[Guest]

Let me complicate the regular balance problem.

All you have is 8 balls and a two pan balance scale (no other weights avaliable).

We all know that it would take 2 weighings, if we have 1 heavy ball and 7 normal balls, to identify the heavy ball.

There are 2 heavy balls and 6 normal balls. How many weighings would it take to identify both heavy balls?

If you are done with 2 then try with 3 heavy balls, then 4.(total 8 balls only)

Need not try with 5/6/7 heavy balls coz the answer would be same as 3/2/1 heavy balls respectively.

We are only looking for the worst case scenarios here.

Note : FYI, all the normal balls weigh the same (lets say 1 unit) and all the heavy balls weigh the same (lets say 1.1 units). the weights mentioned are just for your understanding. No other purpose intended.

[Guest]

2 of 8 balls in 5 tries

3 of 8 balls in 4 tries

Let me complicate the regular balance problem.

All you have is 8 balls and a two pan balance scale (no other weights avaliable).

We all know that it would take 2 weighings, if we have 1 heavy ball and 7 normal balls, to identify the heavy ball.

There are 2 heavy balls and 6 normal balls. How many weighings would it take to identify both heavy balls?

If you are done with 2 then try with 3 heavy balls, then 4.(total 8 balls only)

Need not try with 5/6/7 heavy balls coz the answer would be same as 3/2/1 heavy balls respectively.

We are only looking for the worst case scenarios here.

Note : FYI, all the normal balls weigh the same (lets say 1 unit) and all the heavy balls weigh the same (lets say 1.1 units). the weights mentioned are just for your understanding. No other purpose intended.

[benjer3]

I got 4 for 2/8:

weigh 2 vs 2

-if 1 side heavier, weigh balls on that side against each other

--if balanced, both heavy (2 weighings)

--if 1 heavier, found 1 of the heavy balls; weigh remaining 4 against each other; weigh 2 on heavier side against each other to find other heavy ball (4 weighings)

-if balanced, weigh balls on 1 side against each other

--if 1 heavier, that is 1 heavy ball; find the other by weighing the 2 on the other side against each other (3 weighings)

--if balanced, weigh 2 of remaining against each other

---if balanced, weigh 1 of those vs another of the remaining; heavier side shows which remaining group of 2 is heavier (4 weighings)

---if 1 heavier, weigh that vs another of remaining; balanced means other is other heavy ball, heavier means last ball is other heavy ball (4 weighings)

[Guest]

i would say average 3 ( i knw this aproach is nt correct bt still this is gud)

take 3+3 bals frm 8 of those and weigh them

if they are equal

take 2balls from 1st pan and replace with the remainig two balls from first step

three cases

pan 1 heavy

both equal

pan 2 heavy

pan 1heavy

implies the 2 balls which we kept, are the heavy one

no of weighs(1+1=2) probability of occurence=2/3*1/2=1/6

both equal

implies out of 3 balls in pan 1 the one which we dint replace is the heavy one and the heavy one in second pan can be found by weighing two of those balls

no of weighs(1+1+1=3)

pan 2 heavy

implies two balls which we replace in pan1 contains 1 heavy ball so weigh it to find out and the heavy one in second pan can be found by weighing two of those balls

no of weighs(1+1+1=4) probability of occurence=2/3*1/2=1/6

pan 1 heavy and pan 2 heavy equal probable so we can take average of the giving =(4+2)/2=3

if they are not equal.

implies two heavy balls are on one side.

