Stones game proof by induction
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 $n = k$
Show it is true for $n = 1$
$1 mod 7 = 1$ so p1 losesAssume it is true for $n = k$
$n mod 7 == 0$ OR $n mod 7 == 1$ so p1 loses.
Clearly this is not correct!
induction
add a comment |
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 $n = k$
Show it is true for $n = 1$
$1 mod 7 = 1$ so p1 losesAssume it is true for $n = k$
$n mod 7 == 0$ OR $n mod 7 == 1$ so p1 loses.
Clearly this is not correct!
induction
refer math.meta.stackexchange.com/questions/5020/… for proper formatting
– idea
Nov 24 at 8:50
add a comment |
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 $n = k$
Show it is true for $n = 1$
$1 mod 7 = 1$ so p1 losesAssume it is true for $n = k$
$n mod 7 == 0$ OR $n mod 7 == 1$ so p1 loses.
Clearly this is not correct!
induction
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 $n = k$
Show it is true for $n = 1$
$1 mod 7 = 1$ so p1 losesAssume 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
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
add a comment |
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
add a comment |
1 Answer
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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 $jrightsquigarrow j+1$.
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
add a comment |
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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 $jrightsquigarrow j+1$.
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
add a comment |
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 $jrightsquigarrow j+1$.
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
add a comment |
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 $jrightsquigarrow j+1$.
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 $jrightsquigarrow j+1$.
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
add a comment |
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
add a comment |
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refer math.meta.stackexchange.com/questions/5020/… for proper formatting
– idea
Nov 24 at 8:50