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from itertools import permutations

#  Program to solve IBM's Ponder This puzzle for January 2024.
#  https://research.ibm.com/haifa/ponderthis/challenges/January2024.html
#
#  Using each of the numbers 1–16 exactly once, fill in the blank squares
#  in the grid below, so that the 16 equations are true.
# 
# ┌───┬───┬───┬───┬───┬───┬───┐
# │   │ + │   │ − │   │ − │   │  5 

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;  Inputs: A and B are bit-strings of length N,
;  right-justified in EAX and EBX respectively;
;  N ≤ 30 is in ECX, and A ≠ B.
;
;  Output:  ESI/EDI (a fraction) is the probability
;  that bit-string  A  beats bit-string  B.

Magic:
        PUSH EAX
        PUSH EBX

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from itertools import combinations
from string import ascii_uppercase

states = ["Alabama", "Alaska", "Arizona", "Arkansas", "California", 
          "Colorado", "Connecticut", "Delaware", "Florida", "Georgia",
          "Hawaii", "Idaho", "Illinois", "Indiana", "Iowa",
          "Kansas", "Kentucky", "Louisiana", "Maine", "Maryland",
          "Massachusetts", "Michigan", "Minnesota", "Mississippi", "Missouri",
          "Montana", "Nebraska", "Nevada", "New Hampshire", "New Jersey",
          "New Mexico", "New York", "North Carolina", "North Dakota", "Ohio",

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import numpy as np

#  Compute expected number of turns to complete a snakes & ladders
#  track of length n, given a set of portals (snakes and ladders),
#  which are both represented as lists of pairs (start, end).
#  It's ok to pass lists of [start, end] instead of pairs (start, end).

def E(n, portals):
    portal_map = {k: k for k in range(n)}     # identity map
    portal_map.update(portals)

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• is
• this
• markdown

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#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <string.h>
#include <time.h>

typedef uint8_t byte;
typedef uint32_t quart;
typedef uint64_t pottle;

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from fractions import Fraction
from functools import lru_cache
from itertools import *

def powerset(S):
    return chain.from_iterable(combinations(S, r) for r in range(len(S)+1))

@lru_cache(maxsize=512)
def p(S):                     #  S is a frozenset so it can be cached
    if len(S) == 0:

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	PAGE	58,132

;----------------------------------------------------------------------------
;  Hier sind der neu SETUP komputerprogramm.
;----------------------------------------------------------------------------

	.386P

FALSE	EQU	0

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import random

N = 2018
trials = 10000

def connect(wires):
    loops = 0
    while wires > 1:
        v, w = random.randrange(wires), random.randrange(wires)
        if v == w:

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# Determine how many ways the US election could end in electoral tie.
# Respsonse to https://twitter.com/daveinstpaul/status/792788962230226944

from functools import lru_cache

votes = [9,8,3,3,8,18,11,4,7,6,10,7,55,11,20,9,16,4,7,10,9,3,6,3,3,10,
         11,29,3,38,16,5,6,4,6,3,4,4,13,20,14,12,11,5,5,6,29,10,6,15,3]

@lru_cache(maxsize=10000)
def count_ties(subtotal, votes):