Stones game proof by induction












1














I'm struggling with proof by induction for the following game




Two players called P2 and P2 are playing a game with a starting number
of stones. Player always plays first, and the two players move in
alternating turns. The game's rules are as follows: In a single move,
a player can remove either 2, 3, or 5 stones from the game board. If a
player is unable to make a move, that player loses the game.




A player wins where the number of stones $n$ is $n≥1$, and $n mod 7 = 0$ or $n mod 7 = 1$.



For a proof by induction, I pick $n = 0$ as by base case (here P1 will win)



Do I need to have a second base case for $n = 1$, since if I choose that as the inductive step surely I'm just proving that $2 mod 7 = 1$ which is self-evident.



How do I express these base cases?



My attempt at the proof is as follows:



Base case.:
$n = 0$ (is a multiple of 7)
first player loses



Inductive case:
Assume the theorem is true for $n = k$.
Prove it is true for $n = k$




  1. Show it is true for $n = 1$
    $1 mod 7 = 1$ so p1 loses


  2. Assume it is true for $n = k$
    $n mod 7 == 0$ OR $n mod 7 == 1$ so p1 loses.



Clearly this is not correct!






induction







share|cite|improve this question
























  • refer math.meta.stackexchange.com/questions/5020/… for proper formatting
    – idea
    Nov 24 at 8:50
















1














I'm struggling with proof by induction for the following game




Two players called P2 and P2 are playing a game with a starting number
of stones. Player always plays first, and the two players move in
alternating turns. The game's rules are as follows: In a single move,
a player can remove either 2, 3, or 5 stones from the game board. If a
player is unable to make a move, that player loses the game.




A player wins where the number of stones $n$ is $n≥1$, and $n mod 7 = 0$ or $n mod 7 = 1$.



For a proof by induction, I pick $n = 0$ as by base case (here P1 will win)



Do I need to have a second base case for $n = 1$, since if I choose that as the inductive step surely I'm just proving that $2 mod 7 = 1$ which is self-evident.



How do I express these base cases?



My attempt at the proof is as follows:



Base case.:
$n = 0$ (is a multiple of 7)
first player loses



Inductive case:
Assume the theorem is true for $n = k$.
Prove it is true for $n = k$




  1. Show it is true for $n = 1$
    $1 mod 7 = 1$ so p1 loses


  2. Assume it is true for $n = k$
    $n mod 7 == 0$ OR $n mod 7 == 1$ so p1 loses.



Clearly this is not correct!






induction







share|cite|improve this question
























  • refer math.meta.stackexchange.com/questions/5020/… for proper formatting
    – idea
    Nov 24 at 8:50














1












1








1







I'm struggling with proof by induction for the following game




Two players called P2 and P2 are playing a game with a starting number
of stones. Player always plays first, and the two players move in
alternating turns. The game's rules are as follows: In a single move,
a player can remove either 2, 3, or 5 stones from the game board. If a
player is unable to make a move, that player loses the game.




A player wins where the number of stones $n$ is $n≥1$, and $n mod 7 = 0$ or $n mod 7 = 1$.



For a proof by induction, I pick $n = 0$ as by base case (here P1 will win)



Do I need to have a second base case for $n = 1$, since if I choose that as the inductive step surely I'm just proving that $2 mod 7 = 1$ which is self-evident.



How do I express these base cases?



My attempt at the proof is as follows:



Base case.:
$n = 0$ (is a multiple of 7)
first player loses



Inductive case:
Assume the theorem is true for $n = k$.
Prove it is true for $n = k$




  1. Show it is true for $n = 1$
    $1 mod 7 = 1$ so p1 loses


  2. Assume it is true for $n = k$
    $n mod 7 == 0$ OR $n mod 7 == 1$ so p1 loses.



Clearly this is not correct!






induction







share|cite|improve this question















I'm struggling with proof by induction for the following game




Two players called P2 and P2 are playing a game with a starting number
of stones. Player always plays first, and the two players move in
alternating turns. The game's rules are as follows: In a single move,
a player can remove either 2, 3, or 5 stones from the game board. If a
player is unable to make a move, that player loses the game.




A player wins where the number of stones $n$ is $n≥1$, and $n mod 7 = 0$ or $n mod 7 = 1$.



For a proof by induction, I pick $n = 0$ as by base case (here P1 will win)



Do I need to have a second base case for $n = 1$, since if I choose that as the inductive step surely I'm just proving that $2 mod 7 = 1$ which is self-evident.



