Skip to main content
Mathematics

Return to Answer

Simplified the solution for easy provability
Source Link
blackpen
blackpen
  • 398
  • 3
  • 12

Solution with 40 servants:

Represent the wines using base-4 number system. You will get 5 digits (for 4^5=1024 wines). 
 A-B-C-D-E

(A) For digit A, give wines with values 0,1,2,3 to 4four unique persons respectively. Do similar tests for rest of the digits. There will be 5*4=20 such tests.

(B) Mix wines with AB values {00,1211,2122,33}. Give it to 1one unique person. Do similar tests on all 2-combination of digits. There will be C(5,2)=10 such tests (greenblue graph).

(C) Mix wines with AB values {01,10,2221,32,03}. Give it to 1one unique person. Do similar tests on all 2-combination of digits. There will be C(5,2)=10 such tests (bluegreen graph).

4x4-testsenter image description here

In Step (A), for a given digit, at the most, 2 persons will die. If only one person dies, that digit has one value for both poisonsis resolved. If 2 persons die, it means it has two values (one for each poison).

Let's say two digits D&E got two values each. To link them up, you need to look if the person in Step(B) and Step(C) died for D&E. That is enough to link up the digits.

Explanation:

Look at the two graphs.

Assume D has values (1,3). Together, they connect with 0,1,2,3 (all possible values of E). This is the best case. No matter which two values of E you choose, you can uniquely connect them with values of D.

Assume D has values (1,2). Together, they connect with 0,1,2. One value of E (=3) remains unconnected. This is the worst case. Still, no matter which two values of E you select, at least one of them remains connected with values of D. That is enough for us.

Solution with 40 servants:

Represent the wines using base-4 number system. You will get 5 digits (for 4^5=1024 wines). 
 A-B-C-D-E

(A) For digit A, give wines with values 0,1,2,3 to 4 unique persons respectively. Do similar tests for rest of the digits. There will be 5*4=20 such tests.

(B) Mix wines with AB values {00,12,21,33}. Give it to 1 unique person. Do similar tests on all 2-combination of digits. There will be C(5,2)=10 such tests (green graph).

(C) Mix wines with AB values {01,10,22}. Give it to 1 unique person. Do similar tests on all 2-combination of digits. There will be C(5,2)=10 such tests (blue graph).

4x4-tests

In Step (A), for a given digit, at the most, 2 persons will die. If only one person dies, that digit has one value for both poisons. If 2 persons die, it has two values (one for each poison).

Let's say two digits D&E got two values each. To link them up, you need to look if the person in Step(B) and Step(C) died for D&E. That is enough to link up the digits.

Solution with 40 servants:

Represent the wines using base-4 number system. You will get 5 digits (for 4^5=1024 wines) 
 A-B-C-D-E

(A) For digit A, give wines with values 0,1,2,3 to four unique persons respectively. Do similar tests for rest of the digits. There will be 5*4=20 such tests.

(B) Mix wines with AB values {00,11,22,33}. Give it to one unique person. Do similar tests on all 2-combination of digits. There will be C(5,2)=10 such tests (blue graph).

(C) Mix wines with AB values {10,21,32,03}. Give it to one unique person. Do similar tests on all 2-combination of digits. There will be C(5,2)=10 such tests (green graph).

enter image description here

In Step (A), for a given digit, at the most, 2 persons will die. If only one person dies that digit is resolved. If 2 persons die, it means it has two values (one for each poison).

Let's say two digits D&E got two values each. To link them up, you need to look if the person in Step(B) and Step(C) died for D&E. That is enough to link up the digits.

Explanation:

Look at the two graphs.

Assume D has values (1,3). Together, they connect with 0,1,2,3 (all possible values of E). This is the best case. No matter which two values of E you choose, you can uniquely connect them with values of D.

Assume D has values (1,2). Together, they connect with 0,1,2. One value of E (=3) remains unconnected. This is the worst case. Still, no matter which two values of E you select, at least one of them remains connected with values of D. That is enough for us.

