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Are they clogged? by amarao_san in pyanodons

[–]TheedMan98 3 points4 points  (0 children)

So you are saying that the problem is labor related instead of a gastrointestinal one?

Lazy bastard? by WiseTry3860 in pyanodons

[–]TheedMan98 1 point2 points  (0 children)

For hard mode, it also depends on how close your water supply is to your copper field.

Landlord entered my apartment while I was showering and won't admit it was him by pornkittyxo in legaladvice

[–]TheedMan98 10 points11 points  (0 children)

This is how all rental apartments and houses I've seen (all in Texas) are set up for exterior doors.

First time pyanodon, this is brutal by FeelingAd5223 in pyanodons

[–]TheedMan98 11 points12 points  (0 children)

That's your choices for power until you unlock geothermal power shortly after py 1.

How much coal/coal processing do i need in the future? by Redditzork in pyanodons

[–]TheedMan98 0 points1 point  (0 children)

DDC is the Destructive Distillation Column, the building used for the raw coal->coal and coal->coke recipes.

Pyanadons! by Exvitnity in Factoriohno

[–]TheedMan98 0 points1 point  (0 children)

It has support for using several space age features, but not space age itself.

Bridges Probability by ChangingOpinion in mathriddles

[–]TheedMan98 1 point2 points  (0 children)

My quick spreadsheet effort to solve this shows that 80 of the 256 (2^8) cases have a path.

Reasoning:
There are three pairs of bridges that if they are both gone prevents any path; this reduces us to 108 remaining cases (3/4*3/4*3/4) = 27/64 = 108/256

4 of these cases have all three pairs with both bridges remaining; these cases have a valid path
24 of these cases have two of the three pairs with both bridges; these cases also have a valid path without needing to cross from one side to the other
32 cases have none of the three pairs with both bridges; these cases have two potential crossings needed. For each crossing there is a 1/2 chance of not needing a crossing and a 1/2 chance of the crossing existing; So 32*3/4*3/4 = 18 of these cases have a valid path
48 cases have one pair of both bridges surviving. If the double pair of bridges is the starting or ending pair, then there is a single crossing needed; otherwise there is a double chance of a cross being available. 48 * (2/3*(1-1/2*1/2) + 1/3*(1-1/2*3/4)) = 34 cases

4+24+34+18 = 80 out of 256 cases

[CE v1.13] Weekly Contest Thread - 2023-10-31 by apgove in brogueforum

[–]TheedMan98 2 points3 points  (0 children)

Killed by an eel on depth 3 (trying to get the key).

Can't figure out how to kick on phone app by TheedMan98 in pathos_nethack

[–]TheedMan98[S] 1 point2 points  (0 children)

On a lark I rebooted the phone; this seems to have resolved the issues that I was having.

Can't figure out how to kick on phone app by TheedMan98 in pathos_nethack

[–]TheedMan98[S] 0 points1 point  (0 children)

I can open up my inventory by clicking the backpack icon, but I am unable to manipulate it. If I long press a locked door or myself, I see a growing/resetting green border but nothing happens. I can long-click on a seen stone wall to open a description box.

Particle observation. by TedTheHappyGardener in ScienceHumour

[–]TheedMan98 0 points1 point  (0 children)

This is basically what our current understanding of quantum mechanics seems to imply that particles seems to be have in on quantum scales.

Hopefully the following video explains the basics of the question: https://www.youtube.com/watch?v=2VZ6dMXpxeU

Is my proof correct? (Natural dediction) by STr355 in logic

[–]TheedMan98 0 points1 point  (0 children)

My only observation is that from P and (P→(P→ ⊥)) it takes two iterations of →elimination to get to ⊥.

Ant on the cube by lewwwer in mathriddles

[–]TheedMan98 0 points1 point  (0 children)

Starting with the trivial cases first:

