To the Iceberg, Allons-y!

This is more of a logic/problem solving problem than combinatorics. Here it goes:

1st crossing: You have 3 options. Either 1 penguin, 1 puffin, or 2 puffins cross on the boat. If only 1 animal crosses, then they would end up having to cross back on the boat since the boat can't travel alone. In this case we would be back to where we started. So that leaves the 3rd choice, 2 puffins cross.

2nd crossing: You have 2 options. Either 1 puffin crosses back, or 2 puffins cross back. Its easy to see the 2nd options brings us back to the original problem. So that leaves the 1st choice. 1 puffin crosses back.

3rd crossing: You have 2 options. Either 1 puffin crosses or 1 penguin crosses. Its easy to see the 1st options brings us back to the previous state. That leaves the 2nd choice. 1 penguin crosses.

4th crossing: You have 2 options. Either 1 puffin crosses back or 1 penguin crosses back. Its easy to see the 2nd option brings us back to the previous state. That leaves the 1st choice. 1 puffin crosses back.

5th crossing: You have 3 options. Either 1 penguin, 1 puffin, or 2 puffins cross. Its easy to see the 2nd option brings us back to the previous state. The 1st option would require either one of the penguins to cross back since the boat can't travel alone which would bring us back to the current state. This leaves the 3rd choice. 2 puffins cross.

6th crossing: You have 3 options. Either 1 penguin, 1 puffin, or 2 puffins cross back. Its easy to see the 3rd option brings us back to the previous state. The 1st option would require either 1 of the penguins to cross again, since the boat can't cross alone, which would bring us back to the current state. This leaves the 2nd option, 1 puffin crosses back.

7th crossing: You have 2 options. Either 1 penguin or 1 puffin crosses. By this time it should be obvious why the 2nd option makes no sense. So that means the penguin crosses.

8th crossing: You have 2 options. Either 1 penguin or 1 puffin crosses back. Once again, it should be obvious why the 1st option makes no sense. So that means the puffin crosses back.

9th crossing: Now finally only the 2 puffins need to cross and they can both fit on the boat.

The end.

Sequence: 2 puffins cross $\bullet$ 1 puffin comes back $\bullet$ 1 penguin crosses $\bullet$ 1 puffin comes back $\bullet$ 2 puffins cross $\bullet$ 1 puffin comes back $\bullet$ 1 penguin crosses $\bullet$ 1 puffin comes back $\bullet$ 2 puffin cross

We start with say, 2 P's and 2 p's. It is obvious that a puffin must go first, and have one to go back, so the first thing that happens is the puffins go over. One comes back, and then gives the boat to one penguin. The penguin goes over to the desired shore, and the puffin that was already there rows it back. This took 4 moves to get one penguin over. This must happen once more for the other penguin, so it will take 8 moves. But then it will take an extra trip to get the puffins both to the desired side. So the answer is 9.

Phrase 1: 2puffins cross the river (2puffins//2penguins) (leftside//right side)

Phrase 2: 1puffins cross back (1puffin//2penguins+1puffin)

Phrase 3: 1penguin cross to left side (1puffin+1penguin//1penguin+1puffin)

Phrase 4: 1puffin cross back (1penguin//1penguin+2puffins)

Phrase 5: 2puffins cross the river (1penguin+2puffins//1penguin)

Phrase 6: 1puffin cross back to the right (1penguin+1puffin//1penguin+1puffin)

Phrase 7: 1penguin cross back to the left(2penguins+1puffin//1puffin)

Phrase 8: 1puffin cross back to the right (2penguins//2puffins)

Phrase 9: 2puffins cross the river (2puffins+2penguins//----)

I think the logic is to cross two puffins first then put one puffin back. After that when you put a penguin to the other side, you will have puffin to sail it back. This process occurs twice but you have to bring 2 puffins back at the last phase. 4+4+1=9

let 'pu' be puffins and 'pe' be penguin..

       [ left iceberg]    [ boat direction]           [right iceberg]

state0:.................................................................................. pe1,pe2,pu1,pu2

                             ( pu1,pu2)

state1:.. pu1,pu2......................... <-----------------........................... pe1,pe2

                              ( pu1)

state2: ... pu2..............................-----------------> ........................ ..pu1,pe1,pe2

                              (pe1)

state3: ..... pu2,pe1.................... <------------------ ....................... pu1,pe2

                              (pu2)

state4:.......pe1............................. ----------------->..........................pu1,pu2,pe2

                             (pu1,pu2)

state5:......pe1,pu1,pu2 ...............<----------------............................pe2

                              (pu1)

state6:.....pe1,pu2 .......................... ------------------>.....................pu1, pe2

                             (pe2)

state7: ..... pe1,pe2,pu2.................. <------------------ ................... pu1

                              (pu2)

state8: ....pe1,pe2............................ ------------------->................... pu1,pu2

                              (pu1,pu2)

state9:....pe1,pe2, ........................... <----------------------
.............. pu1,pu2

hence the minimum number of times the rowboat cross the sea is 9.

