# Solution

(related to Problem: "Strand" Patience)

The reader may find a solution quite easy in a little over $200$ moves, but, surprising as it may at first appear, not more than $62$ moves are required. Here is the play: By "4 C up" I mean a transfer of the $4$ of clubs with all the cards that rest on it.

• $1 D$ on space,
• $2 S$ on space,
• $3 D$ on space,
• $2 S$ on $3 D,$
• $1 H$ on $2 S,$
• $2 C$ on space,
• $1 D$ on $2 C,$
• $4 S$ on space,
• $3 H$ on $4 S$ ($9$ moves so far),
• $2 S$ up on $3 H$ ($3$ moves),
• $5 H$ and $5 D$ exchanged, and
• $4 C$ on $5 D$ ($6$ moves),
• $3 D$ on $4 C$ $(1),$
• $6 S$ (with $5 H$) on space $(3),$
• $4 C$ up on $5 H$ $(3),$
• $2 C$ up on $3 D$ $(3),$
• $7 D$ on space $(1),$
• $6 C$ up on $7 D$ $(3),$
• $8 S$ on space $(1),$
• $7 H$ on $8 S$ $(1),$
• $8 C$ on $9 D$ $(1),$
• $7 H$ on $8 C$ $(1),$
• $8 S$ on $9 H$ $(1),$
• $7 H$ on $8 S$ $(1),$
• $7 D$ up on $8 C$ $(5),$
• $4 C$ up on $5 D$ $(9),$
• $6 S$ up on $7 H$ $(3),$
• $4 S$ up on $5 H (7) = 62$ moves in all.

This is my record; perhaps the reader can beat it.

Github: non-Github:
@H-Dudeney

### References

#### Project Gutenberg

1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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