Is there a solution to the problem of tigers crossing the river?

Let the tiger who can drive the boat send the other two tigers across the river, and when they come back, the tiger will get off the boat and change two people to go up, and then let one person and one tiger come back, and when they get to the shore, the person will not come down, let the tiger come down, and the other tiger who will drive the boat will go up and cross with the person to the other side, so that there will be two people and two tigers on the other side, and one of them will drive the boat, and then let one person and the tiger who will drive the boat come back, and let two people go back after coming back, so that there will be three people on the other side and a tiger who can drive the boat. Finally, let the tiger who can drive the boat take the remaining two tigers.

Even if the tiger knows how to sail, it can't be done.

First of all, the first round can be either a human tiger or a tiger tiger.

First of all, there will be no people on the other side of the bank in the first round, otherwise there will be more tigers when you come back. So the first round must have turned out to send a tiger that couldn't sail to the opposite shore (if you left the tiger that could sail on the other side, you couldn't continue).

In the second round, it can't be everyone who goes (otherwise there are 2 tigers left), nor can it be tiger people (otherwise 2 tigers and one person on the other side), only tigers and tigers can pass. Then come back a tiger who can sail a boat (this tiger must be able to sail).

The third round is no longer going on. Suppose that the human tiger passes (the result is 3 tigers on the other side of the river) assumes that everyone goes over (the result is not back, and there are 2 tigers and 1 person on the other side of the river).

Tiger Mother: A b C Tiger Baby: A B C A will row AB across the river and A back; ac crosses the river, a returns; B.B. crosses the river, B.B. comes back; AA crosses the river, CC returns; B.C. crosses the river, A. comes back; ab cross the river, aback; AC crosses the river.

There are three big tigers in A, B and C, and they cross the river with three small tigers in A, B and C. There is only one boat on the river, which can only carry two tigers at a time, regardless of size. How do they cross the river safely?

One thing to note is that the three tiger cubs should not be left alone with a big tiger other than their mother, otherwise they will be eaten by other big tigers. It should be added that a small tiger will also be eaten in the company of two large tigers.

Categories: Entertainment & Leisure >> brain teasers.

Problem description: Three tigresses each take their own children to cross the river, there is only one boat on the river, only two tigers can cross at a time, three big tigers can row, one of the little tigers can also row, the remaining two can't row, if the little tiger is not with the mother, it will be eaten by other big tigers (the little tiger does not eat the little tiger) How to make them all cross the river safely.

Analysis: Let the big tiger be ABC, and the corresponding small tiger be ABC, where C will row.

1. AC crosses the river, C comes back (A little tiger has crossed the river).

2. BC crosses the river, C comes back (AB little tiger has crossed the river) 3, BA crosses the river, BB comes back (AA mother and son have crossed the river) 4, CC crosses the river, AA comes back (CC mother and son have crossed the river) 5, AB crosses the river, C comes back (ABC three big tigers have crossed the river) 6, CA crosses the river, C comes back (ABCA has crossed the river).

7. CB crosses the river, and you're done!

Don't know, right?

Let the three pairs of tigers be AA, BB, CC, the uppercase letters represent the tigress, the lowercase letters represent the little tigers, and the little tiger A can row.

The first AC crosses the river, A comes back, the two banks are AABC and C the second AB crosses the river, A comes back, the two banks are AABC and BC the third BC crosses the river, BB comes back, the two banks are AABB and CC the fourth AA crosses the river, CC comes back, the banks are BBCC and AA the fifth BC crosses the river, A comes back, the banks are ABC and ABC the sixth AB crosses the river, A comes back, the banks are AC and ABBC the seventh AC crosses the river, All six tigers have crossed the river.

One big and one small go first, it's definitely not right, it should be like this.

The little tiger who can row the boat and the other two little tigers go separately, the little tiger who can row the boat comes back, the other two big tigers pass, the big tiger comes over with his cubs, the little tiger who can row the boat and his mother passes, and the other big tiger comes with his cubs.

Two big tigers passed.

The little tigers who can row boats then pick up the little tigers on the other side.

I've verified it, and I'm sure it's correct.

First go past two little tigers, one little tiger comes back, and then two little tigers, the little tigers are on the other side of the river.