To reach the minimum amount of trips you want to maximise the number of birds crossing the sea and minimise the number of birds returning (assuming that you need a bird to pilot the boat). To start with this means you need to get two birds across the sea so that one can return so you haven't wasted a trip (so a penguin must never return or cross to a side where there is not another bird). This means you need to have 1 trip across with 2 puffins and a return trip with 1 puffin. Then to maximise the trips across you must send 1 penguin as they can only go one at a time. Once across the other puffin must return because he is capable of travelling in a 2. The 2 puffins must then cross the sea again and 1 puffin must return after that to allow the final penguin to cross. Once across the puffin on the opposite side must return to pick up the final puffin and take him to the opposite side. This means that the trips go (Assume puffins = p and penguins = P) <--2p, -->p, <--P, -->p, <--2p, -->p, <--P, -->p, <--2p. So a total of 9 trips must be made as a minimum.

Interesting mathcount problem, i solve it with my logic. (imagine the animation while you read my text)

If we said that the land after the sea is 'A land' and the land location of the penguins and puffins started are 'B land'

1) First group to go is the 2's Puffins(to B land) then 1's Puffin return to A land. (2 cross times)

2) 1 of Penguin cross the sea to B land then 1's Puffin return to A land.(2 cross times)

3) 2's Puffin return again to B land then one of them return to A land(2 cross times)

4) Penguin on A land cross to B land then 1's Puffin on B land backs to A land(2 cross times)

5) Last 2's Puffins cross the sea from A land to B land(1 cross time)

So minimum time to use is 9 cross times

P is penguin , p is puffin

Here it is

PPpp

PP > pp

PPp < p

Pp > Pp

Ppp < P

P > Ppp

Pp < Pp

p > PPp

pp < PP

                                          >         PPpp

i have choose the right answer..

9 crossing- write each crossing

1ère traversée : Pour que le bateau puisse revenir et repartir, il faut deux oiseaux dans le bateau, donc 2 macareux. 2ème traversée : Retour de l'un des macareux sui laisse le nateau au 1er pingouin 3ème traversée : traversée du premier pingouin qui reste sur place 4ème traversée : retour du 2ème macareux 5ème traversée : les deux macareux repartent 6ème traversée : retour d'un macareux qui laisse la barque au 2ème pingouin 7ème traversée : traversée du 2ème pingouin qui reste sur place 8ème traversée : retour du 2ème macareux 9ème traversée : les deux macareux repartent et tout le monde est passé

Ans. is 9 -----> 2-1-1-1-2-1-1-1-2("2 and 1" is no. of animals while crossing)

lets assign n for penguin &f for puffin

1 f <-------------> fnn
2. n <-------------> ffn

3 .nf<------------->fn
4. nn<------------->ff
5. nnff<-------------

<--------- 1direction, <----> 2 directions , so total river crossings =9.

in start 2 puffins cross and 1 puffin com back with boat then 1 penguin cross and other puffin also came back with boat then both puffins again cross and again one of them back with boat then 2nd penguin cross and 2nd puffin also came back Lastly both puffins crosses.

i've got d answer as 9 by writing down statements.. but can anyone tel me how it solve it using permutations and combinations???

For minimum number of turns. The only possible way Step 1 : 2 puffins will cross the sea

Step 2: 1 puffin will come back with the boat

Step 3: 1 penguin will cross the sea

Step 4: The puffin on the destination will take the boat and comes back

Step 5: 2 Puffins will cross the sea

Step 6: 1 puffin will come back

Step 7: 1 penguin will cross the sea

Step 8: The puffin on the destination will take the boat and comes back

Step 9: Both the puffins will cross the sea

So Mission completed!!

9

9

1) 2 Puffins cross 2) 1st Puffin returns 3) 1st Penguin crosses 4) 2nd Puffin returns 5) 2 Puffins cross 6) 1st Puffin returns 7) 2nd Penguin crosses 8) 2nd Puffin returns 9) 2 Puffins cross

I made a diagram an always give a point for one crossing. like this

d--------dpp p--------ddp pd--------pd pp--------dd ppdd--------00

2 poffins go over and 1 comes back. 2 trips done. 1 penguin comes over and the other poffin comes back. 4 trips. two poffins go over and one comes back. 6 trips. now we have a poffin and penguin on each side. 1 penguin goes over and a poffin comes back. 8 trips. the remaining two poffins come over and this is 9 trips

1. 2 puffins go to the left
2. 1 puffin comes back
3. puffin gets down and one penguin moves to the left
4. penguin gets down, puffin returns
5. other puffin gets on the boat and both the puffins go left
6. one puffin gets down and the other returns
7. puffin gets down and the penguin moves left
8. penguin gets down and the puffin returns to the right side
9. both the puffins move to the left iceberg. QED. :)

