A little logic game
Author
Discussion

tuscan_v8

Original Poster:

2,496 posts

314 months

Friday 16th December 2005
quotequote all
A = 10 mins
B = 5 mins
C = 2 mins
D = 1 min

We need to cross the bridge, the problem is only 2 persons can cross at a time and one have to go back.

Your task is to take A B C D to other side of the bridge (to attend a concert) in minimum time.

What is the best time you can achieve (the best is 17 mins) tell me how you did it?

I've only managed 19 mins

A+D = 10
<- D = 1
B+D = 5
<- D = 1
C+D = 5
Total = 19 mins

jimothy

5,151 posts

267 months

Friday 16th December 2005
quotequote all
tuscan_v8 said:
A = 10 mins
B = 5 mins
C = 2 mins
D = 1 min

We need to cross the bridge, the problem is only 2 persons can cross at a time and one have to go back.

Your task is to take A B C D to other side of the bridge (to attend a concert) in minimum time.

What is the best time you can achieve (the best is 17 mins) tell me how you did it?

I've only managed 19 mins

A+D = 10
<- D = 1
B+D = 5
<- D = 1
C+D = 5
Total = 19 mins


C+D = 2
<- C = 2
A+B = 10
<- D = 1
C+D = 2
Total = 17 minutes

tuscan_v8

Original Poster:

2,496 posts

314 months

Friday 16th December 2005
quotequote all
That was quick !! know anymore?

nel

4,830 posts

271 months

Friday 16th December 2005
quotequote all
tuscan_v8 said:
I've only managed 19 mins

A+D = 10
<- D = 1
B+D = 5
<- D = 1
C+D = 5
Total = 19 mins


Total = 22 minutes shirley!

Not surprising that you couldn't do it if you can't add up!

omitchell

19,762 posts

265 months

Friday 16th December 2005
quotequote all
jimothy said:
tuscan_v8 said:
A = 10 mins
B = 5 mins
C = 2 mins
D = 1 min


C+D = 2
<- C = 2
A+B = 10
<- D = 1
C+D = 2
Total = 17 minutes


surely c+d = 3 not 2 minutes and a+b = 15 minutes giving you a time of 24 mins? or have i got it completely wrong

Carrera2

8,352 posts

262 months

Friday 16th December 2005
quotequote all
This is an advanced version of the hen/seed/fox problem.

Bear that in mind when trying to solve it.

Einion Yrth

19,575 posts

274 months

Friday 16th December 2005
quotequote all
omitchell said:
jimothy said:
tuscan_v8 said:
A = 10 mins
B = 5 mins
C = 2 mins
D = 1 min


C+D = 2
<- C = 2
A+B = 10
<- D = 1
C+D = 2
Total = 17 minutes


surely c+d = 3 not 2 minutes and a+b = 15 minutes giving you a time of 24 mins? or have i got it completely wrong

any pairing takes the longest time of the two, not the sum of the times.

omitchell

19,762 posts

265 months

Friday 16th December 2005
quotequote all
oh, ok thanks for the clarification

zippee

14,135 posts

264 months

Friday 16th December 2005
quotequote all
omitchell said:
jimothy said:
tuscan_v8 said:
A = 10 mins
B = 5 mins
C = 2 mins
D = 1 min


C+D = 2
<- C = 2
A+B = 10
<- D = 1
C+D = 2
Total = 17 minutes


surely c+d = 3 not 2 minutes and a+b = 15 minutes giving you a time of 24 mins? or have i got it completely wrong


But if c+d are walking across together then it'll be the time taken for both to get across, hence they both do it in 2 minutes. Would be 3 minutes if they were crossing seperately.

wedg1e

27,040 posts

295 months

Friday 16th December 2005
quotequote all
How about if, say, A + B walk across. B gets there in 5 and while A is still on the bridge C sets off. C will overtake A and when C arrives, B sets off back (so still two crossing!) and so on...