to add the previous possibilities and add the current one.

    – Akai shur
    Nov 17 '18 at 20:04



















  • Can you show some attempts and what was not working with these?

    – Willem Van Onsem
    Nov 17 '18 at 19:43






  • 1





    You can here use a form of dynamic programming for this.

    – Willem Van Onsem
    Nov 17 '18 at 19:43











  • This question has been asked here many times before, usually in a different context. Look at this web site for more on this topic of partitions.

    – Rory Daulton
    Nov 17 '18 at 20:03











  • Yes, I currently have a formula that works for me in some cases but not very large numbers. The formula that I am implementing recursively is import math n = 10 p = [0,1,2] for i in range (3, n + 1): p.append ((sum (p) + 1-2) / 2 + 2) print (p) The idea is to add the previous possibilities and add the current one.

    – Akai shur
    Nov 17 '18 at 20:04

















Can you show some attempts and what was not working with these?

– Willem Van Onsem
Nov 17 '18 at 19:43





Can you show some attempts and what was not working with these?

– Willem Van Onsem
Nov 17 '18 at 19:43




1




1





You can here use a form of dynamic programming for this.

– Willem Van Onsem
Nov 17 '18 at 19:43





You can here use a form of dynamic programming for this.

– Willem Van Onsem
Nov 17 '18 at 19:43













This question has been asked here many times before, usually in a different context. Look at this web site for more on this topic of partitions.

– Rory Daulton
Nov 17 '18 at 20:03





This question has been asked here many times before, usually in a different context. Look at this web site for more on this topic of partitions.

– Rory Daulton
Nov 17 '18 at 20:03













Yes, I currently have a formula that works for me in some cases but not very large numbers. The formula that I am implementing recursively is import math n = 10 p = [0,1,2] for i in range (3, n + 1): p.append ((sum (p) + 1-2) / 2 + 2) print (p) The idea is to add the previous possibilities and add the current one.

– Akai shur
Nov 17 '18 at 20:04





Yes, I currently have a formula that works for me in some cases but not very large numbers. The formula that I am implementing recursively is import math n = 10 p = [0,1,2] for i in range (3, n + 1): p.append ((sum (p) + 1-2) / 2 + 2) print (p) The idea is to add the previous possibilities and add the current one.

– Akai shur
Nov 17 '18 at 20:04












1 Answer
1






active

oldest

votes


















2














This is equivalent to finding integer partitions of the number of stones you have. For example:



5 = 1 + 1 + 1 + 1 + 1    -->     X     X   X   X   X

5 = 2 + 1 + 1 + 1        -->     XX    X   X   X

5 = 3 + 1 + 1            -->     XXX   X   X

5 = 2 + 2 + 1            -->     XX    XX  X

5 = 4 + 1                -->     XXXX  X

5 = 3 + 2                -->     XXX   XX

5 = 5                    -->     XXXXX


Each line encodes a possible group configuration, where each integer on the right hand side represents the number of stones in a group.






share|improve this answer
























  • Or the stars and bars problem: en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

    – slider
    Nov 17 '18 at 20:07








  • 1





    @slider They're slightly different: in this case we're distributing identical objects into identical bins, whereas stars+bars is distinct bins.

    – arshajii
    Nov 17 '18 at 20:08











  • Agreed. Although you just need to account for the permutations of the bins.

    – slider
    Nov 17 '18 at 20:12











  • @slider: "just need to account" there is what we use to call "hand-waving". It's not nearly so simple.

    – rici
    Nov 18 '18 at 2:17













  • @rici "It's obvious". Ok I may have slightly underestimated the difficulty.

    – slider
    Nov 18 '18 at 3:23


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














This is equivalent to finding integer partitions of the number of stones you have. For example:



5 = 1 + 1 + 1 + 1 + 1    -->     X     X   X   X   X

5 = 2 + 1 + 1 + 1        -->     XX    X   X   X

5 = 3 + 1 + 1            -->     XXX   X   X

5 = 2 + 2 + 1            -->     XX    XX  X

5 = 4 + 1                -->     XXXX  X

5 = 3 + 2                -->     XXX   XX

5 = 5                    -->     XXXXX


Each line encodes a possible group configuration, where each integer on the right hand side represents the number of stones in a group.






share|improve this answer
























  • Or the stars and bars problem: en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

    – slider
    Nov 17 '18 at 20:07








  • 1





    @slider They're slightly different: in this case we're distributing identical objects into identical bins, whereas stars+bars is distinct bins.

    – arshajii
    Nov 17 '18 at 20:08











  • Agreed. Although you just need to account for the permutations of the bins.

    – slider
    Nov 17 '18 at 20:12











  • @slider: "just need to account" there is what we use to call "hand-waving". It's not nearly so simple.

    – rici
    Nov 18 '18 at 2:17













  • @rici "It's obvious". Ok I may have slightly underestimated the difficulty.

