Henry Ernest Dudeney/Modern Puzzles/203 - The Horse-Shoe Game/Solution
Jump to navigation
Jump to search
Modern Puzzles by Henry Ernest Dudeney: $203$
- The Horse-Shoe Game
- This little game is an interesting companion to our "Noughts and Crosses".
- There are two players.
- One has two white counters, the other two black.
- Playing alternately, each places a counter on a vacant point, where he leaves it.
- When all are played, you slide only, and the player is beaten who is so blocked that he cannot move.
- In the example, Black has just placed his lower counter.
- White now slides his lower one to the centre, and wins.
- Black should have played to the centre himself, and won.
- Now, which player ought to win at this game?
Solution
Every game should end up as a draw.
The only way one player can win is through a blunder made by his opponent.
Proof
This theorem requires a proof. In particular: Martin Gardner tells us that the analysis is in Edouard Lucas' Recreations Mathematiques. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Sources
- 1894: Édouard Lucas: Récréations mathématiques: Volume $\text { 3 }$
- 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $203$. -- The Horse-Shoe Game
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $472$. The Horseshoe Game