Grid Method Example

 
There are two common methods of solving logic problems. This page gives an example of a logic problem being worked out using the "ABC grid", "crosshatch grid" or simply "the grid method". The other method can be called the"table" or "fill-out grid" method. For an example of this puzzle worked out using the table method, click here.
 
 

Three people live in Hopper St. Can you sort out their names, house numbers & their dogs names?

Dogs: Bob, Patch, Toby

Names: Jim, Dick & Mike

Surnames: Black, Turnip & Spade

House No.s: 1,2,3

Clue 1. Dick lives at a higher number than Mr. Spade; neither of these own Bob.

Clue 2. Toby lives at No. 1

Clue 3. Jim Black does not live at No. 3

 
 
 
Jim                  
Dick     X X     X    
Mike                  
1            
2            
3     X      
Bob     X
Patch      
Toby      

Clue 1. Dick lives at a higher number than Mr. Spade; neither of these own Bob.

This clue tells us that Dick cannot be called Mr. Spade. We can now put a X in the grid where Dick intersects with Spade.

Dick lives at a higher No. than Mr. Spade, therefore Dick cannot live at No. 1 and Mr. Spade cannot live at No. 3 and we can put X's where these intersect.

Neither owns Bob and so X's can be placed also following the same principal as above.


 
 
 
Jim                  
Dick     X X     X    
Mike                  
1       X X
2           X
3     X     X
Bob     X
Patch      
Toby      

Clue 2. Toby lives at No. 1

We are now able to place a at the intersection between Toby and No. 1.

This means no other dog can live at No. 1 and Toby cannot be at No. 2 or 3. We can therefore put X's in all these boxes.


 
 
 
Jim X X X X X X
Dick X X X     X X
Mike X X X     X X
1       X X
2           X
3 X   X     X
Bob     X
Patch      
Toby      

Clue 3. Jim Black does not live at No. 3

This tell us Jim's surname is Black and we can place a and X's as before.

Jim does not live at No. 3 and so we can place a X there and also a cross where Black intersects with 3.

Dick has only one space left for a and so he must be called Turnip. We can place a X where Mike intersects with Turnip, meaning Mike must be called Spade and we can place a here.

Spade is not 3 and doesn't own Bob and therefore Mike must be the same. This means Jim owns Bob - place a here. Therefore Jim doesn't own patch or Toby so place X's.

Bob does not live at No. 1 and therefore Jim can't either. Place a X here, which means Jim must live at No. 2. Place a here and X's against Mike and Dick. Dick must therefore live at No. 3 and Mike at No. 1. place here.


 
 
 
Jim X X X X X X
Dick X X X X X X
Mike X X X X X X
1 X X X X
2 X X X X
3 X X X X
Bob X X
Patch X X
Toby X X

Toby & Mike both live at No. 1, so place a here. Place X's for Toby and Dick, and also Mike and Patch. This means Dick must own Patch, so place a here.

This has actually given us all the information we need to answer the puzzle, however we can fill the rest of the grid in from the information given.


 
 

We are then able to complete the table:

Name   Surname   House No.   Dog
Jim   Black   2   Bob

Dick   Turnip   3   Patch

Mike   Spade   1   Toby

I hope the above makes sense to you all. Any further questions or comments, please email The Logic Zone.

 
The Logic Zone
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