She gave a quick knock and slipped into his office. He was wrapping up some old-fashioned paperwork.
"Hi~," she almost sang as she entered.
"Oh, hi, sweetie," he replied, and looked at his monitor. He gathered his papers and bent behind the desk, to file them into the lower drawer, as he said, "Sorry, normally Paul sends me a message when you're on your way up."
"I happened to catch Brian on his way back in," she began.
He shot up like a bolt, just as Brian put the knife to her throat.
"Away from the desk," Brian ordered, "Hands where I can see them. We're going to do this slowly, so that there's no funny business. Grab your monitor by the top and face it this way."
"Brian, look at me," he said. They saw each other, eye-to-eye, and that was all that he needed. The room seemed to darken as a feeling of dread swept over everyone present. But especially Brian. Most especially Brian.
The feeling lifted, but Brian inexplicably threw the knife away, which slid under the couch. He bolted for the door, but recoiled. She went to her husband and clung to him. He held his wife, to reassure her.
"W-What's going--" Brian began.
"Your fillings," came the simple reply.
Eyes visibly widened with dawning terror, the would-be assassin grabbed at the insides of his own mouth with his hands, which muffled his emerging screams.
"Honey," she said, trembling, seeming to look over her shoulder, but with her eyes closed, "what just happened?"
"Brian was let go this morning. I'm sorry, sweetie. In the future, I'll see to it that terminated employees are escorted off of the premises, not just out of the building."
"No, I mean... What did you do to him? How?"
"Well," he began, somewhat awkwardly, still trying to comfort her despite the scene mere feet from them "you remember how you married me for my intelligence, right? Well, I realized that, usually, when people are said to sell their soul, it's just a metaphor for the loss of one's soul, when one profoundly compromises one's ethics. So, it's not exactly like a financial exchange. But, since I was going to take to your religion, I was going to make such a compromise. To minimize my loss, I sold my soul beforehand, to get this power. As for Brian, I made him deathly afraid of metal."
"You... You... sold your soul?" she asked in disbelief, despite ostensibly believing in the existence of souls and the ability to ... lose them? She wasn't sure anymore.
"Of course, sweetie," he replied, still holding her, "I'd do anything for you."
And her trembling only worsened.
Sunday, January 31, 2016
Sudden Short Story 102
Anderson took his box of lollipops off the counter and walked away.
"Have a nice day," said the clerk, but Anderson wasn't interested in exchanging any pleasantries with him or his ilk.
As he walked out, he opened the box to take out a lollipop. He passed by someone who was walking the opposite way, and idly commented, "Can you believe that they put these things in boxes now?"
"Yes, why?" came the reply.
While it might have been someone who was smugly taking his question literally in an attempt to seem witty, the question of "why" was unlikely, and the tone was too... kind.
Anderson ignored the response and kept walking. No matter where he went, it was all well-skinned androids these days. He didn't like them, considering each a walking, talking lie. He unwrapped his lollipop - the one candy that he allowed himself these days - and he suddenly stopped. It wasn't just today, or the past week, or the past month. Was it even in the past year? When was the last time that he'd come across another human being.
Anderson walked more slowly the rest of the way, wondering whether there were any other real humans left. It had been so long.
"Have a nice day," said the clerk, but Anderson wasn't interested in exchanging any pleasantries with him or his ilk.
As he walked out, he opened the box to take out a lollipop. He passed by someone who was walking the opposite way, and idly commented, "Can you believe that they put these things in boxes now?"
"Yes, why?" came the reply.
While it might have been someone who was smugly taking his question literally in an attempt to seem witty, the question of "why" was unlikely, and the tone was too... kind.
Anderson ignored the response and kept walking. No matter where he went, it was all well-skinned androids these days. He didn't like them, considering each a walking, talking lie. He unwrapped his lollipop - the one candy that he allowed himself these days - and he suddenly stopped. It wasn't just today, or the past week, or the past month. Was it even in the past year? When was the last time that he'd come across another human being.
Anderson walked more slowly the rest of the way, wondering whether there were any other real humans left. It had been so long.
Thursday, January 7, 2016
The Self-Referential Number
Recently, singingbanana (who goes by @jamesgrime on twitter) actually managed to post a video. In this case, it was a puzzle. The quest is to find a self-referential number, defined as a 10-digit number (implied to be in base 10) where the 1st digit says how many digits in the number are '0', the 2nd digit says how many digits in the number are '1', and so on.
You might want to try this problem yourself, so consider this your spoiler warning.
Spoilers follow.
The answer that I found was 6,210,001,000. I suspect that this is the only solution, but I haven't proven that yet. Maybe later.
I defined each digit with a letter, such that the number will be a,bcd,efg,hij. Since the number has 10 digits, a+b+c+d+e+f+g+h+i+j=10.
