A CIPHER FOR LOCAL SHERLOCKS

New Zealand Listener, Volume 2, Issue 48, 24 May 1940, Page 16

A CIPHER FOR LOCAL SHERLOCKS

last ‘heard of visiting Nor*way on behalf of the British Government to deliver propaganda lectures prior to the German invasion. Faithful to his instructions, Mr. Macdonnell enthralled the natives with accounts of the lives of. great Englishmen. Among these, he included Sherlock Holmes, who was, he said, at the moment, residing in a quiet country. afea, solving enemy ciphers as fast as the Foreign Office and the Intelligence Department could rush them down to him by special messenger. Who knows but that some similar undying fame might not come to a Listener puzzler? This week, for the first time, a cipher happens on the page. Who can solve it? Who can make up new ones for the puzzlement of other readers? Is there even a Watson among you? Go to it! PROBLEMS Work It Out Quickly! ""A year or so ago," said Mr. Blob the other day, "I could have bought a friend’s magnificent car for £1,026. Next year he was asking £684 for it, and it subsequently came down to £456. It is now going for £304, but I propose to wait for one more reduction in price." If this reduction is consistent with the others, at what price should Blob get the car? ‘a G. MACDONNELL was Cipher ' At last it has happened. A correspondent (C.B.G., of Springfield), sends a cipher for solution. He found it, he Says, in an English magazine, and does not know the answer. Neither do we, but we have hopes that readers will be able to handle it adequately. With imaginary spies all over the country, this sort of thing was bound to happen. Other readers who wish to try their ciphering hands at similar problems, are warned that the Post and Telegraph Department will only transmit material which has a clear meaning in plain English. This is an excellent idea, because it makes us demand that answers be enclosed. Enemy agents, and other idle persons, take notice! Here it is: ‘If you would tread a backward path to happenings of May, The eighth and lesser powers of two os in sequence show the way. DZY TVZOOHU EUVX JBTRUOKAK ALVI IWVQOCYOPP : NKWWEKWOBMGT Shunt Again From a single railway line two sidings curve off from different points to meet in a dead-end. In the dead-end there is

room for one truck only. On each siding line there is one truck. The engine is on the main line. Reverse the positions of the trucks without slipping. -(Problem from Tane). For the Pigs Dimpleton has a house at A, distant 200 yards from a stream BC. These, you will admit, were funny names for houses and streams, but R.E.W. St. C., of Hamilton, says they are right and we must accept them, since he sent the problem. His pigsty also had an unusual name. It was called D, and it was 100 yards from BC on the same side as A. The perpendicular distance from A to the line extended at right angles from BC through D was 500 yards. This, it must be stated, does not mean that D was any further off the grouind than most

pigsties. It simply means what it says, and in mathematics a perpendicular line is a line that does not go round any corners. Now the question is this: What is the shortest distance Dimpleton must walk from his house to draw.water from the stream and water the pigs? It all sounds very complicated, but it’s not really. ANSWERS (Refer to issue of May 10) Match the Matches: Place two matches close together with their heads touching. Slide a third between them to make the third leg of a tripod in such a way that its head comes up to touch both of the others as if they were three mutually adjacent circles. Over these three place three others in the same

position. Now each match is in contact with every other match. -(Problem and answer from R.W.C.). Work for Scissors: This and the other problems sent by R.W.R. for that issue were diversionary, and shall remain so. The Bear: A white bear, for those directions would only fit at the Pole and so it must have been a Polar Bear, which would be white. To the correspondent who suggested it was a Russian Bear, and therefore pink, a brickbat for mentioning such sinister matters. Double Acrostic:

Non-Euclidean Geometry This one came from S8.G.E. (Glenavy), and is thrown in as a sop to our keen mathematicians: Let ABC be a triangle; that is, any triangle, and let the angle A be bisected by AD. Also, let ED be the perpendicular bisector of BC. Let these two lines meet at D. Join DC and DB and from D drop perpendiculars on to AB and AC, to meet these lines in G and F. Then: In the. triangles AGD and AFD, the angle GAD = angle FAD (by construction each is half angle A). The side AD is common. There is a right angle in each, so that the two triangles are equal in all respects. Therefore DG = DF. Similarly, in the triangles BDE and CDE, DE is common to both. BE = EC (because DE is bisector of BC). There is a right angle in each (the included angle). Therefore: DB = DC. This much is obvious: But now, perpend: In the triangles BDG and CDF we have proved that DG = DF and DB = DC. But there is also a right angle in each (i.e. DG and DF are perpendicular to AB and AC); therefore these two triangles are equal in all respects. Therefore BG = CF, But, by the congruency of the triangles AGD and AFD, AG = AF. Therefore, adding these two last results, A.G. + BG = AF + CF. That is, AB = AC, Therefore, it is possible to prove that any triangle is an isoceles triangle. Rugby Risks The other day we were talking Rugby, and conversation turned to the chances of winning the weekly sweepstake, conducted in the Attorney-General’s office with the permission of the Minister of Infernal Affairs, in aid of the Society for Giving Attention to the Graves of Soldiers who died of Cholera in the Crimean War. We wanted to know the odds against winning when the sweepstake was supported by 20 people, and to limit the extent of an otherwise impossible calculation, we decided that it should be assumed that the two football teams between them would score 40 points. Now in that total there are an immense number of possible combinations of scores, and yet the Accounts

Department maintained it would be possible to establish definite odds by a mathematical calculation. It has to be remembered that this number of combinations may be divided among the 20 people taking part in the sweepstake, and, very important, that the number of combinations is limited to a certain extent by the fact that any team at one time can only score three points (for a try or a penalty kick), five points (for a converted try), or four points (for a field goal). To the poor PP, this all sounded very difficult indeed, but he felt obliged to assure the Accounts Department ‘that Listener puzzlers would work it out if it could be worked out at all. So what about it?