How do I express these base cases?



My attempt at the proof is as follows:



Base case.:
$n = 0$ (is a multiple of 7)
first player loses



Inductive case:
Assume the theorem is true for $n = k$.
Prove it is true for $n = k$




  1. Show it is true for $n = 1$
    $1 mod 7 = 1$ so p1 loses


  2. Assume it is true for $n = k$
    $n mod 7 == 0$ OR $n mod 7 == 1$ so p1 loses.



Clearly this is not correct!






induction




induction






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 at 8:54

























asked Nov 24 at 8:44









stevenpcurtis

1062




1062












  • refer math.meta.stackexchange.com/questions/5020/… for proper formatting
    – idea
    Nov 24 at 8:50


















  • refer math.meta.stackexchange.com/questions/5020/… for proper formatting
    – idea
    Nov 24 at 8:50
















refer math.meta.stackexchange.com/questions/5020/… for proper formatting
– idea
Nov 24 at 8:50




refer math.meta.stackexchange.com/questions/5020/… for proper formatting
– idea
Nov 24 at 8:50










1 Answer
1






active

oldest

votes


















1














According to the general theory of such games the set ${mathbb N}_{geq0}$ of positions $n$ is the disjoint union $Wcup L$ of two sets, characterized as follows:




  • If $nin W$ then the next player can enforce a win;

  • If $nin L$ the next player cannot avoid a loss.


From these properties we can surmise that
$$eqalign{nin W&quadRightarrowquad {n-2,n-3,n-5} {rm contains an element of} L ,cr
nin L&quadRightarrowquad {n-2,n-3,n-5}subset W .cr}tag{1}$$
Contrary to what you are writing $0$ and $1$ are losing positions. It follows that ${2,3,4,5,6}subset W$, since from all of these numbers there is a legal move leading to $0$ or $1$. Going stepwise up using $(1)$ one arrives at the conjecture that all numbers $n=7j+k$, $>0leq kleq6$, with $kin{0,1}$ are in $L$, and all numbers $n=7j+k$ with $kin{2,3,4,5,6}$ are in $W$.



To prove this conjecture we do not use induction on $n$, but on $j$. For $j=0$ we have already shown that the claim is true. I leave it to you to perform the step $jrightsquigarrow j+1$.






share|cite|improve this answer





















  • Why would the proof not be for k = k + 1 after math.stackexchange.com/questions/1208641/…
    – stevenpcurtis
    Nov 24 at 13:37










  • Read carefully! Scott's $k$ is my $j$.
    – Christian Blatter
    Nov 24 at 13:55











Your Answer





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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1














According to the general theory of such games the set ${mathbb N}_{geq0}$ of positions $n$ is the disjoint union $Wcup L$ of two sets, characterized as follows:




  • If $nin W$ then the next player can enforce a win;

  • If $nin L$ the next player cannot avoid a loss.


From these properties we can surmise that
$$eqalign{nin W&quadRightarrowquad {n-2,n-3,n-5} {rm contains an element of} L ,cr
nin L&quadRightarrowquad {n-2,n-3,n-5}subset W .cr}tag{1}$$
Contrary to what you are writing $0$ and $1$ are losing positions. It follows that ${2,3,4,5,6}subset W$, since from all of these numbers there is a legal move leading to $0$ or $1$. Going stepwise up using $(1)$ one arrives at the conjecture that all numbers $n=7j+k$, $>0leq kleq6$, with $kin{0,1}$ are in $L$, and all numbers $n=7j+k$ with $kin{2,3,4,5,6}$ are in $W$.



To prove this conjecture we do not use induction on $n$, but on $j$. For $j=0$ we have already shown that the claim is true. I leave it to you to perform the step $jrightsquigarrow j+1$.






share|cite|improve this answer





















  • Why would the proof not be for k = k + 1 after math.stackexchange.com/questions/1208641/…
    – stevenpcurtis
    Nov 24 at 13:37










  • Read carefully! Scott's $k$ is my $j$.
    – Christian Blatter
    Nov 24 at 13:55
















1














According to the general theory of such games the set ${mathbb N}_{geq0}$ of positions $n$ is the disjoint union $Wcup L$ of two sets, characterized as follows:




  • If $nin W$ then the next player can enforce a win;

  • If $nin L$ the next player cannot avoid a loss.


