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Philosopher Bob exercises every morning. He does steps up a flight of emergency stairs leading to the second floor egress door from the courtyard of his apartment high-rise. He first does single steps to the top, descends, runs up the flight, descends, then he takes the steps two at a time and comes down. Set done. He does ten of these for a good aerobic workout. But there’s a reason why he varies his routine in one set, other than for mere variety. Being such a deep thinker, especially when doing rote exercises, he tends to forget mundane things, like how many repetitions he’s done. In fact, he is so deep, he can’t remember past one. If he did the same single-step climbs twice in a row, he would actually lose track of how many he did! What a brilliant guy! But this problem also applies to how many sets he’s done. To help him keep track, when he finishes a set, he can place little white stones (all identical) on a knee-high brick wall surrounding a well landscaped garden near the steps, thus keeping count.

Question: What is the least number of stones Bob needs to put on the brick wall to keep count in a ten-set workout? Explain.

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Assuming that this is a standard brick wall with mortar between individual bricks, he could make do with one stone so long as there were at least 10 bricks. The bricks would be the actual counters, and the stone would mark which set he was on.

If, however, there were less than 10 bricks, he would need to start back at the beginning with two stones once he had done on more set than there were bricks. So he would need 2 stones if there were between 5 and 10 bricks.

Edit: I just realized that the length of the wall may not be provide a recognizable starting brick. In which case he would need one more stone to designate the starting brick. One stone on the starting brick, and the next stone moves to the right or left one brick for each set completed.

Edited by tamjap
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he could make marks with chalk on the wall to have ten spaces and keep moving the stone down each time he does a repitition

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4, using a binary style counting system?

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4, using a binary style counting system?

1,2,4, and 8 would all look very similar, as would 3 and 6. If he had black stones to work as zeros, it would work very

well. But using spaces as zeros would make it difficult to tell the difference between these numbers:

1 = 1

2 = 10

4 = 100

8 = 1000

3 = 11

6 = 110

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Philosopher Bob exercises every morning. He does steps up a flight of emergency stairs leading to the second floor egress door from the courtyard of his apartment high-rise. He first does single steps to the top, descends, runs up the flight, descends, then he takes the steps two at a time and comes down. Set done. He does ten of these for a good aerobic workout. But there’s a reason why he varies his routine in one set, other than for mere variety. Being such a deep thinker, especially when doing rote exercises, he tends to forget mundane things, like how many repetitions he’s done. In fact, he is so deep, he can’t remember past one. If he did the same single-step climbs twice in a row, he would actually lose track of how many he did! What a brilliant guy! But this problem also applies to how many sets he’s done. To help him keep track, when he finishes a set, he can place little white stones (all identical) on a knee-high brick wall surrounding a well landscaped garden near the steps, thus keeping count.

Question: What is the least number of stones Bob needs to put on the brick wall to keep count in a ten-set workout? Explain.

He could draw a rectangle on a paper and divide it into 30 segments (with lines on the paper). He then needs only one stone to keep moving it forward as he goes along completing the sets. The stone should keep the paper from flying off too!

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Also assuming that the wall ledge is not wide enough to have a double row. You could use stone width spacing from the edge of the wall to implement your binary placeholders. Now, you only need to count to 5 as the runner can remember up to 2 sets. So given that, you need 3 stones (being able to count as high as 7 but only needing to go to 5).

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He should only need 5 stones, once the fifth one is placed on the wall, then he would remove one at a time until all are gone. Simple enough.

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One should be fine.

Just put the stone at certain locations on top of a given brick (not in the mortar, so as to be confused as to which brick is selected). I believe 12 positions (4x3) would be easily distinguishable. Start left to right on a front row of 4, then a center row of 4, and then a back row of 4 (not necessarily finished). The four positions in the row would be easily distinguishable by which side of center the brick it is on and how close it is to the center. Since Bob's a smart guy, I'm sure he could distinguish more, but these are fairly simple.

However, noting the exact phrasing of the question, I'd need to change my answer to none. He could just pile the stones up in front of the brick wall (or any other are without the white stones), which would mean he doesn't put any *on* the brick wall. He could even just put the stones in a pocket, etc.

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One should be fine.