A little tiger came back, two big tigers in the past, one big and one small, the mother and son who could row the boat passed, the mother and son who couldn't row came back, the mother and son who could row on the other side of the river, and then the two big tigers came back, there were two little tigers left at the starting point, three big and one small one who could row to the other side, and the following is the little tiger who could row the boat was tired, if I had to process the money, or how about eating the two foodies.

One big and one small past.

Come back a little. Two big pasts.

Come back a lot. Two little pasts.

Come back big and small.

Two little pasts.

Come back a lot. Two big pasts.

Come back a lot. Two big pasts.

When one of them came back alone, the little tiger could row.

One big and one small past, small and then back, one big and one small past, big back, one big and one small past.

Don't blame me for being wrong, I'm only 9 years old!

Together, they crossed the bridge.

Break down the boat, build a bridge and cross it all together.

I didn't say how many boats, if there are many boats, then 1 pair of mother and son, a pair of mother and son can be obtained, the problem is not rigorous.

Cut the boat into three parts, and each one will stand on a pair of mother and son.

Mother and son spend time together. I didn't say it had to be 2 rows.

One big and one small past (shimuzi).

Come back a little. Two big pasts.

Come back a lot. Two little pasts.

Come back big and small.

Two little pasts.

Come back a lot. Two big pasts.

Come back a lot. Two big pasts.

When one of them came back alone, the little tiger could row.

It should be OK.

Let them be complex a, b, c, system a, b, c. A, B, C, and A can row. bai ab crosses the river and a returns;

duzhiac crosses the river, and a returns;

B.B. crosses the river, B.B. comes back; (In this way, the DAO on the other side is only CC) AA crosses the river, CC comes back; (so that there is only AA on the other side) BC crosses the river and A comes back; (So the other side is ABC, and the original shore is ABC!) ab cross the river, aback;

AC crosses the river. (Did all six of them arrive?) )

--make me laugh!

1. Two tigers cross the river first, one tiger stays on the opposite bank, one comes back 2, two tigers cross the river again, two tigers stay on the opposite bank, one comes back 3, two people cross the river, one tiger stays on the other bank, one tiger comes back 4, two people cross the river, three people stay on the other bank, one tiger comes back 5, two tigers cross the river, three people one tiger stays on the other bank, one tiger comes back 6, two tigers cross the river.

6 tigers divided into AABBCC, ABCA will row The steps are as follows:

1. AB crosses the river, A rows back, and B stays on the opposite bank.

2. AC crosses the river, A rows back, C stays on the opposite bank.

3. BC crosses the river, BB comes back again, and CC is on the opposite bank at this time.

4. AA crosses the river, CC and comes back.

5. BC crosses the river, A comes back alone, at this time ABC is on the opposite bank 6, AB crosses the river, A rows back, and B stays on the opposite bank.

7. AC crosses the river and is done.

The next step is to bring back the two tiger cubs who can row boats.

Give them numbing drugs and live as they want.

I'm really sorry, the reply was late.

The index class is used to describe the serial number of a tiger rowing across a river. first is the first tiger in the boat, and second is the second tiger in the boat. (There can only be a maximum of two tigers in the boat).

When the second tiger has an ordinal number of -1, it means that there is no second tiger, i.e., there is only one tiger in the boat.

Private Static ListGetIndex(int len) This method is to select 2 numbers from the number of lens, and how many different possibilities there are in total.

There is no optimization here, what we do first is to find all the subsets of the set a of length len, and then filter out the subset of length 2 of all the subsets of a.

Principle of getting all subsets of a set of length len:

Suppose a set with the number of all its subsets should be 2 3 = 8 (including the empty set), which is a mathematical formula (the number of all subsets of a set is 2 n), it is not proved.

For example, for a set and a set are the same subset, it is not necessary to separate the calculation of the subset of the set.

So it can be assumed that element A appears in the first position by default, element B appears in the second position by default, and element C appears in the third position by default.

With the above assumptions, the problem of calculating all subsets of a set can be reduced to the fact that there is no element at each location, and if there is an element at a location, the element must be known and unique. For example, if there is an element in the first position, then it must be a, it cannot be b or c

The best way to tell in a computer that there is, or isn't, is a binary number. We can represent the entire subset of the set as the following binary number (1 for yes, 0 for none):

Convert these binary numbers into decimal numbers and you'll find that these numbers are 0 7.

Thus it can be summarized that for any set of len length, its entire subset can be represented by a binary number of 0 len-1.

Hopefully, the above explanation will help. -_