• Nim: with 2 heaps

  • Nim: the game with decreasing numbers
  • property:
    • turn-based
    • finite
    • impartial
    • standard winning condition
  • if you can't make a move: you lose
    • 2 players
  • convert: game into a graph

• Conversion

  • each vertex: state
    • if a state $u$ can reach $v$, $u\to v$ state exist
  • the graph: a DAG
    • 👨‍🏫 as the number of states are finite, it
  • a state $u$: winning ($W$) if, starting from $u$, player 1 wins
  • a state $u$: losing ($L$) if, starting from $u$, player 1 wins
  • $W\cup L=V$
    • 👨‍🏫 there is no way that no one wins!

graph LR
  0((W))
  1((W))
  2((L))
  3((L))
  4((W))
  5((W))
  6((L))
  0-->1; 0-->3;
  1-->2; 1-->5;
  3-->1; 3-->5;
  4-->0; 4-->3; 4-->5;
  5-->2; 5-->6;

• rules
  • rule 0: if $v$: no outgoing edges
    • then $v\in L$
  • rule 1: if $\exists u$ s.t. $(v,u)\in E$ and $u\in L$, then $v\in W$
    • optimal players!
  • rule 2: if $\forall u$ s.t. $(v,u)\in E⟹u\in W$, then $v\in L$
• and: apply rules only if all its successors are covered
• aka topological ordering: given a DAG $G=(V,E)$, topological order:
  • permutation $\pi$ of vertices s.t. $\forall e=(u,v)$ then $u$ appears before $v$
  • 👨‍🏫 in our case, we want the reverse: thus, just inverse the order
• theorem: every DAG has a topological order
• $L=\{(x,x)|x\in \mathbb{N}\}$
• $W=\{(x,y)|x\ne y\}$

• Nim with $n$ heaps

  • with $n$ natural numbers: $a_1,a_2,\dots ,a_n$
    • each player: choosing a no. and decrease it in their turn
  • theorem: state of winning: depends on $\oplus$ of all $a_i$
    • $L=\{(a_1,a_2,\dots ,a_n)|{⨁}_{i=1}^n{a}_i=0\}$
      • intuition: you lose when $\forall i,a_i=0$ too
    • $W=\{(a_1,a_2,\dots ,a_n)|{⨁}_{i=1}^n{a}_i\ne 0\}$
  • suppose: ${⨁}_{i=1}^n{a}_i=0$
    • then player: changes $a_j$ to $a_j^{\prime }$
    • e.g. initially: $a_1\oplus a_2\oplus a_3=0$
      • then $a_1\oplus a_2=a_3$
    • similarly, ${⨁}_{i=1}^n{a}_i=0$ means:
      • ${⨁}_{i=1,i\ne j}^n{a}_i=a_j$
  • yet: showing existence of move leading $(a_1,\dots ,a_n)\in W$ to win is hard
    • for ${⨁}_{i=1}^n{a}_i\ne 0$, there exists index $j$ s.t.
    • $({⨁}_{i=1,i\ne j}^n{a}_i)\oplus a_j^{\prime }=0$
    • deduction resulting from XOR: always possible
      • 👨‍🏫 for the first encounter of $1$ in $\oplus$ result, pick a number that is $1$ in that digit

• Nim with 1 heap

  • e.g. when $a_1=5$, its graph $G_5$:

graph LR
    5((5))
    4((4))
    3((3))
    2((2))
    1((1))
    0((0))
    5-->4; 4-->3; 3-->2; 2-->1; 1-->0
    2-->0;
    3-->1; 3-->0;
    4-->2; 4-->1; 4-->0;
    5-->3; 5-->2; 5-->1; 5-->0;

• and for $n$ heap: it's Cartesian product of each $G_{a_i}$
• Sprague–Grundy theorem

• Sprague-Grundy theorem

  • every game: winning conditions can be found, like Nim
  • actually: you can convert any game into Nim!
  • assign a nimber to every vertex in each $G_i$:
    • rule 0: if $u$ has no outgoing edge: $\text{nim}(u)=0$
    • rule 1: otherwise: $\text{nim}(u)=\underset{(u,v)\in E}{\mathrm{mex}}\{\text{nim}(v_i)\}$
      • $\text{mex}\{A\}=\text{min}\{A^c\}$
  • there might be a case: where $(u,v)$ path exist and $\text{nim}(u)<\text{nim}(v)$
    • 👨‍🏫 however, it doesn't really matter: the other player can undo the move!
      • thus, we can ignore this move!
  • $L=\{(v_1,\dots ,v_n)|⨁\text{nim}(v_i)=0\}$
  • $W=\{(v_1,\dots ,v_n)|⨁\text{nim}(v_i)\ne 0\}$
  • and there is a corner case: where neighbors have the same nimber
    • which doesn't exist

• Reverse Nim

  • w/ two natural numbers and two players
  • in each turn, a player can:
    • decrease $a_1$
    • decrease $a_2$
    • decrease both $a_1,a_2$ to $a_1-x,a_2-x$
      • $1\le x\le \min (a_1,a_2)$
      • goal: not giving two equal numbers to opponents
    • 👨‍🏫 exercise: find $L,W$ for this game