Just put the stone at certain locations on top of a given brick (not in the mortar, so as to be confused as to which brick is selected). I believe 12 positions (4x3) would be easily distinguishable. Start left to right on a front row of 4, then a center row of 4, and then a back row of 4 (not necessarily finished). The four positions in the row would be easily distinguishable by which side of center the brick it is on and how close it is to the center. Since Bob's a smart guy, I'm sure he could distinguish more, but these are fairly simple.

However, noting the exact phrasing of the question, I'd need to change my answer to none. He could just pile the stones up in front of the brick wall (or any other are without the white stones), which would mean he doesn't put any *on* the brick wall. He could even just put the stones in a pocket, etc.

One stone, 9 positions on one brick would be even easier to distinguish (3x3) 4 corners, 4 edges and the center. The 10th rep gets no store.

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Philosopher Bob exercises every morning. He does steps up a flight of emergency stairs leading to the second floor egress door from the courtyard of his apartment high-rise. He first does single steps to the top, descends, runs up the flight, descends, then he takes the steps two at a time and comes down. Set done. He does ten of these for a good aerobic workout. But there’s a reason why he varies his routine in one set, other than for mere variety. Being such a deep thinker, especially when doing rote exercises, he tends to forget mundane things, like how many repetitions he’s done. In fact, he is so deep, he can’t remember past one. If he did the same single-step climbs twice in a row, he would actually lose track of how many he did! What a brilliant guy! But this problem also applies to how many sets he’s done. To help him keep track, when he finishes a set, he can place little white stones (all identical) on a knee-high brick wall surrounding a well landscaped garden near the steps, thus keeping count.

Question: What is the least number of stones Bob needs to put on the brick wall to keep count in a ten-set workout? Explain.

I think I need some clarification. One "set" in the work out consists of 3 "laps" up and down the stairs, the 1st "lap" he does single steps (walking), the 2nd "lap" he runs (>2 steps at a time?), the 3rd and final "lap" he does 2 steps at a time (walking). He cannot remember further back than the "lap" he just completed. Is this all correct?

If so, the different "laps" could play a part in his counting. I mean, can I assume he has time to place rocks between laps, or only between sets?

Edited by Louie
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If the bricks used were too small to distinguish easily the placement of a stone on a corner, side or center; then, one could employ the following sequence with three stones.

```Sets Completed - pattern

1 -   1

2 -   11

3 -   111

4 -   1 1

5 -   1 11

6 -   11 1

7 -  1 1 1

8 - 1  1 1

9 - 1 1  1

10 - done```
This method involves moving, placing or removing only a single stone each time and using a total of three stones. All patterns with no space between a pair of '1's indicates the stones are touching. The patterns for sets 8 & 9 exploit the natural ability to see the difference between a single and double space. If these are of issue, then one could take the left stone from set 8, place it between the other two and offset vertically to the top or bottom with the opposite representing the 9th set.
```Optional sets 8 & 9:

8 -  1

1 1

9 - 1 1

1```

However if there is room to stagger the stones vertically as well as horizontally, more interesting orientations could be used, but three stones would still be required.

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Pretty clever solutions here. I like the cheese and beer idea. That method occurred to me as well.

Let me refine:

Bob would not be able to use the bricks in the wall as markers. Let’s say they are too small or uniform or the top is plastered across. In any case, using bricks in the wall as indicators is out.

Bob is too tired and too intent on his metaphysical speculations to worry about spacing or setting up a binary system. Just lumber over to the stones, plop one up and get right back to those steps. Quick and easy.

Bob has two options: either pick up a stone and put it on the wall, or knock it off. Never use the same stone. Never pick up or reposition a stone once placed.

How many on the wall now?

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Bob has two options: either pick up a stone and put it on the wall, or knock it off. Never use the same stone. Never pick up or reposition a stone once placed.

Could you be more specific about using the same stone twice? The way I see it is: Bob takes a single stone from a pile of stones and places in on the brick wall. He can knock down the stone from the wall but not pick it back up. Sooo...

5 stones are used. If there are no stones on the wall he places one there. If there is a stone on the wall he knocks it down. First lap he places a stone on the wall, second lap he knocks it down to the base of the wall. Then repeat this procedure until 5 stones are on the ground. That makes 10 sets.

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Could you be more specific about using the same stone twice? The way I see it is: Bob takes a single stone from a pile of stones and places in on the brick wall. He can knock down the stone from the wall but not pick it back up. Sooo...

5 stones are used. If there are no stones on the wall he places one there. If there is a stone on the wall he knocks it down. First lap he places a stone on the wall, second lap he knocks it down to the base of the wall. Then repeat this procedure until 5 stones are on the ground. That makes 10 sets.