Corrected Trits->Digits
Source Link
blackpen
blackpen
  • 398
  • 3
  • 12

Solution with 40 servants:

Represent the wines using base-4 number system. You will get 5 digits (for 4^5=1024 wines). 
 A-B-C-D-E

(A) For Tritdigit A, give wines with values 0,1,2,3 to 4 unique persons respectively. Do similar tests for rest of the Tritsdigits. There will be 5*4=20 such tests.

(B) Mix wines with AB values {00,12,21,33}. Give it to 1 unique person. Do similar tests on all 2-combination of Tritsdigits. There will be C(5,2)=10 such tests (green graph).

(C) Mix wines with AB values {01,10,22}. Give it to 1 unique person. Do similar tests on all 2-combination of Tritsdigits. There will be C(5,2)=10 such tests (blue graph).

4x4-tests

In Step (A), for a given digit, at the most, 2 persons will die. If only one person dies, that digit has one value for both poisons. If 2 persons die, it has two values (one for each poison).

Let's say two digits D&E got two values each. To link them up, you need to look if the person in Step(B) and Step(C) died for D&E. That is enough to link up the digits.

Solution with 40 servants:

Represent the wines using base-4 number system. You will get 5 digits (for 4^5=1024 wines). 
 A-B-C-D-E

(A) For Trit A, give wines with values 0,1,2,3 to 4 unique persons respectively. Do similar tests for rest of the Trits. There will be 5*4=20 such tests.

(B) Mix wines with AB values {00,12,21,33}. Give it to 1 unique person. Do similar tests on all 2-combination of Trits. There will be C(5,2)=10 such tests (green graph).

(C) Mix wines with AB values {01,10,22}. Give it to 1 unique person. Do similar tests on all 2-combination of Trits. There will be C(5,2)=10 such tests (blue graph).

4x4-tests

In Step (A), for a given digit, at the most, 2 persons will die. If only one person dies, that digit has one value for both poisons. If 2 persons die, it has two values (one for each poison).

Let's say two digits D&E got two values each. To link them up, you need to look if the person in Step(B) and Step(C) died for D&E. That is enough to link up the digits.

Solution with 40 servants:

Represent the wines using base-4 number system. You will get 5 digits (for 4^5=1024 wines). 
 A-B-C-D-E

(A) For digit A, give wines with values 0,1,2,3 to 4 unique persons respectively. Do similar tests for rest of the digits. There will be 5*4=20 such tests.

(B) Mix wines with AB values {00,12,21,33}. Give it to 1 unique person. Do similar tests on all 2-combination of digits. There will be C(5,2)=10 such tests (green graph).

(C) Mix wines with AB values {01,10,22}. Give it to 1 unique person. Do similar tests on all 2-combination of digits. There will be C(5,2)=10 such tests (blue graph).

4x4-tests

In Step (A), for a given digit, at the most, 2 persons will die. If only one person dies, that digit has one value for both poisons. If 2 persons die, it has two values (one for each poison).

Let's say two digits D&E got two values each. To link them up, you need to look if the person in Step(B) and Step(C) died for D&E. That is enough to link up the digits.

Source Link
blackpen
blackpen
  • 398
  • 3
  • 12

Solution with 40 servants:

Represent the wines using base-4 number system. You will get 5 digits (for 4^5=1024 wines). 
 A-B-C-D-E

(A) For Trit A, give wines with values 0,1,2,3 to 4 unique persons respectively. Do similar tests for rest of the Trits. There will be 5*4=20 such tests.

(B) Mix wines with AB values {00,12,21,33}. Give it to 1 unique person. Do similar tests on all 2-combination of Trits. There will be C(5,2)=10 such tests (green graph).

(C) Mix wines with AB values {01,10,22}. Give it to 1 unique person. Do similar tests on all 2-combination of Trits. There will be C(5,2)=10 such tests (blue graph).

4x4-tests

In Step (A), for a given digit, at the most, 2 persons will die. If only one person dies, that digit has one value for both poisons. If 2 persons die, it has two values (one for each poison).

Let's say two digits D&E got two values each. To link them up, you need to look if the person in Step(B) and Step(C) died for D&E. That is enough to link up the digits.