The three groupings permitted on the boat are (1 puffin; 2 puffin; or 1 penguin) [(1) (0)<---------2 puffins (2 penguins left)] [(2) (1 puffin get's off)---------> 1 puffin(2 penguins)] [(3) (1 puffin)<--------- 1 penguin (puffin get's off and 1 penguin left)] [(4) (penguin get's off and puffin get's on) (puffin get's on so both puffins in the boat) ---------> (1 penguin left)] [(5) (1 penguin) <--------- (2 puffins) ( 1 penguin left)] [(6) (1 penguin 1 puffin get's off boat) ---------> (Puffin get's off, penguin get's on)] [(7) (1 penguin 1 puffin) <--------- (penguin) (1 puffin left)] [(8) (penguin get's off boat puffin get's on) ---------> (1 puffin left)] [(9) (2 penguins) <--------- (puffin get's on boat so 2 puffins)] all are now on the other side

first take two puffin(1 time).then come back with 1 puffin(2X),now take 1 penguin(3X) and then return with that one puffin left(4X).after that again take 2 puffin(5X) and as before return back with 1 puffin(6X).take 1 penguin (7X) and return the boat with 1 puffin(8X).at last take the two remaining puffin(9X).

TAKE A PEN AND PAPER AND DO WHAT I SAY FOR BETTER UNDERSTANDING..... :-
FIRST CROSSING - WE SEND TWO PUFFINS. SECOND CROSSING -ONE PUFFIN CAME BACK. THIRD CROSSING - WE SEND A PENGUIN. FOURTH CROSSING - A PUFFIN CAME BACK. FIFTH CROSSING - WE SEND BOTH PUFFINS. SIXTH CROSSING - ONE PUFFIN CAME BACK. SEVENTH CROSSING - WE SEND PENGUIN. EIGHT CROSSING -A PUFFIN CAME BACK. NINTH CROSSING - WE SEND BOTH THE PUFFINS. HOPE ALL LIKE IT :-)...

2 Puffins go 1 return 1 penguin goes 1 puffin returns 2 puffin goes 1 return then 1 penguin goes 1 puffin returns and then finally the puffins go to the other end Min possible crosses=9 :D

so basically, the first trip must be two puffins, because if one penguin, the back trip must be a penguin, rendering the round trip useless.

therefore the first back trip is one puffin, and second trip is one penguin, and second back trip is one puffin,

the third trip must be two puffins due to the first line of reasoning, therefore, the third back trip is one puffin, and the fourth trip is one penguin, the fourth back trip is one puffin, the final trip is two puffins QED

1.2-Puffins Crossed 2.1-Returned 3.1-Penguin Crossed 4.Other Puffin Returned 5.2-Puffins Crossed 6.1-Returned 7.1-Penguin Crossed 8.Other Puffin Returned 9.2-Puffins Crossed

9

Step by Step Solution :

1. <- 2 Puffins | (2 Puffins on opposite).

2. 1 Puffin -> | (1 Puffin on opposite).

3. <- 1 Penguin and 1 Puffin | (2 Puffins and 1 Penguin on opposite).

4. 1 Puffin -> | (1 Puffins and 1 Penguin on opposite).

5. <- 2 Puffins | (2 Puffins and 1 Penguin on opposite).

6. 1 Puffin -> | (1 Puffins and 1 Penguin on opposite).

7. <- 1 Penguin | (1 Puffins and 2 Penguin on opposite).

8. 1 Puffin -> | (2 Penguin on opposite).

9. <- 2 Puffins | (2 Puffins and 2 Penguin on opposite).

:D

2 Pufins cross togetther, 1 puffin returns on Iceberg1 and another stays at iceberg2. 2 journeys..(To and Fro) 1st penguine goes from iceberg 1 -> iceberg2, and 1 pufin brings the boat from iceberg2 -> iceberg1 .. 2 more journeys 2 Pufins cross togetther, 1 puffin returns on Iceberg1 and another stays at iceberg2. 2 journeys..(To and Fro) 2nd penguine goes from iceberg 1 -> iceberg2, and 1 pufin brings boat from iceberg2 . 2 journeys. Both pufins travel together to iceberg 2. 1 journey .. Total 9

at first both the puffins go to the other iceberg and one puffin comes back... then one penguin goes to the other iceberg and the other puffin returns.. then both the puffins go again to the iceberg and one puffin comes back leaving the the other... this time second penguin goes and the puffin is back .... in the last both the puffins come to the iceberg .....hence the the total crssings are 9..

Basically the best strategy is: Both puffins cross the ocean. One puffin goes back. One penguin crosses the ocean. The puffin goes back. Both puffins cross the ocean again and the cycle repeats.

First, two puffins will go to Side 2 and one of them will return. (2 crossings). Then, a penguin will go to side 2, and the puffin from Side 2 would return to side 1 (2 crossings) .The same process will be repeated for second penguin.(4 crossings). Then the two puffins will cross the sea.(1 crossing). Thus, the but will make 2+2+4+1=9 trips.