转换素数

问题描述

这个面试问题。给定两个n位素数，一次将第一个素数转换为第二个改变的一位。中间数字也必须是质数。这需要以最少的步骤数完成（检查素数和更改数字被认为是步骤）

Came across this interview question. Given two n-digit prime numbers, convert the first prime number to the second changing one digit at a time. The intermediate numbers also need to be prime. This needs to be done in the minimum number of steps (checking for primality and changing a digit are considered steps)

将1033转换为8179（1033-> 1733-> 3733-> .......-> 8179）

E.g. convert 1033 to 8179 (1033->1733->3733->.......->8179)

推荐答案

在一个下雨的星期一晚上（无论如何在这里，这是一个不错的挑战！）。可以使用来完成。第一步是创建一个包含所有4位素数的。然后使用Dijkstra的算法找出开始/结束素数之间的最短路径。这是Python的实现：

Nice challenge for a rainy Monday evening (it is here, anyway!). This can be done using Dijkstra's algorithm. The first step is to create a graph containing all 4-digit primes. Then use Dijkstra's algorithm to find the shortest path between the start/end primes. Here's an implementation in Python:

#! /usr/bin/python -tt

# run as: findpath start end

import sys

(start, end) = map(int, sys.argv[1:3])

# http://primes.utm.edu/lists/small/10000.txt
f = open("10000.txt", "r")
lines = f.readlines()
f.close
lines = lines[4:-1] # remove header/footer
all = "".join(lines) # join lines
all = all.split()
all = map(int, all)

# only want the 4-digit primes
fourdigit = [p for p in all if 1000 <= p and p <= 9999]

# returns digits in a number
digits = lambda x: map(int, str(x))

# cache of digits for each prime
digits_for_nums = {}

# returns digits in a number (using cache)
def digits_for_num(x):
    global digits_for_nums
    if x not in digits_for_nums:
        digits_for_nums[x] = digits(x)
    return digits_for_nums[x]

# returns 1 if digits are same, 0 otherwise
diff = lambda pair: 1 if pair[0] == pair[1] else 0

# computes number of identical digits in two numbers
def distance(a, b):
    pair = (a, b)
    pair = map(digits_for_num, pair)
    pair = zip(pair[0], pair[1])
    pair = map(diff, pair)
    same = sum(pair)
    return same

# adjacency list representation of graph of primes
edges = {}

# construct graph
for a in fourdigit:
    edges[a] = []
    for b in fourdigit:
        if distance(a, b) == 3:
            edges[a].append(b)

infinity = sys.maxint

def smallest():
    global dist, Q
    minimum = infinity
    which = None
    for v in Q:
        if dist[v] <= minimum:
            which = v
            minimum = dist[v]
    return which

# Dijkstra's algorithm
dist = {}
previous = {}
Q = edges.keys()
for v in Q:
    dist[v] = infinity
    previous[v] = None
dist[start] = 0
while len(Q) > 0:
    u = smallest()
    if dist[u] == infinity:
        break
    Q.remove(u)
    for v in edges[u]:
        alt = dist[u] + 1
        if alt < dist[v]:
            dist[v] = alt
            previous[v] = u

# get path between start/end nodes
num = end
path = [num]
while num != start:
    num = previous[num]
    path.insert(0, num)
print path

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