The Jungle On The Plot of Land

There is a plot of land that is 16 square acres, arranged in a 4 X 4 grid. There are 16 mammals, two each of the following: beaver, cat, dog, goat, horse, lion, tiger, and walrus. There is one mammal in each acre. The acres are labeled A through P, as shown below.

A B C D
E F G H
I J K L
M N O P

Given the following clues, determine where each mammal is located. The terms "next to" and "connected together" mean vertically, horizontally, or diagonally.

1) A and D do not contain the same mammal, but one of them has a dog.

2) E and O do not contain the same mammal, but one of them has a horse.

3) B and L contain the same mammal, but not a tiger.

4) C and M contain the same mammal, but not a beaver.

5) F, K, N, and P are four different mammals.

6) The tigers are in different columns.

7) The goats are located in two of the following locations: B, C, K, and O.

8) The lions are located in two of the following locations: A, E, F, and M.

9) Cats don't get along with dogs, so neither cat is next to a dog.

10) Tigers have goats on their menu, so each tiger must be next to at least one goat and each goat must be next to at least one tiger.

11) The beavers and walruses live on the same body of water, so all four must be connected together in some fashion.

Click to know how animals are located! 

Oraganised Jungle On The Plot of Land

What was the puzzle?

As per given data - 

There is one mammal in each of 16 acres arranged in 4 x 4 grids. The acres are labeled A through P, as shown below.

A B C D
E F G H
I J K L
M N O P

Given the following clues, determine where each mammal is located. The terms "next to" and "connected together" mean vertically, horizontally, or diagonally.

1) A and D do not contain the same mammal, but one of them has a dog.

2) E and O do not contain the same mammal, but one of them has a horse.

3) B and L contain the same mammal, but not a tiger.

4) C and M contain the same mammal, but not a beaver.

5) F, K, N, and P are four different mammals.

6) The tigers are in different columns.

7) The goats are located in two of the following locations: B, C, K, and O.

8) The lions are located in two of the following locations: A, E, F, and M.

9) Cats don't get along with dogs, so neither cat is next to a dog.

10) Tigers have goats on their menu, so each tiger must be next to at least one goat and each goat must be next to at least one tiger.

11) The beavers and walruses live on the same body of water, so all four must be connected together in some fashion. 

STEPS : 

1] As per Hint 3, blocks B & L contains the same mammal.  As per Hint 4, blocks C & M have the same mammal. So, as per Hint 7 suggests, the goats must be in K & O because if they are in B & C then L & M also supposed to have them but there are total 2 goats only. K - GOAT, O - GOAT

2] If O has a goat then as per Hint 2, E must have a horse.  E - HORSE

3] As per Hint 8, if Lion is located at M then as per Hint 3 the second lion must be in B. But Hint 8 doesn't list B as possible location of lions. Hence, it must not be in M. Also, E is already occupied by horse, hence lions must be at A & F.
A - LION, F - LION

4] So if A has lion, then as per Hint 1, D must have a dog. D - DOG

5] Since C is just near to the D having a dog, as per Hint 9, C can't have cats. As per Hint 10, tigers should be in the 3x3 grids around the goats and goats are located at K and O. But M is out of these grids, hence must not have tigers.

6] And as per Hint 4, C & M having same mammal, but not beaver. As deduced above, it can't be cats or tigers either. With 1 dog already at D, 1 Horse at E, both C & M can't have dogs or horses. Hence, C & M must have walruses (only left mammal out of 8 kind of mammals). C - WALRUS, M - WALRUS

7] As per Hint 11, for walruses (at C & M) to be connected with beavers in some fashion, they must be at G and J. So, G & J have beavers.
G - BEAVER, I - BEAVER

8] Since Hint 3 suggests that B & L are not having tigers, the only animals left in terms of numbers of 2 for to be at B & L are cats. So B & L have cats. B - CAT, L - CAT

9] So far, we have, A - Lion, B - Cat, C - Walrus, E - Horse D - Dog, F - Lion, 
G -  Beaver, J - Beaver, K - Goat, L -  Cat, M - Walrus, O - Goat. Blocks left are  H, I, N, P

10] As per Hint 10 & with goats at K and O, tigers can't be at I. So possible blocks for tigers are H, N and P.

