Problematics | Einstein Bridge

Meet four players from different professions at a game of bridge. Whose cap is purple?

Over the weeks, months and years, Problematics puzzles have gone into various games ranging from cricket and football to cards, dice and Chinese checkers, and occasionally lesser-known games too. One game we have not yet touched, however, is contract bridge. That is because our puzzles belong to a different world from bridge, even though it can be argued that the game’s intricate rules represent a whole puzzle universe in themselves.

For representational purposes only. (Shutterstock)

One way to bring bridge into Problematics is with an Einstein puzzle. For those who have never played or watched bridge, there is no need to go into the detailed rules for this puzzle. All we need to know is: (a) bridge is played by two teams, North & South (one pair of partners) vs East & West (the opposing pair of partners); (b) a “trick” is a set of four cards, one played by each player on the table; (c) this “playing” is preceded by a stage called “bidding”, in which players in turn call out the number of “tricks” their side proposes to win.

#Puzzle 187.1

Four bridge players have just completed bidding when I approach them with a request: “Can I watch your game? There may be a puzzle here for my readers.” They agree, I take a seat alongside them, and observe the way the game progresses

(2) North plays the first card, a heart.

(3) The next card, also a heart, is played by East, an architect.

(4) South plays the third card, another heart, with the game following the traditional clockwise order.

(5) West, who plays the fourth card, is wearing a blue cap.

As bridge players know, these four cards, one played by each contestant, constitute the first “trick”. I note the four cards on the table. All four are hearts, with the trick having been won by the contestant who played the King, the highest ranked among these cards. The other three cards are the 9, 7, and 3 of hearts. I have named the cards in descending order of rank, which is not necessarily the order in which they were played on the table.

“The 9 was played by the banker. The 7 was played by Ms Solanki, who is a chemist,” I note down clues (6) and (7) for Problematics readers.

The game has been paused while I take notes. Mr Sinha gets impatient. “Play the next card (for a new trick),” he tells one of his opponents.

The first card of any new trick is played by the contestant who won the previous trick. In this case, this was the contestant who had played the King of hearts — and who, as we have just noted, is one of Mr Sinha’s opponents. That’s Clue (8) for you.

“Mr Sinha, what do you do for a living?” I want to know. “I am the only architect at this table. That’s your Clue #9,” Mr Sinha replies, peering at my notes.

“Banker, dentist, architect, and…?” I ask one of the players. “(10) I am a dentist,” comes the reply, giving me all four professions.

I gather the two remaining names besides Ms Solanki and Mr Sinha.

(11) One player is Ms Shaikh, an opponent of the banker.

(12) Another player is Mr Sarkar, who played his card immediately after his opponent wearing an orange cap.

All four players, I observe, are wearing very colourful caps. Blue and orange I have already mentioned. (13) One of the other caps is purple. And another is red, worn by one of the opponents of the contestant who played the King of hearts.

Who is what and sitting where, in a cap of which colour? Which card did each one play? As usual, it helps if you send your answers in tabular form.

#Puzzle 187.2

The above was my own puzzle, newly created. The following is a century-old chestnut from Henry Dudeney, which I edited for language but left untouched mathematically. The puzzle must have worked better in the days when solvers didn’t have the internet or AI, but it can be fun even today if you try to solve it on pen and paper before looking for factorisation help online.

A town infested with mice engages a large number of cats. Over a year, the cats kill 1,111,111 mice. Each cat kills an equal number of mice, no fractions involved.

How many cats were at work?

MAILBOX: LAST WEEK’S SOLVERS

#Puzzle 186.1

Hi Kabir,

The four pieces of illegible text specify a sequence of the four towns along the travel path leading to the treasure.

Let us define the locations of various points as the distances from some reference point on line A-B-C-D. For convenience, we can take A as the reference point, zero. With the successive distances between each pair of towns being 5 km, 8 km and 11 km (given), the locations (coordinates) of the four towns along the straight road are 0, 5, 13 and 24 (in km).