No cluttering the ground with stones; they need to go back into the rock pile.

Edited by Shakeepuddn
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Ok with that said, can you still be more specific on not using the same stone twice?

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[quote name='JTZero' date='30 April 2010 - 02:47 AM' timestamp='1272592052

' post='225466']

Ok with that said, can you still be more specific on not using the same stone twice?

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If he's allowed to have two stone piles then he would need a minimum of one stone on the wall. First set he places a stone and for each consecutive set he will replace the stone with another. The replaced stones will go into the new pile. Something tells me this isn't allowed though. From the OP, I can only think of 10 as being the minimum. Am I trying too hard to bend the rules?

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If he's allowed to have two stone piles then he would need a minimum of one stone on the wall. First set he places a stone and for each consecutive set he will replace the stone with another. The replaced stones will go into the new pile. Something tells me this isn't allowed though. From the OP, I can only think of 10 as being the minimum. Am I trying too hard to bend the rules?

Well, there really are no established rules. I was curious if anyone had a better way than I do it. I actually use this method for my own workout and thought it would be a thought provoking puzzler for anyone interested. But I keep having to set boundaries for all you logic masters out there

I have seven

After the fifth set, instead of placing a fifth rock, Bob knocks the four of them down and starts fresh at zero. The "forgetful" clause doesn't apply because Bob has had a good workout and realizes he's already done the first half of ten. One after six, another after seven, one more after eight, and then after nine, knock them off and finish the last set.

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I have seven

After the fifth set, instead of placing a fifth rock, Bob knocks the four of them down and starts fresh at zero. The "forgetful" clause doesn't apply because Bob has had a good workout and realizes he's already done the first half of ten. One after six, another after seven, one more after eight, and then after nine, knock them off and finish the last set.

Using the rule about using a stone more than once, in reality you only need 4 stones. Then you'll reuse 3 of them after the fifth set. Am I wrong?

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Using the rule about using a stone more than once, in reality you only need 4 stones. Then you'll reuse 3 of them after the fifth set. Am I wrong?

No, except for the "reusing" rule that states . . . any stone reused is to be considered a separate stone.

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Pretty clever solutions here. I like the cheese and beer idea. That method occurred to me as well.

Let me refine:

Bob would not be able to use the bricks in the wall as markers. Let’s say they are too small or uniform or the top is plastered across. In any case, using bricks in the wall as indicators is out.

Bob is too tired and too intent on his metaphysical speculations to worry about spacing or setting up a binary system. Just lumber over to the stones, plop one up and get right back to those steps. Quick and easy.

Bob has two options: either pick up a stone and put it on the wall, or knock it off. Never use the same stone. Never pick up or reposition a stone once placed.

How many on the wall now?

Since he must either pick up a stone or remove at least one stone each time, the minimum possible is now 5.
Just how forgetful is he? You say the "forgetful clause" doesn't apply since he's done the first half of the workout, but you don't point out that he can also distinguish other sets. Here's what I see as what you imply he can distinguish from your solution.

2 and 6 (he sees 1 stone)

3 and 7 (he sees 2 stones)

4 and 8 (he sees 3 stones)

5 and 9 (he sees 4 stones)

1 and 10 (he sees no stones)

Yeah, he could probably distinguish between after sets 2 and 6, but 4-8 and 5-9 may be more difficult.

What exactly can he distinguish between?

Here's a solution that only needs 5 stones, but requires the ability to have two piles/groupings (a left pile and a right pile) on the wall.

after set 1 - see (0,0) place rock in the left (only) pile (1,0)

after set 2 - see (1,0) place rock in the left (only) pile (2,0)

after set 3 - see (2,0) place rock in the right pile (2,1)

after set 4 - see (2,1) place rock in the right pile (2,2)

after set 5 - see (2,2) place rock in the right pile (2,3)

after set 6 - see (2,3) remove rock from the left pile (1,3)

after set 7 - see (1,3) remove rock from the left pile (0,3)

after set 8 - see (0,3) remove rock from the right (only) pile (0,2)

after set 9 - see (0,2) remove rock from the right (only) pile (0,1)

after set 10- see (0,1) remove rock from the right (only) pile (0,0)

As I see it, you'd need to distinguish 3-9 and 2-10. From your solution's assumptions, this should be simple enough.

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