11] But as per Hint 6, tigers has to be in different columns and H & P are in the same column. Also, as per Hint 5, N & P can't have same mammals. And since, H & P falls under same column, the tigers must be occupying N and H. 
 H -  TIGER, N - TIGER

12] Since one cat is at L, P can't be a dog. Hence, I must be having a dog and P must be having a horse.  I - DOG, P - HORSE.

SUMMARY : 

A = lion, B = cat, C = walrus, D = dog
E = horse, F = lion, G = beaver, H = tiger
I = dog, J = beaver, K = goat, L = cat
M = walrus, N = tiger, O = goat, P = horse

 

Row Row Row A Tiny RowBoat : Puzzle

Walter, Xavier, Yoshi, and Zeke crossed a river in a tiny rowboat. They all started on the same side of the river. They made three trips from the starting side to the destination side, and two trips from the destination side to the starting side. Here are some facts:

1. On each trip from the starting side of the river to the destination side, two people were in the boat, but only one person rowed the boat, and that person rowed the boat for the entire trip.

2. On each return trip, only one person was in the boat.

3. Walter is the weakest of the group. He could only row the boat if no one else is in it.

4. Xavier is the second weakest of the group. He can only row the boat if he is by himself or if Yoshi, the lightest of the group, is a passenger.

5. Each man rowed the boat at least once. 

Click here for SOLUTION! 

Row Row Row A Tiny RowBoat Puzzle : Solution

What was the puzzle?

We know, Walter, Xavier, Yoshi, and Zeke crossed a river in a tiny rowboat. They all started on the same side of the river. They made three trips from the starting side to the destination side, and two trips from the destination side to the starting side. 

And we have some facts:

1. On each trip from the starting side of the river to the destination side, two people were in the boat, but only one person rowed the boat, and that person rowed the boat for the entire trip.

2. On each return trip, only one person was in the boat.

3. Walter is the weakest of the group. He could only row the boat if no one else is in it.

4. Xavier is the second weakest of the group. He can only row the boat if he is by himself or if Yoshi, the lightest of the group, is a passenger.

5. Each man rowed the boat at least once. 

ANALYSIS :

1] Walter must not had rowed from start to the destination since he is weakest among as per FACT 3. Since, he had to row at least once as per FACT 5, he must had rowed a return tip.


2] The person who had rowed twice, must not had both trips from start to destination. That's because, in that case, he would have had needed third trip in form of return trip to get back to the start once again. So, he must had rowed a return trip at least once.

3] Suppose Walter is the person who rowed the boat twice. Both of his trips must be return trips as concluded in [1] above. 

So if Walter had rowed 2 return trips then each of Xavier, Yoshi and Zeke must have rowed 1 trip from start to the destination. 

If Walter had 'taken' Zeke (while Zeke rowing) across and returned then he would have had to 'take' Yoshi (while Yoshi rowing) across as Xavier being unable to row Walter as per FACT 4. And on returning back to the start after leaving Yoshi across, Walter and Xavier would have had been at the starting point. Now, Walter being unable to row with passenger & Xavier being unable to row Walter, both would have had stuck at the start point.

In short, Walter is not the person for sure who had rowed twice.

4] We know, Walter had one return trip. So the other return trip must have been rowed by someone among Xavier, Yoshi or Zeke. Moreover, the rest of three trips from start to the destination must have rowed by Xavier, Yoshi and Zeke in some order. 

That's how, the one person among Xavier, Yoshi and Zeke, must have rowed twice, one trip from source to destination and other one return trip.

