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The Japanese IQ Test

paperoceans

Une Femme est une femme
Joined
Aug 24, 2009
Messages
834
MBTI Type
ENFP
Enneagram
8w7
"Everybody has to cross the river"

To play, click the link below - then click the blue circle to start.

The following rules apply:
  • Only 2 people on the raft at a time.
  • The Father cannot stay with any of the daughters, without their Mother's presence.
  • The Mother cannot stay with any of the sons, without their Father's presence.
  • The thief (striped shirt) cannot stay with any family member, if the Policeman is not there.
  • Only the Father, the Mother and the Policeman know how to operate the raft.

To move people, click on them. To move the raft, click on the pole on the opposite side of the river. The solution is possible! THIS IS NOT A TRICK!

Link
 

Oaky

Travelling mind
Joined
Jan 15, 2009
Messages
6,180
MBTI Type
INTJ
Enneagram
5w6
Instinctual Variant
sp/so
Well... I Did it. Didn't really find it hard.
 

JTG1984

Permabanned
Joined
Aug 22, 2009
Messages
1,477
MBTI Type
ISFJ
Enneagram
9w1
I call bullshit, tell me how you did it.

Actually nevermind I did it.
 

kyuuei

Emperor/Dictator
Joined
Aug 28, 2008
Messages
13,964
MBTI Type
enfp
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8
I'll put the answer in beige, so you don't have to read it if you don't want to and try to solve it yourself. But I hate it when the answer isn't available too, so I'll oblige the poster above.. It was hard, no joke. I think I played and clicked on it like 400 times :laugh:

Police and thief to the other side.
Police back.
Police and boy to the other side.
Police and thief back.
Dad and boy to the other side.
Dad back.
Dad and mom to the other side.
Mom back.
Police and thief to the other side.
Dad back.
Dad and mom to the other side.
Mom back.
Mom and girl to the other side.
Police and thief back.
Police and girl to the other side.
Police back.
Police and thief to the other side.
 

nzAShadow

New member
Joined
Feb 17, 2008
Messages
64
MBTI Type
INFP
Enneagram
9w1
The way I did it was a little different, so this means there's multiple ways! comon don't cheat JT
 

Oaky

Travelling mind
Joined
Jan 15, 2009
Messages
6,180
MBTI Type
INTJ
Enneagram
5w6
Instinctual Variant
sp/so
Oops. Kyuuei already answered it
 

paperoceans

Une Femme est une femme
Joined
Aug 24, 2009
Messages
834
MBTI Type
ENFP
Enneagram
8w7
Yea, it took me about five minutes to figure it out... either way, it's rather fun so I decided to post it here :D
 

JTG1984

Permabanned
Joined
Aug 22, 2009
Messages
1,477
MBTI Type
ISFJ
Enneagram
9w1
The way I did it was a little different, so this means there's multiple ways! comon don't cheat JT

It took me awhile but I figured it out myself. Thank god because I was getting seriously pissed off.
 

Valiant

Courage is immortality
Joined
Jul 7, 2007
Messages
3,895
MBTI Type
ENTJ
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8w7
Instinctual Variant
sx/so
Nice game. A bit easy for an IQ test, though :p
 

One Day

New member
Joined
Mar 30, 2009
Messages
212
MBTI Type
INfP
Enneagram
5w4
I remember this game from a while ago. Completed it again in about 7-8 minutes. Poor kids have such violent parents... :cry:
 

kyuuei

Emperor/Dictator
Joined
Aug 28, 2008
Messages
13,964
MBTI Type
enfp
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8
Eff all ya'll geniuses. It took me a forever to figure out. What an inefficient system, I'd make their asses swim.
 

Litvyak

No Cigar
Joined
Oct 5, 2008
Messages
1,822
Instinctual Variant
sx/sp
Nice one. I wonder how the fuck does a parent beat a child up while there's a policeman standing two steps away, but that's another story...
 

kyuuei

Emperor/Dictator
Joined
Aug 28, 2008
Messages
13,964
MBTI Type
enfp
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8
:laugh: Good choice, Chosen one. If I were a bit smarter I'd of just consulted the internetz for the answer!
 
P

Phantonym

Guest
To avoid this, I just looked at your answer after 10 minutes, lol.

Ten minutes? Sheesh. I decided they were all violent morons and closed the window in 2. :rolli: Who needs IQ anyway when you can pull the plug. :laugh: I'm too impatient for things like these. :doh:
 

Tamske

Writing...
Joined
Oct 22, 2009
Messages
1,764
MBTI Type
ENTP
I did it.
I think the interesting thing is not IF you did find it, but HOW did you find it.

Friday morning I read the riddle and played a few minutes with the application. Right away, I knew that if there was a solution, there would be two solutions. (Take one, interchange father and sons with mother and daughters and you have the second. That's symmetry.)
I logged off because I had to prepare lessons and give a lesson. On my way to the school I thought about the riddle and after 5 minutes I found something I thought was a solution.
Today I checked my 'solution', finding a kink in it and correcting it, finding a real solution.

Edit: kyuuei's solution with trial-and-error is just another method. You could also make a tree of possible first steps etc...
 

Litvyak

No Cigar
Joined
Oct 5, 2008
Messages
1,822
Instinctual Variant
sx/sp
I think the interesting thing is not IF you did find it, but HOW did you find it.

I think time matters most, just like in 'real' standardized IQ tests (except for the ones that measure the extremes, like... 150+).
 

