Part 95

287.--AN ACROSTIC PUZZLE.

There are twenty-six letters in the alphabet, giving 325 different pairs. Every one of these pairs may be reversed, making 650 ways. But every initial letter may be repeated as the final, producing 26 other ways. The total is therefore 676 different pairs. In other words, the answer is the square of the number of letters in the alphabet.

288.--CHEQUERED BOARD DIVISIONS.

There are 255 different ways of cutting the board into two pieces of exactly the same size and shape. Every way must involve one of the five cuts shown in Diagrams A, B, C, D, and E. To avoid repetitions by reversal and reflection, we need only consider cuts that enter at the points a, b, and c. But the exit must always be at a point in a straight line from the entry through the centre. This is the most important condition to remember. In case B you cannot enter at a, or you will get the cut provided for in E. Similarly in C or D, you must not enter the key-line in the same direction as itself, or you will get A or B. If you are working on A or C and entering at a, you must consider joins at one end only of the key-line, or you will get repetitions. In other cases you must consider joins at both ends of the key; but after leaving a in case D, turn always either to right or left--use one direction only. Figs. 1 and 2 are examples under A; 3 and 4 are examples under B; 5 and 6 come under C;

and 7 is a pretty example of D. Of course, E is a peculiar type, and obviously admits of only one way of cutting, for you clearly cannot enter at b or c.

Here is a table of the results:--

a b c Ways. A = 8 + 17 + 21 = 46 B = 0 + 17 + 21 = 38 C = 15 + 31 + 39 = 85 D = 17 + 29 + 39 = 85 E = 1 + 0 + 0 = 1 -- -- -- --- 41 94 120 255

I have not attempted the task of enumerating the ways of dividing a board 8 x 8--that is, an ordinary chessboard. Whatever the method adopted, the solution would entail considerable labour.

289.--LIONS AND CROWNS.

Here is the solution. It will be seen that each of the four pieces (after making the cuts along the thick lines) is of exactly the same size and shape, and that each piece contains a lion and a crown. Two of the pieces are shaded so as to make the solution quite clear to the eye.

290.--BOARDS WITH AN ODD NUMBER OF SQUARES.

There are fifteen different ways of cutting the 5 x 5 board (with the central square removed) into two pieces of the same size and shape. Limitations of space will not allow me to give diagrams of all these, but I will enable the reader to draw them all out for himself without the slightest difficulty. At whatever point on the edge your cut enters, it must always end at a point on the edge, exactly opposite in a line through the centre of the square. Thus, if you enter at point 1 (see Fig. 1) at the top, you must leave at point 1 at the bottom. Now, 1 and 2 are the only two really different points of entry; if we use any others they will simply produce similar solutions. The directions of the cuts in the following fifteen

solutions are indicated by the numbers on the diagram. The duplication of the numbers can lead to no confusion, since every successive number is contiguous to the previous one. But whichever direction you take from the top downwards you must repeat from the bottom upwards, one direction being an exact reflection of the other.

1, 4, 8. 1, 4, 3, 7, 8. 1, 4, 3, 7, 10, 9. 1, 4, 3, 7, 10, 6, 5, 9. 1, 4, 5, 9. 1, 4, 5, 6, 10, 9. 1, 4, 5, 6, 10, 7, 8. 2, 3, 4, 8. 2, 3, 4, 5, 9. 2, 3, 4, 5, 6, 10, 9. 2, 3, 4, 5, 6, 10, 7, 8. 2, 3, 7, 8. 2, 3, 7, 10, 9. 2, 3, 7, 10, 6, 5, 9. 2, 3, 7, 10, 6, 5, 4, 8.

It will be seen that the fourth direction (1, 4, 3, 7, 10, 6, 5, 9) produces the solution shown in Fig. 2. The thirteenth produces the solution given in propounding the puzzle, where the cut entered at the side instead of at the top. The pieces, however, will be of the same shape if turned over, which, as it was stated in the conditions, would not constitute a different solution.

291.--THE GRAND LAMA'S PROBLEM.

The method of dividing the chessboard so that each of the four parts shall be of exactly the same size and shape, and contain one of the gems, is shown in the diagram. The method of shading the squares is adopted to make the shape of the pieces clear to the eye. Two of the pieces are shaded and two left white.