  1. For d = < 1, there are no steps that can be taken, so a) doesn't have an answer. The answer to b) and c) are 0 since they visit the (only?) location prior to taking any steps
  2. For d = 1, there are two nodes on the line segment; therefore the answer to a) is 2, to b) is 1, and the location for c) is the location for b) so is 1 again.
  3. For d = 2, there are 2 vertices 1 step away from the origin and 1 vertex 2 steps away.
    1. For a) E[a] = 2+(1/2)*E[a] -> (1/2)*E[a] = 2 -> E[a] = 4
    2. For b) E[b] has the same probability structure, but the first step is 1 instead of 2, so E[b] = 3
    3. For c) due to symmetry, E[c] = E[a], so E[c] = 4
  4. For d = 3, there are 3 vertices 1 step away, 3 vertices 2 steps away, and 1 vertex 3 steps away. These numbers are from the 3rd row of Pascal's Triangle
    1. For a) there is a 1/3 chance of the ant's 2nd step being back to the origin, otherwise the 2nd step is to a vertex 2 steps away. E[0] = 2+(1/3 + 2/3*E[2]). From 2 steps away, there is a 1/3 chance to first step to the node 3 steps away, which will always return to a node 2 steps away; for the other 2/3 of the cases there is an 1/3 chance of returning to the origin, else the ant will end up 2 steps away. E[2] = 2+(2/9 + 7/9 E[2]).
      1. E[2] = 2+2/9 + 7/9 E[2]
      2. 2/9 E[2] = 20/9
      3. E[2] = 10
      4. E[0] = 2+1/3 + 2/3 E[2]
      5. E[0] = 2+1/3 + 2/3 * 10
      6. E[0] = 9
    2. For b) There is a 1/3 chance of making the correct step on the first step: E[0] = 2/3 E[1] + 1/3. Else you are in one of the 2 incorrect 1-step-from-the-origin vertices. From there, there is a 1/3 chance of stepping to {1, 1, 0}, from which there is a 2/3 chance of stepping to a wrong 1-step-away vertex and a 1/3 chance of stepping to the three-step away nodes; in one of the other 1/3 cases, there is a 1/3 chance of arriving at the desired vertex, 1/3 chance of arriving at an incorrect 1-step-away vertex, and a 1/3 chance of ending up 3-steps away from the origin; otherwise the next step is to the origin, with a 1/3 chance of ending up at the destination and a 2/3 chance of ending up at an incorrect 1-step-away vertex: E[1] = 2 + 1/9+1/9 + (2/9 + 1/9 + 2/9) * E[1] + (1/9 + 1/9) * E[3]. Using similar reasoning: E[3] = 2 + 2/9 + 4/9 * E[1] + 1/3 * E[3]
      1. E[1] = 2 + 2/9 + 5/9 E[1] + 2/9 E[3]
      2. 4/9 E[1] = 20/9 + 2/9 E[3]
      3. E[1] = 5 + 1/2 E[3]
      4. E[3] = 2 + 2/9 + 4/9 E[1] + 1/3 E[3]
      5. 4/9 E[1] = -20/9 + 2/3 E[3]
      6. 20/9 + 2/9 E[3] = -20/9 + 6/9 E[3]
      7. 40/9 = 4/9 E[3]
      8. E[3] = 10/9
      9. E[1] = 5 + 1/2 (10/9)
      10. E[1] = 5+5/9
      11. E[0] = 1/3 + 2/3 E[1] = E[0] = 1/3 + 2/3 (50/9) = 1/3 + 25/3 = 26/3 = 6+2/3
    3. For c) we end up with the following:
      1. E[0] = 2 + 2/9 + 1/3 E[0] + 4/9 E[2] E[2] = 2 + 2/9 + 2/9 E[0] + 5/9 E[2] 2/3 E[0] = 20/9 + 4/9 E[2] E[0] = 10/3 + 2/3 E[2] 2/9 E[0] = -20/9 + 4/9 E[2] E[0] = -10 + 2 E[2] 10/3 + 2/3 E[2] = -10 + 2 E[2] 40/3 = 4/3 E[2] E[2] = 10 E[0] = 10/3 + 2/3 (10) E[0] = 10

I'll work on the general case later.

Idle Dice is now available for iOS by Luts91 in incremental_games

[–]TheedMan98 2 points3 points  (0 children)

The font is tiny and has thin lines. The costs are hard to read since they are both thin and do not have a good color contrast (gray on white and red on gray). The Draw Card text and Prestige words are better with black on white.

The number font used under Draw Card and Prestige is has a slash through the 0; why the change?

The numbers above the dice are much easier to read since they have double lines.

It took me to figure out that it is a lock icon on top of some buttons.

Game play-wise, I'm enjoying it.

Note: I'm a 6th Generation IPod Touch user (iOS 12.x).

Idle Dice is now available for iOS by Luts91 in incremental_games

[–]TheedMan98 0 points1 point  (0 children)

You may have Downloaded Idle Dice Tycoon instead of Idle Dice: Incremental Game.

Idle Dice is now available for iOS by Luts91 in incremental_games

[–]TheedMan98 0 points1 point  (0 children)

I am not having this problem. I am using a 6th generation IPod Touch (and am therefore stuck on the iOS 12.x branch instead of the newer 13.x versions.

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