Now, let x₁, x₂, x₃ and x₄ be the locations of the four towns, in the sequence to be travelled. These locations (x₁, x₂, x₃, x₄) are a permutation of the numbers (0, 5, 13, 24) in whichever order. Let y₁, y₂ and y₃ be the locations of the points reached, respectively, after the first, second and third legs of travel. These values are defined by the following equations.

y₁ (end of first leg) = x₁ + (x₂ – x₁)/2 = (x₁ + x₂)/2

y₂ (end of second leg) = y₁ + (x₃ – y₁)/3 = (x₁ + x₂ + x₃)/3

y₃ (end of third leg) = y₂ + (x₄ – y₂)/4 = (x₁ + x₂ + x₃ + x₄)/4

But (x₁ + x₂ + x₃ + x₄) is the same as (0 + 5 + 13 + 24) because they are the same locations (regardless of order). So, y₃, the location of the treasure, is (0 + 5 + 13 + 24)/4 = 10.5 km from the reference point A, irrespective of the sequence of the towns chosen. This means the treasure is 10.5 – 5 = 5.5 km ahead of B and 8 – 5.5 = 2.5 km behind C.

Irrespective of the given distances between the towns, the location of the treasure is equal to the average of the locations of the four towns. In fact, the given sequence calculates this average step by step, including the values to be averaged one by one.

— Professor Anshul Kumar, New Delhi

#Puzzle 186.2

The data given leads to some odd fractional results. Let the change from running to walking (senior citizen) and vice versa (youngster) happen after t hours. The remaining time is (3/4 – t) hour. In these two time spans the total distance travelled by the senior citizen and the youngster are respectively:

5t + 2(3/4 – t), or 3t + 3/2; and

4t + 10(3/4 – t), or 15/2 – 6t.

Both these lengths are the distance from one vertex back to it. Equating 3t + 3/2 = 15/2 – 6t, we get t = 2/3 hours. The total distance travelled by each is 3*2/3 + 3/2 = 3.5 km.

Now when they switch after 2/3 hours, the senior citizen is 2/3 x 5 = 3.333 km from the start, and the young man is 4 x 2/3 = 2.667 from the start in the opposite direction. The total of these two distances being more than 3.5 km means they cross before the switch between running and walking.

Let them cross after t₁ hours. The combined distance covered at that point:

5t₁ + 4t₁ = 9t₁ = 7/2, which gives t₁ = 7/18 hours. They meet after the senior citizen has travelled 5 x 7/18 or 35/18 km from one side, and the youngster has travelled 4 x 7/18 or 28/18 km from the other side.

— Kanwarjit Singh, Chief Commissioner of Income-tax, retired

I agree that the data leads to fractional results. One way of simplifying what they represent is (a) 2/3 hours means they meet 40 hours after starting. (b) Each side of the triangle is one-third of 7/2 km, which is 7/6 km or 21/18 km. Since they meet after covering 35/18 and 28/18 km respectively, it means each has completed one side of the triangle, and both are on the third side opposite the starting vertex. The older runner has covered 35/18 – 21/18 = 14/18 of the third side while the youngster has covered 28/18 – 21/18 = 7/18 of that side, meaning they meet at a point that splits the third side in a 2:1 ratio.

For the first puzzle concerning a treasure, a number of readers have observed that regardless of the order in which the towns are mentioned in the instructions, the final site is always 10.5 km from the first town. All of them are counted as correct. Extra credit to Professor Anshul Kumar, Ajay Ashok, Vinod Mahajan, and Yadvendra Somra, who are the only ones who have explained why the order of the towns does not matter.

Solved both puzzles: Professor Anshul Kumar (Delhi), Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Ajay Ashok (Delhi), Shishir Gupta (Indore)

Solved #Puzzle 186.1: Yadvendra Somra (Sonipat), Vinod Mahajan (Delhi), Dr Sunita Gupta (Delhi), Shri Ram Aggarwal (Delhi)

Solved #Puzzle 186.2: Anil Khanna (Ghaziabad), YK Munjal (Delhi)

Problematics will be back next week. Please send in your replies by Friday noon to problematics@hindustantimes.com.