He would drink some very weak poison before coming and give water as his poison to the pharmacist. Then he would survive (weak poison, strong poison, water) and the pharmacist would die (water, strong poison)

Pharmacist would have thought about this. And have water as his poison. He would have (water, water) and sage would have had (weak poison, water, water).

King just got water.

]]>3,4,5,6 needs to be placed in B,D,E,G

B and D cannot be 6

E and G cannot be 3

Since the image is symmetric across vertical line, lets say B is 3. If E is 6, and D and G become 4 and 5 which is not possible. So G is 6. So D is 4 and E is 5.

So, A is 7, B is 3, C is 1, D is 4, E is 5, F is 8 and G is 6.

]]>For 2 weights A and B, you need 1 weighing.

For 3 weights A, B and C, you need 1 weighing to arrange A and B. To place C, you would need 2 more weighings.

For 4 weights, A, B, C and D

Lets arrange A, B and C – It would take 3 weighings. Without loss of generality, lets say it was A>B>C.

To place D, compare D and B. and then depending on gt or lt, compare A and D or B and D.

For 5 weights, A, B, C, D and E.

Lets arrange A, B, C and D – It would take 5 weighings. Without loss of generality, lets say A>B>C>D.

To place E, compare E and B. Depending on result, compare E and A, or E and C. Depending on result of E and C, you might need to do E and D as well.

Number of weighings needed = 8

I do not see how it can be done in 7 in all cases!

]]>For 3 weights A, B and C, you need 1 weighing to arrange A and B. To place C, you would need 2 more weighings.

For 4 weights, A, B, C and D

Lets arrange A, B and C – It would take 3 weighings. Without loss of generality, lets say it was A>B>C.

To place D, compare D and B. and then depending on > or B>C>D.

To place E, compare E and B. Depending on result, compare E and A, or E and C. Depending on result of E and C, you might need to do E and D as well.

Number of weighings needed = 8

I do not see how it can be done in 7 in all cases!

]]>My son Anahat and I worked on it together – with 75% of work done by him for the case of four weights. We started with the case of 3 weights named A, B, and C. It takes three trials on the balance to get it. So let us say we now know their ranking which is A>B>C.

We introduce the fourth weight D and knowing that two trials are left. First, we compare D with B. If D is less than B, we compare again with C to be certain that D is either > than or less than C. This will provide a solution such as A,B,C,D or A,B,D,C. Likewise, D could turn out to be > than B, in which case the argument can be similarly made involving A, leading to solutions such as D>A>B>C or A>D>B>C

Anahat suggested another approach for this case with A/B and C/D as two trials. Say A>B and C>D.

Next, B will be compared with D.

If B is less than D, then we know that the order is A or C >D>B . In the fourth trial C could be compared with A. If A > C, then the answer is A>C>D>B. If C is > than A, we need to compare A with D to see if it is C>A>D>B or C>D>A>B

If B is more than D, the possibilities are A>B>C>D or A>C>B>D or C>A>B>D. Therefore B is compared with C in trial number 4. If it is greater than C, we have found a solution in four trials, else in 5 trials by comparing A with C

Case of five weights

This has been tried by Anahat .

Step 1: C> D

Step 2: A>B

Step 3: A> C

We now know

A>C>D

Step 4 :B> E

Step 5

Compare B with C

If B> C, then A>B>C>D

Step 6,7

Compare E with C and D to get the right answer within 7 steps

If E is greater than C and the previous solution in step 5 was A,B,C,D no further comparison is done and the answer is A,B,E,C,D. But if B is greater than A in Step 5, we need to compare E with A for a final answer in 7 steps

If E is less than C, then from the previous solution in step 5, we need to compare E with D and get the final answer in 7 steps.

If B< C, then

Step 6,7

Compare B with D in Step 6 to get A,C,B,D or A,C,D,B

]]>Trace in image – https://ibb.co/ffOX7a

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