题面摘自 Henry E. Dudeney 公版文本；英文为古腾堡原文整理，中文为本站自译，提示、解答骨架和闲谈保留本站原创结构。
The silent and thoughtful Clerk of Oxenford, of whom it is recorded that "Every farthing that his friends e'er lent, In books and learning was it always spent," was prevailed upon to give his companions a puzzle. He said, "Ofttimes of late have I given much thought to the study of those strange talismans to ward off the plague and such evils that are yclept magic squares, and the secret of such things is very deep and the number of such squares truly great. But the small riddle that I did make yester eve for the purpose of this company is not so hard that any may not find it out with a little patience." He then produced the square shown in the illustration and said that it was desired so to cut it into four pieces (by cuts along the lines) that they would fit together again and form a perfect magic square, in which the four columns, the four rows, and the two long diagonals should add up 34. It will be found that this is a just sufficiently easy puzzle for most people's tastes.
沉默而深思熟虑的奥克森福德书记员，据记载，“他的朋友们借出的每一分钱，总是花在书籍和学习上”，他被说服给他的同伴们出一个谜题。他说道：“我最近常常花很多时间去研究那些用来驱除瘟疫的奇怪护身符和神秘的魔法方块之类的邪恶东西，这些东西的秘密非常深，这样的方块的数量也确实很多。但是我昨天晚上为这家公司的目的而制作的这个小谜语并不难，任何人只要有一点耐心都可能找不到它。”然后他制作了图中所示的正方形，并说需要将其切成四块(沿线切割)，然后将它们重新组合在一起，形成一个完美的魔方，其中四列，四行和两条长对角线加起来应为34。您会发现，对于大多数人的口味来说，这是一个足够简单的拼图。
提示 1
先把每一步允许做什么写成状态表。
提示 2
找一个不会随操作改变的量，或把对象分成互斥类别。
提示 3
检查构造是否覆盖全部对象，而不是只给出一个漂亮例子。
完整解答
固定主人面向门口，就消除了旋转重复。先数其余人任意排列，再减去两位指定朝圣者相邻的情形。相邻时把这两人看成一个双人块，块内部有 2 种顺序；最后用总数减去坏排列。
Dudeney 的趣题常把难点藏在“看起来可以试”的地方。别急着猜答案；先把图、表或状态画出来，再问哪些限制一直没有变。这也是它和 Carroll 逻辑题互补的地方：一个拆句子，一个拆结构。