5] If Xavier had rowed twice, then with one trip he must have taken Yoshi across and in other trip he must have returned as we found the fact in [2] above. So, two return trips 'occupied' by Walter and Xavier, Yoshi wouldn't have got a chance to row which is mandatory.

So, Xavier is not the person who had rowed twice.

6] Suppose Zeke is the person who had rowed twice. He must had rowed Walter across the river to give a chance to row his return trip. After Walter reaching at the start, Xavier must had rowed Yoshi across the river thereby completing his compulsory rowing trip.

Then Zeke must have returned to the start to take Walter across then Yoshi would have been the person who hadn't rowed which is against FACT 5.

Therefore, Zeke must not be the person who rowed twice.

7] So, Yoshi must be the person who rowed twice. 

POSSIBILITY 1 : Zeke took Walter across to give him row trip in return. Then, Xavier rowed Yoshi across and Yoshi returned back to take Walter across.

POSSIBILITY 2 : First Xavier rowed Yoshi across and Yoshi returned back after which Zeke takes Walter across to allow Walter to have return trip, and finally, Yoshi taking Walter across the river.

POSSIBILITY 3 : Yoshi took Walter across the river and Walter returned. Zeke rowed Walter across and Yoshi returned to give Xavier a chance to row him across.

Develop an Unbeatable Strategy!

Consider a two player coin game where each player gets turn one by one. There is a row of even number of coins, and a player on his/her turn can pick a coin from any of the two corners of the row. The player that collects coins with more value wins the game. 

Develop a strategy for the player making the first turn, such he/she never looses the game.

Note : The strategy to pick maximum of two corners may not work. In the following example, first player looses the game when he/she uses strategy to pick maximum of two corners.

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Example :

  18 20 15 30 10 14

First Player picks 18, now row of coins is
  20 15 30 10 14

Second player picks 20, now row of coins is
  15 30 10 14

First Player picks 15, now row of coins is
  30 10 14

Second player picks 30, now row of coins is
  10 14

First Player picks 14, now row of coins is
  10 

Second player picks 10, game over.

The total value collected by second player is more (20 +
30 + 10) compared to first player (18 + 15 + 14).
So the second player wins.
===================================================================
 
This is the unbeatable strategy! 

The Unbeatable Strategy for a Coin Game!

Why strategy needed in the case?

Let's recall the example given in the question.
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Example

  18 20 15 30 10 14

First Player picks 18, now row of coins is
  20 15 30 10 14

Second player picks 20, now row of coins is
  15 30 10 14

First Player picks 15, now row of coins is
  30 10 14

Second player picks 30, now row of coins is
  10 14

First Player picks 14, now row of coins is
  10 

Second player picks 10, game over.

The total value collected by second player is more (20 +
30 + 10) compared to first player (18 + 15 + 14).
So the second player wins.
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Here, it's very clear that the player who chooses coins numbered at even positions wins & the one who chooses odd position loses.

So first player who is going to choose coin first need to be smart. All that he/she need to do is make sum of values of all coin at even position, sum of values of all coins at odd positions and compare them. 

If he finds the sum of values of coins at odd position greater then he should choose 1st coin (odd position) followed by 3rd,5th.....(odd positions) & force other to choose even coins

And if he finds the sum of values of coins at even positions greater then he should choose
last coin (which is at even position as number of coins are even).

For example, in above case, 

  18 20 15 30 10 14

first player calculates 

Sum of odd coins = 18 + 15 + 10 = 43
Sum of even coins = 20 + 30 + 14 = 64.  

Since sum of even coins is greater, he should choose 6th coin (which will be followed by 4th and 2nd) and force other player to choose 5th,3rd and 1st coin.

Even in case second player selects 1st coin after 14 at other corner is selected by first, still he can be forced to choose the coin at 3rd position (odd position) if first player selects 2nd coin.

In short, first player need to make sure that he collects all the coins that are at even positions or odd position whichever has greater sum!   

Note that the total number of coins in the case can't be odd as then distribution of coins among 2 will be unequal.