1 heavy ball on one side and 3 ordinary on other

now jst take two balls from heavy pan weigh it against the remaining 2 balls

3 cases

pan 1 heavy

pan 2 heavy

both equal

pan1 heavy

take one ball from pan1 and keep it in pan2 and from pan2 to pan1

after this if

pan 1 heavy (means the one not exchanged and the remaining from 3 ball of heavy pan of first weigh are two heavy ballS)

no of weighs(1+1+1=3)

 

pan 2 heavy ( means the exchanged ball and the remaining from 3 ball of heavy pan of first weigh are two heavy balls)

no of weighs(1+1+1=3)

both equal ( then balls in the first pan after second weigh are two heavy balls)

no of weighs(1+1+1=3 )

pan2 heavy

take one ball from pan1 and keep it in pan2 and from pan2 to pan1

after this

pan1 heavy ( means the exchanged ball and the remaining from 3 ball of heavy pan of first weigh are two heavy balls)

no of weighs(1+1+1=3)

pan2 heavy (means the one not exchanged and the remaining from 3 ball of heavy pan of first weigh are two heavy ballS)

no of weighs(1+1+1=3)

both equal (not possible)

both equal

take one ball from pan1 and keep it in pan2 and from pan2 to pan1

pan1 heavy (pan 1 conatin now 2 heavy balls)

no of weighs(1+1+1=3)

pan2 heavy (pan 2 contain now 2 heavy balls)

no of weighs(1+1+1=3)

both equal

implies we have exhanged both of same type so jst weigh once more the balls of same pan and u get the answer

no of weighs(1+1+1=4)

PS: a very vague solution . this how it looks when we put thoughts in to words

somebody plz find the better sol

[Guest]

I solved for a 3 pan scale, oops.

Edited May 2, 2011 by Nana7

[Guest]

I believe this uses no more than 5 weighs

1) Weigh 3 (1,2,3) vs 3 (4,5,6).

If they balance, both sides have either 1 or 2 heavy balls. We also know that balls 7 and 8 must balance as well.

2a) Weigh 2 (1,2) from side 1 vs 2 (4,5) from side 2, setting one from each side to the side

3a) If they balance, then weigh the 2 that were only weighed once so far (knowing they must weigh the same) against the 2 not weighed yet.

4a) If they balance, then they are all heavy or light. Weigh one of these 4 vs 1 of the 4 of the other group to see which is heavy or light.

If weighing number 3a did not balance, then they heavy side has 2 heavy balls, and we know that the second weighing above must have 1 heavy ball on each side. Return to the balls weighed in 2a and this time weigh 1 from side 1 vs 1 from side 2. If one is heavy, then weigh the other 2 against each other to find the last heavy ball. If they balanced then they are either both heavy or the other 2 are heavy. Weigh one of these vs a heavy ball to see which is which. This is 5 weighs.

3b) if weigh number 2a did not balance,then we know the 2 just set aside can not balance either, so one of them must be heavy, and it is the opposite of the heavy side we have in weigh 2a. So if side 1 is heavy (in the 2 vs 2 weigh) then ball 6 is heavy. We also know that weigh 2a either has just 1 heavy ball on the heavy side, or 2 on the heavy side and 1 on the light side. This is because weigh number 1 was balanced. We have a known heavy ball, 6, and should weigh it vs 7 (or 8).

4b) If they balance, then balls 6,7,8 are heavy and the last heavy ball is either 1 or 2. If they do not balance, the balls 1,2,6 are heavy and the last heavy ball is either 3 or 4. Weigh ball 6 against one of the last unknowns for the final weighing.

This covers cases where the first weigh was balanced.

If the first weigh did not balance, the possiblities are 2 or 3 heavy balls on the heavy side, 0 or 1 heavy balls on the light side, and 0,1,2 heavy balls for the unweighed balls (7 and 8).

2c) put balls 1,2,3 on one scale and balls 7,8,4 on the other.

If this balances, then balls 7 and 8 are heavy, and the other 2 heavy are from 1,2 or 3. Weigh one of those vs one of those to find which 2 are heavy.

3c) If this does not balance, then 7 and 8 are either both light or have 1 heavy and 1 light. Weigh 7 vs 8.

if they balance, they are light, balls 1,2,3 are all heavy, and the last heavy is from 4,5, or 6. Weigh one of them vs another to find out (ie 4 vs 5, one of them is heavy else 6 is the heavy one)

4c) if they did not balance, we found a heavy ball (say it is 7), know that ball 4 must be light, and that either balls 1,2,3 are heavy or that 2 of them are and the other heavy is ball 5 or 6. Weigh 1,2,8 vs 5,6,7.