From these properties we can surmise that
$$eqalign{nin W&quadRightarrowquad {n-2,n-3,n-5} {rm contains an element of} L ,cr
nin L&quadRightarrowquad {n-2,n-3,n-5}subset W .cr}tag{1}$$
Contrary to what you are writing $0$ and $1$ are losing positions. It follows that ${2,3,4,5,6}subset W$, since from all of these numbers there is a legal move leading to $0$ or $1$. Going stepwise up using $(1)$ one arrives at the conjecture that all numbers $n=7j+k$, $>0leq kleq6$, with $kin{0,1}$ are in $L$, and all numbers $n=7j+k$ with $kin{2,3,4,5,6}$ are in $W$.



To prove this conjecture we do not use induction on $n$, but on $j$. For $j=0$ we have already shown that the claim is true. I leave it to you to perform the step $jrightsquigarrow j+1$.






share|cite|improve this answer





















  • Why would the proof not be for k = k + 1 after math.stackexchange.com/questions/1208641/…
    – stevenpcurtis
    Nov 24 at 13:37










  • Read carefully! Scott's $k$ is my $j$.
    – Christian Blatter
    Nov 24 at 13:55














1












1








1






According to the general theory of such games the set ${mathbb N}_{geq0}$ of positions $n$ is the disjoint union $Wcup L$ of two sets, characterized as follows:




  • If $nin W$ then the next player can enforce a win;

  • If $nin L$ the next player cannot avoid a loss.


From these properties we can surmise that
$$eqalign{nin W&quadRightarrowquad {n-2,n-3,n-5} {rm contains an element of} L ,cr
nin L&quadRightarrowquad {n-2,n-3,n-5}subset W .cr}tag{1}$$
Contrary to what you are writing $0$ and $1$ are losing positions. It follows that ${2,3,4,5,6}subset W$, since from all of these numbers there is a legal move leading to $0$ or $1$. Going stepwise up using $(1)$ one arrives at the conjecture that all numbers $n=7j+k$, $>0leq kleq6$, with $kin{0,1}$ are in $L$, and all numbers $n=7j+k$ with $kin{2,3,4,5,6}$ are in $W$.



To prove this conjecture we do not use induction on $n$, but on $j$. For $j=0$ we have already shown that the claim is true. I leave it to you to perform the step $jrightsquigarrow j+1$.






share|cite|improve this answer












According to the general theory of such games the set ${mathbb N}_{geq0}$ of positions $n$ is the disjoint union $Wcup L$ of two sets, characterized as follows:




  • If $nin W$ then the next player can enforce a win;

  • If $nin L$ the next player cannot avoid a loss.


From these properties we can surmise that
$$eqalign{nin W&quadRightarrowquad {n-2,n-3,n-5} {rm contains an element of} L ,cr
nin L&quadRightarrowquad {n-2,n-3,n-5}subset W .cr}tag{1}$$
Contrary to what you are writing $0$ and $1$ are losing positions. It follows that ${2,3,4,5,6}subset W$, since from all of these numbers there is a legal move leading to $0$ or $1$. Going stepwise up using $(1)$ one arrives at the conjecture that all numbers $n=7j+k$, $>0leq kleq6$, with $kin{0,1}$ are in $L$, and all numbers $n=7j+k$ with $kin{2,3,4,5,6}$ are in $W$.



To prove this conjecture we do not use induction on $n$, but on $j$. For $j=0$ we have already shown that the claim is true. I leave it to you to perform the step $jrightsquigarrow j+1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 24 at 10:33









Christian Blatter

172k7112325




172k7112325












  • Why would the proof not be for k = k + 1 after math.stackexchange.com/questions/1208641/…
    – stevenpcurtis
    Nov 24 at 13:37










  • Read carefully! Scott's $k$ is my $j$.
    – Christian Blatter
    Nov 24 at 13:55


















  • Why would the proof not be for k = k + 1 after math.stackexchange.com/questions/1208641/…
    – stevenpcurtis
    Nov 24 at 13:37










  • Read carefully! Scott's $k$ is my $j$.
    – Christian Blatter
    Nov 24 at 13:55
















Why would the proof not be for k = k + 1 after math.stackexchange.com/questions/1208641/…
– stevenpcurtis
Nov 24 at 13:37




Why would the proof not be for k = k + 1 after math.stackexchange.com/questions/1208641/…
– stevenpcurtis
Nov 24 at 13:37












Read carefully! Scott's $k$ is my $j$.
– Christian Blatter
Nov 24 at 13:55




Read carefully! Scott's $k$ is my $j$.
– Christian Blatter
Nov 24 at 13:55


















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