The reader may find it interesting to compare this puzzle with that of the "Weaver" (No. 14, Canterbury Puzzles).

+===+===+===+===+===+===+===+===+ |:o:| : : : : : : : I...I...+===+===+===+===+===+===+ |:::| o |:::::::::::::::::::::::| I...I...I...+===+===+===+===+...I |:::| |:o:| : : : |:::| I...I...I...I...I===+===+...I...I |:::| |:::| o |:::::::| |:::| I...I...I...+===I===+...I...I...I |:::| |:::::::| |:::| |:::| I...I...+===+===+...+...I...I...I |:::| : : : |:::| |:::| I...+===+===+===+===I...I...I...I |:::::::::::::::::::::::| |:::| +===+===+===+===+===+===+...I...I | : : : : : : |:::| +===+===+===+===+===+===+===+===+

292.--THE ABBOT'S WINDOW.

THE man who was "learned in strange mysteries" pointed out to Father John that the orders of the Lord Abbot of St. Edmondsbury might be easily carried out by blocking up twelve of the lights in the window as shown by the dark squares in the following sketch:--

+===+===+===+===+===+===+===+===+ | : : : : : : : | I...+===+...+...+...+...+===+...I | IIIII : : : IIIII | I...+===+===+...+...+===+===+...I | : IIIII : IIIII : | I...+...+===+===+===+===+...+...I | : : IIIIIIIII : : | I...+...+...+===+===+...+...+...I | : : IIIIIIIII : : | I...+...+===+===+===+===+...+...I | : IIIII : IIIII : | I...+===+===+...+...+===+===+...I | IIIII : : : IIIII | I...+===+...+...+...+...+===+...I | : : : : : : : | +===+===+===+===+===+===+===+===+

Father John held that the four corners should also be darkened, but the sage explained that it was desired to obstruct no more light than was absolutely necessary, and he said, anticipating Lord Dundreary, "A single pane can no more be in a line with itself than one bird can go into a corner and flock in solitude. The Abbot's condition was that no diagonal lines should contain an odd number of lights."

Now, when the holy man saw what had been done he was well pleased, and said, "Truly, Father John, thou art a man of deep wisdom, in that thou hast done that which seemed impossible, and yet withal adorned our window with a device of the cross of St. Andrew, whose name I received from my godfathers and godmothers." Thereafter he slept well and arose refreshed. The window might be seen intact to-day in the monastery of St. Edmondsbury, if it existed, which, alas! the window does not.

293.--THE CHINESE CHESSBOARD.

+===I===+===+===+===I===+===+===+ | |:::: 2 ::::| 3 |:::| 5 |:6:| I...+===+...+===+...I...I...+===I |:::: 1 |:::| ::::| 4 |:::| 7 | I...+===+===+...I===I...I===+===I | |:::: |:::| ::::| 9 |:::| I===I...I===============I...I...I |:::: 11|:::: ::::: 10|:::| 8 | I=======I===I===========I...I...I | ::::: 12|:::: 13::::| |:::| I=======+...I...+===+===|===+===I |:::: 14|:::| |:::| 16::::| 17| I...+...I===I===+...+...+===+...I | ::::| ::::: 15|:::| ::::| I=======+===========+===+=======I |:::: ::::: 18::::: ::::: | +===+===+===+===+===+===+===+===+

+===+===I===I===+===I===+===+===+ | ::::| |:::: |:::| ::::| I...+===I...I=======I...I===+...I |:::| |:::: |:::: |:::| | I...I===I===============I===I...I | |:::: ::::| ::::: |:::| I===I=======I=======I=======I===I |:::| ::::| ::::| ::::| | I...I===+...I...+...I...+===+...I | ::::| |:::: |:::| ::::| I...+===I...+===I===+...I===+...I |:::| |:::: |:::: |:::| | I===I...+=======I=======+...I===I | |:::: ::::| ::::: |:::| I...+=======+...I...+=======+...I |:::: ::::| |:::| ::::: | +===+===+===+===+===+===+===+===+

Eighteen is the maximum number of pieces. I give two solutions. The numbered diagram is so cut that the eighteenth piece has the largest area--eight squares--that is possible under the conditions. The second diagram was prepared under the added condition that no piece should contain more than five squares.