We know 4 and 8 are light and 7 is heavy. If they balance, then we know each side must have 2 heavies. They can not only have 1 heavy each because the only other balls are 3 and 4 and 4 is known to be light. Since 8 is light, then balls 1 and 2 are heavy, and the last heavy is either 5 or 6.

if they did not balance, side 2 must have only 1 heavy, ball 7. Side 1 must therefore have heavy balls 1 and 2. 3 must also be heavy.

Edited May 2, 2011 by Nana7

[k-man]

.. but just to set the target - all cases can be solved in 4 weighings. 4 weighings produce 81 possible outcomes, which is more than enough for any number of heavy balls among 8. 2 heavy balls present 28 possible combinations (just one more than what 3 weighings allow to solve, so you still need 4). 3 heavy balls present 56 combinations and 4 heavy balls present 70 combinations. All are less than 81 outcomes that can be achieved by 4 weighings.

[plainglazed]

I agree and disagree with k-man. there are 28 possible combinations for the 2 heavy balls out of eight and can be done in 4 weighings with one last weighing needed just to compare the last two possible combos as would be expected. there are 56 total combinations for the 3 heavy balls out of eight scenerio and all can also be determined in 4 weighings. for the 70 possibilities of 4 heavy balls out of eight tho, have not been able to do it in four weighings as the math would suggest. listing the combinations after weighing either 2 balls vs 2 balls or 3 balls vs 3 balls initially, have not been able to find a second weighing that will break the outcome into three groups of 9 or less. at best have once again needed one more weighing comparing two last possible combinations.

[plainglazed]

i stand corrected and bow to the k-man. could not visualize if the fact that a balanced scale could mean either an equal number of normal balls and/or an equal number of heavy balls per side might change the number of possibilities.