Then, I put limits on how high each digit could be, just using itself and how many of its thing it represents. For instance, a<=10 (even though 10's not a digit), because 10+10*0<=10. Similarly, b<=5 because 5+5*1<=10, and c<=3 because 3+3*2<=10.
a<=9
b<=5
c<=3
d<=2
e<=2
f<=1
g<=1
h<=1
i<=1
j<=1
Basically, there can't be more than 1 of anything that counts 5s or larger, because having 2 of the thing that it counts (I probably should have come up with symbology for that) would be at least 10, and adding the counter then exceeds 10.
This also applies to the collection, though, so f+g+h+i+j<=1.
This also means that there are at least 4 0s, so 4<=a<=9.
If e=1, then e+4=5, so that would also force f through j to be 0. Thus, e+f+g+h+i+j<=1.
This means that there are at least 5 0s, so 5<=a<=9.
Since a>4, it follows that some member of {f,g,h,i,j} is at least 1, and thus exactly 1. This also means that e=0, so there are no 4s in this number.
Since there's at least one 1, b>=1.
If b were to be 1, though, then there would be at least two 1s, forcing b to be at least 2. This would cause a problem for b, but wait, there's a 2, so c>=1.
Suppose that c=1 and b=2. Using X as a placeholder for a, the number looks like this:
X,21?,0[01][01],[01][01][01]
where only 1 of those [01]s is actually a 1, and the rest 0.
I don't see any 3s, so suppose that d=3. We now know that there are 6 0s. Thus, a=6.
Since there's one 6, g=1. This gives us:
6,210,001,000
QED
P.S.: The writing on this is a bit sloppy, but I'm using plain text and I'm in a hurry. I might edit this later for formality.
UPDATE 2016/Jan/13: So, the morning after I posted this (so the 8th), I thought of a proof that a<7. And now I have a spare moment! :D
Assume that a > 6. Then, hij=001, 010, or 100. Then, b>0.
b=1 and (h=1 xor i=1 xor j=1) ==> b>1 ==> c>0 or d>0. (See above that e+f+g+h+i+j <=1.)
Since a≠0, b≠0, c≠0, and 1 of h, 1, or j ≠0, there are at least 4 non-0 digits, so there are no more than six 0 digits. Thus, a, the number of 0 digits, cannot exceed 6.
QED
You might want to try this problem yourself, so consider this your spoiler warning.
Spoilers follow.
The answer that I found was 6,210,001,000. I suspect that this is the only solution, but I haven't proven that yet. Maybe later.
I defined each digit with a letter, such that the number will be a,bcd,efg,hij. Since the number has 10 digits, a+b+c+d+e+f+g+h+i+j=10.
Then, I put limits on how high each digit could be, just using itself and how many of its thing it represents. For instance, a<=10 (even though 10's not a digit), because 10+10*0<=10. Similarly, b<=5 because 5+5*1<=10, and c<=3 because 3+3*2<=10.
a<=9
b<=5
c<=3
d<=2
e<=2
f<=1
g<=1
h<=1
i<=1
j<=1
Basically, there can't be more than 1 of anything that counts 5s or larger, because having 2 of the thing that it counts (I probably should have come up with symbology for that) would be at least 10, and adding the counter then exceeds 10.
This also applies to the collection, though, so f+g+h+i+j<=1.
This also means that there are at least 4 0s, so 4<=a<=9.
If e=1, then e+4=5, so that would also force f through j to be 0. Thus, e+f+g+h+i+j<=1.
This means that there are at least 5 0s, so 5<=a<=9.
Since a>4, it follows that some member of {f,g,h,i,j} is at least 1, and thus exactly 1. This also means that e=0, so there are no 4s in this number.
Since there's at least one 1, b>=1.
If b were to be 1, though, then there would be at least two 1s, forcing b to be at least 2. This would cause a problem for b, but wait, there's a 2, so c>=1.
Suppose that c=1 and b=2. Using X as a placeholder for a, the number looks like this:
X,21?,0[01][01],[01][01][01]
where only 1 of those [01]s is actually a 1, and the rest 0.
I don't see any 3s, so suppose that d=3. We now know that there are 6 0s. Thus, a=6.
Since there's one 6, g=1. This gives us:
6,210,001,000
QED
P.S.: The writing on this is a bit sloppy, but I'm using plain text and I'm in a hurry. I might edit this later for formality.
UPDATE 2016/Jan/13: So, the morning after I posted this (so the 8th), I thought of a proof that a<7. And now I have a spare moment! :D
Assume that a > 6. Then, hij=001, 010, or 100. Then, b>0.
b=1 and (h=1 xor i=1 xor j=1) ==> b>1 ==> c>0 or d>0. (See above that e+f+g+h+i+j <=1.)
Since a≠0, b≠0, c≠0, and 1 of h, 1, or j ≠0, there are at least 4 non-0 digits, so there are no more than six 0 digits. Thus, a, the number of 0 digits, cannot exceed 6.
QED
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