    – slider
    Nov 18 '18 at 3:23
















2














This is equivalent to finding integer partitions of the number of stones you have. For example:



5 = 1 + 1 + 1 + 1 + 1    -->     X     X   X   X   X

5 = 2 + 1 + 1 + 1        -->     XX    X   X   X

5 = 3 + 1 + 1            -->     XXX   X   X

5 = 2 + 2 + 1            -->     XX    XX  X

5 = 4 + 1                -->     XXXX  X

5 = 3 + 2                -->     XXX   XX

5 = 5                    -->     XXXXX


Each line encodes a possible group configuration, where each integer on the right hand side represents the number of stones in a group.






share|improve this answer
























  • Or the stars and bars problem: en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

    – slider
    Nov 17 '18 at 20:07








  • 1





    @slider They're slightly different: in this case we're distributing identical objects into identical bins, whereas stars+bars is distinct bins.

    – arshajii
    Nov 17 '18 at 20:08











  • Agreed. Although you just need to account for the permutations of the bins.

    – slider
    Nov 17 '18 at 20:12











  • @slider: "just need to account" there is what we use to call "hand-waving". It's not nearly so simple.

    – rici
    Nov 18 '18 at 2:17













  • @rici "It's obvious". Ok I may have slightly underestimated the difficulty.

    – slider
    Nov 18 '18 at 3:23














2












2








2







This is equivalent to finding integer partitions of the number of stones you have. For example:



5 = 1 + 1 + 1 + 1 + 1    -->     X     X   X   X   X

5 = 2 + 1 + 1 + 1        -->     XX    X   X   X

5 = 3 + 1 + 1            -->     XXX   X   X

5 = 2 + 2 + 1            -->     XX    XX  X

5 = 4 + 1                -->     XXXX  X

5 = 3 + 2                -->     XXX   XX

5 = 5                    -->     XXXXX


Each line encodes a possible group configuration, where each integer on the right hand side represents the number of stones in a group.






share|improve this answer













This is equivalent to finding integer partitions of the number of stones you have. For example:



5 = 1 + 1 + 1 + 1 + 1    -->     X     X   X   X   X

5 = 2 + 1 + 1 + 1        -->     XX    X   X   X

5 = 3 + 1 + 1            -->     XXX   X   X

5 = 2 + 2 + 1            -->     XX    XX  X

5 = 4 + 1                -->     XXXX  X

5 = 3 + 2                -->     XXX   XX

5 = 5                    -->     XXXXX


Each line encodes a possible group configuration, where each integer on the right hand side represents the number of stones in a group.







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 17 '18 at 20:05









arshajiiarshajii

101k18182250




101k18182250













  • Or the stars and bars problem: en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

    – slider
    Nov 17 '18 at 20:07








  • 1





    @slider They're slightly different: in this case we're distributing identical objects into identical bins, whereas stars+bars is distinct bins.

    – arshajii
    Nov 17 '18 at 20:08











  • Agreed. Although you just need to account for the permutations of the bins.

    – slider
    Nov 17 '18 at 20:12











  • @slider: "just need to account" there is what we use to call "hand-waving". It's not nearly so simple.

    – rici
    Nov 18 '18 at 2:17













  • @rici "It's obvious". Ok I may have slightly underestimated the difficulty.

    – slider
    Nov 18 '18 at 3:23



















  • Or the stars and bars problem: en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

    – slider
    Nov 17 '18 at 20:07








  • 1





    @slider They're slightly different: in this case we're distributing identical objects into identical bins, whereas stars+bars is distinct bins.

    – arshajii
    Nov 17 '18 at 20:08











  • Agreed. Although you just need to account for the permutations of the bins.

    – slider
    Nov 17 '18 at 20:12











  • @slider: "just need to account" there is what we use to call "hand-waving". It's not nearly so simple.

    – rici
    Nov 18 '18 at 2:17













  • @rici "It's obvious". Ok I may have slightly underestimated the difficulty.

    – slider
    Nov 18 '18 at 3:23

















Or the stars and bars problem: en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

– slider
Nov 17 '18 at 20:07







Or the stars and bars problem: en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

– slider
Nov 17 '18 at 20:07






1




1





@slider They're slightly different: in this case we're distributing identical objects into identical bins, whereas stars+bars is distinct bins.

– arshajii
Nov 17 '18 at 20:08





@slider They're slightly different: in this case we're distributing identical objects into identical bins, whereas stars+bars is distinct bins.

– arshajii
Nov 17 '18 at 20:08













Agreed. Although you just need to account for the permutations of the bins.

– slider
Nov 17 '18 at 20:12





Agreed. Although you just need to account for the permutations of the bins.

– slider
Nov 17 '18 at 20:12













@slider: "just need to account" there is what we use to call "hand-waving". It's not nearly so simple.

– rici
Nov 18 '18 at 2:17







@slider: "just need to account" there is what we use to call "hand-waving". It's not nearly so simple.

– rici
Nov 18 '18 at 2:17















@rici "It's obvious". Ok I may have slightly underestimated the difficulty.

– slider
Nov 18 '18 at 3:23





@rici "It's obvious". Ok I may have slightly underestimated the difficulty.

– slider
Nov 18 '18 at 3:23



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