Matthew_Z

That chalkboard guy
Joined
Jun 15, 2009
Messages
1,256
MBTI Type
xxxx
I just used careful process of elimination. There wasn't much in the way of long dead ends. The rest of this post is how I worked the problem, in invisible font.

[size=-7]Code:
P1 = A parent
p1-1 = the first same-gender child of that parent
p1-2 = the other same-gender child
P2 = The other parent
p2-1 = the first same gender child
p2-2 = the second same gender child

C1 = Police Officer
c1 = Criminal

Step one:
Neither parent can go across alone as that would lead the other parent to attack a child. If the parents go across together, they would have to return together to prevent an attack, which would lead to a useless ad infinitum. The only person capable of going across, then, is C1. Given that he cannot leave c1 with others, he must go across with c1.

Therefore, the first step is:
C1+c1 -->

Step two:
Given that he is the only one capable of operating the boat, C1 must return. Bringing the criminal back would undo the only possible progression, and thus is necessary. C1 must travel back alone.

Therefore, the second step is
<-- C1

Step three:
Given that c1 will attack anyone without the presence of C1, C1 must go. If either parent leaves, a child will be attacked. The problem with both parents leaving is the same as in step 1. C1 must leave with p1-1

Therefore, the third step is:
C1+p1-1 -->

Step 4:
As with Step 2, C1 is the only boat operator. Leaving p1-1 and c1 together is not possible, so the cop must go back with one of them. going with p1-1 would undo the last step, so he must return with c1.

Therefore, the fourth step is:
<-- C1+c1

Step 5:
If the cop were to leave, he would have to leave with the criminal, which would undo the last step. The parents still cannot leave together. As a result, one parent must leave. To prevent "child abuse," P1 must leave with p1-2.

Therefore, the fifth step is:
P1+p1-2 -->

Step 6:
The parent must return as only he can operate the boat. If the parent leaves with either child, he will negate the last step. He must proceed back alone.

Therefore, the sixth step is:
<-- P1

Step 7:
If P1 leaves alone, it will negate the last step. If P2 goes with anyone except P1, child abuse will result. If C1 goes with anyone but c1, the criminal will harm someone. If C1 goes with c1, he would have to come back the same way, which is redundant. By process of elimination, P1 must go across with P2.

Therefore, the seventh step is:
P1+P2 -->

Step 8:
The parents leaving together would undo the last step. P1 going across without P2 would result in child abuse. P2 going across alone, however, would not.

Therefore, the eighth step is:
<-- P2

Step 9:
P2 going across alone would undo the last step. P2 going across with a child would require the step to then be undone or for both parents to return together. However, if both parents then returned together, the only possible step would be C1+c1 -->, which could only then be undone. P2 cannot cross. The only other boat operator is C1, who must cross with the criminal.

Therefore, the ninth step is:
C1+c1 -->

Step 10:
If C1 goes back, he must take the criminal, which would undo the last step. P1 must go across. While P1 could carry p1-2 back with him, the step would then have to be undone or the parents would cross together, at which point the parents would then have to cross back together again to prevent child abuse. The only possible solution here is for P1 to cross alone.

Therefore, the tenth step is:
<-- P1

Step 11:
If P1 one does not leave, P2 must, which would require at least one child remain, to be abused. P1 must leave. Leaving with a child of the opposite gender would be abusive, and leaving alone would undo the last step, so P1 must cross with P2.

Therefore, the eleventh step is:
P1+P2 -->

Step 12:
C1 returning would be redundant, P1 going without P2 would be abusive. P1 going with P2 would then require P1 and P2 to cross the river again to prevent abuse, which be redundant. P2, by process of elimination, must cross. Crossing with anyone other than P1 would result in abuse or criminal attacks, so P2 must cross alone.






Therefore, the twelfth step is:
<-- P2

Step 13:
P2 returning alone would be redundant, so P2 must cross with p2-1.

Therefore, the thirteenth step is:
P2+p2-1 -->

Step 14:
If P2 goes with p2-1, the last step would be undone. If P2 goes with P1, then they must return together to prevent abuse. If P2 goes without P2, abuse results. P2 cannot cross. If P1 goes with anyone except P2, abuse also results. Only C1 can return, and they must so do with the criminal to prevent harm.

Therefore, the fourteenth step is:
<-- C1+c1

Step 15:
C1, being the only boat operator on the same side of the river, must cross. Crossing alone results in harm. Crossing with the criminal is redundant, so C1 must cross with p2-2.

Therefore, the fifteenth step is:
C1+p2-2 -->

Step 16:
If either parent goes across, child abuse results, unless P1 and P2 go together, resulting in a criminal attack. C1 must cross, and without P1 or P2. Taking a child would essentially negate the last step, so C1 must go alone.

Therefore, the sixteenth step is:
<-- C1

Step 17:
C1 going alone would negate the last step, so he must cross with c1.

Therefore, the seventeenth and last step is:
C1+c1 -->
[/size]
 

Little Linguist

Striving for balance
Joined
Jun 23, 2008
Messages
6,880
MBTI Type
xNFP
Instinctual Variant
sx/so
Just think through it logically. Family has to be together, and the thief has to be with the policeman. Duh.
 

Little Linguist

Striving for balance
Joined
Jun 23, 2008
Messages
6,880
MBTI Type
xNFP
Instinctual Variant
sx/so
If you think that was hard, just see what logic puzzles await you in some books. Holy crap. Make your head hurt....With dozens of possibilities and shit if you don't use your logical faculties...Really have to concentrate.
 
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