Technical Interview Brain Teaser - IBM Ponder This - Neighbour Configuration
Source: IBM Ponder This Dec 12 ( http://domino.research.ibm.com/comm/wwwr_ponder.nsf/challenges/December2012.html ) - Mailed to me by Aashay Harlalka (Final Year Student, CSE, IITB)
Problem:
36 people live in a 6x6 grid, and each one of them lives in a separate square of the grid. Each resident's neighbors are those who live in the squares that have a common edge with that resident's square.
Each resident of the grid is assigned a natural number N, such that if a person receives some N>1, then he or she must also have neighbors that have been assigned all of the numbers 1,2,...,N-1.
Find a configuration of the 36 neighbors where the sum of their numbers is at least 90.
As an example, the highest sum we can get in a 3x3 grid is 20:
1 2 1
4 3 4
2 1 2
Update (24 June 2014):
Solution: Available on IBM Research Ponder This - December 2012 Solution
Problem:
36 people live in a 6x6 grid, and each one of them lives in a separate square of the grid. Each resident's neighbors are those who live in the squares that have a common edge with that resident's square.
Each resident of the grid is assigned a natural number N, such that if a person receives some N>1, then he or she must also have neighbors that have been assigned all of the numbers 1,2,...,N-1.
Find a configuration of the 36 neighbors where the sum of their numbers is at least 90.
As an example, the highest sum we can get in a 3x3 grid is 20:
1 2 1
4 3 4
2 1 2
Update (24 June 2014):
Solution: Available on IBM Research Ponder This - December 2012 Solution
3,2,1,2,3,1
ReplyDelete1,5,3,4,5,2
3,4,2,1,1,3
2,1,1,3,4,2
I don't think one can go beyond 86.
ReplyDelete3,1,1,2,2,1
2,5,4,3,5,1
1,3,2,1,4,2
2,4,1,2,3,1
2,5,3,4,5,1
1,1,2,1,2,3
well u ar ealmost right but just in the last row first coloumn and first row last coloumn u just need to change the 1's to 3's and u will get the total to be 90
DeleteNo, I can't do that because we need to have 1 in the neighbour of 2 (Second Last row, first column) and similarly at the other corner as well :-)
DeleteBut, the solution posted below is I think correct.
Well, I thing it could be this:
ReplyDelete1, 2, 2, 1, 3, 2
3, 5, 4, 3, 2, 1
2, 1, 1, 4, 5, 3
3, 5, 4, 2, 1, 2
1, 2, 3, 4, 5, 3
2, 3, 1, 1, 2, 1
1 2 3 2 3 1
ReplyDelete3 5 4 1 5 2
2 1 2 2 4 3
3 4 2 2 1 2
2 5 1 4 5 3
1 3 2 3 2 1
The '3' in the 1st row and 3rd column has not a '1' nearby...
Delete1 2 1 3 4 1
ReplyDelete3 2 3 2 2 4
3 1 4 3 1 3
2 4 5 4 5 2
1 3 4 1 3 1
2 2 1 2 2 1
The '4' in the second last row, the '4' in between the two '5's and the '4' above one '5' do not have a '2' nearby..
ReplyDelete4 2 4 5 1 4
ReplyDelete1 3 1 3 2 3
6 4 5 2 1 5
2 5 6 3 4 6
1 3 1 5 2 3
4 2 3 4 1 4
2 in first row doesnot have a one nearby
DeleteAs posted on http://domino.research.ibm.com/comm/wwwr_ponder.nsf/solutions/December2012.html
ReplyDeleteA solution without 5 (sum = 90):
2 4 1 4 2 1
1 3 2 3 4 1
2 4 4 1 4 2
4 1 3 2 3 4
3 2 4 4 1 1
1 3 1 3 2 3
Optimal solution of 93:
1 3 2 3 1 3
2 5 1 4 5 2
3 4 1 2 3 1
1 2 3 5 4 1
3 5 4 1 2 3
2 1 2 3 4 1
Thanks