...with a maximum of 4 weighings -

label the balls 1,2,3,4,5,6,7,8
1) compare 12 vs 34
if 12 = 34 then the following 26 possibilities contain the four heavy balls
1234 5678
1356 1357 1358 1367 1368 1378
1456 1457 1458 1467 1468 1478
2356 2357 2358 2367 2368 2378
2456 2457 2458 2467 2468 2478
2) compare 126 vs 357
if 126 = 357 then the following 8 possibilities contain the four heavy balls
1356 1367 1458 1478
2356 2367 2458 2478
3) compare 135 vs 248
if 135 = 248 then the four heavy balls must be in one of the following
1458 2367
4) compare 5 vs 7 to see which are the four heavy balls
if 135 > 248 then the four heavy balls must be in one of the following
1356 1367 2356
4) compare 13 vs 56 to see which are the four heavy balls
if 135 < 248 then the four heavy balls must be in one of the following
2478 2458 1478
4) compare 24 vs 78 to see which are the four heavy balls
if 126 > 357 then the following 9 possibilities contain the four heavy balls
1467 1456 2467 2456
1368 1468 2368 2468 1234
3) compare 15 vs 27
if 15 = 27 then the four heavy balls must be in one of the following
2456 1467 1234
4) compare 5 vs 7 to see which are the four heavy balls
if 15 > 27 then the four heavy balls must be in one of the following
1368 1456 1468
4) compare 3 vs 5 to see which are the four heavy balls
if 15 < 27 then the four heavy balls must be in one of the following
2368 2467 2468
4) compare 3 vs 7 to see which are the four heavy balls
if 126 < 357 then the following 9 possibilities contain the four heavy balls
1357 1457 2357 2457
1358 1378 2358 2378 5678
3) compare 15 vs 27
if 15 = 27 then the four heavy balls must be in one of the following
2358 1378 5678
4) compare 5 vs 7 to see which are the four heavy balls
if 15 > 27 then the four heavy balls must be in one of the following
1358 1457 1357
4) compare 13 vs 57 to see which are the four heavy balls
if 15 < 27 then the four heavy balls must be in one of the following
2457 2378 2357
4) compare 4 vs 8 to see which are the four heavy balls
if 12 > 34 then the following 22 possibilities contain the four heavy balls
1567 1568 1578 1678
2567 2568 2578 2678
1235 1236 1237 1238
1245 1246 1247 1248
1256 1257 1258 1267 1268 1278
2) compare 15 vs 27
if 15 = 27 then the following 8 possibilities contain the four heavy balls
1678 2568 1236 1238
1246 1248 1257 1268
3) compare 13 vs 46
if 13 = 46 then the four heavy balls must be in one of the following
1678 1248 1268
4) compare 4 vs 7 to see which are the four heavy balls
if 13 > 46 then the four heavy balls must be in one of the following
1236 1238 1257
4) compare 6 vs 7 to see which are the four heavy balls
if 13 < 46 then the four heavy balls must be in one of the following
1246 2568
4) compare 1 vs 2 to see which are the four heavy balls
if 15 > 27 then the following 7 possibilities contain the four heavy balls
1567 1568 1578
1235 1245 1256 1258
3) compare 6 vs 8
if 6 = 8 then the four heavy balls must be in one of the following
1568 1235 1245
4) compare 3 vs 4 to see which are the four heavy balls
if 6 > 8 then the four heavy balls must be in one of the following
1567 1256
4) compare 2 vs 7 to see which are the four heavy balls
if 6 < 8 then the four heavy balls must be in one of the following
1578 1258
4) compare 2 vs 7 to see which are the four heavy balls
if 15 < 27 the the following 7 possibilities contain the four heavy balls
2567 2578 2678
1237 1247 1267 1278
3) compare 6 vs 8
if 6 = 8 then the four heavy balls must be in one of the following
2678 1237 1247
4) compare 3 vs 4 to see which are the four heavy balls
if 6 > 8 then the four heavy balls must be in one of the following
2567 1267
4) compare 1 vs 5 to see which are the four heavy balls
if 6 < 8 then the four heavy balls must be in one of the following
2578 1278
4) compare 1 vs 5 to see which are the four heavy balls
if 12 < 34 then the following 22 possibilities contain the four heavy balls
3567 3568 3578 3678
4567 4568 4578 4678
3415 3416 3417 3418
3425 3426 3427 3428
3456 3457 3458 3467 3468 3478
2) compare 35 vs 47
if 35 = 47 then the following 8 possibilities contain the four heavy balls
3678 4568 3416 3418
3426 3428 3457 3468
3) compare 13 vs 26
if 13 = 26 then the four heavy balls must be in one of the following
3678 3428 3468
4) compare 2 vs 7 to see which are the four heavy balls
if 13 > 26 then the four heavy balls must be in one of the following
3416 3418 3457
4) compare 6 vs 8 to see which are the four heavy balls
if 13 < 26 then the four heavy balls must be in one of the following
4568 3426
4) compare 3 vs 4 to see which are the four heavy balls
if 35 > 47 then the following 7 possibilities contain the four heavy balls
3567 3568 3578
3415 3425 3456 3458
3) compare 6 vs 8
if 6 = 8 then the four heavy balls must be in one of the following
3568 3415 3425
4) compare 1 vs 8 to see which are the four heavy balls
if 6 > 8 then the four heavy balls must be in one of the following
3567 3456
4) compare 4 vs 7 to see which are the four heavy balls
if 6 < 8 then the four heavy balls must be in one of the following
3578 3458
4) compare 4 vs 7 to see which are the four heavy balls
if 35 < 47 then the following 7 possibilities contain the four heavy balls
4567 4578 4678
3417 3427 3467 3478
3) compare 6 vs 8
if 6 = 8 then the four heavy balls must be in one of the following
4678 3417 3427
4) compare 1 vs 2 to see which are the four heavy balls
if 6 > 8 then the four heavy balls must be in one of the following
4567 3467
4) compare 3 vs 5 to see which are the four heavy balls
if 6 < 8 then the four heavy balls must be in one of the following
4578 3478
4) compare 3 vs 5 to see which are the four heavy balls